Triangle and/or Square? [duplicate]

Question

Explanation of the challenge

Given an input, return what shapes it could be. There are only a couple options.

1. Square
2. Triangle

---

Square

A square is in the following format:

XXXX
XXXX
XXXX
XXXX

So, in this example, the input would be:

16

And the output would be:

1 or "Square"

Triangle

We are only checking for equilateral triangles:

   X
  X X
 X X X
X X X X

So, in this example, the input would be:

10

And the output would be:

2 or "Triangle"

---

Rules for input and output

Input must be a number.

Output can be an integer list or string list as long as you specify how your output works in your post. The output will be either a triangle, square, or both; never neither.

---

Test Cases

36 ->  1,2 or 3 or ["Square","Triangle"] or "Both"

15 ->  2 or "Triangle"

4  ->  1 or "Square"

---

Leaderboard

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Answer 1

Japt, 18 15 11 bytes

[U8*UÄ]®¬v1

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Answer 2

Jelly, 9 7 bytes

×8‘,µÆ²

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Outputs [1,1] for both, [0,1] for square, [1,0] for triangle, [0,0] for neither.

Explanation

×8‘,µÆ² - main link, input a
   ,    - a list of the following two:
×8‘     - a×8+1 (which is square when a is triangle)
        - (implicit) a
    µÆ² - On the previous, check if each is a square

Answer 3

JavaScript (ES7), 40 32 bytes

Returns 1 for square, 2 for triangle, 3 for both or 0 for neither.

n=>!(n**.5%1)+2*!((8*n+1)**.5%1)

• 8 bytes saved in collaboration with ETHproductions and Neil.

---

Try It

Displays "Square/Triangle/Both/Neither" for clarity

f=
n=>!(n**.5%1)+2*!((8*n+1)**.5%1)
i.addEventListener("input",_=>o.innerText=["Neither","Square","Triangle","Both"][f(+i.value)])

<input id=i type=number><pre id=o>

Answer 4

Haskell, 49 48 37 bytes

EDIT:

• -1 byte: @xnor suggested parametrizing on k.
• -11 bytes: And then turned it all inside out...

f takes an integer and returns a list: [3] for square, [2] for triangle, and [2,3] for both (or [] for neither.)

f n=[k|k<-[2,3],0<-scanl(-)n[1,k..n]]

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This uses that triangular numbers are sums 1+2+3+4+...+i while square numbers are sums 1+3+5+7+...+i, thus by starting with n and using scanl(-) to subtract either the range [1,2..n] or the range [1,3..n] consecutively, each of those cases will eventually hit 0.

Answer 5

PHP, 67 bytes

Returns 2 for square, 1 for triangle, 3 for both and 0 for none

<?=(($q=sqrt($i=$argn))==($q^0))*2+(2*$i==($x=sqrt(2*$i)^0)*$x+$x);

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Answer 6

Ohm, 15 7 bytes

D8*≥«Æ²

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EDIT: Saved 8 bytes thanks to this Jelly answer.

Outputs [true, true] if square and triangle, [true, false] for square, and [false, true] for triangle.

Explanation:

D8*≥«Æ²  Main wire, input integer a

D        Duplicate input
 8*≥     Multiply duplicate by 8, increment
    «    Pair
     Ʋ  Check if perfect square

Answer 7

TI-BASIC, 30 23 bytes

-7 for more flexible output requirements.

:Prompt X         //Get input
:{√(X),.5+√(2X+.5 //Compute the square root of X, and the inverse of the triangular number function: X(X-1)/2
:Ans=int(Ans      //Create a list of booleans for whether the values are whole or not, then print

Takes input as an iteger.

This returns a list of integers, either 0 or 1, first for being square, then for being triangular.

Answer 8

Pyth - 16 bytes (possibly 13)

+*2sI@Q2sI@h*8Q2

Prints 2 if square, 1 if triangle, 3 if both, 0 if neither

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Also, 19 18 17

sI@Q2sI@h*8Q2

The first line is 1 if the number is square, 0 if not. The 2nd line is 1 if number is triangle, 0 if not.

