Twenty Four, Again: Difference between revisions
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imported>Kmk21 No edit summary |
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| Line 5: | Line 5: | ||
There are only 10 combinations of parenthesis. | There are only 10 combinations of parenthesis. | ||
(xx)xx | * (xx)xx | ||
x(xx)x | * x(xx)x | ||
xx(xx) | * xx(xx) | ||
(xxx)x | * (xxx)x | ||
((xx)x)x | * ((xx)x)x | ||
(x(xx))x | * (x(xx))x | ||
x(xxx) | * x(xxx) | ||
x((xx)x) | * x((xx)x) | ||
x(x(xx)) | * x(x(xx)) | ||
(xx)(xx) | * (xx)(xx) | ||
4! * 4^3*10 = 15360 | 4! * 4^3*10 = 15360 | ||
Revision as of 02:43, 2 November 2017
there are only 4! permutations of the numbers.
There are only 4^3 combinations of operators.
There are only 10 combinations of parenthesis.
- (xx)xx
- x(xx)x
- xx(xx)
- (xxx)x
- ((xx)x)x
- (x(xx))x
- x(xxx)
- x((xx)x)
- x(x(xx))
- (xx)(xx)
4! * 4^3*10 = 15360
Just iterate through all of them and find the cheapest.
One of the challenges to this problem SHOULD be to actually evaluate the expression...they want you to use a recursive descent parser (which you SHOULD know how to build....), but you can just dump it in java's javascript engine and call "evaluate" which will evaluate the string-based expression for you.