CS102 – Recursion

David DavenportBilkent UniversityAnkara – Turkey

email: david@bilkent.edu.tr

What is Recursion? Method of solving problems

Alternative to iteration All recursive solutions

can be implemented iteratively

Characteristic... recursive methods call themselves

recursive functions defined in terms of (simpler versions) themselves

General recursive case & stopping cases

Factorial N Iterative

N! = N * N-1 * N-2 … 2 * 1

Recursive 1! = 1 - stopping case N! = N * (N-1)! - general case

implement & trace in Java

So, why use recursion? Advantages...

Interesting conceptual framework Intuitive solutions to difficult problems

But, disadvantages... requires more memory & time requires different way of thinking!

Towers of Hanoi The Problem…

Rules only move one disk at a time can’t put a disk on a smaller one

Monks believed world would end once solved for 64 disks!

Takes 580 billion years! at one move per sec.

Print reverse Print first N values in reverse

Simple iterative solution… Recursive?

Hint - what is the simplest case? Write out sequence of cases

& look for previous case within each casegeneralise & note stopping case!

5 7 2 6 3Given... print...

3, 6, 2, 7, 5

Print reverse - solution Solution form

5 7 2 6 3

N’thN-1

To print N values in reverse order if N > 0

Print the N’th value& thenPrint the preceding N-1 values in reverse order

Write Java & trace

Print forward Print set of values in normal order Solution form

5 7 2 6 3

N’thN-1

To print N values in normal order if N > 0

Print the first N-1 values in normal order& thenPrint the N’th value

Write Java & trace

Print – alternative solution Print set of values in order Alternative solution

5 7 2 6 3

To print values btw S & E in order if S <= E

Print the value at S& thenPrint the values btw S+1 & E in order

Write Java & tracefirst S rest E

Summation Find the sum of the first N values

5 7 2 6 3

N’thN-1

To sum first N values of X if N = 0

return 0 else

return N´th value + sum of preceding N-1 values

Summation 2 Find sum of all values before sentinel (-1)

5 7 2 -1 3

first rest

Counting Count number of values before zero

5 7 2 0 3

count no. values before zero starting at S if value at S is zero

return 0 else

return 1 + count no. values starting at S+1

restfirst S

Zero is sentinel value& so is guaranteed to exist.This is a common form of representation for strings!

Common mistakes! (1)

sum = 0;

public void count( int[] X, int S) { if ( X[S] != 0) {

sum++;count( X, S+1);

}}

Problems:• Need another method to getCount()• What happens if called again?

Common mistakes! (2)

public int count( int[] X, int S) { int sum; if ( X[S] == 0)

return sum; else {

sum++;return count( X, S+1);

}}

Problem:• New instance of local variable for each instantiation!

Find Max Find max of first N values

5 7 2 6 3

To find max of N values if N = 1

return N’th value else

return greater of N’th & max of preceding N-1 values

N’thN-1

Think about solution in terms of

these two sub-problems

SEARCHING & SORTING

Sequential search

5 7 2 6 3

Search( first N values of X for target) if N = 0

return not found else if N’th value is target

return found at N’th else

return search( first N-1 values of X for target)

N’thN-1

Think about simplest cases of problem & more

general solution in terms of these two

sub-problems

O(N)O(N)

Binary Search

• Think about using telephone directory• Values must be in order!• Have knowledge of data distribution• If unknown? (number guessing game)

1 2 3 5 6 7 9second halffirst half

middleS E

O(log2N)O(log2N)

Binary Search (continued…)

Search( X btw S & E for target) if S > E

return not found else if ( middle value is target)

return found at middle else if ( target < middle value)

return search( first half of X for target) else

return search( second half of X for target)

First half is S, middle-1

Second half is middle+1, E

Selection sort

5 7 2 6 3

N’thN-1

selectionSort( first N values of X) if N > 1

locOfMax = findmax( first N values of X)exchange N’th value with value at locOfMaxselectionSort( first N-1 values of X)

O(N2)O(N2)

locOfMax

QuickSort

5 7 3 9 2 1 6

pivot

O(Nlog2N)O(Nlog2N)

QuickSort( X btw S & E) if S < E

pivot loc. = partition( X btw S & E)QuickSort( first part of X)

QuickSort( second part of X)

Sort second partSort first part

2 1 3 5 7 9 6

pivot

First part isS to pivot loc.-1Second part ispivot loc.+1 to E

partition

MergeSort

O(Nlog2N)O(Nlog2N)

MergeSort( X btw S & E) if S < E

MergeSort( first half of X) MergeSort( second half of X)

Merge( first & second halves of X)

1 5 7 9 2 3 6

1 2 3 5 6 7 9merge

Merge isthe basis of sequential processing

Sort first halfSort second half

MISC…

Palindrome

R A D A R

lastfirstmiddle

isPalindrome( word) if first >= last

return true else if first char of word != last char of word

return false else

return isPalindrome( middle of word)

BlobCount...

blobCount( x, y) zero

• if x,y outside grid• if x,y is empty

else, one plus blobCount(north)blobCount(east)blobCount(south)blobCount(west)

The plan… Maze & blob counting

Fractals? Recursive-descent-parsing?

…on to data structures (stacks, queues, lists, trees, …)