## Math in manhattan

12

2

I define the following operators:

Manhattan Addition a +M b, for single-digit numbers, is the result of concatenating b onto a. So, a +M b = 10a + b. Therefore, the general operator +M is defined as thus:

a +M b = 10a + b

Manhattan Subtraction a –M b, for single-digit numbers, is the result of removing the last b from a. Therefore, the operator –M is defined as thus in pseudocode:

a –M b = a remove last b

Manhattan Multiplication a ×M b is the result of replacing all instances of b in a with b instances of b. Ergo, ×M is defined in pseudocode as:

a ×M b = a -> s/b/<b copies of b>/g

Manhattan Division a ÷M b is defined in terms of ×M:

1 ÷M b = the first character of b
a ÷M b = a ×M (1 ÷M b)

With all this in mind, create an interpreter that will evaluate infix expressions that use the following operators (i.e., a + b, not a b + or + a b)

+    Addition
-    Subtraction
/    Division
*    Multiplication
*M   Manhattan Multiplication
/M   Manhattan Division
-M   Manhattan Subtraction


Each Manhattan operator has a higher order precedence than their normal counterpart.

Test cases:

> 5 +M 10 + 3
63      // 5*10 + 10 + 3 => 60 + 3
> 10 *M 2
10      // no 2s in 10
> 10 *M 1
10      // one 1 in 10 replaced once
> 23 *M 3
2333    // 23 has one 3, which is replaced with three 3s
> 23 *M 2
223     // 23 has one 2, which is replaced with two 2s
> 232 *M 2
22322   // 232 has two 2s, which are replaced with two 2s
> 232 *M 23
23...(23 times)...232   // ...
> 123 *M 2 * 3
3669    // 1223 * 3 => 3669
> 5 + 3 +M 2
37      // 5 + (3 +M 2) => 5 + 32 => 37
> 150 /M 3
150     // 150 ÷M 3 => 150 ×M 3 => 150
> 150 /M 53
1555550 // 150 ÷M 53 => 150 ×M 5 => 1555550
> 50 -M 0
5
> 500 -M 0
50
> 5234 -M 5
234
> 12 +M 633 *M 3
6333453 // = 12 +M 6333333 = 120 + 6333333 = 6333453


This is a , so the shortest program in bytes wins.

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

# Language Name, N bytes


where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes


If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes


You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes


var QUESTION_ID=59299,OVERRIDE_USER=8478;function answersUrl(e){return"http://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"http://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>

13Why are you using the Unicode symbols × and ÷ instead of ASCII * and /? – ASCIIThenANSI – 2015-10-01T19:47:10.940

1Why does 232 ×M 23 equal 23232? Shouldn't it equal 23 copies of 23 followed by a 2? – senshin – 2015-10-01T21:25:49.207

@senshin Sorry, yes. right. – Conor O'Brien – 2015-10-01T21:33:01.303

1@ASCIIThenANSI I can see why you asked that. The choice is arbitrary. Unless if there is some pressing issue with my choice, I don't think I'll change it. – Conor O'Brien – 2015-10-01T21:33:34.187

4It makes it arbitrarily harder for languages without good Unicode support to participate, which is not very fun, if the challenge isn't about Unicode. – Lynn – 2015-10-01T21:48:04.403

@Mauris Ah. kay then. – Conor O'Brien – 2015-10-01T21:49:23.203

So $a -M b = a$ if $a ≠ b$, $0$ otherwise? Or are $a +M b$ and $a -M b$ defined for arbitrary $a$ and single-digit $b$? Also, should we suppose base 10? – Édouard – 2015-10-08T04:23:59.337

Your definition of subtraction is not related to single digit numbers. Maybe b is single digit but not a – edc65 – 2015-10-08T06:07:59.580

2This question has not received enough attention because is not well specified. You define addition for sngle digit numbers, then your first example have 2 digits numbers. I give up... – edc65 – 2015-10-08T06:25:29.947

@edc65 I clearly defined (fine, specified) what each operator should do. Feel free to give up. – Conor O'Brien – 2015-10-08T11:00:24.197

Do * and / have higher precedence than + and -? What about *M and /M vs +M and -M? – lirtosiast – 2015-10-08T12:54:30.550

How much flexibility do we need in our handling of whitespace? – lirtosiast – 2015-10-08T12:55:46.643

@ThomasKwa I didn't specify it, so whatever works best for your program is fine by me. – Conor O'Brien – 2015-10-08T13:02:30.210

1@edc65 He clearly states a +M b = 10a + b. This is in general, not just for single-digit numbers. The concatenation method was the one that was only for single digit numbers. – mbomb007 – 2015-10-08T13:56:56.243

1Can we assume that -M will always be called with an a that contains at least one b? – lirtosiast – 2015-10-08T14:09:44.287

@ThomasKwa Yes. – Conor O'Brien – 2015-10-08T14:42:06.910

You need some better test cases that test operator precedence, and also a test more than one or two operations at once. – mbomb007 – 2015-10-08T15:24:19.937

@mbomb007 I should, and will do so when the time becomes available. – Conor O'Brien – 2015-10-08T15:27:24.963

Which comes first in order of operations? *M or /M? Same question with +M and -M. They aren't inverses of each other, so I think they need to have a fixed order. Or is it just going to be left-to-right? – mbomb007 – 2015-10-08T15:57:41.033

