Lecture #10

מבוא מורחב1

Lecture #10

Sets (2.3.3)

Huffman Encoding Trees (2.3.4)


Sets

Methods:

(element-of-set? x set)

(adjoin-set x set)

(union-set s1 s2)

(intersection-set s1 s2)

A set is a collection of distinct items.


Implementation

We will see three ways to implement sets.

1. With lists2. With sorted lists3. With trees

And compare the three methods.


First implementation: Lists

Empty set empty list ‘()

Adding an element cons

Set Union append

Set Intersection

But: we also need to remember to remove duplicats.


First implementation: Lists

(define (element-of-set? x set) (cond ((null? set) false) ((equal? x (car set)) true) (else (element-of-set? x (cdr set)))))

equal? : Like eq? for symbols. Works for numbers

Works recursively for compounds.

(eq? (list ‘a ‘b) (list ‘a ‘b)) #f

(equal? (list ‘a ‘b) (list ‘a ‘b)) #t


(define (adjoin-set x set)

(if (element-of-set? x set)

set

(cons x set)))

First Implementation: Adjoin


Intersection

(define (intersection-set set1 set2) (cond ((or (null? set1) (null? set2)) '()) ((element-of-set? (car set1) set2) (cons (car set1) (intersection-set (cdr set1) set2))) (else (intersection-set (cdr set1) set2))))


Union

(define (union-set set1 set2) (cond ((null? set1) set2)) ((not (element-of-set? (car set1) set2)) (cons (car set1) (union-set (cdr set1) set2))) (else (union-set (cdr set1) set2))))

(define (union-set set1 set2) (cond ((null? set1) set2)) (else (adjoin-set (car set1) (union-set (cdr set1) set2)))))


Analysis

Element-of-set

Adjoin-set

Intersection-set

Union-set

(n)

(n2)

(n)

(n2)


Second implementation: Sorted Lists

Empty set empty list ‘()

Adding an element We add the element to the list so that the list is sorted.

Set Union sorted list of union.

Set Intersection sorted list of intersection.


Membership?

(define (element-of-set? x set) (cond ((null? set) false) ((= x (car set)) true) ((< x (car set)) false) (else (element-of-set? x (cdr set)))))

(n) steps.

adjoin-set is similar, try it yourself


Intersection

(define (intersection-set set1 set2) (cond ((or (null? set1) (null? set2)) '()) ((element-of-set? (car set1) set2) (cons (car set1) (intersection-set (cdr set1) set2))) (else (intersection-set (cdr set1) set2))))

Can we do it better ?


Better Intersection

(define (intersection-set set1 set2) (if (or (null? set1) (null? set2)) '() (let ((x1 (car set1)) (x2 (car set2))) (cond ((= x1 x2) (cons x1 (intersection-set (cdr set1) (cdr set2)))) ((< x1 x2) (intersection-set (cdr set1) set2)) ((< x2 x1) (intersection-set set1 (cdr set2)))))))


Example

set1 set2 intersection

(1 3 7 9) (1 4 6 7) (1

(3 7 9) (4 6 7) (1

(7 9) (4 6 7) (1

(7 9) (6 7) (1

(7 9) (7) (1

(9) () (1 7)

Time and space (n)

Union -- similar


Complexity

Element-of-set

Adjoin-set

Intersection-set

Union-set

(n)

(n2)

(n)

(n2)

unsorted sorted

(n)

(n)

(n)

(n)


Version 3: Binary Trees

7

3

1 5

9

12

3

1

5

12

7

9

Balanced Tree Unbalanced Tree

Depth = (log n)


Interface of Binary Trees.

7

3

1 5

9

12 Selectors:

(define (entry tree) (car tree))

(define (left-branch tree) (cadr tree))

(define (right-branch tree) (caddr tree))

Constructor: (define (make-tree entry left right) (list entry left right))


Element-of-set

(define (element-of-set? x set) (cond ((null? set) false) ((= x (entry set)) true) ((< x (entry set)) (element-of-set? x (left-branch set))) ((> x (entry set)) (element-of-set? x (right-branch set)))))

Complexity: (d)

If tree is balanced d log(n)


Complexity

Element-of-set

Adjoin-set

Intersection-set

Union-set

(n)

(n2)

(n)

(n2)

unsorted sorted

(n)

(n)

(n)

(n)

trees

(log(n))

(log(n))

(n log(n))

(n log(n))


Huffman encoding trees


Data Transmission

A B

“sos”

We wish to send information efficiently from A to B


Fixed Length Codes

Represent data as a sequence of 0’s and 1’s

Example: BACADAEAFABBAAAGAH

A 000 B 001 C 010 D 011

E 100 F 101 G 110 H 111

001000010000011000100000101000001001000000000110

000111

This is a fixed length code. Sequence is 18x3=54 bits long.

Can we make the sequence of 0’s and 1’s shorter ?


Variable Length Code

A 0 B 100 C 1010 D 1011

E 1100 F 1101 G 1110 H 1111

Example: BACADAEAFABBAAAGAH

100010100101101100011010100100000111001111

Make use of frequencies. Frequency of A=8, B=3, others 1.

But how do we decode?

42 bits (20% shorter)

24

Prefix code Binary tree

A 0 B 100 C 1010 D 1011

E 1100 F 1101 G 1110 H 1111

Prefix code: No codeword is a prefix of the other

A

B

C D E F G H

0 1

1

1

1

0

0

0

0

00

1

11

25

Decoding Example

10001010A

B

C D E F G H

0 1

1

1

1

0

0

0

0

00

1

11

10001010 B10001010 BA10001010 BAC


Decoding a Message

(define (choose-branch bit branch) (cond ((= bit 0) (left-branch branch)) ((= bit 1) (right-branch branch)) (else (error "bad bit -- CHOOSE-BRANCH" bit))))


Decoding a Message

(define (decode bits Huffman_tree current_branch) (if (null? bits) '() (let ((next-branch (choose-branch (car bits) current_branch))) (if (leaf? next-branch) (cons (symbol-leaf next-branch) (decode (cdr bits) Huffman_tree Huffman_tree)) (decode (cdr bits) Huffman_tree next-branch)))))

28

The cost of a weighted Tree

17

2 2

9

45

2

A

B

C D E F G H

8

3

1 1 1 1 1 1

{G,H}{E,F}{C,D}

{B,C,D} {E,F,G,H}

{B,C,D,E,F,G,H}

{A,B,C,D,E,F,G,H}

29

Huffman Tree = Optimal Length Code

Optimal: no code has better weighted average length

A

B

C D E F G H

0 1

1

1

1

0

0

0

0

00

1

11

8

3

1 1 1 1 1 1

A

D

B C E F G H

0 1

1

1

1

0

0

0

0

00

1

11

8

3 1 1 1 1 1

1


Next Time

• We will describe an efficient algorithm that given the weights, constructs the Huffman tree.

• We will describe the interface it requires and how to implement it.

Lecture #10

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Transcript of Lecture #10