![Mrs Sujata Sathe
1
2 3
8
7
5
2
9
1
1 2 3
1 0 1 8
2 9 0 5
3 2 7 0
A0
Q.Find the shortest path using floyd-warshall algorithm
A1
Now intermediate vertex is 1 i.e so keep the First row and first column as it is and calculate other A[i,j] values using formula
1 2 3
1 0 1 8
2 9 0 5
3 2 0
A1[2,3]= min {A0[2,3], A0[2,1]+A0[1,3] }
=min{5,9+8}=5
5 A1[3,2]= min {A0[3,2], A0[3,1]+A0[1,2] }
=min{7,2+1}=3
3](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-4-320.jpg)
![Mrs Sujata Sathe
A1 1 2 3
1 0 1 8
2 9 0 5
3 2 3 0
Now intermediate vertex is 2 so keep the 2nd row and 2nd column as it is and calculate other A[i,j] values using formula
A2 1 2 3
1 0 1 6
2 9 0 5
3 2 3 0
A2[1,3]= min {A1[1,3], A1[1,2]+A1[2,3] } i =1,j=3 k=2
=min{8,1+5}=6
6
A2[3,1]= min {A1[3,1], A1[3,2]+A1[2,1] } i =3,j=1, k=2
=min{2,3+9}=2
2
Now intermediate vertex is 3 so keep the 3rd row and 3rd column as it is and calculate other A[i,j] values using formula
A3 1 2 3
1 0 6
2 0 5
3 2 3 0
A3[1,2]= min {A2[1,2], A2[1,3]+A2[3,2] } i =1,j=2 ,k=3
=min{1,6+3}=1
1
A3[2,1]= min {A2[2,1], A2[2,3]+A2[3,1] } i =2,j=1, k=3
=min{9, 5+2}=7
7](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-5-320.jpg)
![Mrs Sujata Sathe
1 2 3 4
1 0 4 ∞ 3
2 ∞ 0 2 1
3 5 3 0 ∞
4 1 ∞ 2 0
Q. Use the floyd-warshall algorithm and find all pairs shortest paths for the following adjacency weighted matrix
Cost Matrix =A0
A1
Now intermediate vertex is 1 i.e so keep the First row and first column as it is and calculate other A[i,j] values using formula
1 2 3 4
1 0 4 ∞ 3
2 ∞ 0 2 1
3 5 3 0 8
4 1 5 2 0
A1[2,3]= min {A0[2,3], A0[2,1]+A0[1,3] } i=2, j=3 , k= 1
=min{2,∞ + ∞}=2
A1[2,4]= min {A0[2,4], A0[2,1]+A0[1,4] } i=2, j=4 , k= 1
=min{1,∞ + 3}=1
A1[3,2]= min {A0[3,2], A0[3,1]+A0[1,2] } i=3, j=2 , k= 1
=min{3,5 + 4}=3
A1[3,4]= min {A0[3,4], A0[3,1]+A0[1,4] } i=3, j=4, k= 1
=min{∞,5 + 3}=8
A1[4,2]= min {A0[4,2], A0[4,1]+A0[1,2] } i=4, j=2, k= 1
=min{∞,1 + 4}=5
A1[4,3]= min {A0[4,3], A0[4,1]+A0[1,3] } i=4, j=3, k= 1
=min{2,1 + ∞}=2](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-7-320.jpg)
![Mrs Sujata Sathe
A1 1 2 3 4
1 0 4 ∞ 3
2 ∞ 0 2 1
3 5 3 0 8
4 1 5 2 0
Now intermediate vertex is 2 so keep the 2nd row and 2nd column as it is and calculate other A[i,j] values using formula
A2 1 2 3 4
1 0 4 6 3
2 ∞ 0 2 1
3 5 3 0 4
4 1 5 2 0
A3 1 2 3 4
1 0 4 6 3
2 7 0 2 1
3 5 3 0 4
4 1 5 2 0
Now intermediate vertex is 3 so keep the 3rd row and 3rd column as it is and calculate other A[i,j] values using formula](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-8-320.jpg)
![Mrs Sujata Sathe
A3 1 2 3 4
1 0 4 6 3
2 7 0 2 1
3 5 3 0 4
4 1 5 2 0
