5 kyu
How many strings use all symbols of a given alphabet at least once
74 of 76benjaminzwhite
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Algorithms
Mathematics
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More like a math challenge, however nice.
suggested tag:
combinatoricsFactor translation
@Kacarott - approved, and thanks for translating!
Nice kata. Approved at kyu 5.
Thanks again for the comments, feedback and now approval @dfhwze !
Does this sequence naturally extend with larger 'a' like in your example?
nvm, -1 should have been (-1) :s
Hi @dfhwze , thanks so much for trying my kata.
To answer your question: Yes, but remember if you're going to calculate recursively like this that the multiplications for the various "missing" symbols will probably involve binomial coefficients rather than just linear 4*, 3*, 2* etc.
For example, if we read the part in the description for n=5, a=3 which says:
"So, for each of the 3 choices of F, G, or H there are exactly 2**5 - 2 = 30 forbidden strings in which exactly one of those symbols is missing, but not two of them."
in the above, the "3 choices" here is actually obtained from 3_choose_1 binomial coefficient, because you are choosing 1 symbol to omit from the 3 available.
When you have n=5 and a=4, as you consider the strings missing exactly 1,2,3 symbols, the new terms will involve binomial coefficients like this:
"With a = 4 symbols {F,G,H,I} there are 4_choose_2 = 6 (i.e. NOT just 4) ways of selecting 2 "missing symbols" to not use from these 4 sybmbols."
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I'm not sure what would be best moving forward. Either you update the description with the 5,4 case, or you mark your previous comment as a spoiler. I do think the difficulty level of this kata depends on your decision ;)
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Let's leave it as is. People are free to browse the comments section if they seek some hints.
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