Missing numbers in arithmetic sum

Clash Royale CLAN TAG#URR8PPP

up vote
13
down vote

favorite

Challenge

Giving a valid arithmetic sum with some missing numbers, output the full expression.

Example:

 1#3 123
+ 45# => + 456
-------- --------
 579 579

Input

• Expression format can be an array ["1#3", "45#", "579"], a string "1#3+45#=579", or 3 inputs f("1#3","45#","579")
  

Output

• Same as input


• You don't need to output the result

Notes

• The missing numbers are going to be represented using # or any other constant non-numeric character you want


• Assume result wont have a missing number


• Assume Input/Output consist in 2 terms and a final result


• Assume both term > 0 and result >= 2


• There might be multiple solutions. You can output anyone as long as the sum result match

Test Cases with possibly outputs (pretty format)

 #79 879
+ 44# => + 444
-------- --------
 1323 1323

 5#5 555
+ 3#3 => + 343
-------- --------
 898 898

 # 1
+ # => + 1
-------- --------
 2 2

 ### 998
+ ### => + 1 PD: there are a lot of possible outputs for this one
-------- --------
 999 999


 123 123
+ # => + 1
-------- --------
 124 124


 9 9
+ #6 => + 46
-------- --------
 55 55


 #123651 1123651
+ #98# => + 7981
------------ -----------
 1131632 1131632

---

Standard code-golf rules apply

code-golf arithmetic integer

share|improve this question

edited Aug 30 at 15:44

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

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  Do we need to strip leading zeros?
  – Mnemonic
  Aug 30 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mnemonic not necessarily
  – Luis felipe De jesus Munoz
  Aug 30 at 14:58
  

  
  
  
  


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  can I take the input with the sides around = swapped? e.g. 579=1#3+45#
  – dzaima
  Aug 30 at 15:25
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  "Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  also your added test-case avoids exactly what I was asking for - the leading zeroes
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  

 | 
show 3 more comments

up vote
13
down vote

favorite

Challenge

Giving a valid arithmetic sum with some missing numbers, output the full expression.

Example:

 1#3 123
+ 45# => + 456
-------- --------
 579 579

Input

• Expression format can be an array ["1#3", "45#", "579"], a string "1#3+45#=579", or 3 inputs f("1#3","45#","579")
  

Output

• Same as input


• You don't need to output the result

Notes

• The missing numbers are going to be represented using # or any other constant non-numeric character you want


• Assume result wont have a missing number


• Assume Input/Output consist in 2 terms and a final result


• Assume both term > 0 and result >= 2


• There might be multiple solutions. You can output anyone as long as the sum result match

Test Cases with possibly outputs (pretty format)

 #79 879
+ 44# => + 444
-------- --------
 1323 1323

 5#5 555
+ 3#3 => + 343
-------- --------
 898 898

 # 1
+ # => + 1
-------- --------
 2 2

 ### 998
+ ### => + 1 PD: there are a lot of possible outputs for this one
-------- --------
 999 999


 123 123
+ # => + 1
-------- --------
 124 124


 9 9
+ #6 => + 46
-------- --------
 55 55


 #123651 1123651
+ #98# => + 7981
------------ -----------
 1131632 1131632

---

Standard code-golf rules apply

code-golf arithmetic integer

share|improve this question

edited Aug 30 at 15:44

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do we need to strip leading zeros?
  – Mnemonic
  Aug 30 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mnemonic not necessarily
  – Luis felipe De jesus Munoz
  Aug 30 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  can I take the input with the sides around = swapped? e.g. 579=1#3+45#
  – dzaima
  Aug 30 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  also your added test-case avoids exactly what I was asking for - the leading zeroes
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  

 | 
show 3 more comments

up vote
13
down vote

favorite

up vote
13
down vote

favorite

Challenge

Giving a valid arithmetic sum with some missing numbers, output the full expression.

Example:

 1#3 123
+ 45# => + 456
-------- --------
 579 579

Input

• Expression format can be an array ["1#3", "45#", "579"], a string "1#3+45#=579", or 3 inputs f("1#3","45#","579")
  

Output

• Same as input


• You don't need to output the result

Notes

• The missing numbers are going to be represented using # or any other constant non-numeric character you want


• Assume result wont have a missing number


• Assume Input/Output consist in 2 terms and a final result


• Assume both term > 0 and result >= 2


• There might be multiple solutions. You can output anyone as long as the sum result match

Test Cases with possibly outputs (pretty format)

 #79 879
+ 44# => + 444
-------- --------
 1323 1323

 5#5 555
+ 3#3 => + 343
-------- --------
 898 898

 # 1
+ # => + 1
-------- --------
 2 2

 ### 998
+ ### => + 1 PD: there are a lot of possible outputs for this one
-------- --------
 999 999


 123 123
+ # => + 1
-------- --------
 124 124


 9 9
+ #6 => + 46
-------- --------
 55 55


 #123651 1123651
+ #98# => + 7981
------------ -----------
 1131632 1131632

---

Standard code-golf rules apply

code-golf arithmetic integer

share|improve this question

edited Aug 30 at 15:44

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

Challenge

Giving a valid arithmetic sum with some missing numbers, output the full expression.

Example:

 1#3 123
+ 45# => + 456
-------- --------
 579 579

Input

• Expression format can be an array ["1#3", "45#", "579"], a string "1#3+45#=579", or 3 inputs f("1#3","45#","579")
  

Output

• Same as input


• You don't need to output the result

Notes

• The missing numbers are going to be represented using # or any other constant non-numeric character you want


• Assume result wont have a missing number


• Assume Input/Output consist in 2 terms and a final result


• Assume both term > 0 and result >= 2


• There might be multiple solutions. You can output anyone as long as the sum result match

Test Cases with possibly outputs (pretty format)

 #79 879
+ 44# => + 444
-------- --------
 1323 1323

 5#5 555
+ 3#3 => + 343
-------- --------
 898 898

 # 1
+ # => + 1
-------- --------
 2 2

 ### 998
+ ### => + 1 PD: there are a lot of possible outputs for this one
-------- --------
 999 999


 123 123
+ # => + 1
-------- --------
 124 124


 9 9
+ #6 => + 46
-------- --------
 55 55


 #123651 1123651
+ #98# => + 7981
------------ -----------
 1131632 1131632

---

Standard code-golf rules apply

code-golf arithmetic integer

share|improve this question

edited Aug 30 at 15:44

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

share|improve this question

share|improve this question

edited Aug 30 at 15:44

edited Aug 30 at 15:44

edited Aug 30 at 15:44

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

asked Aug 30 at 14:36

Luis felipe De jesus Munoz

2,9511044

2,9511044

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do we need to strip leading zeros?
  – Mnemonic
  Aug 30 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mnemonic not necessarily
  – Luis felipe De jesus Munoz
  Aug 30 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  can I take the input with the sides around = swapped? e.g. 579=1#3+45#
  – dzaima
  Aug 30 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  also your added test-case avoids exactly what I was asking for - the leading zeroes
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do we need to strip leading zeros?
  – Mnemonic
  Aug 30 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Mnemonic not necessarily
  – Luis felipe De jesus Munoz
  Aug 30 at 14:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  can I take the input with the sides around = swapped? e.g. 579=1#3+45#
  – dzaima
  Aug 30 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  "Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  also your added test-case avoids exactly what I was asking for - the leading zeroes
  – dzaima
  Aug 30 at 15:48
  

  
  
  
  

Do we need to strip leading zeros?
– Mnemonic
Aug 30 at 14:55

Do we need to strip leading zeros?
– Mnemonic
Aug 30 at 14:55

@Mnemonic not necessarily
– Luis felipe De jesus Munoz
Aug 30 at 14:58

@Mnemonic not necessarily
– Luis felipe De jesus Munoz
Aug 30 at 14:58

can I take the input with the sides around = swapped? e.g. 579=1#3+45#
– dzaima
Aug 30 at 15:25

can I take the input with the sides around = swapped? e.g. 579=1#3+45#
– dzaima
Aug 30 at 15:25

1

1

"Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
– dzaima
Aug 30 at 15:48

"Assume both term > 0" does "assume" mean that I have to output both terms > 0 or that I can assume that there's always a solution with both > 0 but output whatever?
– dzaima
Aug 30 at 15:48

1

1

also your added test-case avoids exactly what I was asking for - the leading zeroes
– dzaima
Aug 30 at 15:48

also your added test-case avoids exactly what I was asking for - the leading zeroes
– dzaima
Aug 30 at 15:48

 | 
show 3 more comments

12 Answers
12

active

oldest

votes

up vote
9
down vote

Brachylog, 22 16 bytes

ᵐ²ịᵐ.k+~t

Try it online!