Now intermediate vertex is 4 so keep the 4th row and 4th column as it is and calculate other A[i,j] values using formula
A4
1 2 3 4
1 0 4 5 3
2 2 0 2 1
3 5 3 0 4
4 1 5 2 0](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-9-320.jpg)
![Mrs Sujata Sathe
v4
v3
v1
v2
3
1
6
1
Use bellman –ford algorithm to find shortest path from v1
Source vertex = V1
D[vi]=∞ and D[v1]= 0 0
2
3
1
Edges list
(1,2) (1,3) (2,3) (2,4) (3,4)
Consider the edge (1,2)
d[2]=∞
d[1]+cost[1,2]=0+1=1 so update d[2] = 1
Consider the edge (1,3)
d[3]=∞
d[1]+cost[1,3]=0+3=3, so update d[3] = 3
Relaxtion:if there is an edge from u,v
If d[v]> d[u] +cost[u,v]
Then update d[v]= d[u] +cost[u,v]
Otherwise don’t change
D[v1] D[v2] D[v3
]
D[v4]
0 ∞ ∞ ∞
1 0 1 2 3
2 0 1 2 3
3
Do the relaxation of edges till n-1 times
where n is the no of vertices
D[v1] 0
D[v2] 1
D[v3] 2
D[v4] 3](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-10-320.jpg)
![Mrs Sujata Sathe
1
6
4
5
2
3 7
6
-1
5
5
-2
-2
-1
1
3
3
Edge list
(1,2),(1,3),(1,4),(2,5),(3,2),(3,5),(4,3),(4,6),(5,7),(6,7)
0
1
3
0
3
4
5
Relaxtion:if there is an edge from u,v
If d[v]> d[u] +cost[u,v]
Then update d[v]= d[u] +cost[u,v]
Otherwise don’t change](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-11-320.jpg)
![Mrs Sujata Sathe
Edge list
(1,2),(1,3),(1,4),(2,5),(3,2),(3,5),(4,3),(4,6),(5,7),(6,7)
Iterat
ion
D[v1] D[v2] D[v3] D[v4] D[v5] D[v6] D[v7]
0 0 ∞ ∞ ∞ ∞ ∞ ∞
1 0 3 3 5 5 4 7
2 0 1 3 5 2 4 5
3 0 1 3 5 0 4 3
4 0 1 3 5 0 4 3
D[v1] 0
D[v2] 1
D[v3] 3
D[v4] 5
D[v5] 0
D[v6] 4
D[v7] 3
Solution Set](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-12-320.jpg)
![Mrs Sujata Sathe
b d o
0 1 2
a b c d o
0 1 2 3 4
A
B
int LCS(i,j)
{
if (A[i] = = ‘0’ | | B[j] = = ‘0’)
return 0;
elseif (A[i] = = B[j])
1+ LCS(i+1, j+1);
else
max(LCS(i+1,j),LCS(i,j+1));
}
A[0], B[0]
b ,a
A[1], B[0]
d ,a
A[0], B[1]
b ,b
A[2], B[0]
0 , a
A[1], B[1]
d , b
0
A[2], B[1]
0 ,b
A[1], B[2]
d ,c
0
A[2], B[2]
0 ,c
A[1], B[3]
d ,d
0
A[2], B[4]
0 ,0
1+
1+0=1
1
1
1
1
A[1], B[2]
d ,c
1+
A[1], B[2]
d ,c
A[2], B[2]
0 ,c
A[1], B[3]
d ,d
0
A[2], B[4]
0 ,0
1+
1+0=1
1
1
1
2
2
2
2 2
1 1 1 1
0 0 0 0
a b c d 0
0 1 2 3 4
b 0
d 1
0 2](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-14-320.jpg)
![Mrs Sujata Sathe
S1: {b,d}, S2:{a,b,c,d}
0 0 0 0 0
0 0 1 1 1
0 0 1 1 2
0 1 2 3 4
0
1
2
a b c d
b
d
If s1[i] == s2[j] ( if there is a match)
then 1+ diagonal element
If not
Then max(Prev row, prev col)
d
b
LCS = {b,d}](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-15-320.jpg)
![Mrs Sujata Sathe
W=8, P={1,2,5,6}
W={2,3,4,5}
0 0 0 0 0 0 0 0 0
0 0 1 1 1 1 1 1 1
0 0 1 2 2 3 3 3 3
0 0 1 2 5 5 6 7 7
0 0 1 2 5 6 6 7 8
0 1 2 3 4 5 6 7 8
0
1
2
3
4
V
Pi Wi
1 2
2 3
5 4
