Square-Random-Symmetrical

Clash Royale CLAN TAG#URR8PPP

up vote
18
down vote

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Challenge

Write a program or a function that returns or prints a square-random-symmetrical matrix.

---

Input

N: The size of the matrix i.e 6 x 6

---

Output

The matrix. You can either print it, return it as string (with the newlines) or as a list/array of lists/arrays.

---

Rules

1. You need to use at least N different characters, where N is the size of the square matrix (input). Since we 're using only letter [a, z][A, Z] and digits [0, 9] (and only 1 digit at the time) you can assume that N < 27 and N > 2, that is because at N <= 2 you can't have both letters and digits. Last but not least, every letter/digit must have non-zero probability of occurring (uniform distribution is not a necessity). However, the result must have at least N different letter/digits.




2. The matrix has to be both horizontally and vertically symmetrical.




3. Exactly 2 rows and 2 columns must contain strictly one single-digit number (it's position should be random as well). The rest of rows/cols will contain only letters. Consider letters as [a, z] and [A, Z] and of course single-digit numbers as [0, 9].




4. Just to be easier, you can assume that the case of the letters doesn't matter, as long as the cases are symmetrical which means: a=A, b=B, etc.




5. Every possible output must have a non-zero probability of occurring. The random distribution doesn't need to be uniform.

---

Example

Input: 8

Output:

c r p s s p r c
r k o z z o k r
u t 2 a a 2 t u
y n q z z q n y
y n q z z q n y
u t 2 a a 2 t u
r k o z z o k r
c r p s s p r c

code-golf matrix

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edited Aug 31 at 17:08

asked Aug 29 at 20:54

DimChtz

601111

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  Comments are not for extended discussion; this conversation has been moved to chat.
  – Mego♦
  Aug 31 at 2:30
  

  
  
  
  

add a comment | 

up vote
18
down vote

favorite

2

Challenge

Write a program or a function that returns or prints a square-random-symmetrical matrix.

---

Input

N: The size of the matrix i.e 6 x 6

---

Output

The matrix. You can either print it, return it as string (with the newlines) or as a list/array of lists/arrays.

---

Rules

1. You need to use at least N different characters, where N is the size of the square matrix (input). Since we 're using only letter [a, z][A, Z] and digits [0, 9] (and only 1 digit at the time) you can assume that N < 27 and N > 2, that is because at N <= 2 you can't have both letters and digits. Last but not least, every letter/digit must have non-zero probability of occurring (uniform distribution is not a necessity). However, the result must have at least N different letter/digits.




2. The matrix has to be both horizontally and vertically symmetrical.




3. Exactly 2 rows and 2 columns must contain strictly one single-digit number (it's position should be random as well). The rest of rows/cols will contain only letters. Consider letters as [a, z] and [A, Z] and of course single-digit numbers as [0, 9].




4. Just to be easier, you can assume that the case of the letters doesn't matter, as long as the cases are symmetrical which means: a=A, b=B, etc.




5. Every possible output must have a non-zero probability of occurring. The random distribution doesn't need to be uniform.

---

Example

Input: 8

Output:

c r p s s p r c
r k o z z o k r
u t 2 a a 2 t u
y n q z z q n y
y n q z z q n y
u t 2 a a 2 t u
r k o z z o k r
c r p s s p r c

code-golf matrix

share|improve this question

edited Aug 31 at 17:08

asked Aug 29 at 20:54

DimChtz

601111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Comments are not for extended discussion; this conversation has been moved to chat.
  – Mego♦
  Aug 31 at 2:30
  

  
  
  
  

add a comment | 

up vote
18
down vote

favorite

2

up vote
18
down vote

favorite

2

2

Challenge

Write a program or a function that returns or prints a square-random-symmetrical matrix.

---

Input

N: The size of the matrix i.e 6 x 6

---

Output

The matrix. You can either print it, return it as string (with the newlines) or as a list/array of lists/arrays.

---

Rules

1. You need to use at least N different characters, where N is the size of the square matrix (input). Since we 're using only letter [a, z][A, Z] and digits [0, 9] (and only 1 digit at the time) you can assume that N < 27 and N > 2, that is because at N <= 2 you can't have both letters and digits. Last but not least, every letter/digit must have non-zero probability of occurring (uniform distribution is not a necessity). However, the result must have at least N different letter/digits.




2. The matrix has to be both horizontally and vertically symmetrical.




3. Exactly 2 rows and 2 columns must contain strictly one single-digit number (it's position should be random as well). The rest of rows/cols will contain only letters. Consider letters as [a, z] and [A, Z] and of course single-digit numbers as [0, 9].




4. Just to be easier, you can assume that the case of the letters doesn't matter, as long as the cases are symmetrical which means: a=A, b=B, etc.




5. Every possible output must have a non-zero probability of occurring. The random distribution doesn't need to be uniform.

---

Example

Input: 8

Output:

c r p s s p r c
r k o z z o k r
u t 2 a a 2 t u
y n q z z q n y
y n q z z q n y
u t 2 a a 2 t u
r k o z z o k r
c r p s s p r c

code-golf matrix

share|improve this question

edited Aug 31 at 17:08

asked Aug 29 at 20:54

DimChtz

601111

Challenge

Write a program or a function that returns or prints a square-random-symmetrical matrix.

---

Input

N: The size of the matrix i.e 6 x 6

---

Output

The matrix. You can either print it, return it as string (with the newlines) or as a list/array of lists/arrays.

---

Rules

1. You need to use at least N different characters, where N is the size of the square matrix (input). Since we 're using only letter [a, z][A, Z] and digits [0, 9] (and only 1 digit at the time) you can assume that N < 27 and N > 2, that is because at N <= 2 you can't have both letters and digits. Last but not least, every letter/digit must have non-zero probability of occurring (uniform distribution is not a necessity). However, the result must have at least N different letter/digits.




2. The matrix has to be both horizontally and vertically symmetrical.




3. Exactly 2 rows and 2 columns must contain strictly one single-digit number (it's position should be random as well). The rest of rows/cols will contain only letters. Consider letters as [a, z] and [A, Z] and of course single-digit numbers as [0, 9].




4. Just to be easier, you can assume that the case of the letters doesn't matter, as long as the cases are symmetrical which means: a=A, b=B, etc.




5. Every possible output must have a non-zero probability of occurring. The random distribution doesn't need to be uniform.

---

Example

Input: 8

Output:

c r p s s p r c
r k o z z o k r
u t 2 a a 2 t u
y n q z z q n y
y n q z z q n y
u t 2 a a 2 t u
r k o z z o k r
c r p s s p r c

code-golf matrix

share|improve this question

edited Aug 31 at 17:08

asked Aug 29 at 20:54

DimChtz

601111

share|improve this question

share|improve this question

edited Aug 31 at 17:08

edited Aug 31 at 17:08

edited Aug 31 at 17:08

asked Aug 29 at 20:54

DimChtz

601111

asked Aug 29 at 20:54

DimChtz

601111

asked Aug 29 at 20:54

DimChtz

601111

601111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Comments are not for extended discussion; this conversation has been moved to chat.
  – Mego♦
  Aug 31 at 2:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Comments are not for extended discussion; this conversation has been moved to chat.
  – Mego♦
  Aug 31 at 2:30
  

  
  
  
  

Comments are not for extended discussion; this conversation has been moved to chat.
– Mego♦
Aug 31 at 2:30

Comments are not for extended discussion; this conversation has been moved to chat.
– Mego♦
Aug 31 at 2:30

add a comment | 

14 Answers
14

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oldest

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up vote
4
down vote



accepted

Charcoal, 30 bytes

ＮθＥ⊘⊕θ⭆⊘⊕θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖ＯＯ→↓﹪θ²

Try it online! Link is to verbose version of code. If n is always even, then for 23 bytes:

ＮθＥ⊘θ⭆⊘θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖Ｃ¬

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input $ n $.

Ｅ⊘θ⭆⊘θ‽β

Create an $ frac n 2 $ by $ frac n 2 $ array of random lowercase letters. This prints implicitly as a square.

Ｊ‽⊘θ‽⊘θ

Jump to a random position in the square.

Ｉ‽χ

Print a random digit.

‖Ｃ¬

Reflect horizontally and vertically to complete the matrix.

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answered Aug 30 at 0:07

Neil

75.1k744170

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up vote
11
down vote

R, 124 118 bytes

function(n,i=(n+1)/2,j=n%/%2,m="[<-"(matrix(-letters,i,i),j-1,j-1,0:9-1))cbind(y<-rbind(m,m[j:1,]),y[,j:1])
`-`=sample

Try it online!

In R, things that look like operators are just functions that get special treatment from the parser.

If you redefine an operator (like -) to be some other function, it keeps the special treatment from the parser. Since - is both prefix and infix, and I need to call the sample function with both one and two arguments, I can use

`-`=sample

to get what I want.

So the code -letters is translated to sample(letters), which randomly shuffles the letters built-in. But j-1 is translated to sample(j,1), which randomly samples 1 item from the vector 1:j.

(This behaviour of the sample function depending on the number of parameters and what the first parameter is, is a huge pain in the butt in production code, so I'm happy to find a great use of its perverse nature here!)

Otherwise the code just makes the top left quadrant of the required result, replaces a random element (the j-1,j-1 bit) with a random digit (the 0:9-1 bit), and folds it out for the required symmetry. The i and the j are needed to deal with the even and odd cases.

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edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

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  I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
  – JayCe
  Aug 30 at 23:44
  

  
  
  
  


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  What a fantastic solution and explanation!
  – J.Doe
  Aug 31 at 8:57
  

  
  
  
  

add a comment | 

up vote
6
down vote

Python3, 287 bytes

My first try at golfing something here; I'm sure someone can do far better:

import random as rn, math as m
n=int(input())
x,o=m.ceil(n/2),n%2
c=x-1-o
f=lambda l,n: l.extend((l[::-1], l[:-1][::-1])[o])
q=[rn.sample([chr(i) for i in range(97, 123)],x) for y in range(x)]
q[rn.randint(0,c)][rn.randint(0,c)] = rn.randint(0,9)
for r in q:
 f(r, n)
f(q, n)
print(q)

Try it Online!

Thanks to HyperNeurtrino, Ourous and Heiteria this shrunk down to 193 bytes (see comments). However, TFeld correctly pointed out that multiple calls to sample aren't guaranteeing at least N different characters.

That stuff in mind, try this new version that should guarantee at least N different characters per run.

Python3, 265 260 bytes, at least N distinct characters

from random import *
n=int(input())
x=-(-n//2)
o=n%2
c=x+~o
i=randint
u=[chr(j+97)for j in range(26)]
z,q=u[:],
for y in [1]*x:
 shuffle(z)
 q+=[z[:x]]
 z=z[x:] if len(z[x:])>=x else u[:]
q[i(0,c)][i(0,c)]=i(0,9)
for r in[q]+q:r.extend(r[~o::-1])
print(q)

Try it online!

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edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

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  Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
  – HyperNeutrino
  Aug 30 at 1:56
  

  
  
  
  


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  You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
  – HyperNeutrino
  Aug 30 at 1:59
  

  
  
  
  


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  You can get it down to 236: Try it online!
  – ÎŸurous
  Aug 30 at 4:19
  

  
  
  
  


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  Even further, at 196: Try it online!
  – ÎŸurous
  Aug 30 at 4:58
  
  

  
  
  
  


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  The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
  – TFeld
  Aug 30 at 8:35
  

  
  
  
  

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up vote
3
down vote

APL (Dyalog Classic), 45 44 43 40 bytes

thanks @Adám for -1 byte

26(⎕a,⍺⍴⎕d)[⌈∘⊖⍨⌈∘⌽⍨⍺+@(?⊂⌊⍵÷2)?⍵⍴⍺],⍨

Try it online!

uses ⌈ (max) of the matrix with its reflections to make it symmetric, so it's biased towards the latter part of the alphabet

the digit is chosen uniformly from 0...25 mod 10, so it has a small bias to lower values

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edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

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  ⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
  – Adám
  Aug 30 at 8:59
  

  
  
  
  


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  @Adám clever!­­
  – ngn
  Aug 30 at 9:06
  
  

  
  
  
  


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  Yeah, I just realised.
  – Adám
  Aug 30 at 9:09
  

  
  
  
  


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  If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
  – Adám
  Aug 30 at 9:10
  
  

  
  
  
  


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  @Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
  – ngn
  Aug 30 at 9:11
  

  
  
  
  

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up vote
3
down vote

Japt, 31 bytes (Fixed digit position)

;
/2 c
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv

Try it online!

---

Japt, 41 bytes (Random digit position)

;
/2 c
VÆVÆBö}ÃgMq´VÉ ,MqVÉ @Mq9îêUvÃêUv

Try it online!

---

Explanation

; Change to new vars
/2 c set implicit var V equal to implicit var U / 2 rounded up
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv Main function

VÆ Range from 0 to V and map
 VÆ Range from 0 to V and map
 Bö}Ã return random char from alphabet
 gT0@ map upper-left corner
 Mq9Ã return random number
 ®êUv horizontal mirror
 êUv vertical mirror

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edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

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  Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
  – Kevin Cruijssen
  Aug 30 at 11:50
  
  

  
  
  
  


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  @KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
  – Luis felipe De jesus Munoz
  Aug 30 at 11:54
  

  
  
  
  


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  Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
  – Kevin Cruijssen
  Aug 30 at 12:03
  
  

  
  
  
  

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up vote
2
down vote

Python 2, 259 bytes

from random import*
n=input();c=choice;r=range
w,W=n/2,-~n/2
o=n%2
A=map(chr,r(97,123))
l=[c(r(10))]+sample(A,n)+[c(A)for _ in' '*w*w]
l,e=l[:w*w],l[w*w:W*W]
shuffle(l)
l=[l[w*i:w*-~i]+e[i:i+1]for i in range(w)]+[e[-W:]]
for r in l+l[~o::-1]:print r+r[~o::-1]

Try it online!

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

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  Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
  – Teck-freak
  Sep 3 at 17:09
  

  
  
  
  

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up vote
2
down vote

05AB1E, 29 40 38 bytes

A.rs;ò©n∍9ÝΩ®DnαLʒ®%Ā}<Ωǝ®ô»¹Éi.º.∊ëº∊

+11 bytes to fix the digit being at a random position while still keeping rule 3 in mind for odd inputs..

-2 bytes thanks to @MagicOctopusUrn, changing îï to ò and changing the position of the ».

Try it online of verify some more test cases.

Old (29 27 bytes) answer where the digit positions where always in the corners:

A.rs;ò©n∍¦9ÝΩì®ô»¹Éi.º.∊ëº∊

Try it online or verify some more test cases.

