106 lines
2.6 KiB
Markdown
106 lines
2.6 KiB
Markdown
---
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id: 5900f3a61000cf542c50feb9
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title: 'Problem 58: Spiral primes'
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challengeType: 5
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forumTopicId: 302169
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dashedName: problem-58-spiral-primes
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---
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# --description--
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Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.
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<div style='text-align: center;'>
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<strong><span style='color: red;'>37</span></strong> 36 35 34 33 32 <strong><span style='color: red;'>31</span></strong><br>
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38 <strong><span style='color: red;'>17</span></strong> 16 15 14 <strong><span style='color: red;'>13</span></strong> 30<br>
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39 18 <strong><span style='color: red;'>5</span></strong> 4 <strong><span style='color: red;'>3</span></strong> 12 29<br>
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40 19 6 1 2 11 28<br>
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41 20 <strong><span style='color: red;'>7</span></strong> 8 9 10 27<br>
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42 21 22 23 24 25 26<br>
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<strong><span style='color: red;'>43</span></strong> 44 45 46 47 48 49<br>
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</div>
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It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
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If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the percent of primes along both diagonals first falls below `percent`?
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# --hints--
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`spiralPrimes(50)` should return a number.
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```js
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assert(typeof spiralPrimes(50) === 'number');
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```
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`spiralPrimes(50)` should return `11`.
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```js
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assert.strictEqual(spiralPrimes(50), 11);
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```
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`spiralPrimes(15)` should return `981`.
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```js
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assert.strictEqual(spiralPrimes(15), 981);
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```
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`spiralPrimes(10)` should return `26241`.
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```js
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assert.strictEqual(spiralPrimes(10), 26241);
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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 spiralPrimes(percent) {
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return true;
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}
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spiralPrimes(50);
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```
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# --solutions--
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```js
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function spiralPrimes(percent) {
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function isPrime(n) {
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if (n <= 3) {
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return n > 1;
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} else if (n % 2 === 0 || n % 3 === 0) {
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return false;
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}
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for (let i = 5; i * i <= n; i += 6) {
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if (n % i === 0 || n % (i + 2) === 0) {
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return false;
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}
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}
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return true;
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}
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let totalCount = 1;
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let primesCount = 0;
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let curNumber = 1;
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let curSideLength = 1;
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let ratio = 1;
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const wantedRatio = percent / 100;
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while (ratio >= wantedRatio) {
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curSideLength += 2;
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for (let i = 0; i < 4; i++) {
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curNumber += curSideLength - 1;
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totalCount++;
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if (i !== 3 && isPrime(curNumber)) {
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primesCount++;
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}
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}
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ratio = primesCount / totalCount;
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}
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return curSideLength;
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}
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```
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