6 5
Weights
I
T
E
M
S
V[i,w] = max{ V[i-1,w], V[i-1,w-w[i]]+ P[i]}
V[4,1]=max { V[3. 1], V[3, 1-5] + 6 }
= max{ 0, V[3,-4]+6 }
= 0
Solution set = {0,1,0,1}
8
items X1 X2 X3 X4
0 1 0 1
8-6=2
2-2=0](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-29-320.jpg)
![Mrs Sujata Sathe
A1.A2.A3.A4
5x4 4 x 6 6 X 2 2 x 7
0 120 88
0 48 104
0 84
0
1 2 3 4
1
2
3
4
M
1 1
2 3
3
1 2 3 4
1
2
3
4
S
M[1,1]=0, M[2,2]=0,
M[3,3]=0,M[4,4]
M[1,2]=120
(A1.A2)
5X4 4X6
Cost= 5*4*6=120
M[2,3]=48
A2.A3
4X6 6X2
Cost= 4*6*2=48
M[3,4]=84
A3.A4
6X2 2X7
Cost= 6*2*7=84
M[1,3] = 88 ie A1.A2.A3
A1.(A2.A3) (A1.A2).A3
5X4 4X6 6X2 5X4 4X6 6X2
M[1,1]+M[2,3]+5*4*2 M[1,2]+M[3,3]+
5*6*2
=0+48+40 =120+0+60
=88 =180
M[2,4]= i.e A2.A3.A4
A2.(A3.A4) (A2.A3).A4
4X6 6X2 2X7 4X6 6X2 2X7
M[2,2]+M[3,4]+4*6*7 M[2,3]+M[4,4]+ 4*2*7
=0+84+168 =48+0+56
=252 =104](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-33-320.jpg)
![Mrs Sujata Sathe
0 120 88 158
0 48 104
0 84
0
1 2 3 4
1
2
3
4
M
1 1 3
2 3
3
1 2 3 4
1
2
3
4
S
M[1,4]= min{M[1,1]+ M[2,4]+5*4*7,
M[1,2]+M[3,4]+5*6*7,M[1,3]+M[4,4] +5*2*7}
= {0+104+140, 120+84+210, 88+0+70}
=158](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-34-320.jpg)
![Mrs Sujata Sathe
A1.A2.A3.A4
4x5 5 x 3 3 X 2 2 x 7
0 60 70 126
0 30 100
0 42
0
1 2 3 4
1
2
3
4
M
0 1 1 3
0 2 3
0 3
0
1 2 3 4
1
2
3
4
S
M[1,1]=0, M[2,2]=0, M[3,3]=0,M[4,4]=0,
M[1,2]=60
(A1.A2)
4X5 5X3
Cost= 4*5*3=60
M[2,3]=30
(A2.A3)
M[2,2]+M[3,3]+5X3X2
M[3,4]=
(A3.A4)](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-35-320.jpg)
![Mrs Sujata Sathe
C[i,j] =min {C[i,k-1]+ C[k,j]}+ w[i,j]
i<k≤j
1 2 3 4
10 20 30 40
4 2 6 3
keys
frequency
No
0 4 81 203 263
0 2 103 163
0 6 123
0 3
0
0 1 2 3 4
0
1
2
3
4
i
j
C
C[0,0]= C[1,1]=C[2,2]=C[3,3]=C[4,4]=0
C[0,1]=4
Consider only 1 key i.e10 & Freq=4
C[1,2]=2
C[2,3]=6 C[3,4]=3
C[0,2] means there are 2 keys 10, 20
For j-i=1
1-0=1, C[0,1]
2-1=1, C[1,2]
3-2=1,C[2,3]
4-3=1,C[3,4]
For j-i=2
2-0=2, C[0,2]
3-1=2, C[1,3]
4-2=2, C[2,4]
4
w[0,4]=ΣF(i)
i=1
10
20
20
10
4*1
2*2
2*1
4*2
4*1+2*2= 8 2*1+4*2= 10
C[1,3] means there are 2 keys 20, 30
C[1,3]= min C[1,1]+ C[2,3],
k=2,3 C[1,2]+C[3,3], + w[1,3]
=min{0+6+8, 2+0+8}
=min{14,10}
w[1,3]=2+6=8 Cost of OBST=26 and the root node is 3
3
r[0,4]=3
1 4
r[0,2] r[3,4]
2
r[0,0]
r[1,2]
r[1,1] r[2,2]
r[3,3] r[4,4]
30
10
20
40](https://image.slidesharecdn.com/unit3dynamicprogramming1-240724145818-23f19343/85/unit3_Dynamic_Progrghaiajawzjabagamming1-pptx-39-320.jpg)
unit3_Dynamic_Progrghaiajawzjabagamming1.pptx
- 1. CSC 3504: Design and Analysis of Algorithms-I Mrs. Sujata Sathe Assistant Professor Computer Science Department Fergusson College (Autonomous), Pune Mrs Sujata Sathe
