60
\$\begingroup\$

I love math. But I can't find a single calculator that can multiply correctly. They seem to get everything right except 6*9 (It's the question to life, the universe, and everything! How could they get that wrong?!). So I want you all to write a function for me that can multiply 2 numbers correctly (and 6*9 equals 42 instead of 54. 9*6 equals 54 still).

Oh, and I'm going to have to build the source in Minecraft so... fewest bytes win!

Recap

  • Take 2 numbers as input (type doesn't matter, but only 2 items will be passed, and order must be consistent. So streams, and arrays are ok as long as they preserve the order they where passed in. I.e., a map won't work because it doesn't preserve the order)
  • Output multiple of both numbers except if they are 6 and 9, then output 42 (order matters!)
    • PS. I never was really good with counting, so I think only integers from 0 to 99 are real numbers (type used doesn't matter)
  • Fewest bytes per language wins!

Leaderboard:

var QUESTION_ID=124242,OVERRIDE_USER=61474;function answersUrl(e){return"https://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"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:+r.match(SCORE_REG)[0],language:r.match(LANG_REG)[0].replace(/<\/?[^>]*>/g,"").trim(),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=/\d+((?=$)|(?= Bytes))/i,OVERRIDE_REG=/^Override\s*header:\s*/i;LANG_REG=/^[^,(\n\r]+/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/Sites/codegolf/all.css?v=617d0685f6f3"> <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>

\$\endgroup\$
  • 1
    \$\begingroup\$ @JonathanAllan Yes, added rules for that. \$\endgroup\$ – Tezra Jun 3 '17 at 4:09
  • 73
    \$\begingroup\$ Coincidentally the question id has 42 twice in it. \$\endgroup\$ – Erik the Outgolfer Jun 3 '17 at 6:00
  • 1
    \$\begingroup\$ if only integers up to 99 are real numbers, then e.g. 9 * 11 is a valid prompt but 10 * 10 is not, right? \$\endgroup\$ – Please stop being evil Jun 3 '17 at 8:23
  • 14
    \$\begingroup\$ @EriktheOutgolfer …and the 12 stands for what to subtract if input is 6 and 9. \$\endgroup\$ – Adám Jun 3 '17 at 23:43
  • 11
    \$\begingroup\$ @EriktheOutgolfer Even if you read the ID backwards. \$\endgroup\$ – T. Sar - Reinstate Monica Jun 5 '17 at 12:57

80 Answers 80

65
+100
\$\begingroup\$

Mathematica, 15 bytes

Byte count assumes Windows ANSI encoding (CP-1252).

6±9=42
±n__:=1n

Defines a binary operator ± which solves the problem. We simply define 6±9=42 as a special case which takes precedence and then add a fallback definition which makes ± equal to multiplication. The latter uses a fairly interesting golfing trick. The reason this works is actually quite elaborate and we need to look into sequences. A sequence is similar to what's known as a splat in other languages. It's basically a "list" without any wrapper around it. E.g. f[1, Sequence[2, 3, 4], 5] is really just f[1, 2, 3, 4, 5]. The other important concept is that all operators are just syntactic sugar. In particular, ± can be used as a unary or binary operator and represents the head PlusMinus. So ±x is PlusMinus[x] and a±b is PlusMinus[a,b].

Now we have the definition ±n__. This is shorthand for defining PlusMinus[n__]. But n__ represents an arbitrary sequence of arguments. So this actually adds a definition for binary (and n-ary) usagess of PlusMinus as well. The value of this definition is 1n. How does this multiply the arguments? Well, 1n uses Mathematica's implicit multiplication by juxtaposition so it's equivalent to 1*n. But * is also just shorthand for Times[1,n]. Now, n is sequence of arguments. So if we invoke a±b then this will actually become Times[1,a,b]. And that's just a*b.

I think it's quite neat how this syntax abuse lets us define a binary operator using unary syntax. We could now even do PlusMinus[2,3,4] to compute 24 (which can also be written as ±##&[2,3,4] or 2±Sequence[3,4] but it's just getting crazy at that point).

