CS1022 Computer Programming &

Principles

Lecture 6.2Combinatorics (2)

Plan of lecture• Binomial expansion• Rearrangement theorem• Efficiency of algorithms

2CS1022

Binomial expansion (1)• Numbers C(n, k) arise naturally when we expand

expressions of the form (a b)n algebraically• For example:

(a b)3 (a b) (a b) (a b) aaa aab aba abb baa bab bba bbb a3 3a2b 3ab2 b3

• Terms are from multiplying variables from brackets– Three terms (abb, bab, bba) with one a and two b’s– This is because there are C(3, 2) = 3 ways of selecting

two b’s from the three brackets

3CS1022

Binomial expansion (2)• Coefficients in simplified expansion are– C(3, 0) 1, C(3, 1) 3, C(3, 2) 3, C(3, 3) 3

• To obtain values from C(n, k), we assume 0! = 1 – There is only one way to select nothing from a collection

4CS1022

Binomial expansion (3)• (a b)n expansion contains terms (an – kbk)– Multiplying a from (n – k) brackets and b from remaining

k brackets (k takes values from 0 up to n)– Since there are C(n, k) ways of selecting k brackets, – There are precisely C(n, k) terms of the form (an – kbk), for

k 0, 1, ..., n– Therefore

(a b)n C(n,0)an C(n,1)an – 1b C(n,2)an – 2b2 ... C(n, n)bn

• This formula is called binomial expansion– C(n, k), in this context, is called binomial coefficient

5CS1022

Pascal’s triangle (1)• Binomial coefficients arranged as Pascal’s triangle

C(0, 0)C(1, 0) C(1, 1)

C(2, 0) C(2, 1) C(2, 2) C(3, 0) C(3, 1) C(3, 2) C(3, 3)

C(4, 0) C(4, 1) C(4, 2) C(4, 3) C(4, 4) C(5, 0) C(5, 1) C(5, 2) C(5, 3) C(5, 4) C(5, 5)

... ... ...C(n, 0) C(n, 1) ... C(n, n – 1) C(n, n)

– Entry in row n 1 correspond to coefficients (in order) in the binomial expansion of (a b)n

6CS1022

Pascal’s triangle (2)• If we calculate numerical values we obtain

11 1

1 2 11 3 3 1

1 4 6 4 1 1 5 10 10 5 1

...– Since C(n, 0) C(n, n) 1, the outer edges of Pascal’s

triangle are indeed 1– Vertical symmetry also holds as C(n, k) C(n, n – k)

7CS1022

Rearrangement theorem (1)• Consider the problem of finding rearrangements of

letters in “DEFENDER”– There are 8 letters which can be arranged in 8! Ways– However, the 3 Es are indistinguishable – they can be

permuted in 3! ways without changing the arrangement– Similarly, 2 Ds can be permuted without changing the

arrangement– Therefore, the number of distinguishable arrangements

is just

8CS1022

3360!2!3

!8

Rearrangement theorem (2)• In general, there are

different arrangements of n objects of which– n1 are of type 1

– n2 are of type 2– and so on, up to– nr of type r

9CS1022

!!!

!

21 rnnn

n

Rearrangement theorem (3)• In how many ways can 15 students be divided into 3

tutorial groups with 5 students in each group?• Solution:– There are 15 objects to be arranged into 3 groups of 5– This can be done in

different ways

10CS1022

796,68!5!5!5

!15

Efficiency of algorithms• Issue: design & analysis of efficient algorithms– Two solutions: which one is better?– Is your algorithm the best possible?– What is “better”/“best”?– How can we compare algorithms?

• Measure time and space (memory)– Not size of algorithms (fewer lines not always good)

11CS1022

Towers of Hanoi (1)A “simple” problem/game: the Towers of Hanoi• 3 rods and n disks of different sizes• Start: disks in ascending order on one rod• Finish: disks on another rod

12CS1022

Rules:• Only one disk to be moved at a time•Move: take upper disk from a rod &

slide it onto another rod, on top of any other disks• No disk may be placed on top of a

smaller disk.

Towers of Hanoi (2)

13CS1022

Animation of solution for 4 disks:

Towers of Hanoi (3)• Many algorithms to solve this problem:

14CS1022

Alternate between smallest and next-smallest disksFor an even number of disks:

make the legal move between pegs A and Bmake the legal move between pegs A and Cmake the legal move between pegs B and C

repeat until completeFor an odd number of disks:

make the legal move between pegs A and Cmake the legal move between pegs A and Bmake the legal move between pegs B and C

repeat until complete To move n discs from peg A to peg C:move n − 1 discs from A to Bmove disc n from A to Cmove n − 1 discs from B to C

Towers of Hanoi (4)• How can we compare solutions?• What about counting the number of moves?– The fewer moves, the better!