- 2. Mrs Sujata Sathe Unit-III Dynamic Programming Strategy used in DP, difference between DP and DandC Principle of optimality (Tabulation and Memoization) Matrix chain multiplication Problem Example 1:Matrix Chain multiplication Example 2:Matrix Chain multiplication Shortest paths theory Floyd Warshall Algorithm Examples of calculating shortest path using Floyd Warshall Algorithm Analysis of Floyd Warshall Algorithm Bellman - ford algorithm Examples of calculating shortest path using Bellman - ford Algorithm Concept of String editing and its usage Example 1: String Editing Example 2: String Editing Knapsack problem strategy 0/1 Knapsack problem Examples of knapsack Longest common Subsequence Example 1: LCS Example 2: LCS Travelling Salesman Problem Example 1: TSP Example 2: TSP Example 3: TSP Optimal Binary Search Trees Example 1: OBST Example 2: OBST Content
- 3. Mrs Sujata Sathe Finding all pairs shortest path 1->2 1->3 2->1 2->3 3->1 3->2
- 4. Mrs Sujata Sathe 1 2 3 8 7 5 2 9 1 1 2 3 1 0 1 8 2 9 0 5 3 2 7 0 A0 Q.Find the shortest path using floyd-warshall algorithm A1 Now intermediate vertex is 1 i.e so keep the First row and first column as it is and calculate other A[i,j] values using formula 1 2 3 1 0 1 8 2 9 0 5 3 2 0 A1[2,3]= min {A0[2,3], A0[2,1]+A0[1,3] } =min{5,9+8}=5 5 A1[3,2]= min {A0[3,2], A0[3,1]+A0[1,2] } =min{7,2+1}=3 3
- 5. Mrs Sujata Sathe A1 1 2 3 1 0 1 8 2 9 0 5 3 2 3 0 Now intermediate vertex is 2 so keep the 2nd row and 2nd column as it is and calculate other A[i,j] values using formula A2 1 2 3 1 0 1 6 2 9 0 5 3 2 3 0 A2[1,3]= min {A1[1,3], A1[1,2]+A1[2,3] } i =1,j=3 k=2 =min{8,1+5}=6 6 A2[3,1]= min {A1[3,1], A1[3,2]+A1[2,1] } i =3,j=1, k=2 =min{2,3+9}=2 2 Now intermediate vertex is 3 so keep the 3rd row and 3rd column as it is and calculate other A[i,j] values using formula A3 1 2 3 1 0 6 2 0 5 3 2 3 0 A3[1,2]= min {A2[1,2], A2[1,3]+A2[3,2] } i =1,j=2 ,k=3 =min{1,6+3}=1 1 A3[2,1]= min {A2[2,1], A2[2,3]+A2[3,1] } i =2,j=1, k=3 =min{9, 5+2}=7 7
- 7. Mrs Sujata Sathe 1 2 3 4 1 0 4 ∞ 3 2 ∞ 0 2 1 3 5 3 0 ∞ 4 1 ∞ 2 0 Q. Use the floyd-warshall algorithm and find all pairs shortest paths for the following adjacency weighted matrix Cost Matrix =A0 A1 Now intermediate vertex is 1 i.e so keep the First row and first column as it is and calculate other A[i,j] values using formula 1 2 3 4 1 0 4 ∞ 3 2 ∞ 0 2 1 3 5 3 0 8 4 1 5 2 0 A1[2,3]= min {A0[2,3], A0[2,1]+A0[1,3] } i=2, j=3 , k= 1 =min{2,∞ + ∞}=2 A1[2,4]= min {A0[2,4], A0[2,1]+A0[1,4] } i=2, j=4 , k= 1 =min{1,∞ + 3}=1 A1[3,2]= min {A0[3,2], A0[3,1]+A0[1,2] } i=3, j=2 , k= 1 =min{3,5 + 4}=3 A1[3,4]= min {A0[3,4], A0[3,1]+A0[1,4] } i=3, j=4, k= 1 =min{∞,5 + 3}=8 A1[4,2]= min {A0[4,2], A0[4,1]+A0[1,2] } i=4, j=2, k= 1 =min{∞,1 + 4}=5 A1[4,3]= min {A0[4,3], A0[4,1]+A0[1,3] } i=4, j=3, k= 1 =min{2,1 + ∞}=2
