Chop off the matrix to get the desired sum

Clash Royale CLAN TAG#URR8PPP

up vote
21
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Definition

Given a matrix $M$ of non-negative integers and a non-negative integer $k$, we define $F_k$ as the "chop-off" function that removes all rows and all columns in $M$ that contain $k$.

Example:

$$beginalignM=pmatrixcolorred6&colorred1&colorwhitebbox[red,1pt]5\1&2&colorred8\colorred9&colorred8&colorwhitebbox[red,1pt]5\6&0&colorred4\\F_5(M)=pmatrix1&2\6&0endalign$$

Your task

Given $M$ and a target sum $S$, your task is to find all possible values of $k$ such that the sum of the remaining elements in $F_k(M)$ is equal to $S$.

Example:

Given the above matrix $M$ and $S=9$:

• $k=5$ is a solution, because $F_5(M)=pmatrix1&2\6&0$ and $1+2+6+0=9$


• $k=1$ is the only other possible solution: $F_1(M)=pmatrix5\4$ and $5+4=9$

So the expected output would be $1,5$.

Clarifications and rules

• The input is guaranteed to admit at least one solution.


• The sum of the elements in the original matrix is guaranteed to be greater than $S$.


• You may assume $S>0$. It means that an empty matrix will never lead to a solution.


• The values of $k$ may be printed or returned in any order and in any reasonable, unambiguous format.


• You are allowed not to deduplicate the output (e.g. $[1,1,5,5]$ or $[1,5,1,5]$ are considered valid answers for the above example).


• This is code-golf.

Test cases

M = [[6,1,5],[1,2,8],[9,8,5],[6,0,4]]
S = 9
Solution = 1,5

M = [[7,2],[1,4]]
S = 7
Solution = 4

M = [[12,5,2,3],[17,11,18,8]]
S = 43
Solution = 5

M = [[7,12],[10,5],[0,13]]
S = 17
Solution = 0,13

M = [[1,1,0,1],[2,0,0,2],[2,0,1,0]]
S = 1
Solution = 2

M = [[57,8,33,84],[84,78,19,14],[43,14,81,30]]
S = 236
Solution = 19,43,57

M = [[2,5,8],[3,5,8],[10,8,5],[10,6,7],[10,6,4]]
S = 49
Solution = 2,3,4,7

M = [[5,4,0],[3,0,4],[8,2,2]]
S = 8
Solution = 0,2,3,4,5,8

code-golf array-manipulation matrix

share|improve this question

edited Sep 6 at 6:46

asked Sep 5 at 9:08

Arnauld

63.7k580268

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  Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
  – Shaggy
  Sep 5 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaggy Yes. That looks reasonable.
  – Arnauld
  Sep 5 at 16:33
  

  
  
  
  

add a comment | 

up vote
21
down vote

favorite

Definition

Given a matrix $M$ of non-negative integers and a non-negative integer $k$, we define $F_k$ as the "chop-off" function that removes all rows and all columns in $M$ that contain $k$.

Example:

$$beginalignM=pmatrixcolorred6&colorred1&colorwhitebbox[red,1pt]5\1&2&colorred8\colorred9&colorred8&colorwhitebbox[red,1pt]5\6&0&colorred4\\F_5(M)=pmatrix1&2\6&0endalign$$

Your task

Given $M$ and a target sum $S$, your task is to find all possible values of $k$ such that the sum of the remaining elements in $F_k(M)$ is equal to $S$.

Example:

Given the above matrix $M$ and $S=9$:

• $k=5$ is a solution, because $F_5(M)=pmatrix1&2\6&0$ and $1+2+6+0=9$


• $k=1$ is the only other possible solution: $F_1(M)=pmatrix5\4$ and $5+4=9$

So the expected output would be $1,5$.

Clarifications and rules

• The input is guaranteed to admit at least one solution.


• The sum of the elements in the original matrix is guaranteed to be greater than $S$.


• You may assume $S>0$. It means that an empty matrix will never lead to a solution.


• The values of $k$ may be printed or returned in any order and in any reasonable, unambiguous format.


• You are allowed not to deduplicate the output (e.g. $[1,1,5,5]$ or $[1,5,1,5]$ are considered valid answers for the above example).


• This is code-golf.

Test cases

M = [[6,1,5],[1,2,8],[9,8,5],[6,0,4]]
S = 9
Solution = 1,5

M = [[7,2],[1,4]]
S = 7
Solution = 4

M = [[12,5,2,3],[17,11,18,8]]
S = 43
Solution = 5

M = [[7,12],[10,5],[0,13]]
S = 17
Solution = 0,13

M = [[1,1,0,1],[2,0,0,2],[2,0,1,0]]
S = 1
Solution = 2

M = [[57,8,33,84],[84,78,19,14],[43,14,81,30]]
S = 236
Solution = 19,43,57

M = [[2,5,8],[3,5,8],[10,8,5],[10,6,7],[10,6,4]]
S = 49
Solution = 2,3,4,7

M = [[5,4,0],[3,0,4],[8,2,2]]
S = 8
Solution = 0,2,3,4,5,8

code-golf array-manipulation matrix

share|improve this question

edited Sep 6 at 6:46

asked Sep 5 at 9:08

Arnauld

63.7k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
  – Shaggy
  Sep 5 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaggy Yes. That looks reasonable.
  – Arnauld
  Sep 5 at 16:33
  

  
  
  
  

add a comment | 

up vote
21
down vote

favorite

up vote
21
down vote

favorite

Definition

Given a matrix $M$ of non-negative integers and a non-negative integer $k$, we define $F_k$ as the "chop-off" function that removes all rows and all columns in $M$ that contain $k$.

Example:

$$beginalignM=pmatrixcolorred6&colorred1&colorwhitebbox[red,1pt]5\1&2&colorred8\colorred9&colorred8&colorwhitebbox[red,1pt]5\6&0&colorred4\\F_5(M)=pmatrix1&2\6&0endalign$$

Your task

Given $M$ and a target sum $S$, your task is to find all possible values of $k$ such that the sum of the remaining elements in $F_k(M)$ is equal to $S$.

Example:

Given the above matrix $M$ and $S=9$:

• $k=5$ is a solution, because $F_5(M)=pmatrix1&2\6&0$ and $1+2+6+0=9$


• $k=1$ is the only other possible solution: $F_1(M)=pmatrix5\4$ and $5+4=9$

So the expected output would be $1,5$.

Clarifications and rules

• The input is guaranteed to admit at least one solution.


• The sum of the elements in the original matrix is guaranteed to be greater than $S$.


• You may assume $S>0$. It means that an empty matrix will never lead to a solution.


• The values of $k$ may be printed or returned in any order and in any reasonable, unambiguous format.


• You are allowed not to deduplicate the output (e.g. $[1,1,5,5]$ or $[1,5,1,5]$ are considered valid answers for the above example).


• This is code-golf.

Test cases

M = [[6,1,5],[1,2,8],[9,8,5],[6,0,4]]
S = 9
Solution = 1,5

M = [[7,2],[1,4]]
S = 7
Solution = 4

M = [[12,5,2,3],[17,11,18,8]]
S = 43
Solution = 5

M = [[7,12],[10,5],[0,13]]
S = 17
Solution = 0,13

M = [[1,1,0,1],[2,0,0,2],[2,0,1,0]]
S = 1
Solution = 2

M = [[57,8,33,84],[84,78,19,14],[43,14,81,30]]
S = 236
Solution = 19,43,57

M = [[2,5,8],[3,5,8],[10,8,5],[10,6,7],[10,6,4]]
S = 49
Solution = 2,3,4,7

M = [[5,4,0],[3,0,4],[8,2,2]]
S = 8
Solution = 0,2,3,4,5,8

code-golf array-manipulation matrix

share|improve this question

edited Sep 6 at 6:46

asked Sep 5 at 9:08

Arnauld

63.7k580268

Definition

Given a matrix $M$ of non-negative integers and a non-negative integer $k$, we define $F_k$ as the "chop-off" function that removes all rows and all columns in $M$ that contain $k$.

Example:

$$beginalignM=pmatrixcolorred6&colorred1&colorwhitebbox[red,1pt]5\1&2&colorred8\colorred9&colorred8&colorwhitebbox[red,1pt]5\6&0&colorred4\\F_5(M)=pmatrix1&2\6&0endalign$$

Your task

Given $M$ and a target sum $S$, your task is to find all possible values of $k$ such that the sum of the remaining elements in $F_k(M)$ is equal to $S$.

Example:

Given the above matrix $M$ and $S=9$:

• $k=5$ is a solution, because $F_5(M)=pmatrix1&2\6&0$ and $1+2+6+0=9$


• $k=1$ is the only other possible solution: $F_1(M)=pmatrix5\4$ and $5+4=9$

So the expected output would be $1,5$.

Clarifications and rules

• The input is guaranteed to admit at least one solution.


• The sum of the elements in the original matrix is guaranteed to be greater than $S$.


• You may assume $S>0$. It means that an empty matrix will never lead to a solution.


• The values of $k$ may be printed or returned in any order and in any reasonable, unambiguous format.


• You are allowed not to deduplicate the output (e.g. $[1,1,5,5]$ or $[1,5,1,5]$ are considered valid answers for the above example).


• This is code-golf.

Test cases

M = [[6,1,5],[1,2,8],[9,8,5],[6,0,4]]
S = 9
Solution = 1,5

M = [[7,2],[1,4]]
S = 7
Solution = 4

M = [[12,5,2,3],[17,11,18,8]]
S = 43
Solution = 5

M = [[7,12],[10,5],[0,13]]
S = 17
Solution = 0,13

M = [[1,1,0,1],[2,0,0,2],[2,0,1,0]]
S = 1
Solution = 2

M = [[57,8,33,84],[84,78,19,14],[43,14,81,30]]
S = 236
Solution = 19,43,57

M = [[2,5,8],[3,5,8],[10,8,5],[10,6,7],[10,6,4]]
S = 49
Solution = 2,3,4,7

M = [[5,4,0],[3,0,4],[8,2,2]]
S = 8
Solution = 0,2,3,4,5,8

code-golf array-manipulation matrix

share|improve this question

edited Sep 6 at 6:46

asked Sep 5 at 9:08

Arnauld

63.7k580268

share|improve this question

share|improve this question

edited Sep 6 at 6:46

edited Sep 6 at 6:46

edited Sep 6 at 6:46

asked Sep 5 at 9:08

Arnauld

63.7k580268

asked Sep 5 at 9:08

Arnauld

63.7k580268

asked Sep 5 at 9:08

Arnauld

63.7k580268

63.7k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
  – Shaggy
  Sep 5 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaggy Yes. That looks reasonable.
  – Arnauld
  Sep 5 at 16:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
  – Shaggy
  Sep 5 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaggy Yes. That looks reasonable.
  – Arnauld
  Sep 5 at 16:33
  

  
  
  
  

Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
– Shaggy
Sep 5 at 16:28

Would retaining the original structure of the input array (e.g., [[1,5],[1],[5],] for the first test case) be a valid means of output?
– Shaggy
Sep 5 at 16:28

@Shaggy Yes. That looks reasonable.
– Arnauld
Sep 5 at 16:33

@Shaggy Yes. That looks reasonable.
– Arnauld
Sep 5 at 16:33

add a comment | 

14 Answers
14

active

oldest

votes

up vote
10
down vote

K (ngn/k), 39 bytes

a@&y=x+//x*(&/'b)&:&/b:~x=y/:a:,/x

Try it online!

thanks @Adám for this explanation:

… function, x is M and y is S

 ,/x flatten M (these are the k candidates)

 a: assign to a

 x…/: apply the following function to each while using M as fixed left argument (x):

  x=y Boolean matrix indicating where elements of M equal the current k candidate

  ~ negate that

  b: assign that to b

  &/ AND reduction (finds columns without that k)

  (…)&: AND that with each of the following:

   &/'b AND reduction of each (finds rows without that k)

  x* multiply M by that

  +// grand sum

 y= list of Booleans indicating where S equals those sums

 & indices of Trues

 a@ use that to index into the elements (the k candidates)

share|improve this answer

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

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  Feel free to correct the explanation.
  – Adám
  Sep 5 at 10:11
  

  
  
  
  


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  The dangers of copy-paste explanation…
  – Adám
  Sep 5 at 10:15
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

APL (Dyalog Unicode), 35 33 28 bytes^SBCS

-7 thanks to ngn.

