module4_dynamic programming_2022.pdf

Dynamic Programming
Module 4
Shiwani Gupta 1

Dynamic Programming
 Dynamic Programming is applied to optimization problems.
 Solves problems with overlapping subproblems.
 Each subproblem solved only once and result recorded in a table.
Shiwani Gupta 2

Divide n Conquer vs Dynamic
Programming
 Subproblems are independent
 Recomputations performed
 Less Efficient due to rework
 Recursive Method (Top Down
Approach of problem Solving)
 Splits its input at specific
deterministic points usually in
middle
 Subproblems are overlapping
 No need to recompute
 More Efficient
 Iterative Method (Bottom Up
Approach of problem Solving)
 Splits its input at every possible
split point
Shiwani Gupta 3

Greedy vs Dynamic Programming
 Used to obtain optimum solution
 Picks optimum solution from a set
of feasible solutions
 Optimum selection is without
revising previously generated
selections
 Only one decision sequence is
ever generated
 No guarantee of getting optimum
solution
 Also obtains optimum solution
 No special set of feasible
solutions
 Considers all possible
sequences to obtain possible
solution
 Many decision sequences may
be generated
 Guarantees optimal solution
using Principle of optimality
In an optimal sequence of decisions or choices, each subsequence must
also be optimal. Shiwani Gupta 4

Principle of optimality
 Principle of optimality: Suppose that in solving a problem, we have to
make a sequence of decisions D1, D2, …, Dn. If this sequence is
optimal, then the last k decisions, 1  k  n must be optimal.
 e.g. the shortest path problem
If i, i1, i2, …, j is a shortest path from i to j, then i1, i2, …, j must be a
shortest path from i1 to j
Shiwani Gupta 5

Applications of Dynamic Programming
 Multistage Graphs
 Single Source Shortest Path
 All Pair Shortest Path
 Optimal Binary Search Tree
 0/1 Knapsack Problem
 Traveling Salesman Problem
 Longest Common Subsequence
 Flow Shop Scheduling
Shiwani Gupta 6

Multistage Graph
 Multistage Graph G=(V,E) is a directed graph whose vertices are
partitioned into k stages, k ≥ 2
 To find a shortest path from source to sink
 Apply the greedy method :
the shortest path from S to T : 1 + 2 + 5 = 8
S A B T
3
4
5
2 7
1
5 6
Shiwani Gupta 7

The shortest path in Multistage Graphs
The greedy method can not be applied to this case:
(S, A, D, T) 1+4+18 = 23.
The real shortest path is: (S, C, F, T) 5+2+2 = 9.
We obtain min. path at each current stage by considering path length
of each vertex obtained in earlier stage.
S T
13
2
B E
9
A D
4
C F
2
1
5
11
5
16
18
2
Shiwani Gupta 8

 Dynamic programming approach (forward approach):
 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}
S T
2
B
A
C
1
5
d(C, T)
d(B, T)
d(A, T)
A
T
4
E
D
11
d(E, T)
d(D, T)
 d(A,T) = min{4+d(D,T), 11+d(E,T)} = min{4+18, 11+13} = 22.
Forward approach
(backward reasoning)
Shiwani Gupta 9

 d(B, T) = min{9+d(D, T), 5+d(E, T), 16+d(F, T)}
= min{9+18, 5+13, 16+2} = 18.
 d(C, T) = min{ 2+d(F, T) } = 2+2 = 4
 d(S, T) = min{1+d(A, T), 2+d(B, T), 5+d(C, T)}
= min{1+22, 2+18, 5+4} = 9.
 The above way of reasoning is called backward reasoning.
B T
5
E
D
F
9
16
d(F, T )
d(E, T )
d(D, T)
Dynamic programming approach
Shiwani Gupta 10