Explanation:

A # Take the lowercase alphabet
 .r # Randomly shuffle it
 # i.e. "abcdefghijklmnopqrstuvwxyz" → "uovqxrcijfgyzlbpmhatnkwsed"
s # Swap so the (implicit) input is at the top of the stack
 ; # Halve the input
 # i.e. 7 → 3.5
 ò # Bankers rounding to the nearest integer
 # i.e. 3.5 → 4
 © # And save this number in the register
 n # Take its square
 # i.e. 4 → 16
 ∍ # Shorten the shuffled alphabet to that length
 # i.e. "uovqxrcijfgyzlbpmhatnkwsed" and 16 → "uovqxrcijfgyzlbp"
9ÝΩ # Take a random digit in the range [0,9]
 # i.e. 3
 ®Dnα # Take the difference between the saved number and its square:
 # i.e. 4 and 16 → 12
 L # Create a list in the range [1,n]
 # i.e. 12 → [1,2,3,4,5,6,7,8,9,10,11,12]
ʒ } # Filter this list by:
 ®%Ā # Remove any number that's divisible by the number we've saved
 # i.e. [1,2,3,4,5,6,7,8,9,10,11,12] and 4 → [1,2,3,5,6,7,9,10,11]
 < # Decrease each by 1 (to make it 0-indexed)
 # i.e. [1,2,3,5,6,7,9,10,11] → [0,1,2,3,5,6,7,9,10]
 Ω # Take a random item from this list
 # i.e. [0,1,2,3,5,6,7,9,10] → 6
 ǝ # Replace the character at this (0-indexed) position with the digit
 # i.e. "uovqxrcijfgyzlbp" and 3 and 6 → "uovqxr3ijfgyzlbp"
®ô # Split the string into parts of length equal to the number we've saved
 # i.e. "uovqxr3ijfgyzlbp" and 4 → ["uovq","xr3i","jfgy","zlbp"]
 » # Join them by new-lines (this is done implicitly in the legacy version)
 # i.e. ["uovq","xr3i","jfgy","zlbp"] → "uovqnxr3injfgynzlbp"
 ¹Éi # If the input is odd:
 # i.e. 7 → 1 (truthy)
 .º # Intersect mirror the individual items
 # i.e. "uovqnxr3injfgynzlbp"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 .∊ # And intersect vertically mirror the whole thing
 # i.e. "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblznjfgygfjnxr3i3rxnuovqvou"
 ë # Else (input was even):
 º∊ # Do the same, but with non-intersecting mirrors

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edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

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  You could also save 2 bytes with the legacy version since it doesn't require »
  – Emigna
  Aug 30 at 13:17
  

  
  
  
  


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  @Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
  – Kevin Cruijssen
  Aug 30 at 14:50
  

  
  
  
  


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  @MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
  – Kevin Cruijssen
  Aug 30 at 15:32
  

  
  
  
  


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  @KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
  – Magic Octopus Urn
  Aug 30 at 15:36
  

  
  
  
  


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  @MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
  – Kevin Cruijssen
  Aug 30 at 15:43
  

  
  
  
  

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up vote
2
down vote

C (gcc), 198 197 bytes

Saved 1 byte thanks to ceilingcat (4 bytes actually but I lost 3 bytes fixing the bug which caused the program to print regular pattern for $N=26$.

#define A(x)(x<n/2?x:n-1-x)
#define R rand()
S(n,x,y)

Try it online!

Explanation:

// Coordinate conversion for symmetry
#define A (x) (x < n / 2 ? x : n - 1 - x)
// Get a random and seed
#define R rand()

S (n, x, y)

share|improve this answer

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

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  Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
  – ceilingcat
  Sep 4 at 23:11
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

JavaScript (ES6), 213 209 206 bytes

n=>(a=,F=(x=y=d=c=0,R=k=>Math.random()*k|0,g=y=>(r=a[y]=a[y]||)[x]=r[n+~x]=v.toString(36))=>y<n/2?F(g(y,R[v=R(m=~-n/2)<!d&x<m&y<m?R(d=10):R(26)+10]=R[v]||++c,g(n+~y))&&++x<n/2?x:+!++y,R):!d|c<n?F():a)()

Try it online!

Commented

n => ( // n = input
 a = , // a = output matrix
 F = ( // F = main recursive function taking:
 x = y = // (x, y) = current coordinates
 d = c = 0, // d = digit flag; c = distinct character counter
 R = k => // R() = helper function to get a random value in [0,k[
 Math.random() * k | 0, // also used to store characters
 g = y => // g() = helper function to update the matrix
 (r = a[y] = a[y] || )[x] // with horizontal symmetry
 = r[n + ~x] = v.toString(36) // using the base-36 representation of v
 ) => //
 y < n / 2 ? // if we haven't reached the middle row(s) of the matrix:
 F( // do a recursive call to F():
 g( // invoke g() ...
 y, // ... on the current row
 R[v = // compute v = next value to be inserted
 R(m = ~-n/2) < !d & // we may insert a digit if no digit has been
 x < m & // inserted so far and the current coordinates are
 y < m ? // compatible: 2 distinct rows / 2 distinct columns
 R(d = 10) // if so, pick v in [0, 9] and update d
 : // else:
 R(26) + 10 // pick v in [10, 35] for a letter
 ] = R[v] || ++c, // set this character as used; update c accordingly
 g(n + ~y) // invoke g() on the mirror row
 ) && // end of outer call to g()
 ++x < n / 2 ? // if we haven't reached the middle column(s):
 x // use x + 1
 : // else
 +!++y, // increment y and reset x to 0
 R // explicitly pass R, as it is used for storage
 ) // end of recursive call to F()
 : // else:
 !d | c < n ? F() : a // either return the matrix or try again if it's invalid
)() // initial call to F()

share|improve this answer

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

add a comment | 

up vote
1
down vote

Clean, 346 312 bytes

will golf more tomorrow

import StdEnv,Data.List,Math.Random,System.Time,System._Unsafe
$n#q=twice(transpose oq=zipWith((++)o reverse o drop(n-n/2*2))q q)[[(['a'..'z']++['0'..'9'])!!(c rem 36)\c<-genRandInt(toInt(accUnsafe(time)))]%(i*n/2,i*n/2+(n-1)/2)\i<-[1..(n+1)/2]]
|length(nub(flatten q))>=n&&sum[1\c<-q|any isDigit c]==2=q= $n

Try it online!

share|improve this answer

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

add a comment | 

up vote
1
down vote

Python 3, 197 bytes

As mentioned by @Emigna, doesn't work for odd values of N (I didn't understand the question properly)

from random import*
def m(N):M=N//2;E=reversed;R=range;B=[randint(48,57),*(sample(R(97,123),N)*N)][:M*M];shuffle(B);r=R(M);m=[k+[*E(k)]for k in[[chr(B.pop())for i in r]for j in r]];m+=E(m);return m

Try it online!

I do think the calls to randint() + sample() + shuffle() are too much, and getting rid of in-place shuffling would be great :)

I'm pretty sure this part (that selects the letters & digit) could be golfed a bit more.

share|improve this answer

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doesn't seem correct for odd N.
  – Emigna
  Aug 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
  – etene
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  These are some examples of odd symmetrical matrices.
  – Emigna
  Aug 30 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
  – etene
  Aug 30 at 13:59
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 2, 275 266 bytes

from random import*
def f(n):
 R=range;C=choice;A=map(chr,R(97,123));b=N=n-n/2;c=`C(R(10))`;s=[c]+sample(A,n-1)+[C(A)for i in R(N*N-n)]
 while b:shuffle(s);i=s.index(c);b=n%2>(i<N*N-N>N-1>i%N)
 a=[r+r[~(n%2)::-1]for r in[s[i::N]for i in R(N)]];return a+a[~(n%2)::-1]

Try it online!

Returns the array as a list of lists of characters. To satisfy Rule 1, we set up a pool of characters:

s = [c] # the unique digit...
 + sample(A,n-1) # then sample without replacement `n-1` chars in a-z, 
 # so we have `n` distinct chars
 + [C(A)for i in R(N*N-n)] # and fill out the rest with any in a-z

The next tricky bit is rule 3: there must be exactly 2 columns and rows having a digit; this means for n odd, that the chosen digit may not appear in the middle column or middle row. Since we construct the array using a twice reflected square sub array s, that is accomplished here by using:

while b: # to save a couple bytes, `b` is initialized 
 # to `N`, which is greater than 0.
 shuffle(s) # shuffle at least once...
 i = s.index(c) # c is the unique digit used
 b = n%2 
 > # if n is even, 0>(any boolean) will be false,
 # so exit the loop; otherwise n odd, and we are
 # evaluating '1 > some boolean', which is equivalent 
 # to 'not (some boolean)'
 (i<N*N-N # i is not the last column of s...
 > # shortcut for ' and ', since N*N-N is always > N-1
 N-1>i%N) # is not the last row of s

i.e., shuffle at least once; and then, if n is odd, keep looping if the digit is in the last column or the last row of s.

share|improve this answer

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

add a comment | 

up vote
1
down vote

Pyth, 48 bytes

L+b_<b/Q2JmO/Q2 2jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K

Try it out online here.

The program is in 3 parts - definition of palindromisation function, choosing location of numeric, and main function.

Implicit: Q=eval(input()), T=10, G=lower case alphabet

L+b_<b/Q2 Palindromisation function
L Define a function, y(b)
 /Q2 Half input number, rounding down
 <b Take that many elements from the start of the sequence
 _ Reverse them
 +b Prepend the unaltered sequence

JmO/Q2 2 Choose numeric location
 O/Q2 Choose a random number between 0 and half input number
 m 2 Do the above twice, wrap in array
J Assign to variable J

jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K Main function
 cQ2 Divide input number by 2
 .E Round up
 K Assign the above to K
 .SG Shuffle the alphabet
 sm Q Do the above Q times, concatenate
 c K Chop the above into segments of length K
 < K Take the first K of the above
 .e Map (element, index) as (b,k) using:
 qhJk Does k equal first element of J?
 W If so...
 X b Replace in b...
 eJ ...at position <last element of J>...
 OT ...a random int less than 10
 Otherwise, b without replacement
 y Apply palindromisation to the result of the above
 y Palindromise the set of lines
j Join on newlines, implicit print

Using several shuffled alphabets should ensure that the number of unique characters is always more then the input number.

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

add a comment | 

up vote
1
down vote

Python 2/Python 3, 227 bytes

from random import*
def m(N):n=N-N//2;r=range;C=choice;c=n*[chr(i+97)for i in r(26)];shuffle(c);c[C([i for i in r(n*(N-n))if(i+1)%n+1-N%2])]=`C(r(10))`;R=[c[i*n:i*n+n]+c[i*n:i*n+n-N%2][::-1]for i in r(n)];return R+R[::-1][N%2:]

ungolfing a bit:

from random import * # get 'choice' and 'shuffle'
def matrix(N):
 n = ceil(N/2) # get the size of the base block
 # get a shuffleable lowercase alphabet
 c = [chr(i+97)for i in range(26)]
 c = n*c # make it large enough to fill the base-block
 shuffle(c) # randomize it
 digit = choice('1234567890') # get random digit string
 ## this is only needed as to prevent uneven side-length matrices
 # from having centerline digits.
 allowed_indices = [i for i in range( # get all allowed indices
 n*(N-n)) # skip those, that are in an unmirrored center-line
 if(i+1)%n # only use those that are not in the center column
 +1-N%2] # exept if there is no center column
 index = choice(allowed_indices) # get random index
 c[index]=digit # replace one field at random with a random digit
 ## 
 R=
 for i in range(n):
 r = c[i*n:i*n+n] # chop to chunks sized fit for the base block
 R.append(r+r[::-1][N%2:]) # mirror skipping the center line
 return R+R[::-1][N%2:] # mirror skipping the center line and return

Older, almost correct versions below:

Python2, Python3, 161 bytes

from random import *
N=26
n=N-N//2
c=[chr(i+97)for i in range(26)]
R=[ r+r[::-1][N%2:]for r in[(shuffle(c),c[:n])[1]for i in range(n)]]
R+=R[::-1][N%2:]
print(R)

It seems N differing elements is only almost guarranteed.

Python 2/Python 3, 170 bytes

from random import*
def m(N):n=N-N//2;r=range;c=n*[chr(i+97)for i in r(26)][:n*n];shuffle(c);R=[_+_[::-1][N%2:]for _ in[c[i*n:i*n+n]for i in r(n)]];return R+R[::-1][N%2:]

It seems I forgot rule 3. Also somehow the [:n*n] slipped in.

share|improve this answer

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
  – Chas Brown
  Sep 2 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
  – Teck-freak
  Sep 3 at 0:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
  – Chas Brown
  Sep 3 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I litterally ran it over 10000 times without any error.
  – Teck-freak
  Sep 3 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
  – Teck-freak
  Sep 3 at 15:11
  

  
  
  
  

 | 
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14 Answers
14

active

oldest

votes

14 Answers
14

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Charcoal, 30 bytes

ＮθＥ⊘⊕θ⭆⊘⊕θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖ＯＯ→↓﹪θ²

Try it online! Link is to verbose version of code. If n is always even, then for 23 bytes:

ＮθＥ⊘θ⭆⊘θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖Ｃ¬

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input $ n $.

Ｅ⊘θ⭆⊘θ‽β

Create an $ frac n 2 $ by $ frac n 2 $ array of random lowercase letters. This prints implicitly as a square.

Ｊ‽⊘θ‽⊘θ

Jump to a random position in the square.

Ｉ‽χ

Print a random digit.

‖Ｃ¬

Reflect horizontally and vertically to complete the matrix.

share|improve this answer

answered Aug 30 at 0:07

Neil

75.1k744170

add a comment | 

up vote
4
down vote



accepted

Charcoal, 30 bytes

ＮθＥ⊘⊕θ⭆⊘⊕θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖ＯＯ→↓﹪θ²

Try it online! Link is to verbose version of code. If n is always even, then for 23 bytes:

ＮθＥ⊘θ⭆⊘θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖Ｃ¬

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input $ n $.

Ｅ⊘θ⭆⊘θ‽β

Create an $ frac n 2 $ by $ frac n 2 $ array of random lowercase letters. This prints implicitly as a square.

Ｊ‽⊘θ‽⊘θ

Jump to a random position in the square.

Ｉ‽χ

Print a random digit.

‖Ｃ¬

Reflect horizontally and vertically to complete the matrix.

share|improve this answer

answered Aug 30 at 0:07

Neil

75.1k744170

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Charcoal, 30 bytes

ＮθＥ⊘⊕θ⭆⊘⊕θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖ＯＯ→↓﹪θ²

Try it online! Link is to verbose version of code. If n is always even, then for 23 bytes:

ＮθＥ⊘θ⭆⊘θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖Ｃ¬

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input $ n $.