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  • \$\begingroup\$ I particularly like how this is basically a patch for my 'calculator' :3 For fun, does this work for * too? :3 \$\endgroup\$ – Tezra Jun 3 '17 at 4:14
  • 3
    \$\begingroup\$ @Tezra well, * is a built-in operator, so you'd have to Unprotect it to add further definitions, but Unprotect[Times];6*9=42 should work (can't test right now though). \$\endgroup\$ – Martin Ender Jun 3 '17 at 5:24
  • 1
    \$\begingroup\$ Doing this to the * operator is just so evil.... I love it! >:3 \$\endgroup\$ – Tezra Jun 3 '17 at 15:51
  • 1
    \$\begingroup\$ How can I upvote when there are exactly 42 others?!? Here's my "future +1" to be awarded after someone else breaks it! :-) \$\endgroup\$ – The Vee Jun 5 '17 at 15:11
  • 1
    \$\begingroup\$ @MartinEnder Aww; But it's the shortest Mathematica one, and my favorite so far. :3 \$\endgroup\$ – Tezra Jun 5 '17 at 15:17
58
\$\begingroup\$

Haskell, 14 bytes

6&9=42
a&b=a*b

Try it online!

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25
\$\begingroup\$

C, 32 31 29 28 bytes

-2 thanks to Digital Trauma
-1 thanks to musicman523

#define f(a,b)a^6|b^9?a*b:42

Pretty simple. Declares a macro function f that takes two arguments, a and b.
If a is 6 and b is 9, return 42. Otherwise return a x b.

Try it online!

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  • 2
    \$\begingroup\$ Use ^ instead of == and adjust the logic a bit: #define f(a,b)a^6||b^9?a*b:42 - saves 2 bytes. \$\endgroup\$ – Digital Trauma Jun 2 '17 at 21:51
  • \$\begingroup\$ @DigitalTrauma Cheers :D \$\endgroup\$ – MD XF Jun 2 '17 at 21:54
  • 1
    \$\begingroup\$ I think you can use | instead of || to save another byte, since it still has lower precedence than ^ \$\endgroup\$ – musicman523 Jun 2 '17 at 22:07
  • \$\begingroup\$ @musicman523 Thanks! Editing. \$\endgroup\$ – MD XF Jun 2 '17 at 22:09
  • 1
    \$\begingroup\$ You should update your shortC version to take these changes as well \$\endgroup\$ – musicman523 Jun 2 '17 at 22:21
17
\$\begingroup\$

JavaScript (ES6), 20 bytes

x=>y=>x-6|y-9?x*y:42

Explanation:

Iff x==6 and y==9, x-6|y-9 will be 0 (falsy), and 42 will be the result.

Snippet:

f=

x=>y=>x-6|y-9?x*y:42

console.log(f(6)(9));
console.log(f(9)(6));

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  • 4
    \$\begingroup\$ Very nicely done; wish I'd thought of it. +1 \$\endgroup\$ – Shaggy Jun 2 '17 at 21:55
14
\$\begingroup\$

Python 2, 30 29 bytes

Thanks to Jonathan Allan for saving a byte!

lambda x,y:x*[y,7][6==x==y-3]

Try it online!

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  • 2
    \$\begingroup\$ Save a byte using the fact that 6*7 is 42: lambda x,y:x*[y,7][6==x==y-3] \$\endgroup\$ – Jonathan Allan Jun 2 '17 at 22:18
  • \$\begingroup\$ @JonathanAllan Ohh, that's neat! Thanks! :) \$\endgroup\$ – Adnan Jun 3 '17 at 8:13
  • \$\begingroup\$ This solution also works in Python 3 \$\endgroup\$ – AMK Aug 22 '17 at 20:18
  • \$\begingroup\$ Exactly what I got! Can't seem to find any way to golf it further. \$\endgroup\$ – FlipTack Oct 24 '17 at 11:17
12
\$\begingroup\$

05AB1E, 15 11 9 bytes

-4 bytes thanks to @Emigna

-2 bytes thanks to @Adnan

P¹69SQi42

Try it online!

