1 recursion. 2 basic recursions break a problem into smaller identical problems –each recursive...

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Recursion

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Basic Recursions

• Break a problem into smaller identical problems– Each recursive call solves an identical but smaller

problem.• Stop the break-down process at a special case whose

solution is obvious, (termed a base case)– Each recursive call tests the base case to

eventually stop.

– Otherwise, we fall into an infinite recursion.

Basic Recursions

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Example 1: The Sum of the First NPositive Integers• Definition:

sum(n) = n + (n-1) + (n-2) + … + 1 for any integer n > 0• Recursive relation;

sum(n) = n + [(n-1) + (n-2) + … + 1]= n + sum(n-1)

Looks so nice, but how about n == 1?sum(1) = 1 + sum(0), but the argument to sum( ) must be positive

• Final recursive definition:sum(n) = 1 if n = 1 (Base case)

= n + sum(n-1) if n > 1 (Recursive call)

Basic Recursions

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Recursive Definition of sum(n)int sum(int n){ if (n == 1) // base case return 1; else return n + sum(n-1); // recursive call A}

How should I test this function’s correctness?

Use the box method:

A box has each function’s local environment ex. cout << sum(3);

1. Arguments2. Local variables3. Place holder for the value returned by a called function4. A return value

n = 3

A: sum(n-1) = ?

return ?

Basic Recursions

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Box trace of sum(3)

n = 3A: sum(n-1)=?return ?


n = 1

return 1

n = 3A: sum(n-1)= ?return ?


n = 1

return 1

Each box corresponds to a function’s activation record or stack.

cout << sum(3);

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Basic Recursions

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Example 2: The Factorial of n• Definition:

factorial(n) = n * (n-1) * (n-2) * … * 1 for any integer n > 0factorial(0) = 1

• Recursive relation;factorial(n) = n * [(n-1) * (n-2) * … * 1]

= n * factorial(n-1)Looks so nice, but how about n = 1?factorial(1) = 1 * factorial(0) but the argument to factorial( ) must be positive

• Final recursive definition:factorial(n) = 1 if n = 1

= n *factorial(n-1) if n > 1

Basic Recursions

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Recursive Definition of facotrial(n)int fact(int n){ if (n == 1) // base case return 1; else return n * fact(n-1); // recursive call A}

Trace its execution using the box method:

ex. cout << fact(3);

n = 3

A: fact(n-1) = ?

return ?

Basic Recursions

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Box trace of fact(3)

n = 3A: fact(n-1)=?return ?


n = 1

return 1

n = 3A: fact(n-1)= ?return ?


n = 1

return 1

cout << fact(3);

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Basic Recursions

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Precondition of fact(n)

• Precondition: n >= 1• If the precondition is violated, what happen?

– Fact(0) calls fact(-1), which calls fact(-2), ….• For robustness, given a negative number or 0, fact should stop a

recursive call.

Always correct to recursions

int fact(int n){ if (n <= 1) // base case for n = 1 or bad args. return 1; else return n * fact(n-1); // recursive call A}

Basic Recursions

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Example 3: Printing Numbers in Any Base

How to convert a decimal to a hexadecimal

Dividend Remainder16) 1234567890(10)……..…16) 7716049(10)………616) 482253(10)………116) 30140(10)…13(D)16) 1883(10)…12(C)16) 117(10)…11(B)16) 7(10)……....516) 0(10)……....7

1. Divide by a base(16)

2. Print each reminder from the bottom:

75BCD16(16)

Basic Recursions

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Recursive Definition of printNum(n,base)void printNum(int n, int base){ static string DIGIT_TABLE = “0123456789abcdef”;

if (n >= base) // recursive call A printNum( n / base, base ); // else n < base base case cout << DIGIT_TALBE[ n % base ];}

Trace its execution using the box method:

ex. cout << print(123456789, 16);

n = 123456789, base = 16

A: fact(n-1) returns nothingPrint: n % base = ?

return nothing

Basic Recursions

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Example 4: Binary Search

• The search divides a list into two small sub-lists till a sub-list is no more divisible.

• Each step needs the top/tail indexes and a target item.

First half

First half

An entire sorted list

First half Second half

Second half

Can we call binarySearch( ) recursively?

Basic Recursions

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Binary Search (Recursive Version)

int binarySearch( const int array[], int top, int tail, int target ) {

int mid = (top + tail)/2;if ( target == array[mid] )

return mid;else if ( target < array[mid] )

return binarySearch( array, top, mid-1, target );else if ( target > array[mid] )

return binarySearch( array, mid+1, tail, target );}

Question: if it does not find the target, how can we stop this program?

