79d9012432
* fix(curriculum): tests quotes * fix(curriculum): fill seed-teardown * fix(curriculum): fix tests and remove unneeded seed-teardown
78 lines
2.6 KiB
Markdown
78 lines
2.6 KiB
Markdown
---
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id: 5900f4b41000cf542c50ffc7
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challengeType: 5
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title: 'Problem 328: Lowest-cost Search'
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---
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## Description
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<section id='description'>
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We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions.
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Each number (question) we ask, has a cost equal to the number asked and we get one of three possible answers: "Your guess is lower than the hidden number", or
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"Yes, that's it!", or
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"Your guess is higher than the hidden number".
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Given the value of n, an optimal strategy minimizes the total cost (i.e. the sum of all the questions asked) for the worst possible case. E.g.
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If n=3, the best we can do is obviously to ask the number "2". The answer will immediately lead us to find the hidden number (at a total cost = 2).
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If n=8, we might decide to use a "binary search" type of strategy: Our first question would be "4" and if the hidden number is higher than 4 we will need one or two additional questions.
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Let our second question be "6". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8.
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Thus, our third question will be "7" and the total cost for this worst-case scenario will be 4+6+7=17.
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We can improve considerably the worst-case cost for n=8, by asking "5" as our first question.
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If we are told that the hidden number is higher than 5, our second question will be "7", then we'll know for certain what the hidden number is (for a total cost of 5+7=12).
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If we are told that the hidden number is lower than 5, our second question will be "3" and if the hidden number is lower than 3 our third question will be "1", giving a total cost of 5+3+1=9.
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Since 12>9, the worst-case cost for this strategy is 12. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy.
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So, in fact, we have just described an optimal strategy for n=8.
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Let C(n) be the worst-case cost achieved by an optimal strategy for n, as described above.
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Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12.
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Similarly, C(100) = 400 and C(n) = 17575.
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Find C(n).
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>euler328()</code> should return 260511850222.
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testString: assert.strictEqual(euler328(), 260511850222, '<code>euler328()</code> should return 260511850222.');
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function euler328() {
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// Good luck!
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return true;
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}
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euler328();
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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</section>
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