-6 bytes thanks to @Fatelize

Explanation

ᵐ²ịᵐ.k+~t
 ᵐ² # for each letter in each string
 Ṣ∧Ị∋ # if " " return a digit; else input
 | #
 ịᵐ # cast each string to number
 k+ # the sum of all but the last one
 ~t # is equal to the last one
 . # output that list

share|improve this answer

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

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  ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
  – Fatalize
  Aug 31 at 6:57
  
  

  
  
  
  


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  Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
  – Fatalize
  Aug 31 at 6:59
  

  
  
  
  

add a comment | 

up vote
8
down vote

JavaScript (ES6), 74 57 bytes

Takes input as (a)(b)(result), where a and b are strings with . for unknown digits and result is an integer. Returns an array of 2 integers.

a=>b=>F=(c,n)=>`$r=[c,n]`.match(`^`+[a,b])?r:F(c-1,-~n)

Try it online!

Commented

a => b => // a, b = term patterns (e.g. a = ".79", b = "44.")
 F = (c, // c = expected result (e.g. 1323)
 n) => // n = guessed value of b, initially undefined
 `$r = [c, n]` // we coerce r = [c, n] to a string (e.g. "879,444")
 // if n is still undefined, this gives just c followed by a comma
 .match(`^` + [a, b]) // we coerce [a, b] to a string, prefixed with "^" (e.g. "^.79,44.")
 ? // this is implicitly turned into a regular expression; if matching:
 r // return r
 : // else:
 F(c - 1, -~n) // decrement c, increment n and do a recursive call

share|improve this answer

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

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  Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
  – Kevin Cruijssen
  Aug 31 at 15:59
  
  

  
  
  
  


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  @KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
  – Arnauld
  Aug 31 at 16:08
  

  
  
  
  


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  Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
  – Kevin Cruijssen
  Aug 31 at 16:38
  
  

  
  
  
  


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  @KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
  – Arnauld
  Aug 31 at 17:05
  

  
  
  
  

add a comment | 

up vote
7
down vote

Matlab, 143 134 132 119 115 bytes

function[m]=f(x,y,r),p=@(v)str2num(strrep(v,'#',char(randi([48,57]))));m=[1,1];while sum(m)-r,m=[p(x),p(y)];end;end

-4 bytes thanks to @Luismendo

Try it Online

---

Pretty big and pretty stupid. It simply replaces all # with random digits until it finds the correct ones.

share|improve this answer

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

add a comment | 

up vote
4
down vote

R, 67 51 bytes

Rock simple and scales horribly, just grep all the sum combinations. Use "." for unknown digits. It won't find the same answer as test case number 4, but it will give a possible answer, which follows the letter of the rules as given.

-16 bytes by grepping after forming the output and replacing paste with the ? operator.

function(x,y,z,`?`=paste)grep(x?y,1:z?z:1-1,v=T)[1]

Try it online!

share|improve this answer

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

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  Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
  – JayCe
  Aug 31 at 0:44
  

  
  
  
  


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  I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
  – JayCe
  Aug 31 at 14:26
  

  
  
  
  


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  I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
  – J.Doe
  Aug 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
3
down vote

Charcoal, 32 bytes

Ｆ²⊞υ0Ｗ⁻ζΣＩυ≔Ｅ⟦θη⟧⭆κ⎇⁼μ#‽χμυ←Ｅυ⮌ι

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

Ｆ²⊞υ0

Push two string 0s to the predefined empty list u to get the while loop going.

Ｗ⁻ζΣＩυ

Repeat while the sum of casting the values in u to integer is not equal to the desired result.

≔Ｅ⟦θη⟧

Create an array of the two inputs and map over it.

⭆κ⎇⁼μ#‽χμυ

Replace each # with a random digit and assign the result back to u.

←Ｅυ⮌ι

Print the result right justified. (Left justified would be just υ for a 4-byte saving.)

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

add a comment | 

up vote
3
down vote

Jelly, 20 bytes

ØDṛċ¡V€)ŒpḌð€ŒpS⁼¥ƇḢ

Try it online!

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
3
down vote

05AB1E (legacy), 23 20 bytes

[²³«εð9ÝΩ:}²gôJDO¹Q#

-3 bytes thanks to @Emigna.

Unknown digits are spaces ( ). Input order should be: expected result; longest string; shortest string.

Try it online.

Explanation:

[ # Start an infinite loop
 ²³« # Take the second and third inputs, and merge them together
 # i.e. " 79" and " 4 " → " 79 4 "
 ε } # Map each character to:
 ð : # Replace a space with:
 9ÝΩ # A random digit in the range [0,9]
 # i.e. " 79 4 " → ['3','7','9','2','4','3']
 # i.e. " 79 4 " → ['5','7','9','7','4','4']
²g # Get the length of the second input
 # i.e. " 79" → 3
 ô # And split it into two numbers again
 # i.e. ['3','7','9','2','4','3'] and 3 → [['3','7','9'],['2','4','3']]
 # i.e. ['5','7','9','7','4','4'] and 3 → [['5','7','9'],['7','4','4']]
 J # Join each list together to a single number
 # i.e. [['3','7','9'],['2','4','3']] → [379,243]
 # i.e. [['5','7','9'],['7','4','4']] → [579,744]
 D # Duplicate this list
 O # Sum the list
 # i.e. [379,243] → 622
 # i.e. [579,744] → 1323
 ¹Q# # If it's equal to the first input: stop the infinite loop
 # (and output the duplicate list implicitly)
 # i.e. 1323 and 622 → 0 (falsey) → continue the loop
 # i.e. 1323 and 1323 → 1 (truthy) → stop the loop and output [579,744]

share|improve this answer

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

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  Replace saves 3 over the if.
  – Emigna
  Aug 31 at 6:10
  

  
  
  
  


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  @Emigna Ah, of course. Thanks!
  – Kevin Cruijssen
  Aug 31 at 6:52
  

  
  
  
  

add a comment | 

up vote
3
down vote

Perl 6, 81 74 bytes

-7 bytes thanks to nwellnhof!

first try S/=/==/.EVAL,map $^a;S:g[#]=$a[$++],[X] ^10 xx.comb('#')

Try it online!

Anonymous code block that takes input as a string containing an arithmetical expression e.g. "12#+45#=579". Substitutes each # with possible permutations of digits, substitutes the= with == and finds the first valid result.

Explanation:

 # Anonymous code block 
 first # Find the first of:
 ^10 # The range of 0 to 9
 xx.comb('#') # Multiplied by the number #s in the code
 ,[X] # The cross-product of these lists
 map # Map each crossproduct to:
 $^a;.trans: "#"=>$a[$++] # The given string with each # translated to each element in the list
 try S/=/==/.EVAL, # Find which of these is true when = are changed to == and it is eval'd

share|improve this answer

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

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  You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
  – nwellnhof
  Sep 1 at 13:05
  

  
  
  
  


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  @nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
  – Jo King
  Sep 1 at 13:13
  

  
  
  
  

add a comment | 

up vote
2
down vote

APL (Dyalog Unicode), 22 bytes

'#'⎕R⍕?10t⍣⍎⍺t∘←

Try it online!

share|improve this answer

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

add a comment | 

up vote
2
down vote

Java 10, 203 197 bytes

(a,b,c)->int A=0,B=0,l=a.length();for(a+=b,b="";A+B!=c;A=c.valueOf(b.substring(0,l)),B=c.valueOf(b.substring(l)),b="")for(var t:a.toCharArray())b+=t<36?(int)(Math.random()*9)+"":t;return A+"+"+B;

Try it online.

Explanation:

(a,b,c)-> // Method with 2 Strings & integer parameters and String return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.length(); // Length of the first String-input
 for(a+=b, // Concat the second String-input to the first
 b=""; // Reuse `b`, and start it as an empty String
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=c.valueOf(b.substring(0,l)),
 // Set `A` to the first String-part as integer
 B=c.valueOf(n.substring(l)),
 // Set `B` to the second String-part as integer
 b="") // Reset `b` to an empty String
 for(var t:a.toCharArray())
 // Inner loop over the characters of the concatted String inputs
 b+=t<36? // If the current character is a '#':
 (int)(Math.random()*9)+""
 // Append a random digit to `b`
 : // Else (it already is a digit):
 t; // Append this digit as is to `b`
 return A+"+"+B; // After the loop, return `A` and `B` as result

share|improve this answer

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
1
down vote

C# .NET, 225 220 196 bytes

(a,b,c)=>int A=0,B=0,l=a.Length;for(a+=b,b="";A+B!=c;A=int.Parse(b.Substring(0,l)),B=int.Parse(b.Substring(l)),b="")foreach(var t in a)b+=(t<36?new System.Random().Next(10):t-48)+"";return(A,B);

Port of my Java 10 answer.

(I'm very rusty in C# .NET golfing, so can defintely be golfed..)

-3 bytes implicitly thanks to @user82593 and this new C# tip he added.

-29 bytes thanks to @hvd.

Try it online.