Anonymous infix lambda. Takes S as left argument and M as right argument.

⍵[⍸⍺=(+/∘,⍵×∧/∘.∧∧⌿)¨⍵≠⊂⍵]

Try it online!

… "dfn", ⍺ and ⍵ are left and right arguments (S and M) respectively:

 ⍵[…] index M with the following coordinates:

  ⊂⍵ enclose M to treat it as a single element

  ⍵= compare each element (i.e. k candidate) of M to that entire M

  (…)¨ apply the following tacit function to each:

   ∧⌿ vertical AND reduction (finds columns without that k candidate)

…∘.∧ Cartesian Boolean product with:

    ∧/ horizontal AND reduction (finds rows without that k candidate)

   ⍵× multiply M with that mask

   +/∘, sum the flattened matrix

  ⍺= Boolean indicating where S equals those sums

  ⍸ indices where that's true

share|improve this answer

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

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  M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
  – ngn
  Sep 5 at 10:06
  

  
  
  
  


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  @ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
  – Adám
  Sep 5 at 10:24
  
  

  
  
  
  


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  passing ⍵ as ⍺ in the inner dfn is equally confusing to me
  – ngn
  Sep 5 at 10:28
  

  
  
  
  


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  ⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
  – ngn
  Sep 5 at 10:35
  
  

  
  
  
  


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  @ngn Yeah, I wanted to do something like that. Thanks!
  – Adám
  Sep 5 at 11:59
  

  
  
  
  

add a comment | 

up vote
6
down vote

R, 78 73 bytes

function(m,s)m[Map(function(y)sum(m[(w=-which(m==y,T))[,1],w[,2]]),m)==s]

Try it online!

Does not sort or deduplicate the output.

Credit to J.Doe & Giuseppe for -5 bytes.

share|improve this answer

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

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  76 bytes
  – J.Doe
  Sep 5 at 12:55
  

  
  
  
  


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  4
  

  
  
  
  
  
  

  
  @J.Doe 73 bytes
  – Giuseppe
  Sep 5 at 12:56
  
  

  
  
  
  

add a comment | 

up vote
5
down vote

Jelly, 20 19 17 15 14 bytes

pZnⱮFȦ€€ḋFẹƓịF

This is a monadic link that takes M as argument and reads S from STDIN.

Try it online!

How it works

pZnⱮFȦ€€ḋFẹƓịF Main link. Argument: M

 Z Zip; transpose the rows and columns of M.
p Take the Cartesian product of M and its transpose, yielding all pairs
 (r, c) of rows and columns of M.
 F Flatten; yield the elements of M.
 nâ±® Not equal map; for each element e of M, compare the elements of the
 pairs (r, c) with e.
 Ȧ€€ All each each; for each array of Booleans corresponding to an (r, c)
 pair, test if all of them are true.
 F Flatten; yield the elements of M.
 ḋ Take the dot product of each list of resulting Booleans and the
 elements of M.
 Ɠ Read an integer S from STDIN.
 ẹ Find all indices of S in the dot products.
 F Flatten; yield the elements of M.
 ị Retrieve the elements of the right at the indices from the left.

share|improve this answer

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

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  Mark my words lol :) Nice answer, +1
  – Mr. Xcoder
  Sep 5 at 17:26
  

  
  
  
  

add a comment | 

up vote
5
down vote

Haskell, 88 86 84 77 bytes

• -2 bytes thanks to BWO
  


• -7 bytes thanks to Tesseract
  

m!s=[k|k<-m>>=id,s==sum[x|r<-m,all(/=k)r,(i,x)<-zip[0..]r,all((/=k).(!!i))m]]

Verify all testcases.

Explanation

m ! s = -- function !, taking m and s as input
 [k | -- the list of all k's such that
 k <- m >>= id, -- * k is an entry of m
 s == sum -- * s equals the sum of
 [x | -- the list of x's such that
 r <- m, -- * r is a row of m
 all (/= k) r, -- * r does not contain k
 (i, x) <- zip [0 ..] r, -- * i is a valid column index; also let x = r[i]
 all ((/= k) . (!! i)) m -- * none of the rows contain k at index i
 ]
 ]

share|improve this answer

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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  Should that say “function f”?
  – Quintec
  Sep 5 at 11:47
  

  
  
  
  


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  @Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
  – Delfad0r
  Sep 5 at 11:51
  

  
  
  
  

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up vote
5
down vote

Pyth,  27 23 22 21  20 bytes

fqvzss.DRsxLTQ-I#TQs

Test suite!

Does not deduplicate.

How it works?

fqvzss.DRsxLTQ-I#TQs Full program.
f s Flatten M and keep only those elements T which satisfy:
 qvzss.DRsxLTQ-I#TQ The filtering function. Breakdown:
 -I#TQ Discard the rows that contain T. More elaborate explanation:
 # Q |-> In M, keep only those elements that are...
 I |-> Invariant under (equal to themselves after...)
 - T |-> Removing T.
 Let's call the result of this expression CR (chopped rows).
 xLTQ Map over the rows M and retrieve all indices of T.
 s Collect indices in 1D list (flatten). Call this I.
 .DR For each row left in CR, remove the elements at indices in I.
 ss Sum all the elements of this matrix flattened.
 qvz And then finally check whether they equal S.

share|improve this answer

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

add a comment | 

up vote
4
down vote

Python 2, 114 108 bytes

lambda m,s:a for a in sum(m,)if s==sum(v for l in m for i,v in enumerate(l)ifa-set(l)-set(zip(*m)[i]))

Try it online!

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

add a comment | 

up vote
4
down vote

Perl 6, 80 bytes

$^s;@^m[*;*].grep($s==sum @m[($!=*.grep(:k,!*.grep($^v)))(@m);[$!([Z] @m)]])

Try it online!

share|improve this answer

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

add a comment | 

up vote
3
down vote

05AB1E, 21 bytes

²˜ʒQεZ+}øεZ<~}ø_*OO¹Q

Try it online!

Only after I've written this answer have I seen Kevin's. I believe this is substantially different, so I am posting it separately. My intuition says that the optimal byte count is around 18, so I'll have to revisit this and see what else I can do. With the current code, it is impossible to write a test suite but I have verified all test cases myself and the results are correct.

Cropping Algorithm

To better illustrate the technique used, let's grab an example, with the element to crop $k=5$:

$$M=left(beginmatrix6&1&bbox[green,1pt]colorwhite5\1&2&8\9&8&bbox[green,1pt]colorwhite5\6&0&4endmatrixright)$$

It first generates the matrix of equality with $k$:

$$left(beginmatrix0&0&1\0&0&0\0&0&1\0&0&0endmatrixright)$$

Then, it goes through $M$ and for each row $R$, it adds $max(R)$ to each element in $R$, highlighting the necessary rows while still preserving the uniqueness of the horizontal positions of the searched element:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0endmatrixright)$$

Then, it goes through the transpose of $M$ and for each column $C$, it performs the operation $(max(C)-1)space ||space cspaceforallspace cin C:$ (where $||$ is 05AB1E's bitwise OR – addition should work too, so replace ~ with + if you want to test that too), which results in:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1endmatrixright)$$

Finally, it maps $0$ to $1$ and all other integers to $0$ and performs element-wise multiplication with $M$:

$$left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0\colorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0endmatrixright):longrightarrow:left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&2&colorgreen0\colorgreen0&colorgreen0&colorgreen0\6&0&colorgreen0endmatrixright)$$

After which the sum of the resulting matrix is computed.

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edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

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  1
  

  
  
  
  
  
  

  
  Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
  – Kevin Cruijssen
  Sep 5 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
  – Mr. Xcoder
  Sep 5 at 13:24
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

05AB1E (legacy), 27 26 bytes

˜ʒ©¹ε®å_}¹ζʒ®å_}ζ‚ζ€€OPOIQ

Does not sort nor uniquify the result.

Only works in the legacy (for now), because sum-each seems to be doing some odd things when part of the inner lists are integers and other are lists..

Try it online or verify all test cases.

Explanation:

˜ # Flatten the (implicit) matrix-input
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [6,1,5,1,2,8,9,8,5,6,0,4]
 ʒ # Filter this list by:
 © # Store the current value in a register-variable
 ¹ # Take the matrix-input
 ε } # Map it to:
 ®å_ # 0 if the current number is in this row, 1 if not
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] and 6 → [0,1,1,0]
 ¹ # Take the matrix-input again
 ζ # Swap its rows and columns
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [[6,1,9,6],[1,2,8,0],[5,8,5,4]]
 ʒ } # Filter it by:
 ®å_ # Only keep the inner lists that does not contain the current number
 # i.e. [[6,1,9,6],[1,2,8,0],[5,8,5,4]] and 6 → [[1,2,8,0],[5,8,5,4]]
 # i.e. [[1,2,2],[1,0,0],[0,0,1],[1,2,0]] and 1 → 
 ζ # After filtering, swap it's rows and columns back again
 # i.e. [[1,2,8,0],[5,8,5,4]] → [[1,5],[2,8],[8,5],[0,4]]
 ‚ζ # Pair both lists together and zip them
 # i.e. [0,1,1,0] and [[1,5],[2,8],[8,5],[0,4]]
 # → [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # i.e. [0,1,0] and → [[0,' '],[1,' '],[0,' ']]
 € # Map each inner list / value to:
 €O # Sum each
 # i.e. [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # → [[0,6],[1,10],[1,13],[0,4]]
 # i.e. [[0,' '],[1,' '],[0,' ']]
 # → [[0,0],[1,0],[0,0]]
 # (NOTE: For most test cases just `O` instead of `€€O` would be enough,
 # but not if we removed ALL zipped inner lists for a number, like the 
 # second example above with input [[1,1,0,1],[2,0,0,2],[2,0,1,0]] and 1)
 P # Now take the product of each inner list
 # i.e. [[0,6],[1,10],[1,13],[0,4]] → [0,10,13,0]
 O # Then take the sum of those
 # i.e. [0,10,13,0] → 23
 IQ # And only keep those that are equal to the number-input
 # i.e. 23 and 9 → 0 (falsey), so it's removed from the flattened input

share|improve this answer

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

add a comment | 

up vote
1
down vote

Jelly, 22 bytes

=+Ṁ$€Z$⁺’¬×⁸FS
F³ç=⁹ƲƇ

Try it online!

