It's Hip to be Square

Question 1

Challenge

So, um, it seems that, while we have plenty of challenges that work with square numbers or numbers of other shapes, we don't have one 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 as long as it's permitted by standard I/O rules.
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

Question 2

@Neil I realized my mistake. I retract that suggestion, and instead offer 18014398509481982 (2**54-2), which is representable with a double, and causes answers that use sqrt to fail.

Question 3

@Mego I'm probably wrong or just misunderstanding what you're saying, but I'm sure 2**54-2 is still larger than a double can safely handle, at least in JavaScript 18014398509481982 > 9007199254740991

Question 4

@Mego I think the limiting value is 9007199515875288. It's not the square of 94906267, because that's not representable in a double, but if you take its square root, then you get that integer as the result.

Question 5

@Tom Type 2**54-2 into a JS console, and compare what you get with 18014398509481982 (the exact value). JS outputs the exact value, therefore 2**54-2 is representable with a double. If that still doesn't convince you, take the binary data 0100001101001111111111111111111111111111111111111111111111111111, interpret it as a IEEE-754 double-precision float, and see what value you get.

Question 6

Sorry, guys, stepped away for lunch and ... well, that escalated! And there I thought this would be a nice, simple challenge! Would adding a rule that you need not handle inputs that lead to floating point inaccuracies in your chosen language cover it?

Question 7

Ohm, 2 bytes

Æ2

Uses CP-437 encoding.

Explanation

Implicit Input -> Perfect square built-in -> Implicit Output...

Question 8

Java 8, 20 bytes

n->Math.sqrt(n)%1==0

Input is an int.

Try it here.

Question 9

Not debatable: the question explicitly says "Given an integer n (where n>=0)". The shortest answer is the best. Edit: won't +1 until the shortest answer isn't the first :p

Question 10

@OlivierGrégoire Hmm, that's a good way to look at it. But you still wouldn't know whether it's an int, long, short. And with questions where they ask for an integer but the input format is flexible, I sometimes use a String input to save some bytes. Personally I think using n-> is fine, and you should just state what the type is, but apparently not everyone agrees with this. On the other hand, coming from a Java 7 answer history, going from int c(int n){return ...;} to (int n)->... makes more sense than n->... (even though I personally prefer the second since shorter of course).

Question 11

@OlivierGrégoire Ok, I've changed it. After reading the discussion in this answer, I came to the conclusion that stating the input is an integer in Java, is no difference than stating the input is a list of two Strings in CJam or a cell array of strings in MATL.

Question 12

nvm, I'm stupid :P

Question 13

Add++, (削除) 24 (削除ここまで) (削除) 13 (削除ここまで) 11 bytes

+?
S
%1
N
O

Try it online!

I removed the clunky function at the top and rewrote it into the body of the question to remove 11 bytes.

As the first section is already explained below, let's only find out how the new part works

S Square root
%1 Modulo by 1. Produced 0 for integers and a decimal for floats
N Logical NOT

Old version, 24 bytes

D,i,@,1@%!
+?
^.5
$i,x
O

Try it online!

The function at the top (D,i,@,1@%!) is the main part of the program, so let's go into more detail.

D, Create a function...
 i, ...called i...
 @, ...that takes 1 argument (for this example, let's say 3.162 (root 10))
 1 push 1 to the stack; STACK = [1, 3.162]
 @ reverse the stack; STACK = [3.162, 1]
 % modulo the stack; STACK = [0.162]
 ! logical NOT; STACK = [False]
+? Add the input to accumulator (x)
^.5 Square root (exponent by 0.5)
$i,x Apply function i to x
O Output the result

Question 14

Python 3, (削除) 28 27 (削除ここまで) 25 bytes

Thanks to @mdahmoune for 1 byte: compare int of root squared with original
2 bytes saved: lambda shortened

lambda x:int(x**.5)**2==x

Try it online!

Question 15

what about f=lambda x:int(x**.5)**2==x 27bytes

Question 16

CJam, 12 bytes

{_),2f##)!!}

Try it online!

Question 17

JavaScript on NodeJS & Chrome, 51 bytes

// OLD: t=n=>{i=Number.isSafeInteger;return i(n)&&i(n**.5)}
i=Number.isSafeInteger;t=n=>i(n)&&i(n**.5) 
// TestCases:
let l=console.log;l(`t(1): ${t(1)}; t(64): ${t(64)}; t(88): ${t(88)};`)

Try it online!

Question 18

Welcome to PPCG. You don't need to include the t= in your byte count nor is input validation required for this challenge.

Question 19

Also works for Firefox

Question 20

@rickhitchcock does have a point - input validation should not be required! Quote: "You need not handle inputs greater than what your chosen language can natively handle nor which would lead to floating point inaccuracies."

Question 21

Python, (削除) 53 (削除ここまで) (削除) 50 (削除ここまで) (削除) 49 (削除ここまで) (削除) 48 (削除ここまで) (削除) 49 (削除ここまで) 48 bytes

This should in theory work for an input of any size. Returns True if the given number is a square, False otherwise.

f=lambda n,x=0:x<=n if~-(x<=n!=x*x)else f(n,x+1)

Try it online!

Explanation:

f= # assign a name so we can call it
 lambda n,x=0: # counter variable x
 x<=n # counter bigger than input?
 if~-( ) # "negate" inner condition
 x<=n # counter not bigger
 n!=x*x # and n not square of x
 else f(n,x+1) # else recurse

The condition is just a de-Morgan'd if x>n or n==x**2, i.e. we return if the counter is bigger than the input, or we found a proof for squareness.

Saved 1 byte thanks to Gábor Fekete.

Question 22

Your outputs are the wrong way 'round ;)

Question 23

Technically, my output is "one of two consistent truthy/falsey values". But if you deem this invalid, I can correct it at the cost of one byte.

