* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
113 lines
4.3 KiB
Markdown
113 lines
4.3 KiB
Markdown
---
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id: 5a7dad05be01840e1778a0d1
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title: Fractran
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challengeType: 3
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videoUrl: ''
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dashedName: fractran
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---
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# --description--
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<div class='rosetta'><p class='rosetta__paragraph'> <span class='rosetta__text--bold'><a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/FRACTRAN' title='wp:FRACTRAN'>FRACTRAN</a></span>是由数学家<a class='rosetta__link--wiki' href='https://en.wikipedia.org/wiki/John Horton Conway' title='wp:John Horton Conway'>John Horton Conway</a>发明的图灵完备的深奥编程语言。 </p><br><p class='rosetta__paragraph'> FRACTRAN程序是正分数$ P =(f_1,f_2,\ ldots,f_m)$的有序列表,以及初始正整数输入$ n $。 </p><br><p class='rosetta__paragraph'>该程序通过更新整数$ n $来运行,如下所示: </p><br><ul class='rosetta__unordered-list'><li class='rosetta__list-item--unordered'>对于第一个分数$ f_i $,在$ nf_i $为整数的列表中,用$ nf_i $替换$ n $; </li><li class='rosetta__list-item--unordered'>重复此规则,直到列表中没有分数乘以$ n $时产生整数,然后停止。 </li></ul><br><p class='rosetta__paragraph'>康威为FRACTRAN提供了素数计划: </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 17/91 $,$ 78/85 $,$ 19/51 $,$ 23/38 $,$ 29/33 $,$ 77/29 $,$ 95/23 $,$ 77/19 $,$ 1/17 $,$ 11/13 $, $ 13/11 $ $,$ 15/14 $,$ 15/2 $,$ 55/1 $</span> </p><br><p class='rosetta__paragraph'>从$ n = 2 $开始,此FRACTRAN程序将$ n $更改为$ 15 = 2 \ times(15/2)$,然后$ 825 = 15 \ times(55/1)$,生成以下整数序列: </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 2 $,$ 15 $,$ 825 $,$ 725 $,$ 1925 $,$ 2275 $,$ 425 $,$ 390 $,$ 330 $,$ 290 $,$ 770 $,$ \ ldots $</span> </p><br><p class='rosetta__paragraph'> 2之后,此序列包含以下2的幂: </p><br><p class='rosetta__paragraph'> <span class='rosetta__text--indented'>$ 2 ^ 2 = 4 $,$ 2 ^ 3 = 8 $,$ 2 ^ 5 = 32 $,$ 2 ^ 7 = 128 $,$ 2 ^ {11} = 2048 $,$ 2 ^ {13} = 8192 $,$ 2 ^ {17 } = 131072 $,$ 2 ^ {19} = 524288 $,$ \ ldots $</span> </p><br><p class='rosetta__paragraph'>这是2的主要权力。 </p><br><dl class='rosetta__description-list'><dt class='rosetta__description-title'>任务: </dt></dl><p class='rosetta__paragraph'>编写一个函数,将fractran程序作为字符串参数,并将程序的前10个数字作为数组返回。如果结果没有10个数字,则按原样返回数字。 </p></div>
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# --hints--
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`fractran`应该是一个功能。
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```js
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assert(typeof fractran == 'function');
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```
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`fractran(""+tests[0]+"")`应该返回一个数组。
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```js
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assert(Array.isArray(fractran('3/2, 1/3')));
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```
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`fractran(""+tests[0]+"")`应返回`"+JSON.stringify(results[0])+"` 。
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```js
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assert.deepEqual(fractran('3/2, 1/3'), [2, 3, 1]);
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```
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`fractran(""+tests[1]+"")`应返回`"+JSON.stringify(results[1])+"` 。
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```js
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assert.deepEqual(fractran('3/2, 5/3, 1/5'), [2, 3, 5, 1]);
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```
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`fractran(""+tests[2]+"")`应返回`"+JSON.stringify(results[2])+"` 。
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```js
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assert.deepEqual(fractran('3/2, 6/3'), [2, 3, 6, 9, 18, 27, 54, 81, 162, 243]);
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```
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`fractran(""+tests[3]+"")`应返回`"+JSON.stringify(results[3])+"` 。
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```js
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assert.deepEqual(fractran('2/7, 7/2'), [2, 7, 2, 7, 2, 7, 2, 7, 2, 7]);
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```
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`fractran(""+tests[4]+"")`应返回`"+JSON.stringify(results[4])+"` 。
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```js
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assert.deepEqual(
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fractran(
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'17/91, 78/85, 19/51, 23/38, 29/33, 77/29, 95/23, 77/19, 1/17, 11/13, 13/11, 15/14, 15/2, 55/1'
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),
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[2, 15, 825, 725, 1925, 2275, 425, 390, 330, 290]
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);
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```
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# --seed--
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## --seed-contents--
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```js
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function fractran(progStr) {
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}
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```
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# --solutions--
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```js
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function fractran(progStr){
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var num = new Array();
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var den = new Array();
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var val ;
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var out="";
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function compile(prog){
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var regex = /\s*(\d*)\s*\/\s*(\d*)\s*(.*)/m;
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while(regex.test(prog)){
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num.push(regex.exec(prog)[1]);
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den.push(regex.exec(prog)[2]);
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prog = regex.exec(prog)[3];
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}
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}
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function step(val){
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var i=0;
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while(i<den.length && val%den[i] != 0) i++;
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return num[i]*val/den[i];
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}
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var seq=[]
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function exec(val){
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var i = 0;
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while(val && i<limit){
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seq.push(val)
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val = step(val);
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i ++;
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}
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}
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// Main
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compile(progStr);
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var limit = 10;
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exec(2);
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return seq;
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}
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```
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