Download Polynomial Time Hierarchy

Lecture 4
Divide-and-Conquer
What’s Self-Reducibility?
• A problem can be reduced to several
“same” problems with smaller inputs.
Divide and Conquer
• Divide the problem into subproblems.
• Conquer the subproblems by solving
them recursively.
• Combine the solutions to subproblems
into the solution for original problem.
Tree Structure
problem
subproblem
subsubproblem
subproblem
subsubproblem
subsubproblem
subsubproblem
Insertion Sort
5,2,4,6,1,3
5,2,4,6,1
3
5,2,4,6
1
5,2,4
6
5,2
5
4
2
key

5, 2, 4, 6, 1, 3

2, 5, 4, 6, 1, 3

2, 5, 4, 6, 1, 3

2, 4, 5, 6, 1, 3

2, 4, 5, 6, 1, 3

2, 4, 5, 6, 1 , 3

2, 4, 5, 1 , 6, 3

2, 4, 1 , 5, 6, 3

2, 1 , 4, 5, 6, 3

1 , 2, 4, 5, 6, 3

1, 2, 4, 5, 6, 3

1, 2, 4, 5, 3, 6

1, 2, 4, 3, 5, 6

1, 2, 3, 4, 5, 6
Merge Sort
Let Merge - Sort ( A, p, r ) be a procedure
which sorts elements in subarray A[ p...r ].
Main Program
begin
Merge - Sort ( A,1, n);
end
Procedure
Merge - Sort ( A, p, r )
if p  r
then begin q  ( p  r ) / 2;
Merge - Sort ( A, p, q );
Merge - Sort ( A, q  1, r );
Merge ( A, p, q, r );
end - then.
divide
conquer
combine
Heapsort
16
14
7
8
2
10
4
1
9
3
Heapsort
16,14,10,8,7,9,3,2,4,1
1,14,10,8,7,9,3,2,4
16
Heapsort
16,14,10,8,7,9,3,2,4,1
14,8,10,4,7,9,3,2,1
16
Quicksort
Quicksort ( A, p, r )
if p  r
the n begin
q  Partition ( A, p, r ); divide
Quicksort ( A, p, q  1);
conquer
Quicksort ( A, q  1, r );
end - then combine is simplified
Selection with O(n) Comparisons
Algorithm Select :
1. Divide n elements into n/5  groups of 5 elements, possibly
except one of less than 5 elements.
divide
2. Find the median of each group by insertion sort. In case,
the exceptiona l one has two medians, take the larger one.
3. Use Select recursivel y to find the median x of n/5  group
medians.
conquer
4. Exchange x with the last element in input array and apply
Partition subroutine . Let k be the number of elements on the
low side of the partition.
5. Use Select recursivel y to find the ith smallest element on the
low side if i  k , or the (i  k ) th smallest element on the high
side if i  k .
conquer
Remark on Divide and Conquer
Key Point:
1. Self - reducibili ty!!!
2. A recursive tree structure exists between problem
and subproblem s.
Merge Sort
Quick Sort
T (n)  2T (n / 2)  (n)
1
E[T (n)]  (E[T (0)]  E[T (n  1)]  c1n)
n
1
 (E[T (1)]  E[T (n  2)]  c1n)
n

1
 (E[T (n  1)]  E[T (0)]  c1n)
n
Selection
T (n)  T (n / 5)  T (7n / 10  3)  O(n)
Analysis requires to solve a recurrence!
Maximum Subarray
Problem
Given a sequence of integers a1 , a2 ,..., an ,
find a consecutuv e subsequenc e ai , ai 1 ,..., a j
to maximize the sume ai  ai 1    a j .
Max - Subarray( A,low, hig h)
begin
Max - Subarray( low,mid );
Max - Subarray( mid  1,high );
Find - Max - Crossing - Subarray( A,low,mid,high );
return the max one among above three subsequenc es;
end.
Find  Max  Cross  Subarray can be done in O(n)
time. Therefore,
T (n)  2T ( n / 2)  O(n)
T (n)  O(n log n).
Closest pair of points
Given n points in the Euclidean plane, find
a pair of points to minimize the distance between
them.
Divide all points into two half parts based on
x  coordinate s.
Find the closest pair of points in each part.
Suppose 1 and  2 are distances of closest pairs
in two parts, respective ly.
Let   min( 1 ,  2 ).
Matrix Multiplication
C  AB
Step 1
 C11 C12   A11

  
 C21 C22   A21
A12  B11

A22  B21
B12 

B22 
n/2
n/2
Step 2
Step 3
S1  B12  B22
S 6  B11  B22
S 2  A11  A12
S 7  A12  A22
S3  A21  A22
S8  B21  B22
S 4  B21  B11
S9  A11  A21
S5  A11  A22
S10  B11  B12
P1  A11  S1
P5  S5  S 6
P2  S 2  B22
P6  S 7  S8
P3  S3  B11
P7  S9  S10
P4  A22  S 4
( n )
2
7  T ( n / 2)
Step 4
C11  P5  P4  P2  P6
C12  P1  P2
C21  P3  P4
C22  P5  P1  P3  P7
if n  1,
(1)
T ( n)  
2
7

T
(
n
/
2
)


(
n
) if n  1.

By Master Theorem,
T (n)  O(n log2 7 ).
( n )
2
Master Theorem
Let T (n)  aT (n / b)  f (n) where constants
a  1, b  1, and n / b means n / b  or n / b .
If f (n)  O(n logb a  ) for some constant   0,
then T (n)  (n logb a ).
If f (n)  (n logb a ), then T (n)  (n logb a lg n).
If f (n)  (n
logb a 
) for some constant   0,
and if af (n / b)  cf (n) for a constant c  1 and
sufficient ly large n, then T (n)  ( f (n)).