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Function Definition by Cases and RecursionPowerPoint Presentation

Function Definition by Cases and Recursion

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Definitions Revisited

- A definition
- double :: Int -> Int
- double x = 2*x
- makes a true statement about the function defined,
- (whatever x is, then double x and 2*x are equal)
- gives a way of computing calls of the function.

Quiz

Given the definition

x :: Int

x*x = 4

Is x equal to 2?

NO! This is not a valid Haskell definition.

It makes a true statement about x, but it does not

give a way of computing x.

Computing with Definitions

- A function call is computed by
- replacing the call with a copy of the right hand side,
- with the argument names replaced by the actual arguments.

double :: Int -> Int

double x = 2*x

double 8 2*8

16

Evaluation Order

There may be more than one way to evaluate an expression:

double 8

double (3+5)

2*8

16

2*(3+5)

You can use any order of evaluation; they all give the same result. Haskell chooses a suitable one; you don’t need to know which.

Sharing Evaluation

double :: Int -> Int

double x = x+x

double (3*5)

Is it more work to

evaluate the expression

in this order?

(3*5)+(3*5)

double 15

15+(3*5)

15+15

30

Sharing Evaluation

double :: Int -> Int

double x = x+x

double (3*5)

NO!

Haskell `remembers´

that both occurrences

of 3*5 are really the

same, and evaluates

both in one step.

(3*5)+(3*5)

double 15

15+15

30

Definition by Cases

Often programs must make decisions, and compute different results in different cases.

Example: Define max x y to return the maximum of its

two arguments.

If x <= y, then max x y should be y.

If x>y, then max x y should be x.

The Type of Booleans

We make a decision by asking: does a condition hold?

(e.g. Does x<=y hold?)

A condition is either true or false: this is a piece of data, a value!

We introduce a new basic type with two values, named after the mathematician George Boole:

True, False :: Bool

Constants begin with a capital letter.

Some Operators Producing Booleans

Note two equals signs, to

avoid confusion with a

definition.

2 <= 3 True

2 > 3 False

2 < 3 True

2 == 3 False

2 /= 3 True

Not equals.

Functions Returning Booleans

Functions can return boolean results (or any other type).

Example:

inOrder :: Int -> Int -> Int -> Bool

inOrder x y z = x <= y && y <= z

a && b is True if

both a and b are True.

Using Booleans to Define Functions by Cases

max :: Int -> Int -> Int

max x y | x <= y = y

max x y | x > y = x

max :: Int -> Int -> Int

max x y | x <= y = y

| x > y = x

OR

A guard: an expression of

type Bool.

If the guard is True,

the equation applies.

Evaluation with Guards

- To evaluate a function call,
- evaluate each guard in turn until one is True,
- replace the call with a copy of the right hand side following the true guard.

max 4 2 ?? 4 <= 2 False

?? 4 > 2 True

4

max :: Int -> Int -> Int

max x y | x <= y = y

| x > y = x

Is max Correct?

Programming is a very error prone process; programs are rarely correct `first time´.

A large part of the cost of software development goes on finding and correcting errors.

It is essential to test software: try it on a variety of inputs and see if the output is correct.

Choosing Test Data

Test data should be chosen carefully, to include `difficult´ cases that might induce a failure.

The max function should be tested at least with x<y, x==y, x>y, and probably combinations of positive and negative arguments.

Choose enough test examples so that every case in your program is used at least once!

Dijkstra on Testing

”Testing can never demonstrate the absence of errors in software, only their presence”

Edsger W. Dijkstra

(but it is very good at the latter).

Specifications

What do we mean by `max is correct´?

A specification formulates properties we expect max to satisfy.

Property: x <= max x y

Property: y <= max x y

Why Formulate Specifications?

- Helps us clarify what max is supposed to do.
- Can help in testing.
- Enables us to prove programs correct.

Specifications and Testing

We can define function to check whether properties hold.

prop_Max :: Int -> Int -> Bool

prop_Max x y = x <= max x y && y <= max x y

If prop_Max always returns True, then the specification is satisfied.

We can test max on many inputs without needing to inspect the results by hand.

Testing with QuickCheck

QuickCheck is a tool to help you test your programs.

Main> quickCheck prop_Max

OK, passed 100 tests

quickCheck generates random values to test your property thoroughly.

Testing with QuickCheck (2)

- What if we make a mistake?
- max x y | x <= y = x
- | x > y = y

Main> quickCheck prop_Max

Falsifiable, after 0 tests

1

0

Specifications and Proofs

From the definition of max:

x <= y ==> max x y = y

x > y ==> max x y = x

Theorem: x <= max x y

Proof: Consider two cases:

Case x <= y: y = max x y, so x <= max x y.

Case x > y: max x y = x and x <= x, so x <= max x y.

Formal Methods

- Proofs are costly and also error-prone, but can guarantee correctness.
- Thorough testing is the most common method today.
- Customers for safety critical software demand proofs today.
- Proofs of correctness will play a growing role, thanks to
- automatic tools to help with proving,
- demand for better quality software.