Backward approach
(forward reasoning)
 d(S,T) = min{d(S, D)+d(D, T),d(S,E)+d(E,T), d(S, F)+d(F, T)}
= min{d(S, D)+18,d(S,E)+13, d(S, F)+2}
 d(S,D) = min{d(S, A)+d(A, D),d(S, B)+d(B, D)}
= min{d(S, A)+4,d(S, B)+9}
Shiwani Gupta 11
S T
13
2
B E
9
A D
4
C F
2
1
5
11
5
16
18
2

Backward approach
(forward reasoning)
 d(S,E) = min{d(S, A)+d(A, E),d(S, B)+d(B, E)}
= min{d(S, A)+11,d(S, B)+5}
 d(S,F) = min{d(S, B)+d(B, F),d(S, C)+d(C, F)}
= min{d(S, B)+16,d(S, C)+2}
 d(S,A) = 1; d(S, B) = 2;d(S, C) = 5
d(S,D)= min{ 1+4, 2+9 } = 5 d(S,E)= min{ 1+11, 2+5 } = 7
d(S,F)= min{ 2+16, 5+2 } = 7 d(S,T) = min{ 5+18, 7+13, 7+2 } = 9
Shiwani Gupta 12
S T
13
2
B E
9
A D
4
C F
2
1
5
11
5
16
18
2

Shortest Path Problems
 Single-source (single-destination). Find a shortest path
from a given source (vertex s) to each of the vertices.
 Dijkstra’s Algorithm - Greedy
 Bellman-Ford Algorithm – Dynamic Programming
 Single-pair. Given two vertices, find a shortest path
between them. Solution to single-source problem solves
this problem efficiently, too.
 All-pairs. Find shortest-paths for every pair of vertices.
Dynamic programming algorithm.
 Floyd Warshall’s Algorithm – Dynamic Programming
 Applications
 static/dynamic network routing
 robot motion planning
 map/route generation in traffic
Shiwani Gupta 14

Bellman-Ford Algorithm
 Dijkstra’s doesn’t work when there are negative edges
 Bellman-Ford algorithm detects negative cycles (returns false) or
returns the shortest path-tree
Algorithm BellmanFord (vertices, edges, source)
for(each vertex v) //loop to initialize graph
if (v is source) then
v.distance ← 0
else
v.distance ← infinity
v.pred ← null
for (i← 1 to tot_vertices-1) //O(n)
for (each edge uv) //O(nm)
u ← uv.source
v ← uv.destination Shiwani Gupta 15

Bellman-Ford Algorithm
//newly obtained min dist
if( v.distance > u.distance + uv.weight ) then
v.distance = u.distance + uv.weight //relax edge
v.pred ← u
for (each edge uv)
u ← uv.source
v ← uv.destination
if( v.distance > u.distance + uv.weight ) then
write(“Graph has negative weights”)
return false
Time Complexity: O (nm)
Shiwani Gupta 16

Bellman-Ford Example
5
¥ ¥
¥ ¥
0
s
z
y
6
7
8
-3
7
2
9
-2
x
t
-4
6 ¥
7 ¥
0
s
z
y
6
7
8
-3
7
2
9
-2
x
t
-4
5
6 4
7 2
0
s
z
y
6
7
8
-3
7
2
9
-2
x
t
-4
5
2 4
7 2
0
s
z
y
6
7
8
-3
7
2
9
-2
x
t
-4
5
2 4
7 -2
0
s
z
y
6
7
8
-3
7
2
9
-2
x
t
-4
5
Shiwani Gupta 17

All Pair Shortest Path
 Find the distance between every pair of vertices in a
weighted directed graph G.
 We can make n calls to Dijkstra’s algorithm (if no negative
edges), which takes O(n(nlog n+m)) time or O(n3) time.
 Likewise, n calls to Bellman-Ford would take O(n2m) time.
 We can achieve O(n3) time using dynamic programming
 Floyd Warshall’s Algorithm (Generalization of Dijkstra’s)
Shiwani Gupta 19

