10
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Definition

The chain rule with two functions state that:

D[f(g(x))] = f'(g(x)) * g'(x)

Or, alternatively:

D[f1(f2(x))] = f1'(f2(x)) * f2'(x)

The chain rule with three functions state that:

D[f(g(h(x)))] = f'(g(h(x))) * g'(h(x)) * h'(x)

Or, alternatively:

D[f1(f2(f3(x)))] = f1'(f2(f3(x))) * f2'(f3(x)) * f3'(x)

Et cetera.

Task

  • Given an integer between 2 and 21, output the chain rule with that many functions, either in the first form or in the second form.
  • Please specify if you are using the second form.

Specs

  • The format of the string must be exactly that stated above, with:
    1. all the spaces kept intact
    2. a capitalized D
    3. a square bracket immediately following D
    4. the asterisk kept intact
  • One extra trailing space (U+0020) is allowed.
  • Leading zeros in the function names in the second form (e.g. f01 instead of f1) is allowed.

Testcases

If you use the first form:

input output
2     D[f(g(x))] = f'(g(x)) * g'(x)
3     D[f(g(h(x)))] = f'(g(h(x))) * g'(h(x)) * h'(x)

If you use the second form:

input output
2     D[f1(f2(x))] = f1'(f2(x)) * f2'(x)
3     D[f1(f2(f3(x)))] = f1'(f2(f3(x))) * f2'(f3(x)) * f3'(x)

Leaderboard

var QUESTION_ID=86652,OVERRIDE_USER=48934;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>

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  • \$\begingroup\$ Do function names have to be lowercase? \$\endgroup\$ – betseg Jul 26 '16 at 23:21
  • \$\begingroup\$ @betseg Yes of course. \$\endgroup\$ – Leaky Nun Jul 26 '16 at 23:22
6
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Python 2, 79 bytes

f=lambda n:0**n*"D[x] ="or f(n-1).replace("x","f%d(x)"%n)+1%n*" *"+" f%d'(x)"%n

Outputs with numbered functions.

Builds the output by repeatedly replacing each x with fn(x), then appending * fn'(x). The * is omitted for n==1.

Compare to the iterate program (92 bytes):

r="D[x] = ";n=0
exec'n+=1;r=r.replace("x","f%d(x)"%n)+"f%d\'(x) * "%n;'*input()
print r[:-3]

96 bytes:

n=input();r='';s='x'
while n:s='f%d(%%s)'%n%s;r=" * f%d'"%n+s[2:]+r;n-=1
print"D[%s] = "%s+r[3:]

Outputs with numbered functions.

Accumulates the nested function f1(f2(f3(x))) in s and the right-hand-side expression in r. The string formatting is clunky; f-strings from 3.6 would do better.

| improve this answer | |
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5
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Sesos, 49 bytes

0000000: 2ac992 63fb92 245fb6 6c57be 255bbe 2cc9bf 6d49da  *..c..$_.lW.%[.,..mI.
0000015: 025e7f fdced0 fd67f8 fcde33 b6a7b2 643d4f 65597e  .^.....g...3...d=OeY~
000002a: f77a72 dd73cf fe                                  .zr.s..

Try it online

Disassembled

set numin
add 68   ; 'D'
put
sub 7    ; '=' - 'D'
fwd 1
add 32   ; ' '
fwd 1
add 91   ; '['
put
add 2    ; ']' - '['
fwd 1
add 102  ; 'f'
fwd 1
add 40   ; '('
fwd 3
add 120  ; 'x'
rwd 2
get
jmp
    jmp
        fwd 1
        add 1
        rwd 3
        put
        add 1
        fwd 1
        put
        fwd 1
        sub 1
    jnz
    sub 1
    rwd 1
    add 1    ; ')' - '('
    fwd 3
    put
    rwd 1
    jmp
        rwd 3
        sub 1
        fwd 1
        put
        fwd 1
        add 1
        fwd 1
        sub 1
    jnz
    rwd 4
    put
    get
    add 41   ; ')'
    rwd 1
    put
    rwd 1
    put
    get
    add 42   ; '*'
    fwd 1
    put
    fwd 2
    put
    add 1
    fwd 1
    sub 2    ; '\'' - ')'
    put
    add 1    ; '(' - '\''
    put
    fwd 1
jnz
fwd 2
put
rwd 5
put
| improve this answer | |
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3
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JavaScript (ES6), 89 bytes

f=n=>--n?f(n).replace(/x/g,`${c=(n+15).toString(36)}(x)`)+` * ${c}'(x)`:`D[f(x)] = f'(x)`