- 8. Mrs Sujata Sathe A1 1 2 3 4 1 0 4 ∞ 3 2 ∞ 0 2 1 3 5 3 0 8 4 1 5 2 0 Now intermediate vertex is 2 so keep the 2nd row and 2nd column as it is and calculate other A[i,j] values using formula A2 1 2 3 4 1 0 4 6 3 2 ∞ 0 2 1 3 5 3 0 4 4 1 5 2 0 A3 1 2 3 4 1 0 4 6 3 2 7 0 2 1 3 5 3 0 4 4 1 5 2 0 Now intermediate vertex is 3 so keep the 3rd row and 3rd column as it is and calculate other A[i,j] values using formula
- 9. Mrs Sujata Sathe A3 1 2 3 4 1 0 4 6 3 2 7 0 2 1 3 5 3 0 4 4 1 5 2 0 Now intermediate vertex is 4 so keep the 4th row and 4th column as it is and calculate other A[i,j] values using formula A4 1 2 3 4 1 0 4 5 3 2 2 0 2 1 3 5 3 0 4 4 1 5 2 0
- 10. Mrs Sujata Sathe v4 v3 v1 v2 3 1 6 1 Use bellman –ford algorithm to find shortest path from v1 Source vertex = V1 D[vi]=∞ and D[v1]= 0 0 2 3 1 Edges list (1,2) (1,3) (2,3) (2,4) (3,4) Consider the edge (1,2) d[2]=∞ d[1]+cost[1,2]=0+1=1 so update d[2] = 1 Consider the edge (1,3) d[3]=∞ d[1]+cost[1,3]=0+3=3, so update d[3] = 3 Relaxtion:if there is an edge from u,v If d[v]> d[u] +cost[u,v] Then update d[v]= d[u] +cost[u,v] Otherwise don’t change D[v1] D[v2] D[v3 ] D[v4] 0 ∞ ∞ ∞ 1 0 1 2 3 2 0 1 2 3 3 Do the relaxation of edges till n-1 times where n is the no of vertices D[v1] 0 D[v2] 1 D[v3] 2 D[v4] 3
- 11. Mrs Sujata Sathe 1 6 4 5 2 3 7 6 -1 5 5 -2 -2 -1 1 3 3 Edge list (1,2),(1,3),(1,4),(2,5),(3,2),(3,5),(4,3),(4,6),(5,7),(6,7) 0 1 3 0 3 4 5 Relaxtion:if there is an edge from u,v If d[v]> d[u] +cost[u,v] Then update d[v]= d[u] +cost[u,v] Otherwise don’t change
- 12. Mrs Sujata Sathe Edge list (1,2),(1,3),(1,4),(2,5),(3,2),(3,5),(4,3),(4,6),(5,7),(6,7) Iterat ion D[v1] D[v2] D[v3] D[v4] D[v5] D[v6] D[v7] 0 0 ∞ ∞ ∞ ∞ ∞ ∞ 1 0 3 3 5 5 4 7 2 0 1 3 5 2 4 5 3 0 1 3 5 0 4 3 4 0 1 3 5 0 4 3 D[v1] 0 D[v2] 1 D[v3] 3 D[v4] 5 D[v5] 0 D[v6] 4 D[v7] 3 Solution Set
- 13. Mrs Sujata Sathe Longest Common Subsequence Str 1 : a c d p s t r i u x t m n Str 2 :g m n u x p i c s n Sequence : m n n u x n p i n S1 = {B, C, D, A, A, C, D} S2 = {A, C, D, B, A, C} Seq: {A,A,C }, {C, D,A,C }, {B,A,C},{D,A,C},{B,C}, {A.C}..... LCS={C,D,A,C}
- 14. Mrs Sujata Sathe b d o 0 1 2 a b c d o 0 1 2 3 4 A B int LCS(i,j) { if (A[i] = = ‘0’ | | B[j] = = ‘0’) return 0; elseif (A[i] = = B[j]) 1+ LCS(i+1, j+1); else max(LCS(i+1,j),LCS(i,j+1)); } A[0], B[0] b ,a A[1], B[0] d ,a A[0], B[1] b ,b A[2], B[0] 0 , a A[1], B[1] d , b 0 A[2], B[1] 0 ,b A[1], B[2] d ,c 0 A[2], B[2] 0 ,c A[1], B[3] d ,d 0 A[2], B[4] 0 ,0 1+ 1+0=1 1 1 1 1 A[1], B[2] d ,c 1+ A[1], B[2] d ,c A[2], B[2] 0 ,c A[1], B[3] d ,d 0 A[2], B[4] 0 ,0 1+ 1+0=1 1 1 1 2 2 2 2 2 1 1 1 1 0 0 0 0 a b c d 0 0 1 2 3 4 b 0 d 1 0 2
