1. Page 1 Recursion and Exhaustion Yip Lik Yau

2. Page 2 Why Exhaustion?  Many problems can be solved by brute force in a reasonable amount of time if the problem size is small  Some problems have no known “fast” algorithm  Example: Traveling Salesman, Hamiltonian Path, Graph Isomorphism…  Estimating the time needed for brute force let us decide whether to use exhaustion or not  Example: O(2 N ) is okay for N~30, O(N!) is okay for N~13  Knowing what types of problems are “hard” prevents you from wasting time in competitions

3. Page 3 Problem: Generating Permutations  Given N, generate all permutations of integers 1..N  When N = 3: 123 132 213 231 312 321

4. Page 4 Problem: Generating Permutations  Nested For loop is out of the question  We need a more advanced method – recursion  Outline of algorithm: var used: array[0..100] of boolean; n: integer; procedure run(stage : integer); begin if (level = n) do_something; else for i = 1 to n do if not used[i] begin used[i] = true; run(stage+1); used[i] = false; end;

5. Page 5 Problem: Generating Permutations  Search Tree 1 23 2 13 3 12 323121

6. Page 6 Be Jewelled!  ``Be Jewelled!'' is a popular Palm game. You are given 81 jewels arranged in a 9x9 matrix. Two adjacent jewels are allowed to be swapped if and only if, after the swapping, some jewels can be counted for scoring.  You can make one such swapping and get score. The score is calculated according to the following rules:  1. If there are any three or more jewels of the same kind lined up horizontally, these jewels will be counted for scoring.  2. If there are any three or more jewels of the same kind lined up vertically, these jewels will be counted for scoring.  If a jewel fulfils both rule 1 and 2, this jewel will count only once for scoring. The score gained = 5*3^(number of scoring jewelled-3)

7. Page 7 Be Jewelled! PQPQPQ QPXYXP PYYXYQ QYQYQP PQPYPQ QPQPQP PQPQPQ QPXXXP PYYYYQ QYQYQP PQPYPQ QPQPQP The jewels in red will be counted for score: 5*3^(9-3)=3645 Swapping the jewels in red.

8. Page 8 General Idea of Recursion  To solve a problem, break it into smaller, similar instances  Suppose we can solve the smaller instances, then combining these results can solve the original problem  For the algorithm to terminate, we also need to know how to solve the base case

9. Page 9 The N-Queens Problem  In a NxN chessboard, place N queens so that no two of them can attack each other  One brute force algorithm is to find all permutations of 1..N  Example for N=3: permutation is 132, then we place the queens at (1,1), (2,3) and (3,2)  The do_something at base case is to check whether 2 queens are attacking  Optimization: when placing each queen, we can immediately check if it has violated the constraint  Cuts off useless branches quickly Q Q Q Q Q Q Q Q

10. Page 10 Estimating Time Complexity of Exhaustion  The time complexity for the previous 2 algorithms are easy  First stage has N choices, second stage has N-1 choices, …  Total time = N*(N-1)*(N-2)*…*2*1 = N!  If still not convinced, look at the search tree  How about other problems?  Start at the top-left corner, move to the bottom-right corner  Each step can go down or right only  Maximize sum of numbers in the path 15425 69281 47614 24312

11. Page 11 Two useful mathematical concepts  Permutations (n P r) : number of ways to arrange r out of n objects with ordering  5 books, but only 3 slots on a bookshelf, how many permutations?  The first slot has 5 choices  The second slot has 4 choices  The third slot has 3 choices  Total number of permutations = 5*4*3  Formula : n P r = n! / (n-r)!

12. Page 12 Two useful mathematical concepts  Combinations (n C r) : number of ways to choose r out of n objects without ordering  Mark Six has 49 numbers, draw 6 of them, how many combinations?  The first ball has 49 choices  The second ball has 48 choices ……  Total number of combinations = 49*48*47*46*45*44  However, 1 3 5 10 11 12 is just the same as 12 11 10 5 1 3, 10 5 1 3 11 12 or any of 6! permutations  Therefore need to divide it by 6!  Formula : n C r = n! / (n-r)! r! (n P r / r!)  Property : n C r = n C (n-r)  Choosing r items is equal to choosing which n-r items to leave out

13. Page 13 Estimating Time Complexity of Exhaustion  For the equation x 1 + x 2 + … + x k = s, where x i >=0, how many possible solutions?  Answer = (s+k-1) C (k-1), why?  N people sit around a round table, how many permutations? (rotations of a permutation are considered the same)  Answer = (N-1)!, why?  Return to the previous problem  To get from the top-left to bottom-right, we need to make 7 moves (4 right, 3 down)  This is equivalent to choosing 4 out of 7 items  Answer = 7 C 4 (or generally, (width + height – 2) C height )

14. Page 14 Lexicographical Ordering  The objects we permutate may have some “ordering”  Integers : 1 < 2 < 3 < … < N  Alphabets : a < b < c < … < z  If such an ordering exists, we can extend it to an ordering between permutations  123 < 213 < 321  bdca < cabd < cbad < dcab  This is called “lexicographical ordering” or “dictionary ordering”

15. Page 15 Lexicographical Ordering  Three natural questions to ask……  Given a permutation, determine its “next” permutation  231 -> 312  Find the Nth permutation (Deranking)  3 rd permutation of 1,2,3 = 213  Given a permutation, determine its position (Ranking)  132 -> position = 2