Or 75 bytes using the second form:

f=n=>n>1?f(n-1).replace(/x/g,`f${n}(x)`)+` * f${n}'(x)`:`D[f1(x)] = f1'(x)`

Or 82/64 bytes if I'm allowed an extra 1 * term:

f=n=>n?f(n-1).replace(/x/g,`${c=(n+14).toString(36)}(x)`)+` * ${c}'(x)`:`D[x] = 1`
f=n=>n?f(n-1).replace(/x/g,`f${n}(x)`)+` * f${n}'(x)`:`D[x] = 1`
| improve this answer | |
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  • 1
    \$\begingroup\$ Why don't you put the 73-byte version as your main? \$\endgroup\$ – Leaky Nun Jul 26 '16 at 23:02
  • 1
    \$\begingroup\$ If you're being recursive, you need the f= or else your program won't work. \$\endgroup\$ – Value Ink Jul 26 '16 at 23:21
  • 2
    \$\begingroup\$ @KevinLau-notKenny Bah, I always forget to do that. \$\endgroup\$ – Neil Jul 27 '16 at 0:09
1
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Ruby, 72 bytes

Second form. Port to Ruby from @Neil answer.

f=->n{n>1?f[n-1].gsub(?x,"f#{n}(x)")+" * f#{n}'(x)":"D[f1(x)] = f1'(x)"}
| improve this answer | |
\$\endgroup\$
1
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Julia, 66 bytes

!x=x>1?replace(!~-x,"x","f$x(x)")*" * f$x'(x)":"D[f1(x)] = f1'(x)"

Port of @Neil's ES6 answer. Uses the second form.

| improve this answer | |
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0
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Python 2, 174 bytes

i=input()+1
a=lambda x:'('.join(['f%d'%e for e in range(x,i)])+'(x'+')'*(i-x)
print'D[%s] = %s'%(a(1),' * '.join(["f%d'%s%s)"%(e+1,'('*(i-e-2>0),a(e+2))for e in range(i-1)]))

Lots of room left to be golfed here. Uses the second form (f1, f2, etc.).

| improve this answer | |
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0
\$\begingroup\$

JavaScript (ES6), 194 bytes

n=>{s="D[";for(i=1;i<=n;i++)s+="f"+i+"(";s+="x";s+=")".repeat(n);s+="] = ";for(i=1;i<=n;i++){s+="f"+i+"'(";for(j=i+1;j<=n;j++)s+="f"+j+"(";s+="x";s+=")".repeat(n-i+1);if(i-n)s+=" * ";}return s;}

31 bytes saved thanks to @LeakyNun.

It's an anonymous lambda function. I'm sure there's some way to shorten this...

Ungolfed

var chain = function(n) {
    var str = "D["; // derivative symbol
    for (var i = 1; i <= n; i++) {
        str += "f"+i+"("; // put in n functions, each labeled f1(, f2(, etc.
    }
    str += "x"; // add in the function input, usually "x"
    for (var i = 0; i < n; i++) {
        str += ")"; // add in n end parentheses
    }
    str += "] = "; // add in the end bracket and the equals operator
    for (var i = 1; i <= n; i++) {
        str += "f"+i+"'("; // add in all n of the outer functions with the prime operator
        for (var j = i+1; j <= n; j++) {
            str += "f"+j+"("; // add in all of the inner functions
        }
        str += "x"; // add in the input, "x"
        for (var j = 1; j <= n; j++) {
            str += ")"; // close the parentheses
        }
        if (i !== n) {
            str += " * "; // the multiplication of all of the outer primed functions
        }
    }
    return str;
};
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Change str into s. \$\endgroup\$ – Leaky Nun Jul 26 '16 at 23:25
  • \$\begingroup\$ Ok, I forgot to change it. Lol \$\endgroup\$ – Drew Christensen Jul 26 '16 at 23:27
  • \$\begingroup\$ Change for(j=i;j<=n;j++)s+=")"; to s+=")".repeat(n-i+1); \$\endgroup\$ – Leaky Nun Jul 27 '16 at 0:30

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