Difference between revisions of "99 questions/Solutions/49"
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(another solution using list comprehension) |
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</haskell> |
</haskell> |
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| + | A similar solution using list comprehension to build the sub-lists: |
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| − | (The [http://en.wikipedia.org/wiki/Gray_code Wikipedia article] seems to approve this.) |
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| + | <haskell> |
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| + | gray :: Int -> [String] |
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| + | gray 0 = [""] |
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| + | gray n = [ '0' : x | x <- prev ] ++ [ '1' : x | x <- reverse prev ] |
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| + | where prev = gray (n-1) |
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| + | </haskell> |
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| + | |||
| ⚫ | The Gray code can be recursively defined in the way that for determining the gray code of n we take the Gray code of n-1, prepend a 0 to each word, take the Gray code for n-1 again, reverse it and prepend a 1 to each word. At last we have to append these two lists. For more see [http://en.wikipedia.org/wiki/Gray_code the Wikipedia article]. |
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Revision as of 21:58, 17 July 2010
(**) Gray codes.
An n-bit Gray code is a sequence of n-bit strings constructed according to certain rules. Find out the construction rules and write a predicate with the following specification:
% gray(N,C) :- C is the N-bit Gray code
Solution:
gray :: Int -> [String]
gray 0 = [""]
gray n = let xs = gray (n-1) in map ('0':) xs ++ map ('1':) (reverse xs)
A similar solution using list comprehension to build the sub-lists:
gray :: Int -> [String]
gray 0 = [""]
gray n = [ '0' : x | x <- prev ] ++ [ '1' : x | x <- reverse prev ]
where prev = gray (n-1)
The Gray code can be recursively defined in the way that for determining the gray code of n we take the Gray code of n-1, prepend a 0 to each word, take the Gray code for n-1 again, reverse it and prepend a 1 to each word. At last we have to append these two lists. For more see the Wikipedia article.