cs 180 programming i aditya p. mathur department of computer sciences purdue university march 29-31,...

CS 180 Programming I

Aditya P. MathurDepartment of Computer SciencesPurdue UniversityMarch 29-31, 2004

Last update: March 30, 2004

Recursion

Recursion: CS 180 Spring 2004 Aditya P. Mathur 2004 2 of 59

Learning Objectives

How to use recursion for solving problems?

How do recursive programs work?

What is recursion?

Recursion versus iteration


What is Recursion?

It is independent of programming language used. You can use recursion in Java, C, C++, or almost any of the well known languages.

A program that uses recursion is known as a recursive program.

We will learn how to code recursive solutions in Java.

A technique for solving problems.


Iterative solutions: Summation

Problem: Given n integers, find their sum.

Iterative solution: a1, a2, … an

sum=0

sum=0+a1

sum=0+a1+a2

sum=0+a1+a2+…an

.

.

.

sum=0;for(int i=1; i<=n; i++){

sum=sum+a[i];}


Iterative solutions: Factorial

Problem: Given n>0, find !n (factorial of n).

Iterative solution: product=1

product=1*2

product=1*2*3

product=1*2*3…*n

.

.

.

product=1;for(int i=2; i<=n; i++){

product=product*i;}


Iterative solutions: Fibonacci Series

Problem: Given n>=0, find Fibonacci(n).

Definition: Fib(0)=0; Fib(1)=1; n>1, Fib(n)=Fib(n-1)+Fib(n-2)

Fibonacci series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55....


Iterative solutions: Fibonacci Series

public int fibonacci(int n) { int fib, fibPrev, fibPrevPrev; if (n==0) return(0); if (n==1) return(1); fibPrevPrev=0; fibPrev=1; fib=fibPrev+fibPrevPrev; for (int i=2; i<=n; i++) { System.out.println("Fib("+i+")="+fib); fibPrevPrev=fibPrev; fibPrev=fib; fib=fibPrev+fibPrevPrev; } return(fib); }

fib=0

fib=1

fib=0+1

fib=0+1+1+2+3+5....

.

.

.

fib=0+1+1


Recursive solutions: Background: What is tree?

17

12 29

18 32

30 45

Root node

Leaf nodes Internal nodes

Parent nodeChild node

We will see that it is easy to understand how recursionworks with the help of a tree.


Recursive solutions: Summation


Recursive solution:(a1+ a2+ … +an)

a1+ (a2+ … +an)

a2+ (a3+ … +an)

a3+ (a4+....+an)...

a(n-1)+ (an)

(a(n-1)+an)

winding phaseunwinding phase


Recursive solutions: Summation


Recursive solution:

(2+3+4)public int sum(int []A, int low, int high){

if(low= =high)return(A[low]);if(low<high)

{ return(A[low]+sum(A,low+1,high));

}}

2+ (3+4)

3+ (4)

=7

=9

Recursive callDo you see any loop here?

=4


Recursive solutions: Factorial

Problem: Given n>0, find !n (factorial of n).

Recursive solution:

public int fact(int n){ if (n= =0|| n= =1) return(1); return(n*fact(n-1));}

!4

4* !3

3* !2

=6

=24

2* !1

=2

=1

Recursive call

Again, no loop!


Recursive solutions: Fibonacci Series

Problem: Given n>=0, find Fibonacci(n).

Definition: Fib(0)=0; Fib(1)=1; Fib(n)=Fib(n-1)+Fib(n-2)

public int fib(int n) { if (n= =0 ) return(0); if (n= =1 ) return(1); return(fib(n-2)+fib(n-1)); }

fib(4)

fib(2)+ fib(3)

fib(1)+ fib(2)

=3

fib(0)+ fib(1)

fib(0)+ fib(1)

=2

=1

=1

=1

=0

=1

=0 =1

Fibonacci series: 0, 1, 1, 2, 3, 5....Recursive calls

Once again, no loop!


Developing a recursive solution: Base case

Step 1: Identify a base case.