How it works

P          # multiply input
 ¹         # push first number
  69       # the number 69
    S      # split per character
     Q     # equality for both inputs
       i42 # if so, print 42
           # otherwise print product
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  • \$\begingroup\$ You could save 4 bytes with ‚D96SQi42ëP \$\endgroup\$ – Emigna Jun 2 '17 at 21:23
  • \$\begingroup\$ @Emigna huh... Why the , at the beginning? \$\endgroup\$ – Neil A. Jun 2 '17 at 21:26
  • \$\begingroup\$ Pairing the 2 inputs to compare only once as a list. \$\endgroup\$ – Emigna Jun 2 '17 at 21:45
  • \$\begingroup\$ I suppose 6Qs9Q* would have worked as well for the same byte count. \$\endgroup\$ – Emigna Jun 2 '17 at 21:46
  • \$\begingroup\$ Changing the input format saves 2 bytes: P¹69SQi42 \$\endgroup\$ – Adnan Jun 3 '17 at 8:33
10
\$\begingroup\$

Java (OpenJDK 8), 24 22 bytes

-2 bytes thanks to @OlivierGrégoire

a->b->a==6&b==9?42:a*b

Try it online!

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  • 3
    \$\begingroup\$ Welcome to PPCG! I don't know much about Java, but could you remove the System.out.println() call and just let the function return the result? \$\endgroup\$ – ETHproductions Jun 2 '17 at 23:04
  • 2
    \$\begingroup\$ @LưuVĩnhPhúc not in Java, because I'd have to write (a^6|b^9)==0 since there is no implicit "different than 0" comparison. The resulting code snippet would be 27 bytes long. Anyways, thanks for the suggestion, and please tell me if I got your tip wrong. \$\endgroup\$ – Bashful Beluga Jun 4 '17 at 2:59
  • 1
    \$\begingroup\$ @Riker nope, it does not work like that in java. E.g. the snippet int a = 5; if (a) do_some_stuff(); else do_other_stuff(); gives a Type mismatch: cannot convert from int to boolean compilation error. They must be made explicitly with boolean values; refer to SO and ORACLE. \$\endgroup\$ – Bashful Beluga Jun 4 '17 at 4:10
  • 3
    \$\begingroup\$ You can use currying to spare one byte, and you can get rid of the semicolon as it's not part of the lambda to spare another byte: a->b->a==6&b==9?42:a*b. \$\endgroup\$ – Olivier Grégoire Jun 4 '17 at 16:14
  • 1
    \$\begingroup\$ Just a note why the 0 is not false. Java is type safe so 0 is an integer not a boolean and unsafe typecasting is not allowed so you can't use falsy values \$\endgroup\$ – Martin Barker Jun 6 '17 at 14:52
6
\$\begingroup\$

Ruby, 24 bytes

->a,b{a==6&&b==9?42:a*b}
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  • \$\begingroup\$ a^6|b^9<1 might work as boolean. Hard to test on my smartphone. \$\endgroup\$ – Eric Duminil Jun 2 '17 at 22:53
  • 1
    \$\begingroup\$ @EricDuminil Unfortunately that expression is parsed as (((a^6)|b)^9), i.e. a.^(6).|(b).^(9), so it won't work correctly. a-6|b-9==0 would work, but that's no shorter. \$\endgroup\$ – Ventero Jun 3 '17 at 1:05
  • \$\begingroup\$ @Ventero: I didn't think about that. Thanks. a,b==6,9 would be nice, but it also doesn't work. \$\endgroup\$ – Eric Duminil Jun 3 '17 at 8:38
6
\$\begingroup\$

Brain-Flak, 158 154 148 140 138 126 bytes

(({}<>)(((([()()()]<>)){})<({}{}({}))>{(()()()){}(<{}>)}{}))([()]{()(<{}>)}{})(({<{}>{}((<>))}{}){}<{}>{<({}[()])><>({})<>}{})

Try it online!

Explanation

This code is pretty simple. We make copies of the top two items on the stack, we subtract 6 from one and 9 from the other. We then take the not of the two values. We and those values, multiply the result by 12. Multiply the inputs and subtract the two results.

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  • 2
    \$\begingroup\$ You might want to capitalize not and and (or backtick them), reading your description rather tripped me up. \$\endgroup\$ – MD XF Jun 2 '17 at 21:39
6
\$\begingroup\$

Factorio, 661 bytes, 6 combinators with 9 connections

There is one constant combinator set to output A and B. Change these to set the input.