Explanation:

(a,b,c)=> // Method with 2 string & int parameters and int-tuple return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.Length; // Length of the first string-input
 for(a+=b, // Concat the second string-input to the first
 b=""; // Reuse `b`, and start it as an empty string
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=int.Parse(b.Substring(0,l)),
 // Set `A` to the first string-part as integer
 B=int.Parse(b.Substring(l)),
 // Set `B` to the second string-part as integer
 b="") // Reset `b` to an empty string
 foreach(var t in a)
 // Inner loop over the characters of the concatted string inputs
 b+=(t<36? // If the current character is a '#':
 new System.Random().Next(10)
 // Use a random digit
 : // Else (it already is a digit):
 t-48) // Use this digit as is
 +""; // And convert it to a string so it can be appended to the string
 return(A,B); // After the loop, return `A` and `B` in a tuple as result

share|improve this answer

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the regular using System; instead, it is shorter than namespace System.
  – hvd
  Sep 1 at 8:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
  – Kevin Cruijssen
  Sep 1 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
  – hvd
  Sep 1 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
  – Kevin Cruijssen
  Sep 1 at 11:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 3, 121 155 152 149 bytes

import re
def f(i,k=0,S=re.sub):s=S('#','%s',i)%(*list('%0*d'%(i.count('#'),k)),);print(s)if eval(S('=','==',S('\b0*([1-9])','\1',s)))else f(i,k+1)

Try it online!

+34 New solution with regex to circumvent the fact that python doesn't support numbers with leading zeroes.

-3 thanks to @Jonathan Frech

---

The old solution doesn't work if # is the first character in any number (because eval doesn't accept leading zeroes) and is therefore invalid :(

def f(i,k=0):
 s=i.replace('#','%s')%(*list('%0*d'%(i.count('#'),k)),)
 print(s)if eval(s.replace('=','=='))else f(i,k+1)

Try it online!

share|improve this answer

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm afraid this submission is invalid for the reason stated in the post.
  – Erik the Outgolfer
  Aug 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Since your function does not contain any compound statements you can condense it to only one line.
  – Jonathan Frech
  Aug 30 at 23:21
  

  
  
  
  

add a comment | 

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12 Answers
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up vote
9
down vote

Brachylog, 22 16 bytes

ᵐ²ịᵐ.k+~t

Try it online!

-6 bytes thanks to @Fatelize

Explanation

ᵐ²ịᵐ.k+~t
 ᵐ² # for each letter in each string
 Ṣ∧Ị∋ # if " " return a digit; else input
 | #
 ịᵐ # cast each string to number
 k+ # the sum of all but the last one
 ~t # is equal to the last one
 . # output that list

share|improve this answer

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
  – Fatalize
  Aug 31 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
  – Fatalize
  Aug 31 at 6:59
  

  
  
  
  

add a comment | 

up vote
9
down vote

Brachylog, 22 16 bytes

ᵐ²ịᵐ.k+~t

Try it online!

-6 bytes thanks to @Fatelize

Explanation

ᵐ²ịᵐ.k+~t
 ᵐ² # for each letter in each string
 Ṣ∧Ị∋ # if " " return a digit; else input
 | #
 ịᵐ # cast each string to number
 k+ # the sum of all but the last one
 ~t # is equal to the last one
 . # output that list

share|improve this answer

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
  – Fatalize
  Aug 31 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
  – Fatalize
  Aug 31 at 6:59
  

  
  
  
  

add a comment | 

up vote
9
down vote

up vote
9
down vote

Brachylog, 22 16 bytes

ᵐ²ịᵐ.k+~t

Try it online!

-6 bytes thanks to @Fatelize

Explanation

ᵐ²ịᵐ.k+~t
 ᵐ² # for each letter in each string
 Ṣ∧Ị∋ # if " " return a digit; else input
 | #
 ịᵐ # cast each string to number
 k+ # the sum of all but the last one
 ~t # is equal to the last one
 . # output that list

share|improve this answer

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

Brachylog, 22 16 bytes

ᵐ²ịᵐ.k+~t

Try it online!

-6 bytes thanks to @Fatelize

Explanation

ᵐ²ịᵐ.k+~t
 ᵐ² # for each letter in each string
 Ṣ∧Ị∋ # if " " return a digit; else input
 | #
 ịᵐ # cast each string to number
 k+ # the sum of all but the last one
 ~t # is equal to the last one
 . # output that list

share|improve this answer

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

share|improve this answer

share|improve this answer

edited Aug 31 at 8:07

edited Aug 31 at 8:07

edited Aug 31 at 8:07

answered Aug 30 at 23:08

Kroppeb

88628

answered Aug 30 at 23:08

Kroppeb

88628

answered Aug 30 at 23:08

Kroppeb

88628

88628

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
  – Fatalize
  Aug 31 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
  – Fatalize
  Aug 31 at 6:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
  – Fatalize
  Aug 31 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
  – Fatalize
  Aug 31 at 6:59
  

  
  
  
  

1

1

ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
– Fatalize
Aug 31 at 6:57

ᵐ²á»‹áµ.k+~t is 4 bytes shorter. I'm not sure why you did something this convoluted in your map.
– Fatalize
Aug 31 at 6:57

Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
– Fatalize
Aug 31 at 6:59

Since we can use any non-numeric character, you should use e.g. space with á¹¢ instead of "#" which will save two more bytes.
– Fatalize
Aug 31 at 6:59

add a comment | 

up vote
8
down vote

JavaScript (ES6), 74 57 bytes

Takes input as (a)(b)(result), where a and b are strings with . for unknown digits and result is an integer. Returns an array of 2 integers.

a=>b=>F=(c,n)=>`$r=[c,n]`.match(`^`+[a,b])?r:F(c-1,-~n)

Try it online!

Commented

a => b => // a, b = term patterns (e.g. a = ".79", b = "44.")
 F = (c, // c = expected result (e.g. 1323)
 n) => // n = guessed value of b, initially undefined
 `$r = [c, n]` // we coerce r = [c, n] to a string (e.g. "879,444")
 // if n is still undefined, this gives just c followed by a comma
 .match(`^` + [a, b]) // we coerce [a, b] to a string, prefixed with "^" (e.g. "^.79,44.")
 ? // this is implicitly turned into a regular expression; if matching:
 r // return r
 : // else:
 F(c - 1, -~n) // decrement c, increment n and do a recursive call

share|improve this answer

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
  – Kevin Cruijssen
  Aug 31 at 15:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
  – Arnauld
  Aug 31 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
  – Kevin Cruijssen
  Aug 31 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
  – Arnauld
  Aug 31 at 17:05
  

  
  
  
  

add a comment | 

up vote
8
down vote

JavaScript (ES6), 74 57 bytes

Takes input as (a)(b)(result), where a and b are strings with . for unknown digits and result is an integer. Returns an array of 2 integers.

a=>b=>F=(c,n)=>`$r=[c,n]`.match(`^`+[a,b])?r:F(c-1,-~n)

Try it online!

Commented

a => b => // a, b = term patterns (e.g. a = ".79", b = "44.")
 F = (c, // c = expected result (e.g. 1323)
 n) => // n = guessed value of b, initially undefined
 `$r = [c, n]` // we coerce r = [c, n] to a string (e.g. "879,444")
 // if n is still undefined, this gives just c followed by a comma
 .match(`^` + [a, b]) // we coerce [a, b] to a string, prefixed with "^" (e.g. "^.79,44.")
 ? // this is implicitly turned into a regular expression; if matching:
 r // return r
 : // else:
 F(c - 1, -~n) // decrement c, increment n and do a recursive call

share|improve this answer

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
  – Kevin Cruijssen
  Aug 31 at 15:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
  – Arnauld
  Aug 31 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
  – Kevin Cruijssen
  Aug 31 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
  – Arnauld
  Aug 31 at 17:05
  

  
  
  
  

add a comment | 

up vote
8
down vote

up vote
8
down vote

JavaScript (ES6), 74 57 bytes

Takes input as (a)(b)(result), where a and b are strings with . for unknown digits and result is an integer. Returns an array of 2 integers.

a=>b=>F=(c,n)=>`$r=[c,n]`.match(`^`+[a,b])?r:F(c-1,-~n)

Try it online!

Commented

a => b => // a, b = term patterns (e.g. a = ".79", b = "44.")
 F = (c, // c = expected result (e.g. 1323)
 n) => // n = guessed value of b, initially undefined
 `$r = [c, n]` // we coerce r = [c, n] to a string (e.g. "879,444")
 // if n is still undefined, this gives just c followed by a comma
 .match(`^` + [a, b]) // we coerce [a, b] to a string, prefixed with "^" (e.g. "^.79,44.")
 ? // this is implicitly turned into a regular expression; if matching:
 r // return r
 : // else:
 F(c - 1, -~n) // decrement c, increment n and do a recursive call

share|improve this answer

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

JavaScript (ES6), 74 57 bytes

Takes input as (a)(b)(result), where a and b are strings with . for unknown digits and result is an integer. Returns an array of 2 integers.

a=>b=>F=(c,n)=>`$r=[c,n]`.match(`^`+[a,b])?r:F(c-1,-~n)

Try it online!