This is the case where the solution is derived without recursion.

In most cases it is a simple solution.

The winding phase terminates when the base case is encountered; unwinding begins.


Base case: Examples

Summation: (a1+a2+...an) , low=1, high=n

For n=1, sum(A, low, high)=alow

Factorial: n=0 !n=1 n=1 !n=1

Fibonacci: n=0 fib(0)=0 n=1 fib(1)=1


Recursive case

Step 2: Identify the recursive, or the general, case.

This is the case where the solution is derived by invoking the same method recursively until a base case encountered.

A recursive call continues the winding phase.

Did you notice the Divide and Conquer strategyin our examples?


Recursive case: Examples

Summation: For n>1, low=1, high=n. (low<high), sum(A, low, high)=alow+sum(A, low+1, high)

Factorial: For n>1, fact(n)=n*fact(n-1)

Fibonacci: For n>1, fib(n)=fin(n-2)+fib(n-1)


Recursive and base cases: summary

Summation:

A=(a1, a2,...an), n>0, low=1, high=n low=high, sum(A, low, high)=alow low<high, sum(A, low, high)=alow+sum(A, low+1, high)


Recursive and base cases: summary

Factorial: n=0, !n=1 n=1, !n=1 n>1, fact(n)=n*fact(n-1)

Fibonacci: n=0, fib(n)=0 n=1, fib(n)=1

n>1, fib(n)=fin(n-2)+fib(n-1)


Loops and recursion

In fact a program may use recursion and iteration and hence will be composed of loops as well as recursive calls.

In our examples, there was no loop (e.g. while-do, for).

This does not imply that programs with recursion are free of loops created using while-do, for, etc.


Difficulty in writing a recursive program

Some find it difficult to write a recursive program than to write an iterative program.

The difficulty arises in identifying the base and the recursive cases.

In most problems, it is easy to identify the base case. The difficulty lies in identifying the recursive , or the “general” case.

Suggestion: Focus on identifying the base and general cases beforeyou start writing a recursive program.


Recursion and “lazy evaluation”

Note that in our examples, the intermediate solutions are “postponed.”

Thus to find (2+3+4) we first added 3 to 4 and then added 2 to the sum.

Similarly, to find !4, we first found !3 and then multiplied it by 4. To find !3 we found !2 and multiplied it by 3. To find !2 we found !1 and multiplied it by 2. !1 is a base case.

This strategy of postponing the application of operationsis also known as “lazy evaluation.”


“lazy evaluation” requires memory

In each of our three examples, when we make a recursive call, we postponed a computation.

For example to find !4, we first found !3 but remembered that !3 is to be multiplied by 4 to get !4. Similarly, to find !3, we remembered (a) 3 is to be multiplied by !2 to get !3 and (b) that 4 is to be multiplied by !3.

Java does this “remembering” for us automatically by using a stack.


Stack

Upon each recursive call, Java interpreter pushes, whatever is to be remembered, on a stack.

The stack “grows” in the winding phase as more and more data is added to it.

During the unwinding phase, Java interpreter removes data rom the top of the stack and performs the desired operation.


Quiz (March 29, 2004)

What is Fibonacci(6)?


Stack Example: Summation: Winding

(2+3+4)

Data stack empty before sum( ) is calledlow=1, high=3

2 pushed on the stackright before the first recursive call.low=2, high=3

2

(2+(3+4))

3 pushed on the stackright before the second recursive call. low=3, high=3

2

3

(2+(3+(4)))


Stack Example: Summation: Unwind

(2+(3+4))

4 pushed on the stackright after return from the last call

2

3

4

4+3 pushed on the stackright after another return.

2

7

(2+7)

9 pushed on the stackright after return to mainmain gets the result from the stack.

(9)

9


Iteration versus recursion

Recursion is often more expensive than iteration in terms of execution time and stack memory required.

For some problems recursion offers an easy and natural solution. Iterative solutions might be difficult and unnatural.

Four examples follow: Towers of Hanoi Merge sort Tree traversal Shortest distance between two cities.