-6 bytes using the general approach from Mr. Xcoder's 05AB1E answer.

share|improve this answer

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

add a comment | 

up vote
1
down vote

Charcoal, 33 bytes

ＦθＦι⊞υκＩΦυ⁼ηΣＥθ∧¬№λιΣＥλ∧¬№Ｅθ§πξιν

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

ＦθＦι⊞υκ

Flatten the first input array q into the predefined list u.

 υ Flattened array
 Φ Filter elements
 θ Input array
 ï¼¥ Map over rows
 ι Current element
 λ Current row
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 λ Current row
 ï¼¥ Map over columns
 θ Input array
 ï¼¥ Map over rows
 π Inner row
 ξ Column index
 § Inner element
 ι Current element
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 ν Current element
 Σ Sum
 Σ Sum
 η Second input
 ⁼ Equals
Ｉ Cast to string
 Implicitly print each result on its own line

For each element of the list, sum the array, but if the row contains the element then use 0 instead of its sum, and when summing rows that don't contain the element, if the column contains the element then use 0 instead of the column's value. This is very slightly golfier than filtering the elements out as Charcoal can't sum an empty list.

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

add a comment | 

up vote
1
down vote

Clean, 92 bytes

import StdEnv
$m s=[c\r<-m,c<-r|sum[b\a<-m|all((<>)c)a,b<-a&x<-[0..]|all(u=u!!x<>c)m]==s]

Try it online!

Explained:

$ m s // the function $ of `m` and `s`
 = [ // is equal to
 c // the value `c`
 \ r <- m // for every row `r` in `m`
 , c <- r // for every value `c` in `r`
 | sum [ // where the sum of
 b // the value `b`
 \ a <- m // for every row `a` in `m`
 | all ((<>)c) a // where `c` isn't in `a`
 , b <- a // for every value `b` in `a`
 & x <- [0..] // with every column index `x` from zero
 | all (u = u!!x <> c) m // where `c` isn't in column `x`
 ] == s // equals `s`
 ]

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

add a comment | 

up vote
0
down vote

MATLAB - 78 bytes

Compacted:

function f(M,s);for k=1:9;if sum(sum(M(~sum(M==k,2),~sum(M==k))))==s;k,end;end

And in a fully developped version:

function getthesum(M,s)

for k=1:9 % For each single digit
 x = M==k ; % Index elements equal to "k"
 N = M( ~sum(x,2) , ~sum(x) ) ; % New matrix with only the appropriate rows/columns
 if sum(sum(N))==s % sum rows and columns and compare to "s"
 k % display "k" in console if "k" is valid
 end
end

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  k could be any element of the matrix.
  – Dennis♦
  Sep 5 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
  – Hoki
  Sep 6 at 11:04
  
  

  
  
  
  

add a comment | 

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14 Answers
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up vote
10
down vote

K (ngn/k), 39 bytes

a@&y=x+//x*(&/'b)&:&/b:~x=y/:a:,/x

Try it online!

thanks @Adám for this explanation:

… function, x is M and y is S

 ,/x flatten M (these are the k candidates)

 a: assign to a

 x…/: apply the following function to each while using M as fixed left argument (x):

  x=y Boolean matrix indicating where elements of M equal the current k candidate

  ~ negate that

  b: assign that to b

  &/ AND reduction (finds columns without that k)

  (…)&: AND that with each of the following:

   &/'b AND reduction of each (finds rows without that k)

  x* multiply M by that

  +// grand sum

 y= list of Booleans indicating where S equals those sums

 & indices of Trues

 a@ use that to index into the elements (the k candidates)

share|improve this answer

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Feel free to correct the explanation.
  – Adám
  Sep 5 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dangers of copy-paste explanation…
  – Adám
  Sep 5 at 10:15
  
  

  
  
  
  

add a comment | 

up vote
10
down vote

K (ngn/k), 39 bytes

a@&y=x+//x*(&/'b)&:&/b:~x=y/:a:,/x

Try it online!

thanks @Adám for this explanation:

… function, x is M and y is S

 ,/x flatten M (these are the k candidates)

 a: assign to a

 x…/: apply the following function to each while using M as fixed left argument (x):

  x=y Boolean matrix indicating where elements of M equal the current k candidate

  ~ negate that

  b: assign that to b

  &/ AND reduction (finds columns without that k)

  (…)&: AND that with each of the following:

   &/'b AND reduction of each (finds rows without that k)

  x* multiply M by that

  +// grand sum

 y= list of Booleans indicating where S equals those sums

 & indices of Trues

 a@ use that to index into the elements (the k candidates)

share|improve this answer

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Feel free to correct the explanation.
  – Adám
  Sep 5 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dangers of copy-paste explanation…
  – Adám
  Sep 5 at 10:15
  
  

  
  
  
  

add a comment | 

up vote
10
down vote

up vote
10
down vote

K (ngn/k), 39 bytes

a@&y=x+//x*(&/'b)&:&/b:~x=y/:a:,/x

Try it online!

thanks @Adám for this explanation:

… function, x is M and y is S

 ,/x flatten M (these are the k candidates)

 a: assign to a

 x…/: apply the following function to each while using M as fixed left argument (x):

  x=y Boolean matrix indicating where elements of M equal the current k candidate

  ~ negate that

  b: assign that to b

  &/ AND reduction (finds columns without that k)

  (…)&: AND that with each of the following:

   &/'b AND reduction of each (finds rows without that k)

  x* multiply M by that

  +// grand sum

 y= list of Booleans indicating where S equals those sums

 & indices of Trues

 a@ use that to index into the elements (the k candidates)

share|improve this answer

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

K (ngn/k), 39 bytes

a@&y=x+//x*(&/'b)&:&/b:~x=y/:a:,/x

Try it online!

thanks @Adám for this explanation:

… function, x is M and y is S

 ,/x flatten M (these are the k candidates)

 a: assign to a

 x…/: apply the following function to each while using M as fixed left argument (x):

  x=y Boolean matrix indicating where elements of M equal the current k candidate

  ~ negate that

  b: assign that to b

  &/ AND reduction (finds columns without that k)

  (…)&: AND that with each of the following:

   &/'b AND reduction of each (finds rows without that k)

  x* multiply M by that

  +// grand sum

 y= list of Booleans indicating where S equals those sums

 & indices of Trues

 a@ use that to index into the elements (the k candidates)

share|improve this answer

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

share|improve this answer

share|improve this answer

edited Sep 5 at 21:51

edited Sep 5 at 21:51

edited Sep 5 at 21:51

answered Sep 5 at 9:27

ngn

6,13812256

answered Sep 5 at 9:27

ngn

6,13812256

answered Sep 5 at 9:27

ngn

6,13812256

6,13812256

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Feel free to correct the explanation.
  – Adám
  Sep 5 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dangers of copy-paste explanation…
  – Adám
  Sep 5 at 10:15
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Feel free to correct the explanation.
  – Adám
  Sep 5 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The dangers of copy-paste explanation…
  – Adám
  Sep 5 at 10:15
  
  

  
  
  
  

Feel free to correct the explanation.
– Adám
Sep 5 at 10:11

Feel free to correct the explanation.
– Adám
Sep 5 at 10:11

The dangers of copy-paste explanation…
– Adám
Sep 5 at 10:15

The dangers of copy-paste explanation…
– Adám
Sep 5 at 10:15

add a comment | 

up vote
6
down vote

APL (Dyalog Unicode), 35 33 28 bytes^SBCS

-7 thanks to ngn.

Anonymous infix lambda. Takes S as left argument and M as right argument.

⍵[⍸⍺=(+/∘,⍵×∧/∘.∧∧⌿)¨⍵≠⊂⍵]

Try it online!

… "dfn", ⍺ and ⍵ are left and right arguments (S and M) respectively:

 ⍵[…] index M with the following coordinates:

  ⊂⍵ enclose M to treat it as a single element

  ⍵= compare each element (i.e. k candidate) of M to that entire M

  (…)¨ apply the following tacit function to each:

   ∧⌿ vertical AND reduction (finds columns without that k candidate)

…∘.∧ Cartesian Boolean product with:

    ∧/ horizontal AND reduction (finds rows without that k candidate)

   ⍵× multiply M with that mask

   +/∘, sum the flattened matrix

  ⍺= Boolean indicating where S equals those sums

  ⍸ indices where that's true

share|improve this answer

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
  – ngn
  Sep 5 at 10:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
  – Adám
  Sep 5 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  passing ⍵ as ⍺ in the inner dfn is equally confusing to me
  – ngn
  Sep 5 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
  – ngn
  Sep 5 at 10:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Yeah, I wanted to do something like that. Thanks!
  – Adám
  Sep 5 at 11:59
  

  
  
  
  

add a comment | 

up vote
6
down vote

APL (Dyalog Unicode), 35 33 28 bytes^SBCS

-7 thanks to ngn.

Anonymous infix lambda. Takes S as left argument and M as right argument.

⍵[⍸⍺=(+/∘,⍵×∧/∘.∧∧⌿)¨⍵≠⊂⍵]

Try it online!

… "dfn", ⍺ and ⍵ are left and right arguments (S and M) respectively:

 ⍵[…] index M with the following coordinates:

  ⊂⍵ enclose M to treat it as a single element

  ⍵= compare each element (i.e. k candidate) of M to that entire M

  (…)¨ apply the following tacit function to each:

   ∧⌿ vertical AND reduction (finds columns without that k candidate)

…∘.∧ Cartesian Boolean product with:

    ∧/ horizontal AND reduction (finds rows without that k candidate)

   ⍵× multiply M with that mask

   +/∘, sum the flattened matrix

  ⍺= Boolean indicating where S equals those sums

  ⍸ indices where that's true

share|improve this answer

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
  – ngn
  Sep 5 at 10:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
  – Adám
  Sep 5 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  passing ⍵ as ⍺ in the inner dfn is equally confusing to me
  – ngn
  Sep 5 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
  – ngn
  Sep 5 at 10:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Yeah, I wanted to do something like that. Thanks!
  – Adám
  Sep 5 at 11:59
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

APL (Dyalog Unicode), 35 33 28 bytes^SBCS

-7 thanks to ngn.

Anonymous infix lambda. Takes S as left argument and M as right argument.

⍵[⍸⍺=(+/∘,⍵×∧/∘.∧∧⌿)¨⍵≠⊂⍵]

Try it online!

