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Ari’s class: PHO 211 – 2/7/06
1. heapsort (I've already introduced min- and max- heaps)
2. information-theoretic lower-bound on comparison-based
sorting
3. counting sort (aka bin sort)
4. radix sort
5. bucket sort
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Priority Heaps

Basic Heap ADT:
Data is a[i] i = 1,...,N
 Methods: Insert (key), Delete(key), DeleteMin, Build and Sort


Q: When is a tree an array? A: complete tree
Parent a[i]  a[2i] = left child & a[2i+1] = right child
 Child a[j]:  a[ j/2 ] = parent (integer division).


Build Heap is O(N) and DeleteMin is O(log(N)) 

Heap sort by deleting min over and over is O(Nlog(N)).
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Min Heap Order
depth =0/ height = 3
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depth =1/height = 2
depth =2/height=1
24
height =4
9
2
23
10
25
Partial
Ordering
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3
18
4
120
depth = 3/heigth = 0
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Algebraic Details for TH(N) vs TD(N)
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Week 4: Lecture 7, Sept 23, 2002
Proof of
(Nlog(N))
Decision Tree
a, b, c
a<b
a<b, c
b<a, c
a<c
b<c
a< b ,a<c
c< a < b
b <a,c<a
c < b <a
c< a
b<c
a<b< c
a< c < b
b< a < c
b<c< a
Binary decisions: 3! = 6 possible outcomes. Longest path: log(3!)
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Lower Bound Theorem for Camparision Sort
Proof: Compute the maximum depth D of decision tree?



Need N! leaves to get all possible outcomes of a sorting routine.
Each level at most doubles:
1  2  4  8      2D
Consequently for D levels:
 T ( N )  ( D)  (log 2 ( N!))  ( N log 2 ( N ))
Information  log 2 ( N!)  N log 2 ( N )
Number of bits to encode any (initial) state is information ( - Entropy)
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Bisection Search of Sorted List

int a[0], a[1],a[2],a[3],.... a[m],....
a[2],a[N-1]
i
j
i= 0; j= N-1; m = N/2
while(b!=a[m] && i!=j ){
if(b>a[m]) i = m;
if(b<a[m]) j = m;
m = (j-i)/2 + i;}
if(b==a[m])) “found it” else “not found”
T(N) = T(N/2) + c0

Choose
mid point
T(N) » Log(N)
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Dictionary: Sorted and Uniform

int a[0], a[1],a[2],a[3],.... a[m],....
i
Dictionary: Same code EXCEPT
estimate location of b
x = fractional distance (0<x<1)
x = (b-a[i])/(a[j] – a[i]) ;
m = x (j-i) + i ;
T(N) = T(N1/2) + c0

a[2],a[N-1]

j
m
T(N) » Log(Log(N))
Extra Knowledge Helps: % Error » 1/N1/2
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O(N): Bin, Bucket & Radix

BIN Sort – make histogram:


N integers 0 < a[i] < M in the range v= 0,...,M-1.
Count number of occurrences in a[i]
for(v=0; v<M; v++) bin[ v ] =0;
for(i=0;i<N; i++) bin[a[i]] ++;
j=0;
for(v=0; v< M; v++) {
for(i=0; i<bin[v]; i++)
a[ j ] = v; j++; }

O(M + N) so if M » N it is O(N)
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Bucket Sort

Choose B Buckets as bins for high digits of a[i]


place N numbers in a[i] in buckets
Sort average of N/B elements in each bucket.
Linked list:
Bucket:
0
1
2
3
4
5
6
 O( N + B*(N/B log(N/B) ) = O( N + N log(N/B))
7
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B = O(N)
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Radix Sort (IBM Card Sorter!)

Represent integers in a[i] in base B: n0 + n1 B + n2 B2 + .... + np BP

Sort into buckets by low digits first: n0, then n1, etc.
Queues: B= 10
Example: 64, 8, 216,
Bucket: 0
1
2
#1
0
1
512
#2
8
1
0
216
512
729
27
125
#3
64
27
8
1
0
125
216
3
343
4
64
512,
0,
1, 343, 125
5
6
7
8
9
125
216
27
8
729
343
343
27, 729,
64
512
729
O( N P ) where BP = M or P= log(M) / log(B) = O(1)
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End Ari’s Class
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