Fix: Problem 39: Integer right triangles (#38145)
* fix: correct test and add solution I also changed the seed to report the results of an easier example to the user, since just evaluating the function mostly wastes time. * fix: use a better solution * fix: credit original authorpull/38187/head
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@ -24,8 +24,8 @@ For which value of p ≤ n, is the number of solutions maximised?
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tests:
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- text: <code>intRightTriangles(500)</code> should return 420.
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testString: assert(intRightTriangles(500) == 420);
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- text: <code>intRightTriangles(800)</code> should return 420.
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testString: assert(intRightTriangles(800) == 420);
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- text: <code>intRightTriangles(800)</code> should return 720.
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testString: assert(intRightTriangles(800) == 720);
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- text: <code>intRightTriangles(900)</code> should return 840.
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testString: assert(intRightTriangles(900) == 840);
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- text: <code>intRightTriangles(1000)</code> should return 840.
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@ -46,7 +46,7 @@ function intRightTriangles(n) {
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return n;
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}
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intRightTriangles(1000);
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console.log(intRightTriangles(500)); // 420
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```
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</div>
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@ -59,7 +59,34 @@ intRightTriangles(1000);
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<section id='solution'>
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```js
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// solution required
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// Original idea for this solution came from
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// https://www.xarg.org/puzzle/project-euler/problem-39/
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function intRightTriangles(n) {
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// store the number of triangles with a given perimeter
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let triangles = {};
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// a is the shortest side
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for (let a = 3; a < n / 3; a++)
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// o is the opposite side and is at least as long as a
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for (let o = a; o < n / 2; o++) {
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let h = Math.sqrt(a * a + o * o); // hypotenuse
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let p = a + o + h; // perimeter
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if ((h % 1) === 0 && p <= n) {
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triangles[p] = (triangles[p] || 0) + 1;
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}
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}
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let max = 0, maxp = null;
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for (let p in triangles) {
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if (max < triangles[p]) {
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max = triangles[p];
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maxp = p;
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}
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}
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return maxp;
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}
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```
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</section>
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