Question 24

I'll edit the question to clarify, I see now that my phrasing in the rules was open to "interpretation", although it was clearer in the intro.

Question 25

@Shaggy Yes, done.

Question 26

Use x*x instead of x**2, saves 1 byte :)

Question 27

Actually, 6 bytes

;ur♂2c

Try it online!

-2 bytes from Erik the Outgolfer

Explanation:

;ur♂2c
;ur range(n+1) ([0, n])
 ♂2 square each element
 c does the list contain the input?

This takes a while for large inputs - TIO will timeout.

Question 28

Can't you replace íub with c?

Question 29

@EriktheOutgolfer You're absolutely right I believe, and I have a lot of answers to update because I didn't think about that.

Question 30

I actually tried to outgolf you, but then I thought the algorithm was way too similar to post, so I converted it to a golfing tip instead...

Question 31

C (gcc), (削除) 66 (削除ここまで) (削除) 43 (削除ここまで) 42 bytes

f(float n){return!(sqrt(n)-(int)sqrt(n));}

Try it online!

Thanks to TheLethalCoder for the tip!

@hvd Thanks for saving a byte!

Question 32

Can you just do: bool f(float n){return !sqrt(n)-(int)sqrt(n)>0;} or similar? Been a while since I've used C.

Question 33

@TheLethalCoder yea absolutely man. :) i would have to incluede stdbool.h though, therefore int as return type would be shorter.

Question 34

There's no need for a space between return and !.

Question 35

Suggest f(n) instead of f(float n) and (n=sqrt(n)) instead of (int)sqrt(n).

Question 36

PowerShell, 26 bytes

!([math]::Sqrt("$args")%1)

Try it online!

Question 37

MIPS, 112 bytes

main:li $v0,7
syscall
sqrt.d $f0,$f0
mfc1 $a0,$f0
li $v0,1
beqz $a0,t
li $a0,0
syscall
b e
t:li $a0,1
syscall
e:

Try it online!

Outputs 1 if the input is square, 0 if not.

Explanation

main:li $v0,7 #Start of program. Load service 7 (read input as float to $f0). 
 #Input can be an integer, but MIPS will interpret it as a float.
syscall #Execute.
sqrt.d $f0,$f0 #Overwrite $f0 with its square root, stored as a double.
mfcl $a0,$f0 #Move $f0 to $a0.
li $v0,1 #Load service 1 (print int from $a0).
beqz $a0,t #Branch to label 't' if $a0 = 0. Otherwise continue.
#If input is non-square...
li $a0,0 #Load 0 into $a0.
syscall #Execute (print $a0).
b e #Branch to label 'e'.
#If input is square...
t:li $a0,1 #Start of label 't'. Load 1 into $a0.
syscall #Execute (print $a0).
e: #Start of label 'e'. Used to avoid executing 't' when input isn't square.

A double in MIPS is 16 hexes. It shares its address with a float containing its low-order 8 hexes ($f0 in this case). The high-order hexes are stored in the next register ($f1), also as a float.

 float double 
$f0 0000 0000 1111 1111 0000 0000
$f1 1111 1111

Taking the square root of a non-square number requires the entire double in order to be stored, meaning the high and low floats are populated. The square root of a square number only needs a few hexes from the double to be stored, and it is stored specifically in its high-order hexes. This means the low float is left at 0.

If the low float equals 0, the input is a square number.

Question 38

you have a bit of superfluous whitespace

Question 39

91?

Question 40

Whitespace, 95 bytes

[S S S N
_Push_0][S N
S _Duplicate_top][T N
T T _Read_STDIN_as_integer][N
S S N
_Create_Label_LOOP][S N
S _Duplicate_top][S N
S _Duplicate_top][T S S N
_Multiply_top_two][S S S N
_Push_0][T T T _Retrieve_input][S N
T _Swap_top_two][T S S T _Subtract][S N
S _Duplicate_top][N
T S S N
_If_0_jump_to_Label_TRUE][N
T T T N
_If_negative_jump_to_Label_FALSE][S S S T N
_Push_1][T S S S _Add][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_TRUE][S S S T N
_Push_1][T N
S T _Print_as_integer][N
N
N
_Stop_program][N
S S T N
_Create_Label_FALSE][S S S N
_Push_0][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs 1/0 for truthy/falsey respectively. A few bytes could be saved if something like 00/0 for truthy/falsey is allowed instead.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Read STDIN as integer, and store it in the heap
Integer i = 0
Start LOOP:
 Integer temp = i*i - input from heap
 If(temp == 0):
 Call function TRUE
 If(temp < 0):
 Call function FALSE
 i = i + 1
 Go to next iteration of LOOP
function TRUE:
 Print 1 as integer
 Stop program
function FALSE:
 Print 0 as integer
 (implicitly stop the program with an error)

Here a port of this approach in Java:

int func(int input){
 for(int i=0; /*infinite loop*/; i++){
 int m = input - i*i;
 if(m==0) return 1;
 if(m<1) return 0;
 }
}

Try it online.

Question 41

Rockstar, 55 bytes

First time posting a solution to one of my own challenges, if you'll forgive me the indulgence.

Outputs 1 for truthy or nothing for falsey.

listen to N
X's0
while N-X
let X be+1
if X*X is N
say 1

Try it here (Code will need to be pasted in)

Question 42

For a Rockstar answer, anything can be forgiven.

Question 43

V (vim), 34 bytes

C<C-r>=sqrt(<C-r>")
<esc>:s/.\+\.0\n
:s/.\+/1

Try it online!

Returns empty string for a square, 1 for non-square.

