Problem 1873. Break it up! Break it up!

Created by James

Solve Later

{"relationships":[{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/document","targetMode":"","relationshipId":"rId1","target":"/matlab/document.xml"},{"relationshipType":"http://schemas.mathworks.com/matlab/code/2013/relationships/output","targetMode":"","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\"?>\n<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>You have N pennies. Write a Matlab script that will reveal how many different ways you can break up those pennies. For example, with 5 pennies, there are seven different ways you can divide them:</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(1,1,1,1,1)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(2,1,1,1)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(2,2,1)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(3,2)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(3,1,1)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(4,1)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"ListParagraph\"/><w:numPr><w:numId w:val=\"1\"/></w:numPr></w:pPr><w:r><w:t>(5)</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>The order of the coins does not matter, so (2,2,1) is considered the same combination as (2,1,2). You can assume that N will always be a positive integer.</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>"}]}

Solve

Solution Stats

89 Solutions
40 Solvers

Last Solution submitted on Jul 13, 2022

Last 200 Solutions

Problem Comments

4 Comments

4 Comments

Show 1 older comment

Tim on 17 Sep 2013

Test case p(300)=9253082936723604 exceeds the maximum integer that can be represented exactly as a double precision number (namely, 2^53=9007199254740992); the fact that some solutions work for this case is just luck.

James on 18 Sep 2013

Changed to p(299), which is <2^53. Thanks for the heads up.

James on 18 Sep 2013

Interestingly, p(299) had some issues as well. Changed again to p(199), which seems to be working OK.

ChrisR on 23 Apr 2022

If I ever want to learn something, I dial up a problem from James.

Solution Comments

Solution 8712630

1 Comment

1 Comment

Aditi Singh on 12 Jul 2022

Great

Solution 7282685

1 Comment

1 Comment

Dyuman Joshi on 21 Dec 2021

any input greater than 700 and this code will fail to compile in Cody's time limit

Solution 938319

1 Comment

1 Comment

Peng Liu on 18 Aug 2016

Test 2: function y = partitions(x) works!

Solution 938316

7 Comments

7 Comments

Show 4 older comments

Peng Liu on 18 Aug 2016

Test 1: function y = partitions(n) fails

Aditya Jain on 18 Aug 2016

isn't it suppose to be partitions(x)?

Peng Liu on 18 Aug 2016

But this is just function definition.

Peng Liu on 18 Aug 2016

Either partitions(x) or partitions(n) should not matter in the function definition.

James on 19 Aug 2016

Strange. The problem you were having had something to do with the error checking suite that was above the solutions. Your solution passed the error checks, but failed to actually work. Now that I have removed the error checks, your solution with partitions(n) passes. I may have to muck around with this a bit later on.

James on 19 Aug 2016

The problem was with the following line in the error checking:

st = sprintf('function y = partitions(x)\n%%%s', st);

This automatically changed the first line of your script to use partitions(x) instead of partitions(n). That was done to prevent folks from using the "function ans=blahblahblah(x)" technique when that still worked, That was just a holdover that happened to trip you up. Now that that line (and the rest of the suite0 has been removed, partitions(n) should work just fine.

Peng Liu on 19 Aug 2016

@James oops! I was so careless that I didn't notice that single-line error checking (I thought your check was intended to forbid the usage of functions such as eval, str2num, etc, and thus I completely ignored that when I was solving the problem). Now I realized this is not a strange behavior or bug of Cody. Thanks for your clarification.

Problem Recent Solvers40

Problem Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Problem 1873. Break it up! Break it up!

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers40

Suggested Problems

More from this Author80

Problem Tags

Community Treasure Hunt

Problem 1873. Break it up! Break it up!

Solution Stats

Last 200 Solutions

Problem Comments

Solution Comments

Problem Recent Solvers40

Suggested Problems

More from this Author80

Problem Tags

Community Treasure Hunt

Players