Daa unit 2

Unit-2
Algorithms Using
Divide-and-conquer

Introduction
 Divide-and-conquer is a top-down
technique for designing algorithms
that consists of dividing the problem
into smaller sub problems hoping
that the solutions of the sub
problems are easier to find and then
composing the partial solutions into
the solution of the original problem.

Divide-and-conquer
 1. Divide: Break the problem into
sub-problems of same type.
2. Conquer: Recursively solve
these sub-problems.
3. Combine: Combine the solution
sub-problems.

Divide-and-conquer
 For Example: Assuming that each
divide step creates two sub-
problems

Divide-and-conquer
 For Example: If we can divide the
problem into more than two, it
looks like this

Divide-and-conquer Detail
Divide/Break
 This step involves breaking the problem
into smaller sub-problems.
 Sub-problems should represent a part of
the original problem.
 This step generally takes a recursive
approach to divide the problem until no
sub-problem is further divisible.
 At this stage, sub-problems become
atomic in nature but still represent some
part of the actual problem.

Conquer/Solve
 This step receives a lot of smaller sub-
problems to be solved.
 Generally, at this level, the problems are
considered 'solved' on their own.

Merge/Combine
 When the smaller sub-problems are
solved, this stage recursively combines
them until they formulate a solution of
the original problem.
 This algorithmic approach works
recursively and conquer & merge steps
works so close that they appear as one.

Standard Algorithms
based on D & C
 The following algorithms are based on
divide-and-conquer algorithm design
paradigm.
 Merge Sort
 Quick Sort
 Binary Search
 Strassen's Matrix Multiplication
 Closest pair (points)
 Cooley–Tukey Fast Fourier Transform
(FFT) algorithm

Advantages of D & C
1. Solving difficult problems:
Divide and conquer is a powerful tool for solving
conceptually difficult problems: all it requires is a
way of breaking the problem into sub-problems,
of solving the trivial cases and of combining sub-
problems to the original problem.
2. Parallelism:
Divide and conquer algorithms are naturally
adapted for execution in multi-processor
machines, especially shared-memory systems
where the communication of data between
processors does not need to be planned in
advance, because distinct sub-problems can be
executed on different processors.

Advantages of D & C
3. Memory Access:
Divide-and-conquer algorithms naturally tend to
make efficient use of memory caches. The reason
is that once a sub-problem is small enough, it and
all its sub-problems can, in principle, be solved
within the cache, without accessing the slower
main memory.
4. Roundoff control:
In computations with rounded arithmetic, e.g.
with floating point numbers, a divide-and-conquer
algorithm may yield more accurate results than a
superficially equivalent iterative method.

Advantages of D & C
 For solving difficult problems like Tower
Of Hanoi, divide & conquer is a powerful
tool
 Results in efficient algorithms.
 Divide & Conquer algorithms are adapted
foe execution in multi-processor
machines
 Results in algorithms that use memory
cache efficiently.

Limitations of D & C
 Recursion is slow.
 Very simple problem may be more
complicated than an iterative approach.
Example: adding n numbers etc

Example of Multiplication of
N digit integers.
 Multiplication can be perform using divide
and conquer technique.
 First we know the decimal system of
number which are shown as under.
Number is 3754
3*103 + 7*102 + 5*101 + 4*100

Example of Multiplication of
N digit integers.
 2345 * 6789
2*103 + 3*102 + 4*101 + 5*100
6*103 + 7*102 + 8*101 + 9*100
4 3 2 1

Closest-Pair Problem:
Divide and Conquer
 Brute force approach requires comparing every point
with every other point
 Given n points, we must perform 1 + 2 + 3 + … + n-
2 + n-1 comparisons.
 Brute force  O(n2)
 The Divide and Conquer algorithm yields  O(n log
n)
 Reminder: if n = 1,000,000 then
 n2 = 1,000,000,000,000 whereas
 n log n = 20,000,000
2
)1(1
1
nn
k
n
k