Explanation

C<C-r>=sqrt(<C-r>")
C cut the input line, enter insert mode
 <C-r>= evaluate the following math command:
 sqrt( ) square root of
 <C-r>" cut input line in register "
<esc>:s/.\+\.0\n/
<esc> exit insert mode
 :s/.\+\.0\n/ remove instance of <digits>.0 
 this removes a perfect square root.
:s/.\+/1 replace any other non-newline chars left with
 a single 1.

Question 44

JavaScript, 44 bytes

n=>{for(i=0n;i<=n;){if(i++**2n==n)return 1}}

Obligatory bigger but BigInt answer.

Returns 1 if a square, or undefined when not a square.

+4 bytes if you want to actually get your answer today:

n=>{for(i=0n;i**2n<=n;){if(i++**2n==n)return 1}}

Question 45

@emanresuA could you elaborate on how the first one would be but not the second? because the only thing I can think of is i not fitting in your memory at some point, but that can happen for both

Question 46

Ah, never mind, I misread it as not having a condition on the loop. Sorry

Question 47

Knight, (削除) 32 (削除ここまで) (削除) 33 (削除ここまで) (削除) 32 (削除ここまで) 30 bytes

;=i+1=xP;W!>0=i-i 1|-^i 2xQ1Q0

Try it online!

Takes input from standard in, outputs by error code (1 if input is square, 0 otherwise).

Expanded code:

;=i +1 (=x PROMPT)
;WHILE !> 0(=i -i 1)
 | (-(^i 2) x)
 QUIT 1
QUIT 0

+1 bytes from @Jonah because I somehow didn't notice this failed on input 1
-2 bytes from @Jonah by using short circuit evaluation of | instead of IF

Question 48

I think Q?0x can just be Q0 because if you get there we know it's not square.

Question 49

@Jonah if the input is 0, then the while loop will never run, so it will go straight to the "false" case -- I need to check for that.

Question 50

I see. Looks like current answer fails for 1.

Question 51

-2 with ;=i+1=xP;W!>0=i-i 1|-^i 2xQ1Q0

Question 52

Pyth, 5 bytes

/^R2h

Try it here!

Question 53

An alternative 5-byte solution that doesn't work for larger numbers (e.g. 2**127 - 1) would be sI@Q2, but I have to admit that I am quite fond of your right map answer.

Question 54

QBIC, (削除) 16 (削除ここまで) 18 bytes

[:|~b/a=a|_Xq}?b=0

Added two bytes for 0-case

This runs through i = 1 ... n to see if n / i == i. Prints 1 if such an i is found, prints -1 for N=0 and 0 in all other cases. Both 1 and -1 are considered truthy in QBasic (-1 being the actual value for true, but IF (n) only is false on n=0).

Question 55

Can you add a TIO (or equivalent)?

Question 56

@Shaggy unfortunately not. My attempts to have QBasic / QBIC / DosBox run on a webpage have been unfruitful, to put it mildly... I've considered running it from CGI, but QBasic is just too powerful to do that, that's asking for exploits... I'm open to suggestions.

Question 57

I just wanted to check if it returns true for 0.

Question 58

@Shaggy it would print 0 for N=0. It never enters the FOR loop and jumps straight to the last statement. I'll fix that

Question 59

J, 8 bytes

(=~<.)%:

Explanation:

%: square root
=~ is the argument equal to itself
<. floor of
=~<. a J hook, which modifies the right argument by applying <.
so, "is the floor of the square root equal to itself?"

Note: If we want to save the above to a variable as a verb, we must do, eg:

issq=.(=~<.)@%:

Question 60

Could you add a TIO (or equivalent) to this, please?

Question 61

I tried it, but it's not producing any output. Not sure if the J interpreter is down or if I'm missing something, but if you paste the following test into jconsole: (=~<.)%: 1 4 9 16 ,: 2 3 7 21 it will return a table with a row of 1s on top of a row of 0s, as expected. I tried putting the same into the "Code" section of TIO and then hitting run, but I get no output.

Question 62

C, 33 bytes

#define f(x)sqrt(x)==(int)sqrt(x)

Takes an integer x. Checks if the square root of x is the square root of x rounded to an integer.

Try it online!

Question 63

Pari/GP, 8 bytes

issquare

Try it online!

Question 64

Pyke, 3 bytes

,$P

Try it here!

, - sqrt(input)
 $ - float(^)
 P - is_int(^)

This could be two bytes (and is in older versions) if Pyke didn't helpfully automatically cast the results of sqrt to an integer if it's a square number.

Linnea Gräf 3,07917 silver badges35 bronze badges · Answer 1 · 2017-06-08 18:48:40Z

3

\$\begingroup\$

Ohm, 2 bytes

Æ2

Uses CP-437 encoding.

Explanation

Implicit Input -> Perfect square built-in -> Implicit Output...

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edited Jun 17, 2020 at 9:04

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answered Jun 8, 2017 at 18:48

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Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 2 · 2017-06-08 12:00:38Z

3

\$\begingroup\$

Java 8, 20 bytes

n->Math.sqrt(n)%1==0

Input is an int.

Try it here.

