Problem 2635. Higher! Lower!

Created by James

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","relationshipId":"rId2","target":"/matlab/output.xml"}],"parts":[{"partUri":"/matlab/document.xml","relationship":[],"contentType":"application/vnd.mathworks.matlab.code.document+xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\"?><w:document xmlns:w=\"http://schemas.openxmlformats.org/wordprocessingml/2006/main\"><w:body><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>My kids have started to play the \"I'm thinking of a number between X and Y\" game in the car on trips to and from their swim lessons. I'm trying to teach them the best way to arrive at the correct answer, so I could use a little MATLAB help here to show them what to do...</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>Using the bisection method (split X and Y, round down if there's a remainder), guess what the number is. Your output should be all of guesses it took for you to get to that number, with the last number in that trail being the number that I'm thinking of.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>For example, I'm thinking of a number (8) between 1 and 10.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>First guess is (1+10)/2 rounded down, or 5. Five is lower than 8, so your next guess should be (5+10)/2, rounding down to 7. Seven is lower than 8, so your next guess should be (7+10)/2, rounding down to 8.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>Since this is the correct answer, your output should be [5 7 8].</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/><w:jc w:val=\"left\"/></w:pPr><w:r><w:t>The input to the function will be the low bound, the high bound, and the number I'm thinking of. You can assume that they're not cheating, and that the correct answer will be between the high and low bounds. The answer can be equal to the high or low bound, however.</w:t></w:r></w:p></w:body></w:document>"},{"partUri":"/matlab/output.xml","contentType":"text/xml","content":"<?xml version=\"1.0\" encoding=\"UTF-8\" standalone=\"no\" ?><embeddedOutputs><metaData><evaluationState>manual</evaluationState><layoutState>code</layoutState><outputStatus>ready</outputStatus></metaData><outputArray type=\"array\"/><regionArray type=\"array\"/></embeddedOutputs>"}]}

My kids have started to play the "I'm thinking of a number between X and Y" game in the car on trips to and from their swim lessons. I'm trying to teach them the best way to arrive at the correct answer, so I could use a little MATLAB help here to show them what to do...Using the bisection method (split X and Y, round down if there's a remainder), guess what the number is. Your output should be all of guesses it took for you to get to that number, with the last number in that trail being the number that I'm thinking of.For example, I'm thinking of a number (8) between 1 and 10.First guess is (1+10)/2 rounded down, or 5.
Five is lower than 8, so your next guess should be (5+10)/2, rounding down to 7.
Seven is lower than 8, so your next guess should be (7+10)/2, rounding down to 8.Since this is the correct answer, your output should be [5 7 8].The input to the function will be the low bound, the high bound, and the number I'm thinking of. You can assume that they're not cheating, and that the correct answer will be between the high and low bounds. The answer can be equal to the high or low bound, however.

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94 Solutions
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Last Solution submitted on Jul 03, 2020

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7 Comments

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Tim on 17 Oct 2014

Test suite should have assert(isequal(...)).

Guillaume on 19 Oct 2014

It's an interesting problem. Unfortunately without any assert, the test suite does not test anything. So any syntactically correct answer succeeds even if it doesn't solve the problem.

James on 20 Oct 2014

Fixed, and (hopefully) rescored. Thanks for the heads up.

Alfonso Nieto-Castanon on 6 Jan 2015

your example ([5 7 8]) is not really following the optimal bisection approach. After you choose 5, the new bounds are not 5 and 10, they are 6 and 10, so your next choice should be 8 ((6+10)/2) instead of 7...

Gergely Patay on 7 Jul 2015

The misplaced comment line in the first test case breaks the test suite (at least for "conventional" solutions).

Michael Jarboe on 4 Jun 2019

Emphasized Alfonso Nieto-Castanon Comments "your example ([5 7 8]) is not really following the optimal bisection approach. After you choose 5, the new bounds are not 5 and 10, they are 6 and 10, so your next choice should be 8 ((6+10)/2) instead of 7..."

Nikolaos Nikolaou on 9 Sep 2019

test suit is broken plz fix since its a budge problem

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Problem 2635. Higher! Lower!

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Problem 2635. Higher! Lower!

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