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 author
pull/38187/head
Oliver Eyton-Williams 2020-02-08 16:39:32 +01:00 committed by GitHub
parent 158188924b
commit a7075a579c
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1 changed files with 31 additions and 4 deletions

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@ -24,8 +24,8 @@ For which value of p ≤ n, is the number of solutions maximised?
tests:
- text: <code>intRightTriangles(500)</code> should return 420.
testString: assert(intRightTriangles(500) == 420);
- text: <code>intRightTriangles(800)</code> should return 420.
testString: assert(intRightTriangles(800) == 420);
- text: <code>intRightTriangles(800)</code> should return 720.
testString: assert(intRightTriangles(800) == 720);
- text: <code>intRightTriangles(900)</code> should return 840.
testString: assert(intRightTriangles(900) == 840);
- text: <code>intRightTriangles(1000)</code> should return 840.
@ -46,7 +46,7 @@ function intRightTriangles(n) {
return n;
}
intRightTriangles(1000);
console.log(intRightTriangles(500)); // 420
```
</div>
@ -59,7 +59,34 @@ intRightTriangles(1000);
<section id='solution'>
```js
// solution required
// Original idea for this solution came from
// https://www.xarg.org/puzzle/project-euler/problem-39/
function intRightTriangles(n) {
// store the number of triangles with a given perimeter
let triangles = {};
// a is the shortest side
for (let a = 3; a < n / 3; a++)
// o is the opposite side and is at least as long as a
for (let o = a; o < n / 2; o++) {
let h = Math.sqrt(a * a + o * o); // hypotenuse
let p = a + o + h; // perimeter
if ((h % 1) === 0 && p <= n) {
triangles[p] = (triangles[p] || 0) + 1;
}
}
let max = 0, maxp = null;
for (let p in triangles) {
if (max < triangles[p]) {
max = triangles[p];
maxp = p;
}
}
return maxp;
}
```
</section>