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Exercises for Week 5: Recursive Datatypes

Here are some exercises designed to help you practice programming with recursive datatypes.

If you do not have time to do all these exercises, don't worry. The exercises are intended to provide enough work to keep the most experienced students busy. If you do all exercises marked with an (*) you have probably understood this week's material.

Good luck!

0 (*). Expression and Integer Trees


From Thompson's book, read Chapter 14, sections 14.2 to 14.4. Alternatively, you can take a look at the datatype Expr for expressions with variables from the lecture: ExprVar.hs.

Among other things, the book introduces a datatype Expr

data Expr = Lit Int
          | Add Expr Expr
          | Sub Expr Expr
which is similar to the type I used in the lecture. Given an expression we want to evaluate it. This is done using eval defined by
eval :: Expr -> Int
eval (Lit n)     = n
eval (Add e1 e2) = eval e1 + eval e2
eval (Sub e1 e2) = eval e1 - eval e2
We also want be able to show an expression

showExpr :: Expr -> String
showExpr (Lit n)     = show n
showExpr (Add e1 e2) = "(" ++ showExpr e1 ++ "+" ++ showExpr e2 ++")"
showExpr (Sub e1 e2) = "(" ++ showExpr e1 ++ "-" ++ showExpr e2 ++")"

A. Give calculations of

eval (Num 67)
eval (Add (Sub (Lit 3) (Lit 1)) (Lit 3))
showExpr (Add (Lit 67) (Lit -34))

B(*). Define the function

size :: Expr -> Int
which counts the number of operators in an expression.

C(*). Add the operations of multiplication and integer division to the type Expr, and redefine the functions eval,showExpr, and size to include the new cases. What does your eval do when you divide by zero? Write one version of eval with the result type Maybe Int.

D. Instead of adding extra constructors to the Expr datatype as in C it is possible to factor the definitions

data Expr = Lit Int
          | Op Ops Expr Expr

data Ops = Add | Sub | Mul | Div
Show how the functions eval,showExpr, and size are defined for this type. How would you add yet another extra operation Mod for remainder on integer division?

E. In Haskell back-quotes allow us to use constructor in infix form (indeed any function) like in (Lit 3) `Add` (Lit 15). However, if this expression is shown it appears in prefix form as Add (Lit 3) (Lit 15).

It is also possible to use infix operators for constructor names where the first character must be a ':'. We can, e.g., redefine Expr as

data Expr = Lit Int
          | Expr :+: Expr
          | Expr :-: Expr
Redefine the above functions using this datatype with infix constructors.

Integer Trees

A tree of integers is either nil or given by combining a value and two sub-trees. In Haskell we can introduce the datatype of integer trees NTree by

data NTree = NilT
           | Node Int NTree NTree
We can sum the nodes and calculate the depths of a tree as follows
sumTree :: NTree -> Int
sumTree NilT           = 0
sumTree (Node n t1 t2) = n + sumTree t1 + sumTree t2

depth :: NTree -> Int
depth NilT           = 0
depth (Node n t1 t2) = 1 + max (depth t1) (depth t2)

A. Give a calculation of

sumTree (Node 3 (Node 4 NilT NilT) NilT)
depth (Node 3 (Node 4 NilT NilT) NilT)

B(*). Define functions to return the left- and right-hand sub-trees of an NTree.

C(*). Define a function to decide whether a number is an element of an NTree.

D. Define functions to find the maximum and minimum values held in an NTree.

E(*). A tree is reflected by swapping left and right sub-trees, recursively. Define a function to reflect an NTree. What is the result of reflecting twice? Write a QuickCheck property for that!

(In order to be able to test your properties, you have to make NTree an instance of Arbitrary.)

F. Define functions

collapse, sort :: NTree -> [Int]
which turn a tree into a list. The function collapse should enumerate the left sub-tree, then the value at the node and finally the right sub-tree; sort should sort the elements in ascending order. For example,
collapse (Node 3 (Node 4 NilT NilT) NilT) = [4,3]
sort (Node 3 (Node 4 NilT NilT) NilT) = [3,4]
Write a suitable QuickCheck property for your functions!

1 (*). File Systems

A file either contains data or is a directory. A directory contains other files (which may themselves be directories) along with a name for each one.