… "dfn", ⍺ and ⍵ are left and right arguments (S and M) respectively:

 ⍵[…] index M with the following coordinates:

  ⊂⍵ enclose M to treat it as a single element

  ⍵= compare each element (i.e. k candidate) of M to that entire M

  (…)¨ apply the following tacit function to each:

   ∧⌿ vertical AND reduction (finds columns without that k candidate)

…∘.∧ Cartesian Boolean product with:

    ∧/ horizontal AND reduction (finds rows without that k candidate)

   ⍵× multiply M with that mask

   +/∘, sum the flattened matrix

  ⍺= Boolean indicating where S equals those sums

  ⍸ indices where that's true

share|improve this answer

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

APL (Dyalog Unicode), 35 33 28 bytes^SBCS

-7 thanks to ngn.

Anonymous infix lambda. Takes S as left argument and M as right argument.

⍵[⍸⍺=(+/∘,⍵×∧/∘.∧∧⌿)¨⍵≠⊂⍵]

Try it online!

… "dfn", ⍺ and ⍵ are left and right arguments (S and M) respectively:

 ⍵[…] index M with the following coordinates:

  ⊂⍵ enclose M to treat it as a single element

  ⍵= compare each element (i.e. k candidate) of M to that entire M

  (…)¨ apply the following tacit function to each:

   ∧⌿ vertical AND reduction (finds columns without that k candidate)

…∘.∧ Cartesian Boolean product with:

    ∧/ horizontal AND reduction (finds rows without that k candidate)

   ⍵× multiply M with that mask

   +/∘, sum the flattened matrix

  ⍺= Boolean indicating where S equals those sums

  ⍸ indices where that's true

share|improve this answer

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

share|improve this answer

share|improve this answer

edited Sep 5 at 12:04

edited Sep 5 at 12:04

edited Sep 5 at 12:04

answered Sep 5 at 9:42

Adám

27.7k268185

answered Sep 5 at 9:42

Adám

27.7k268185

answered Sep 5 at 9:42

Adám

27.7k268185

27.7k268185

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
  – ngn
  Sep 5 at 10:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
  – Adám
  Sep 5 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  passing ⍵ as ⍺ in the inner dfn is equally confusing to me
  – ngn
  Sep 5 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
  – ngn
  Sep 5 at 10:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Yeah, I wanted to do something like that. Thanks!
  – Adám
  Sep 5 at 11:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
  – ngn
  Sep 5 at 10:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
  – Adám
  Sep 5 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  passing ⍵ as ⍺ in the inner dfn is equally confusing to me
  – ngn
  Sep 5 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
  – ngn
  Sep 5 at 10:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ngn Yeah, I wanted to do something like that. Thanks!
  – Adám
  Sep 5 at 11:59
  

  
  
  
  

1

1

M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
– ngn
Sep 5 at 10:06

M[⍸⍺=+/,(∧⌿d)/M⌿⍨∧/d←M≠⍵¨M←⍵]
– ngn
Sep 5 at 10:06

@ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
– Adám
Sep 5 at 10:24

@ngn Thanks. I won't use the global, though, as it makes order of evaluation confusing: — How can you index into M when it hasn't been created yet?
– Adám
Sep 5 at 10:24

passing ⍵ as ⍺ in the inner dfn is equally confusing to me
– ngn
Sep 5 at 10:28

passing ⍵ as ⍺ in the inner dfn is equally confusing to me
– ngn
Sep 5 at 10:28

⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
– ngn
Sep 5 at 10:35

⍵[⍸⍺=+/¨(,⍵×∧/∘.∧∧⌿)¨âµâ‰ âŠ‚⍵]
– ngn
Sep 5 at 10:35

@ngn Yeah, I wanted to do something like that. Thanks!
– Adám
Sep 5 at 11:59

@ngn Yeah, I wanted to do something like that. Thanks!
– Adám
Sep 5 at 11:59

add a comment | 

up vote
6
down vote

R, 78 73 bytes

function(m,s)m[Map(function(y)sum(m[(w=-which(m==y,T))[,1],w[,2]]),m)==s]

Try it online!

Does not sort or deduplicate the output.

Credit to J.Doe & Giuseppe for -5 bytes.

share|improve this answer

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  76 bytes
  – J.Doe
  Sep 5 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @J.Doe 73 bytes
  – Giuseppe
  Sep 5 at 12:56
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

R, 78 73 bytes

function(m,s)m[Map(function(y)sum(m[(w=-which(m==y,T))[,1],w[,2]]),m)==s]

Try it online!

Does not sort or deduplicate the output.

Credit to J.Doe & Giuseppe for -5 bytes.

share|improve this answer

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  76 bytes
  – J.Doe
  Sep 5 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @J.Doe 73 bytes
  – Giuseppe
  Sep 5 at 12:56
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

R, 78 73 bytes

function(m,s)m[Map(function(y)sum(m[(w=-which(m==y,T))[,1],w[,2]]),m)==s]

Try it online!

Does not sort or deduplicate the output.

Credit to J.Doe & Giuseppe for -5 bytes.

share|improve this answer

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

R, 78 73 bytes

function(m,s)m[Map(function(y)sum(m[(w=-which(m==y,T))[,1],w[,2]]),m)==s]

Try it online!

Does not sort or deduplicate the output.

Credit to J.Doe & Giuseppe for -5 bytes.

share|improve this answer

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

share|improve this answer

share|improve this answer

edited Sep 5 at 13:03

edited Sep 5 at 13:03

edited Sep 5 at 13:03

answered Sep 5 at 11:40

Kirill L.

2,4061116

answered Sep 5 at 11:40

Kirill L.

2,4061116

answered Sep 5 at 11:40

Kirill L.

2,4061116

2,4061116

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  76 bytes
  – J.Doe
  Sep 5 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @J.Doe 73 bytes
  – Giuseppe
  Sep 5 at 12:56
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  76 bytes
  – J.Doe
  Sep 5 at 12:55
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  @J.Doe 73 bytes
  – Giuseppe
  Sep 5 at 12:56
  
  

  
  
  
  

4

4

76 bytes
– J.Doe
Sep 5 at 12:55

76 bytes
– J.Doe
Sep 5 at 12:55

4

4

@J.Doe 73 bytes
– Giuseppe
Sep 5 at 12:56

@J.Doe 73 bytes
– Giuseppe
Sep 5 at 12:56

add a comment | 

up vote
5
down vote

Jelly, 20 19 17 15 14 bytes

pZnⱮFȦ€€ḋFẹƓịF

This is a monadic link that takes M as argument and reads S from STDIN.

Try it online!

How it works

pZnⱮFȦ€€ḋFẹƓịF Main link. Argument: M

 Z Zip; transpose the rows and columns of M.
p Take the Cartesian product of M and its transpose, yielding all pairs
 (r, c) of rows and columns of M.
 F Flatten; yield the elements of M.
 nâ±® Not equal map; for each element e of M, compare the elements of the
 pairs (r, c) with e.
 Ȧ€€ All each each; for each array of Booleans corresponding to an (r, c)
 pair, test if all of them are true.
 F Flatten; yield the elements of M.
 ḋ Take the dot product of each list of resulting Booleans and the
 elements of M.
 Ɠ Read an integer S from STDIN.
 ẹ Find all indices of S in the dot products.
 F Flatten; yield the elements of M.
 ị Retrieve the elements of the right at the indices from the left.

share|improve this answer

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Mark my words lol :) Nice answer, +1
  – Mr. Xcoder
  Sep 5 at 17:26
  

  
  
  
  

add a comment | 

up vote
5
down vote

Jelly, 20 19 17 15 14 bytes

pZnⱮFȦ€€ḋFẹƓịF

This is a monadic link that takes M as argument and reads S from STDIN.

Try it online!

How it works

pZnⱮFȦ€€ḋFẹƓịF Main link. Argument: M

 Z Zip; transpose the rows and columns of M.
p Take the Cartesian product of M and its transpose, yielding all pairs
 (r, c) of rows and columns of M.
 F Flatten; yield the elements of M.
 nâ±® Not equal map; for each element e of M, compare the elements of the
 pairs (r, c) with e.
 Ȧ€€ All each each; for each array of Booleans corresponding to an (r, c)
 pair, test if all of them are true.
 F Flatten; yield the elements of M.
 ḋ Take the dot product of each list of resulting Booleans and the
 elements of M.
 Ɠ Read an integer S from STDIN.
 ẹ Find all indices of S in the dot products.
 F Flatten; yield the elements of M.
 ị Retrieve the elements of the right at the indices from the left.

share|improve this answer

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Mark my words lol :) Nice answer, +1
  – Mr. Xcoder
  Sep 5 at 17:26
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

Jelly, 20 19 17 15 14 bytes

pZnⱮFȦ€€ḋFẹƓịF

This is a monadic link that takes M as argument and reads S from STDIN.

Try it online!

How it works

pZnⱮFȦ€€ḋFẹƓịF Main link. Argument: M

 Z Zip; transpose the rows and columns of M.
p Take the Cartesian product of M and its transpose, yielding all pairs
 (r, c) of rows and columns of M.
 F Flatten; yield the elements of M.
 nâ±® Not equal map; for each element e of M, compare the elements of the
 pairs (r, c) with e.
 Ȧ€€ All each each; for each array of Booleans corresponding to an (r, c)
 pair, test if all of them are true.
 F Flatten; yield the elements of M.
 ḋ Take the dot product of each list of resulting Booleans and the
 elements of M.
 Ɠ Read an integer S from STDIN.
 ẹ Find all indices of S in the dot products.
 F Flatten; yield the elements of M.
 ị Retrieve the elements of the right at the indices from the left.

share|improve this answer

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

Jelly, 20 19 17 15 14 bytes

pZnⱮFȦ€€ḋFẹƓịF

This is a monadic link that takes M as argument and reads S from STDIN.

Try it online!

How it works

pZnⱮFȦ€€ḋFẹƓịF Main link. Argument: M

 Z Zip; transpose the rows and columns of M.
p Take the Cartesian product of M and its transpose, yielding all pairs
 (r, c) of rows and columns of M.
 F Flatten; yield the elements of M.
 nâ±® Not equal map; for each element e of M, compare the elements of the
 pairs (r, c) with e.
 Ȧ€€ All each each; for each array of Booleans corresponding to an (r, c)
 pair, test if all of them are true.
 F Flatten; yield the elements of M.
 ḋ Take the dot product of each list of resulting Booleans and the
 elements of M.
 Ɠ Read an integer S from STDIN.
 ẹ Find all indices of S in the dot products.
 F Flatten; yield the elements of M.
 ị Retrieve the elements of the right at the indices from the left.

share|improve this answer

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

share|improve this answer

share|improve this answer

edited Sep 5 at 17:31

edited Sep 5 at 17:31

edited Sep 5 at 17:31

answered Sep 5 at 15:37

Dennis♦

182k32291722

answered Sep 5 at 15:37

Dennis♦

182k32291722

answered Sep 5 at 15:37

Dennis♦

182k32291722

182k32291722

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Mark my words lol :) Nice answer, +1
  – Mr. Xcoder
  Sep 5 at 17:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Mark my words lol :) Nice answer, +1
  – Mr. Xcoder
  Sep 5 at 17:26
  

  
  
  
  

Mark my words lol :) Nice answer, +1
– Mr. Xcoder
Sep 5 at 17:26

Mark my words lol :) Nice answer, +1
– Mr. Xcoder
Sep 5 at 17:26

add a comment | 

up vote
5
down vote

Haskell, 88 86 84 77 bytes

• -2 bytes thanks to BWO
  


• -7 bytes thanks to Tesseract
  

m!s=[k|k<-m>>=id,s==sum[x|r<-m,all(/=k)r,(i,x)<-zip[0..]r,all((/=k).(!!i))m]]

Verify all testcases.

