Recursion Big Picture Recursion is a technique for solving problems - - PDF document

SLIDE 1

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Recursive Methods That Return Values

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Recursion Big Picture

• Recursion is a technique for solving problems

(akin to Divide, Conquer, and Glue)

• Recursion is effective when sub-problems have

the same “shape” as the original problem

L - 3

PictureWorld Example

Suppose we have the following two methods:

public Picture triangles(Color c1, Color c2) public Picture LL(Picture p) p

Goal

How can we generate this picture? Should we use recursion?

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PictureWorld Example

Suppose we have the following two methods:

public Picture triangles(Color c1, Color c2) public Picture LL(Picture p) p

Goal

If we had a method which generated this picture, then maybe we could create our goal picture...

How can we solve this problem using a smaller version of the same problem?

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PictureWorld Example

Suppose we have the following two methods:

public Picture triangles(Color c1, Color c2) public Picture LL(Picture p) p

Goal

If we had a method which generated this picture, then maybe we could create our goal picture...

triangleDesign(3,Color.red,Color.blue)

Assume we have such a method... a leap of faith!

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PictureWorld Example

Suppose we have the following two methods:

public Picture triangles(Color c1, Color c2) public Picture LL(Picture p) p

Goal

Put this picture in the lower left corner... If we had a method which generated this picture, then maybe we could create our goal picture...

triangleDesign(3,Color.red,Color.blue) LL(triangleDesign( 3,Color.red,Color.blue))

SLIDE 2

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PictureWorld Example

Suppose we have the following two methods:

public Picture triangles(Color c1, Color c2) public Picture LL(Picture p) p

Goal

Overlay

LL(triangleDesign( 3,Color.red,Color.blue)) triangles(Color.red,Color.blue) L - 8

triangleDesign Method

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JEM

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

4 rtw

Execution Land

red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

3 rtw red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

1 rtw red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

2 rtw red blue

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JEM

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

4 rtw

Execution Land

red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

3 rtw red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

1 rtw red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

2 rtw red blue return

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JEM

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

4 rtw

Execution Land

red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

3 rtw red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

2 rtw red blue return p2 p1 p1 p2 ? ?

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JEM

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

4 rtw

Execution Land

red blue

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

3 rtw red blue return p2 p1 p1 p2 ? ?

SLIDE 3

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JEM

if (steps <= 1) { // base case return triangles(c1, c2); } else { // recursive case Picture p1 = triangleDesign(steps-1, c1, c2); Picture p2 = triangles(c1, c2); return overlay(LL(p1), p2); }

this steps c1 c2

triangleDesign

4 rtw

Execution Land

red blue return p2 p1 p1 p2 ? ?

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Comments

• We now have recursion with returned values,

yippee!!!

• In this case, the number of times to recurse is

determined by a counter (steps) - explicit.

• In this case, there are things left to do in method

after recursive call: non-tail recursion.

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Buggle Example

Given: Suppose we have a random size grid with a random number of bagels in the first row. Problem: Write a method eatBagels such that

• Buggle walks up to wall eating bagels
• Method returns number of bagels which Buggle has eaten

return 5

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eatBagels Method

public int eatBagels() { }

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Exercise 1

Problem: Write a method bagelForward such that

• The method has one integer parameter
• Buggle moves forward the given number of steps dropping

bagels at each cell along the way.

becky.bagelForward(6);

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Exercise 2

Problem: Write a method distanceToWall such that

• The method returns an integer representing how far the

Buggle is from the wall

• The Buggle returns to the cell it started from before the

method was invoked

becky.brushUp(); becky.dropInt(becky.distanceToWall());

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