Problem 1307. RPN Calculator for simple arithmetic expressions

Created by Muthu Annamalai

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>Reverse-Polish-Notation (RPN) is a machine friendly form of calculating expressions. Example, to evaluate, (1+2)*4 + 5 - 3 you enter the sequence into the calculator as, '5 1 2 + 4 * + 3 -' and obtain the result.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>For this challenge write a RPN calculator for simple arithmetic expressions, '+','-','*','%','/'</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Note: Chosen interpretations of operators, illustrated from first few test cases, are as follows: rpn('a','b','-') is interpreted as 'a-b', while rpn('a','b','/') is interpreted as 'b/a'. and rpz('a','b','%') which is interpreted as 'b%a'.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>with operator '%' 'b % a' being mod(b,a) operator.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Following your 'stack' solution to</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://www.mathworks.com/matlabcentral/cody/problems/1303-is-the-paranthesis-sequence-balanced\"><w:r><w:t>http://www.mathworks.com/matlabcentral/cody/problems/1303-is-the-paranthesis-sequence-balanced</w:t></w:r></w:hyperlink><w:r><w:t> try using a expression, and a value stack to work this problem.</w:t></w:r></w:p><w:p><w:pPr><w:pStyle w:val=\"text\"/></w:pPr><w:r><w:t>Hint:</w:t></w:r><w:r><w:t> </w:t></w:r><w:hyperlink w:docLocation=\"http://en.wikipedia.org/wiki/Reverse_Polish_notation\"><w:r><w:t>http://en.wikipedia.org/wiki/Reverse_Polish_notation</w:t></w:r></w:hyperlink></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>"}]}

Reverse-Polish-Notation (RPN) is a machine friendly form of calculating expressions. Example, to evaluate, (1+2)*4 + 5 - 3 you enter the sequence into the calculator as, '5 1 2 + 4 * + 3 -' and obtain the result.For this challenge write a RPN calculator for simple arithmetic expressions, '+','-','*','%','/'Note: Chosen interpretations of operators, illustrated from first few test cases, are as follows: rpn('a','b','-') is interpreted as 'a-b', while rpn('a','b','/') is interpreted as 'b/a'. and rpz('a','b','%') which is interpreted as 'b%a'.with operator '%' 'b % a' being mod(b,a) operator.Following your 'stack' solution to <a href = "http://www.mathworks.com/matlabcentral/cody/problems/1303-is-the-paranthesis-sequence-balanced">http://www.mathworks.com/matlabcentral/cody/problems/1303-is-the-paranthesis-sequence-balanced</a> try using a expression, and a value stack to work this problem.Hint: <a href = "http://en.wikipedia.org/wiki/Reverse_Polish_notation">http://en.wikipedia.org/wiki/Reverse_Polish_notation</a>

Solve

Solution Stats

45 Solutions
7 Solvers

Last Solution submitted on Aug 29, 2020

Last 200 Solutions

Problem Comments

4 Comments

4 Comments

Show 1 older comment

Alfonso Nieto-Castanon on 1 Mar 2013

in your test suite it seems odd that rpz('a','b','-') is interpreted as 'a-b' while rpz('a','b','/') is interpreted as 'b/a' (instead of 'a/b'). Same goes for rpz('a','b','%') which is interpreted as 'b%a' instead of 'a%b'. Last, your last test case rpn(z(1),'+',z(2),'*',z(1),z(3),'-',z(4),z(5),'%','*') is plain weird... it should probably read rpn(z(1),z(2),'+',z(1),'*',z(3),'-',z(4),z(5),'%','*') instead...

Muthu Annamalai on 2 Mar 2013

Thank you for your comments; I will update the description and fix last test case.

Muthu Annamalai on 22 Mar 2013

I will retire the last test case for the problem.

Baki on 6 Nov 2016

I have an rpn solver but it only accepts strings

Problem Recent Solvers7

Problem Tags

rpn stack polish-notation expression evaluation

Community Treasure Hunt

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

Start Hunting!

Problem 1307. RPN Calculator for simple arithmetic expressions

Solution Stats

Last 200 Solutions

Problem Comments

Problem Recent Solvers7

Suggested Problems

More from this Author10

Problem Tags

Community Treasure Hunt

Problem 1307. RPN Calculator for simple arithmetic expressions

Solution Stats

Last 200 Solutions

Problem Comments

Problem Recent Solvers7

Suggested Problems

More from this Author10

Problem Tags

Community Treasure Hunt

Players