Explanation

m ! s = -- function !, taking m and s as input
 [k | -- the list of all k's such that
 k <- m >>= id, -- * k is an entry of m
 s == sum -- * s equals the sum of
 [x | -- the list of x's such that
 r <- m, -- * r is a row of m
 all (/= k) r, -- * r does not contain k
 (i, x) <- zip [0 ..] r, -- * i is a valid column index; also let x = r[i]
 all ((/= k) . (!! i)) m -- * none of the rows contain k at index i
 ]
 ]

share|improve this answer

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should that say “function f”?
  – Quintec
  Sep 5 at 11:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
  – Delfad0r
  Sep 5 at 11:51
  

  
  
  
  

add a comment | 

up vote
5
down vote

Haskell, 88 86 84 77 bytes

• -2 bytes thanks to BWO
  


• -7 bytes thanks to Tesseract
  

m!s=[k|k<-m>>=id,s==sum[x|r<-m,all(/=k)r,(i,x)<-zip[0..]r,all((/=k).(!!i))m]]

Verify all testcases.

Explanation

m ! s = -- function !, taking m and s as input
 [k | -- the list of all k's such that
 k <- m >>= id, -- * k is an entry of m
 s == sum -- * s equals the sum of
 [x | -- the list of x's such that
 r <- m, -- * r is a row of m
 all (/= k) r, -- * r does not contain k
 (i, x) <- zip [0 ..] r, -- * i is a valid column index; also let x = r[i]
 all ((/= k) . (!! i)) m -- * none of the rows contain k at index i
 ]
 ]

share|improve this answer

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should that say “function f”?
  – Quintec
  Sep 5 at 11:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
  – Delfad0r
  Sep 5 at 11:51
  

  
  
  
  

add a comment | 

up vote
5
down vote

up vote
5
down vote

Haskell, 88 86 84 77 bytes

• -2 bytes thanks to BWO
  


• -7 bytes thanks to Tesseract
  

m!s=[k|k<-m>>=id,s==sum[x|r<-m,all(/=k)r,(i,x)<-zip[0..]r,all((/=k).(!!i))m]]

Verify all testcases.

Explanation

m ! s = -- function !, taking m and s as input
 [k | -- the list of all k's such that
 k <- m >>= id, -- * k is an entry of m
 s == sum -- * s equals the sum of
 [x | -- the list of x's such that
 r <- m, -- * r is a row of m
 all (/= k) r, -- * r does not contain k
 (i, x) <- zip [0 ..] r, -- * i is a valid column index; also let x = r[i]
 all ((/= k) . (!! i)) m -- * none of the rows contain k at index i
 ]
 ]

share|improve this answer

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Haskell, 88 86 84 77 bytes

• -2 bytes thanks to BWO
  


• -7 bytes thanks to Tesseract
  

m!s=[k|k<-m>>=id,s==sum[x|r<-m,all(/=k)r,(i,x)<-zip[0..]r,all((/=k).(!!i))m]]

Verify all testcases.

Explanation

m ! s = -- function !, taking m and s as input
 [k | -- the list of all k's such that
 k <- m >>= id, -- * k is an entry of m
 s == sum -- * s equals the sum of
 [x | -- the list of x's such that
 r <- m, -- * r is a row of m
 all (/= k) r, -- * r does not contain k
 (i, x) <- zip [0 ..] r, -- * i is a valid column index; also let x = r[i]
 all ((/= k) . (!! i)) m -- * none of the rows contain k at index i
 ]
 ]

share|improve this answer

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this answer

share|improve this answer

edited Sep 5 at 18:42

edited Sep 5 at 18:42

edited Sep 5 at 18:42

answered Sep 5 at 10:49

Delfad0r

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Sep 5 at 10:49

Delfad0r

914

answered Sep 5 at 10:49

Delfad0r

914

914

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Delfad0r is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should that say “function f”?
  – Quintec
  Sep 5 at 11:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
  – Delfad0r
  Sep 5 at 11:51
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Should that say “function f”?
  – Quintec
  Sep 5 at 11:47
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
  – Delfad0r
  Sep 5 at 11:51
  

  
  
  
  

Should that say “function f”?
– Quintec
Sep 5 at 11:47

Should that say “function f”?
– Quintec
Sep 5 at 11:47

1

1

@Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
– Delfad0r
Sep 5 at 11:51

@Quintec Indeed it should have, but I changed it to "function !" to save 2 bytes thanks to BWO
– Delfad0r
Sep 5 at 11:51

add a comment | 

up vote
5
down vote

Pyth,  27 23 22 21  20 bytes

fqvzss.DRsxLTQ-I#TQs

Test suite!

Does not deduplicate.

How it works?

fqvzss.DRsxLTQ-I#TQs Full program.
f s Flatten M and keep only those elements T which satisfy:
 qvzss.DRsxLTQ-I#TQ The filtering function. Breakdown:
 -I#TQ Discard the rows that contain T. More elaborate explanation:
 # Q |-> In M, keep only those elements that are...
 I |-> Invariant under (equal to themselves after...)
 - T |-> Removing T.
 Let's call the result of this expression CR (chopped rows).
 xLTQ Map over the rows M and retrieve all indices of T.
 s Collect indices in 1D list (flatten). Call this I.
 .DR For each row left in CR, remove the elements at indices in I.
 ss Sum all the elements of this matrix flattened.
 qvz And then finally check whether they equal S.

share|improve this answer

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

add a comment | 

up vote
5
down vote

Pyth,  27 23 22 21  20 bytes

fqvzss.DRsxLTQ-I#TQs

Test suite!

Does not deduplicate.

How it works?

fqvzss.DRsxLTQ-I#TQs Full program.
f s Flatten M and keep only those elements T which satisfy:
 qvzss.DRsxLTQ-I#TQ The filtering function. Breakdown:
 -I#TQ Discard the rows that contain T. More elaborate explanation:
 # Q |-> In M, keep only those elements that are...
 I |-> Invariant under (equal to themselves after...)
 - T |-> Removing T.
 Let's call the result of this expression CR (chopped rows).
 xLTQ Map over the rows M and retrieve all indices of T.
 s Collect indices in 1D list (flatten). Call this I.
 .DR For each row left in CR, remove the elements at indices in I.
 ss Sum all the elements of this matrix flattened.
 qvz And then finally check whether they equal S.

share|improve this answer

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

add a comment | 

up vote
5
down vote

up vote
5
down vote

Pyth,  27 23 22 21  20 bytes

fqvzss.DRsxLTQ-I#TQs

Test suite!

Does not deduplicate.

How it works?

fqvzss.DRsxLTQ-I#TQs Full program.
f s Flatten M and keep only those elements T which satisfy:
 qvzss.DRsxLTQ-I#TQ The filtering function. Breakdown:
 -I#TQ Discard the rows that contain T. More elaborate explanation:
 # Q |-> In M, keep only those elements that are...
 I |-> Invariant under (equal to themselves after...)
 - T |-> Removing T.
 Let's call the result of this expression CR (chopped rows).
 xLTQ Map over the rows M and retrieve all indices of T.
 s Collect indices in 1D list (flatten). Call this I.
 .DR For each row left in CR, remove the elements at indices in I.
 ss Sum all the elements of this matrix flattened.
 qvz And then finally check whether they equal S.

share|improve this answer

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

Pyth,  27 23 22 21  20 bytes

fqvzss.DRsxLTQ-I#TQs

Test suite!

Does not deduplicate.

How it works?

fqvzss.DRsxLTQ-I#TQs Full program.
f s Flatten M and keep only those elements T which satisfy:
 qvzss.DRsxLTQ-I#TQ The filtering function. Breakdown:
 -I#TQ Discard the rows that contain T. More elaborate explanation:
 # Q |-> In M, keep only those elements that are...
 I |-> Invariant under (equal to themselves after...)
 - T |-> Removing T.
 Let's call the result of this expression CR (chopped rows).
 xLTQ Map over the rows M and retrieve all indices of T.
 s Collect indices in 1D list (flatten). Call this I.
 .DR For each row left in CR, remove the elements at indices in I.
 ss Sum all the elements of this matrix flattened.
 qvz And then finally check whether they equal S.

share|improve this answer

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

share|improve this answer

share|improve this answer

edited Sep 5 at 20:09

edited Sep 5 at 20:09

edited Sep 5 at 20:09

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

answered Sep 5 at 10:03

Mr. Xcoder

30.3k758193

30.3k758193

add a comment | 

add a comment | 

up vote
4
down vote

Python 2, 114 108 bytes

lambda m,s:a for a in sum(m,)if s==sum(v for l in m for i,v in enumerate(l)ifa-set(l)-set(zip(*m)[i]))

Try it online!

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

add a comment | 

up vote
4
down vote

Python 2, 114 108 bytes

lambda m,s:a for a in sum(m,)if s==sum(v for l in m for i,v in enumerate(l)ifa-set(l)-set(zip(*m)[i]))

Try it online!

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

add a comment | 

up vote
4
down vote

up vote
4
down vote

Python 2, 114 108 bytes

lambda m,s:a for a in sum(m,)if s==sum(v for l in m for i,v in enumerate(l)ifa-set(l)-set(zip(*m)[i]))

Try it online!

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

Python 2, 114 108 bytes

lambda m,s:a for a in sum(m,)if s==sum(v for l in m for i,v in enumerate(l)ifa-set(l)-set(zip(*m)[i]))

Try it online!