Commented

a => b => // a, b = term patterns (e.g. a = ".79", b = "44.")
 F = (c, // c = expected result (e.g. 1323)
 n) => // n = guessed value of b, initially undefined
 `$r = [c, n]` // we coerce r = [c, n] to a string (e.g. "879,444")
 // if n is still undefined, this gives just c followed by a comma
 .match(`^` + [a, b]) // we coerce [a, b] to a string, prefixed with "^" (e.g. "^.79,44.")
 ? // this is implicitly turned into a regular expression; if matching:
 r // return r
 : // else:
 F(c - 1, -~n) // decrement c, increment n and do a recursive call

share|improve this answer

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

share|improve this answer

share|improve this answer

edited Aug 30 at 19:19

edited Aug 30 at 19:19

edited Aug 30 at 19:19

answered Aug 30 at 15:23

Arnauld

63.6k580268

answered Aug 30 at 15:23

Arnauld

63.6k580268

answered Aug 30 at 15:23

Arnauld

63.6k580268

63.6k580268

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
  – Kevin Cruijssen
  Aug 31 at 15:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
  – Arnauld
  Aug 31 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
  – Kevin Cruijssen
  Aug 31 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
  – Arnauld
  Aug 31 at 17:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
  – Kevin Cruijssen
  Aug 31 at 15:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
  – Arnauld
  Aug 31 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
  – Kevin Cruijssen
  Aug 31 at 16:38
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
  – Arnauld
  Aug 31 at 17:05
  

  
  
  
  

Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
– Kevin Cruijssen
Aug 31 at 15:59

Ah, so that's what's happening. I tried to understand your code without explanation yesterday (and I'm bad at JS), but didn't understand why -~n couldn't be just n+1 and how F=(c,n)=> was used. Now that you added an explanation it all makes sense. c is the third input, n is undefined (and ~undefined becomes -1 unlike undefined+1). All clear now (and not something I can port to Java unfortunately, which was mainly why I tried to understand it xD). PS: Already upvoted yesterday, so I just upvoted one of your other answers (which I didn't upvote already, not a lot available..) ;p
– Kevin Cruijssen
Aug 31 at 15:59

@KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
– Arnauld
Aug 31 at 16:08

@KevinCruijssen FWIW, I wrote a tip about this some times ago. But yeah ... that's a JS thing and is probably not portable to many other languages.
– Arnauld
Aug 31 at 16:08

Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
– Kevin Cruijssen
Aug 31 at 16:38

Well, I might be able to semi-port it but just creating a recursive second method, and using a ternary-if to check for null, manually converting it to -1. However, Java has a (very) limited recursive StackOverflow limit, so using a recursive method with randomness hoping it would end up correct within about 1024 recursive calls isn't going to work anyway in Java. Ah well. I've upvoted your tip. Have a nice weekend! :)
– Kevin Cruijssen
Aug 31 at 16:38

@KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
– Arnauld
Aug 31 at 17:05

@KevinCruijssen My first JS attempt was doing exactly that: trying random values with a recursive function. And it was usually doing significantly less iterations than the one using a counter. Fun fact: even for ###+###=999, your odds are 1 in 1000. So with 1024 iterations, you should succeed slightly more often than you fail. :)
– Arnauld
Aug 31 at 17:05

add a comment | 

up vote
7
down vote

Matlab, 143 134 132 119 115 bytes

function[m]=f(x,y,r),p=@(v)str2num(strrep(v,'#',char(randi([48,57]))));m=[1,1];while sum(m)-r,m=[p(x),p(y)];end;end

-4 bytes thanks to @Luismendo

Try it Online

---

Pretty big and pretty stupid. It simply replaces all # with random digits until it finds the correct ones.

share|improve this answer

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

add a comment | 

up vote
7
down vote

Matlab, 143 134 132 119 115 bytes

function[m]=f(x,y,r),p=@(v)str2num(strrep(v,'#',char(randi([48,57]))));m=[1,1];while sum(m)-r,m=[p(x),p(y)];end;end

-4 bytes thanks to @Luismendo

Try it Online

---

Pretty big and pretty stupid. It simply replaces all # with random digits until it finds the correct ones.

share|improve this answer

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

add a comment | 

up vote
7
down vote

up vote
7
down vote

Matlab, 143 134 132 119 115 bytes

function[m]=f(x,y,r),p=@(v)str2num(strrep(v,'#',char(randi([48,57]))));m=[1,1];while sum(m)-r,m=[p(x),p(y)];end;end

-4 bytes thanks to @Luismendo

Try it Online

---

Pretty big and pretty stupid. It simply replaces all # with random digits until it finds the correct ones.

share|improve this answer

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

Matlab, 143 134 132 119 115 bytes

function[m]=f(x,y,r),p=@(v)str2num(strrep(v,'#',char(randi([48,57]))));m=[1,1];while sum(m)-r,m=[p(x),p(y)];end;end

-4 bytes thanks to @Luismendo

Try it Online

---

Pretty big and pretty stupid. It simply replaces all # with random digits until it finds the correct ones.

share|improve this answer

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

share|improve this answer

share|improve this answer

edited Aug 31 at 0:50

edited Aug 31 at 0:50

edited Aug 31 at 0:50

answered Aug 30 at 15:18

DimChtz

601111

answered Aug 30 at 15:18

DimChtz

601111

answered Aug 30 at 15:18

DimChtz

601111

601111

add a comment | 

add a comment | 

up vote
4
down vote

R, 67 51 bytes

Rock simple and scales horribly, just grep all the sum combinations. Use "." for unknown digits. It won't find the same answer as test case number 4, but it will give a possible answer, which follows the letter of the rules as given.

-16 bytes by grepping after forming the output and replacing paste with the ? operator.

function(x,y,z,`?`=paste)grep(x?y,1:z?z:1-1,v=T)[1]

Try it online!

share|improve this answer

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
  – JayCe
  Aug 31 at 0:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
  – JayCe
  Aug 31 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
  – J.Doe
  Aug 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
4
down vote

R, 67 51 bytes

Rock simple and scales horribly, just grep all the sum combinations. Use "." for unknown digits. It won't find the same answer as test case number 4, but it will give a possible answer, which follows the letter of the rules as given.

-16 bytes by grepping after forming the output and replacing paste with the ? operator.

function(x,y,z,`?`=paste)grep(x?y,1:z?z:1-1,v=T)[1]

Try it online!

share|improve this answer

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
  – JayCe
  Aug 31 at 0:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
  – JayCe
  Aug 31 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
  – J.Doe
  Aug 31 at 14:31
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

R, 67 51 bytes

Rock simple and scales horribly, just grep all the sum combinations. Use "." for unknown digits. It won't find the same answer as test case number 4, but it will give a possible answer, which follows the letter of the rules as given.

-16 bytes by grepping after forming the output and replacing paste with the ? operator.

function(x,y,z,`?`=paste)grep(x?y,1:z?z:1-1,v=T)[1]

Try it online!

share|improve this answer

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

R, 67 51 bytes

Rock simple and scales horribly, just grep all the sum combinations. Use "." for unknown digits. It won't find the same answer as test case number 4, but it will give a possible answer, which follows the letter of the rules as given.

-16 bytes by grepping after forming the output and replacing paste with the ? operator.

function(x,y,z,`?`=paste)grep(x?y,1:z?z:1-1,v=T)[1]

Try it online!

share|improve this answer

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

share|improve this answer

share|improve this answer

edited Aug 31 at 14:42

edited Aug 31 at 14:42

edited Aug 31 at 14:42

answered Aug 30 at 18:01

J.Doe

5718

answered Aug 30 at 18:01

J.Doe

5718

answered Aug 30 at 18:01

J.Doe

5718

5718

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
  – JayCe
  Aug 31 at 0:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
  – JayCe
  Aug 31 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
  – J.Doe
  Aug 31 at 14:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
  – JayCe
  Aug 31 at 0:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
  – JayCe
  Aug 31 at 14:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
  – J.Doe
  Aug 31 at 14:31
  

  
  
  
  

1

1

Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
– JayCe
Aug 31 at 0:44

Awesome idea, I would never have thought of that. You can save a few bytes by using * in place of grepl : tio.run/##PYzLCoMwEEX3/…
– JayCe
Aug 31 at 0:44

1

1

I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
– JayCe
Aug 31 at 14:26

I was looking for various operators and you come up with ?... I think this is a first. by the way, I forgot if I already told you but we're trying to get R nominated for september language of the month - you can upvote if not done already.
– JayCe
Aug 31 at 14:26

I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
– J.Doe
Aug 31 at 14:31

I could've chosen anything with low precedence. Feels like there should be a better way to get the match...
– J.Doe
Aug 31 at 14:31

add a comment | 

up vote
3
down vote

Charcoal, 32 bytes

Ｆ²⊞υ0Ｗ⁻ζΣＩυ≔Ｅ⟦θη⟧⭆κ⎇⁼μ#‽χμυ←Ｅυ⮌ι

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

Ｆ²⊞υ0

Push two string 0s to the predefined empty list u to get the while loop going.