- 15. Mrs Sujata Sathe S1: {b,d}, S2:{a,b,c,d} 0 0 0 0 0 0 0 1 1 1 0 0 1 1 2 0 1 2 3 4 0 1 2 a b c d b d If s1[i] == s2[j] ( if there is a match) then 1+ diagonal element If not Then max(Prev row, prev col) d b LCS = {b,d}
- 16. Mrs Sujata Sathe S1: {c,b,d,a }, S2:{a,c,a,d,b} 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 1 1 2 0 0 1 1 2 2 0 1 1 2 2 2 0 1 2 3 4 5 a c a d b 0 1 2 3 4 c b d a b c Lcs: c,b
- 17. Mrs Sujata Sathe String Editing Transform X= abcdg to Y= aecbf and cost of inserting and deleting is 1 and that of replacing is 2 1) Delete all char from X, insert the char from Y,Tot Oper= 5 + 5 =10 2) Replace b with e, replace d with b, replace g with f= Tot operation =6
- 18. Mrs Sujata Sathe Transform X= aabab to Y= babb and cost of inserting and deleting is 1 and that of replacing is 2 1) Delete all char from X, insert the char from Y, Tot Oper= 5 + 4 =9 2) Delete first two chars from X, Then insert the char b at the end , Tot operation =2+1=3
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- 22. Mrs Sujata Sathe Q.Transform X= adceg to Y=abcfg Rows=tot chars from x+1 Cols=tot chars from y+1 0 1 2 3 4 5 1 0 1 2 3 4 2 1 1 2 3 4 3 2 2 1 2 3 4 3 3 2 2 3 5 4 4 3 3 2 null a b c f g null a d c e g Insert deletion Replace a d c e g a b c f g 1. If r ≠ c 2. If r = c just copy the diagonal element REPLACE REMOVE INSERT MIN(DIAG, UP,LEFT)+1 REPLACE REMOVE INSERT Copy diagonal ele
- 23. Mrs Sujata Sathe Q. Find the edit (Levenshtein) distance between the two strings A= sitting, and B= kitten. 0 1 2 3 4 5 6 1 1 2 3 4 5 6 2 2 1 2 3 4 5 3 3 2 1 2 3 4 4 4 3 2 1 2 3 5 5 4 3 2 2 3 6 6 5 4 3 3 2 7 7 6 5 4 4 3 null k i t t e n null s i t t i n g s i t t i n g k i t t e n We did 2 replacements and 1 deletion
- 24. Mrs Sujata Sathe Travelling Salesman Problem Given a graph G=(V,E) which is directed weightED graph, Let us assume that verter1 is the starting node. The problem involves computation of the minimum cost path that starting from vertex 1 visits all other vertices exactly once and returns back to the starting vertex. g(i,S) = min { Cik + g(k,S-{k}) k€S g(1,{2,3,4}) = min { C12 + g(2, {3,4}), C13 + g(3,{2,4}), C14 + g(4, {2,3})} k€{2,3,4} Base case: g(v,ɸ)= Cv1 1 3 4 2