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edited Jun 17, 2020 at 9:04

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answered Jun 8, 2017 at 12:00

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Kevin Cruijssen

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\$\begingroup\$ Not debatable: the question explicitly says "Given an integer n (where n>=0)". The shortest answer is the best. Edit: won't +1 until the shortest answer isn't the first :p \$\endgroup\$

Olivier Grégoire
– Olivier Grégoire

2017年06月08日 13:03:59 +00:00
Commented Jun 8, 2017 at 13:03
\$\begingroup\$ @OlivierGrégoire Hmm, that's a good way to look at it. But you still wouldn't know whether it's an int, long, short. And with questions where they ask for an integer but the input format is flexible, I sometimes use a String input to save some bytes. Personally I think using n-> is fine, and you should just state what the type is, but apparently not everyone agrees with this. On the other hand, coming from a Java 7 answer history, going from int c(int n){return ...;} to (int n)->... makes more sense than n->... (even though I personally prefer the second since shorter of course). \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2017年06月08日 13:10:06 +00:00
Commented Jun 8, 2017 at 13:10
2

\$\begingroup\$ @OlivierGrégoire Ok, I've changed it. After reading the discussion in this answer, I came to the conclusion that stating the input is an integer in Java, is no difference than stating the input is a list of two Strings in CJam or a cell array of strings in MATL. \$\endgroup\$

Kevin Cruijssen
– Kevin Cruijssen

2017年06月09日 08:05:10 +00:00
Commented Jun 9, 2017 at 8:05
\$\begingroup\$ nvm, I'm stupid :P \$\endgroup\$

0xff
– 0xff

2021年10月16日 12:55:36 +00:00
Commented Oct 16, 2021 at 12:55

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score 3 · Answer 3 · 2017-06-08 14:10:18Z

Add++, (削除) 24 (削除ここまで) (削除) 13 (削除ここまで) 11 bytes

+?
S
%1
N
O

Try it online!

I removed the clunky function at the top and rewrote it into the body of the question to remove 11 bytes.

As the first section is already explained below, let's only find out how the new part works

S Square root
%1 Modulo by 1. Produced 0 for integers and a decimal for floats
N Logical NOT

Old version, 24 bytes

D,i,@,1@%!
+?
^.5
$i,x
O

Try it online!

The function at the top (D,i,@,1@%!) is the main part of the program, so let's go into more detail.

D, Create a function...
 i, ...called i...
 @, ...that takes 1 argument (for this example, let's say 3.162 (root 10))
 1 push 1 to the stack; STACK = [1, 3.162]
 @ reverse the stack; STACK = [3.162, 1]
 % modulo the stack; STACK = [0.162]
 ! logical NOT; STACK = [False]
+? Add the input to accumulator (x)
^.5 Square root (exponent by 0.5)
$i,x Apply function i to x
O Output the result

0xffcourse 2,7993 gold badges15 silver badges27 bronze badges · Answer 4 · 2017-06-15 10:31:39Z

3

\$\begingroup\$

Python 3, (削除) 28 27 (削除ここまで) 25 bytes

Thanks to @mdahmoune for 1 byte: compare int of root squared with original
2 bytes saved: lambda shortened

lambda x:int(x**.5)**2==x

Try it online!

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edited Jul 3, 2017 at 16:06

answered Jun 15, 2017 at 10:31

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0xffcourse

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1

\$\begingroup\$ what about f=lambda x:int(x**.5)**2==x 27bytes \$\endgroup\$

mdahmoune
– mdahmoune

2017年07月03日 15:30:36 +00:00
Commented Jul 3, 2017 at 15:30

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Erik the Outgolfer 40.8k5 gold badges46 silver badges125 bronze badges · Answer 5 · 2017-06-08 12:07:17Z

2

\$\begingroup\$

CJam, 12 bytes

{_),2f##)!!}

Try it online!

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answered Jun 8, 2017 at 12:07

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Erik the Outgolfer

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BogdanBiv 1234 bronze badges · Answer 6 · 2017-06-08 13:49:13Z

2

\$\begingroup\$

JavaScript on NodeJS & Chrome, 51 bytes

// OLD: t=n=>{i=Number.isSafeInteger;return i(n)&&i(n**.5)}
i=Number.isSafeInteger;t=n=>i(n)&&i(n**.5) 
// TestCases:
let l=console.log;l(`t(1): ${t(1)}; t(64): ${t(64)}; t(88): ${t(88)};`)

Try it online!

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edited Jun 8, 2017 at 14:01

answered Jun 8, 2017 at 13:49

BogdanBiv's user avatar

BogdanBiv

1234 bronze badges

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3

\$\begingroup\$ Welcome to PPCG. You don't need to include the t= in your byte count nor is input validation required for this challenge. \$\endgroup\$

Shaggy
– Shaggy

2017年06月08日 14:00:49 +00:00
Commented Jun 8, 2017 at 14:00
\$\begingroup\$ Also works for Firefox \$\endgroup\$

WORNG ALL
– WORNG ALL

2017年07月03日 14:57:53 +00:00
Commented Jul 3, 2017 at 14:57
\$\begingroup\$ @rickhitchcock does have a point - input validation should not be required! Quote: "You need not handle inputs greater than what your chosen language can natively handle nor which would lead to floating point inaccuracies." \$\endgroup\$

BogdanBiv
– BogdanBiv

2017年07月10日 13:27:37 +00:00
Commented Jul 10, 2017 at 13:27

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L3viathan 3,44116 silver badges26 bronze badges · Answer 7 · 2017-06-08 12:35:52Z

Python, (削除) 53 (削除ここまで) (削除) 50 (削除ここまで) (削除) 49 (削除ここまで) (削除) 48 (削除ここまで) (削除) 49 (削除ここまで) 48 bytes

This should in theory work for an input of any size. Returns True if the given number is a square, False otherwise.

f=lambda n,x=0:x<=n if~-(x<=n!=x*x)else f(n,x+1)

Try it online!

Explanation:

f= # assign a name so we can call it
 lambda n,x=0: # counter variable x
 x<=n # counter bigger than input?
 if~-( ) # "negate" inner condition
 x<=n # counter not bigger
 n!=x*x # and n not square of x
 else f(n,x+1) # else recurse

The condition is just a de-Morgan'd if x>n or n==x**2, i.e. we return if the counter is bigger than the input, or we found a proof for squareness.

Saved 1 byte thanks to Gábor Fekete.