Blueprint string (0.15.18):

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

The output is signal Z and is to be taken from the top and bottom deciders.

Screenshot

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  • \$\begingroup\$ waaa... first time i ever see a factorio answer in ppcg xD \$\endgroup\$ – V. Courtois Sep 12 '17 at 21:52
  • 2
    \$\begingroup\$ Not golfy enough. \$\endgroup\$ – jimmy23013 Apr 1 '18 at 14:04
6
\$\begingroup\$

Jelly, 8 7 bytes

Vf96SạP

Input is as an array of two integers: first the right operand, then the left one.

Try it online!

How it works

Vf96SạP  Main link. Argument: [b, a]

V        Cast [b, a] to string, then eval the resulting string.
         For [b, a] = [9, 6], this yields 96.
 f96     Filter with 96, yielding [96] if V returned 96, [] otherwise.
    S    Take the sum, yielding either 96 or 0.
      P  Compute the product of [b, a], yielding ba = ab.
     ạ   Compute the absolute difference of the results to both sides.
         When the sum is 0, this simply yields the product.
         However, when [b, a] = [9, 6], this yields 96 - 54 = 42.
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  • \$\begingroup\$ This has like -1 degree of freedom. How do these coincidences even occur to you? \$\endgroup\$ – lirtosiast Nov 29 '18 at 8:46
5
\$\begingroup\$

Factorio, 581 bytes, 3 combinators with 4 connections

Blueprint string (0.16.36):

0eNqllNtu4jAQht9lLldmldNCFWkvto/RCkUhGWAkYkfOGDVCefeOnV1Km7ACemPJ9vibf04+webgsLWkGfITUGV0B/nrCTra6fLgz7hvEXIgxgYU6LLxO2/HpeZFZZoN6ZKNhUEB6RrfII+HtQLUTEw44sKmL7RrNmjF4AyqsaIa7SVHQWs6eWq0dy+46OcvBT3ki1hc1GSxGi8T5XWwNYdig/vySPJYXvxFFnJXB0znT7dkOy4mYR3JspOTs6DRYoFHtD3vSe98XP/CFZ9xtsqe0mW29KdNW9qgOYffgjCOW3eHk+eR3fai1WkuttY0BWlhQM7W4TC61mPAIYzYLxbry6yS7FKxJFs54rANFdhZRP3VMBnWQk08ZvZ+ChpExqSCyX9bYVLCRfxRwbmabenAaK+03rX0/RnT5z7VJbroQnUH7HkGlq7OsDFtc8WYzWJ8WxbTs4rSEu8bZKpuGoXopkn4gH5vGEKiO/SMO5vbtCgDEjTCjwcm5AWGO4ZgknX16Tq7OhRfHiZXypU91PTRd6ZYdIjo8PnmF3+1AvmfuuBq+bRKYmnWKM2G4R1hAPnz

The bottom left constant combinator should be set to output A and B as input. The output is signal Z from the bottom right arithmetic combinator.

enter image description here

Top left: 2147483640 A, 2147483637 B
Top right: If everything = 2147483646 output B, input count
Bottom left: (input) A, (input) B
Bottom right: A * B -> Z
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5
\$\begingroup\$

MATL, 11 bytes

[BE]=?42}Gp

Input is an array with the two numbers.

Try it online!

Explanation

[BE]    % Push array [6, 9]
=       % Implicit input: array of two numbers. Compare with [6, 9] element-wise
?       % If the two entries are true
  42    %   Push 42
}       % Else
  G     %   Push input
  p     %   Product of array
        % Implicit end. Implicit display
\$\endgroup\$
4
\$\begingroup\$

GW-BASIC, 55 bytes

1INPUT A:INPUT B
2IF A=6THEN IF B=9THEN ?"42":END
3?A*B

Output:

output

The first machine at pcjs has IBM BASIC, which is practically the same thing. To test this, head over there, hit Run on the machine, Press Enter-Enter and type BASICA to get into BASIC mode. Then enter the source code (it will automatically prettyprint for you), type RUN, input two integers, and done!