Danger!

If your base case is incorrect, or coded incorrectly, you might end up with a program that has infinite recursion!


Towers of Hanoi

Source Peg (A) Intermediate Peg (B) Destination Peg (C)


Towers of Hanoi: The Problem

Source Peg (A) has N disks stacked vertically in the order of their size as shown. (N>0).

The problem is find the shortest sequence of moves that will transfer all N disks from the Source Peg (A) to the Destination Peg (C).

The Intermediate Peg (B) can be used for temporary placement of disks.

A larger disk should NEVER be placed on top of a smaller disk.

Only one Disk can be moved from the top of one Peg and placed on top of another peg in one move.


Towers of Hanoi: Base and Recursive Cases

Base case: N=1

Solution: moveDisk (A, C). This moves one disk from A to C.

Recursive case: N>1

Solution: moveDisks(N-1, A, B, C): Move N-1 disks from A to B via C.moveDisk(A, C): This moves one disk from A to C. moveDisks(N-1, B, C, A): Move N-1 disks from B to C via A.

from to via


Towers of Hanoi: Recursive case


moveDisks(N-1, A, B, C): Move N-1 disks from A to B via C.moveDisk(A, C): This moves one disk from A to C. moveDisks(N-1, B, C, A): Move N-1 disks from B to C via A.




moveDisks(N-1, A, B, C): Move N-1 disks from A to B via C.moveDisk(A, C): This moves one disk from A to C. moveDisks(N-1, B, C, A): Move N-1 disks from B to C via A.




. moveDisks(N-1, A, B, C): Move N-1 disks from A to B via C.moveDisk(A, C): This moves one disk from A to C. moveDisks(N-1, B, C, A): Move N-1 disks from B to C via A.


Towers of Hanoi: A Java Program

private void moveDisk(String from, String to) { System.out.println(“Move disk from:"+ from+" to: "+to); }


Towers of Hanoi: A Java Program

private void moveDisks(int disks, String from, String to, String intermediate) { if(disks==1) { moveDisk(from, to); } else { moveDisks(disks-1, from, intermediate, to); moveDisk(from, to); moveDisks(disks-1, intermediate, to, from); } }


Towers of Hanoi: Sample Execution

moveDisks(3, “A”, “B”, “C”);

Move: 1 Move disk from:Peg A to: Peg CMove: 2 Move disk from:Peg A to: Peg BMove: 3 Move disk from:Peg C to: Peg BMove: 4 Move disk from:Peg A to: Peg CMove: 5 Move disk from:Peg B to: Peg AMove: 6 Move disk from:Peg B to: Peg CMove: 7 Move disk from:Peg A to: Peg C


Merge sort

A method for sorting an array of data.

A=[a1 a2 a3...an]

Divide array into two halves: front and tail.

front=[a1 a2...a(n/2)]

tail=[a(n/2) a(n/2+1).....an]

Sort front and tail and merge the two in to A.


Merge sort: Base and recursive cases

n=1: A is already sorted. Do nothing!

n>1: Divide array into two halves: front and tail.

front=[a1 a2...a(n/2)]

tail=[a(n/2) a(n/2+1).....an]

sort(front) and sort(tail); do this recursively. Merge sorted front and tail into A.

sort(A[]);


Merge sort: Illustration

sort([14 2 0 ])

wind unwind

merge([2 14], [0])

[0 2 14]

merge([14],[2])[2 14]sort([14 2]) sort([0])front tail

sort([2])sort([14])

front tail


Problem: Tree traversal: Infix order

17

12 29

18 32

30 45

Traverse a binary tree in infix order and print the data in its nodes.

Infix order

Traverse the left subtree, print root, traverse the right subtree.

Infix order: 12, 17, 18, 29, 30, 32, 45


Problem: Tree traversal: Prefix order

*

a +

b /

c d

Traverse a binary tree in prefix order and print the data in its nodes.

Prefix order

Print root, traverse the left subtree, traverse the right subtree.