Basic Recursions

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Binary Search Revisited

In the next recursive call,top = i + 1;tail = i;thus top > tail!!

Base cases: (1) if (target == array[mid]) return mid;(2) if top > tail return –1;

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top tail

i i+1

mid = (top + tail)/2 = (i + i+1)/2 = itop = mid + 1 = i + 1

Basic Recursions

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Binary Search (Correct Recursive Version)

int binarySearch( const int array[], int top, int tail, int target ) {

int mid = (top + tail)/2;if ( target == array[mid] )

return mid; // find itelse if ( top > tail )

return -1; // cannot find itelse if ( target < array[mid] )

return binarySearch( array, top, mid-1, target );else if ( target > array[mid] )

return binarySearch( array, mid+1, tail, target );}

Basic Recursions

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Babies

Example 5: Multiplying Mice

Quite difficult to keep track of the

population explosion.

Month 1

Month 2

Month 4

Month 5

Month 3

Month 6

Month 7

Recursive relation:#babies in Month 7= #mice in Month 5#adults in Month 7 = #mice in Month 6

Basic Recursions

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The Fibonacci Sequence• mice(n) = mice(n-1) + mice(n-2)• Then, what are base cases?

• Recursive definition:

mice(1) = mice(0) + mice(-1) can’t happen.Thus, mice(1) = 1;

mice(2) = mice(1) + mice(0) can’t happen, either.Thus, mice(2) = 1;

mice(n) = 1 if n = 1 or 2mice(n-1) + mice(n-2) if n > 2

Basic Recursions

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Tracing mice(n)

m(6)Return m(5)+m(4)




m(2)Return 1

m(1)Return 1


m(1)Return 1

m(2)Return 1

m(1)Return 1


m(2)Return 1


1 1m(2)Return 1

m(2)Return 1

2 1

1 12

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11 1

23

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Basic Recursions

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Recursive Applications

• Discrete Mathematics (Combinatorics, Puzzles, Coding Theory)– Tower of Hanoi and Gray Code

• Divide and Conquer– Mergesort, Convex Hall, and Fast Fourier Transform

• Backtrack– 8 Queens, Maze and Classic Chess Program

• Fractal Figures– Koch, Sierpinski Allowhead, Gosper, Hilbert, and

Dragon curves


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Towers of Hanoi

A B C

Find how to move dishes from A to B using C.Restrictions:


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Solution

A B C

A B C

A B C

A B C

if we could move n-1 dishesfrom A to C via B,

we could move a dish, (i.e.,the last one) from A to B (via C),

and thereafter, we would move n-1Dishes from C to B via A!

move(n, A, B, C)

move(n-1, A, C, B)

move(1, A, B, C)

move(n-1, C, B, A)

In order to move n dishes fromA to B via C,


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Example Code#include <iostream>#include <stdlib.h>using namespace std;

void move( int n, char orig, char dest, char temp ) { if ( n == 1 ) cout << "move disk 1 from " << orig << " to " << dest << endl; else { move( n - 1, orig, temp, dest ); cout << "move disk " << n << " from " << orig << " to " << dest << endl; move( n - 1, temp, dest, orig ); }}

int main( int argc, char* argv[] ) { int nDishes = atoi( argv[1] ); move( nDishes, 'A', 'B', 'C' );

cout << "completed" << endl; return 0;}

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Tracing move(n, A, B, C)move(3, A, B, C)

move(2, A, C, B) move(1, A, B, C) move(2, C, B, A)

move(1, A, B, C)

move(1, A, C, B)

move(1, B, C, A)

move(1, C, A, B)

move(1, C, B, A)

move(1, A, B, C)

Locally, A is A, C is B, B is C. Locally C is A, B is B, A is C.

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Divide-and-Conquer AlgorithmsComputational Geometry – Finding a Convex Hall

• Divide:– Divide a problem into two

sub-problems recursively so that the sub-problems are calculated by themselves.

• Conquer:– Form a new solution from

the solutions answered by the sub-problems


The convex hull problemTrivial algorithm: O(n2)

Divide & conquer: O(nlogn)

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BacktrackTrace every combination until encountering a solution

• The eight queens problem– Place 8 queens on a 8 * 8

chessboard so that no queen can attack any other queen.

– Place a queen from the column 0 to 7 as checking if the queens placed on the former columns can attack the current queen.


Q

Q

Q

Q

Q

Q

Q

Q

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BacktrackThe Eight Queen Problem

• A Naïve Solution


Q

Q

Q

Q ??