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

share|improve this answer

share|improve this answer

answered Sep 5 at 10:04

TFeld

11.3k2833

answered Sep 5 at 10:04

TFeld

11.3k2833

answered Sep 5 at 10:04

TFeld

11.3k2833

11.3k2833

add a comment | 

add a comment | 

up vote
4
down vote

Perl 6, 80 bytes

$^s;@^m[*;*].grep($s==sum @m[($!=*.grep(:k,!*.grep($^v)))(@m);[$!([Z] @m)]])

Try it online!

share|improve this answer

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

add a comment | 

up vote
4
down vote

Perl 6, 80 bytes

$^s;@^m[*;*].grep($s==sum @m[($!=*.grep(:k,!*.grep($^v)))(@m);[$!([Z] @m)]])

Try it online!

share|improve this answer

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

add a comment | 

up vote
4
down vote

up vote
4
down vote

Perl 6, 80 bytes

$^s;@^m[*;*].grep($s==sum @m[($!=*.grep(:k,!*.grep($^v)))(@m);[$!([Z] @m)]])

Try it online!

share|improve this answer

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

Perl 6, 80 bytes

$^s;@^m[*;*].grep($s==sum @m[($!=*.grep(:k,!*.grep($^v)))(@m);[$!([Z] @m)]])

Try it online!

share|improve this answer

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

share|improve this answer

share|improve this answer

edited Sep 5 at 11:44

edited Sep 5 at 11:44

edited Sep 5 at 11:44

answered Sep 5 at 11:30

nwellnhof

3,503714

answered Sep 5 at 11:30

nwellnhof

3,503714

answered Sep 5 at 11:30

nwellnhof

3,503714

3,503714

add a comment | 

add a comment | 

up vote
3
down vote

05AB1E, 21 bytes

²˜ʒQεZ+}øεZ<~}ø_*OO¹Q

Try it online!

Only after I've written this answer have I seen Kevin's. I believe this is substantially different, so I am posting it separately. My intuition says that the optimal byte count is around 18, so I'll have to revisit this and see what else I can do. With the current code, it is impossible to write a test suite but I have verified all test cases myself and the results are correct.

Cropping Algorithm

To better illustrate the technique used, let's grab an example, with the element to crop $k=5$:

$$M=left(beginmatrix6&1&bbox[green,1pt]colorwhite5\1&2&8\9&8&bbox[green,1pt]colorwhite5\6&0&4endmatrixright)$$

It first generates the matrix of equality with $k$:

$$left(beginmatrix0&0&1\0&0&0\0&0&1\0&0&0endmatrixright)$$

Then, it goes through $M$ and for each row $R$, it adds $max(R)$ to each element in $R$, highlighting the necessary rows while still preserving the uniqueness of the horizontal positions of the searched element:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0endmatrixright)$$

Then, it goes through the transpose of $M$ and for each column $C$, it performs the operation $(max(C)-1)space ||space cspaceforallspace cin C:$ (where $||$ is 05AB1E's bitwise OR – addition should work too, so replace ~ with + if you want to test that too), which results in:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1endmatrixright)$$

Finally, it maps $0$ to $1$ and all other integers to $0$ and performs element-wise multiplication with $M$:

$$left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0\colorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0endmatrixright):longrightarrow:left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&2&colorgreen0\colorgreen0&colorgreen0&colorgreen0\6&0&colorgreen0endmatrixright)$$

After which the sum of the resulting matrix is computed.

share|improve this answer

edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
  – Kevin Cruijssen
  Sep 5 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
  – Mr. Xcoder
  Sep 5 at 13:24
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

05AB1E, 21 bytes

²˜ʒQεZ+}øεZ<~}ø_*OO¹Q

Try it online!

Only after I've written this answer have I seen Kevin's. I believe this is substantially different, so I am posting it separately. My intuition says that the optimal byte count is around 18, so I'll have to revisit this and see what else I can do. With the current code, it is impossible to write a test suite but I have verified all test cases myself and the results are correct.

Cropping Algorithm

To better illustrate the technique used, let's grab an example, with the element to crop $k=5$:

$$M=left(beginmatrix6&1&bbox[green,1pt]colorwhite5\1&2&8\9&8&bbox[green,1pt]colorwhite5\6&0&4endmatrixright)$$

It first generates the matrix of equality with $k$:

$$left(beginmatrix0&0&1\0&0&0\0&0&1\0&0&0endmatrixright)$$

Then, it goes through $M$ and for each row $R$, it adds $max(R)$ to each element in $R$, highlighting the necessary rows while still preserving the uniqueness of the horizontal positions of the searched element:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0endmatrixright)$$

Then, it goes through the transpose of $M$ and for each column $C$, it performs the operation $(max(C)-1)space ||space cspaceforallspace cin C:$ (where $||$ is 05AB1E's bitwise OR – addition should work too, so replace ~ with + if you want to test that too), which results in:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1endmatrixright)$$

Finally, it maps $0$ to $1$ and all other integers to $0$ and performs element-wise multiplication with $M$:

$$left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0\colorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0endmatrixright):longrightarrow:left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&2&colorgreen0\colorgreen0&colorgreen0&colorgreen0\6&0&colorgreen0endmatrixright)$$

After which the sum of the resulting matrix is computed.

share|improve this answer

edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
  – Kevin Cruijssen
  Sep 5 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
  – Mr. Xcoder
  Sep 5 at 13:24
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

05AB1E, 21 bytes

²˜ʒQεZ+}øεZ<~}ø_*OO¹Q

Try it online!

Only after I've written this answer have I seen Kevin's. I believe this is substantially different, so I am posting it separately. My intuition says that the optimal byte count is around 18, so I'll have to revisit this and see what else I can do. With the current code, it is impossible to write a test suite but I have verified all test cases myself and the results are correct.

Cropping Algorithm

To better illustrate the technique used, let's grab an example, with the element to crop $k=5$:

$$M=left(beginmatrix6&1&bbox[green,1pt]colorwhite5\1&2&8\9&8&bbox[green,1pt]colorwhite5\6&0&4endmatrixright)$$

It first generates the matrix of equality with $k$:

$$left(beginmatrix0&0&1\0&0&0\0&0&1\0&0&0endmatrixright)$$

Then, it goes through $M$ and for each row $R$, it adds $max(R)$ to each element in $R$, highlighting the necessary rows while still preserving the uniqueness of the horizontal positions of the searched element:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0endmatrixright)$$

Then, it goes through the transpose of $M$ and for each column $C$, it performs the operation $(max(C)-1)space ||space cspaceforallspace cin C:$ (where $||$ is 05AB1E's bitwise OR – addition should work too, so replace ~ with + if you want to test that too), which results in:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1endmatrixright)$$

Finally, it maps $0$ to $1$ and all other integers to $0$ and performs element-wise multiplication with $M$:

$$left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0\colorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0endmatrixright):longrightarrow:left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&2&colorgreen0\colorgreen0&colorgreen0&colorgreen0\6&0&colorgreen0endmatrixright)$$

After which the sum of the resulting matrix is computed.

share|improve this answer

edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

05AB1E, 21 bytes

²˜ʒQεZ+}øεZ<~}ø_*OO¹Q

Try it online!

Only after I've written this answer have I seen Kevin's. I believe this is substantially different, so I am posting it separately. My intuition says that the optimal byte count is around 18, so I'll have to revisit this and see what else I can do. With the current code, it is impossible to write a test suite but I have verified all test cases myself and the results are correct.

Cropping Algorithm

To better illustrate the technique used, let's grab an example, with the element to crop $k=5$:

$$M=left(beginmatrix6&1&bbox[green,1pt]colorwhite5\1&2&8\9&8&bbox[green,1pt]colorwhite5\6&0&4endmatrixright)$$

It first generates the matrix of equality with $k$:

$$left(beginmatrix0&0&1\0&0&0\0&0&1\0&0&0endmatrixright)$$

Then, it goes through $M$ and for each row $R$, it adds $max(R)$ to each element in $R$, highlighting the necessary rows while still preserving the uniqueness of the horizontal positions of the searched element:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite2\0&0&0endmatrixright)$$

Then, it goes through the transpose of $M$ and for each column $C$, it performs the operation $(max(C)-1)space ||space cspaceforallspace cin C:$ (where $||$ is 05AB1E's bitwise OR – addition should work too, so replace ~ with + if you want to test that too), which results in:

$$left(beginmatrixcolorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1\colorgreen1&colorgreen1&bbox[green,1pt]colorwhite3\0&0&colorgreen1endmatrixright)$$

Finally, it maps $0$ to $1$ and all other integers to $0$ and performs element-wise multiplication with $M$:

$$left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0\colorgreen0&colorgreen0&colorgreen0\1&1&colorgreen0endmatrixright):longrightarrow:left(beginmatrixcolorgreen0&colorgreen0&colorgreen0\1&2&colorgreen0\colorgreen0&colorgreen0&colorgreen0\6&0&colorgreen0endmatrixright)$$

After which the sum of the resulting matrix is computed.

share|improve this answer

edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

share|improve this answer

share|improve this answer

edited Sep 5 at 13:27

edited Sep 5 at 13:27

edited Sep 5 at 13:27

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

answered Sep 5 at 13:11

Mr. Xcoder

30.3k758193

30.3k758193

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
  – Kevin Cruijssen
  Sep 5 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
  – Mr. Xcoder
  Sep 5 at 13:24
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
  – Kevin Cruijssen
  Sep 5 at 13:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
  – Mr. Xcoder
  Sep 5 at 13:24
  
  

  
  
  
  

1

1

Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
– Kevin Cruijssen
Sep 5 at 13:21

Nice answer! I knew mine would be golfable for sure. I was already happy enough to have it working, including the annoying case of [[1,1,0,1],[2,0,0,2],[2,0,1,0]] which screwed me over for number 1 (which removes every column..) I indeed had slightly below 20 in my head as well as possibility. Too bad there are barely any builtins for matrices, despite the added product ones recently. As for the 1|2 (1 2~ in 05AB1E synthax) resulting in 3, this is because the logical OR acts like a binary OR when numbers other than 0/1 are involved (I think/assume).
– Kevin Cruijssen
Sep 5 at 13:21

@KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
– Mr. Xcoder
Sep 5 at 13:24

@KevinCruijssen Oh you are right! Then, the docs should write bitwise OR, not logical OR. I am going to have to correct that soon. Anyway, bitwise OR should work as well I think. It can be replaced by + anyway I guess, so I hope there will be no issues with it :)
– Mr. Xcoder
Sep 5 at 13:24

add a comment | 

up vote
2
down vote

05AB1E (legacy), 27 26 bytes

˜ʒ©¹ε®å_}¹ζʒ®å_}ζ‚ζ€€OPOIQ

Does not sort nor uniquify the result.

Only works in the legacy (for now), because sum-each seems to be doing some odd things when part of the inner lists are integers and other are lists..

Try it online or verify all test cases.

