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Challenges

It's Hip to be Square

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Challenge

A catalogue type challenge that simply asks: Given an integer n (where n>=0) as input return a truthy value if n is a perfect square or a falsey value if not.


Rules

  • You may take input by any reasonable, convenient means.
  • You need not handle inputs greater than what your chosen language can natively handle nor which would lead to floating point inaccuracies.
  • Output should be one of two consistent truthy/falsey values (e.g., true or false, 1 or 0) - truthy if the input is a perfect square, falsey if it's not.
  • This is code-golf so lowest byte count wins.

Test Cases

Input:  0
Output: true

Input:  1
Output: true

Input:  64
Output: true

Input:  88
Output: false

Input:  2147483647
Output: false
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20 answers

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Lua 5.4, 9 bytes

n^.5%1==0

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The entire compressed code is 33 bytes:

function f(n)return n^.5%1==0 end

Try it here.

It's not clear whether the rules allow the first form (i.e. only the relevant expression) or the second form (i.e. the entire piece of code that does the check). I see some use the former, some use the latter.

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Python 3, 20 19 bytes

lambda n:n**.5%1==0

-1 byte thanks to @shaggy : 0.5 -> .5

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1 comment thread

Welcome to Code Golf :) You can save a byte with `.5` instead of `0.5`. (2 comments)
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APL (Dyalog Extended), 3 bytes

√∊…

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Is the square root in the range?

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Fails for `0`, which should be truthy. I think you'll need `√∊0,⍳`, unless APL has a 0-based range bu... (3 comments)
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Haskell, 24 bytes

Probably the optimal solution. Credits go to nimi from PPCG.

f n=elem n$map(^2)[0..n]

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My original solution, 25 bytes

f a=any((==a).(^2))[0..a]

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You should be able to get this down to 24 by mapping and squaring the range and then checking if it c... (3 comments)
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Ruby, 16 bytes

->n{n**0.5%1==0}

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+3
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Prolog (SWI), 42 bytes

R-S:-S is R*R;R<S,M is R+1,M-S.
f(S):-0-S.

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Japt, 6 3 bytes

Cut in half thanks to Shaggy!

¬v1

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Golfed thanks to Shaggy's interpreter auto-golf feature.

¬v1
¬      //Square root
 v1    //Is that an integer?
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You can reduce that by half (7 comments)
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Husk, 5 bytes

±£Θİ□

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similar to haskell, checks if the number is in the infinite list of squares.

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This fails for `0`, which should return a truthy value. (4 comments)
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Vyxal, 2 bytes

∆²

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Woooooooo built-in elements ftw.

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General (1 comment)
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C (gcc), 57 64 bytes

x,s;main(){scanf("%d",&x);s=sqrt(x);printf("%d",s*s==x);}

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Strictly speaking, both outputs here are truthy as they're both non-empty strings. (4 comments)
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Python 3, 46 bytes

Saved 8 bytes thanks to Shaggy!

lambda n:[n-i*i or exit(1)for i in range(1+n)]

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Python 3.8 (pre-release), 50 54 52 bytes

Fixed a silly mistake thanks to Shaggy and saved 2 bytes!

for i in range(1+(x:=int(input()))):x-i*i or exit(1)

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Here's a stupid solution that outputs using the exit code (1 if it's a perfect square, 0 otherwise).

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2 comment threads

Incorrect output (6 comments)
Cheaper to set x outside the loop (2 comments)
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C (gcc), 40 bytes

i;f(n){for(i=1;n>0;n-=i,i+=2);return!n;}

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PHP, 69 37 bytes

<?=($y=sqrt($x=fgets(STDIN)))*$y==$x;

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Golfed 32 bytes thanks to @Shaggy's advice.

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Does nothing (1 comment)
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Sclipting, (UTF-16) 8 bytes

根❶圜同

Takes the square root, pushes a copy, rounds the copy, and compares. Basically, checks that the square root is an integer.

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Can you add a [TIO](https://tio.run/) or equivalent, please? (4 comments)
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C (clang), 189 181 176 173 bytes

i,n,x,y;main(){scanf("%i",&x);y=1,n=10001;int z[n];for(i=0;i<n;i++,y++){z[i]=pow(y,2);}for(i=0;i<n;i++){if(x==z[i]){puts("True");break;}if(x!=z[i]&&i==n-1){puts("False");}}}

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If libraries also count in the code, the total byte count will be 226 218 213 210.

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This seems to return `false` for all squares greater than `4`. Also, we're not restricted to using TI... (4 comments)
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JavaScript (V8), 13 bytes

x=>x**.5%1==0

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C (gcc), 37 bytes

f(n,p){while(++p*p<n);return p*p==n;}

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The solution is based on the simple fact that:

$$\forall n \in \mathbb{N}, n \text{ is a perfect square} \Longleftrightarrow \exists p \in \mathbb{N}, p \le n \text{ / } p^2=n$$

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J, 6 bytes

0=1|%:

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Embed ESCR, 28 characters

[= [exp [rnd [sqrt n]] 2] n]

The number to test is in N. There are 4 nested functions. From inner to outer: SQRT takes the square root of N. This produces a floating point result. RND rounds that to the nearest integer. EXP is used to raise that integer to the power of 2 (square it). This produces an integer since both inputs are integers. "=" compares the squared result to N. The result is a boolean value, which is either "TRUE" or "FALSE" in its text representation.

Here is a complete program illustrating the above applied to input values 0 to 20:

loop with n from 0 to 20
  show n ": " [= [exp [rnd [sqrt n]] 2] n]
  endloop

Its output is:

0: TRUE
1: TRUE
2: FALSE
3: FALSE
4: TRUE
5: FALSE
6: FALSE
7: FALSE
8: FALSE
9: TRUE
10: FALSE
11: FALSE
12: FALSE
13: FALSE
14: FALSE
15: FALSE
16: TRUE
17: FALSE
18: FALSE
19: FALSE
20: FALSE
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dc, 9 bytes

_?dvd*-^p

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Comparisons are expensive in dc, so you have to get a bit creative. I came up with $0^{n-\lfloor\sqrt{n}\rfloor^2}$.

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