Queensland University of Technology CRICOS No. 00213J ITB001 Problem Solving and Programming Lecture 4 Faculty of Information Technology Queensland University

Queensland University of Technology

CRICOS No. 00213J

ITB001Problem Solving and Programming

Lecture 4

Faculty of Information Technology

Queensland University of Technology

CRICOS No. 00213J

a university for the worldrealR 2

Aims of this lecture

• Last week we saw how parameterised procedures allow us to reuse solutions

• We look now at how to break a large problem down so that it can be solved by repeated application of small steps

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References

• Concrete Abstractions: An Introduction to Computer Science Using Scheme, Sections 2.1–2.3

• Structure and Interpretation of Computer Programs, Sections 1.1.7–1.1.8, and 1.2

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An illustration

• How can we make a paper chain with n links?

1. Line up at least n people

2. Ask the person on the end for a chain of length n

3. For each person who is asked for a chain of length i, where i exceeds 1:• Ask the next person for a chain of length i 1• When the chain is returned, add one link to it and

give it to the person who asked– If asked for a chain of length 1, create a single link

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What happened?

• Each person had to join one link only• Each person did exactly the same thing (except the

person who created the single link)• When asked to make a chain, each person first got

someone else to do most of the work (except the person who created the single link)

• We took a problem of size n and broke it into n problems of size 1

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The recursion strategy

• This is an example of a powerful problem-solving strategy: Do nearly all the work first, then do the little bit that’s

left over

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A simple example of problem solving using the recursion strategy

• In last week’s exercises you wrote a small procedure called gst to calculate the Goods and Services Tax paid on a given price– Example: (gst 110) returns 10

• Now imagine that you have been asked to write a procedure called total-gst to calculate the total tax paid on a list of prices– Example: (total-gst (list 150 25 45))

should return 20

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Total GST: Analysing the problem

• Before attempting to solve the problem, we should summarise what we know:– We know that the desired result is the sum of the

taxes paid on each item in the given list– We know how to extract the first item in a list (using first) and get the remaining items (using rest)

– We already have a procedure, gst, capable of calculating the tax for a single item

• We should also consider some ‘extreme’ examples, such as what result to return if given an empty list

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Total GST: Designing a solution

• One way to solve this problem (but not the only one) is to calculate the tax paid for each item in the list and sum the results– To express this solution recursively, consider that the

desired result is the tax paid on the first item in the list plus the total tax paid for all the remaining items

– In other words, if we already know the total tax paid for all items in the list except the first, then all we need to do is calculate the tax for the first item and add it

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Total GST: Designing a solution

• Action refinement:

1. Return the sum of the tax paid for a list of items

1.1 If the list is empty return zero (Base Case)

1.2 Otherwise return the tax paid for the first item in the list added to the tax paid for the rest of the items (Recursive Case)

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Total GST: Implementation

• The following Scheme procedure calculates taxes using this recursive strategy: ; Sums the tax paid for a list of items (define [total-gst prices] ; If the list is empty return zero (if (empty? prices) 0 ; Add the tax paid for the first item to ; the tax paid for the rest of the items (+ (gst (first prices)) (total-gst (rest prices)))))

• Example:– (total-gst (list 5 15 6 7)) returns 3

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The process for calculating taxes

• The main steps in calculating the total tax paid on two items costing $22 and $55 are as follows:

1. (total-gst (list 22 55))2. (+ 2 (total-gst (list 55)))3. (+ 2 (+ 5 (total-gst (list))))4. (+ 2 (+ 5 0))5. (+ 2 5)6. 7

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Calculating factorials

• As another example of the recursive problem-solving strategy, consider calculating factorials, written ‘n!’

• The number of ways that n objects can be arranged is:

n! n (n 1) 1• By definition 0! 1 because zero objects can be

arranged in one way only

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• For example, 3 objects can be arranged in 3 2 1 6 ways

• But notice that 4 objects can be arranged in 4 3 2 1 24 ways, which is equal to 4 3!

• In general, therefore, if asked to calculate n! we can solve the problem by getting someone else to calculate (n 1)! and by then multiplying this by n

• If asked to calculate 0! we just return 1

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• The following Scheme procedure calculates factorials using this recursive strategy: (define [factorial n] (if (zero? n) 1 (* n (factorial (- n 1)))))

• Example:– (factorial 3) returns 6

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The process for calculating factorials

• The main steps in calculating 3! are as follows:

1. (factorial 3)2. (* 3 (factorial 2))3. (* 3 (* 2 (factorial 1)))4. (* 3 (* 2 (* 1 (factorial 0))))5. (* 3 (* 2 (* 1 1)))6. (* 3 (* 2 1))7. (* 3 2)8. 6

