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**CS1022 Computer Programming & Principles**

Lecture 6.2 Combinatorics (2)

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**Plan of lecture Binomial expansion Rearrangement theorem**

Efficiency of algorithms CS1022

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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 CS1022

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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 CS1022

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

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 CS1022

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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 CS1022

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**Pascal’s triangle (2) If we calculate numerical values we obtain 1 1 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) CS1022

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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 CS1022

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**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 CS1022

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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 different ways CS1022

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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) CS1022

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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. CS1022

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Towers of Hanoi (2) Animation of solution for 4 disks: CS1022

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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: 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 CS1022

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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 CS1022

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**584,942,417.36 years 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, sec 307,445,734,561, min 5,124,095,576, hrs 213,503,982, days 584,942, years CS1022

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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) CS1022

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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 CS1022

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**Searching in a dictionary (2)**

Performance (in terms of comparisons) n 1 + log2n 8 4 64 7 n 1 + log2n 8 4 n 1 + log2n n 1 + log2n 8 4 64 7 218 250,000 19 log22k= k CS1022

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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 B C D E n 3n2 2n2 + 4n n3 2n 1 1ms 3ms 6ms 2ms 10 10ms 300ms 240ms 1s 1.024s 100 100ms 30s 20.4s 0.28h 41017 centuries 1000 1000ms 0.83h 0.56h 11.6 days 10176 centuries A B C D E n 3n2 2n2 + 4n n3 2n 1 1ms 3ms 6ms 2ms 10 10ms 300ms 240ms 1s 1.024s 100 100ms 30s 20.4s 0.28h 41017 centuries 1000 1000ms 0.83h 0.56h 11.6 days 10176 centuries CS1022

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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 CS1022

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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 CS1022

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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: 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 CS1022

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**Hierarchy of functions (2)**

Growth of functions as a graph CS1022

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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 CS1022

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Further reading R. Haggarty. “Discrete Mathematics for Computing”. Pearson Education Ltd (Chapter 6) Wikipedia Analysis of Wikipedia 100 solutions to the “Tower of Hanoi” problem CS1022

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