Ｅ⊘θ⭆⊘θ‽β

Create an $ frac n 2 $ by $ frac n 2 $ array of random lowercase letters. This prints implicitly as a square.

Ｊ‽⊘θ‽⊘θ

Jump to a random position in the square.

Ｉ‽χ

Print a random digit.

‖Ｃ¬

Reflect horizontally and vertically to complete the matrix.

share|improve this answer

answered Aug 30 at 0:07

Neil

75.1k744170

Charcoal, 30 bytes

ＮθＥ⊘⊕θ⭆⊘⊕θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖ＯＯ→↓﹪θ²

Try it online! Link is to verbose version of code. If n is always even, then for 23 bytes:

ＮθＥ⊘θ⭆⊘θ‽βＪ‽⊘θ‽⊘θＩ‽χ‖Ｃ¬

Try it online! Link is to verbose version of code. Explanation:

Ｎθ

Input $ n $.

Ｅ⊘θ⭆⊘θ‽β

Create an $ frac n 2 $ by $ frac n 2 $ array of random lowercase letters. This prints implicitly as a square.

Ｊ‽⊘θ‽⊘θ

Jump to a random position in the square.

Ｉ‽χ

Print a random digit.

‖Ｃ¬

Reflect horizontally and vertically to complete the matrix.

share|improve this answer

answered Aug 30 at 0:07

Neil

75.1k744170

share|improve this answer

share|improve this answer

answered Aug 30 at 0:07

Neil

75.1k744170

answered Aug 30 at 0:07

Neil

75.1k744170

answered Aug 30 at 0:07

Neil

75.1k744170

75.1k744170

add a comment | 

add a comment | 

up vote
11
down vote

R, 124 118 bytes

function(n,i=(n+1)/2,j=n%/%2,m="[<-"(matrix(-letters,i,i),j-1,j-1,0:9-1))cbind(y<-rbind(m,m[j:1,]),y[,j:1])
`-`=sample

Try it online!

In R, things that look like operators are just functions that get special treatment from the parser.

If you redefine an operator (like -) to be some other function, it keeps the special treatment from the parser. Since - is both prefix and infix, and I need to call the sample function with both one and two arguments, I can use

`-`=sample

to get what I want.

So the code -letters is translated to sample(letters), which randomly shuffles the letters built-in. But j-1 is translated to sample(j,1), which randomly samples 1 item from the vector 1:j.

(This behaviour of the sample function depending on the number of parameters and what the first parameter is, is a huge pain in the butt in production code, so I'm happy to find a great use of its perverse nature here!)

Otherwise the code just makes the top left quadrant of the required result, replaces a random element (the j-1,j-1 bit) with a random digit (the 0:9-1 bit), and folds it out for the required symmetry. The i and the j are needed to deal with the even and odd cases.

share|improve this answer

edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
  – JayCe
  Aug 30 at 23:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What a fantastic solution and explanation!
  – J.Doe
  Aug 31 at 8:57
  

  
  
  
  

add a comment | 

up vote
11
down vote

R, 124 118 bytes

function(n,i=(n+1)/2,j=n%/%2,m="[<-"(matrix(-letters,i,i),j-1,j-1,0:9-1))cbind(y<-rbind(m,m[j:1,]),y[,j:1])
`-`=sample

Try it online!

In R, things that look like operators are just functions that get special treatment from the parser.

If you redefine an operator (like -) to be some other function, it keeps the special treatment from the parser. Since - is both prefix and infix, and I need to call the sample function with both one and two arguments, I can use

`-`=sample

to get what I want.

So the code -letters is translated to sample(letters), which randomly shuffles the letters built-in. But j-1 is translated to sample(j,1), which randomly samples 1 item from the vector 1:j.

(This behaviour of the sample function depending on the number of parameters and what the first parameter is, is a huge pain in the butt in production code, so I'm happy to find a great use of its perverse nature here!)

Otherwise the code just makes the top left quadrant of the required result, replaces a random element (the j-1,j-1 bit) with a random digit (the 0:9-1 bit), and folds it out for the required symmetry. The i and the j are needed to deal with the even and odd cases.

share|improve this answer

edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
  – JayCe
  Aug 30 at 23:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What a fantastic solution and explanation!
  – J.Doe
  Aug 31 at 8:57
  

  
  
  
  

add a comment | 

up vote
11
down vote

up vote
11
down vote

R, 124 118 bytes

function(n,i=(n+1)/2,j=n%/%2,m="[<-"(matrix(-letters,i,i),j-1,j-1,0:9-1))cbind(y<-rbind(m,m[j:1,]),y[,j:1])
`-`=sample

Try it online!

In R, things that look like operators are just functions that get special treatment from the parser.

If you redefine an operator (like -) to be some other function, it keeps the special treatment from the parser. Since - is both prefix and infix, and I need to call the sample function with both one and two arguments, I can use

`-`=sample

to get what I want.

So the code -letters is translated to sample(letters), which randomly shuffles the letters built-in. But j-1 is translated to sample(j,1), which randomly samples 1 item from the vector 1:j.

(This behaviour of the sample function depending on the number of parameters and what the first parameter is, is a huge pain in the butt in production code, so I'm happy to find a great use of its perverse nature here!)

Otherwise the code just makes the top left quadrant of the required result, replaces a random element (the j-1,j-1 bit) with a random digit (the 0:9-1 bit), and folds it out for the required symmetry. The i and the j are needed to deal with the even and odd cases.

share|improve this answer

edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

R, 124 118 bytes

function(n,i=(n+1)/2,j=n%/%2,m="[<-"(matrix(-letters,i,i),j-1,j-1,0:9-1))cbind(y<-rbind(m,m[j:1,]),y[,j:1])
`-`=sample

Try it online!

In R, things that look like operators are just functions that get special treatment from the parser.

If you redefine an operator (like -) to be some other function, it keeps the special treatment from the parser. Since - is both prefix and infix, and I need to call the sample function with both one and two arguments, I can use

`-`=sample

to get what I want.

So the code -letters is translated to sample(letters), which randomly shuffles the letters built-in. But j-1 is translated to sample(j,1), which randomly samples 1 item from the vector 1:j.

(This behaviour of the sample function depending on the number of parameters and what the first parameter is, is a huge pain in the butt in production code, so I'm happy to find a great use of its perverse nature here!)

Otherwise the code just makes the top left quadrant of the required result, replaces a random element (the j-1,j-1 bit) with a random digit (the 0:9-1 bit), and folds it out for the required symmetry. The i and the j are needed to deal with the even and odd cases.

share|improve this answer

edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

share|improve this answer

share|improve this answer

edited Aug 31 at 14:49

edited Aug 31 at 14:49

edited Aug 31 at 14:49

answered Aug 30 at 20:05

ngm

2,07920

answered Aug 30 at 20:05

ngm

2,07920

answered Aug 30 at 20:05

ngm

2,07920

2,07920

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
  – JayCe
  Aug 30 at 23:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What a fantastic solution and explanation!
  – J.Doe
  Aug 31 at 8:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
  – JayCe
  Aug 30 at 23:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What a fantastic solution and explanation!
  – J.Doe
  Aug 31 at 8:57
  

  
  
  
  

I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
– JayCe
Aug 30 at 23:44

I wish I could +2 for the great explanation and also editing the related R golfing tip answer. You can save a few more bytes
– JayCe
Aug 30 at 23:44

What a fantastic solution and explanation!
– J.Doe
Aug 31 at 8:57

What a fantastic solution and explanation!
– J.Doe
Aug 31 at 8:57

add a comment | 

up vote
6
down vote

Python3, 287 bytes

My first try at golfing something here; I'm sure someone can do far better:

import random as rn, math as m
n=int(input())
x,o=m.ceil(n/2),n%2
c=x-1-o
f=lambda l,n: l.extend((l[::-1], l[:-1][::-1])[o])
q=[rn.sample([chr(i) for i in range(97, 123)],x) for y in range(x)]
q[rn.randint(0,c)][rn.randint(0,c)] = rn.randint(0,9)
for r in q:
 f(r, n)
f(q, n)
print(q)

Try it Online!

Thanks to HyperNeurtrino, Ourous and Heiteria this shrunk down to 193 bytes (see comments). However, TFeld correctly pointed out that multiple calls to sample aren't guaranteeing at least N different characters.

That stuff in mind, try this new version that should guarantee at least N different characters per run.

Python3, 265 260 bytes, at least N distinct characters

from random import *
n=int(input())
x=-(-n//2)
o=n%2
c=x+~o
i=randint
u=[chr(j+97)for j in range(26)]
z,q=u[:],
for y in [1]*x:
 shuffle(z)
 q+=[z[:x]]
 z=z[x:] if len(z[x:])>=x else u[:]
q[i(0,c)][i(0,c)]=i(0,9)
for r in[q]+q:r.extend(r[~o::-1])
print(q)

Try it online!

share|improve this answer

edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
  – HyperNeutrino
  Aug 30 at 1:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
  – HyperNeutrino
  Aug 30 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can get it down to 236: Try it online!
  – ÎŸurous
  Aug 30 at 4:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even further, at 196: Try it online!
  – ÎŸurous
  Aug 30 at 4:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
  – TFeld
  Aug 30 at 8:35
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

Python3, 287 bytes

My first try at golfing something here; I'm sure someone can do far better:

import random as rn, math as m
n=int(input())
x,o=m.ceil(n/2),n%2
c=x-1-o
f=lambda l,n: l.extend((l[::-1], l[:-1][::-1])[o])
q=[rn.sample([chr(i) for i in range(97, 123)],x) for y in range(x)]
q[rn.randint(0,c)][rn.randint(0,c)] = rn.randint(0,9)
for r in q:
 f(r, n)
f(q, n)
print(q)

Try it Online!

Thanks to HyperNeurtrino, Ourous and Heiteria this shrunk down to 193 bytes (see comments). However, TFeld correctly pointed out that multiple calls to sample aren't guaranteeing at least N different characters.

That stuff in mind, try this new version that should guarantee at least N different characters per run.

Python3, 265 260 bytes, at least N distinct characters

from random import *
n=int(input())
x=-(-n//2)
o=n%2
c=x+~o
i=randint
u=[chr(j+97)for j in range(26)]
z,q=u[:],
for y in [1]*x:
 shuffle(z)
 q+=[z[:x]]
 z=z[x:] if len(z[x:])>=x else u[:]
q[i(0,c)][i(0,c)]=i(0,9)
for r in[q]+q:r.extend(r[~o::-1])
print(q)

Try it online!

share|improve this answer

edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
  – HyperNeutrino
  Aug 30 at 1:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
  – HyperNeutrino
  Aug 30 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can get it down to 236: Try it online!
  – ÎŸurous
  Aug 30 at 4:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even further, at 196: Try it online!
  – ÎŸurous
  Aug 30 at 4:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
  – TFeld
  Aug 30 at 8:35
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

up vote
6
down vote

Python3, 287 bytes

My first try at golfing something here; I'm sure someone can do far better:

import random as rn, math as m
n=int(input())
x,o=m.ceil(n/2),n%2
c=x-1-o
f=lambda l,n: l.extend((l[::-1], l[:-1][::-1])[o])
q=[rn.sample([chr(i) for i in range(97, 123)],x) for y in range(x)]
q[rn.randint(0,c)][rn.randint(0,c)] = rn.randint(0,9)
for r in q:
 f(r, n)
f(q, n)
print(q)

Try it Online!

Thanks to HyperNeurtrino, Ourous and Heiteria this shrunk down to 193 bytes (see comments). However, TFeld correctly pointed out that multiple calls to sample aren't guaranteeing at least N different characters.

That stuff in mind, try this new version that should guarantee at least N different characters per run.

Python3, 265 260 bytes, at least N distinct characters

from random import *
n=int(input())
x=-(-n//2)
o=n%2
c=x+~o
i=randint
u=[chr(j+97)for j in range(26)]
z,q=u[:],
for y in [1]*x:
 shuffle(z)
 q+=[z[:x]]
 z=z[x:] if len(z[x:])>=x else u[:]
q[i(0,c)][i(0,c)]=i(0,9)
for r in[q]+q:r.extend(r[~o::-1])
print(q)

Try it online!

share|improve this answer

edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

Python3, 287 bytes

My first try at golfing something here; I'm sure someone can do far better:

import random as rn, math as m
n=int(input())
x,o=m.ceil(n/2),n%2
c=x-1-o
f=lambda l,n: l.extend((l[::-1], l[:-1][::-1])[o])
q=[rn.sample([chr(i) for i in range(97, 123)],x) for y in range(x)]
q[rn.randint(0,c)][rn.randint(0,c)] = rn.randint(0,9)
for r in q:
 f(r, n)
f(q, n)
print(q)

Try it Online!

Thanks to HyperNeurtrino, Ourous and Heiteria this shrunk down to 193 bytes (see comments). However, TFeld correctly pointed out that multiple calls to sample aren't guaranteeing at least N different characters.

That stuff in mind, try this new version that should guarantee at least N different characters per run.