Floyd Warshall’s Algorithm
Representation of the Input
We assume that the input is represented by a weight matrix
W= (wij); i,j in E that is defined by
wij= 0 if i=j
wij= w(i,j) if ij and (i,j) in E
wij=  if ij and (i,j) not in E
Representation of the Output
If the graph has n vertices, we return a distance matrix (dij), where dij the
length of the path from i to j.
Intermediate Vertices
Without loss of generality, we will assume that V={1,2,…,n}, i.e., that
the vertices of the graph are numbered from 1 to n.
Given a path p=(v1, v2,…, vm) in the graph, we will call the vertices vk
with index k in {2,…,m-1} the intermediate vertices of p.
Shiwani Gupta 20

The key to the Floyd-Warshall algorithm is the following definition:
Let dij
(k) denote the length of the shortest path from i to j such that all
intermediate vertices are contained in the set {1,…,k}.
Consider a shortest path p from i to j such that the intermediate vertices
are from the set {1,…,k}.
 If the vertex k is not an intermediate vertex on p, then dij
(k) = dij
(k-1)
If the vertex k is an intermediate vertex on p, then dij
(k) = dik
(k-1) + dkj
(k-
1)
Interestingly, in either case, the subpaths contain merely nodes from
{1,…,k-1}.
Floyd Warshall’s Algorithm
Shiwani Gupta 21

Recursive Formulation
Therefore, we can conclude that dij
(k) = min{dij
(k-1) , dik
(k-1) + dkj
(k-1)}
If we do not use intermediate nodes, i.e.,
when k=0, then dij
(0) = wij
If k>0, then dij
(k) = min{dij
(k-1) , dik
(k-1) + dkj
(k-1)}
Shiwani Gupta 22

Floyd Warshall algorithm
Algorithm AllPair(G) //assumes vertices 1,…,n
for all vertex pairs (i,j)
if i = j
D0[i,i]  0
else if (i,j) is an edge in G
D0[i,j]  weight of edge (i,j)
else
D0[i,j]  + 
for k  1 to n do
for i  1 to n do
for j  1 to n do
Dk[i,j]  min{Dk-1[i,j], Dk-1[i,k]+Dk-1[k,j]}
return Dn
Can reduce compexity to O (n2) by using single array
Shiwani Gupta 23

Example of Floyd-Warshall
Shiwani Gupta 28

The 0/1 Knapsack Problem
 Given: A set S of n items, with each item i having
 wi - a positive weight
 bi - a positive benefit
 Goal: Choose items with maximum total benefit but with
weight at most W.
 If we are not allowed to take fractional amounts, then this is
the 0/1 knapsack problem.
 In this case, we let T denote the set of items we take
 Objective: maximize
 Constraint:

T
i
i
b



T
i
i W
w
Shiwani Gupta 30

 Given: A set S of n items, with each item i having
 bi - a positive “benefit”
 wi - a positive “weight”
 Goal: Choose items with maximum total benefit but with weight at
most W.
Example
Weight:
Benefit:
1 2 3 4 5
4 in 2 in 2 in 6 in 2 in
$20 $3 $6 $25 $80
Items:
box of width 9 in
Solution:
• item 5 ($80, 2 in)
• item 1 ($20, 4in)
• item 3 ($6, 2in)
“knapsack”
Shiwani Gupta 31

0/1 Knapsack Algorithm, First Attempt
 Sk: Set of items numbered 1 to k.
 Define B[k] = best selection from Sk.
 Problem: does not have subproblem optimality:
 Consider set S={(3,2),(5,4),(8,5),(4,3),(10,9)} of
(benefit, weight) pairs and total weight W = 20
Best for S4:
Best for S5:
Shiwani Gupta 32

0/1 Knapsack Algorithm, Second Attempt
 Sk: Set of items numbered 1 to k.
 Define B[k,w] to be the best selection from Sk with weight at most w
 Good news: this does have subproblem optimality.
i.e., the best subset of Sk with weight at most w is either
 the best subset of Sk-1 with weight at most w or
 the best subset of Sk-1 with weight at most w-wk plus item k










else
}
]
,
1
[
],
,
1
[
max{
if
]
,
1
[
]
,
[
k
k
k
b
w
w
k
B
w
k
B
w
w
w
k
B
w
k
B
Shiwani Gupta 33
https://youtu.be/nLmhmB6NzcM 19:21

