Difference between revisions of "99 questions/46 to 50"
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−  [[99_Haskell_exercisesBack to 99 Haskell exercises]] 

−  
__NOTOC__ 
__NOTOC__ 

−  These are Haskell translations of [http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L99_NinetyNine_Lisp_Problems.html Ninety Nine Lisp Problems]. 

+  This is part of [[H99:_NinetyNine_Haskell_ProblemsNinetyNine Haskell Problems]], based on [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p99/ NinetyNine Prolog 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. 

== Logic and Codes == 
== Logic and Codes == 

Line 16:  Line 12:  
Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables. 
Now, write a predicate table/3 which prints the truth table of a given logical expression in two variables. 

+  
+  Example: 

<pre> 
<pre> 

−  Example: 

(table A B (and A (or A B))) 
(table A B (and A (or A B))) 

true true true 
true true true 

Line 24:  Line 21:  
fail true fail 
fail true fail 

fail fail fail 
fail fail fail 

+  </pre> 

Example in Haskell: 
Example in Haskell: 

−  > table2 (\a b > (and' a (or' a b)) 

+  
+  <haskell> 

+  λ> table (\a b > (and' a (or' a b))) 

True True True 
True True True 

True False True 
True False True 

False True False 
False True False 

False False False 
False False False 

−  </pre> 

−  
−  Solution: 

−  <haskell> 

−  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]] 

</haskell> 
</haskell> 

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

+  [[99 questions/Solutions/46  Solutions]] 

−  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 :) 

−  
== Problem 47 == 
== Problem 47 == 

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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. 
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. 

−  <pre> 

Example: 
Example: 

+  
+  <pre> 

* (table A B (A and (A or not B))) 
* (table A B (A and (A or not B))) 

true true true 
true true true 

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fail true fail 
fail true fail 

fail fail fail 
fail fail fail 

+  </pre> 

Example in Haskell: 
Example in Haskell: 

−  > table2 (\a b > a `and'` (a `or'` not b)) 

+  
+  <haskell> 

+  λ> table2 (\a b > a `and'` (a `or'` not b)) 

True True True 
True True True 

True False True 
True False True 

False True False 
False True False 

False False False 
False False False 

−  </pre> 

−  
−  Solution: 

−  <haskell> 

−   functions as in solution 46 

−  infixl 4 `or'` 

−  infixl 6 `and'` 

−   "not" has fixity 9 by default 

</haskell> 
</haskell> 

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

+  [[99 questions/Solutions/47  Solutions]] 

−  <pre> 

−  logical not 

−  equality 

−  and 

−  xor 

−  or 

−  </pre> 

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

== Problem 48 == 
== Problem 48 == 

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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. 
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. 

−  <pre> 

Example: 
Example: 

+  
+  <pre> 

* (table (A,B,C) (A and (B or C) equ A and B or A and C)) 
* (table (A,B,C) (A and (B or C) equ A and B or A and C)) 

true true true true 
true true true true 

Line 127:  Line 81:  
fail fail true true 
fail fail true true 

fail fail fail true 
fail fail fail true 

+  </pre> 

Example in Haskell: 
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 

−  </pre> 

−  Solution: 

<haskell> 
<haskell> 

−   functions as in solution 46 

+  λ> tablen 3 (\[a,b,c] > a `and'` (b `or'` c) `equ'` a `and'` b `or'` a `and'` c) 

−  infixl 
+   infixl 3 `equ'` 
−  +  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 

−  table3 :: (Bool > Bool > Bool > Bool) > IO () 

+   infixl 7 `equ'` 

−  table3 f = putStrLn . unlines $ [show a ++ " " ++ show b ++ " " ++ show c ++ " " ++ show (f a b c) 

+  True True True True 

−   a < [True, False], b < [True, False], c < [True, False]] 

+  True True False True 

+  True False True True 

+  True False False False 

+  False True True False 

+  False True False False 

+  False False True False 

+  False False False False 

</haskell> 
</haskell> 

−  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. 

+  [[99 questions/Solutions/48  Solutions]] 

−  
+  
== Problem 49 == 
== Problem 49 == 

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

+  
<pre> 
<pre> 

n = 1: C(1) = ['0','1']. 
n = 1: C(1) = ['0','1']. 

Line 170:  Line 121:  
Find out the construction rules and write a predicate with the following specification: 
Find out the construction rules and write a predicate with the following specification: 

+  <pre> 

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

+  </pre> 

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

−  <pre> 

Example in Haskell: 
Example in Haskell: 

−  P49> gray 3 

−  ["000","001","011","010","110","111","101","100"] 

−  </pre> 

−  Solution: 

<haskell> 
<haskell> 

−  gray :: Int > [String] 

+  λ> gray 3 

−  gray 0 = [""] 

+  ["000","001","011","010","110","111","101","100"] 

−  gray n = let xs = gray (n1) in map ('0':) xs ++ map ('1':) (reverse xs) 

</haskell> 
</haskell> 

−  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 n1, prepend a 0 to each word, take the Gray code for n1 again, reverse it and prepend a 1 to each word. At last we have to append these two lists. 

+  [[99 questions/Solutions/49  Solutions]] 

−  (The [http://en.wikipedia.org/wiki/Gray_code Wikipedia article] seems to approve this.) 

+  
== Problem 50 == 
== Problem 50 == 

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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: 
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: 

<pre> 
<pre> 

−  *Exercises> huffman [('a',45),('b',13),('c',12),('d',16),('e',9),('f',5)] 

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

−  [('a',"0"),('b',"101"),('c',"100"),('d',"111"),('e',"1101"),('f',"1100")] 

</pre> 
</pre> 

−  Solution: 

+  Example in Haskell: 

<haskell> 
<haskell> 

−  import Data.List 

+  λ> 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")] 

−  data HTree = Leaf Char  Branch HTree HTree 

+  </haskell> 

−  deriving Show 

−  huffman :: (Ord a, Num a) => [(Char,a)] > [(Char,[Char])] 

+  [[99 questions/Solutions/50  Solutions]] 

−  huffman freq = sortBy cmpFst $ serialize $ 

−  htree $ sortBy cmpFst $ [(w, Leaf c)  (c,w) < freq] 

−  where htree [(_, t)] = t 

−  htree ((wx,tx):(wy,ty):rest) = 

−  htree $ insertBy cmpFst (wx + wy, Branch tx ty) rest 

−  cmpFst x y = compare (fst x) (fst y) 

−  serialize (Branch l r) = 

−  [(c, '0':code)  (c, code) < serialize l] ++ 

−  [(c, '1':code)  (c, code) < serialize r] 

−  serialize (Leaf c) = [(c, "")] 

−  </haskell> 

[[Category:Tutorials]] 
[[Category:Tutorials]] 
Latest revision as of 09:29, 9 February 2019
This is part of NinetyNine Haskell Problems, based on NinetyNine Prolog Problems.
Logic and Codes
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:
λ> table (\a b > (and' a (or' a b)))
True True True
True False True
False True False
False False False
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:
λ> table2 (\a b > a `and'` (a `or'` not b))
True True True
True False True
False True False
False False False
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:
λ> tablen 3 (\[a,b,c] > a `and'` (b `or'` c) `equ'` a `and'` b `or'` a `and'` c)
 infixl 3 `equ'`
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
 infixl 7 `equ'`
True True True True
True True False True
True False True True
True False False False
False True True False
False True False False
False False True False
False False False False
Problem 49
(**) Gray codes.
An nbit Gray code is a sequence of nbit 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 Nbit 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:
λ> gray 3
["000","001","011","010","110","111","101","100"]
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:
λ> 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")]