99 lines
2.2 KiB
Markdown
99 lines
2.2 KiB
Markdown
---
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id: 5900f3a51000cf542c50feb8
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title: 'Problem 57: Square root convergents'
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challengeType: 5
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forumTopicId: 302168
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dashedName: problem-57-square-root-convergents
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---
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# --description--
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It is possible to show that the square root of two can be expressed as an infinite continued fraction.
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<div style='text-align: center;'>$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$</div>
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By expanding this for the first four iterations, we get:
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$1 + \\frac 1 2 = \\frac 32 = 1.5$
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$1 + \\frac 1 {2 + \\frac 1 2} = \\frac 7 5 = 1.4$
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$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 2}} = \\frac {17}{12} = 1.41666 \\dots$
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$1 + \\frac 1 {2 + \\frac 1 {2+\\frac 1 {2+\\frac 1 2}}} = \\frac {41}{29} = 1.41379 \\dots$
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The next three expansions are $\\frac {99}{70}$, $\\frac {239}{169}$, and $\\frac {577}{408}$, but the eighth expansion, $\\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
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In the first `n` expansions, how many fractions contain a numerator with more digits than denominator?
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# --hints--
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`squareRootConvergents(10)` should return a number.
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```js
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assert(typeof squareRootConvergents(10) === 'number');
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```
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`squareRootConvergents(10)` should return 1.
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```js
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assert.strictEqual(squareRootConvergents(10), 1);
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```
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`squareRootConvergents(100)` should return 15.
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```js
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assert.strictEqual(squareRootConvergents(100), 15);
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```
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`squareRootConvergents(1000)` should return 153.
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```js
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assert.strictEqual(squareRootConvergents(1000), 153);
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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 squareRootConvergents(n) {
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return true;
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}
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squareRootConvergents(1000);
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```
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# --solutions--
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```js
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function squareRootConvergents(n) {
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function countDigits(number) {
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let counter = 0;
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while (number > 0) {
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counter++;
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number = number / 10n;
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}
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return counter;
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}
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// Use BigInt as integer won't handle all cases
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let numerator = 3n;
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let denominator = 2n;
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let moreDigitsInNumerator = 0;
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for (let i = 2; i <= n; i++) {
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[numerator, denominator] = [
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numerator + 2n * denominator,
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denominator + numerator
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];
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if (countDigits(numerator) > countDigits(denominator)) {
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moreDigitsInNumerator++;
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}
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}
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return moreDigitsInNumerator;
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}
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```
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