Python3, 265 260 bytes, at least N distinct characters

from random import *
n=int(input())
x=-(-n//2)
o=n%2
c=x+~o
i=randint
u=[chr(j+97)for j in range(26)]
z,q=u[:],
for y in [1]*x:
 shuffle(z)
 q+=[z[:x]]
 z=z[x:] if len(z[x:])>=x else u[:]
q[i(0,c)][i(0,c)]=i(0,9)
for r in[q]+q:r.extend(r[~o::-1])
print(q)

Try it online!

share|improve this answer

edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

share|improve this answer

share|improve this answer

edited Aug 30 at 20:58

edited Aug 30 at 20:58

edited Aug 30 at 20:58

answered Aug 30 at 1:51

souldeux

1614

answered Aug 30 at 1:51

souldeux

1614

answered Aug 30 at 1:51

souldeux

1614

1614

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
  – HyperNeutrino
  Aug 30 at 1:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
  – HyperNeutrino
  Aug 30 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can get it down to 236: Try it online!
  – ÎŸurous
  Aug 30 at 4:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even further, at 196: Try it online!
  – ÎŸurous
  Aug 30 at 4:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
  – TFeld
  Aug 30 at 8:35
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
  – HyperNeutrino
  Aug 30 at 1:56
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
  – HyperNeutrino
  Aug 30 at 1:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can get it down to 236: Try it online!
  – ÎŸurous
  Aug 30 at 4:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Even further, at 196: Try it online!
  – ÎŸurous
  Aug 30 at 4:58
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
  – TFeld
  Aug 30 at 8:35
  

  
  
  
  

1

1

Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
– HyperNeutrino
Aug 30 at 1:56

Welcome to PPCG! You can golf a few of the whitespaces out; there's no need to put spaces between symbols and symbols and letters. a[:-1][::-1] is fundamentally equivalent to a[:-2::-1], and you can import random as r instead of rn, and you can move the for loop into an inline expression. Try It Online!
– HyperNeutrino
Aug 30 at 1:56

2

2

You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
– HyperNeutrino
Aug 30 at 1:59

You can remove the math import by using -(-a // 2) instead of math.ceil(a / 2) which is basically negative floor-div of the negative (effectively ceiling). tio.run/##XY7LagMxDEX3/…
– HyperNeutrino
Aug 30 at 1:59

1

1

You can get it down to 236: Try it online!
– ÎŸurous
Aug 30 at 4:19

You can get it down to 236: Try it online!
– ÎŸurous
Aug 30 at 4:19

1

1

Even further, at 196: Try it online!
– ÎŸurous
Aug 30 at 4:58

Even further, at 196: Try it online!
– ÎŸurous
Aug 30 at 4:58

1

1

The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
– TFeld
Aug 30 at 8:35

The multiple sample()s don't guarantee that you get at least N different characters. I managed to get [['g', 'x', 'x', 'g'], [7, 'x', 'x', 7], [7, 'x', 'x', 7], ['g', 'x', 'x', 'g']] for N=4, which only has 3 distinct chars
– TFeld
Aug 30 at 8:35

 | 
show 5 more comments

up vote
3
down vote

APL (Dyalog Classic), 45 44 43 40 bytes

thanks @Adám for -1 byte

26(⎕a,⍺⍴⎕d)[⌈∘⊖⍨⌈∘⌽⍨⍺+@(?⊂⌊⍵÷2)?⍵⍴⍺],⍨

Try it online!

uses ⌈ (max) of the matrix with its reflections to make it symmetric, so it's biased towards the latter part of the alphabet

the digit is chosen uniformly from 0...25 mod 10, so it has a small bias to lower values

share|improve this answer

edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
  – Adám
  Aug 30 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám clever!­­
  – ngn
  Aug 30 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, I just realised.
  – Adám
  Aug 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
  – Adám
  Aug 30 at 9:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
  – ngn
  Aug 30 at 9:11
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote

APL (Dyalog Classic), 45 44 43 40 bytes

thanks @Adám for -1 byte

26(⎕a,⍺⍴⎕d)[⌈∘⊖⍨⌈∘⌽⍨⍺+@(?⊂⌊⍵÷2)?⍵⍴⍺],⍨

Try it online!

uses ⌈ (max) of the matrix with its reflections to make it symmetric, so it's biased towards the latter part of the alphabet

the digit is chosen uniformly from 0...25 mod 10, so it has a small bias to lower values

share|improve this answer

edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
  – Adám
  Aug 30 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám clever!­­
  – ngn
  Aug 30 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, I just realised.
  – Adám
  Aug 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
  – Adám
  Aug 30 at 9:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
  – ngn
  Aug 30 at 9:11
  

  
  
  
  

 | 
show 1 more comment

up vote
3
down vote

up vote
3
down vote

APL (Dyalog Classic), 45 44 43 40 bytes

thanks @Adám for -1 byte

26(⎕a,⍺⍴⎕d)[⌈∘⊖⍨⌈∘⌽⍨⍺+@(?⊂⌊⍵÷2)?⍵⍴⍺],⍨

Try it online!

uses ⌈ (max) of the matrix with its reflections to make it symmetric, so it's biased towards the latter part of the alphabet

the digit is chosen uniformly from 0...25 mod 10, so it has a small bias to lower values

share|improve this answer

edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

APL (Dyalog Classic), 45 44 43 40 bytes

thanks @Adám for -1 byte

26(⎕a,⍺⍴⎕d)[⌈∘⊖⍨⌈∘⌽⍨⍺+@(?⊂⌊⍵÷2)?⍵⍴⍺],⍨

Try it online!

uses ⌈ (max) of the matrix with its reflections to make it symmetric, so it's biased towards the latter part of the alphabet

the digit is chosen uniformly from 0...25 mod 10, so it has a small bias to lower values

share|improve this answer

edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

share|improve this answer

share|improve this answer

edited Aug 30 at 9:29

edited Aug 30 at 9:29

edited Aug 30 at 9:29

answered Aug 30 at 8:54

ngn

6,11812256

answered Aug 30 at 8:54

ngn

6,11812256

answered Aug 30 at 8:54

ngn

6,11812256

6,11812256

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
  – Adám
  Aug 30 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám clever!­­
  – ngn
  Aug 30 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, I just realised.
  – Adám
  Aug 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
  – Adám
  Aug 30 at 9:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
  – ngn
  Aug 30 at 9:11
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
  – Adám
  Aug 30 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám clever!­­
  – ngn
  Aug 30 at 9:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yeah, I just realised.
  – Adám
  Aug 30 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
  – Adám
  Aug 30 at 9:10
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
  – ngn
  Aug 30 at 9:11
  

  
  
  
  

1

1

⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
– Adám
Aug 30 at 8:59

⌊2⍴⍵÷2)?⍵ ⍵⍴26]} → ⌊⍺⍵÷2)?⍺⍵⍴26]}⍨
– Adám
Aug 30 at 8:59

@Adám clever!­­
– ngn
Aug 30 at 9:06

@Adám clever!­­
– ngn
Aug 30 at 9:06

Yeah, I just realised.
– Adám
Aug 30 at 9:09

Yeah, I just realised.
– Adám
Aug 30 at 9:09

If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
– Adám
Aug 30 at 9:10

If I'm not mistaken, you can change ⌊⍺⍵÷2 → ⍺⍵.
– Adám
Aug 30 at 9:10

@Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
– ngn
Aug 30 at 9:11

@Adám I can't - if N is odd, the digit might end up in the centre and there'd be only 1 row/column containing it
– ngn
Aug 30 at 9:11

 | 
show 1 more comment

up vote
3
down vote

Japt, 31 bytes (Fixed digit position)

;
/2 c
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv

Try it online!

---

Japt, 41 bytes (Random digit position)

;
/2 c
VÆVÆBö}ÃgMq´VÉ ,MqVÉ @Mq9îêUvÃêUv

Try it online!

---

Explanation

; Change to new vars
/2 c set implicit var V equal to implicit var U / 2 rounded up
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv Main function

VÆ Range from 0 to V and map
 VÆ Range from 0 to V and map
 Bö}Ã return random char from alphabet
 gT0@ map upper-left corner
 Mq9Ã return random number
 ®êUv horizontal mirror
 êUv vertical mirror

share|improve this answer

edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
  – Kevin Cruijssen
  Aug 30 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
  – Luis felipe De jesus Munoz
  Aug 30 at 11:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
  – Kevin Cruijssen
  Aug 30 at 12:03
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

Japt, 31 bytes (Fixed digit position)

;
/2 c
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv

Try it online!

---

Japt, 41 bytes (Random digit position)

;
/2 c
VÆVÆBö}ÃgMq´VÉ ,MqVÉ @Mq9îêUvÃêUv

Try it online!

---

Explanation

; Change to new vars
/2 c set implicit var V equal to implicit var U / 2 rounded up
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv Main function

VÆ Range from 0 to V and map
 VÆ Range from 0 to V and map
 Bö}Ã return random char from alphabet
 gT0@ map upper-left corner
 Mq9Ã return random number
 ®êUv horizontal mirror
 êUv vertical mirror

share|improve this answer

edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
  – Kevin Cruijssen
  Aug 30 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
  – Luis felipe De jesus Munoz
  Aug 30 at 11:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
  – Kevin Cruijssen
  Aug 30 at 12:03
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Japt, 31 bytes (Fixed digit position)

;
/2 c
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv

Try it online!

---

Japt, 41 bytes (Random digit position)

;
/2 c
VÆVÆBö}ÃgMq´VÉ ,MqVÉ @Mq9îêUvÃêUv

Try it online!

---

Explanation

; Change to new vars
/2 c set implicit var V equal to implicit var U / 2 rounded up
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv Main function

VÆ Range from 0 to V and map
 VÆ Range from 0 to V and map
 Bö}Ã return random char from alphabet
 gT0@ map upper-left corner
 Mq9Ã return random number
 ®êUv horizontal mirror
 êUv vertical mirror

share|improve this answer

edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

Japt, 31 bytes (Fixed digit position)

;
/2 c
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv

Try it online!

---

Japt, 41 bytes (Random digit position)

;
/2 c
VÆVÆBö}ÃgMq´VÉ ,MqVÉ @Mq9îêUvÃêUv

Try it online!

---

Explanation

; Change to new vars
/2 c set implicit var V equal to implicit var U / 2 rounded up
VÆVÆBö}ÃgT0@Mq9îêUvÃêUv Main function

VÆ Range from 0 to V and map
 VÆ Range from 0 to V and map
 Bö}Ã return random char from alphabet
 gT0@ map upper-left corner
 Mq9Ã return random number
 ®êUv horizontal mirror
 êUv vertical mirror

share|improve this answer

edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

share|improve this answer

share|improve this answer

edited Aug 30 at 12:35

edited Aug 30 at 12:35

edited Aug 30 at 12:35

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

answered Aug 29 at 22:38

Luis felipe De jesus Munoz

2,9511044

2,9511044

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
  – Kevin Cruijssen
  Aug 30 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
  – Luis felipe De jesus Munoz
  Aug 30 at 11:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
  – Kevin Cruijssen
  Aug 30 at 12:03
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
  – Kevin Cruijssen
  Aug 30 at 11:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
  – Luis felipe De jesus Munoz
  Aug 30 at 11:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
  – Kevin Cruijssen
  Aug 30 at 12:03
  
  

  
  
  
  

Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
– Kevin Cruijssen
Aug 30 at 11:50

Your digits are currently always inserted at the same place. Based on the challenge, the position of the digits should be random as well (and may not be in the middle row and/or column for odd inputs due to rule 4).
– Kevin Cruijssen
Aug 30 at 11:50

@KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
– Luis felipe De jesus Munoz
Aug 30 at 11:54

@KevinCruijssen I dont see where the challenge say the number position must be random as well, I'll ask OP for clarification though
– Luis felipe De jesus Munoz
Aug 30 at 11:54

1

1

Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
– Kevin Cruijssen
Aug 30 at 12:03

Ah, you're indeed right. I saw it's random in all other answer, so I might have falsely assumed it's mandatory. We'll see what OP says. I actually hope fixed it allowed, it would make it a lot easier to fix that problem for my prepared answer.. ;)
– Kevin Cruijssen
Aug 30 at 12:03

add a comment | 

up vote
2
down vote

Python 2, 259 bytes

from random import*
n=input();c=choice;r=range
w,W=n/2,-~n/2
o=n%2
A=map(chr,r(97,123))
l=[c(r(10))]+sample(A,n)+[c(A)for _ in' '*w*w]
l,e=l[:w*w],l[w*w:W*W]
shuffle(l)
l=[l[w*i:w*-~i]+e[i:i+1]for i in range(w)]+[e[-W:]]
for r in l+l[~o::-1]:print r+r[~o::-1]

Try it online!

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
  – Teck-freak
  Sep 3 at 17:09
  

  
  
  
  

add a comment | 

up vote
2
down vote

Python 2, 259 bytes

from random import*
n=input();c=choice;r=range
w,W=n/2,-~n/2
o=n%2
A=map(chr,r(97,123))
l=[c(r(10))]+sample(A,n)+[c(A)for _ in' '*w*w]
l,e=l[:w*w],l[w*w:W*W]
shuffle(l)
l=[l[w*i:w*-~i]+e[i:i+1]for i in range(w)]+[e[-W:]]
for r in l+l[~o::-1]:print r+r[~o::-1]

Try it online!

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
  – Teck-freak
  Sep 3 at 17:09
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python 2, 259 bytes

from random import*
n=input();c=choice;r=range
w,W=n/2,-~n/2
o=n%2
A=map(chr,r(97,123))
l=[c(r(10))]+sample(A,n)+[c(A)for _ in' '*w*w]
l,e=l[:w*w],l[w*w:W*W]
shuffle(l)
l=[l[w*i:w*-~i]+e[i:i+1]for i in range(w)]+[e[-W:]]
for r in l+l[~o::-1]:print r+r[~o::-1]

Try it online!

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

Python 2, 259 bytes

from random import*
n=input();c=choice;r=range
w,W=n/2,-~n/2
o=n%2
A=map(chr,r(97,123))
l=[c(r(10))]+sample(A,n)+[c(A)for _ in' '*w*w]
l,e=l[:w*w],l[w*w:W*W]
shuffle(l)
l=[l[w*i:w*-~i]+e[i:i+1]for i in range(w)]+[e[-W:]]
for r in l+l[~o::-1]:print r+r[~o::-1]

Try it online!

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

share|improve this answer

share|improve this answer

answered Aug 30 at 9:20

TFeld

11.2k2833

answered Aug 30 at 9:20

TFeld

11.2k2833

answered Aug 30 at 9:20

TFeld

11.2k2833

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
  – Teck-freak
  Sep 3 at 17:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
  – Teck-freak
  Sep 3 at 17:09
  

  
  
  
  

Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
– Teck-freak
Sep 3 at 17:09

Is using ints directly allowed? Cool idea on the ~ by the way. I was thinking of that too, but I'm not yet really used to it.
– Teck-freak
Sep 3 at 17:09

add a comment | 

up vote
2
down vote

05AB1E, 29 40 38 bytes

A.rs;ò©n∍9ÝΩ®DnαLʒ®%Ā}<Ωǝ®ô»¹Éi.º.∊ëº∊

+11 bytes to fix the digit being at a random position while still keeping rule 3 in mind for odd inputs..

-2 bytes thanks to @MagicOctopusUrn, changing îï to ò and changing the position of the ».

Try it online of verify some more test cases.

Old (29 27 bytes) answer where the digit positions where always in the corners:

A.rs;ò©n∍¦9ÝΩì®ô»¹Éi.º.∊ëº∊

Try it online or verify some more test cases.