A. Design a data type to represent the contents of a directory. Ignore the contents of files: you are just trying to represent file names and the way they are organised into directories here.

B. Define a function to search for a given file name in a directory. You should return a path leading to a file with the given name. Thus if your directory contains a, b, and c, and b is a directory containing x and y, then searching for x should produce b/x.

2. Exercises on Propositional Logic

A proposition is a boolean formula of one of the following forms:

where p and q are propositions. For example, p | ~p is a proposition.

A. Design a data type Prop to represent propositions.

B. Define a function

vars :: Prop -> [String]

which returns a list of the variables in a proposition. Make sure each variable appears only once in the list you return.

Suppose you are given a list of variable names and their values, of type Bool, for example, [("p",True),("q",False)]. Define a function

truthValue :: Prop -> [(String,Bool)] -> Bool

which determines whether the proposition is true when the variables have the values given.

C. Define a function

tautology :: Prop -> Bool

which returns true if the proposition holds for all values of the variables appearing in it.

Congratulations! You have implemented a simple theorem prover.

3 (*). Exercises on type Expr from the Lecture

(This exercise consists of writing QuickCheck properties, and checking them using QuickCheck. Define the properties in the groups exercises, but you can leave the QuickChecking of them until a later time, if you do not have access to a computer in your group room.)

Take a look at the datatype Expr for expressions with variables from the lecture: ExprVar.hs.

In the lecture, I showed a function for differentiating expressions, called diff (Swedish: "derivat").

  diff :: Expr -> Name -> Expr
  diff (Num n)   x = Num 0
  diff (Add a b) x = Add (diff a x) (diff b x)
  diff (Mul a b) x = Add (Mul a (diff b x)) (Mul b (diff a x))
  diff (Var y)   x
    | x == y       = Num 1
    | otherwise    = Num 0

A. Define a property that checks that, for each expression e, the derivative of e to x does not contain more variables than e.

Does the opposite hold also?

The result of the diff function often contains expressions that can be enormously simplified.

B.(*) Define a function simplify that, given an expression e, creates an expression that is equivalent to e, but simplified. Examples of simplifications you could do are:

  • 2+3 --> 5

  • 2*x+6+5*x+6 --> 7*x+12

  • 0*x+-2+5*y+3 --> 5*y+1
  • You can decide yourself how ambitious you want to be!

    We have made a small start for you in the file ExprVar.hs

    Hint: This exercise is more open-ended.

    You can try to define simplify recursively. You will however notice that it is difficult to accomplish many simplifications. Take a look at the function assoc in chapter 14 in Thompson's book:

    assoc :: Expr -> Expr
    assoc (Add (Add e1 e2) e3) = assoc (Add e1 (Add e2 e3))
    assoc (Add e1          e2) = Add (assoc e1) (assoc e2)
    assoc (Sub e1 e2)          = Sub (assoc e1) (assoc e2)
    assoc (Lit n)              = Lit n

    (A different, but more ambitious, approach is to design a new type, that represents a normal form for expressions, for example polynomials. Your simplify function could simply transform an expression into a polynomial, and back into expressions again. How would you model polynomials over multiple variables as a type in Haskell?)

    Make sure that the following property holds for your simplify function: For each expression e, evaluating it in an environment generates the same result before and after simplifying:

      prop_SimplifyCorrect e (Env env) =
        eval env e == eval env (simplify e)

    C. Define a property:

      prop_SimplifyNoJunk :: Expr -> Bool
    that checks that the result of simplification does not "leave any junk". In other words, the result of simplification should for example not have subexpressions of the form:

  • Add (Num n) (Num m)

  • Mul (Num 0) b

  • Add a a
  • And so forth. What is allowed as "junk" and what is not of course depends on your simplification function, and what you expect from it.

    This property boils down to defining a function noJunk:: Expr -> Bool. We have already made a small start for you in ExprVar.hs.

    D. Define a property:

      prop_SimplifyDiff :: Expr -> Bool
    that checks that differentiating an expression and then simplifying it should have the same result as simplifying it first, then deriving, and then simplifying.

    Do you expect it to hold? Does it actually hold for your simplification function?