CS1022 Computer Programming & Principles

1. CS1022Computer Programming & Principles Lecture 2 Combinatorics

2. Plan of lecture • Binomial expansion • Rearrangement theorem • Efficiency of algorithms CS1022

3. 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

4. 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

5. Binomial expansion (3) • (ab)n expansion contains terms (an –kbk) • Multiplying afrom (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

6. 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

7. Pascal’s triangle (2) • If we calculate numerical values we obtain 1 1 1 1 2 1 1 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) CS1022

8. 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

9. 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

10. 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

11. 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

12. 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

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

14. 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 CS1022

15. 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

16. 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 CS1022

17. 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

18. 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

19. Searching in a dictionary (2) Performance (in terms of comparisons) log22k= k CS1022

20. 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 CS1022

21. 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 mostg(n), written as O(g(n)), • If there is a positive constant c, |f(n)|  c|g(n)|, n N CS1022

22. 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 CS1022

23. 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 CS1022

24. Hierarchy of functions (2) • Growth of functions as a graph CS1022

25. 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

26. 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 CS1022