Explanation:

A # Take the lowercase alphabet
 .r # Randomly shuffle it
 # i.e. "abcdefghijklmnopqrstuvwxyz" → "uovqxrcijfgyzlbpmhatnkwsed"
s # Swap so the (implicit) input is at the top of the stack
 ; # Halve the input
 # i.e. 7 → 3.5
 ò # Bankers rounding to the nearest integer
 # i.e. 3.5 → 4
 © # And save this number in the register
 n # Take its square
 # i.e. 4 → 16
 ∍ # Shorten the shuffled alphabet to that length
 # i.e. "uovqxrcijfgyzlbpmhatnkwsed" and 16 → "uovqxrcijfgyzlbp"
9ÝΩ # Take a random digit in the range [0,9]
 # i.e. 3
 ®Dnα # Take the difference between the saved number and its square:
 # i.e. 4 and 16 → 12
 L # Create a list in the range [1,n]
 # i.e. 12 → [1,2,3,4,5,6,7,8,9,10,11,12]
ʒ } # Filter this list by:
 ®%Ā # Remove any number that's divisible by the number we've saved
 # i.e. [1,2,3,4,5,6,7,8,9,10,11,12] and 4 → [1,2,3,5,6,7,9,10,11]
 < # Decrease each by 1 (to make it 0-indexed)
 # i.e. [1,2,3,5,6,7,9,10,11] → [0,1,2,3,5,6,7,9,10]
 Ω # Take a random item from this list
 # i.e. [0,1,2,3,5,6,7,9,10] → 6
 ǝ # Replace the character at this (0-indexed) position with the digit
 # i.e. "uovqxrcijfgyzlbp" and 3 and 6 → "uovqxr3ijfgyzlbp"
®ô # Split the string into parts of length equal to the number we've saved
 # i.e. "uovqxr3ijfgyzlbp" and 4 → ["uovq","xr3i","jfgy","zlbp"]
 » # Join them by new-lines (this is done implicitly in the legacy version)
 # i.e. ["uovq","xr3i","jfgy","zlbp"] → "uovqnxr3injfgynzlbp"
 ¹Éi # If the input is odd:
 # i.e. 7 → 1 (truthy)
 .º # Intersect mirror the individual items
 # i.e. "uovqnxr3injfgynzlbp"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 .∊ # And intersect vertically mirror the whole thing
 # i.e. "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblznjfgygfjnxr3i3rxnuovqvou"
 ë # Else (input was even):
 º∊ # Do the same, but with non-intersecting mirrors

share|improve this answer

edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also save 2 bytes with the legacy version since it doesn't require »
  – Emigna
  Aug 30 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
  – Kevin Cruijssen
  Aug 30 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
  – Kevin Cruijssen
  Aug 30 at 15:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
  – Magic Octopus Urn
  Aug 30 at 15:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
  – Kevin Cruijssen
  Aug 30 at 15:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

05AB1E, 29 40 38 bytes

A.rs;ò©n∍9ÝΩ®DnαLʒ®%Ā}<Ωǝ®ô»¹Éi.º.∊ëº∊

+11 bytes to fix the digit being at a random position while still keeping rule 3 in mind for odd inputs..

-2 bytes thanks to @MagicOctopusUrn, changing îï to ò and changing the position of the ».

Try it online of verify some more test cases.

Old (29 27 bytes) answer where the digit positions where always in the corners:

A.rs;ò©n∍¦9ÝΩì®ô»¹Éi.º.∊ëº∊

Try it online or verify some more test cases.

Explanation:

A # Take the lowercase alphabet
 .r # Randomly shuffle it
 # i.e. "abcdefghijklmnopqrstuvwxyz" → "uovqxrcijfgyzlbpmhatnkwsed"
s # Swap so the (implicit) input is at the top of the stack
 ; # Halve the input
 # i.e. 7 → 3.5
 ò # Bankers rounding to the nearest integer
 # i.e. 3.5 → 4
 © # And save this number in the register
 n # Take its square
 # i.e. 4 → 16
 ∍ # Shorten the shuffled alphabet to that length
 # i.e. "uovqxrcijfgyzlbpmhatnkwsed" and 16 → "uovqxrcijfgyzlbp"
9ÝΩ # Take a random digit in the range [0,9]
 # i.e. 3
 ®Dnα # Take the difference between the saved number and its square:
 # i.e. 4 and 16 → 12
 L # Create a list in the range [1,n]
 # i.e. 12 → [1,2,3,4,5,6,7,8,9,10,11,12]
ʒ } # Filter this list by:
 ®%Ā # Remove any number that's divisible by the number we've saved
 # i.e. [1,2,3,4,5,6,7,8,9,10,11,12] and 4 → [1,2,3,5,6,7,9,10,11]
 < # Decrease each by 1 (to make it 0-indexed)
 # i.e. [1,2,3,5,6,7,9,10,11] → [0,1,2,3,5,6,7,9,10]
 Ω # Take a random item from this list
 # i.e. [0,1,2,3,5,6,7,9,10] → 6
 ǝ # Replace the character at this (0-indexed) position with the digit
 # i.e. "uovqxrcijfgyzlbp" and 3 and 6 → "uovqxr3ijfgyzlbp"
®ô # Split the string into parts of length equal to the number we've saved
 # i.e. "uovqxr3ijfgyzlbp" and 4 → ["uovq","xr3i","jfgy","zlbp"]
 » # Join them by new-lines (this is done implicitly in the legacy version)
 # i.e. ["uovq","xr3i","jfgy","zlbp"] → "uovqnxr3injfgynzlbp"
 ¹Éi # If the input is odd:
 # i.e. 7 → 1 (truthy)
 .º # Intersect mirror the individual items
 # i.e. "uovqnxr3injfgynzlbp"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 .∊ # And intersect vertically mirror the whole thing
 # i.e. "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblznjfgygfjnxr3i3rxnuovqvou"
 ë # Else (input was even):
 º∊ # Do the same, but with non-intersecting mirrors

share|improve this answer

edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also save 2 bytes with the legacy version since it doesn't require »
  – Emigna
  Aug 30 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
  – Kevin Cruijssen
  Aug 30 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
  – Kevin Cruijssen
  Aug 30 at 15:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
  – Magic Octopus Urn
  Aug 30 at 15:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
  – Kevin Cruijssen
  Aug 30 at 15:43
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

05AB1E, 29 40 38 bytes

A.rs;ò©n∍9ÝΩ®DnαLʒ®%Ā}<Ωǝ®ô»¹Éi.º.∊ëº∊

+11 bytes to fix the digit being at a random position while still keeping rule 3 in mind for odd inputs..

-2 bytes thanks to @MagicOctopusUrn, changing îï to ò and changing the position of the ».

Try it online of verify some more test cases.

Old (29 27 bytes) answer where the digit positions where always in the corners:

A.rs;ò©n∍¦9ÝΩì®ô»¹Éi.º.∊ëº∊

Try it online or verify some more test cases.

Explanation:

A # Take the lowercase alphabet
 .r # Randomly shuffle it
 # i.e. "abcdefghijklmnopqrstuvwxyz" → "uovqxrcijfgyzlbpmhatnkwsed"
s # Swap so the (implicit) input is at the top of the stack
 ; # Halve the input
 # i.e. 7 → 3.5
 ò # Bankers rounding to the nearest integer
 # i.e. 3.5 → 4
 © # And save this number in the register
 n # Take its square
 # i.e. 4 → 16
 ∍ # Shorten the shuffled alphabet to that length
 # i.e. "uovqxrcijfgyzlbpmhatnkwsed" and 16 → "uovqxrcijfgyzlbp"
9ÝΩ # Take a random digit in the range [0,9]
 # i.e. 3
 ®Dnα # Take the difference between the saved number and its square:
 # i.e. 4 and 16 → 12
 L # Create a list in the range [1,n]
 # i.e. 12 → [1,2,3,4,5,6,7,8,9,10,11,12]
ʒ } # Filter this list by:
 ®%Ā # Remove any number that's divisible by the number we've saved
 # i.e. [1,2,3,4,5,6,7,8,9,10,11,12] and 4 → [1,2,3,5,6,7,9,10,11]
 < # Decrease each by 1 (to make it 0-indexed)
 # i.e. [1,2,3,5,6,7,9,10,11] → [0,1,2,3,5,6,7,9,10]
 Ω # Take a random item from this list
 # i.e. [0,1,2,3,5,6,7,9,10] → 6
 ǝ # Replace the character at this (0-indexed) position with the digit
 # i.e. "uovqxrcijfgyzlbp" and 3 and 6 → "uovqxr3ijfgyzlbp"
®ô # Split the string into parts of length equal to the number we've saved
 # i.e. "uovqxr3ijfgyzlbp" and 4 → ["uovq","xr3i","jfgy","zlbp"]
 » # Join them by new-lines (this is done implicitly in the legacy version)
 # i.e. ["uovq","xr3i","jfgy","zlbp"] → "uovqnxr3injfgynzlbp"
 ¹Éi # If the input is odd:
 # i.e. 7 → 1 (truthy)
 .º # Intersect mirror the individual items
 # i.e. "uovqnxr3injfgynzlbp"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 .∊ # And intersect vertically mirror the whole thing
 # i.e. "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblznjfgygfjnxr3i3rxnuovqvou"
 ë # Else (input was even):
 º∊ # Do the same, but with non-intersecting mirrors

share|improve this answer

edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

05AB1E, 29 40 38 bytes

A.rs;ò©n∍9ÝΩ®DnαLʒ®%Ā}<Ωǝ®ô»¹Éi.º.∊ëº∊

+11 bytes to fix the digit being at a random position while still keeping rule 3 in mind for odd inputs..

-2 bytes thanks to @MagicOctopusUrn, changing îï to ò and changing the position of the ».

Try it online of verify some more test cases.

Old (29 27 bytes) answer where the digit positions where always in the corners:

A.rs;ò©n∍¦9ÝΩì®ô»¹Éi.º.∊ëº∊

Try it online or verify some more test cases.

Explanation:

A # Take the lowercase alphabet
 .r # Randomly shuffle it
 # i.e. "abcdefghijklmnopqrstuvwxyz" → "uovqxrcijfgyzlbpmhatnkwsed"
s # Swap so the (implicit) input is at the top of the stack
 ; # Halve the input
 # i.e. 7 → 3.5
 ò # Bankers rounding to the nearest integer
 # i.e. 3.5 → 4
 © # And save this number in the register
 n # Take its square
 # i.e. 4 → 16
 ∍ # Shorten the shuffled alphabet to that length
 # i.e. "uovqxrcijfgyzlbpmhatnkwsed" and 16 → "uovqxrcijfgyzlbp"
9ÝΩ # Take a random digit in the range [0,9]
 # i.e. 3
 ®Dnα # Take the difference between the saved number and its square:
 # i.e. 4 and 16 → 12
 L # Create a list in the range [1,n]
 # i.e. 12 → [1,2,3,4,5,6,7,8,9,10,11,12]
ʒ } # Filter this list by:
 ®%Ā # Remove any number that's divisible by the number we've saved
 # i.e. [1,2,3,4,5,6,7,8,9,10,11,12] and 4 → [1,2,3,5,6,7,9,10,11]
 < # Decrease each by 1 (to make it 0-indexed)
 # i.e. [1,2,3,5,6,7,9,10,11] → [0,1,2,3,5,6,7,9,10]
 Ω # Take a random item from this list
 # i.e. [0,1,2,3,5,6,7,9,10] → 6
 ǝ # Replace the character at this (0-indexed) position with the digit
 # i.e. "uovqxrcijfgyzlbp" and 3 and 6 → "uovqxr3ijfgyzlbp"
®ô # Split the string into parts of length equal to the number we've saved
 # i.e. "uovqxr3ijfgyzlbp" and 4 → ["uovq","xr3i","jfgy","zlbp"]
 » # Join them by new-lines (this is done implicitly in the legacy version)
 # i.e. ["uovq","xr3i","jfgy","zlbp"] → "uovqnxr3injfgynzlbp"
 ¹Éi # If the input is odd:
 # i.e. 7 → 1 (truthy)
 .º # Intersect mirror the individual items
 # i.e. "uovqnxr3injfgynzlbp"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 .∊ # And intersect vertically mirror the whole thing
 # i.e. "uovqvounxr3i3rxnjfgygfjnzlbpblz"
 # → "uovqvounxr3i3rxnjfgygfjnzlbpblznjfgygfjnxr3i3rxnuovqvou"
 ë # Else (input was even):
 º∊ # Do the same, but with non-intersecting mirrors

share|improve this answer

edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

share|improve this answer

share|improve this answer

edited Aug 30 at 15:49

edited Aug 30 at 15:49

edited Aug 30 at 15:49

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

answered Aug 30 at 12:25

Kevin Cruijssen

29.5k550162

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also save 2 bytes with the legacy version since it doesn't require »
  – Emigna
  Aug 30 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
  – Kevin Cruijssen
  Aug 30 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
  – Kevin Cruijssen
  Aug 30 at 15:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
  – Magic Octopus Urn
  Aug 30 at 15:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
  – Kevin Cruijssen
  Aug 30 at 15:43
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You could also save 2 bytes with the legacy version since it doesn't require »
  – Emigna
  Aug 30 at 13:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
  – Kevin Cruijssen
  Aug 30 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
  – Kevin Cruijssen
  Aug 30 at 15:32
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
  – Magic Octopus Urn
  Aug 30 at 15:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
  – Kevin Cruijssen
  Aug 30 at 15:43
  

  
  
  
  

You could also save 2 bytes with the legacy version since it doesn't require »
– Emigna
Aug 30 at 13:17

You could also save 2 bytes with the legacy version since it doesn't require »
– Emigna
Aug 30 at 13:17

@Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
– Kevin Cruijssen
Aug 30 at 14:50

@Emigna Verified with OP, and the position should indeed be random as well. Fixed for +11 bytes due to rule 3 with odd inputs.. >.> And 3 bytes could have been saved in the legacy because the ï was done implicitely as well. Unfortunately this doesn't apply to the 40-byte version because ∊ would insert instead of replace.
– Kevin Cruijssen
Aug 30 at 14:50

@MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
– Kevin Cruijssen
Aug 30 at 15:32

@MagicOctopusUrn The TIO you linked still contained my 29 bytes answer instead of 28, do you have the correct link? As for the failing for 2, the input is guaranteed to be 3 <= N <= 26.
– Kevin Cruijssen
Aug 30 at 15:32

1

1

@KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
– Magic Octopus Urn
Aug 30 at 15:36

@KevinCruijssen you're right, I'm a moron, here's the one I was working out: Try it online!
– Magic Octopus Urn
Aug 30 at 15:36

@MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
– Kevin Cruijssen
Aug 30 at 15:43

@MagicOctopusUrn Oh, didn't knew about that bankers rounding. That saves a byte in my current answer as well! :D And first appending a random digit and only then shuffling is a pretty smart approach as well. Not sure if it's 100% valid though, since you will always have the first n letters of the alphabet, instead of n random letters of the alphabet. And first joining by newlines and only then doing the mirrors saves a byte as well in mine. Thanks for -2 bytes! :) PS: One byte can be saved in your 28-byter by removing the trailing }. :)
– Kevin Cruijssen
Aug 30 at 15:43

add a comment | 

up vote
2
down vote

C (gcc), 198 197 bytes

Saved 1 byte thanks to ceilingcat (4 bytes actually but I lost 3 bytes fixing the bug which caused the program to print regular pattern for $N=26$.

