# 99 questions/Solutions/49

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(another solution using list comprehension) |
|||

Line 15: | Line 15: | ||

</haskell> | </haskell> | ||

− | + | A similar solution using list comprehension to build the sub-lists: | |

− | + | ||

+ | <haskell> | ||

+ | gray :: Int -> [String] | ||

+ | gray 0 = [""] | ||

+ | gray n = [ '0' : x | x <- prev ] ++ [ '1' : x | x <- reverse prev ] | ||

+ | where prev = gray (n-1) | ||

+ | </haskell> | ||

+ | |||

+ | 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]. |

## 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.