Ｗ⁻ζΣＩυ

Repeat while the sum of casting the values in u to integer is not equal to the desired result.

≔Ｅ⟦θη⟧

Create an array of the two inputs and map over it.

⭆κ⎇⁼μ#‽χμυ

Replace each # with a random digit and assign the result back to u.

←Ｅυ⮌ι

Print the result right justified. (Left justified would be just υ for a 4-byte saving.)

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

add a comment | 

up vote
3
down vote

Charcoal, 32 bytes

Ｆ²⊞υ0Ｗ⁻ζΣＩυ≔Ｅ⟦θη⟧⭆κ⎇⁼μ#‽χμυ←Ｅυ⮌ι

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

Ｆ²⊞υ0

Push two string 0s to the predefined empty list u to get the while loop going.

Ｗ⁻ζΣＩυ

Repeat while the sum of casting the values in u to integer is not equal to the desired result.

≔Ｅ⟦θη⟧

Create an array of the two inputs and map over it.

⭆κ⎇⁼μ#‽χμυ

Replace each # with a random digit and assign the result back to u.

←Ｅυ⮌ι

Print the result right justified. (Left justified would be just υ for a 4-byte saving.)

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

add a comment | 

up vote
3
down vote

up vote
3
down vote

Charcoal, 32 bytes

Ｆ²⊞υ0Ｗ⁻ζΣＩυ≔Ｅ⟦θη⟧⭆κ⎇⁼μ#‽χμυ←Ｅυ⮌ι

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

Ｆ²⊞υ0

Push two string 0s to the predefined empty list u to get the while loop going.

Ｗ⁻ζΣＩυ

Repeat while the sum of casting the values in u to integer is not equal to the desired result.

≔Ｅ⟦θη⟧

Create an array of the two inputs and map over it.

⭆κ⎇⁼μ#‽χμυ

Replace each # with a random digit and assign the result back to u.

←Ｅυ⮌ι

Print the result right justified. (Left justified would be just υ for a 4-byte saving.)

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

Charcoal, 32 bytes

Ｆ²⊞υ0Ｗ⁻ζΣＩυ≔Ｅ⟦θη⟧⭆κ⎇⁼μ#‽χμυ←Ｅυ⮌ι

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

Ｆ²⊞υ0

Push two string 0s to the predefined empty list u to get the while loop going.

Ｗ⁻ζΣＩυ

Repeat while the sum of casting the values in u to integer is not equal to the desired result.

≔Ｅ⟦θη⟧

Create an array of the two inputs and map over it.

⭆κ⎇⁼μ#‽χμυ

Replace each # with a random digit and assign the result back to u.

←Ｅυ⮌ι

Print the result right justified. (Left justified would be just υ for a 4-byte saving.)

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

share|improve this answer

share|improve this answer

answered Aug 30 at 16:16

Neil

75.1k744170

answered Aug 30 at 16:16

Neil

75.1k744170

answered Aug 30 at 16:16

Neil

75.1k744170

75.1k744170

add a comment | 

add a comment | 

up vote
3
down vote

Jelly, 20 bytes

ØDṛċ¡V€)ŒpḌð€ŒpS⁼¥ƇḢ

Try it online!

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
3
down vote

Jelly, 20 bytes

ØDṛċ¡V€)ŒpḌð€ŒpS⁼¥ƇḢ

Try it online!

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
3
down vote

up vote
3
down vote

Jelly, 20 bytes

ØDṛċ¡V€)ŒpḌð€ŒpS⁼¥ƇḢ

Try it online!

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

Jelly, 20 bytes

ØDṛċ¡V€)ŒpḌð€ŒpS⁼¥ƇḢ

Try it online!

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

share|improve this answer

share|improve this answer

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

answered Aug 30 at 16:30

Erik the Outgolfer

29.4k42698

29.4k42698

add a comment | 

add a comment | 

up vote
3
down vote

05AB1E (legacy), 23 20 bytes

[²³«εð9ÝΩ:}²gôJDO¹Q#

-3 bytes thanks to @Emigna.

Unknown digits are spaces ( ). Input order should be: expected result; longest string; shortest string.

Try it online.

Explanation:

[ # Start an infinite loop
 ²³« # Take the second and third inputs, and merge them together
 # i.e. " 79" and " 4 " → " 79 4 "
 ε } # Map each character to:
 ð : # Replace a space with:
 9ÝΩ # A random digit in the range [0,9]
 # i.e. " 79 4 " → ['3','7','9','2','4','3']
 # i.e. " 79 4 " → ['5','7','9','7','4','4']
²g # Get the length of the second input
 # i.e. " 79" → 3
 ô # And split it into two numbers again
 # i.e. ['3','7','9','2','4','3'] and 3 → [['3','7','9'],['2','4','3']]
 # i.e. ['5','7','9','7','4','4'] and 3 → [['5','7','9'],['7','4','4']]
 J # Join each list together to a single number
 # i.e. [['3','7','9'],['2','4','3']] → [379,243]
 # i.e. [['5','7','9'],['7','4','4']] → [579,744]
 D # Duplicate this list
 O # Sum the list
 # i.e. [379,243] → 622
 # i.e. [579,744] → 1323
 ¹Q# # If it's equal to the first input: stop the infinite loop
 # (and output the duplicate list implicitly)
 # i.e. 1323 and 622 → 0 (falsey) → continue the loop
 # i.e. 1323 and 1323 → 1 (truthy) → stop the loop and output [579,744]

share|improve this answer

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Replace saves 3 over the if.
  – Emigna
  Aug 31 at 6:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, of course. Thanks!
  – Kevin Cruijssen
  Aug 31 at 6:52
  

  
  
  
  

add a comment | 

up vote
3
down vote

05AB1E (legacy), 23 20 bytes

[²³«εð9ÝΩ:}²gôJDO¹Q#

-3 bytes thanks to @Emigna.

Unknown digits are spaces ( ). Input order should be: expected result; longest string; shortest string.

Try it online.

Explanation:

[ # Start an infinite loop
 ²³« # Take the second and third inputs, and merge them together
 # i.e. " 79" and " 4 " → " 79 4 "
 ε } # Map each character to:
 ð : # Replace a space with:
 9ÝΩ # A random digit in the range [0,9]
 # i.e. " 79 4 " → ['3','7','9','2','4','3']
 # i.e. " 79 4 " → ['5','7','9','7','4','4']
²g # Get the length of the second input
 # i.e. " 79" → 3
 ô # And split it into two numbers again
 # i.e. ['3','7','9','2','4','3'] and 3 → [['3','7','9'],['2','4','3']]
 # i.e. ['5','7','9','7','4','4'] and 3 → [['5','7','9'],['7','4','4']]
 J # Join each list together to a single number
 # i.e. [['3','7','9'],['2','4','3']] → [379,243]
 # i.e. [['5','7','9'],['7','4','4']] → [579,744]
 D # Duplicate this list
 O # Sum the list
 # i.e. [379,243] → 622
 # i.e. [579,744] → 1323
 ¹Q# # If it's equal to the first input: stop the infinite loop
 # (and output the duplicate list implicitly)
 # i.e. 1323 and 622 → 0 (falsey) → continue the loop
 # i.e. 1323 and 1323 → 1 (truthy) → stop the loop and output [579,744]

share|improve this answer

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Replace saves 3 over the if.
  – Emigna
  Aug 31 at 6:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, of course. Thanks!
  – Kevin Cruijssen
  Aug 31 at 6:52
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

05AB1E (legacy), 23 20 bytes

[²³«εð9ÝΩ:}²gôJDO¹Q#

-3 bytes thanks to @Emigna.

Unknown digits are spaces ( ). Input order should be: expected result; longest string; shortest string.

Try it online.

Explanation:

[ # Start an infinite loop
 ²³« # Take the second and third inputs, and merge them together
 # i.e. " 79" and " 4 " → " 79 4 "
 ε } # Map each character to:
 ð : # Replace a space with:
 9ÝΩ # A random digit in the range [0,9]
 # i.e. " 79 4 " → ['3','7','9','2','4','3']
 # i.e. " 79 4 " → ['5','7','9','7','4','4']
²g # Get the length of the second input
 # i.e. " 79" → 3
 ô # And split it into two numbers again
 # i.e. ['3','7','9','2','4','3'] and 3 → [['3','7','9'],['2','4','3']]
 # i.e. ['5','7','9','7','4','4'] and 3 → [['5','7','9'],['7','4','4']]
 J # Join each list together to a single number
 # i.e. [['3','7','9'],['2','4','3']] → [379,243]
 # i.e. [['5','7','9'],['7','4','4']] → [579,744]
 D # Duplicate this list
 O # Sum the list
 # i.e. [379,243] → 622
 # i.e. [579,744] → 1323
 ¹Q# # If it's equal to the first input: stop the infinite loop
 # (and output the duplicate list implicitly)
 # i.e. 1323 and 622 → 0 (falsey) → continue the loop
 # i.e. 1323 and 1323 → 1 (truthy) → stop the loop and output [579,744]

share|improve this answer

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

05AB1E (legacy), 23 20 bytes

[²³«εð9ÝΩ:}²gôJDO¹Q#

-3 bytes thanks to @Emigna.