Quiz

- Define
- abs x to return the absolute value of x (e.g. abs 2 = 2, abs (-3) = 3.
- sign x to return 1 if x is positive, and -1 if x is negative.
- State (and prove?) a property relating abs and sign.

Quiz Answer

abs x | x <= 0 = -x

| x > 0 = x

sign x | x < 0 = -1

| x > 0 = 1

| x == 0 = 0

Property: x == sign x * abs x

Did you consider this case?

This can also be written

sign 0 = 0

Recursion

Problem: define fac :: Int -> Int

fac n = 1 * 2 * … * n

What if we already know the value of fac (n-1)?

Then fac n = 1 * 2 * … * (n-1) * n

= fac (n-1) * n

A Table of Factorials

Must start somewhere:

we know that fac 0 = 1.

n fac n

0 1

1 1

2 2

3 6

4 24

...

So fac 1 = 1 * 1.

So fac 2 = 1 * 2.

So fac 3 = 2 * 3.

A Recursive Definition of Factorial

Base case.

fac :: Int -> Int

fac 0 = 1

fac n | n > 0 = fac (n-1) * n

Recursive case.

Evaluating Factorials

fac :: Int -> Int

fac 0 = 1

fac n | n > 0 = fac (n-1) * n

fac 4 ?? 4 == 0 False

?? 4 > 0 True

fac (4-1) * 4

fac 3 * 4

fac 2 * 3 * 4

fac 1 * 2 * 3 * 4

fac 0 * 1 * 2 * 3 * 4

1 * 1 * 2 * 3 * 4

24

There is No Magic!

What if we define

fac :: Int -> Int

fac n = div (fac (n+1)) (n+1) ?

fac 4 div (fac 5) 5

div (div (fac 6) 6) 5

div (div (div (fac 7) 7) 6) 5

...

A true statement.

Not a useful

definition.

Primitive Recursion

- Define
- f n in terms of f (n-1), for n > 0.
- f 0 separately.
- What if I already know the value of f (n-1)?
- Can I compute f n from it?

Quiz

Define a function power so that

power x n == x * x * … * x

n times

(Of course, power x n == x^n, but you should define power without using ^).

Quiz

Define a function power so that

power x n == x * x * … * x

n times

Don’t forget the base case!

power x 0 = 1

power x n | n > 0 = power x (n-1) * x

Since this equals

(x * x * … * x) * x

n-1 times

General Recursion

What if I know the values of f x for all x less than n?

Can I compute f n from them?

Example

x^(2*n) == (x*x)^n

x^(2*n+1) == (x*x)^n * x

Power Using General Recursion

Base case is still needed.

power :: Int -> Int -> Int

power x 0 = 1

power x n

| n `mod` 2 == 0 = power (x*x) (n `div` 2)

| n `mod` 2 == 1 = power (x*x) (n `div` 2) * x

Two recursive cases.

Why might this definition of power be preferred?

Comparing the Versions

First Version

power 3 5

power 3 4 * 3

power 3 3 * 3 * 3

power 3 2 * 3 * 3 * 3

power 3 1 * 3 * 3 * 3 * 3

power 3 0 * 3 * 3 * 3 * 3 * 3

1 * 3 * 3 * 3 * 3 * 3

243

Second Version

power 3 5

power 9 2 * 3

power 81 1 * 3

power 81 0 * 81 * 3

1 * 81 * 3

243

4 function calls,

4 multiplications.

6 function calls,

5 multiplications.

A More Difficult Example

Define prime :: Int -> Bool, so that prime n is True if n is a prime number.

What if we know whether (n-1) is prime?

What if we know whether each smaller number is prime?

NO HELP!

Generalise the Problem!

n is prime means No k in the range 2<=k<n

divides n.

Generalisation

Replace 2 by a variable.

Define

factors m n == True if Some k in the range m<=k<n

divides n.

So prime n = not (factors 2 n)

not x is True if x is False,

and vice versa.

Recursive Decomposition

Problem: Does any k in the range m<=k<n divide n?

What if we know whether any k in a smaller range divides n?

Some k in the range m<=k<n divides n

if m divides n,

or some k in the range m+1<=k<n divides n.

Recursive Solution

factors :: Int -> Int -> Bool

factors m n

| m == n = False

| m < n = divides m n || factors (m+1) n

divides :: Int -> Int -> Bool

divides m n = n `mod` m == 0

There is no k in the

range n<=k<n.

x || y is True

if x is True or y is True.

What is Getting Smaller?

The range m<=k<n contains n-m elements. Call this the problem size.

factors m n

| m == n = False

| m < n = divides m n || factors (m+1) n

Base case: n-m == 0

Recursive case: n-(m+1) == (n-m)-1

The problem size gets smaller in each call, until it reaches zero. So recursion terminates.

Lessons

- Recursion lets us decompose a problem into smaller subproblems of the same kind -- a powerful problem solving tool in any programming language!
- A more general problem may be easier to solve recursively than a `simpler´ one, because the recursive calls can do more.
- To ensure termination, define a `problem size´ which must be greater than zero in the recursive cases, and decreases by at least one in each recursive call.

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