Technically, my output is "one of two consistent truthy/falsey values". But if you deem this invalid, I can correct it at the cost of one byte.
I'll edit the question to clarify, I see now that my phrasing in the rules was open to "interpretation", although it was clearer in the intro.

user45941 · Answer 8 · 2017-06-08 11:50:08Z

2

\$\begingroup\$

Actually, 6 bytes

;ur♂2c

Try it online!

-2 bytes from Erik the Outgolfer

Explanation:

;ur♂2c
;ur range(n+1) ([0, n])
 ♂2 square each element
 c does the list contain the input?

This takes a while for large inputs - TIO will timeout.

Share

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edited Jun 10, 2017 at 11:10

answered Jun 8, 2017 at 11:50

user45941

\$\endgroup\$

3

\$\begingroup\$ Can't you replace íub with c? \$\endgroup\$

Erik the Outgolfer
– Erik the Outgolfer

2017年06月09日 15:09:16 +00:00
Commented Jun 9, 2017 at 15:09
\$\begingroup\$ @EriktheOutgolfer You're absolutely right I believe, and I have a lot of answers to update because I didn't think about that. \$\endgroup\$

user45941
– user45941

2017年06月10日 01:30:12 +00:00
Commented Jun 10, 2017 at 1:30
\$\begingroup\$ I actually tried to outgolf you, but then I thought the algorithm was way too similar to post, so I converted it to a golfing tip instead... \$\endgroup\$

Erik the Outgolfer
– Erik the Outgolfer

2017年06月10日 10:09:36 +00:00
Commented Jun 10, 2017 at 10:09

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Abel Tom 1,1701 gold badge7 silver badges10 bronze badges · Answer 9 · 2017-06-08 12:00:45Z

2

\$\begingroup\$

C (gcc), (削除) 66 (削除ここまで) (削除) 43 (削除ここまで) 42 bytes

f(float n){return!(sqrt(n)-(int)sqrt(n));}

Try it online!

Thanks to TheLethalCoder for the tip!

@hvd Thanks for saving a byte!

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edited Jun 11, 2017 at 15:19

answered Jun 8, 2017 at 12:00

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Abel Tom

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4

\$\begingroup\$ Can you just do: bool f(float n){return !sqrt(n)-(int)sqrt(n)>0;} or similar? Been a while since I've used C. \$\endgroup\$

TheLethalCoder
– TheLethalCoder

2017年06月08日 12:21:05 +00:00
Commented Jun 8, 2017 at 12:21
\$\begingroup\$ @TheLethalCoder yea absolutely man. :) i would have to incluede stdbool.h though, therefore int as return type would be shorter. \$\endgroup\$

Abel Tom
– Abel Tom

2017年06月08日 12:26:38 +00:00
Commented Jun 8, 2017 at 12:26
\$\begingroup\$ There's no need for a space between return and !. \$\endgroup\$

hvd
– hvd

2017年06月11日 13:54:33 +00:00
Commented Jun 11, 2017 at 13:54
1

\$\begingroup\$ Suggest f(n) instead of f(float n) and (n=sqrt(n)) instead of (int)sqrt(n). \$\endgroup\$

ceilingcat
– ceilingcat

2017年11月02日 04:33:24 +00:00
Commented Nov 2, 2017 at 4:33

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mazzy 7,2292 gold badges13 silver badges23 bronze badges · Answer 10 · 2019-03-03 06:24:04Z

2

\$\begingroup\$

PowerShell, 26 bytes

!([math]::Sqrt("$args")%1)

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answered Mar 3, 2019 at 6:24

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Bsoned 314 bronze badges · Answer 11 · 2019-03-06 03:47:37Z

MIPS, 112 bytes

main:li $v0,7
syscall
sqrt.d $f0,$f0
mfc1 $a0,$f0
li $v0,1
beqz $a0,t
li $a0,0
syscall
b e
t:li $a0,1
syscall
e:

Try it online!

Outputs 1 if the input is square, 0 if not.

Explanation

main:li $v0,7 #Start of program. Load service 7 (read input as float to $f0). 
 #Input can be an integer, but MIPS will interpret it as a float.
syscall #Execute.
sqrt.d $f0,$f0 #Overwrite $f0 with its square root, stored as a double.
mfcl $a0,$f0 #Move $f0 to $a0.
li $v0,1 #Load service 1 (print int from $a0).
beqz $a0,t #Branch to label 't' if $a0 = 0. Otherwise continue.
#If input is non-square...
li $a0,0 #Load 0 into $a0.
syscall #Execute (print $a0).
b e #Branch to label 'e'.
#If input is square...
t:li $a0,1 #Start of label 't'. Load 1 into $a0.
syscall #Execute (print $a0).
e: #Start of label 'e'. Used to avoid executing 't' when input isn't square.

A double in MIPS is 16 hexes. It shares its address with a float containing its low-order 8 hexes ($f0 in this case). The high-order hexes are stored in the next register ($f1), also as a float.

 float double 
$f0 0000 0000 1111 1111 0000 0000
$f1 1111 1111

Taking the square root of a non-square number requires the entire double in order to be stored, meaning the high and low floats are populated. The square root of a square number only needs a few hexes from the double to be stored, and it is stored specifically in its high-order hexes. This means the low float is left at 0.

If the low float equals 0, the input is a square number.