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Procedures and theprocesses they generate

• The form of a recursive procedure influences the efficiency of the corresponding computational process– When making paper chains, we had to wait until our

neighbours had produced their part before we could add a link

– When calculating taxes we had to wait until the subtotal for all other items was calculated before adding the tax on the first item

– In the factorial process each multiplication is deferred until the recursive procedure application returns

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Base cases

• Observations:– When making paper chains, the person who created a

single link behaved differently than everyone else– When calculating taxes we just returned zero if given

an empty list of prices– In the factorial procedure, no multiplication was

needed to calculate 0! (which equals 1 by convention)• These are called the base cases of our solution

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Recursive cases

• Observations:– When making paper chains, everyone except the

‘base case’ person added a link to an existing chain– When calculating taxes, we applied the gst

procedure to each price in the list– When calculating factorials, every number except for

the base case was multiplied by another number• These are called the recursive cases of our solution

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Designing recursive solutions

• Each recursive case should ‘break off’ a piece of the given problem, to create a subproblem that has the same form as the original problem– This ensures that the recursive procedure can make

use of itself– When asked to make a paper chain, each person

(except the last) asked their neighbour to also make a (shorter) chain

– When calculating the total tax, we made use of a subtotal

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Designing recursive solutions

• Each recursive case should create a subproblem that is closer to the base case than the given problem– This ensures that the recursive procedure eventually

terminates– When calculating the tax for a list, we first found the

subtotal for a shorter list, stopping when the list was empty

– When calculating n!, the factorial procedure’s recursive case decremented n, taking it closer to the base case of zero

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Exercise 4-1: Multiplicationby repeated addition

• Define a procedure called mult which accepts two non-negative numbers m and n and returns the value m n

• Challenge: You may use the pre-defined addition procedure ‘+’ but not the multiplication procedure ‘*’

• Hints:– Consider changing parameter n in each recursive

step– What is the base case value of n?– What do you need to do in each recursive step?

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Recursive definitions in mathematics

• Recursively-defined procedures express computations in the same way as in mathematics

• The following mathematical definition of factorials has a base case and a recursive case, just like our procedure

1, if n 0

n (n 1)!, otherwisen!

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“Like a circle in a spiral, Like a wheel within a wheel”

• But isn’t this definition circular?• No, it spirals towards the base case

– Each recursive step ‘breaks off’ a piece of the problem until only the base case is left

• Truly circular definitions don’t help us find solutions n! = n!

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Multiple base and recursive cases

• To solve more complicated problems, we may need more than just one base case and one recursive case

• Multiple base cases allow the computation to terminate in different ways

• Multiple recursive cases allow us to do different things at each step

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Exercise 4-2: Number ofoccurrences of an item in a list

• Write a recursive procedure called how-often with two parameters, a list items and a particular item x, which returns the number of occurrences of x in items

• Hint: You may need more than one recursive case• Example:

– (how-often (list 1 2 3 4 2 5 6 2 7) 2) returns 3

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“My head’s spinning”

• Trying to think about how a recursive procedure is evaluated is confusing

• Instead, consider just one layer of evaluation• For other layers use ‘wishful thinking’ and assume they

work as required– Since every layer is the same, this is sufficient to

show that your solution is correct– (This is the basis of inductive proofs in mathematics)

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Wishful thinking

• For instance, to understand the factorial procedure, just assume that the recursive application ‘(factorial (- n 1))’ returns the desired value (n 1)!

(define [factorial n] (if (zero? n) 1 (* n (n 1)!)))

• Now we can easily see that this procedure returns 1 if parameter n is zero and otherwise returns n (n 1)! which is n!, as desired

This is not a legal Scheme expression

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Deciding how big a step to take

• So far it has been obvious what constitutes a ‘step’ required to solve a problem– For constructing paper chains it was sensible to make

‘add one link’ a single step– For calculating total GST it was easiest to calculate

the tax on one price at a time– For calculating factorials it seemed reasonable to

decrease the number by one at each step• But for some problems the hard part is deciding how to

take a step

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Exercise 4-3: Number of digits

• Define a procedure called num-digits which, given a non-negative number n, returns the number of digits in n

• Hint: You will need to use the pre-defined integer division procedure quotient to make progress towards the base case, e.g., (quotient 487 10) returns 48

• Examples:– (num-digits 487) returns 3– (num-digits 6549) returns 4

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Exercise 4-3: Number of digits

• Extra challenge: Make your procedure work for negative numbers too

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Advanced topic: Tree recursion

• So far, all of our solutions have required only one recursive step at a time

• However, some problems require more than one recursive procedure application at each step

• This is known as tree recursion, due to the shape of the process it generates– Such processes are typically very inefficient