The same as normal: As * and / are evaluated first, from left-to-right, so are *M and /M. – Conor O'Brien – 2015-10-08T17:05:09.557

@CᴏɴᴏʀO'Bʀɪᴇɴ You’ve left several questions about your specs unanswered. First of all, your operations are not defined on numbers, but on their representations. I suppose that you want us to use base 10 representations at all time, but it’s not in the specs. If I get it right, 2 -M 1 = 2 if we suppose base 10, but 1 if we suppose base 2. – Édouard – 2015-10-09T16:10:39.017

@CᴏɴᴏʀO'Bʀɪᴇɴ Second, you define +M and -M for single digits, then you say “in general”. Does that mean “for any integer”? Because, then, I’m not sure whether 123 -M 12 = 3 and I have no idea what 123 -M 13 is. – Édouard – 2015-10-09T16:16:56.677

@Édouard Yes, in general applies to all integers. 123 -M 12 = 3 and 123 -M 13 = 123. If there is no instance of b in a, then a remains unchanged. – Conor O'Brien – 2015-10-09T16:18:50.570

You should add 4342343 -M 3443423 as a test case. – Adám – 2016-06-30T06:53:46.363

5

# Dyalog APL, 104817993 75 bytes

Edit: Now handles 4342343 -M 3443423 correctly.

M←{⍎(5|⌊⍺⍺2)⊃'⍺×M⍣(⍺≠1)⍎⊃b'(b⎕R(⍵⍴'&')⊢a)'10⊥⍺⍵'(('(.*)',b←⍕⍵)⎕R'\1'⊢a←⍕⍺)}


### Background

This extends APL to include the Manhattan operator. An operator in APL terminology is a modifier of functions (e.g. ÷). An example of an operator is ⍨ which modifies functions to swap their arguments so 3 = 2 ÷⍨ 6. So too, M modifies the basic arithmetic functions to be their Manhattan relatives. Note that since the resulting language is an extension of APL, APL's strict right-to-left precedence remains.

### Explanation

The overarching structure is M←{⍎(5|⌊⍺⍺2)⊃} which applies the function (+ or - or × or ÷) to 2 and uses the result to chose which string to evaluate. The strings are:

3 for -M: (('(.*)',b←⍕⍵)⎕R'\1'⊢a←⍕⍺)
regex remove the last occurrence of b (string rep. of right arg.) in a (string rep. of left arg.)

2 for +M: '10⊥⍺⍵'
evaluate the arguments as base-10 digits

1 for ×M: (b⎕R(⍵⍴'&')⊢a)
replace occurrences of b with b ampersands (i.e. regex for the

0 for ÷M: '⍺×M⍣(⍺≠1)⍎⊃b'
⍎⊃b first digit of b
⍺×M⍣(⍺≠1) apply ⍺ ×M if ⍺ ≠ 1

⊃ out of of the above four strings, pick number:

(5|⌊⍺⍺2) mod-5 of the floor of the function applied to 2, namely:
3 = 5 | ⌊-2
2 = 5 | ⌊+2
1 = 5 | ⌊×2 because ×2 ⇔ sgn(2) ⇔ 1
0 = 5 | ⌊÷2 because ÷2 ⇔ 1 ÷ 2 ⇔ 0.5

### Lots of thanks to my dear friend ngn for amazing shavings.

1This is fine. It fits what I had desired. – Conor O'Brien – 2015-10-12T21:03:53.883

Great, I'll edit the post then. – Adám – 2015-10-12T21:15:53.280

@CᴏɴᴏʀO'Bʀɪᴇɴ I might have lost out on the bonus, but it sure is the shortest now. – Adám – 2016-06-29T09:47:00.070

@CᴏɴᴏʀO'Bʀɪᴇɴ Forgot? I just edited today, making it shorter than the accepted one. – Adám – 2016-06-29T20:02:20.847

Oops again. I thought you were reminding me to accept – Conor O'Brien – 2016-06-29T20:09:45.053

12

# Perl, 10099 98 bytes

97 bytes code + 1 byte command line

s/ |.*\K(\d)(\d*)-M\1|\+M/\2/g+s/(\d+)\*M(.)/$1=~s@$2@$&x$&@erg/e+s#/(M.)\d+#*\1#&&redo,\$\=eval}{


Usage example:

echo "123 *M 2 * 3 + 150 /M 53" | perl -p entry.pl


If it makes your code shorter, you only have to use *M for xM and /M for <div>M. – Conor O'Brien – 2015-10-01T21:53:22.133

Congrats on the bounty! – Conor O'Brien – 2015-10-13T15:51:28.267

7

# Python, 644 bytes

import operator as o,re
x,q,t,r,w='*/+-M';mm,md,ma,ms='*M /M +M -M'.split()
n=lambda x:x
a=lambda a,b:str(10*int(a)+int(b))
v=lambda a,b:a[::-1].replace(b,'',1)[::-1]
m=lambda a,b:a.replace(b,b*int(b))
d=lambda a,b:m(a,b[0])if a>0 else b[0]
def u(s):z=r'\d+ (?:\{0}{2}|\{1}{2}) \d+';return re.sub(z.format(t,r,''),p,re.sub(z.format(t,r,w),p,re.sub(z.format(x,q,''),p,re.sub(z.format(x,q,w),p,re.sub(r'$$(.*)$$',u,s)))))

I don't see a c or f. – RK. – 2016-06-29T16:17:31.150