\$\begingroup\$ you have a bit of superfluous whitespace \$\endgroup\$

ASCII-only
– ASCII-only

2019年03月13日 03:49:07 +00:00
Commented Mar 13, 2019 at 3:49
\$\begingroup\$ 91? \$\endgroup\$

ASCII-only
– ASCII-only

2019年03月13日 04:17:31 +00:00
Commented Mar 13, 2019 at 4:17

Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 12 · 2019-08-27 10:26:33Z

Whitespace, 95 bytes

[S S S N
_Push_0][S N
S _Duplicate_top][T N
T T _Read_STDIN_as_integer][N
S S N
_Create_Label_LOOP][S N
S _Duplicate_top][S N
S _Duplicate_top][T S S N
_Multiply_top_two][S S S N
_Push_0][T T T _Retrieve_input][S N
T _Swap_top_two][T S S T _Subtract][S N
S _Duplicate_top][N
T S S N
_If_0_jump_to_Label_TRUE][N
T T T N
_If_negative_jump_to_Label_FALSE][S S S T N
_Push_1][T S S S _Add][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_TRUE][S S S T N
_Push_1][T N
S T _Print_as_integer][N
N
N
_Stop_program][N
S S T N
_Create_Label_FALSE][S S S N
_Push_0][T N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Outputs 1/0 for truthy/falsey respectively. A few bytes could be saved if something like 00/0 for truthy/falsey is allowed instead.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Read STDIN as integer, and store it in the heap
Integer i = 0
Start LOOP:
 Integer temp = i*i - input from heap
 If(temp == 0):
 Call function TRUE
 If(temp < 0):
 Call function FALSE
 i = i + 1
 Go to next iteration of LOOP
function TRUE:
 Print 1 as integer
 Stop program
function FALSE:
 Print 0 as integer
 (implicitly stop the program with an error)

Here a port of this approach in Java:

int func(int input){
 for(int i=0; /*infinite loop*/; i++){
 int m = input - i*i;
 if(m==0) return 1;
 if(m<1) return 0;
 }
}

Try it online.

Shaggy 45k4 gold badges39 silver badges95 bronze badges · Answer 13 · 2020-09-24 21:10:20Z

Rockstar, 55 bytes

First time posting a solution to one of my own challenges, if you'll forgive me the indulgence.

Outputs 1 for truthy or nothing for falsey.

listen to N
X's0
while N-X
let X be+1
if X*X is N
say 1

Try it here (Code will need to be pasted in)

1

\$\begingroup\$ For a Rockstar answer, anything can be forgiven. \$\endgroup\$

user
– user

2021年03月22日 00:57:40 +00:00
Commented Mar 22, 2021 at 0:57

Razetime 27.6k3 gold badges31 silver badges77 bronze badges · Answer 14 · 2021-04-10 11:08:39Z

V (vim), 34 bytes

C<C-r>=sqrt(<C-r>")
<esc>:s/.\+\.0\n
:s/.\+/1

Try it online!

Returns empty string for a square, 1 for non-square.

Explanation

C<C-r>=sqrt(<C-r>")
C cut the input line, enter insert mode
 <C-r>= evaluate the following math command:
 sqrt( ) square root of
 <C-r>" cut input line in register "
<esc>:s/.\+\.0\n/
<esc> exit insert mode
 :s/.\+\.0\n/ remove instance of <digits>.0 
 this removes a perfect square root.
:s/.\+/1 replace any other non-newline chars left with
 a single 1.

Leaf 5214 silver badges7 bronze badges · Answer 15 · 2022-08-04 15:32:38Z

2

\$\begingroup\$

JavaScript, 44 bytes

n=>{for(i=0n;i<=n;){if(i++**2n==n)return 1}}

Obligatory bigger but BigInt answer.

Returns 1 if a square, or undefined when not a square.

+4 bytes if you want to actually get your answer today:

n=>{for(i=0n;i**2n<=n;){if(i++**2n==n)return 1}}

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answered Aug 4, 2022 at 15:32

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Leaf

5214 silver badges7 bronze badges

\$\endgroup\$

2

\$\begingroup\$ @emanresuA could you elaborate on how the first one would be but not the second? because the only thing I can think of is i not fitting in your memory at some point, but that can happen for both \$\endgroup\$

Leaf
– Leaf

2024年03月07日 09:49:54 +00:00
Commented Mar 7, 2024 at 9:49
1

\$\begingroup\$ Ah, never mind, I misread it as not having a condition on the loop. Sorry \$\endgroup\$

emanresu A
– emanresu A

2024年03月07日 18:36:42 +00:00
Commented Mar 7, 2024 at 18:36

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97.100.97.109 7,46513 silver badges48 bronze badges · Answer 16 · 2022-08-04 15:40:44Z

2

\$\begingroup\$

Knight, (削除) 32 (削除ここまで) (削除) 33 (削除ここまで) (削除) 32 (削除ここまで) 30 bytes

;=i+1=xP;W!>0=i-i 1|-^i 2xQ1Q0

Try it online!

Takes input from standard in, outputs by error code (1 if input is square, 0 otherwise).

Expanded code:

;=i +1 (=x PROMPT)
;WHILE !> 0(=i -i 1)
 | (-(^i 2) x)
 QUIT 1
QUIT 0

+1 bytes from @Jonah because I somehow didn't notice this failed on input 1
-2 bytes from @Jonah by using short circuit evaluation of | instead of IF

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edited Aug 6, 2022 at 21:18

answered Aug 4, 2022 at 15:40

97.100.97.109's user avatar

97.100.97.109

7,46513 silver badges48 bronze badges

\$\endgroup\$

4

\$\begingroup\$ I think Q?0x can just be Q0 because if you get there we know it's not square. \$\endgroup\$

Jonah
– Jonah

2022年08月06日 20:29:38 +00:00
Commented Aug 6, 2022 at 20:29
1

\$\begingroup\$ @Jonah if the input is 0, then the while loop will never run, so it will go straight to the "false" case -- I need to check for that. \$\endgroup\$

97.100.97.109
– 97.100.97.109

2022年08月06日 20:32:43 +00:00
Commented Aug 6, 2022 at 20:32
\$\begingroup\$ I see. Looks like current answer fails for 1. \$\endgroup\$

Jonah
– Jonah

2022年08月06日 20:36:38 +00:00
Commented Aug 6, 2022 at 20:36
1

\$\begingroup\$ -2 with ;=i+1=xP;W!>0=i-i 1|-^i 2xQ1Q0 \$\endgroup\$

Jonah
– Jonah

2022年08月06日 21:14:36 +00:00
Commented Aug 6, 2022 at 21:14

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Erik the Outgolfer 40.8k5 gold badges46 silver badges125 bronze badges · Answer 17 · 2017-06-08 11:59:04Z

1

\$\begingroup\$

Pyth, 5 bytes

/^R2h

Try it here!

