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Subject:
Design and analysis of algorithm( DAA )
Notes prepared By:
Ahmad Sir
9867597554
All the best…
Notes Prepared by- Ahmad Sir
: 9867597554
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UNIT -III
Divide and Conquer Strategies & Greedy
Approach
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: 9867597554
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1. Divide and Conquer strategies:
1. Binary Search :
Algorithm
bsearch (a[i], key, low, high,)
{
mid =( low + high) /2
if ( low > high ) then
return(1)
else if ( a[ mid] = key ) then
return(mid)
else if( a[ mid] < key )
return(bsearch (a[i], key, mid+1, high,))
else
return((bsearch (a[i], key, low, mid-1))
}
Complexity: O(log2n)
This analysis is to determine average no of comparison for successful and unsuccessful search.
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: 9867597554
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2. Quick Sort :
Algorithm
Quick Sort(a, low, high)
{
low = 0
,high = n-1
if (low < high )
mid = partition ( a, low, high)
Quick Sort ( a, low, mid-1)
Quick Sort ( a, mid+1, high)
}
Algorithm partition ( a, low, high)
{
i = low, j = high
while ( a[i] < = key && I < high )
i++
while ( a[j] > key && j < low )
j- if( i< j) then
interchange a[i] & a[j]
else
interchange a[j] & key
}
Complexity:
Best case & Average case: n log2n
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: 9867597554
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Worst case O (n2)
3. Merge Sort:
Algorithm Mergesort( a, low, high)
{
if ( low < high)
{
mid =( low + high) /2
mergesort( a, low, mid)
mergesort( a, mid+1, high)
mergesort( a, low, high)
}
}
Complexity: n log2n
Where ‘n’ is the no of comparisons and log2n is the hight of tree.
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: 9867597554
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4. Min_max :
Algorithm
min_max( low, high, min, max)
{
if ( low = high) then
{
min = max = a [low]
else if ( low = high -1) then
else if ( a[low] < a [high]) then
min = a[low]
max = a[high]
else
min = a[high]
max = a[low]
}
else
mid =( low + high) /2
min_max( low, mid, min, max)
min_max(mid+1, high, min1, max1)
if ( min1 < min) then
min = min1
if( max1> max) then
max = max1
}
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: 9867597554
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Complexity: 2 log2n
5.Heap sort:
Algorithm
Heapsort(A)
{
Build_ heap (A)
For i = n to 1 do
swap ( A[1], A[i] )
n- heapify(A, 1)
Build Max_heap(A)
Heap-size(A) = length [A]
For i int(length[A]/2) to 1 do
Max_heap(A, i)
}
Complexity: O(n logn)
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: 9867597554
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6.Multiplication Algorithm:
Stasseens matrix multiplication:
A
A = ( 𝟏𝟏
A𝟐𝟏
A𝟏𝟐
)
A𝟐𝟐
B= (
B𝟏𝟏
B𝟐𝟏
B𝟏𝟐
)
B𝟐𝟐
S1 = B12 – B22
S2 = A11 + A12
S3 = A21 + A22
S4 = B21 – B11
S5 = A11 + A22
S6 = B11 + B22
S7 = A12 – A22
S8 = B21 + B22
S9 = A11 – A21
S10 = B11 + B12
P1 = S1 ∙A11
P2 = S2 ∙B22
P3 = S3 ∙B11
P4 = S4 ∙A22
C11 = - P2 + P4 + P5 + P6
C12 = P1 + P2
C21 = P3 + P4
C11 = P1 – P3 + P5 - P7
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: 9867597554
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A
( 𝟏𝟏
A𝟐𝟏
A𝟏𝟐
B
) . ( 𝟏𝟏
B𝟐𝟏
A𝟐𝟐
B𝟏𝟐
C
) = ( 𝟏𝟏
B𝟐𝟐
C𝟐𝟏
C𝟏𝟐
)
C𝟐𝟐
Greedy Methods:
1. Knapsack Problem:
Assumption:
1. P[1….n] profit array
2. W[1…n] weight array
Capacity = total capacity
Weight = 0
x[1…n] is the array that contains three types of values
1 = complete
0 = reject