#define A(x)(x<n/2?x:n-1-x)
#define R rand()
S(n,x,y)

Try it online!

Explanation:

// Coordinate conversion for symmetry
#define A (x) (x < n / 2 ? x : n - 1 - x)
// Get a random and seed
#define R rand()

S (n, x, y)

share|improve this answer

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
  – ceilingcat
  Sep 4 at 23:11
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

C (gcc), 198 197 bytes

Saved 1 byte thanks to ceilingcat (4 bytes actually but I lost 3 bytes fixing the bug which caused the program to print regular pattern for $N=26$.

#define A(x)(x<n/2?x:n-1-x)
#define R rand()
S(n,x,y)

Try it online!

Explanation:

// Coordinate conversion for symmetry
#define A (x) (x < n / 2 ? x : n - 1 - x)
// Get a random and seed
#define R rand()

S (n, x, y)

share|improve this answer

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
  – ceilingcat
  Sep 4 at 23:11
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

C (gcc), 198 197 bytes

Saved 1 byte thanks to ceilingcat (4 bytes actually but I lost 3 bytes fixing the bug which caused the program to print regular pattern for $N=26$.

#define A(x)(x<n/2?x:n-1-x)
#define R rand()
S(n,x,y)

Try it online!

Explanation:

// Coordinate conversion for symmetry
#define A (x) (x < n / 2 ? x : n - 1 - x)
// Get a random and seed
#define R rand()

S (n, x, y)

share|improve this answer

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

C (gcc), 198 197 bytes

Saved 1 byte thanks to ceilingcat (4 bytes actually but I lost 3 bytes fixing the bug which caused the program to print regular pattern for $N=26$.

#define A(x)(x<n/2?x:n-1-x)
#define R rand()
S(n,x,y)

Try it online!

Explanation:

// Coordinate conversion for symmetry
#define A (x) (x < n / 2 ? x : n - 1 - x)
// Get a random and seed
#define R rand()

S (n, x, y)

share|improve this answer

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

share|improve this answer

share|improve this answer

edited Sep 3 at 5:44

edited Sep 3 at 5:44

edited Sep 3 at 5:44

answered Aug 31 at 14:48

Max Yekhlakov

2415

answered Aug 31 at 14:48

Max Yekhlakov

2415

answered Aug 31 at 14:48

Max Yekhlakov

2415

2415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
  – ceilingcat
  Sep 4 at 23:11
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
  – ceilingcat
  Sep 4 at 23:11
  
  

  
  
  
  

Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
– ceilingcat
Sep 4 at 23:11

Suggest for(s[R%y+R%y*n]=48+R%10; instead of s[R%y+n*(R%y)]=48+R%10;for(;
– ceilingcat
Sep 4 at 23:11

add a comment | 

up vote
1
down vote

JavaScript (ES6), 213 209 206 bytes

n=>(a=,F=(x=y=d=c=0,R=k=>Math.random()*k|0,g=y=>(r=a[y]=a[y]||)[x]=r[n+~x]=v.toString(36))=>y<n/2?F(g(y,R[v=R(m=~-n/2)<!d&x<m&y<m?R(d=10):R(26)+10]=R[v]||++c,g(n+~y))&&++x<n/2?x:+!++y,R):!d|c<n?F():a)()

Try it online!

Commented

n => ( // n = input
 a = , // a = output matrix
 F = ( // F = main recursive function taking:
 x = y = // (x, y) = current coordinates
 d = c = 0, // d = digit flag; c = distinct character counter
 R = k => // R() = helper function to get a random value in [0,k[
 Math.random() * k | 0, // also used to store characters
 g = y => // g() = helper function to update the matrix
 (r = a[y] = a[y] || )[x] // with horizontal symmetry
 = r[n + ~x] = v.toString(36) // using the base-36 representation of v
 ) => //
 y < n / 2 ? // if we haven't reached the middle row(s) of the matrix:
 F( // do a recursive call to F():
 g( // invoke g() ...
 y, // ... on the current row
 R[v = // compute v = next value to be inserted
 R(m = ~-n/2) < !d & // we may insert a digit if no digit has been
 x < m & // inserted so far and the current coordinates are
 y < m ? // compatible: 2 distinct rows / 2 distinct columns
 R(d = 10) // if so, pick v in [0, 9] and update d
 : // else:
 R(26) + 10 // pick v in [10, 35] for a letter
 ] = R[v] || ++c, // set this character as used; update c accordingly
 g(n + ~y) // invoke g() on the mirror row
 ) && // end of outer call to g()
 ++x < n / 2 ? // if we haven't reached the middle column(s):
 x // use x + 1
 : // else
 +!++y, // increment y and reset x to 0
 R // explicitly pass R, as it is used for storage
 ) // end of recursive call to F()
 : // else:
 !d | c < n ? F() : a // either return the matrix or try again if it's invalid
)() // initial call to F()

share|improve this answer

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

add a comment | 

up vote
1
down vote

JavaScript (ES6), 213 209 206 bytes

n=>(a=,F=(x=y=d=c=0,R=k=>Math.random()*k|0,g=y=>(r=a[y]=a[y]||)[x]=r[n+~x]=v.toString(36))=>y<n/2?F(g(y,R[v=R(m=~-n/2)<!d&x<m&y<m?R(d=10):R(26)+10]=R[v]||++c,g(n+~y))&&++x<n/2?x:+!++y,R):!d|c<n?F():a)()

Try it online!

Commented

n => ( // n = input
 a = , // a = output matrix
 F = ( // F = main recursive function taking:
 x = y = // (x, y) = current coordinates
 d = c = 0, // d = digit flag; c = distinct character counter
 R = k => // R() = helper function to get a random value in [0,k[
 Math.random() * k | 0, // also used to store characters
 g = y => // g() = helper function to update the matrix
 (r = a[y] = a[y] || )[x] // with horizontal symmetry
 = r[n + ~x] = v.toString(36) // using the base-36 representation of v
 ) => //
 y < n / 2 ? // if we haven't reached the middle row(s) of the matrix:
 F( // do a recursive call to F():
 g( // invoke g() ...
 y, // ... on the current row
 R[v = // compute v = next value to be inserted
 R(m = ~-n/2) < !d & // we may insert a digit if no digit has been
 x < m & // inserted so far and the current coordinates are
 y < m ? // compatible: 2 distinct rows / 2 distinct columns
 R(d = 10) // if so, pick v in [0, 9] and update d
 : // else:
 R(26) + 10 // pick v in [10, 35] for a letter
 ] = R[v] || ++c, // set this character as used; update c accordingly
 g(n + ~y) // invoke g() on the mirror row
 ) && // end of outer call to g()
 ++x < n / 2 ? // if we haven't reached the middle column(s):
 x // use x + 1
 : // else
 +!++y, // increment y and reset x to 0
 R // explicitly pass R, as it is used for storage
 ) // end of recursive call to F()
 : // else:
 !d | c < n ? F() : a // either return the matrix or try again if it's invalid
)() // initial call to F()

share|improve this answer

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (ES6), 213 209 206 bytes

n=>(a=,F=(x=y=d=c=0,R=k=>Math.random()*k|0,g=y=>(r=a[y]=a[y]||)[x]=r[n+~x]=v.toString(36))=>y<n/2?F(g(y,R[v=R(m=~-n/2)<!d&x<m&y<m?R(d=10):R(26)+10]=R[v]||++c,g(n+~y))&&++x<n/2?x:+!++y,R):!d|c<n?F():a)()

Try it online!

Commented

n => ( // n = input
 a = , // a = output matrix
 F = ( // F = main recursive function taking:
 x = y = // (x, y) = current coordinates
 d = c = 0, // d = digit flag; c = distinct character counter
 R = k => // R() = helper function to get a random value in [0,k[
 Math.random() * k | 0, // also used to store characters
 g = y => // g() = helper function to update the matrix
 (r = a[y] = a[y] || )[x] // with horizontal symmetry
 = r[n + ~x] = v.toString(36) // using the base-36 representation of v
 ) => //
 y < n / 2 ? // if we haven't reached the middle row(s) of the matrix:
 F( // do a recursive call to F():
 g( // invoke g() ...
 y, // ... on the current row
 R[v = // compute v = next value to be inserted
 R(m = ~-n/2) < !d & // we may insert a digit if no digit has been
 x < m & // inserted so far and the current coordinates are
 y < m ? // compatible: 2 distinct rows / 2 distinct columns
 R(d = 10) // if so, pick v in [0, 9] and update d
 : // else:
 R(26) + 10 // pick v in [10, 35] for a letter
 ] = R[v] || ++c, // set this character as used; update c accordingly
 g(n + ~y) // invoke g() on the mirror row
 ) && // end of outer call to g()
 ++x < n / 2 ? // if we haven't reached the middle column(s):
 x // use x + 1
 : // else
 +!++y, // increment y and reset x to 0
 R // explicitly pass R, as it is used for storage
 ) // end of recursive call to F()
 : // else:
 !d | c < n ? F() : a // either return the matrix or try again if it's invalid
)() // initial call to F()

share|improve this answer

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

JavaScript (ES6), 213 209 206 bytes

n=>(a=,F=(x=y=d=c=0,R=k=>Math.random()*k|0,g=y=>(r=a[y]=a[y]||)[x]=r[n+~x]=v.toString(36))=>y<n/2?F(g(y,R[v=R(m=~-n/2)<!d&x<m&y<m?R(d=10):R(26)+10]=R[v]||++c,g(n+~y))&&++x<n/2?x:+!++y,R):!d|c<n?F():a)()

Try it online!

Commented

n => ( // n = input
 a = , // a = output matrix
 F = ( // F = main recursive function taking:
 x = y = // (x, y) = current coordinates
 d = c = 0, // d = digit flag; c = distinct character counter
 R = k => // R() = helper function to get a random value in [0,k[
 Math.random() * k | 0, // also used to store characters
 g = y => // g() = helper function to update the matrix
 (r = a[y] = a[y] || )[x] // with horizontal symmetry
 = r[n + ~x] = v.toString(36) // using the base-36 representation of v
 ) => //
 y < n / 2 ? // if we haven't reached the middle row(s) of the matrix:
 F( // do a recursive call to F():
 g( // invoke g() ...
 y, // ... on the current row
 R[v = // compute v = next value to be inserted
 R(m = ~-n/2) < !d & // we may insert a digit if no digit has been
 x < m & // inserted so far and the current coordinates are
 y < m ? // compatible: 2 distinct rows / 2 distinct columns
 R(d = 10) // if so, pick v in [0, 9] and update d
 : // else:
 R(26) + 10 // pick v in [10, 35] for a letter
 ] = R[v] || ++c, // set this character as used; update c accordingly
 g(n + ~y) // invoke g() on the mirror row
 ) && // end of outer call to g()
 ++x < n / 2 ? // if we haven't reached the middle column(s):
 x // use x + 1
 : // else
 +!++y, // increment y and reset x to 0
 R // explicitly pass R, as it is used for storage
 ) // end of recursive call to F()
 : // else:
 !d | c < n ? F() : a // either return the matrix or try again if it's invalid
)() // initial call to F()

share|improve this answer

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

share|improve this answer

share|improve this answer

edited Aug 30 at 10:26

edited Aug 30 at 10:26

edited Aug 30 at 10:26

answered Aug 30 at 8:29

Arnauld

63.6k580268

answered Aug 30 at 8:29

Arnauld

63.6k580268

answered Aug 30 at 8:29

Arnauld

63.6k580268

63.6k580268

add a comment | 

add a comment | 

up vote
1
down vote

Clean, 346 312 bytes

will golf more tomorrow

import StdEnv,Data.List,Math.Random,System.Time,System._Unsafe
$n#q=twice(transpose oq=zipWith((++)o reverse o drop(n-n/2*2))q q)[[(['a'..'z']++['0'..'9'])!!(c rem 36)\c<-genRandInt(toInt(accUnsafe(time)))]%(i*n/2,i*n/2+(n-1)/2)\i<-[1..(n+1)/2]]
|length(nub(flatten q))>=n&&sum[1\c<-q|any isDigit c]==2=q= $n

Try it online!

share|improve this answer

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

add a comment | 

up vote
1
down vote

Clean, 346 312 bytes

will golf more tomorrow

import StdEnv,Data.List,Math.Random,System.Time,System._Unsafe
$n#q=twice(transpose oq=zipWith((++)o reverse o drop(n-n/2*2))q q)[[(['a'..'z']++['0'..'9'])!!(c rem 36)\c<-genRandInt(toInt(accUnsafe(time)))]%(i*n/2,i*n/2+(n-1)/2)\i<-[1..(n+1)/2]]
|length(nub(flatten q))>=n&&sum[1\c<-q|any isDigit c]==2=q= $n

Try it online!

share|improve this answer

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

add a comment | 

up vote
1
down vote

up vote
1
down vote

Clean, 346 312 bytes

will golf more tomorrow

import StdEnv,Data.List,Math.Random,System.Time,System._Unsafe
$n#q=twice(transpose oq=zipWith((++)o reverse o drop(n-n/2*2))q q)[[(['a'..'z']++['0'..'9'])!!(c rem 36)\c<-genRandInt(toInt(accUnsafe(time)))]%(i*n/2,i*n/2+(n-1)/2)\i<-[1..(n+1)/2]]
|length(nub(flatten q))>=n&&sum[1\c<-q|any isDigit c]==2=q= $n

Try it online!

share|improve this answer

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

Clean, 346 312 bytes

will golf more tomorrow

import StdEnv,Data.List,Math.Random,System.Time,System._Unsafe
$n#q=twice(transpose oq=zipWith((++)o reverse o drop(n-n/2*2))q q)[[(['a'..'z']++['0'..'9'])!!(c rem 36)\c<-genRandInt(toInt(accUnsafe(time)))]%(i*n/2,i*n/2+(n-1)/2)\i<-[1..(n+1)/2]]
|length(nub(flatten q))>=n&&sum[1\c<-q|any isDigit c]==2=q= $n

Try it online!

share|improve this answer

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

share|improve this answer

share|improve this answer

edited Aug 30 at 12:26

edited Aug 30 at 12:26

edited Aug 30 at 12:26

answered Aug 30 at 1:36

Οurous

5,2031931

answered Aug 30 at 1:36

Οurous

5,2031931

answered Aug 30 at 1:36

Οurous

5,2031931

5,2031931

add a comment | 

add a comment | 

up vote
1
down vote

Python 3, 197 bytes

As mentioned by @Emigna, doesn't work for odd values of N (I didn't understand the question properly)

from random import*
def m(N):M=N//2;E=reversed;R=range;B=[randint(48,57),*(sample(R(97,123),N)*N)][:M*M];shuffle(B);r=R(M);m=[k+[*E(k)]for k in[[chr(B.pop())for i in r]for j in r]];m+=E(m);return m

Try it online!