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answered Jun 8, 2017 at 11:59

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Erik the Outgolfer

40.8k5 gold badges46 silver badges125 bronze badges

\$\endgroup\$

1

\$\begingroup\$ An alternative 5-byte solution that doesn't work for larger numbers (e.g. 2**127 - 1) would be sI@Q2, but I have to admit that I am quite fond of your right map answer. \$\endgroup\$

notjagan
– notjagan

2017年06月08日 15:19:25 +00:00
Commented Jun 8, 2017 at 15:19

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steenbergh 8,1741 gold badge27 silver badges43 bronze badges · Answer 18 · 2017-06-08 12:55:25Z

1

\$\begingroup\$

QBIC, (削除) 16 (削除ここまで) 18 bytes

[:|~b/a=a|_Xq}?b=0

Added two bytes for 0-case

This runs through i = 1 ... n to see if n / i == i. Prints 1 if such an i is found, prints -1 for N=0 and 0 in all other cases. Both 1 and -1 are considered truthy in QBasic (-1 being the actual value for true, but IF (n) only is false on n=0).

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edited Jun 8, 2017 at 13:31

answered Jun 8, 2017 at 12:55

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steenbergh

8,1741 gold badge27 silver badges43 bronze badges

\$\endgroup\$

4

\$\begingroup\$ Can you add a TIO (or equivalent)? \$\endgroup\$

Shaggy
– Shaggy

2017年06月08日 13:21:06 +00:00
Commented Jun 8, 2017 at 13:21
\$\begingroup\$ @Shaggy unfortunately not. My attempts to have QBasic / QBIC / DosBox run on a webpage have been unfruitful, to put it mildly... I've considered running it from CGI, but QBasic is just too powerful to do that, that's asking for exploits... I'm open to suggestions. \$\endgroup\$

steenbergh
– steenbergh

2017年06月08日 13:23:40 +00:00
Commented Jun 8, 2017 at 13:23
\$\begingroup\$ I just wanted to check if it returns true for 0. \$\endgroup\$

Shaggy
– Shaggy

2017年06月08日 13:25:28 +00:00
Commented Jun 8, 2017 at 13:25
\$\begingroup\$ @Shaggy it would print 0 for N=0. It never enters the FOR loop and jumps straight to the last statement. I'll fix that \$\endgroup\$

steenbergh
– steenbergh

2017年06月08日 13:28:52 +00:00
Commented Jun 8, 2017 at 13:28

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Jonah 34.1k4 gold badges41 silver badges95 bronze badges · Answer 19 · 2017-06-08 15:31:45Z

1

\$\begingroup\$

J, 8 bytes

(=~<.)%:

Explanation:

%: square root
=~ is the argument equal to itself
<. floor of
=~<. a J hook, which modifies the right argument by applying <.
so, "is the floor of the square root equal to itself?"

Note: If we want to save the above to a variable as a verb, we must do, eg:

issq=.(=~<.)@%:

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edited Jun 17, 2020 at 9:04

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1

answered Jun 8, 2017 at 15:31

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Jonah

34.1k4 gold badges41 silver badges95 bronze badges

\$\endgroup\$

2

\$\begingroup\$ Could you add a TIO (or equivalent) to this, please? \$\endgroup\$

Shaggy
– Shaggy

2017年06月09日 08:19:47 +00:00
Commented Jun 9, 2017 at 8:19
\$\begingroup\$ I tried it, but it's not producing any output. Not sure if the J interpreter is down or if I'm missing something, but if you paste the following test into jconsole: (=~<.)%: 1 4 9 16 ,: 2 3 7 21 it will return a table with a row of 1s on top of a row of 0s, as expected. I tried putting the same into the "Code" section of TIO and then hitting run, but I get no output. \$\endgroup\$

Jonah
– Jonah

2017年06月09日 18:24:05 +00:00
Commented Jun 9, 2017 at 18:24

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MD XF 14.2k5 gold badges69 silver badges107 bronze badges · Answer 20 · 2017-06-09 03:27:11Z

C, 33 bytes

#define f(x)sqrt(x)==(int)sqrt(x)

Takes an integer x. Checks if the square root of x is the square root of x rounded to an integer.

Try it online!

alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 21 · 2017-06-10 17:17:39Z

1

\$\begingroup\$

Pari/GP, 8 bytes

issquare

Try it online!

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answered Jun 10, 2017 at 17:17

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alephalpha

51.9k7 gold badges75 silver badges196 bronze badges

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Blue 28.8k8 gold badges53 silver badges101 bronze badges · Answer 22 · 2017-06-11 15:46:06Z

Pyke, 3 bytes

,$P

Try it here!

, - sqrt(input)
 $ - float(^)
 P - is_int(^)

This could be two bytes (and is in older versions) if Pyke didn't helpfully automatically cast the results of sqrt to an integer if it's a square number.

ASCII-only 5,17922 silver badges43 bronze badges · Answer 23 · 2017-06-19 02:45:30Z

1

\$\begingroup\$

Charcoal, 8 bytes

¬%XN·5¦1

Try it online!

