CS1022
Computer Programming &
Principles
Lecture 2
Combinatorics
Plan of lecture
• Binomial expansion
• Rearrangement theorem
• Efficiency of algorithms
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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
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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
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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
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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
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Pascal’s triangle (2)
• If we calculate numerical values we obtain
1
1
1
1
1
2
3
1
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)
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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
8!
 3360
3!2!
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Rearrangement theorem (2)
• In general, there are
n!
n1!n 2! nr!
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
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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
15!
 68,796
5!5!5!
different ways
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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)
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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
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.
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Towers of Hanoi (2)
Animation of solution for 4 disks:
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Towers of Hanoi (3)
• Many algorithms to solve this problem:
Alternate between smallest and next-smallest disks
For an even number of disks:
 make the legal move between pegs A and B
 make the legal move between pegs A and C
 make the legal move between pegs B and C
repeat until complete
For an odd number of disks:
 make the legal move between pegs A and C
 make the legal move between pegs A and B
 make 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 B
 move disc n from A to C
 move n − 1 discs from B to C
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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
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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 sec
307,445,734,561,825.87 min
5,124,095,576,030.43 hrs
213,503,982,334.60 days
584,942,417.36 years
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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)
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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
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Searching in a dictionary (2)
Performance (in terms of comparisons)
n
8
64
218  250,000
1 + log2n
4
7
19
log2
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k
2=
k
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Comparing algorithms (1)
• 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
A
n
1
1ms
10
10ms
100 100ms
1000 1000ms
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B
C
D
E
3n2 2n2 + 4n
2n
n3
3ms
6ms
1ms
2ms
300ms 240ms
1s
1.024s
30s
20.4s
0.28h 41017 centuries
0.83h 0.56h 11.6 days 10176 centuries
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Comparing algorithms (2)
• 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
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Comparing algorithms (3)
• 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
C
3n2 2n2 + 4n
1
3ms
6ms
10 300ms 240ms
100
30s
20.4s
1000 0.83h 0.56h
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Hierarchy of functions (1)
• We can define a hierarchy of functions
– Each has greater order of magnitude than predecessors
• Example of hierarchy:
1
log n
n n2... nk...
2n




constant logarithmic polynomial exponential
• Left-to-right: greater order of magnitude
– As n increases, the value of latter functions increases
more rapidly
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Hierarchy of functions (2)
• Growth of functions as a graph
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Complexity of functions
• 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
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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
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