0/1 Knapsack Algorithm
Consider set S={(1,1),(2,2),(4,3),(2,2),(5,5)} of (benefit, weight)
pairs and total weight W = 10
Shiwani Gupta 34

Trace back to find the items picked
Shiwani Gupta 35

 Each diagonal arrow corresponds to adding one item into the bag
 Pick items 2,3,5
 {(2,2),(4,3),(5,5)} are what you will take away
Shiwani Gupta 36

Algorithm 01Knapsack(B,w,n,W):
Input: set S of n items with benefit Bi and weight wi; maximum weight W
Output: benefit of best subset of S with weight at most W
for w  0 to W do
B[0,w]  0
for i  1 to n do
for w  0 to W do
if wi ≤ w then
B[i,w]  max{B[i-1,w],B[i]+B[i-1,w-wi]}
keep[i,w]=1 //if keeping ith file in knapsack
else
B[i,w]  B[i-1,w]
keep[i,w]=0
return B[n,W] //max computing time of files that can fit into storage
K=W
for i = n downto 1
if keep[i,k]==1
output i
k=k-w[i]
Running time: O(nW).
Shiwani Gupta 37

Shiwani Gupta 38
Weights: {3, 4, 6, 5}
Profits: {2, 3, 1, 4}
The weight of the knapsack is 8 kg

Travelling Salesman Problem
Let G be a directed graph denoted by (V,E). Edges are given along with
their cost Cij. A tour for the graph should be such that all the vertices
should be visited only once terminating at source vertex with tour of
minimum cost (sum of cost of edges on tour).
Dynamic Programming Approach
• Let function g(1,V-{1}) is total length of tour terminating at 1.
• Let d[i,j] be shortest path b/w two vertices i and j.
• Principle of optimality: Path Vi, Vi+1, … Vj must be optimal for all
paths beginning at Vi and ending at Vj and passing through all
intermediate vertices {Vi+1, … Vj-1} once.
• g(1,V-{1}) = min {c1k + g(k,V-{1,k}) }
• g(i,S) = min {c1j + g(j,S-{j}) } where j є S and i є S
• g(i,Ф) = ci1 , 1≤i ≤n
2≤k ≤n
j є S
Shiwani Gupta 39
https://youtu.be/XaXsJJh-Q5Y

Travelling Salesman Problem
1 2 3 4
1 0 10 15 20
2 5 0 9 10
3 6 13 0 12
4 8 8 9 0
g(2, Ф) = c21 = 5; g(3, Ф) = c31 = 6; g(4, Ф) = c41 = 8
• K=1
Set{2}
g(3,{2})= c32 + g(2,Ф) = c32 + c21= 13+5=18; p(3,{2})=2
g(4,{2})= c42 + g(2,Ф) = c42 + c21= 8+5=13; p(4,{2})=2
Set{3}
g(2,{3})= c23 + g(3,Ф) = c23 + c31= 9+6=15; p(2,{3})=3
g(4,{3})= c43 + g(3,Ф) = c43 + c31= 9+6=15; p(4,{3})=3
Set{4}
g(2,{4})= c24 + g(4,Ф) = c24 + c41= 10+8=18; p(2,{4})=4
g(3,{4})= c34 + g(4,Ф) = c34 + c41= 12+8=20; p(3,{4})=4
• k=2
Set{3,4}: g(2,{3,4}) = min {c23 + g(3,{4}), c24 + g(4,{3})}=25; p(2,{3,4})=4
Set{2,4}: g(3,{2,4}) = min {c32 + g(2,{4}), c34 + g(4,{2})}=25; p(3,{2,4})=4
Set{2,3}: g(4,{2,3}) = min {c42 + g(2,{3}), c43 + g(3,{2})}=23; p(4,{2,3})=2
• k=3
Set{2,3,4}: f= 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} = 35; p(1,{2,3,4})=2
ans: 12431
40