- 25. Mrs Sujata Sathe 0 10 15 20 5 0 9 10 6 13 0 12 8 8 9 0 1 2 3 4 1 2 3 4 So the starting vertex is 1. g(i,S) = min { Cik + g(k,S-{k}) k€S g(1,{2,3,4}) = min { C12 + g(2, {3,4}), C13 + g(3,{2,4}), C14 + g(4, {2,3})} k€{2,3,4} 1 3 2 4 C12 + g(2, {3,4}) 10+ g(2,{3,4}) C13 + g(3,{2,4}) 15 + g(3,{2,4}) C14 + g(4, {2,3}) 20 + g(4, {2,3}) 3 4 C23 + g(3, {4}) 9 + g(3,{4}) 9+20=29 C24 + g(4, {3}) 10 + g(4,{3})= 10+15=25 4 C34 + g(4, ɸ) 12 + 8 =20 3 C43 + g(3, ɸ) 9 + 6 =15 START NODE 2 4 C32 + g(2, {4}) 13 + g(2,{4})= 13+18=31 C34 + g(4, {2}) 12 + g(4,{2})= 12+13=25 4 C24 + g(4, ɸ) 10+ 8 = 18 2 C42 + g(2, ɸ) 8 + 5 = 13 2 3 C42 + g(2, {3}) 8 + g(2, {3})= 8+15=23 C43 + g(3, {2}) 9 + g(3, {2})= 9+18=27 3 C23 + g(3, ɸ) 9 + 6= 15 2 C32 + g(2, ɸ) 13 + 5= 18 g(1,{2,3,4}) = min { C12 + g(2, {3,4}), C13 + g(3,{2,4}), C14 + g(4, {2,3})} = min{ 10+25, 15+25, 20+23} =min{35, 40, 43} =35
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- 27. Mrs Sujata Sathe 0 10 15 20 5 0 9 10 6 13 0 12 8 8 9 0 0 10 15 20 5 0 9 10 6 13 0 12 8 8 9 0 1 2 3 4 1 2 3 4
- 28. Mrs Sujata Sathe 0/1 Knapsack Problem W is the capacity of of the Bag, wi and pi is the weight and profit associated with each each item xi. The total profit earned is Σxi*pi Solution set is X={x1,x2…..xn} Q given below is the weight and profit of the items and capacity of the bag is 15. Solve the problem to maximize the profit Solution set {0/1,0/1,0/1} x1 x2 x3 weights 10 5 7 profits 90 110 80
- 29. Mrs Sujata Sathe W=8, P={1,2,5,6} W={2,3,4,5} 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 1 2 2 3 3 3 3 0 0 1 2 5 5 6 7 7 0 0 1 2 5 6 6 7 8 0 1 2 3 4 5 6 7 8 0 1 2 3 4 V Pi Wi 1 2 2 3 5 4 6 5 Weights I T E M S V[i,w] = max{ V[i-1,w], V[i-1,w-w[i]]+ P[i]} V[4,1]=max { V[3. 1], V[3, 1-5] + 6 } = max{ 0, V[3,-4]+6 } = 0 Solution set = {0,1,0,1} 8 items X1 X2 X3 X4 0 1 0 1 8-6=2 2-2=0
- 30. Mrs Sujata Sathe Q2. Capacity=3, Profit={1,6,4} Weight={1,2,4} Q3. Capacity=12, Profit={14,20,24, 30} Weight={2,4,6,10}
- 31. Mrs Sujata Sathe Matrix Chain Multiplication A X B ≠ BX A…not commutative A X B m*n x*y Rule for matrix multiplication is that the col of First matrix should be equal to the row of the second matrix. (n= x) The dimensions of the resultant matrix = m*y Given three matrices A, B, and C, you can perform matrix multiplication as A.(B.C), (A.B). C A X B m*n x*y Given a set of matrices {A1,A2,…An} with the dimensions {p0*p1,p1*p2,….,pn-1*pn}, find the optimal order of matrices such that the cost of multiplication is optimal. A:5X4, B:4X3 = Resultant Matrix (AXB): 5X3 = R A:5X4 B:4X3 R: 5X3 The multiplication cost of the resultant matrix = 5*4*3=60
- 32. Mrs Sujata Sathe Q. Consider the following three matrices A:2X3 ; B: 3X4; C: 4X3 What are the possible orderings? What is the optimal order? (A XB)X C= (2*3*4)+(2*4*3)=24+24=48 AX(BXC)= (3*4*3)+(?) (A1.A2).(A3.A4) A1.(A2.A3).A4 .. .