Unknown digits are spaces ( ). Input order should be: expected result; longest string; shortest string.

Try it online.

Explanation:

[ # Start an infinite loop
 ²³« # Take the second and third inputs, and merge them together
 # i.e. " 79" and " 4 " → " 79 4 "
 ε } # Map each character to:
 ð : # Replace a space with:
 9ÝΩ # A random digit in the range [0,9]
 # i.e. " 79 4 " → ['3','7','9','2','4','3']
 # i.e. " 79 4 " → ['5','7','9','7','4','4']
²g # Get the length of the second input
 # i.e. " 79" → 3
 ô # And split it into two numbers again
 # i.e. ['3','7','9','2','4','3'] and 3 → [['3','7','9'],['2','4','3']]
 # i.e. ['5','7','9','7','4','4'] and 3 → [['5','7','9'],['7','4','4']]
 J # Join each list together to a single number
 # i.e. [['3','7','9'],['2','4','3']] → [379,243]
 # i.e. [['5','7','9'],['7','4','4']] → [579,744]
 D # Duplicate this list
 O # Sum the list
 # i.e. [379,243] → 622
 # i.e. [579,744] → 1323
 ¹Q# # If it's equal to the first input: stop the infinite loop
 # (and output the duplicate list implicitly)
 # i.e. 1323 and 622 → 0 (falsey) → continue the loop
 # i.e. 1323 and 1323 → 1 (truthy) → stop the loop and output [579,744]

share|improve this answer

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

share|improve this answer

share|improve this answer

edited Aug 31 at 6:51

edited Aug 31 at 6:51

edited Aug 31 at 6:51

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

answered Aug 30 at 16:50

Kevin Cruijssen

29.5k550162

29.5k550162

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Replace saves 3 over the if.
  – Emigna
  Aug 31 at 6:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, of course. Thanks!
  – Kevin Cruijssen
  Aug 31 at 6:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Replace saves 3 over the if.
  – Emigna
  Aug 31 at 6:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Emigna Ah, of course. Thanks!
  – Kevin Cruijssen
  Aug 31 at 6:52
  

  
  
  
  

1

1

Replace saves 3 over the if.
– Emigna
Aug 31 at 6:10

Replace saves 3 over the if.
– Emigna
Aug 31 at 6:10

@Emigna Ah, of course. Thanks!
– Kevin Cruijssen
Aug 31 at 6:52

@Emigna Ah, of course. Thanks!
– Kevin Cruijssen
Aug 31 at 6:52

add a comment | 

up vote
3
down vote

Perl 6, 81 74 bytes

-7 bytes thanks to nwellnhof!

first try S/=/==/.EVAL,map $^a;S:g[#]=$a[$++],[X] ^10 xx.comb('#')

Try it online!

Anonymous code block that takes input as a string containing an arithmetical expression e.g. "12#+45#=579". Substitutes each # with possible permutations of digits, substitutes the= with == and finds the first valid result.

Explanation:

 # Anonymous code block 
 first # Find the first of:
 ^10 # The range of 0 to 9
 xx.comb('#') # Multiplied by the number #s in the code
 ,[X] # The cross-product of these lists
 map # Map each crossproduct to:
 $^a;.trans: "#"=>$a[$++] # The given string with each # translated to each element in the list
 try S/=/==/.EVAL, # Find which of these is true when = are changed to == and it is eval'd

share|improve this answer

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
  – nwellnhof
  Sep 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
  – Jo King
  Sep 1 at 13:13
  

  
  
  
  

add a comment | 

up vote
3
down vote

Perl 6, 81 74 bytes

-7 bytes thanks to nwellnhof!

first try S/=/==/.EVAL,map $^a;S:g[#]=$a[$++],[X] ^10 xx.comb('#')

Try it online!

Anonymous code block that takes input as a string containing an arithmetical expression e.g. "12#+45#=579". Substitutes each # with possible permutations of digits, substitutes the= with == and finds the first valid result.

Explanation:

 # Anonymous code block 
 first # Find the first of:
 ^10 # The range of 0 to 9
 xx.comb('#') # Multiplied by the number #s in the code
 ,[X] # The cross-product of these lists
 map # Map each crossproduct to:
 $^a;.trans: "#"=>$a[$++] # The given string with each # translated to each element in the list
 try S/=/==/.EVAL, # Find which of these is true when = are changed to == and it is eval'd

share|improve this answer

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
  – nwellnhof
  Sep 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
  – Jo King
  Sep 1 at 13:13
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Perl 6, 81 74 bytes

-7 bytes thanks to nwellnhof!

first try S/=/==/.EVAL,map $^a;S:g[#]=$a[$++],[X] ^10 xx.comb('#')

Try it online!

Anonymous code block that takes input as a string containing an arithmetical expression e.g. "12#+45#=579". Substitutes each # with possible permutations of digits, substitutes the= with == and finds the first valid result.

Explanation:

 # Anonymous code block 
 first # Find the first of:
 ^10 # The range of 0 to 9
 xx.comb('#') # Multiplied by the number #s in the code
 ,[X] # The cross-product of these lists
 map # Map each crossproduct to:
 $^a;.trans: "#"=>$a[$++] # The given string with each # translated to each element in the list
 try S/=/==/.EVAL, # Find which of these is true when = are changed to == and it is eval'd

share|improve this answer

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

Perl 6, 81 74 bytes

-7 bytes thanks to nwellnhof!

first try S/=/==/.EVAL,map $^a;S:g[#]=$a[$++],[X] ^10 xx.comb('#')

Try it online!

Anonymous code block that takes input as a string containing an arithmetical expression e.g. "12#+45#=579". Substitutes each # with possible permutations of digits, substitutes the= with == and finds the first valid result.

Explanation:

 # Anonymous code block 
 first # Find the first of:
 ^10 # The range of 0 to 9
 xx.comb('#') # Multiplied by the number #s in the code
 ,[X] # The cross-product of these lists
 map # Map each crossproduct to:
 $^a;.trans: "#"=>$a[$++] # The given string with each # translated to each element in the list
 try S/=/==/.EVAL, # Find which of these is true when = are changed to == and it is eval'd

share|improve this answer

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

share|improve this answer

share|improve this answer

edited Sep 1 at 13:13

edited Sep 1 at 13:13

edited Sep 1 at 13:13

answered Aug 30 at 21:18

Jo King

14.9k24084

answered Aug 30 at 21:18

Jo King

14.9k24084

answered Aug 30 at 21:18

Jo King

14.9k24084

14.9k24084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
  – nwellnhof
  Sep 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
  – Jo King
  Sep 1 at 13:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
  – nwellnhof
  Sep 1 at 13:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
  – Jo King
  Sep 1 at 13:13
  

  
  
  
  

You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
– nwellnhof
Sep 1 at 13:05

You can use S:g[#]=$a[$++] instead of trans for 74 bytes.
– nwellnhof
Sep 1 at 13:05

@nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
– Jo King
Sep 1 at 13:13

@nwellnhof I didn't realise you could use S/// in that sort of syntax! Thanks!
– Jo King
Sep 1 at 13:13

add a comment | 

up vote
2
down vote

APL (Dyalog Unicode), 22 bytes

'#'⎕R⍕?10t⍣⍎⍺t∘←

Try it online!

share|improve this answer

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

add a comment | 

up vote
2
down vote

APL (Dyalog Unicode), 22 bytes

'#'⎕R⍕?10t⍣⍎⍺t∘←

Try it online!

share|improve this answer

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

add a comment | 

up vote
2
down vote

up vote
2
down vote

APL (Dyalog Unicode), 22 bytes

'#'⎕R⍕?10t⍣⍎⍺t∘←

Try it online!

share|improve this answer

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

APL (Dyalog Unicode), 22 bytes

'#'⎕R⍕?10t⍣⍎⍺t∘←

Try it online!

share|improve this answer

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

share|improve this answer

share|improve this answer

edited Aug 30 at 17:10

edited Aug 30 at 17:10

edited Aug 30 at 17:10

answered Aug 30 at 17:01

dzaima

12.8k21652

answered Aug 30 at 17:01

dzaima

12.8k21652

answered Aug 30 at 17:01

dzaima

12.8k21652

12.8k21652

add a comment | 

add a comment | 

up vote
2
down vote

Java 10, 203 197 bytes

(a,b,c)->int A=0,B=0,l=a.length();for(a+=b,b="";A+B!=c;A=c.valueOf(b.substring(0,l)),B=c.valueOf(b.substring(l)),b="")for(var t:a.toCharArray())b+=t<36?(int)(Math.random()*9)+"":t;return A+"+"+B;

Try it online.

