# 99 questions/46 to 50

These are Haskell translations of Ninety Nine Lisp Problems.

If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in lisp>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.

## Problem 46

Define predicates and/2, or/2, nand/2, nor/2, xor/2, impl/2 and equ/2 (for logical equivalence) which succeed or fail according to the result of their respective operations; e.g. and(A,B) will succeed, if and only if both A and B succeed.

A logical expression in two variables can then be written as in the following example: and(or(A,B),nand(A,B)).

Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables.

```Example:
(table A B (and A (or A B)))
true true true
true fail true
fail true fail
fail fail fail

> table (\a b -> (and' a (or' a b))
True True True
True False True
False True False
False False False
```

Solution:

```and' True True = True
and' _    _    = False

or' True _    = True
or' _    True = True
or' _    _    = False

not' True  = False
not' False = True

nor'  = not' . or'
nand' = not' . and'

xor' True  False = True
xor' False True  = True
xor' _     _     = False

impl' a b = (not' a) `or'` b

equ' True  True  = True
equ' False False = True
equ' _     _     = False

table f = putStrLn . unlines \$ [show a ++ " " ++ show b ++ " " ++ show (f a b)
| a <- [True, False], b <- [True, False]
```

The implementations of the logic functions are quite verbose and can be shortened in places (like "equ' = (==)").

The table function in Lisp supposedly uses lisps symbol handling to substitute variables on the fly in the expression. I chose passing a binary function instead because parsing an expression would be more verbose in haskell than in Lisp

## Problem 47

<Problem description>

```Example:
<example in lisp>

```

Solution:

```<solution in haskell>
```

<description of implementation>

## Problem 48

<Problem description>

```Example:
<example in lisp>

```

Solution:

```<solution in haskell>
```

<description of implementation>

## Problem 49

An n-bit Gray code is a sequence of n-bit strings constructed according to certain rules. For example,

```n = 1: C(1) = ['0','1'].
n = 2: C(2) = ['00','01','11','10'].
n = 3: C(3) = ['000','001','011','010',´110´,´111´,´101´,´100´].
```

Find out the construction rules and write a predicate with the following specification:

% gray(N,C) :- C is the N-bit Gray code

Can you apply the method of "result caching" in order to make the predicate more efficient, when it is to be used repeatedly?

```Example in Haskell:
P49> gray 3
["000","001","011","010","110","111","101","100"]
```

Solution:

```gray :: Int -> [String]
gray 1     = ["0", "1"]
gray (n+1) = let xs = gray n in map ('0':) xs ++ map ('1':) (reverse xs)
```

It seems that 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 perpend a 1 to each word. At last we have to append these two lists. (Wikipedia seems to approve this.)

Instead of the equation for `gray 1 = ...` we could also use

```gray 0     = [""]
```

what leads to the same results.

## Problem 50

<Problem description>

```Example:
<example in lisp>

```<solution in haskell>