# Difference between revisions of "99 questions/Solutions/49"

(another solutions using folds) |
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</haskell> |
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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). |
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). |
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+ | |||

+ | Here is another solution using folds that yields the specified result. |
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+ | <haskell> |
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+ | gray' :: Integral a => a -> [String] |
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+ | gray' 0 = [""] |
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+ | gray' n = gl $ gr where gl y = foldr (\s acc -> ('0':s):acc) y gp |
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+ | gr = foldl (\acc s -> ('1':s):acc) [] gp |
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+ | gp = gray' (n-1) |
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+ | </haskell> |
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+ | |||

## Revision as of 04:28, 31 May 2021

(**) 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 the space which is occupied by the list itself:

```
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).

Here is another solution using folds that yields the specified result.

```
gray' :: Integral a => a -> [String]
gray' 0 = [""]
gray' n = gl $ gr where gl y = foldr (\s acc -> ('0':s):acc) y gp
gr = foldl (\acc s -> ('1':s):acc) [] gp
gp = gray' (n-1)
```