# 99 questions/46 to 50

### From HaskellWiki

This is part of Ninety-Nine Haskell Problems, based on Ninety-Nine Prolog Problems.

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## 1 Logic and Codes

## 2 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 Example in Haskell: > table2 (\a b -> (and' a (or' a b)) True True True True False True False True False False False False

Solution:

not' :: Bool -> Bool not' True = False not' False = True and',or',nor',nand',xor',impl',equ' :: Bool -> Bool -> Bool and' True True = True and' _ _ = False or' True _ = True or' _ True = True or' _ _ = False nor' a b = not' $ or' a b nand' a b = not' $ and' a b 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 table2 :: (Bool -> Bool -> Bool) -> IO () table2 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 Lisp's 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 it is in Lisp. Template Haskell could also be used :)

## 3 Problem 47

(*) Truth tables for logical expressions (2).

Continue problem P46 by defining and/2, or/2, etc as being operators. This allows to write the logical expression in the more natural way, as in the example: A and (A or not B). Define operator precedence as usual; i.e. as in Java.

Example: * (table A B (A and (A or not B))) true true true true fail true fail true fail fail fail fail Example in Haskell: > table2 (\a b -> a `and'` (a `or'` not b)) True True True True False True False True False False False False

Solution:

-- functions as in solution 46 infixl 4 `or'` infixl 6 `and'` -- "not" has fixity 9 by default

Java operator precedence (descending) as far as I could fathom it:

logical not equality and xor or

Using "not" as a non-operator is a little evil, but then again these problems were designed for languages other than haskell :)

## 4 Problem 48

(**) Truth tables for logical expressions (3).

Generalize problem P47 in such a way that the logical expression may contain any number of logical variables. Define table/2 in a way that table(List,Expr) prints the truth table for the expression Expr, which contains the logical variables enumerated in List.

Example: * (table (A,B,C) (A and (B or C) equ A and B or A and C)) true true true true true true fail true true fail true true true fail fail true fail true true true fail true fail true fail fail true true fail fail fail true Example in Haskell: > table3 (\a b c -> a `and'` (b `or'` c) `equ'` a `and'` b `or'` a `and'` c) True True True True True True False True True False True True True False False True False True True True False True False True False False True True False False False True

Solution:

-- functions as in solution 46 infixl 4 `or'` infixl 4 `nor'` infixl 5 `xor'` infixl 6 `and'` infixl 6 `nand'` infixl 3 `equ'` -- was 7, changing it to 3 got me the same results as in the original question :( table3 :: (Bool -> Bool -> Bool -> Bool) -> IO () table3 f = putStrLn . unlines $ [show a ++ " " ++ show b ++ " " ++ show c ++ " " ++ show (f a b c) | a <- [True, False], b <- [True, False], c <- [True, False]]

Using individual table functions for different numbers of variables is even more ugly, but anything else would be a bit of a pain in haskell AFAIK.

## 5 Problem 49

(**) Gray codes.

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 0 = [""] gray n = let xs = gray (n-1) 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 prepend a 1 to each word. At last we have to append these two lists. (The Wikipedia article seems to approve this.)

## 6 Problem 50

(***) Huffman codes.

We suppose a set of symbols with their frequencies, given as a list of fr(S,F) terms. Example: [fr(a,45),fr(b,13),fr(c,12),fr(d,16),fr(e,9),fr(f,5)]. Our objective is to construct a list hc(S,C) terms, where C is the Huffman code word for the symbol S. In our example, the result could be Hs = [hc(a,'0'), hc(b,'101'), hc(c,'100'), hc(d,'111'), hc(e,'1101'), hc(f,'1100')] [hc(a,'01'),...etc.]. The task shall be performed by the predicate huffman/2 defined as follows:

% huffman(Fs,Hs) :- Hs is the Huffman code table for the frequency table Fs

Example in Haskell:

*Exercises> huffman [('a',45),('b',13),('c',12),('d',16),('e',9),('f',5)] [('a',"0"),('b',"101"),('c',"100"),('d',"111"),('e',"1101"),('f',"1100")]

Solution:

import Data.List data HTree a = Leaf a | Branch (HTree a) (HTree a) deriving Show huffman :: (Ord a, Ord w, Num w) => [(a,w)] -> [(a,[Char])] huffman freq = sortBy (comparing fst) $ serialize $ htree $ sortBy (comparing fst) $ [(w, Leaf x) | (x,w) <- freq] where htree [(_, t)] = t htree ((w1,t1):(w2,t2):wts) = htree $ insertBy (comparing fst) (w1 + w2, Branch t1 t2) wts comparing f x y = compare (f x) (f y) serialize (Branch l r) = [(x, '0':code) | (x, code) <- serialize l] ++ [(x, '1':code) | (x, code) <- serialize r] serialize (Leaf x) = [(x, "")]

The argument to `htree` is a list of (weight, tree) pairs, in order of increasing weight.
The implementation could be made more efficient by using a priority queue instead of an ordered list.