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Fibonacci numbers

• Fibonacci numbers model population growth in societies where deaths are rare (e.g., in bacterial colonies)

• They consist of the following sequence in which each number is the sum of its two predecessors

0, 1, 1, 2, 3, 5, 8, 13, 21, …• The nth Fibonacci number can be expressed by:

0, if n 0

1, if n 1

Fn - 1 Fn - 2, otherwise

Fn

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Exercise 4-4: Fibonacci numbers

• Fill in the blanks to create a procedure for calculating the nth (counting from zero) Fibonacci number: (define [fib n] (cond [(…) 0] [(…) 1] [else (+ (…) (…))]))

• Examples:– (fib 7) returns 13– (fib 8) returns 21– Try some bigger arguments—what happens?

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Further Exercises

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Exercise 4-5: The TwelveDays of Christmas

On the first day of Christmas,

My true love gave to me,

A partridge in a pear tree.

On the second day of Christmas,


Two turtle doves,

And a partridge in a pear tree.

On the third day of Christmas,


Three French hens,

Two turtle doves,

And a partridge in a pear tree.

Etcetera, ad nauseam!

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Exercise 4-5: The TwelveDays of Christmas

1. Define a procedure presents-on-day which, given a number n, returns the number of presents my true love gave me on the nth day of Christmas

2. Use your procedure presents-on-day to define a procedure presents-upto-day which returns the total number of presents my true love gave me up until a given day

3. Use your procedure presents-upto-day to find out how many presents I got over Christmas

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Exercise 4-6: Permutations

• The number of ways m objects can be arranged in n spaces is m! (m n)!– For example, given 52 playing cards we can deal 52!

47! 311,875,200 different 5-card hands, assuming the order of cards in the hand is significant

• Although we could calculate m! (m n)! by applying our factorial procedure twice and then dividing, this would not be very efficient because we would calculate term (m n)! twice

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Exercise 4-6: Permutations

• Define a recursive procedure permutations which, given two numbers m and n, where m n, calculates m! (m n)! efficiently

• Hint: In the playing card example, notice that the result is equal to 52 51 50 49 48

• Examples:– (permutations 6 6) returns 720– (permutations 6 1) returns 6– (permutations 6 3) returns 120– (permutations 4 2) returns 12

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Exercise 4-7: Length of longest word

• Write a procedure called longest-word-length which accepts a list of strings representing words and returns the length of the longest word in the list (or the special value 1 if the list is empty)

• Hint: The pre-defined procedure string-length returns the length of a string

• Example:– (longest-word-length

(list "See" "Spot" "run")) returns 4

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Exercise 4-8: Summing digits

• Write a procedure sum-digits that computes the sum of the digits in a given number

• Hint: The pre-defined procedure remainder returns the remainder after integer division, e.g., (remainder 472 10) returns 2

• Example:– (sum-digits 472) returns 13

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Exercise 4-9: Counting letters

• Write a procedure count-letters which, when given a string comprising letters and space characters only, returns the number of letters

• Hint: Count-letters should convert the string into a list and use another procedure to do the counting– The pre-defined procedure string->list turns a

string into a list of characters– The space character is written #\space

• Example:– (count-letters "My name is Fred")

returns 12

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Exercise 4-10: Quotient byrepeated subtraction

• Write a procedure quot which, given a non-negative numerator n and a positive denominator d, computes the integral part of n d by repeatedly subtracting d from n

• Hint: n d 0 when n d• Example:

– (quot 7 2) returns 3

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Exercise 4-11: Generalised quotient

• Extend your quot procedure to allow for the possibility that either, or both, of n and d are negative

• Hint: n d (n d) when d is negative n d (n d) when n is negative

• Example: (quot 7 -2) returns -3

• (This is an interesting example where the various recursive cases do quite different things)

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Exercise 4-12: Square rootsby Newton’s method

• For further study, look at the example of calculating square roots by Newton’s method in Structure and Interpretation of Computer Programs, Section 1.1.7

• This is a good example of a large problem involving recursion and several procedures

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Exercise 4-13: Extractingalphabetic characters from text

• Define a procedure extract-alphas which takes a string of text and returns a string containing the alphabetic characters only

• Hint: You will need to convert the string into a list and back again using procedures string->list and list->string

• Hint: Pre-defined predicate char-alphabetic? can be used to test if a character is alphabetic or not

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Exercise 4-13: Extractingalphabetic characters from text

• Example:– (extract-alphas "Hello, World!") returns "HelloWorld"

Documents

Queensland University of Technology CRICOS No. 00213J ITB001 Problem Solving and Programming Lecture 4 Faculty of Information Technology Queensland University