I do think the calls to randint() + sample() + shuffle() are too much, and getting rid of in-place shuffling would be great :)

I'm pretty sure this part (that selects the letters & digit) could be golfed a bit more.

share|improve this answer

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doesn't seem correct for odd N.
  – Emigna
  Aug 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
  – etene
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  These are some examples of odd symmetrical matrices.
  – Emigna
  Aug 30 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
  – etene
  Aug 30 at 13:59
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 3, 197 bytes

As mentioned by @Emigna, doesn't work for odd values of N (I didn't understand the question properly)

from random import*
def m(N):M=N//2;E=reversed;R=range;B=[randint(48,57),*(sample(R(97,123),N)*N)][:M*M];shuffle(B);r=R(M);m=[k+[*E(k)]for k in[[chr(B.pop())for i in r]for j in r]];m+=E(m);return m

Try it online!

I do think the calls to randint() + sample() + shuffle() are too much, and getting rid of in-place shuffling would be great :)

I'm pretty sure this part (that selects the letters & digit) could be golfed a bit more.

share|improve this answer

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doesn't seem correct for odd N.
  – Emigna
  Aug 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
  – etene
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  These are some examples of odd symmetrical matrices.
  – Emigna
  Aug 30 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
  – etene
  Aug 30 at 13:59
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 3, 197 bytes

As mentioned by @Emigna, doesn't work for odd values of N (I didn't understand the question properly)

from random import*
def m(N):M=N//2;E=reversed;R=range;B=[randint(48,57),*(sample(R(97,123),N)*N)][:M*M];shuffle(B);r=R(M);m=[k+[*E(k)]for k in[[chr(B.pop())for i in r]for j in r]];m+=E(m);return m

Try it online!

I do think the calls to randint() + sample() + shuffle() are too much, and getting rid of in-place shuffling would be great :)

I'm pretty sure this part (that selects the letters & digit) could be golfed a bit more.

share|improve this answer

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

Python 3, 197 bytes

As mentioned by @Emigna, doesn't work for odd values of N (I didn't understand the question properly)

from random import*
def m(N):M=N//2;E=reversed;R=range;B=[randint(48,57),*(sample(R(97,123),N)*N)][:M*M];shuffle(B);r=R(M);m=[k+[*E(k)]for k in[[chr(B.pop())for i in r]for j in r]];m+=E(m);return m

Try it online!

I do think the calls to randint() + sample() + shuffle() are too much, and getting rid of in-place shuffling would be great :)

I'm pretty sure this part (that selects the letters & digit) could be golfed a bit more.

share|improve this answer

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

share|improve this answer

share|improve this answer

edited Aug 30 at 14:02

edited Aug 30 at 14:02

edited Aug 30 at 14:02

answered Aug 30 at 12:59

etene

4487

answered Aug 30 at 12:59

etene

4487

answered Aug 30 at 12:59

etene

4487

4487

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doesn't seem correct for odd N.
  – Emigna
  Aug 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
  – etene
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  These are some examples of odd symmetrical matrices.
  – Emigna
  Aug 30 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
  – etene
  Aug 30 at 13:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Doesn't seem correct for odd N.
  – Emigna
  Aug 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
  – etene
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  These are some examples of odd symmetrical matrices.
  – Emigna
  Aug 30 at 13:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
  – etene
  Aug 30 at 13:59
  

  
  
  
  

Doesn't seem correct for odd N.
– Emigna
Aug 30 at 13:50

Doesn't seem correct for odd N.
– Emigna
Aug 30 at 13:50

Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
– etene
Aug 30 at 13:51

Damn, I had just assumed N would always be even since I don't get how the matrix could be symmetrical if it's odd !
– etene
Aug 30 at 13:51

1

1

These are some examples of odd symmetrical matrices.
– Emigna
Aug 30 at 13:57

These are some examples of odd symmetrical matrices.
– Emigna
Aug 30 at 13:57

Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
– etene
Aug 30 at 13:59

Okay, thanks, I hadn't seen it that way ! Well I guess my answer is worthless as is then.
– etene
Aug 30 at 13:59

add a comment | 

up vote
1
down vote

Python 2, 275 266 bytes

from random import*
def f(n):
 R=range;C=choice;A=map(chr,R(97,123));b=N=n-n/2;c=`C(R(10))`;s=[c]+sample(A,n-1)+[C(A)for i in R(N*N-n)]
 while b:shuffle(s);i=s.index(c);b=n%2>(i<N*N-N>N-1>i%N)
 a=[r+r[~(n%2)::-1]for r in[s[i::N]for i in R(N)]];return a+a[~(n%2)::-1]

Try it online!

Returns the array as a list of lists of characters. To satisfy Rule 1, we set up a pool of characters:

s = [c] # the unique digit...
 + sample(A,n-1) # then sample without replacement `n-1` chars in a-z, 
 # so we have `n` distinct chars
 + [C(A)for i in R(N*N-n)] # and fill out the rest with any in a-z

The next tricky bit is rule 3: there must be exactly 2 columns and rows having a digit; this means for n odd, that the chosen digit may not appear in the middle column or middle row. Since we construct the array using a twice reflected square sub array s, that is accomplished here by using:

while b: # to save a couple bytes, `b` is initialized 
 # to `N`, which is greater than 0.
 shuffle(s) # shuffle at least once...
 i = s.index(c) # c is the unique digit used
 b = n%2 
 > # if n is even, 0>(any boolean) will be false,
 # so exit the loop; otherwise n odd, and we are
 # evaluating '1 > some boolean', which is equivalent 
 # to 'not (some boolean)'
 (i<N*N-N # i is not the last column of s...
 > # shortcut for ' and ', since N*N-N is always > N-1
 N-1>i%N) # is not the last row of s

i.e., shuffle at least once; and then, if n is odd, keep looping if the digit is in the last column or the last row of s.

share|improve this answer

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

add a comment | 

up vote
1
down vote

Python 2, 275 266 bytes

from random import*
def f(n):
 R=range;C=choice;A=map(chr,R(97,123));b=N=n-n/2;c=`C(R(10))`;s=[c]+sample(A,n-1)+[C(A)for i in R(N*N-n)]
 while b:shuffle(s);i=s.index(c);b=n%2>(i<N*N-N>N-1>i%N)
 a=[r+r[~(n%2)::-1]for r in[s[i::N]for i in R(N)]];return a+a[~(n%2)::-1]

Try it online!

Returns the array as a list of lists of characters. To satisfy Rule 1, we set up a pool of characters:

s = [c] # the unique digit...
 + sample(A,n-1) # then sample without replacement `n-1` chars in a-z, 
 # so we have `n` distinct chars
 + [C(A)for i in R(N*N-n)] # and fill out the rest with any in a-z

The next tricky bit is rule 3: there must be exactly 2 columns and rows having a digit; this means for n odd, that the chosen digit may not appear in the middle column or middle row. Since we construct the array using a twice reflected square sub array s, that is accomplished here by using:

while b: # to save a couple bytes, `b` is initialized 
 # to `N`, which is greater than 0.
 shuffle(s) # shuffle at least once...
 i = s.index(c) # c is the unique digit used
 b = n%2 
 > # if n is even, 0>(any boolean) will be false,
 # so exit the loop; otherwise n odd, and we are
 # evaluating '1 > some boolean', which is equivalent 
 # to 'not (some boolean)'
 (i<N*N-N # i is not the last column of s...
 > # shortcut for ' and ', since N*N-N is always > N-1
 N-1>i%N) # is not the last row of s

i.e., shuffle at least once; and then, if n is odd, keep looping if the digit is in the last column or the last row of s.

share|improve this answer

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 2, 275 266 bytes

from random import*
def f(n):
 R=range;C=choice;A=map(chr,R(97,123));b=N=n-n/2;c=`C(R(10))`;s=[c]+sample(A,n-1)+[C(A)for i in R(N*N-n)]
 while b:shuffle(s);i=s.index(c);b=n%2>(i<N*N-N>N-1>i%N)
 a=[r+r[~(n%2)::-1]for r in[s[i::N]for i in R(N)]];return a+a[~(n%2)::-1]

Try it online!

Returns the array as a list of lists of characters. To satisfy Rule 1, we set up a pool of characters:

s = [c] # the unique digit...
 + sample(A,n-1) # then sample without replacement `n-1` chars in a-z, 
 # so we have `n` distinct chars
 + [C(A)for i in R(N*N-n)] # and fill out the rest with any in a-z

The next tricky bit is rule 3: there must be exactly 2 columns and rows having a digit; this means for n odd, that the chosen digit may not appear in the middle column or middle row. Since we construct the array using a twice reflected square sub array s, that is accomplished here by using:

while b: # to save a couple bytes, `b` is initialized 
 # to `N`, which is greater than 0.
 shuffle(s) # shuffle at least once...
 i = s.index(c) # c is the unique digit used
 b = n%2 
 > # if n is even, 0>(any boolean) will be false,
 # so exit the loop; otherwise n odd, and we are
 # evaluating '1 > some boolean', which is equivalent 
 # to 'not (some boolean)'
 (i<N*N-N # i is not the last column of s...
 > # shortcut for ' and ', since N*N-N is always > N-1
 N-1>i%N) # is not the last row of s

i.e., shuffle at least once; and then, if n is odd, keep looping if the digit is in the last column or the last row of s.

share|improve this answer

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

Python 2, 275 266 bytes

from random import*
def f(n):
 R=range;C=choice;A=map(chr,R(97,123));b=N=n-n/2;c=`C(R(10))`;s=[c]+sample(A,n-1)+[C(A)for i in R(N*N-n)]
 while b:shuffle(s);i=s.index(c);b=n%2>(i<N*N-N>N-1>i%N)
 a=[r+r[~(n%2)::-1]for r in[s[i::N]for i in R(N)]];return a+a[~(n%2)::-1]

Try it online!

Returns the array as a list of lists of characters. To satisfy Rule 1, we set up a pool of characters:

s = [c] # the unique digit...
 + sample(A,n-1) # then sample without replacement `n-1` chars in a-z, 
 # so we have `n` distinct chars
 + [C(A)for i in R(N*N-n)] # and fill out the rest with any in a-z

The next tricky bit is rule 3: there must be exactly 2 columns and rows having a digit; this means for n odd, that the chosen digit may not appear in the middle column or middle row. Since we construct the array using a twice reflected square sub array s, that is accomplished here by using:

while b: # to save a couple bytes, `b` is initialized 
 # to `N`, which is greater than 0.
 shuffle(s) # shuffle at least once...
 i = s.index(c) # c is the unique digit used
 b = n%2 
 > # if n is even, 0>(any boolean) will be false,
 # so exit the loop; otherwise n odd, and we are
 # evaluating '1 > some boolean', which is equivalent 
 # to 'not (some boolean)'
 (i<N*N-N # i is not the last column of s...
 > # shortcut for ' and ', since N*N-N is always > N-1
 N-1>i%N) # is not the last row of s

i.e., shuffle at least once; and then, if n is odd, keep looping if the digit is in the last column or the last row of s.

share|improve this answer

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

share|improve this answer

share|improve this answer

edited Sep 3 at 6:19

edited Sep 3 at 6:19

edited Sep 3 at 6:19

answered Sep 3 at 4:58

Chas Brown

4,1361319

answered Sep 3 at 4:58

Chas Brown

4,1361319

answered Sep 3 at 4:58

Chas Brown

4,1361319

4,1361319

add a comment | 

add a comment | 

up vote
1
down vote

Pyth, 48 bytes

L+b_<b/Q2JmO/Q2 2jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K

Try it out online here.

The program is in 3 parts - definition of palindromisation function, choosing location of numeric, and main function.

Implicit: Q=eval(input()), T=10, G=lower case alphabet

L+b_<b/Q2 Palindromisation function
L Define a function, y(b)
 /Q2 Half input number, rounding down
 <b Take that many elements from the start of the sequence
 _ Reverse them
 +b Prepend the unaltered sequence

JmO/Q2 2 Choose numeric location
 O/Q2 Choose a random number between 0 and half input number
 m 2 Do the above twice, wrap in array
J Assign to variable J

jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K Main function
 cQ2 Divide input number by 2
 .E Round up
 K Assign the above to K
 .SG Shuffle the alphabet
 sm Q Do the above Q times, concatenate
 c K Chop the above into segments of length K
 < K Take the first K of the above
 .e Map (element, index) as (b,k) using:
 qhJk Does k equal first element of J?
 W If so...
 X b Replace in b...
 eJ ...at position <last element of J>...
 OT ...a random int less than 10
 Otherwise, b without replacement
 y Apply palindromisation to the result of the above
 y Palindromise the set of lines
j Join on newlines, implicit print

Using several shuffled alphabets should ensure that the number of unique characters is always more then the input number.

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

add a comment | 

up vote
1
down vote

Pyth, 48 bytes

L+b_<b/Q2JmO/Q2 2jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K

Try it out online here.

The program is in 3 parts - definition of palindromisation function, choosing location of numeric, and main function.

Implicit: Q=eval(input()), T=10, G=lower case alphabet

L+b_<b/Q2 Palindromisation function
L Define a function, y(b)
 /Q2 Half input number, rounding down
 <b Take that many elements from the start of the sequence
 _ Reverse them
 +b Prepend the unaltered sequence

JmO/Q2 2 Choose numeric location
 O/Q2 Choose a random number between 0 and half input number
 m 2 Do the above twice, wrap in array
J Assign to variable J

jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K Main function
 cQ2 Divide input number by 2
 .E Round up
 K Assign the above to K
 .SG Shuffle the alphabet
 sm Q Do the above Q times, concatenate
 c K Chop the above into segments of length K
 < K Take the first K of the above
 .e Map (element, index) as (b,k) using:
 qhJk Does k equal first element of J?
 W If so...
 X b Replace in b...
 eJ ...at position <last element of J>...
 OT ...a random int less than 10
 Otherwise, b without replacement
 y Apply palindromisation to the result of the above
 y Palindromise the set of lines
j Join on newlines, implicit print

Using several shuffled alphabets should ensure that the number of unique characters is always more then the input number.

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

add a comment | 

up vote
1
down vote

up vote
1
down vote

Pyth, 48 bytes

L+b_<b/Q2JmO/Q2 2jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K

Try it out online here.

