99 questions/54A to 60
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(Fix bug in P57) 
DonStewart (Talk  contribs) (expand on why it is trivially true) 

Line 23:  Line 23:  
</haskell> 
</haskell> 

−  The type system ensures that all terms of type <haskell>Tree a</haskell> are binary trees. Hence, it is redundant to introduce a predicate to check this property. 
+  The type system ensures that all terms of type <haskell>Tree a</haskell> 
−  +  are binary trees: it is just not possible to construct an invalid tree 

+  using with this type. Hence, it is redundant to introduce a predicate to 

+  check this property  the istree predicate is trivially true, for 

+  anything of type <hask>Tree a</hask>. 

+  
+  <haskell> 

+  istree :: Tree a > Bool 

+  istree _ = True 

+  </haskell> 

+  
+  Running this: 

+  
+  <haskell> 

+  *M> istree Leaf 

+  True 

+  
+  *M> istree (Branch 1 Leaf Leaf) 

+  True 

+  
+  *M> istree (Branch 1 Leaf (Branch 2 Leaf Leaf)) 

+  True 

+  </haskell> 

+  
== Problem 55 == 
== Problem 55 == 

Revision as of 23:18, 13 December 2006
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.
1 Problem 54A
(*) Check whether a given term represents a binary tree
Example:
* (istree (a (b nil nil) nil)) T * (istree (a (b nil nil))) NIL
The typical solution in Haskell is to introduce an algebraic data type:
data Tree a = Leaf  Branch a (Tree a) (Tree a) deriving Show
Tree a
are binary trees: it is just not possible to construct an invalid tree using with this type. Hence, it is redundant to introduce a predicate to check this property  the istree predicate is trivially true, for
anything of typeistree :: Tree a > Bool istree _ = True
Running this:
*M> istree Leaf True *M> istree (Branch 1 Leaf Leaf) True *M> istree (Branch 1 Leaf (Branch 2 Leaf Leaf)) True
2 Problem 55
(**) Construct completely balanced binary trees In a completely balanced binary tree, the following property holds for every node: The number of nodes in its left subtree and the number of nodes in its right subtree are almost equal, which means their difference is not greater than one.
Write a function cbaltree to construct completely balanced binary trees for a given number of nodes. The predicate should generate all solutions via backtracking. Put the letter 'x' as information into all nodes of the tree. Example:
* cbaltree(4,T). T = t(x, t(x, nil, nil), t(x, nil, t(x, nil, nil))) ; T = t(x, t(x, nil, nil), t(x, t(x, nil, nil), nil)) ; etc......No
Example in Haskell:
*Main> cbalTree 4 [Branch 'x' (Branch 'x' Leaf Leaf) (Branch 'x' Leaf (Branch 'x' Leaf Leaf)),Branch 'x' (Branch 'x' Leaf Leaf) (Branch 'x' (Branch 'x' Leaf Leaf) Leaf),Branch 'x' (Branch 'x' Leaf (Branch 'x' Leaf Leaf)) (Branch 'x' Leaf Leaf),Branch 'x' (Branch 'x' (Branch 'x' Leaf Leaf) Leaf) (Branch 'x' Leaf Leaf)]
Solution:
cbalTree 0 = [Leaf] cbalTree n = [Branch 'x' l r  i < [q .. q + r], l < cbalTree i, r < cbalTree (n  i  1)] where (q, r) = quotRem (n1) 2
Here we use the list monad to enumerate all the trees, in a style that is more natural than standard backtracking.
3 Problem 56
(**) Symmetric binary trees Let us call a binary tree symmetric if you can draw a vertical line through the root node and then the right subtree is the mirror image of the left subtree. Write a predicate symmetric/1 to check whether a given binary tree is symmetric. Hint: Write a predicate mirror/2 first to check whether one tree is the mirror image of another. We are only interested in the structure, not in the contents of the nodes.
Example in Haskell:
*Main> symmetric (Branch 'x' (Branch 'x' Leaf Leaf) Leaf) False *Main> symmetric (Branch 'x' (Branch 'x' Leaf Leaf) (Branch 'x' Leaf Leaf)) True
Solution:
mirror Leaf Leaf = True mirror (Branch _ a b) (Branch _ x y) = mirror a y && mirror b x mirror _ _ = False symmetric Leaf = True symmetric (Branch _ l r) = mirror l r
4 Problem 57
(**) Binary search trees (dictionaries) Use the predicate add/3, developed in chapter 4 of the course, to write a predicate to construct a binary search tree from a list of integer numbers. Example:
* construct([3,2,5,7,1],T). T = t(3, t(2, t(1, nil, nil), nil), t(5, nil, t(7, nil, nil)))
Then use this predicate to test the solution of the problem P56. Example:
* testsymmetric([5,3,18,1,4,12,21]). Yes * testsymmetric([3,2,5,7,1]). No
Example in Haskell:
*Main> construct [3, 2, 5, 7, 1] Branch 3 (Branch 2 (Branch 1 Leaf Leaf) (Branch 5 Leaf Leaf)) (Branch 7 Leaf Leaf) *Main> symmetric . construct $ [5, 3, 18, 1, 4, 12, 21] True *Main> symmetric . construct $ [3, 2, 5, 7, 1] True
Solution:
add :: Ord a => a > Tree a > Tree a add x Leaf = Branch x Leaf Leaf add x t@(Branch y l r) = case compare x y of LT > Branch y (add x l) r GT > Branch y l (add x r) EQ > t construct xs = foldl (flip add) Leaf xs
Here, the definition of construct is trivial, because the pattern of accumulating from the left is captured by the standard function foldl.
5 Problem 58
(**) Generateandtest paradigm Apply the generateandtest paradigm to construct all symmetric, completely balanced binary trees with a given number of nodes. Example:
* symcbaltrees(5,Ts). Ts = [t(x, t(x, nil, t(x, nil, nil)), t(x, t(x, nil, nil), nil)), t(x, t(x, t(x, nil, nil), nil), t(x, nil, t(x, nil, nil)))]
Example in Haskell:
*Main> symCbalTrees 5 [Branch 'x' (Branch 'x' Leaf (Branch 'x' Leaf Leaf)) (Branch 'x' (Branch 'x' Leaf Leaf) Leaf),Branch 'x' (Branch 'x' (Branch 'x' Leaf Leaf) Leaf) (Branch 'x' Leaf (Branch 'x' Leaf Leaf))]
Solution:
symCbalTrees = filter symmetric . cbalTree
6 Problem 59
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>
7 Problem 60
<Problem description>
Example: <example in lisp> Example in Haskell: <example in Haskell>
Solution:
<solution in haskell>
<description of implementation>