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  • 3
    \$\begingroup\$ Are you sure the bytecount is correct? GW-BASIC uses an encoding in which some words, like INPUT, are encoded in fewer bytes than the characters that make them up would suggest. The count therefore seems high to me. \$\endgroup\$ – user62131 Jun 2 '17 at 21:33
  • \$\begingroup\$ @ais523 Well, I ran it through wc, and I got 55... Copy-pasted into my emulator and it had the expected behavior. \$\endgroup\$ – MD XF Jun 2 '17 at 21:34
  • 3
    \$\begingroup\$ Right, my point is that you're probably scoring your submission higher than it needs to be. Get GW-BASIC to save the file, and then look at the size of the resulting file on disk; it should be smaller. \$\endgroup\$ – user62131 Jun 2 '17 at 21:35
  • \$\begingroup\$ @ais523 Saved as OUT.BAS: i.stack.imgur.com/32eH1.png Bytecount is the middle value. \$\endgroup\$ – MD XF Jun 2 '17 at 21:38
  • \$\begingroup\$ OK, I wasn't expecting that, but I guess it's a wash in this situation. (Or perhaps there's more than one save format?) \$\endgroup\$ – user62131 Jun 2 '17 at 21:45
4
\$\begingroup\$

R, 33 bytes

function(a,b)`if`(a-6|b-9,a*b,42)

Returns a function.

Try it online!

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4
\$\begingroup\$

Check, 34 33 bytes

.:+&>#v
#>42#v#9-!\>6-!*?
 d* ##p

Check is my new esolang. It uses a combination of 2D and 1D semantics.

Input is two numbers passed through command line arguments.

Explanation

The stack starts with the command line arguments on it. Let's call the arguments a and b.

The first part, .:+&, essentially duplicates the stack, leaving it as a, b, a, b. > pushes 0 to the stack (it is part of a numeric literal completed by 9).

# switches to 2D semantics, and v redirects the IP downwards. The IP immediately runs into a #, which switches back to 1D semantics again.

9-! checks whether b is equal to 9 (by subtracting 9 and taking the logical NOT). \>6-! then checks whether a is equal to 6. The stack now contains a, b, 1, 1 if and only if b==9 and a==6. Multiplying with * takes the logical AND of these two values, giving a, b, 1 if the inputs were 6 and 9, and a, b, 0 otherwise.

After this, the IP runs into a ?. This will switch to 2D mode if the top stack value is nonzero, and otherwise will continue in 1D mode.

If the top stack value was 1, this means that the other stack values are 6 and 9, so we push 42 to the stack with >42 and then move down to the second # on the last line.

If the top stack value was 0, then execution moves down to the next line. d removes the 0 (as ? does not do so), and then we multiply the two inputs with *. The ## switches in and out of 2D mode, doing nothing.

The branches have now joined again. The stack either contains 6, 9, 1, 42, or a*b. p prints the top stack value and then the program ends, discarding the rest of the stack.

\$\endgroup\$
  • \$\begingroup\$ This looks like a nifty language! \$\endgroup\$ – Not a tree Jun 3 '17 at 13:12
3
\$\begingroup\$

JavaScript (ES6), 25 bytes

x=>y=>[x*y,42][x==6&y==9]
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3
\$\begingroup\$

Python 3, 36 33 bytes

lambda x,y:42if x==6==y-3else x*y

Try it online!

\$\endgroup\$
  • 1
    \$\begingroup\$ Alternate version, same size: lambda x:x[0]*x[1]if x!=(6,9)else 42. The only difference is the input type. \$\endgroup\$ – musicman523 Jun 2 '17 at 21:46
  • \$\begingroup\$ Nevermind - your edited version is shorter :) \$\endgroup\$ – musicman523 Jun 2 '17 at 21:58
  • \$\begingroup\$ The code you posted doesn't actually work, so I switched it to the working version on TIO that you linked to. \$\endgroup\$ – ETHproductions Jun 2 '17 at 22:49
  • \$\begingroup\$ It works for me: In [1]: f=lambda x,y:42if 6==x==y-3else x*y In [2]: f(6,9) Out[2]: 42 In [3]: f(9,6) Out[3]: 54 @ETHproductions \$\endgroup\$ – Martmists Jun 2 '17 at 22:56
  • \$\begingroup\$ @Martmists You were missing a space then, because your code was f=lambda x,y:42if6==x==y-3else x*y \$\endgroup\$ – ETHproductions Jun 2 '17 at 22:58
3
\$\begingroup\$

APL (Dyalog), 10 bytes

×-12×6 9≡,

Try it online!