The program is in 3 parts - definition of palindromisation function, choosing location of numeric, and main function.

Implicit: Q=eval(input()), T=10, G=lower case alphabet

L+b_<b/Q2 Palindromisation function
L Define a function, y(b)
 /Q2 Half input number, rounding down
 <b Take that many elements from the start of the sequence
 _ Reverse them
 +b Prepend the unaltered sequence

JmO/Q2 2 Choose numeric location
 O/Q2 Choose a random number between 0 and half input number
 m 2 Do the above twice, wrap in array
J Assign to variable J

jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K Main function
 cQ2 Divide input number by 2
 .E Round up
 K Assign the above to K
 .SG Shuffle the alphabet
 sm Q Do the above Q times, concatenate
 c K Chop the above into segments of length K
 < K Take the first K of the above
 .e Map (element, index) as (b,k) using:
 qhJk Does k equal first element of J?
 W If so...
 X b Replace in b...
 eJ ...at position <last element of J>...
 OT ...a random int less than 10
 Otherwise, b without replacement
 y Apply palindromisation to the result of the above
 y Palindromise the set of lines
j Join on newlines, implicit print

Using several shuffled alphabets should ensure that the number of unique characters is always more then the input number.

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

Pyth, 48 bytes

L+b_<b/Q2JmO/Q2 2jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K

Try it out online here.

The program is in 3 parts - definition of palindromisation function, choosing location of numeric, and main function.

Implicit: Q=eval(input()), T=10, G=lower case alphabet

L+b_<b/Q2 Palindromisation function
L Define a function, y(b)
 /Q2 Half input number, rounding down
 <b Take that many elements from the start of the sequence
 _ Reverse them
 +b Prepend the unaltered sequence

JmO/Q2 2 Choose numeric location
 O/Q2 Choose a random number between 0 and half input number
 m 2 Do the above twice, wrap in array
J Assign to variable J

jy.eyXWqhJkbeJOT<csm.SGQK.EcQ2K Main function
 cQ2 Divide input number by 2
 .E Round up
 K Assign the above to K
 .SG Shuffle the alphabet
 sm Q Do the above Q times, concatenate
 c K Chop the above into segments of length K
 < K Take the first K of the above
 .e Map (element, index) as (b,k) using:
 qhJk Does k equal first element of J?
 W If so...
 X b Replace in b...
 eJ ...at position <last element of J>...
 OT ...a random int less than 10
 Otherwise, b without replacement
 y Apply palindromisation to the result of the above
 y Palindromise the set of lines
j Join on newlines, implicit print

Using several shuffled alphabets should ensure that the number of unique characters is always more then the input number.

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

share|improve this answer

share|improve this answer

answered Sep 3 at 9:38

Sok

2,851722

answered Sep 3 at 9:38

Sok

2,851722

answered Sep 3 at 9:38

Sok

2,851722

2,851722

add a comment | 

add a comment | 

up vote
1
down vote

Python 2/Python 3, 227 bytes

from random import*
def m(N):n=N-N//2;r=range;C=choice;c=n*[chr(i+97)for i in r(26)];shuffle(c);c[C([i for i in r(n*(N-n))if(i+1)%n+1-N%2])]=`C(r(10))`;R=[c[i*n:i*n+n]+c[i*n:i*n+n-N%2][::-1]for i in r(n)];return R+R[::-1][N%2:]

ungolfing a bit:

from random import * # get 'choice' and 'shuffle'
def matrix(N):
 n = ceil(N/2) # get the size of the base block
 # get a shuffleable lowercase alphabet
 c = [chr(i+97)for i in range(26)]
 c = n*c # make it large enough to fill the base-block
 shuffle(c) # randomize it
 digit = choice('1234567890') # get random digit string
 ## this is only needed as to prevent uneven side-length matrices
 # from having centerline digits.
 allowed_indices = [i for i in range( # get all allowed indices
 n*(N-n)) # skip those, that are in an unmirrored center-line
 if(i+1)%n # only use those that are not in the center column
 +1-N%2] # exept if there is no center column
 index = choice(allowed_indices) # get random index
 c[index]=digit # replace one field at random with a random digit
 ## 
 R=
 for i in range(n):
 r = c[i*n:i*n+n] # chop to chunks sized fit for the base block
 R.append(r+r[::-1][N%2:]) # mirror skipping the center line
 return R+R[::-1][N%2:] # mirror skipping the center line and return

Older, almost correct versions below:

Python2, Python3, 161 bytes

from random import *
N=26
n=N-N//2
c=[chr(i+97)for i in range(26)]
R=[ r+r[::-1][N%2:]for r in[(shuffle(c),c[:n])[1]for i in range(n)]]
R+=R[::-1][N%2:]
print(R)

It seems N differing elements is only almost guarranteed.

Python 2/Python 3, 170 bytes

from random import*
def m(N):n=N-N//2;r=range;c=n*[chr(i+97)for i in r(26)][:n*n];shuffle(c);R=[_+_[::-1][N%2:]for _ in[c[i*n:i*n+n]for i in r(n)]];return R+R[::-1][N%2:]

It seems I forgot rule 3. Also somehow the [:n*n] slipped in.

share|improve this answer

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
  – Chas Brown
  Sep 2 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
  – Teck-freak
  Sep 3 at 0:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
  – Chas Brown
  Sep 3 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I litterally ran it over 10000 times without any error.
  – Teck-freak
  Sep 3 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
  – Teck-freak
  Sep 3 at 15:11
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

Python 2/Python 3, 227 bytes

from random import*
def m(N):n=N-N//2;r=range;C=choice;c=n*[chr(i+97)for i in r(26)];shuffle(c);c[C([i for i in r(n*(N-n))if(i+1)%n+1-N%2])]=`C(r(10))`;R=[c[i*n:i*n+n]+c[i*n:i*n+n-N%2][::-1]for i in r(n)];return R+R[::-1][N%2:]

ungolfing a bit:

from random import * # get 'choice' and 'shuffle'
def matrix(N):
 n = ceil(N/2) # get the size of the base block
 # get a shuffleable lowercase alphabet
 c = [chr(i+97)for i in range(26)]
 c = n*c # make it large enough to fill the base-block
 shuffle(c) # randomize it
 digit = choice('1234567890') # get random digit string
 ## this is only needed as to prevent uneven side-length matrices
 # from having centerline digits.
 allowed_indices = [i for i in range( # get all allowed indices
 n*(N-n)) # skip those, that are in an unmirrored center-line
 if(i+1)%n # only use those that are not in the center column
 +1-N%2] # exept if there is no center column
 index = choice(allowed_indices) # get random index
 c[index]=digit # replace one field at random with a random digit
 ## 
 R=
 for i in range(n):
 r = c[i*n:i*n+n] # chop to chunks sized fit for the base block
 R.append(r+r[::-1][N%2:]) # mirror skipping the center line
 return R+R[::-1][N%2:] # mirror skipping the center line and return

Older, almost correct versions below:

Python2, Python3, 161 bytes

from random import *
N=26
n=N-N//2
c=[chr(i+97)for i in range(26)]
R=[ r+r[::-1][N%2:]for r in[(shuffle(c),c[:n])[1]for i in range(n)]]
R+=R[::-1][N%2:]
print(R)

It seems N differing elements is only almost guarranteed.

Python 2/Python 3, 170 bytes

from random import*
def m(N):n=N-N//2;r=range;c=n*[chr(i+97)for i in r(26)][:n*n];shuffle(c);R=[_+_[::-1][N%2:]for _ in[c[i*n:i*n+n]for i in r(n)]];return R+R[::-1][N%2:]

It seems I forgot rule 3. Also somehow the [:n*n] slipped in.

share|improve this answer

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
  – Chas Brown
  Sep 2 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
  – Teck-freak
  Sep 3 at 0:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
  – Chas Brown
  Sep 3 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I litterally ran it over 10000 times without any error.
  – Teck-freak
  Sep 3 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
  – Teck-freak
  Sep 3 at 15:11
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

up vote
1
down vote

Python 2/Python 3, 227 bytes

from random import*
def m(N):n=N-N//2;r=range;C=choice;c=n*[chr(i+97)for i in r(26)];shuffle(c);c[C([i for i in r(n*(N-n))if(i+1)%n+1-N%2])]=`C(r(10))`;R=[c[i*n:i*n+n]+c[i*n:i*n+n-N%2][::-1]for i in r(n)];return R+R[::-1][N%2:]

ungolfing a bit:

from random import * # get 'choice' and 'shuffle'
def matrix(N):
 n = ceil(N/2) # get the size of the base block
 # get a shuffleable lowercase alphabet
 c = [chr(i+97)for i in range(26)]
 c = n*c # make it large enough to fill the base-block
 shuffle(c) # randomize it
 digit = choice('1234567890') # get random digit string
 ## this is only needed as to prevent uneven side-length matrices
 # from having centerline digits.
 allowed_indices = [i for i in range( # get all allowed indices
 n*(N-n)) # skip those, that are in an unmirrored center-line
 if(i+1)%n # only use those that are not in the center column
 +1-N%2] # exept if there is no center column
 index = choice(allowed_indices) # get random index
 c[index]=digit # replace one field at random with a random digit
 ## 
 R=
 for i in range(n):
 r = c[i*n:i*n+n] # chop to chunks sized fit for the base block
 R.append(r+r[::-1][N%2:]) # mirror skipping the center line
 return R+R[::-1][N%2:] # mirror skipping the center line and return

Older, almost correct versions below:

Python2, Python3, 161 bytes

from random import *
N=26
n=N-N//2
c=[chr(i+97)for i in range(26)]
R=[ r+r[::-1][N%2:]for r in[(shuffle(c),c[:n])[1]for i in range(n)]]
R+=R[::-1][N%2:]
print(R)

It seems N differing elements is only almost guarranteed.

Python 2/Python 3, 170 bytes

from random import*
def m(N):n=N-N//2;r=range;c=n*[chr(i+97)for i in r(26)][:n*n];shuffle(c);R=[_+_[::-1][N%2:]for _ in[c[i*n:i*n+n]for i in r(n)]];return R+R[::-1][N%2:]

It seems I forgot rule 3. Also somehow the [:n*n] slipped in.

share|improve this answer

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

Python 2/Python 3, 227 bytes

from random import*
def m(N):n=N-N//2;r=range;C=choice;c=n*[chr(i+97)for i in r(26)];shuffle(c);c[C([i for i in r(n*(N-n))if(i+1)%n+1-N%2])]=`C(r(10))`;R=[c[i*n:i*n+n]+c[i*n:i*n+n-N%2][::-1]for i in r(n)];return R+R[::-1][N%2:]

ungolfing a bit:

from random import * # get 'choice' and 'shuffle'
def matrix(N):
 n = ceil(N/2) # get the size of the base block
 # get a shuffleable lowercase alphabet
 c = [chr(i+97)for i in range(26)]
 c = n*c # make it large enough to fill the base-block
 shuffle(c) # randomize it
 digit = choice('1234567890') # get random digit string
 ## this is only needed as to prevent uneven side-length matrices
 # from having centerline digits.
 allowed_indices = [i for i in range( # get all allowed indices
 n*(N-n)) # skip those, that are in an unmirrored center-line
 if(i+1)%n # only use those that are not in the center column
 +1-N%2] # exept if there is no center column
 index = choice(allowed_indices) # get random index
 c[index]=digit # replace one field at random with a random digit
 ## 
 R=
 for i in range(n):
 r = c[i*n:i*n+n] # chop to chunks sized fit for the base block
 R.append(r+r[::-1][N%2:]) # mirror skipping the center line
 return R+R[::-1][N%2:] # mirror skipping the center line and return

Older, almost correct versions below:

Python2, Python3, 161 bytes

from random import *
N=26
n=N-N//2
c=[chr(i+97)for i in range(26)]
R=[ r+r[::-1][N%2:]for r in[(shuffle(c),c[:n])[1]for i in range(n)]]
R+=R[::-1][N%2:]
print(R)

It seems N differing elements is only almost guarranteed.

Python 2/Python 3, 170 bytes

from random import*
def m(N):n=N-N//2;r=range;c=n*[chr(i+97)for i in r(26)][:n*n];shuffle(c);R=[_+_[::-1][N%2:]for _ in[c[i*n:i*n+n]for i in r(n)]];return R+R[::-1][N%2:]

It seems I forgot rule 3. Also somehow the [:n*n] slipped in.

share|improve this answer

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

share|improve this answer

share|improve this answer

edited Sep 3 at 16:57

edited Sep 3 at 16:57

edited Sep 3 at 16:57

answered Aug 31 at 22:38

Teck-freak

1314

answered Aug 31 at 22:38

Teck-freak

1314

answered Aug 31 at 22:38

Teck-freak

1314

1314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
  – Chas Brown
  Sep 2 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
  – Teck-freak
  Sep 3 at 0:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
  – Chas Brown
  Sep 3 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I litterally ran it over 10000 times without any error.
  – Teck-freak
  Sep 3 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
  – Teck-freak
  Sep 3 at 15:11
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
  – Chas Brown
  Sep 2 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
  – Teck-freak
  Sep 3 at 0:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
  – Chas Brown
  Sep 3 at 5:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I litterally ran it over 10000 times without any error.
  – Teck-freak
  Sep 3 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
  – Teck-freak
  Sep 3 at 15:11
  

  
  
  
  

Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
– Chas Brown
Sep 2 at 22:30

Your answer is very clever in the way it constructs the symmetric matrix, but you have not satisfied rule 3 (as you don't have any digits in your result), nor rule 5 (e.g., if n = 3, you will never have an output containing a 'z', so not every output is possible).
– Chas Brown
Sep 2 at 22:30

Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
– Teck-freak
Sep 3 at 0:57

Well pickle me and ... you're correct @ChasBrown ! Well, the [:n*n] is a remainder from a different approach and frankly it shouldn't be there. But you're correct about rule three. I'll have to correct it. Give me a bit.
– Teck-freak
Sep 3 at 0:57

Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
– Chas Brown
Sep 3 at 5:01

Tried your solution here, but there was an index error... BTW, TryItOnline is super handy here at PPCG! (Also, this problem is way trickier than I thought at first...)
– Chas Brown
Sep 3 at 5:01

I litterally ran it over 10000 times without any error.
– Teck-freak
Sep 3 at 15:06

I litterally ran it over 10000 times without any error.
– Teck-freak
Sep 3 at 15:06

found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
– Teck-freak
Sep 3 at 15:11

found it. a ':' was missing. I copied it directly from my script, but it must have gotten lost. it should be "... :-1][N%2:]for i ..." instead of "... :-1][N%2]for i ...".
– Teck-freak
Sep 3 at 15:11

 | 
show 1 more comment

 

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