Longest Common Subsequence
 Given two strings, find a longest subsequence that they share
 substring vs. subsequence of a string
 Substring: the characters in a substring of S must occur
contiguously in S
 Subsequence: the characters can be interspersed with gaps.
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC
Shiwani Gupta 32

Longest common subsequence
Sequences x1,…,xn, and y1,…,ym
LCS(i,j) = length of a longest common
subsequence of x1,…,xi and y1,…,yj
if xi  yj then
LCS(i,j) = max (LCS(i-1,j),LCS(i,j-1))
if xi = yj then
LCS(i,j) = 1 + LCS(i-1,j-1)
Running time = O(mn)
Shiwani Gupta 33

Let’s give a score M an alignment in this way,
M=sum s(xi,yj), where
xi is the i character in the first aligned sequence
yj is the j character in the second aligned sequence
s(xi,yj)= 1 if xi= yj
s(xi,yj)= 0 if xi≠yj or any of them is a gap
The score for alignment:
ababc.
abd.cb
M=s(a,a)+s(b,b)+s(a,d)+s(b,.)+s(c,c)+s(.,b)=3
To find the longest common subsequence between sequences S1 and
S2 is to find the alignment that maximizes score M.
Shiwani Gupta 34

Score matrix M Trace back table B
M7,11=6 (lower right corner of Score matrix)
This tells us that the best alignment has a score of 6
What is the best alignment?
Shiwani Gupta 35

We need to use trace back table to find out the best alignment,
which has a score of 6
(1) Find the path from
lower right corner to
upper left corner
Thus, the optimal alignment is
The longest common subsequence is
G.A.T.C.G.A
Shiwani Gupta 36

Algorithm LCS (string A, string B) {
Input strings A and B
Output the longest common subsequence of A and B
M: Score Matrix
B: trace back table (use letter a, b, c for )
n=A.length()
m=B.length()
// fill in M and B
for (i=0;i<m+1;i++)
for (j=0;j<n+1;j++)
if (i==0) || (j==0)
then M(i,j)=0;
else if (A[i]==B[j])
M(i,j)=M[i-1,j-1]+1
{update the entry in trace table B}
else
M(i,j)=max {M[i-1,j], M[i,j-1]}
{update the entry in trace table B}
then use trace back table B to print out the optimal alignment
Running time = O(mn)

Applications of LCS
 Molecular Biology: DNA sequences (genes) can be
represented as sequences of four letters ACGT, corresponding
to the four submolecules forming DNA. When biologists find
new sequences, they typically want to know what other
sequences it is most similar to. One way of computing how
similar two sequences are is to find the length of their longest
common subsequence.
 File Compression: The Unix program "diff" is used to
compare two different versions of the same file, to determine
what changes have been made to the file. It works by finding a
longest common subsequence of the lines of the two files; any
line in the subsequence has not been changed, so what it
displays is the remaining set of lines that have changed. In this
instance of the problem we should think of each line of a file
as being a single complicated character in a string.

Dynamic Programming to LCS
 Optimal Substructure
 Overlapping subproblems
 The method of dynamic programming reduces the number
of function calls. It stores the result of each function call so
that it can be used in future calls without the need for
redundant calls.
 In the above dynamic algorithm, the results obtained from
each comparison between elements of X and the elements
of Y are stored in a table so that they can be used in future
computations.
 So, the time taken by a dynamic approach is the time taken
to fill the table (ie. O(mn)).
Shiwani Gupta 39

Task
 Determine LCS of <0100110101> and <10101101>.
 Find LCS of following strings X=“ABCBDAB”
Y=“BDCABA”
Shiwani Gupta 40

module4_dynamic programming_2022.pdf

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