Explanation

¬ Not
 % ¦1 Modulo 1
 X ·5 To the power of .5
 N Next input as number,

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answered Jun 19, 2017 at 2:45

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ASCII-only

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\$\endgroup\$

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space junk 3152 silver badges6 bronze badges · Answer 24 · 2017-06-28 09:59:56Z

05AB1E, (削除) 5 (削除ここまで) 3 bytes

t.ï

Try it online!

~~(削除) Longer than @Erik's but just wanted to give it a shot. (削除ここまで)~~

Now shorter than Erik's but fails for large numbers...

Explanation

t # square roots the input
 .ï # checks if number on stack is an int
 # implicit output of result (0 or 1)

jacoblaw 2661 silver badge4 bronze badges · Answer 25 · 2017-06-09 20:02:47Z

Python 3, (削除) 39 (削除ここまで) 38 Bytes

lambda n:n in(i*i for i in range(n+1))

@mathmandan I had the same idea, and this implementation is 1 byte shorter. I wanted to comment on your post but do not yet have 50 reputation. I hope you see this!

This is just brute force, and I did not get it to complete 2147483647 in a reasonable amount of time.

Thanks @DJMcMayhem for suggesting i remove the space after in

Try it Online

Welcome to the site! Just FYI, you could remove a space after in

Wernisch 3,03411 silver badges23 bronze badges · Answer 26 · 2017-06-09 09:40:31Z

1

\$\begingroup\$

Excel, (削除) 18 (削除ここまで) 16 bytes

=MOD(A1^0.5,1)=0

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edited Jul 10, 2017 at 10:36

answered Jun 9, 2017 at 9:40

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Wernisch

3,03411 silver badges23 bronze badges

\$\endgroup\$

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Renzo 2,63816 silver badges22 bronze badges · Answer 27 · 2017-07-10 10:57:21Z

1

\$\begingroup\$

Common Lisp, 30 bytes

(defun g(x)(=(mod(sqrt x)1)0))

Try it online!

or, with the same length,

(defun g(x)(integerp(sqrt x)))

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answered Jul 10, 2017 at 10:57

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Renzo

2,63816 silver badges22 bronze badges

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Sok 6,2341 gold badge19 silver badges30 bronze badges · Answer 28 · 2017-07-10 13:44:47Z

1

\$\begingroup\$

><>, 22+3 = 25 bytes

01-\:*=n;
(?!1円+::*{:}

Try it online!

Input is expected on the stack at program start, so +3 bytes for the -v flag.

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answered Jul 10, 2017 at 13:44

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Sok

6,2341 gold badge19 silver badges30 bronze badges

\$\endgroup\$

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G B 23.4k1 gold badge24 silver badges55 bronze badges · Answer 29 · 2018-01-12 07:14:32Z

1

\$\begingroup\$

Ruby, 16 bytes

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

Trivial solution.

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edited Jun 17, 2020 at 9:04

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answered Jan 12, 2018 at 7:14

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DLosc 40.7k6 gold badges87 silver badges142 bronze badges · Answer 30 · 2018-01-12 09:06:28Z

tinylisp, 67 bytes

(load library
(d S(q((n)(contains?(map(q((x)(* x x)))(c 0(1to n)))n

Try it online!

Generates a list of numbers from 0 through n, maps a lambda function that squares each one, and checks if n is in the resulting list of squares. Wildly inefficient, of course.

Here's a 77-byte solution that doesn't use the library and runs an order of magnitude faster:

(d _(q((x n s)(i(e n s)1(i(l n s)0(_(a x 1)n(a s(a x(a x 1
(d S(q((n)(_ 0 n 0

Try it online!

This one uses a helper function _ which tracks a counter x and its square s. At each level of recursion, we return success if s equals n and failure if s is greater than n; otherwise, if s is still less than n, we recurse, incrementing x and calculating the next s by the formula (x+1)^2 = x^2 + x + x + 1.

It's Hip to be Square

Challenge

Rules

Test Cases

84 Answers 84

Ohm, 2 bytes

Explanation

Java 8, 20 bytes

Add++, (削除) 24 (削除ここまで) (削除) 13 (削除ここまで) 11 bytes

Old version, 24 bytes

Python 3, (削除) 28 27 (削除ここまで) 25 bytes

CJam, 12 bytes

JavaScript on NodeJS & Chrome, 51 bytes

Python, (削除) 53 (削除ここまで) (削除) 50 (削除ここまで) (削除) 49 (削除ここまで) (削除) 48 (削除ここまで) (削除) 49 (削除ここまで) 48 bytes

Explanation:

Actually, 6 bytes

C (gcc), (削除) 66 (削除ここまで) (削除) 43 (削除ここまで) 42 bytes

PowerShell, 26 bytes

MIPS, 112 bytes

Explanation

Whitespace, 95 bytes

Rockstar, 55 bytes

V (vim), 34 bytes

Explanation

JavaScript, 44 bytes

Knight, (削除) 32 (削除ここまで) (削除) 33 (削除ここまで) (削除) 32 (削除ここまで) 30 bytes

Pyth, 5 bytes

QBIC, (削除) 16 (削除ここまで) 18 bytes

J, 8 bytes

C, 33 bytes

Pari/GP, 8 bytes

Pyke, 3 bytes

Charcoal, 8 bytes

Explanation

05AB1E, (削除) 5 (削除ここまで) 3 bytes

Explanation

Python 3, (削除) 39 (削除ここまで) 38 Bytes

Excel, (削除) 18 (削除ここまで) 16 bytes

Common Lisp, 30 bytes

><>, 22+3 = 25 bytes

Ruby, 16 bytes

tinylisp, 67 bytes

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