Explanation:

(a,b,c)-> // Method with 2 Strings & integer parameters and String return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.length(); // Length of the first String-input
 for(a+=b, // Concat the second String-input to the first
 b=""; // Reuse `b`, and start it as an empty String
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=c.valueOf(b.substring(0,l)),
 // Set `A` to the first String-part as integer
 B=c.valueOf(n.substring(l)),
 // Set `B` to the second String-part as integer
 b="") // Reset `b` to an empty String
 for(var t:a.toCharArray())
 // Inner loop over the characters of the concatted String inputs
 b+=t<36? // If the current character is a '#':
 (int)(Math.random()*9)+""
 // Append a random digit to `b`
 : // Else (it already is a digit):
 t; // Append this digit as is to `b`
 return A+"+"+B; // After the loop, return `A` and `B` as result

share|improve this answer

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
2
down vote

Java 10, 203 197 bytes

(a,b,c)->int A=0,B=0,l=a.length();for(a+=b,b="";A+B!=c;A=c.valueOf(b.substring(0,l)),B=c.valueOf(b.substring(l)),b="")for(var t:a.toCharArray())b+=t<36?(int)(Math.random()*9)+"":t;return A+"+"+B;

Try it online.

Explanation:

(a,b,c)-> // Method with 2 Strings & integer parameters and String return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.length(); // Length of the first String-input
 for(a+=b, // Concat the second String-input to the first
 b=""; // Reuse `b`, and start it as an empty String
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=c.valueOf(b.substring(0,l)),
 // Set `A` to the first String-part as integer
 B=c.valueOf(n.substring(l)),
 // Set `B` to the second String-part as integer
 b="") // Reset `b` to an empty String
 for(var t:a.toCharArray())
 // Inner loop over the characters of the concatted String inputs
 b+=t<36? // If the current character is a '#':
 (int)(Math.random()*9)+""
 // Append a random digit to `b`
 : // Else (it already is a digit):
 t; // Append this digit as is to `b`
 return A+"+"+B; // After the loop, return `A` and `B` as result

share|improve this answer

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
2
down vote

up vote
2
down vote

Java 10, 203 197 bytes

(a,b,c)->int A=0,B=0,l=a.length();for(a+=b,b="";A+B!=c;A=c.valueOf(b.substring(0,l)),B=c.valueOf(b.substring(l)),b="")for(var t:a.toCharArray())b+=t<36?(int)(Math.random()*9)+"":t;return A+"+"+B;

Try it online.

Explanation:

(a,b,c)-> // Method with 2 Strings & integer parameters and String return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.length(); // Length of the first String-input
 for(a+=b, // Concat the second String-input to the first
 b=""; // Reuse `b`, and start it as an empty String
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=c.valueOf(b.substring(0,l)),
 // Set `A` to the first String-part as integer
 B=c.valueOf(n.substring(l)),
 // Set `B` to the second String-part as integer
 b="") // Reset `b` to an empty String
 for(var t:a.toCharArray())
 // Inner loop over the characters of the concatted String inputs
 b+=t<36? // If the current character is a '#':
 (int)(Math.random()*9)+""
 // Append a random digit to `b`
 : // Else (it already is a digit):
 t; // Append this digit as is to `b`
 return A+"+"+B; // After the loop, return `A` and `B` as result

share|improve this answer

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

Java 10, 203 197 bytes

(a,b,c)->int A=0,B=0,l=a.length();for(a+=b,b="";A+B!=c;A=c.valueOf(b.substring(0,l)),B=c.valueOf(b.substring(l)),b="")for(var t:a.toCharArray())b+=t<36?(int)(Math.random()*9)+"":t;return A+"+"+B;

Try it online.

Explanation:

(a,b,c)-> // Method with 2 Strings & integer parameters and String return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.length(); // Length of the first String-input
 for(a+=b, // Concat the second String-input to the first
 b=""; // Reuse `b`, and start it as an empty String
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=c.valueOf(b.substring(0,l)),
 // Set `A` to the first String-part as integer
 B=c.valueOf(n.substring(l)),
 // Set `B` to the second String-part as integer
 b="") // Reset `b` to an empty String
 for(var t:a.toCharArray())
 // Inner loop over the characters of the concatted String inputs
 b+=t<36? // If the current character is a '#':
 (int)(Math.random()*9)+""
 // Append a random digit to `b`
 : // Else (it already is a digit):
 t; // Append this digit as is to `b`
 return A+"+"+B; // After the loop, return `A` and `B` as result

share|improve this answer

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

share|improve this answer

share|improve this answer

edited Aug 31 at 15:46

edited Aug 31 at 15:46

edited Aug 31 at 15:46

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

answered Aug 30 at 18:01

Kevin Cruijssen

29.5k550162

29.5k550162

add a comment | 

add a comment | 

up vote
1
down vote

C# .NET, 225 220 196 bytes

(a,b,c)=>int A=0,B=0,l=a.Length;for(a+=b,b="";A+B!=c;A=int.Parse(b.Substring(0,l)),B=int.Parse(b.Substring(l)),b="")foreach(var t in a)b+=(t<36?new System.Random().Next(10):t-48)+"";return(A,B);

Port of my Java 10 answer.

(I'm very rusty in C# .NET golfing, so can defintely be golfed..)

-3 bytes implicitly thanks to @user82593 and this new C# tip he added.

-29 bytes thanks to @hvd.

Try it online.

Explanation:

(a,b,c)=> // Method with 2 string & int parameters and int-tuple return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.Length; // Length of the first string-input
 for(a+=b, // Concat the second string-input to the first
 b=""; // Reuse `b`, and start it as an empty string
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=int.Parse(b.Substring(0,l)),
 // Set `A` to the first string-part as integer
 B=int.Parse(b.Substring(l)),
 // Set `B` to the second string-part as integer
 b="") // Reset `b` to an empty string
 foreach(var t in a)
 // Inner loop over the characters of the concatted string inputs
 b+=(t<36? // If the current character is a '#':
 new System.Random().Next(10)
 // Use a random digit
 : // Else (it already is a digit):
 t-48) // Use this digit as is
 +""; // And convert it to a string so it can be appended to the string
 return(A,B); // After the loop, return `A` and `B` in a tuple as result

share|improve this answer

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the regular using System; instead, it is shorter than namespace System.
  – hvd
  Sep 1 at 8:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
  – Kevin Cruijssen
  Sep 1 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
  – hvd
  Sep 1 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
  – Kevin Cruijssen
  Sep 1 at 11:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

C# .NET, 225 220 196 bytes

(a,b,c)=>int A=0,B=0,l=a.Length;for(a+=b,b="";A+B!=c;A=int.Parse(b.Substring(0,l)),B=int.Parse(b.Substring(l)),b="")foreach(var t in a)b+=(t<36?new System.Random().Next(10):t-48)+"";return(A,B);

Port of my Java 10 answer.

(I'm very rusty in C# .NET golfing, so can defintely be golfed..)

-3 bytes implicitly thanks to @user82593 and this new C# tip he added.

-29 bytes thanks to @hvd.

Try it online.

Explanation:

(a,b,c)=> // Method with 2 string & int parameters and int-tuple return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.Length; // Length of the first string-input
 for(a+=b, // Concat the second string-input to the first
 b=""; // Reuse `b`, and start it as an empty string
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=int.Parse(b.Substring(0,l)),
 // Set `A` to the first string-part as integer
 B=int.Parse(b.Substring(l)),
 // Set `B` to the second string-part as integer
 b="") // Reset `b` to an empty string
 foreach(var t in a)
 // Inner loop over the characters of the concatted string inputs
 b+=(t<36? // If the current character is a '#':
 new System.Random().Next(10)
 // Use a random digit
 : // Else (it already is a digit):
 t-48) // Use this digit as is
 +""; // And convert it to a string so it can be appended to the string
 return(A,B); // After the loop, return `A` and `B` in a tuple as result

share|improve this answer

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the regular using System; instead, it is shorter than namespace System.
  – hvd
  Sep 1 at 8:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
  – Kevin Cruijssen
  Sep 1 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
  – hvd
  Sep 1 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
  – Kevin Cruijssen
  Sep 1 at 11:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

C# .NET, 225 220 196 bytes

(a,b,c)=>int A=0,B=0,l=a.Length;for(a+=b,b="";A+B!=c;A=int.Parse(b.Substring(0,l)),B=int.Parse(b.Substring(l)),b="")foreach(var t in a)b+=(t<36?new System.Random().Next(10):t-48)+"";return(A,B);

Port of my Java 10 answer.

(I'm very rusty in C# .NET golfing, so can defintely be golfed..)

-3 bytes implicitly thanks to @user82593 and this new C# tip he added.

-29 bytes thanks to @hvd.

Try it online.

