Functional Programming -- Exercises 5 | TDA452 & DIT142 | LP2 | HT2015 | [Home] |

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!

**Expressions**

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**

which is similar to the type I used in the lecture. Given an expression we want to evaluate it. This is done usingdata Expr = Lit Int | Add Expr Expr | Sub Expr Expr

We also want be able to show an expressioneval :: Expr -> Int eval (Lit n) = n eval (Add e1 e2) = eval e1 + eval e2 eval (Sub e1 e2) = eval e1 - eval e2

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

**B(*).** Define the function

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

**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

Show how the functionsdata Expr = Lit Int | Op Ops Expr Expr data Ops = Add | Sub | Mul | Div

**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

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

**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

We can sum the nodes and calculate the depths of a tree as followsdata NTree = NilT | Node Int NTree NTree

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

which turn a tree into a list. The functioncollapse, sort :: NTree -> [Int]

Write a suitable QuickCheck property for your functions!collapse (Node 3 (Node 4 NilT NilT) NilT) = [4,3] sort (Node 3 (Node 4 NilT NilT) NilT) = [3,4]

**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.

- a variable name (a string)
- p & q
*(and)* - p | q
*(or)* - ~p
*(not)*

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.

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:

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 :: Expr -> Env -> Property prop_SimplifyCorrect e (Env env) = eval env e === eval env (simplify e)

**C.** Define a property:

prop_SimplifyNoJunk :: Expr -> Boolthat 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: 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 -> Boolthat 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?