# 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"


var QUESTION_ID=118960,OVERRIDE_USER=20260;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/118980/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

• @ScottMilner In the post I specified you could use your own output method.
– Neil
May 4 '17 at 21:45
• The output needs to be ["Square"], as in a singleton list of the string Square? Or the string ["Square"]? Even just the string Square seems needlessly specific.
– xnor
May 4 '17 at 21:45
• Possibly a dupe of Polygonal Numbers!, which asks to list all k for a which a number is k-gonal. This restricts to k=3 and k=4.
– xnor
May 4 '17 at 21:56
• Is it acceptable to output a 1 or 0 to indicate whether it's a square, and another 1 or 0 to indicate if it's a triangle? May 4 '17 at 22:38
• Is [True, False] an OK output format (first one for triangle, second for square)?
– xnor
May 4 '17 at 23:10

# Jelly, 9 7 bytes

×8‘,µÆ²


Try it online!

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


# Japt, 18 15 11 bytes

[U8*UÄ]®¬v1


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• You can save a few bytes with shortcuts: [U8*UÄ]®¬u1 ¥0 Also, u1 ¥0 can be reduced to v1 to save a couple bytes :-) May 4 '17 at 23:18

# 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)


## 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>

• Nice one! I think you can remove both pairs of parentheses around exponentiations (that is, n=>(n**.5%1?0:1)+((8*n+1)**.5%1?0:2). You can save another couple bytes with !(n**.5%1) May 4 '17 at 23:20
• In that case, 2*!((8*n+1)**.5%1) might save a byte too.
– Neil
May 4 '17 at 23:52

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.

• Nice idea with the scan. I think it would help to write the two options as [1,k..n] for k<-[2,3].
– xnor
May 4 '17 at 23:30
• One weird possibility is f n=[k|k<-[2,3],0<-scanl(-)n[1,k..n]].
– xnor
May 4 '17 at 23:40
• @xnor That's starting to get pretty different, sure you don't want to post it yourself? May 4 '17 at 23:46
• @xnor Someone should do it, so I'm taking silence for a no ;) May 5 '17 at 12:01

# PHP, 67 bytes

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

(^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.

# 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


# 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.

• Why len(input)? May 4 '17 at 22:24
• You can change this to Python 2 and remove the len (which I think should be int). May 4 '17 at 22:40
• @Svetlana The specs were changed after I answered. Thanks for notifying me. May 4 '17 at 23:38
• @pycoder ^ Thanks as well May 4 '17 at 23:38
• Why is this getting downvotes? May 5 '17 at 1:09

# 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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