Introduction to Functional Programming – Exercises Week 2: "Recursion and Datatypes" | TDA555 / DIT440, LP1 2016 |

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Introduction to Functional Programming – Exercises Week 2: "Recursion and Datatypes" | TDA555 / DIT440, LP1 2016 |

Home | Schedule | Labs | Lectures | Exercises | Exam | About | FAQ | Fire | Forum | TimeEdit | YouTube | Links |

Here are some exercises designed to help you practice writing and reasoning
about recursive definitions and datatypes.
## 1. (*) The Maximum Function

## 2. Sum of squares.

## 3. (*) The Towers of Hanoi

## 4. Fibonacci Numbers

## 5. Factors.

## 6. (*) Defining Types

## 7. (*) Replication

Define a function that replicates a given word n times

The following exercises use*recursion over lists* which we have not yet covered in the lectures. Interested students can read about it in LYAH here and here.
You can also use the ## 8. (*) Multiplying list elements

## 9. Avoiding Duplicates

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!

Complete the following function definition:

-- maxi x y returns the maximum of x and y

(We call this function maxi, because there is a standard function max which does the same thing. Of course, you should not use it.)

Begin by writing a type signature for maxi, and a left hand side "equal to undefined". Before you write the definition, think of at least one property that you will use for testing your code.

You will need to consider two cases: what are they? Write the left hand sides, and make sure GHCi accepts them.

Define a function which computes the sum of the squares of the numbers from 1 to n.

-- sumsq n returns 1*1 + 2*2 + ... + n*n

*Hint:* use recursion to compute sumsq n from sumsq (n-1); **do not**
use a formula (such as the one below) to compute this without recursion.

Some say that sumsq n is equal to n (n+1) (2n + 1) / 6.

The Towers of Hanoi is an ancient puzzle, consisting of a collection of rings of different sizes, and three posts mounted on a base. At the beginning all the rings are on the left-most post as shown, and the goal is to move them all to the rightmost post, by moving one ring at a time from one post to another. But, at no time may a larger ring be placed on top of a smaller one!

Can you find a strategy for solving the puzzle based on recursion? That is, if you already know how to move n-1 rings from one post to another, can you find a way to move n rings?

If you try out your strategy, you will quickly discover that quite a large number of moves are needed to solve the puzzle with, say, five rings. Can you define a Haskell function

hanoi n

which computes the number of moves needed to move n rings from one post to another using your strategy? How many moves would be needed to solve the puzzle with ten rings? Legend has it that the original version of the puzzle has 32 rings, and is being solved at this very moment by Bhuddist monks in a monastery. When the puzzle is complete, the world will end.

**More difficult** (only do this if you want a challenge): Now
suppose we add a fourth post to the puzzle. Can you think of a strategy
which makes use of the fourth post? Define a Haskell function to compute the
number of moves your new strategy needs to solve the puzzle. How many moves are
now required to solve the puzzle with ten rings? How much closer would the end
of the world be if the monks added a fourth post to their puzzle?

Can you explain this behaviour? *Hint*: what kind of recursion do your
two strategies use?

The *Fibonacci numbers* are defined by

so the sequence of values begins 1, 1, 2, 3, 5, 8...

Write a recursive function

-- fib n computes the nth Fibonacci number

based on the mathematical definition above. Use it to compute the 10th, 15th, 20th, 25th, and 30th Fibonacci number. What do you notice? Can you explain this behaviour?

**More Difficult:** There is a faster way to compute Fibonacci numbers.
Suppose we define a function fibAux which satisfies the property

fibAux i (fib n) (fib (n+1)) == fib (n+i)

Notice that this is *not* a definition, it is a property!

*From this property* can you see how we might *redefine* fib by
using fibAux? (Hint: try setting n to 0 in the property).

It is possible to *derive* a recursive definition of fibAux from this
property just using algebra: the two cases we want are

fibAux 0 a b = ...

fibAux i a b | i>0 = ...

See if you can use the property to figure out what the right hand sides should be.

Use the new version of fib to compute the 10th, 15th, 20th, 25th, and 30th Fibonacci number. What do you notice?

Calculate fibAux 4 1 1 by hand. If you have programmed before, then observe the way the values of i, a, and b change in successive recursive calls. Does it remind you of anything?

A prime number p has only two factors, 1 and p itself. A composite number has more than two factors. Define a function

smallestFactor n

which returns the smallest factor of n larger than one. For example,

smallestFactor 14 ==
2

smallestFactor 15 == 3

*Hint:* write smallestFactor using an auxiliary function nextFactor k n which returns the
smallest factor of n larger than k. You can define smallestFactor using
nextFactor, and nextFactor by recursion.

Now define

numFactors n

which computes the number of factors of n in the range 1..n.

Define a data type Month to represent months, and a function

daysInMonth :: Month -> Integer -> Integer

which computes the number of days in a month, given also the year. (You can ignore leap centuries and the like: just assume that every fourth year is a leap year).

Define a data type Date, containing a year, month, and day, and a function

validDate :: Date -> Bool

that returns True if the day in the date lies between 1 and the number of days in the month.

repli :: Integer -> String -> StringExample:

*Main> repli 3 "apa" "apaapaapa"As always, try to make the base case as small as possible!

Use the following function to concatenate two strings:

(++) :: String -> String -> StringExample:

*Main> "flyg" ++ "plan" "flygplan"

The following exercises use

`List`

type defined in the lecture on lists. `List`

is basically the same as the built-in lists, only the syntax is slightly different.

Define a function

multiply :: Num a => [a] -> a

which multiplies together all the elements of a list. (Think: what should its value be for the empty list?). For example

Main> multiply [1,2,3,4,5]

120

(This is actually a standard function, called product).

If you do this for the `List`

data type in the lecture, your function will have the type:

multiply :: List Integer -> Integer

In many situations, lists should not contain *duplicate* elements. For
example, a pack of cards should not contain the same card twice. Define a
function

duplicates :: Eq a => [a] -> Bool

which returns True if its argument contains duplicate elements.

Main> duplicates [1,2,3,4,5]

False

Main> duplicates [1,2,3,2]

True

*Hint:* the standard function elem, which tests whether an element
occurs in a list, is helpful here.

One way to *ensure* a list contains no duplicates is to start with a
list that might contain duplicate elements, and remove them. Define a function

removeDuplicates :: Eq a => [a] -> [a]

which returns a list containing the same elements as its argument, but without duplicates. Test it using the following property:

prop_duplicatesRemoved :: [Integer] -> Bool

prop_duplicatesRemoved xs = not (duplicates (removeDuplicates
xs))

Does this property guarantee that removeDuplicates behaves correctly? If not, what is missing?

(removeDuplicates is actually a standard function, called nub).