Explanation:

˜ # Flatten the (implicit) matrix-input
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [6,1,5,1,2,8,9,8,5,6,0,4]
 ʒ # Filter this list by:
 © # Store the current value in a register-variable
 ¹ # Take the matrix-input
 ε } # Map it to:
 ®å_ # 0 if the current number is in this row, 1 if not
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] and 6 → [0,1,1,0]
 ¹ # Take the matrix-input again
 ζ # Swap its rows and columns
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [[6,1,9,6],[1,2,8,0],[5,8,5,4]]
 ʒ } # Filter it by:
 ®å_ # Only keep the inner lists that does not contain the current number
 # i.e. [[6,1,9,6],[1,2,8,0],[5,8,5,4]] and 6 → [[1,2,8,0],[5,8,5,4]]
 # i.e. [[1,2,2],[1,0,0],[0,0,1],[1,2,0]] and 1 → 
 ζ # After filtering, swap it's rows and columns back again
 # i.e. [[1,2,8,0],[5,8,5,4]] → [[1,5],[2,8],[8,5],[0,4]]
 ‚ζ # Pair both lists together and zip them
 # i.e. [0,1,1,0] and [[1,5],[2,8],[8,5],[0,4]]
 # → [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # i.e. [0,1,0] and → [[0,' '],[1,' '],[0,' ']]
 € # Map each inner list / value to:
 €O # Sum each
 # i.e. [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # → [[0,6],[1,10],[1,13],[0,4]]
 # i.e. [[0,' '],[1,' '],[0,' ']]
 # → [[0,0],[1,0],[0,0]]
 # (NOTE: For most test cases just `O` instead of `€€O` would be enough,
 # but not if we removed ALL zipped inner lists for a number, like the 
 # second example above with input [[1,1,0,1],[2,0,0,2],[2,0,1,0]] and 1)
 P # Now take the product of each inner list
 # i.e. [[0,6],[1,10],[1,13],[0,4]] → [0,10,13,0]
 O # Then take the sum of those
 # i.e. [0,10,13,0] → 23
 IQ # And only keep those that are equal to the number-input
 # i.e. 23 and 9 → 0 (falsey), so it's removed from the flattened input

share|improve this answer

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

add a comment | 

up vote
2
down vote

05AB1E (legacy), 27 26 bytes

˜ʒ©¹ε®å_}¹ζʒ®å_}ζ‚ζ€€OPOIQ

Does not sort nor uniquify the result.

Only works in the legacy (for now), because sum-each seems to be doing some odd things when part of the inner lists are integers and other are lists..

Try it online or verify all test cases.

Explanation:

˜ # Flatten the (implicit) matrix-input
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [6,1,5,1,2,8,9,8,5,6,0,4]
 ʒ # Filter this list by:
 © # Store the current value in a register-variable
 ¹ # Take the matrix-input
 ε } # Map it to:
 ®å_ # 0 if the current number is in this row, 1 if not
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] and 6 → [0,1,1,0]
 ¹ # Take the matrix-input again
 ζ # Swap its rows and columns
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [[6,1,9,6],[1,2,8,0],[5,8,5,4]]
 ʒ } # Filter it by:
 ®å_ # Only keep the inner lists that does not contain the current number
 # i.e. [[6,1,9,6],[1,2,8,0],[5,8,5,4]] and 6 → [[1,2,8,0],[5,8,5,4]]
 # i.e. [[1,2,2],[1,0,0],[0,0,1],[1,2,0]] and 1 → 
 ζ # After filtering, swap it's rows and columns back again
 # i.e. [[1,2,8,0],[5,8,5,4]] → [[1,5],[2,8],[8,5],[0,4]]
 ‚ζ # Pair both lists together and zip them
 # i.e. [0,1,1,0] and [[1,5],[2,8],[8,5],[0,4]]
 # → [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # i.e. [0,1,0] and → [[0,' '],[1,' '],[0,' ']]
 € # Map each inner list / value to:
 €O # Sum each
 # i.e. [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # → [[0,6],[1,10],[1,13],[0,4]]
 # i.e. [[0,' '],[1,' '],[0,' ']]
 # → [[0,0],[1,0],[0,0]]
 # (NOTE: For most test cases just `O` instead of `€€O` would be enough,
 # but not if we removed ALL zipped inner lists for a number, like the 
 # second example above with input [[1,1,0,1],[2,0,0,2],[2,0,1,0]] and 1)
 P # Now take the product of each inner list
 # i.e. [[0,6],[1,10],[1,13],[0,4]] → [0,10,13,0]
 O # Then take the sum of those
 # i.e. [0,10,13,0] → 23
 IQ # And only keep those that are equal to the number-input
 # i.e. 23 and 9 → 0 (falsey), so it's removed from the flattened input

share|improve this answer

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

add a comment | 

up vote
2
down vote

up vote
2
down vote

05AB1E (legacy), 27 26 bytes

˜ʒ©¹ε®å_}¹ζʒ®å_}ζ‚ζ€€OPOIQ

Does not sort nor uniquify the result.

Only works in the legacy (for now), because sum-each seems to be doing some odd things when part of the inner lists are integers and other are lists..

Try it online or verify all test cases.

Explanation:

˜ # Flatten the (implicit) matrix-input
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [6,1,5,1,2,8,9,8,5,6,0,4]
 ʒ # Filter this list by:
 © # Store the current value in a register-variable
 ¹ # Take the matrix-input
 ε } # Map it to:
 ®å_ # 0 if the current number is in this row, 1 if not
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] and 6 → [0,1,1,0]
 ¹ # Take the matrix-input again
 ζ # Swap its rows and columns
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [[6,1,9,6],[1,2,8,0],[5,8,5,4]]
 ʒ } # Filter it by:
 ®å_ # Only keep the inner lists that does not contain the current number
 # i.e. [[6,1,9,6],[1,2,8,0],[5,8,5,4]] and 6 → [[1,2,8,0],[5,8,5,4]]
 # i.e. [[1,2,2],[1,0,0],[0,0,1],[1,2,0]] and 1 → 
 ζ # After filtering, swap it's rows and columns back again
 # i.e. [[1,2,8,0],[5,8,5,4]] → [[1,5],[2,8],[8,5],[0,4]]
 ‚ζ # Pair both lists together and zip them
 # i.e. [0,1,1,0] and [[1,5],[2,8],[8,5],[0,4]]
 # → [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # i.e. [0,1,0] and → [[0,' '],[1,' '],[0,' ']]
 € # Map each inner list / value to:
 €O # Sum each
 # i.e. [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # → [[0,6],[1,10],[1,13],[0,4]]
 # i.e. [[0,' '],[1,' '],[0,' ']]
 # → [[0,0],[1,0],[0,0]]
 # (NOTE: For most test cases just `O` instead of `€€O` would be enough,
 # but not if we removed ALL zipped inner lists for a number, like the 
 # second example above with input [[1,1,0,1],[2,0,0,2],[2,0,1,0]] and 1)
 P # Now take the product of each inner list
 # i.e. [[0,6],[1,10],[1,13],[0,4]] → [0,10,13,0]
 O # Then take the sum of those
 # i.e. [0,10,13,0] → 23
 IQ # And only keep those that are equal to the number-input
 # i.e. 23 and 9 → 0 (falsey), so it's removed from the flattened input

share|improve this answer

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

05AB1E (legacy), 27 26 bytes

˜ʒ©¹ε®å_}¹ζʒ®å_}ζ‚ζ€€OPOIQ

Does not sort nor uniquify the result.

Only works in the legacy (for now), because sum-each seems to be doing some odd things when part of the inner lists are integers and other are lists..

Try it online or verify all test cases.

Explanation:

˜ # Flatten the (implicit) matrix-input
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [6,1,5,1,2,8,9,8,5,6,0,4]
 ʒ # Filter this list by:
 © # Store the current value in a register-variable
 ¹ # Take the matrix-input
 ε } # Map it to:
 ®å_ # 0 if the current number is in this row, 1 if not
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] and 6 → [0,1,1,0]
 ¹ # Take the matrix-input again
 ζ # Swap its rows and columns
 # i.e. [[6,1,5],[1,2,8],[9,8,5],[6,0,4]] → [[6,1,9,6],[1,2,8,0],[5,8,5,4]]
 ʒ } # Filter it by:
 ®å_ # Only keep the inner lists that does not contain the current number
 # i.e. [[6,1,9,6],[1,2,8,0],[5,8,5,4]] and 6 → [[1,2,8,0],[5,8,5,4]]
 # i.e. [[1,2,2],[1,0,0],[0,0,1],[1,2,0]] and 1 → 
 ζ # After filtering, swap it's rows and columns back again
 # i.e. [[1,2,8,0],[5,8,5,4]] → [[1,5],[2,8],[8,5],[0,4]]
 ‚ζ # Pair both lists together and zip them
 # i.e. [0,1,1,0] and [[1,5],[2,8],[8,5],[0,4]]
 # → [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # i.e. [0,1,0] and → [[0,' '],[1,' '],[0,' ']]
 € # Map each inner list / value to:
 €O # Sum each
 # i.e. [[0,[1,5]],[1,[2,8]],[1,[8,5]],[0,[0,4]]]
 # → [[0,6],[1,10],[1,13],[0,4]]
 # i.e. [[0,' '],[1,' '],[0,' ']]
 # → [[0,0],[1,0],[0,0]]
 # (NOTE: For most test cases just `O` instead of `€€O` would be enough,
 # but not if we removed ALL zipped inner lists for a number, like the 
 # second example above with input [[1,1,0,1],[2,0,0,2],[2,0,1,0]] and 1)
 P # Now take the product of each inner list
 # i.e. [[0,6],[1,10],[1,13],[0,4]] → [0,10,13,0]
 O # Then take the sum of those
 # i.e. [0,10,13,0] → 23
 IQ # And only keep those that are equal to the number-input
 # i.e. 23 and 9 → 0 (falsey), so it's removed from the flattened input

share|improve this answer

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

share|improve this answer

share|improve this answer

edited Sep 5 at 13:06

edited Sep 5 at 13:06

edited Sep 5 at 13:06

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

answered Sep 5 at 11:58

Kevin Cruijssen

29.6k551164

29.6k551164

add a comment | 

add a comment | 

up vote
1
down vote

Jelly, 22 bytes

=+Ṁ$€Z$⁺’¬×⁸FS
F³ç=⁹ƲƇ

Try it online!

-6 bytes using the general approach from Mr. Xcoder's 05AB1E answer.

share|improve this answer

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

add a comment | 

up vote
1
down vote

Jelly, 22 bytes

=+Ṁ$€Z$⁺’¬×⁸FS
F³ç=⁹ƲƇ

Try it online!

-6 bytes using the general approach from Mr. Xcoder's 05AB1E answer.

share|improve this answer

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

add a comment | 

up vote
1
down vote

up vote
1
down vote

Jelly, 22 bytes

=+Ṁ$€Z$⁺’¬×⁸FS
F³ç=⁹ƲƇ

Try it online!

-6 bytes using the general approach from Mr. Xcoder's 05AB1E answer.

share|improve this answer

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

Jelly, 22 bytes

=+Ṁ$€Z$⁺’¬×⁸FS
F³ç=⁹ƲƇ

Try it online!

-6 bytes using the general approach from Mr. Xcoder's 05AB1E answer.

share|improve this answer

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

share|improve this answer

share|improve this answer

edited Sep 5 at 14:50

edited Sep 5 at 14:50

edited Sep 5 at 14:50

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

answered Sep 5 at 14:26

HyperNeutrino

18.6k437147

18.6k437147

add a comment | 

add a comment | 

up vote
1
down vote

Charcoal, 33 bytes

ＦθＦι⊞υκＩΦυ⁼ηΣＥθ∧¬№λιΣＥλ∧¬№Ｅθ§πξιν

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

ＦθＦι⊞υκ

Flatten the first input array q into the predefined list u.