Prefix order: *a(+b(/cd))


Problem: Tree traversal: Postfix order

*

a +

b /

c d

Traverse a binary tree in prefix order and print the data in its nodes.

Postfix order

Traverse the left subtree, traverse the right subtree, print root.

Postfix order: a b c d / + *


Problem: Tree traversal: Infix order

Write a Java program to traverse a binary tree in infix order and print the data in its nodes.

17

12 29

18 32

30 45

head

Node

leftlink

data rightlink


Problem: Tree traversal: Base and Recursive Cases

Base case: head=nullOutput: None

Recursive case: head!=nullOutput:

traverse(head.leftLink( ))print head.data( )traverse(head.rightLink( ))

17

12 29

18 32

30 45

head

156

6 17 18 29 30 32 451512


Shortest Path: The Problem

We are given a network of cities and links (roads) that connect them.

Links are assumed to be directional and have distances associated with them.

Given two cities C1 and C2 find the shortest path from C1 to C2 in terms of the distance if a path exists otherwise notify that there is no such path.

shortestPath(C1, C2) returns Path. A Path has path which is a non-empty string of cities and a distance, if it exists. Else the path is an empty string.


Shortest Path: The Network

1

4

5

2

3

650

50

12040

80

130

70

45

60 70

shortestPath(1, 6): path: 1-->4-->6 distance=120shortestPath(3, 6): path: “” distance=undefined


Base case:

Base Case 1:There are one or more direct links from C1 to C2.

Solution:

Find link with the shortest distance.

shortestPath(1,2): Path: 1-->2. Distance: 50.

12

60

50


Base case:

Base Case 2:There are no outgoing links from C1.

Solution:

Path does not exist.

3

shortestPath(3, 6): Path: “” distance: undefined.


Base case:

Base Case 3:C1 and C2 are the same

Solution:

shortestPath(4, 4): Path: “4-->4” distance: 0.


Recursive case

There is no direct link from C1 to C2.

if (count==0) then return(“”);else

return(paths[j]) such that paths[j].distance() is the smallest amongst all elements of paths[];

Solution:

count=0;

for (each outgoing link from C1)

{ Path p=shortestPath(tail(link), C2);

if(p!=“”)paths[count++]=addPaths((link(C1, C2), p);}

12

60

50

440


Recursive case: Examine links from 1

shortestPath(1, 6): There is no direct link from 1 to 6. Hence we find shortestPath(2, 6), shortestPath(2, 6), and shortestPath(4, 6) corresponding to the three outgoing links from 1.

1

4

5

2

3

650

50

12040

80

130

70

45

60 70



shortestPath(2, 6): There is no direct link from 2 to 6. Hence we find shortestPath(3, 6) corresponding to the lone outgoing link from 2.

1

4

5

2

3

650

50

12040

80

130

70

45

60 70



shortestPath(3, 6): There is no link from 3 and hence return empty path (implying that no path was found) This also happens for the other link from 1 to 2 (with distance=60).

130

1

4

5

2

3

650

50

12040

80

70

45

60 70



shortestPath(4, 6): There is a direct link from 4 to 6 and hence return path with string “4-->6” and distance=80. shortstPath(3, 6) returns an empty path.

1

4

5

2

3

650

50

12040

80

130

70

45

60 70


Recursive case: Get back to 1

append(link(1, 4), shortestPath(4,6)): We get the path “1-->4-->6” and a distance of 120. Done!

1

4

5

2

3

650

50

12040

80

130

70

45

60 70


Issues (Been to MapQuest?)

How does one handle cycles?

1

4

5

2

3

650

50

12040

80

130

70

45

60 70

How to handle bi-directional links?...etc. etc.


Quiz (March 31, 2004)

Saturn

Jupiter Neptune

Uranus Earth

head

In what order will the planets in the following tree be listed if the tree is traversed in infix order?


Bye!

Have funwith

Recursion!

cs 180 programming i aditya p. mathur department of computer sciences purdue university march 29-31,...

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