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Comments on Operation "Find The Operator" [released]

Post

Operation "Find The Operator" [released]

+3
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Challenge

Make a program that takes input of 3 non-negative integers: a result and 2 other values that once calculated results to the 3rd value.

  • The program must figure out how to get the 2 first numbers to equivalence itself with the 3rd. It could be either addition, subtraction, multiplication, integer division, modulo and exponentiation.
  • You can't change the places of the 2 variables when building the equation, meaning you can't get some value by intertwining in the methods excluding addition and multiplication.
  • The returning value is the operator that makes the equation true (it's best to use some distinguishable character to separate it from the other methods).
  • Use whatever characters you want to use to determine the method that makes the equation truthy.
  • If multiple operators make the equation truthy, then output them in any order.
  • What makes it true? If an equation exists within the 2 numbers that results to the value of the 3rd.
  • What to return when it's false? Something but not nothing (whitespaces and newlines aren't allowed as such output).
  • This is code-golf so the shortest program wins!

Test Cases

# Input (#, #, #)
// Output (+-*/%^ or .)

2 3 5
+

7 3 2
/

9 8 4
.

1 2 2
*

2 2 4
+ * ^

14 6 8
-

5 0 2
.

0 0 1
^

8 4 0
%

0 0 0
+ - *
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7 comment threads

Output type (3 comments)
Correction for test case, suggestion of another (1 comment)
In case multiple operators match (1 comment)
Challenge name idea (1 comment)
General (1 comment)
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General
Shaggy‭ wrote about 3 years ago

Can we use our language's native operators (e.g., p in Japt for exponentiation)? I'm guessing from your use of . in the test cases we can. Also, you need to mention that / is floor/integer division.