Explanation:

(a,b,c)=> // Method with 2 string & int parameters and int-tuple return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.Length; // Length of the first string-input
 for(a+=b, // Concat the second string-input to the first
 b=""; // Reuse `b`, and start it as an empty string
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=int.Parse(b.Substring(0,l)),
 // Set `A` to the first string-part as integer
 B=int.Parse(b.Substring(l)),
 // Set `B` to the second string-part as integer
 b="") // Reset `b` to an empty string
 foreach(var t in a)
 // Inner loop over the characters of the concatted string inputs
 b+=(t<36? // If the current character is a '#':
 new System.Random().Next(10)
 // Use a random digit
 : // Else (it already is a digit):
 t-48) // Use this digit as is
 +""; // And convert it to a string so it can be appended to the string
 return(A,B); // After the loop, return `A` and `B` in a tuple as result

share|improve this answer

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

C# .NET, 225 220 196 bytes

(a,b,c)=>int A=0,B=0,l=a.Length;for(a+=b,b="";A+B!=c;A=int.Parse(b.Substring(0,l)),B=int.Parse(b.Substring(l)),b="")foreach(var t in a)b+=(t<36?new System.Random().Next(10):t-48)+"";return(A,B);

Port of my Java 10 answer.

(I'm very rusty in C# .NET golfing, so can defintely be golfed..)

-3 bytes implicitly thanks to @user82593 and this new C# tip he added.

-29 bytes thanks to @hvd.

Try it online.

Explanation:

(a,b,c)=> // Method with 2 string & int parameters and int-tuple return-type
 int A=0,B=0, // Result-integers, starting both at 0
 l=a.Length; // Length of the first string-input
 for(a+=b, // Concat the second string-input to the first
 b=""; // Reuse `b`, and start it as an empty string
 A+B!=c // Loop as long as `A+B` isn't equal to the integer-input
 ; // After every iteration:
 A=int.Parse(b.Substring(0,l)),
 // Set `A` to the first string-part as integer
 B=int.Parse(b.Substring(l)),
 // Set `B` to the second string-part as integer
 b="") // Reset `b` to an empty string
 foreach(var t in a)
 // Inner loop over the characters of the concatted string inputs
 b+=(t<36? // If the current character is a '#':
 new System.Random().Next(10)
 // Use a random digit
 : // Else (it already is a digit):
 t-48) // Use this digit as is
 +""; // And convert it to a string so it can be appended to the string
 return(A,B); // After the loop, return `A` and `B` in a tuple as result

share|improve this answer

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

share|improve this answer

share|improve this answer

edited Sep 1 at 11:48

edited Sep 1 at 11:48

edited Sep 1 at 11:48

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

answered Aug 31 at 15:43

Kevin Cruijssen

29.5k550162

29.5k550162

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the regular using System; instead, it is shorter than namespace System.
  – hvd
  Sep 1 at 8:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
  – Kevin Cruijssen
  Sep 1 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
  – hvd
  Sep 1 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
  – Kevin Cruijssen
  Sep 1 at 11:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can use the regular using System; instead, it is shorter than namespace System.
  – hvd
  Sep 1 at 8:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
  – Kevin Cruijssen
  Sep 1 at 11:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
  – hvd
  Sep 1 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
  – Kevin Cruijssen
  Sep 1 at 11:49
  

  
  
  
  

You can use the regular using System; instead, it is shorter than namespace System.
– hvd
Sep 1 at 8:01

You can use the regular using System; instead, it is shorter than namespace System.
– hvd
Sep 1 at 8:01

@hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
– Kevin Cruijssen
Sep 1 at 11:01

@hvd That was it!.. I haven't done C# in years, lol.. I tried using System.*; similar as imports in Java, but that didn't work. Forgot I had to remove the .* part.. Thanks for the -5 bytes.
– Kevin Cruijssen
Sep 1 at 11:01

1

1

Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
– hvd
Sep 1 at 11:05

Re-reading it now, that was actually a sub-optimal suggestion. You can write int.Parse (-4), use new System.Random() (+7) and drop using System; (-13) to save another 10 bytes. :) Also, you do not need .ToCharArray(), that takes off 14 more bytes.
– hvd
Sep 1 at 11:05

@hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
– Kevin Cruijssen
Sep 1 at 11:49

@hvd Thanks! Not sure how I forgot about int.Parse vs System.Int32.Parse... It's basically the same as System.String and string.. And didn't knew it was possible to loop over the characters without the .ToCharArray(). Thanks for another -24 bytes. :D
– Kevin Cruijssen
Sep 1 at 11:49

add a comment | 

up vote
1
down vote

Python 3, 121 155 152 149 bytes

import re
def f(i,k=0,S=re.sub):s=S('#','%s',i)%(*list('%0*d'%(i.count('#'),k)),);print(s)if eval(S('=','==',S('\b0*([1-9])','\1',s)))else f(i,k+1)

Try it online!

+34 New solution with regex to circumvent the fact that python doesn't support numbers with leading zeroes.

-3 thanks to @Jonathan Frech

---

The old solution doesn't work if # is the first character in any number (because eval doesn't accept leading zeroes) and is therefore invalid :(

def f(i,k=0):
 s=i.replace('#','%s')%(*list('%0*d'%(i.count('#'),k)),)
 print(s)if eval(s.replace('=','=='))else f(i,k+1)

Try it online!

share|improve this answer

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm afraid this submission is invalid for the reason stated in the post.
  – Erik the Outgolfer
  Aug 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Since your function does not contain any compound statements you can condense it to only one line.
  – Jonathan Frech
  Aug 30 at 23:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 3, 121 155 152 149 bytes

import re
def f(i,k=0,S=re.sub):s=S('#','%s',i)%(*list('%0*d'%(i.count('#'),k)),);print(s)if eval(S('=','==',S('\b0*([1-9])','\1',s)))else f(i,k+1)

Try it online!

+34 New solution with regex to circumvent the fact that python doesn't support numbers with leading zeroes.

-3 thanks to @Jonathan Frech

---

The old solution doesn't work if # is the first character in any number (because eval doesn't accept leading zeroes) and is therefore invalid :(

def f(i,k=0):
 s=i.replace('#','%s')%(*list('%0*d'%(i.count('#'),k)),)
 print(s)if eval(s.replace('=','=='))else f(i,k+1)

Try it online!

share|improve this answer

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm afraid this submission is invalid for the reason stated in the post.
  – Erik the Outgolfer
  Aug 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Since your function does not contain any compound statements you can condense it to only one line.
  – Jonathan Frech
  Aug 30 at 23:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 3, 121 155 152 149 bytes

import re
def f(i,k=0,S=re.sub):s=S('#','%s',i)%(*list('%0*d'%(i.count('#'),k)),);print(s)if eval(S('=','==',S('\b0*([1-9])','\1',s)))else f(i,k+1)

Try it online!

+34 New solution with regex to circumvent the fact that python doesn't support numbers with leading zeroes.

-3 thanks to @Jonathan Frech

---

The old solution doesn't work if # is the first character in any number (because eval doesn't accept leading zeroes) and is therefore invalid :(

def f(i,k=0):
 s=i.replace('#','%s')%(*list('%0*d'%(i.count('#'),k)),)
 print(s)if eval(s.replace('=','=='))else f(i,k+1)

Try it online!

share|improve this answer

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

Python 3, 121 155 152 149 bytes

import re
def f(i,k=0,S=re.sub):s=S('#','%s',i)%(*list('%0*d'%(i.count('#'),k)),);print(s)if eval(S('=','==',S('\b0*([1-9])','\1',s)))else f(i,k+1)

Try it online!

+34 New solution with regex to circumvent the fact that python doesn't support numbers with leading zeroes.

-3 thanks to @Jonathan Frech

---

The old solution doesn't work if # is the first character in any number (because eval doesn't accept leading zeroes) and is therefore invalid :(

def f(i,k=0):
 s=i.replace('#','%s')%(*list('%0*d'%(i.count('#'),k)),)
 print(s)if eval(s.replace('=','=='))else f(i,k+1)

Try it online!

share|improve this answer

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

share|improve this answer

share|improve this answer

edited Sep 6 at 13:07

edited Sep 6 at 13:07

edited Sep 6 at 13:07

answered Aug 30 at 17:02

Heiteira

913

answered Aug 30 at 17:02

Heiteira

913

answered Aug 30 at 17:02

Heiteira

913

913

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm afraid this submission is invalid for the reason stated in the post.
  – Erik the Outgolfer
  Aug 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Since your function does not contain any compound statements you can condense it to only one line.
  – Jonathan Frech
  Aug 30 at 23:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I'm afraid this submission is invalid for the reason stated in the post.
  – Erik the Outgolfer
  Aug 30 at 17:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Since your function does not contain any compound statements you can condense it to only one line.
  – Jonathan Frech
  Aug 30 at 23:21
  

  
  
  
  

1

1

I'm afraid this submission is invalid for the reason stated in the post.
– Erik the Outgolfer
Aug 30 at 17:10

I'm afraid this submission is invalid for the reason stated in the post.
– Erik the Outgolfer
Aug 30 at 17:10

2

2

Since your function does not contain any compound statements you can condense it to only one line.
– Jonathan Frech
Aug 30 at 23:21

Since your function does not contain any compound statements you can condense it to only one line.
– Jonathan Frech
Aug 30 at 23:21

add a comment | 

 

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