- 33. Mrs Sujata Sathe A1.A2.A3.A4 5x4 4 x 6 6 X 2 2 x 7 0 120 88 0 48 104 0 84 0 1 2 3 4 1 2 3 4 M 1 1 2 3 3 1 2 3 4 1 2 3 4 S M[1,1]=0, M[2,2]=0, M[3,3]=0,M[4,4] M[1,2]=120 (A1.A2) 5X4 4X6 Cost= 5*4*6=120 M[2,3]=48 A2.A3 4X6 6X2 Cost= 4*6*2=48 M[3,4]=84 A3.A4 6X2 2X7 Cost= 6*2*7=84 M[1,3] = 88 ie A1.A2.A3 A1.(A2.A3) (A1.A2).A3 5X4 4X6 6X2 5X4 4X6 6X2 M[1,1]+M[2,3]+5*4*2 M[1,2]+M[3,3]+ 5*6*2 =0+48+40 =120+0+60 =88 =180 M[2,4]= i.e A2.A3.A4 A2.(A3.A4) (A2.A3).A4 4X6 6X2 2X7 4X6 6X2 2X7 M[2,2]+M[3,4]+4*6*7 M[2,3]+M[4,4]+ 4*2*7 =0+84+168 =48+0+56 =252 =104
- 34. Mrs Sujata Sathe 0 120 88 158 0 48 104 0 84 0 1 2 3 4 1 2 3 4 M 1 1 3 2 3 3 1 2 3 4 1 2 3 4 S M[1,4]= min{M[1,1]+ M[2,4]+5*4*7, M[1,2]+M[3,4]+5*6*7,M[1,3]+M[4,4] +5*2*7} = {0+104+140, 120+84+210, 88+0+70} =158
- 35. Mrs Sujata Sathe A1.A2.A3.A4 4x5 5 x 3 3 X 2 2 x 7 0 60 70 126 0 30 100 0 42 0 1 2 3 4 1 2 3 4 M 0 1 1 3 0 2 3 0 3 0 1 2 3 4 1 2 3 4 S M[1,1]=0, M[2,2]=0, M[3,3]=0,M[4,4]=0, M[1,2]=60 (A1.A2) 4X5 5X3 Cost= 4*5*3=60 M[2,3]=30 (A2.A3) M[2,2]+M[3,3]+5X3X2 M[3,4]= (A3.A4)
- 36. Mrs Sujata Sathe A1.A2.A3.A4.A5 4x5 5 x 3 3 X 2 2 x 7 7X2 0 0 0 0 0 1 2 3 4 5 1 2 3 4 5 M 0 0 0 0 0 1 2 3 4 5 1 2 3 4 5 S
- 37. Mrs Sujata Sathe Optimal Binary Search Tree A BST is a binary tree that possesses the following properties 1. Each node of a BST has one key, 2. All the nodes on the LHS of a BST is less than or equal to the keys stored in the root. 3. All the nodes on the RHS of a BST are greater than or equal to the keys stored in the root
- 38. Mrs Sujata Sathe Optimal Binary Search Tree Keys = 10,20,30,40,50,60,70 40 60 20 10 30 50 70 Keys = 10,20,30 Frequency=3,2,5 10 20 30 30 10 20 10 30 20 30 10 10 30 20 20 To find the node 50, how many comparisons are done 3 comparisons To find element 29 4comp 3*1 2*2 5*3 1+2+3/3= 6/3=2 1 2 3 1+2+3/3= 6/3=2 1 2 2 1+2+2/3= 5/3=<2 1+2+3/3= 6/3=2 1+2+3/3= 6/3=2 3+4+15= 22 3*1+5*2+2*3= 19 2*1+3*2+5*2= 17 5*1+3*2+2*3= 17 5*1+2*2+3*3= 18
- 39. Mrs Sujata Sathe C[i,j] =min {C[i,k-1]+ C[k,j]}+ w[i,j] i<k≤j 1 2 3 4 10 20 30 40 4 2 6 3 keys frequency No 0 4 81 203 263 0 2 103 163 0 6 123 0 3 0 0 1 2 3 4 0 1 2 3 4 i j C C[0,0]= C[1,1]=C[2,2]=C[3,3]=C[4,4]=0 C[0,1]=4 Consider only 1 key i.e10 & Freq=4 C[1,2]=2 C[2,3]=6 C[3,4]=3 C[0,2] means there are 2 keys 10, 20 For j-i=1 1-0=1, C[0,1] 2-1=1, C[1,2] 3-2=1,C[2,3] 4-3=1,C[3,4] For j-i=2 2-0=2, C[0,2] 3-1=2, C[1,3] 4-2=2, C[2,4] 4 w[0,4]=ΣF(i) i=1 10 20 20 10 4*1 2*2 2*1 4*2 4*1+2*2= 8 2*1+4*2= 10 C[1,3] means there are 2 keys 20, 30 C[1,3]= min C[1,1]+ C[2,3], k=2,3 C[1,2]+C[3,3], + w[1,3] =min{0+6+8, 2+0+8} =min{14,10} w[1,3]=2+6=8 Cost of OBST=26 and the root node is 3 3 r[0,4]=3 1 4 r[0,2] r[3,4] 2 r[0,0] r[1,2] r[1,1] r[2,2] r[3,3] r[4,4] 30 10 20 40