Recursion and Exhaustion

1. Recursion and Exhaustion Yip Lik Yau

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(2N) 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. Problem: Generating Permutations • Given N, generate all permutations of integers 1..N • When N = 3: 123 132 213 231 312 321

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; end;

5. Problem: Generating Permutations • Search Tree 1 2 3 2 3 1 3 1 2 3 2 3 1 2 1

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. Be Jewelled! The jewels in red will be counted for score: 5*3^(9-3)=3645 Swapping the jewels in red.

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. 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

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

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. 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. Estimating Time Complexity of Exhaustion • For the equation x1 + x2 + … + xk = s, where xi>=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. 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. Lexicographical Ordering • Three natural questions to ask…… • Given a permutation, determine its “next” permutation • 231 -> 312 • Find the Nth permutation (Deranking) • 3rd permutation of 1,2,3 = 213 • Given a permutation, determine its position (Ranking) • 132 -> position = 2