× the product (of the arguments)

- minus

12× twelve times

6 9≡ whether (6,9) is identical to

, the concatenation (of the arguments)

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  • \$\begingroup\$ Oh wow I just saw this and my J answer is exactly the same as this :/ except one byte longer \$\endgroup\$ – Conor O'Brien Jun 5 '17 at 0:36
  • \$\begingroup\$ @ConorO'Brien Makes sense. J and tacit APL are mostly equivalent, save for J's multi-char primitives (and needing Cap for a final atop). \$\endgroup\$ – Adám Jun 5 '17 at 0:39
3
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R, 41 I think I don't know how to count bytes I am new :D

function(a,b){

if(a==6&b==9){42} else {a*b}

}

I define a funtion whose arguments are a and b in this order. If a equals to 6 and b equals to 9 it returns 42, otherwise, a times b

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  • \$\begingroup\$ Welcome to the site! :) \$\endgroup\$ – DJMcMayhem Jun 6 '17 at 16:07
  • \$\begingroup\$ Welcome. Take into account that newlines and spaces count too. In your case, if you remove newlines and spaces (which you can) it's only 41. \$\endgroup\$ – Masclins Jun 12 '17 at 14:27
  • 1
    \$\begingroup\$ You can cut two bytes by using ifelse(a==6&b==9,42,a*b) \$\endgroup\$ – Masclins Jun 12 '17 at 14:33
  • \$\begingroup\$ You can cut the whole thing down to 33 bytes as function(a,b)`if`(a-6|b-9,a*b,42). \$\endgroup\$ – rturnbull Jun 12 '17 at 14:45
  • \$\begingroup\$ This is only 41 bytes after you remove the unneeded whitespace, unil then, it's 47 bytes. \$\endgroup\$ – Pavel Jul 11 '17 at 1:24
3
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SPL, 356 bytes

a.Ajax,.Puck,.Act I:.Scene I:.[Enter Ajax and Puck]Ajax:Listen to your heart!Puck:Listen to your heart!Are you as big as the sum of a big big big cat and a cat?If so, am I as big as the sum of a big big cat and a big cat?If so, you are as big as the product of I and the sum of I and a cat.If not, you are as big as the product of you and I.Open your heart

With newlines and spaces:

a.                       *Title*
Ajax,.                   *Declare variable Ajax*
Puck,.                   *Declare variable Puck*
Act I:.
Scene I:.
[Enter Ajax and Puck]
Ajax: Listen to your heart!                  *Set Puck's value to user input*
Puck: Listen to your heart!                  *Set Ajax's value to user input*
      Are you as big as the sum of a big 
       big big cat and a cat?                *Is Ajax=9?* 
      If so, am I as big as the sum of a 
       big big cat and a big cat?            *Is Puck=6?* 
      If so, you are as big as the product 
       of I and the sum of I and a cat.      *If so, set Ajax=42* 
      If not, you are as big as the product 
       of you and I.                         *If not set Ajax=(Ajax)(Puck)*
      Open your heart                        *Print Ajax's value*
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3
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Japt, 13 11 12 bytes

¥6&V¥9?42:N×

Try it online

  • 2 1 bytes saved thanks to obarakon.
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  • 1
    \$\begingroup\$ N¬¥69?42:N× for 11 bytes \$\endgroup\$ – Oliver Jun 3 '17 at 1:54
  • \$\begingroup\$ Nice one, @obarakon. \$\endgroup\$ – Shaggy Jun 3 '17 at 12:44
3
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Standard ML (MLton), 22 20 bytes

Saved 2 bytes thanks to @Laikoni!

fun$6 $9=42| $x y=x*y

Try it online!

This is the kind of thing SML is meant for, which is why it beats shortC and Python.