 υ Flattened array
 Φ Filter elements
 θ Input array
 ï¼¥ Map over rows
 ι Current element
 λ Current row
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 λ Current row
 ï¼¥ Map over columns
 θ Input array
 ï¼¥ Map over rows
 π Inner row
 ξ Column index
 § Inner element
 ι Current element
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 ν Current element
 Σ Sum
 Σ Sum
 η Second input
 ⁼ Equals
Ｉ Cast to string
 Implicitly print each result on its own line

For each element of the list, sum the array, but if the row contains the element then use 0 instead of its sum, and when summing rows that don't contain the element, if the column contains the element then use 0 instead of the column's value. This is very slightly golfier than filtering the elements out as Charcoal can't sum an empty list.

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

add a comment | 

up vote
1
down vote

Charcoal, 33 bytes

ＦθＦι⊞υκＩΦυ⁼ηΣＥθ∧¬№λιΣＥλ∧¬№Ｅθ§πξιν

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

ＦθＦι⊞υκ

Flatten the first input array q into the predefined list u.

 υ Flattened array
 Φ Filter elements
 θ Input array
 ï¼¥ Map over rows
 ι Current element
 λ Current row
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 λ Current row
 ï¼¥ Map over columns
 θ Input array
 ï¼¥ Map over rows
 π Inner row
 ξ Column index
 § Inner element
 ι Current element
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 ν Current element
 Σ Sum
 Σ Sum
 η Second input
 ⁼ Equals
Ｉ Cast to string
 Implicitly print each result on its own line

For each element of the list, sum the array, but if the row contains the element then use 0 instead of its sum, and when summing rows that don't contain the element, if the column contains the element then use 0 instead of the column's value. This is very slightly golfier than filtering the elements out as Charcoal can't sum an empty list.

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

add a comment | 

up vote
1
down vote

up vote
1
down vote

Charcoal, 33 bytes

ＦθＦι⊞υκＩΦυ⁼ηΣＥθ∧¬№λιΣＥλ∧¬№Ｅθ§πξιν

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

ＦθＦι⊞υκ

Flatten the first input array q into the predefined list u.

 υ Flattened array
 Φ Filter elements
 θ Input array
 ï¼¥ Map over rows
 ι Current element
 λ Current row
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 λ Current row
 ï¼¥ Map over columns
 θ Input array
 ï¼¥ Map over rows
 π Inner row
 ξ Column index
 § Inner element
 ι Current element
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 ν Current element
 Σ Sum
 Σ Sum
 η Second input
 ⁼ Equals
Ｉ Cast to string
 Implicitly print each result on its own line

For each element of the list, sum the array, but if the row contains the element then use 0 instead of its sum, and when summing rows that don't contain the element, if the column contains the element then use 0 instead of the column's value. This is very slightly golfier than filtering the elements out as Charcoal can't sum an empty list.

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

Charcoal, 33 bytes

ＦθＦι⊞υκＩΦυ⁼ηΣＥθ∧¬№λιΣＥλ∧¬№Ｅθ§πξιν

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

ＦθＦι⊞υκ

Flatten the first input array q into the predefined list u.

 υ Flattened array
 Φ Filter elements
 θ Input array
 ï¼¥ Map over rows
 ι Current element
 λ Current row
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 λ Current row
 ï¼¥ Map over columns
 θ Input array
 ï¼¥ Map over rows
 π Inner row
 ξ Column index
 § Inner element
 ι Current element
 № Count matching elements
 ¬ Logical Not
 ∧ Logical And
 ν Current element
 Σ Sum
 Σ Sum
 η Second input
 ⁼ Equals
Ｉ Cast to string
 Implicitly print each result on its own line

For each element of the list, sum the array, but if the row contains the element then use 0 instead of its sum, and when summing rows that don't contain the element, if the column contains the element then use 0 instead of the column's value. This is very slightly golfier than filtering the elements out as Charcoal can't sum an empty list.

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

share|improve this answer

share|improve this answer

answered Sep 5 at 15:59

Neil

75.1k744170

answered Sep 5 at 15:59

Neil

75.1k744170

answered Sep 5 at 15:59

Neil

75.1k744170

75.1k744170

add a comment | 

add a comment | 

up vote
1
down vote

Clean, 92 bytes

import StdEnv
$m s=[c\r<-m,c<-r|sum[b\a<-m|all((<>)c)a,b<-a&x<-[0..]|all(u=u!!x<>c)m]==s]

Try it online!

Explained:

$ m s // the function $ of `m` and `s`
 = [ // is equal to
 c // the value `c`
 \ r <- m // for every row `r` in `m`
 , c <- r // for every value `c` in `r`
 | sum [ // where the sum of
 b // the value `b`
 \ a <- m // for every row `a` in `m`
 | all ((<>)c) a // where `c` isn't in `a`
 , b <- a // for every value `b` in `a`
 & x <- [0..] // with every column index `x` from zero
 | all (u = u!!x <> c) m // where `c` isn't in column `x`
 ] == s // equals `s`
 ]

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

add a comment | 

up vote
1
down vote

Clean, 92 bytes

import StdEnv
$m s=[c\r<-m,c<-r|sum[b\a<-m|all((<>)c)a,b<-a&x<-[0..]|all(u=u!!x<>c)m]==s]

Try it online!

Explained:

$ m s // the function $ of `m` and `s`
 = [ // is equal to
 c // the value `c`
 \ r <- m // for every row `r` in `m`
 , c <- r // for every value `c` in `r`
 | sum [ // where the sum of
 b // the value `b`
 \ a <- m // for every row `a` in `m`
 | all ((<>)c) a // where `c` isn't in `a`
 , b <- a // for every value `b` in `a`
 & x <- [0..] // with every column index `x` from zero
 | all (u = u!!x <> c) m // where `c` isn't in column `x`
 ] == s // equals `s`
 ]

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

add a comment | 

up vote
1
down vote

up vote
1
down vote

Clean, 92 bytes

import StdEnv
$m s=[c\r<-m,c<-r|sum[b\a<-m|all((<>)c)a,b<-a&x<-[0..]|all(u=u!!x<>c)m]==s]

Try it online!

Explained:

$ m s // the function $ of `m` and `s`
 = [ // is equal to
 c // the value `c`
 \ r <- m // for every row `r` in `m`
 , c <- r // for every value `c` in `r`
 | sum [ // where the sum of
 b // the value `b`
 \ a <- m // for every row `a` in `m`
 | all ((<>)c) a // where `c` isn't in `a`
 , b <- a // for every value `b` in `a`
 & x <- [0..] // with every column index `x` from zero
 | all (u = u!!x <> c) m // where `c` isn't in column `x`
 ] == s // equals `s`
 ]

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

Clean, 92 bytes

import StdEnv
$m s=[c\r<-m,c<-r|sum[b\a<-m|all((<>)c)a,b<-a&x<-[0..]|all(u=u!!x<>c)m]==s]

Try it online!

Explained:

$ m s // the function $ of `m` and `s`
 = [ // is equal to
 c // the value `c`
 \ r <- m // for every row `r` in `m`
 , c <- r // for every value `c` in `r`
 | sum [ // where the sum of
 b // the value `b`
 \ a <- m // for every row `a` in `m`
 | all ((<>)c) a // where `c` isn't in `a`
 , b <- a // for every value `b` in `a`
 & x <- [0..] // with every column index `x` from zero
 | all (u = u!!x <> c) m // where `c` isn't in column `x`
 ] == s // equals `s`
 ]

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

share|improve this answer

share|improve this answer

answered Sep 5 at 23:05

Οurous

5,2131931

answered Sep 5 at 23:05

Οurous

5,2131931

answered Sep 5 at 23:05

Οurous

5,2131931

5,2131931

add a comment | 

add a comment | 

up vote
0
down vote

MATLAB - 78 bytes

Compacted:

function f(M,s);for k=1:9;if sum(sum(M(~sum(M==k,2),~sum(M==k))))==s;k,end;end

And in a fully developped version:

function getthesum(M,s)

for k=1:9 % For each single digit
 x = M==k ; % Index elements equal to "k"
 N = M( ~sum(x,2) , ~sum(x) ) ; % New matrix with only the appropriate rows/columns
 if sum(sum(N))==s % sum rows and columns and compare to "s"
 k % display "k" in console if "k" is valid
 end
end

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  k could be any element of the matrix.
  – Dennis♦
  Sep 5 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
  – Hoki
  Sep 6 at 11:04
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

MATLAB - 78 bytes

Compacted:

function f(M,s);for k=1:9;if sum(sum(M(~sum(M==k,2),~sum(M==k))))==s;k,end;end

And in a fully developped version:

function getthesum(M,s)

for k=1:9 % For each single digit
 x = M==k ; % Index elements equal to "k"
 N = M( ~sum(x,2) , ~sum(x) ) ; % New matrix with only the appropriate rows/columns
 if sum(sum(N))==s % sum rows and columns and compare to "s"
 k % display "k" in console if "k" is valid
 end
end

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  k could be any element of the matrix.
  – Dennis♦
  Sep 5 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
  – Hoki
  Sep 6 at 11:04
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

MATLAB - 78 bytes

Compacted:

function f(M,s);for k=1:9;if sum(sum(M(~sum(M==k,2),~sum(M==k))))==s;k,end;end

And in a fully developped version:

function getthesum(M,s)

for k=1:9 % For each single digit
 x = M==k ; % Index elements equal to "k"
 N = M( ~sum(x,2) , ~sum(x) ) ; % New matrix with only the appropriate rows/columns
 if sum(sum(N))==s % sum rows and columns and compare to "s"
 k % display "k" in console if "k" is valid
 end
end

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

MATLAB - 78 bytes

Compacted:

function f(M,s);for k=1:9;if sum(sum(M(~sum(M==k,2),~sum(M==k))))==s;k,end;end

And in a fully developped version:

function getthesum(M,s)

for k=1:9 % For each single digit
 x = M==k ; % Index elements equal to "k"
 N = M( ~sum(x,2) , ~sum(x) ) ; % New matrix with only the appropriate rows/columns
 if sum(sum(N))==s % sum rows and columns and compare to "s"
 k % display "k" in console if "k" is valid
 end
end

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

share|improve this answer

share|improve this answer

answered Sep 5 at 18:22

Hoki

24127

answered Sep 5 at 18:22

Hoki

24127

answered Sep 5 at 18:22

Hoki

24127

24127

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  k could be any element of the matrix.
  – Dennis♦
  Sep 5 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
  – Hoki
  Sep 6 at 11:04
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  k could be any element of the matrix.
  – Dennis♦
  Sep 5 at 19:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
  – Hoki
  Sep 6 at 11:04
  
  

  
  
  
  

1

1

k could be any element of the matrix.
– Dennis♦
Sep 5 at 19:12

k could be any element of the matrix.
– Dennis♦
Sep 5 at 19:12

@Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
– Hoki
Sep 6 at 11:04

@Dennis, oops that's right ... My bad, i'll correct it later today. Thanks for pointing it out.
– Hoki
Sep 6 at 11:04

add a comment | 

 

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