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Recursion and Exhaustion

Yip Lik Yau

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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

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Problem: Generating Permutations

Given N, generate all permutations of integers 1..N

When N = 3:123132213231312321

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Problem: Generating Permutations Nested For loop is out of the question We need a more advanced method – recursion Outline of algorithm:

varused : array[0..100] of boolean;n : integer;

procedure run(stage : integer);begin

if (level = n)do_something;

else for i = 1 to n doif not used[i] begin

used[i] = true;run(stage+1);used[i] = false;

end;end;

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Problem: Generating Permutations Search Tree

1

2 3

2

1 3

3

1 2

3 2 3 1 2 1

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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)

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Be Jewelled!

P Q P Q P Q

Q P X Y X P

P Y Y X Y Q

Q Y Q Y Q P

P Q P Y P Q

Q P Q P Q P

P Q P Q P Q

Q P X X X P

P Y Y Y Y Q

Q Y Q Y Q P

P Q P Y P Q

Q P Q P Q P

The jewels in red will be counted for score:5*3^(9-3)=3645

Swapping the jewels in red.

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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

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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

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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

1 5 4 2 5

6 9 2 8 1

4 7 6 1 4

2 4 3 1 2

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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)!

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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

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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 )

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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”

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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