The old version looked much nicer. :P

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  • \$\begingroup\$ 20 bytes: Try it online! \$\endgroup\$ – Laikoni Jun 23 '17 at 13:29
  • \$\begingroup\$ @Laikoni Wow, I had no idea you could use $ as an identifier! Why doesn't this compile if you remove the space between | and $? \$\endgroup\$ – musicman523 Jun 23 '17 at 14:14
  • \$\begingroup\$ SML distinguishes alphanumeric and symbolic identifiers, which can be quite handy for golfing. |$ is parsed as a single symbolic identifier, so everything breaks. I plan to write a tips question for SML soon and add an answer about those two types of identifiers. \$\endgroup\$ – Laikoni Jun 23 '17 at 15:21
2
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Pyth, 12 bytes

-*FQ*12q(6 9

Try it online

Explanation

 -*FQ*12q(6 9
  *FQ             Take the product
        q(6 9)Q   Check if the (implicit) input is (6, 9)
 -   *12          If so, subtract 12
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  • \$\begingroup\$ Clever solution. I tried it with ternary statements in 15: AQ?&q6Gq9G42*GH \$\endgroup\$ – Tornado547 Dec 14 '17 at 2:54
2
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Retina, 36 bytes

^6 9$
6 7
\d+
$*
1(?=1* (1+))|.
$1
1

Try it online! Standard unary multiplication, just alters the input to handle the special case.

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2
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Jelly, 10 bytes

⁼6,9ȧ42ȯ⁸P

A monadic link taking a list of the two numbers.

Try it online!

How?

⁼6,9ȧ42ȯ⁸P - Link: list of numbers [a,b]
 6,9       - 6 paired with 9, [6,9]
⁼          - equals? (non-vectorising) (1 or 0)
     42    - literal answer, 42
    ȧ      - logical and               (42 or 0)
        ⁸  - link's left argument, [a,b]
       ȯ   - logical or                (42 or [a,b])
         P - product                   (42 or a*b)
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  • \$\begingroup\$ You could've just used ?, like I did. ;) \$\endgroup\$ – Erik the Outgolfer Jun 3 '17 at 5:52
  • \$\begingroup\$ Ah because , is special in that it is part of the literal regex pattern, so 6,9 is parsed as a single token and the quick $ can combine it with . Did you reason that, or just try it and notice that it worked? \$\endgroup\$ – Jonathan Allan Jun 3 '17 at 19:53
  • 1
    \$\begingroup\$ I reasoned that. \$\endgroup\$ – Erik the Outgolfer Jun 4 '17 at 6:52
2
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Jelly, 9 bytes

42P⁼6,9$?

Try it online!

Takes list of numbers as input.

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2
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S.I.L.O.S, 81 67 bytes

readIO
J=i
readIO
a=(J-6)^2+(i-9)^2
a/a
a-1
a*12
x=J*i+a
printInt x

Try it online!

In some sense addition functions as an interesting NAND gate in SILOS.

-14 bytes thanks to @Leaky Nun

Essentially we create a number "a" which is 0 (falsy) iff j is 6 and i=9, then we divide it by itself subtract one and multiply it by 12 in order to add to our product.

If "a" was 1 after subtracting one and multiplying, it becomes a no-op, however in the case where a is 0, 0/0 silently throws an error (which is auto-magically caught) a becomes 0, and then becomes -1 and we end up subtracting 12 from our product.

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  • \$\begingroup\$ 69 bytes \$\endgroup\$ – Leaky Nun Jun 3 '17 at 3:02
  • \$\begingroup\$ 67 bytes \$\endgroup\$ – Leaky Nun Jun 3 '17 at 3:06
  • \$\begingroup\$ @LeakyNun ooh, that's clever. \$\endgroup\$ – Rohan Jhunjhunwala Jun 3 '17 at 11:09
  • \$\begingroup\$ Actually, 0/0 becomes 0. \$\endgroup\$ – Leaky Nun Jun 3 '17 at 12:43
  • \$\begingroup\$ @LeakyNun I meant to say becomes 0, and then decremented. Fixing. \$\endgroup\$ – Rohan Jhunjhunwala Jun 3 '17 at 13:14
2
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Convex, 16 14 13 bytes

_6 9¶=\:*42¶=

Try it online!

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2
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shortC, 23 bytes

Df(a,b)a==6&b==9?42:a*b

Try it online!

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