Q

Q

for ( int col = 0; col < 7; col++ ) { for ( int row = 0; row < 7; row++ ) { if ( safeLocation( table, row, col ) ) { table[row][col] = true; break; } } // oops! No safe row to place a new queen // I got to go back to the previous column!}

• Then, how can I go back to the previous column? Just col--?

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• A Recursive Solution:


Q

Q

Q

Q ??

Q

Q

[0,0] -> [0,1] NO [1,1] NO [2,1] -> [0,2] NO [1,2] NO [2,2] NO [3,2] NO [4,2] -> [0,3] NO [1,3] -> [0,4] NO [1,4] NO [2,4] NO [3,4] -> [0,5] NO [1,5] NO [2,5] NO [3,5] NO [4,5] NO [5,5] NO [6,5] NO [7,5] NO [4,4] NO [5,4] NO [6,4] NO [7,4] -> [0,5]

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bool addQueen( bool t[SIZE][SIZE], int col ) { if ( col >= SIZE ) return true; // all cols have been examined for ( int row = 0; row < SIZE; row++ ) { if ( safeLocation( t, row, col ) ) { // this row may be a candidate t[row][col] = true; // place a new queen; if ( addQueen( t, col + 1 ) )

return true; // all the following cols were filled else

t[row][col] = false; // A wrong position. Try the next row } } return false; // all rows examined, but no candidates}

int main( ) { bool table[SIZE][SIZE]; // there are no matrix templates in STL

init( table ); // all table[i][j] = false; if ( addQueen( table, 0 ) ) print( table ); else cout << "no solution" << endl;}

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Fractal CurvesDrawing a Ruler


// java code to draw a rulervoid drawRuler( Graphics g, int left, int right, int level ){ if ( level < 1 ) return;

int mid = ( left + right ) / 2;

g.drawLine( mid, 80, mid, 80 – level * 5 );

drawRuler( g, left, mid – 1, level – 1 ); drawRuler( g, mid + 1, right, level – 1 );}

mid at level 3

mid at level 2 mid at level 2

mid at level 1 mid at level 1 mid at level 1 mid at level 1

Initial left Initial right

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Fractal CurvesDrawing a Koch Curve using a Turtle


class Turtle {public: Turtle( float initX=0.0, float initY=0.0, float initAngle=0.0 ) { ~Turtle( ); void draw( float d ); // draw a line from the current position for distance d void move( float d ); // simply move from the current position for distance d void turn( float a ); // turn left by angle a

private: float angle; // presented in degree but not in radian oftream out;};

int main( ) { Turtle t;

t.draw( 10 ); t.turn( 45 ); t.move( 10 ); t.draw( 10 );}

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Fractal CurvesDrawing a Koch Curve using a Turtle


class Koch : Turtle {public: Koch( float initX=0.0, float initY=0.0, float initAngle=0.0) : Turtle( initX, initY, initAngle ) { ) ~Koch( ); curve( int level, float length ) { if ( level > 1 ) { curve( level – 1, length ); turn( 60 ); curve( level – 1, length ); turn( -120 ); curve( level – 1, length ); turn( 60 ); curve( level – 1, length ); } draw( length ); }};

int main( ) { Koch k(100, 100, 0); k.curve( 3, 10 );}

Draw the level-3 Koch curve here

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Recursion and Efficiency

• Programmability– Capable of breaking down a complex problem– Based on mathematical induction

• Efficiency– The overhead associated with function calls

(Bookkeeping the current stack and creating a new one)

– The inherent inefficiency of some algorithm(Repeating the same computation)

Efficiency

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Efficiency of mice(n)





m(2)Return 1

m(1)Return 1


m(1)Return 1

m(2)Return 1

m(1)Return 1


m(2)Return 1


m(2)Return 1

m(2)Return 1

O(2n)

Efficiency

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Iterative Solution of mice(n)int iterativeMice(int n){

int previous = 1;int current = 1;int next = 1;for (int i = 3; i <= n; ++i) {

next = current + previous;previous = current;current = next;

}return next;

}

Efficiency of mice(n):O(n-2)=O(n)

Efficiency

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Recursion versus Iteration

• Discover a recursive solution if it is easier.• Convert it to iteration if an iterative solution is more efficient.

– When a recursion incurs more iterations– When each iteration includes a few computation

Problems Recursion IterationFibonacci sequence O(2n) O(n)Sum/Fact/printNum O(n) O(n)

function call overhead

Binary search O(log n) O(log n)Tower of Hanoi O(2n)Fractal figures (Koch) O(4n) Can you program it?

N queens O(Nn)

Efficiency

1 recursion. 2 basic recursions break a problem into smaller identical problems –each recursive...

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