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Answer 9

JavaScript (ES7), 52 bytes

Returns 1 for square, 2 for triangle, 3 for both (and 0 for none, although irrelevant).

f=(n,t=0,s=(n**.5|0)**2==n)=>t<n?f(n-++t,t,s):2*!n|s

Demo

f=(n,t=0,s=(n**.5|0)**2==n)=>t<n?f(n-++t,t,s):2*!n|s

console.log(f(7))
console.log(f(15))
console.log(f(25))
console.log(f(36))

Answer 10

Python 2, 39 bytes

lambda n:[`x**.5%1==0`for x in 8*n+1,n]

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Outputs like ["True", "False"] for triangle then square. Probably could be shorter if allowed output format is clarified.

Answer 11

Batch, 120 bytes

@set/ar=i=j=k=0
:g
@if %j%==%1 set/ar+=1
@if %k%==%1 set/ar+=2
@set/aj+=i+=1,k=i*i
@if %j% leq %1 goto g
@echo %r%

Outputs 1 for triangular numbers, 2 for squares, 3 for both, 0 for neither.

Answer 12

Python 3, 49 42 bytes

lambda i:(i**.5%1==0)+2*((1+8*i)**.5%1==0)

Returns:

• 1 for Triangle
• 2 for Square
• 3 for Both

Answer 13

C#, 61 bytes

s=n=>System.Math.Sqrt(n)%1==0;n=>(s(n)?1:0)+(s((8*n)+1)?2:0);

Could be a lot shorter if there was (or I knew) an alternative to System.Math.Sqrt.

Formatted version with test cases:

Func<int, bool> s = n => System.Math.Sqrt(n) % 1 == 0;

Func<int, int> f = n => (s(n) ? 1 : 0) + (s((8 * n) + 1) ? 2 : 0);

Console.WriteLine(f(36));
Console.WriteLine(f(15));
Console.WriteLine(f(4));
Console.WriteLine(f(2));

Answer 14

Retina, 25 bytes

*M`(^1|1\1)+$
(^1|11\1)+$

Takes input in unary, outputs t and s on separate lines, where t and s are 0 or 1 indicating whether the shape is a triangle and/or square, respectively.

Try it online! (Contains an additional \ to put both outputs on the same line to make the test cases more easily distinguishable.)

Explanation

Triangular numbers are the sums of consecutive integers starting from one. Square numbers are the ums of consecutive odd integers starting from one. Both of those are fairly easy to match with forward references.

*M`(^1|1\1)+$

This prints 0 or 1 depending on whether the given regex matches, but it doesn't actually modify the string (the * indicates a dry run). The regex either matches the initial one with ^1 or it matches exactly one more 1 than in the previous iteration with 1\1 (the \1 refers to what (^1|1\1) matched last time).

(^1|11\1)+$

This is basically the same, except that we don't need a dry run and M is implicit. To go up by odd integers instead of all integers, we simply add two 1s on each iteration with 11\1.

Answer 15

MATL, 11 10 bytes

t:UmGt:Ysm

Outputs two values 0 or 1 separated by newline. The first indicates if the input is square, and the second indicates if it is triangular.

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Explanation

Consider input 4 as an example.

t     % Implicit input. Duplicate
      % STACK: 4, 4
:     % Range
      % STACK: 4, [1 2 3 4]
U     % Square, elementwise. This gives square numbers
      % STACK: 4, [1 4 9 16]
m     % Ismember
      % STACK: 1
Gt    % Push input again. Duplicate
      % STACK: 1, 4, 4
:     % Range
      % STACK: 1, 4, [1 2 3 4]
Ys    % Cumulative sum. This gives triangular numbers
      % STACK: 1, 4, [1 3 6 10]
m     % Ismember. Implicitly display
      % STACK 1, 0

Answer 16

Python 3, 72 67 bytes

x=input()
k=int((x*2)**.5)
print((x==int(x**.5)**2)+2*(2*x==k*k+k))

A bit longer than wanted. This is a full program that takes input from STDIN and outputs to STDOUT. STDERR should be empty as long as there isn't an I/O Exception and the input isn't invalid.

Answer 17

Octave, 38 bytes

@(x)~mod(x^.5,1)+2*any(x==cumsum(1:x))

Outputs 1 for square, 2 for triangular, 3 for both, 0 for neither.

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