• Instead of using a stop-clock, we count how often– The most frequent instruction/move is performed OR– The most expensive instruction/move is performed

• We simplify our analysis, so we don’t consider– Computer, operating system, or language we might use

15CS1022

Towers of Hanoi (5)• 64 disks require 264 moves • That’s 18,446,744,073,709,551,616 moves– If we carry out one move per millisecond, we have

18,446,744,073,709,551,616 ms

18,446,744,073,709,551.616 sec307,445,734,561,825.87 min

5,124,095,576,030.43 hrs213,503,982,334.60 days

584,942,417.36 years16CS1022

Estimating time• We don’t measure time with a stop-clock!– It’s an algorithm, so it won’t run...– Implement it and then time it (not good either)– Some algorithms take too long (centuries!)

• We estimate the time the algorithm takes as a function of the number of values processed– Simplest way: count, for an input of size n, the number

of elementary operations carried out• Important: same considerations about time apply to

space needed (memory)

17CS1022

Searching in a dictionary (1)• Search for word X in a dictionary with n words • Simple solution: sequential search– Check if X is the 1st word, if not check if it’s the 2nd, etc.– Worst case: X is not in dictionary (n comparisons)

• Another solution: binary search– Check if X is the “middle” word (words are ordered)– If not, decide if X could be on the first or second half – Dictionary always “half the size”– Worst case: 1 + log2n comparisons

18CS1022

n 1 + log2n

Searching in a dictionary (2)Performance (in terms of comparisons)

19CS1022

log22k= k

n 1 + log2n8 4n 1 + log2n8 4

64 7

n 1 + log2n8 4

64 7218 250,000 19

A B C D En 3n2 2n2 + 4n n3 2n

1 1ms 3ms 6ms 1ms 2ms10 10ms 300ms 240ms 1s 1.024s100 100ms 30s 20.4s 0.28h 41017 centuries1000 1000ms 0.83h 0.56h 11.6 days 10176 centuries

A B C D En 3n2 2n2 + 4n n3 2n

1 1ms 3ms 6ms 1ms 2ms10 10ms 300ms 240ms 1s 1.024s100 100ms 30s 20.4s 0.28h 41017 centuries1000 1000ms 0.83h 0.56h 11.6 days 10176 centuries

Comparing algorithms (1)

20CS1022

• Suppose we have a choice of algorithms A, B, C, D, E require n, 3n2, 2n2 + 4n, n3, 2n

elementary operations (respectively)– Each operation takes 1 millisecond

• Find overall running time for n = 1, 10, 100 & 1000

Comparing algorithms (2)

21CS1022

• Table shows huge difference in times– Formulae with powers of n (polynomial functions)– Formulae with powering by n (exponential functions)

• When the same highest power of n is involved, running times are comparable– For instance, B and C in the previous table

• If f(n) and g(n) measure efficiency of 2 algorithms– They are called time-complexity functions

• f(n) is of order at most g(n), written as O(g(n)),– If there is a positive constant c, |f(n)| c|g(n)|, n N

Comparing algorithms (3)

22CS1022

• Suppose two functions f(n) and g(n), such that |f(n)| c1|g(n)| and |g(n)| c2|f(n)|

one is of order at most the other• They have same order of magnitude• Their times are comparable– They perform, when n is big enough, similarly

B C3n2 2n2 + 4n

1 3ms 6ms10 300ms 240ms100 30s 20.4s1000 0.83h 0.56h

Hierarchy of functions (1)

23CS1022

• We can define a hierarchy of functions– Each has greater order of magnitude than predecessors

• Example of hierarchy:

• Left-to-right: greater order of magnitude– As n increases, the value of latter functions increases

more rapidly

1 log n n n2... nk... 2n

constant

logarithmic

polynomial

exponential

Hierarchy of functions (2)

24CS1022

• Growth of functions as a graph

Complexity of functions

25CS1022

• We compare algorithms based on the complexity of the function describing their performance

• We “simplify” a function to work out which curve (or which class in the hierarchy) “bounds” it

• Example: f(n) = 9n + 3n6 + 7 log n – Constants do not affect magnitudes, so

9n is O(n) 3n6 = O(n6) 7 log n = O(log n)– Since n and log n occur earlier than n6 in hierarchy, we

say that 9n and 7 log n are in O(n6)– Hence f(n) is O(n6), as the fastest growing term is 3n6

– The function increases no faster than function n6

Further reading• R. Haggarty. “Discrete Mathematics for

Computing”. Pearson Education Ltd. 2002. (Chapter 6)

• Combinatorics @ Wikipedia• Analysis of algorithms @ Wikipedia• 100 solutions to the “Tower of Hanoi”

problem

26CS1022