Fraction = partially
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: 9867597554
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Algorithm
knapsack(P, W, capacity, weight)
Step1. For i = 1 to n do
x[i] = 0
weight = 0
Step2. while ( weight < capacity) do
{
Select the best object with max (Pi/ Wi) ratio, let its index = i
if (weight + W[i] <= capacity ) then
{
x[i] = 1
weight = weight +W[i]
}
else
{
x[i] =
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦−𝑤𝑒𝑖𝑔ℎ𝑡
𝑊[𝑖]
weight = capacity
}
Step3. Profit = 0
For i = 1 to n do
Profit = profit + [x[i]×P[i]]
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}
Complexity: O( nlog n)
2.Application to job sequencing with deadlines problem:
job_scheduling(d[0….n] : j[0…n])
Algorithm
{
d[0] = 0
j[0] = 0
k =1 , j[1] = 1
for i = 2 to n do
{
r=k
while( (d[ j[r]) > max (d[i],r))
r = r-1
if( d[i] > r ) then
for m = k to r + 1 do
j[r +1] = i
k ++
}
}
All the best…
Notes Prepared by- Ahmad Sir
: 9867597554
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3.Minimum cost spanning trees:
1.Kruskal’s algorithm:
Assumption:
1. G is a graph which represented in the form of cost matrix of size n×n
2. minimum cost = 0
3. there are two functions used in algorithm
1. Find:
These function will find the component in which the vertex are present.
2. Merge:
These function will merge the two component if they are different.
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Algorithm
kruskal(cost, G, n, mincost: t)
{
Step1. Create component as per no of vertices in the graph
// n: component
Step2.
i=1
mincost = 0
while( i < = n-1) do
select the best edge and it be (u,v)
x = Find(u)
y = Find(v)
if (x ≠ y)
t[i, 1] = u
t[i, 2] = v
t[i, 3] = cost[u,v]
mincost = cost[u,v] + mincost
Merge[x,y]
i = i+1
}
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Complexity: Elog V
2. Prim’s algorithm:
Algorithm prims(G, n, cost, near, mincost:t)
{
Step1. Consider t
Let cost[k, l] be the min value in cost matrix
t[1,1] = k
t[1, 2] = l
t[1,3] = cost[k, l]
// consider near array
For i = 1 to n do
{
if( cost [k, i] < cost[l, i] )
near[i] = k
else
near[i] = l
}
near[k] = 0
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near[l] = 0
step2. // refer t
for i = 2 to n-1
{
Let j be index in near array such that near[j] ≠0 & cost[ near[j], j] is min
t[i, 1] = near[j]
t[i, 2] = j
t[i, 3] = cost[near[j], j]
mincost = mincost + cost[near[j], j]
near[j] = 0
for k = 1 to n do
{
if(near[k] ≠ 0) and ( cost[k, j] < cost[k.near[k])
near[k] = j
}
}
Complexity: E + log V
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Notes Prepared by- Ahmad Sir
: 9867597554
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Difference between Kruskal’s and Prim’s algorithm:
Kruskal’s
S.N.
Prim’s
1.
In this, edge is selected with minimum
value.
In this, edge is selected with minimum value
and connectivity.
2.
Graph generated with isolated and will
required connecting edge
Graph generated will be connected but no
inter path require.
3.
Complexity is E log V
Complexity is E+ log V
4. Single Source Shortest path(SSSP):
Algorithm SSSP( cost, n, dist)
Step1. For i = 1 to n do
{
S[i] = 0
dist[i] = cost[v,i]
}
Step2.
S[v] = 0
dist[v] = 0
Step3. for i =2 to n do
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{
Determine n-1 path from vertex ‘v’ and let the vertices ‘u’ select a vertex such that
S[i] ≠ 1 and dist[u] = min
for ( each ‘w’ adjacent to vertex ‘u’)
{
if( dist[w] > dist[u] + cost[u, w]) then
dist[w] = dist[u] + cost[u,w]
}
}
Complexity: O(n2)
All the best…
Notes Prepared by- Ahmad Sir
: 9867597554
Page 17