Closest-Pair Algorithm
Given: A set of points in 2-D

Step 1: Sort the points in one D

Lets sort based on the X-axis
O(n log n) using quicksort or mergesort
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4

Step 2: Split the points, i.e.,
Draw a line at the mid-point between 7
and 8
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
Sub-Problem
1
Sub-Problem
2

Advantage: Normally, we’d have to
compare each of the 14 points with every
other point.
(n-1)n/2 = 13*14/2 = 91 comparisons
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
Sub-Problem
1
Sub-Problem
2

Advantage: Now, we have two sub-
problems of half the size. Thus, we have to
do 6*7/2 comparisons twice, which is 42
comparisons
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
d1
d2
Sub-Problem
1
Sub-Problem
2
solution d = min(d1, d2)

Advantage: With just one split we cut the
number of comparisons in half. Obviously,
we gain an even greater advantage if we
split the sub-problems.
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
d1
d2
Sub-Problem
1
Sub-Problem
2
d = min(d1, d2)

Problem: However, what if the closest two
points are each from different sub-
problems?
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
d1
d2
Sub-Problem
1
Sub-Problem
2

Here is an example where we have to
compare points from sub-problem 1 to the
points in sub-problem 2.
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
d1
d2
Sub-Problem
1
Sub-Problem
2

However, we only have to compare points
inside the following “strip.”
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
d1
d2
Sub-Problem
1
Sub-Problem
2
dd
d = min(d1, d2)

Step 3: In fact we can continue to split
until each sub-problem is trivial, i.e., takes
one comparison.
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4

Finally: The solution to each sub-problem
is combined until the final solution is
obtained
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4

Finally: On the last step the ‘strip’ will
likely be very small. Thus, combining the
two largest sub-problems won’t require
much work.
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4

1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3 1
4
 In this example, it takes 22 comparisons to
find the closets-pair.
 The brute force algorithm would have taken
91 comparisons.
 But, the real advantage occurs when there are
millions of points.

Closest-Pair Algo
Float DELTA-LEFT, DELTA-RIGHT;
Float DELTA;
If n= 2 then
return distance form p(1) to p(2);
Else
P-LEFT <- (p(1),p(2),p(n/2));
P-RIGHT <- (p(n/2 + 1),p(n/2 + 2),p(n));
DELTA-LEFT <- Closest_pair(P-LEFT,n2);
DELTA-RIGHT <- Closest_pair(P-RIGHT,n2);
DELTA<- minimum(DELTA-LEFT,DELTA-RIGHT)

Closest-Pair Algo
For I in 1---s do
for j in i+1..s do
if(|x[i] – x[j] | > DELTA and
|y[i] – y[j]|> DELTA ) then
exit;
end
if(distance(q[I],q[j] <DELTA)) then
DELTA <- distance (q[i],q[j]);
end
end

Closest-Pair Algo
Array a[]
Array b[]
Integer h,i,j,k;
h<- low;
i <- low;
j=<- mid + 1;
While( h<= mid and j<= high) do
if(a[h] <= a[j]) then
b[i]<- a[h]
h<-h+1

Closest-Pair Algo
Else
b[i]<- a[j]
j<- j+1;
end
i<- i+1;
End
If(h>mid) then
for(k<-j to high) do
b[i] <- a[k];
i<-i+1;

Closest-Pair Algo
Else
for(k<- to mid) do
b[i] <- a[k]
i=i+1;
End
For(k<-low to high)
a[k]<- b[k]
end

Timing Analysis
 D&C algorithm running time in
mainly affected by 3 factors
 The number of sub-instance(α)
into which a problem is split.
 The ratio of initial problem size to
sub problem size(ß)
 The number of steps required to
divide the initial instance and to
combine sub-solutions expressed
as a function of the input size n.

Timing Analysis
 P is D & C Where α sub instance each of
size n/ß
 Let Tp(n) donete the number of steps
taken by P on instances of size n.
Tp(n0) = constant (recursive-base);
Tp(n)= αTp (n/ß)+y(n);
α is number
of sub
instance
ß is number
of sub size
y is constant

Daa unit 2

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