# 99 questions/Solutions/49

### From HaskellWiki

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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]. | 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]. | ||

− | Another completely different solution (using folds) that is way more efficient, because it needs just | + | Another completely different solution (using folds) that is way more efficient, because it needs just O(1) (constant) space: |

<haskell> | <haskell> | ||

gray :: Integral a => a -> [String] | gray :: Integral a => a -> [String] |

## Revision as of 10:47, 16 December 2012

(**) 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.

Another completely different solution (using folds) that is way more efficient, because it needs just O(1) (constant) space:

gray :: Integral a => a -> [String] gray 0 = [""] gray n = foldr (\s acc -> ("0" ++ s):("1" ++ s):acc) [] $ gray (n-1)

The key is that the algorithm just crawls one time over the input-list and uses the (++) operator in a way that it just has a running time of O(1).