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Dynamic [email protected], [email protected]
www.mbchandak.com
Tutorial &Practice on Longest Common Sub-sequence
Optimal Binary Search Tree
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Characteristics of Dynamic Programming• Developed by Richard Bellman in 1950.• To provide accurate solution with the help of series of decisions. • Define a small part of problem and find out the optimal solution to
small part.• Enlarge the small part to next version and again find optimal
solution.• Continue till problem is solved.• Find the complete solution by collecting the solution of optimal
problems in bottom up manner.
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Characteristics• Types of problems and solution strategies.• 1. Problem can be solved in stages.• 2. Each problem has number of states• 3. The decision at a stage updates the state at the stage into the
state for next stage.• 4. Given the current state, the optimal decision for the remaining
stages is independent of decisions made in previous states.• 5. There is recursive relationship between the value of decision at a
stage and the value of optimum decisions at previous stages.
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Characteristics• How memory requirement is reduce:• How recursion overheads are converted into advantage• - By storing the results of previous n-2 computations in computation
of n-1 stage and will be used for computation of “n” stage.• Not very fast, but very accurate• It belongs to smart recursion class, in which recursion results and
decisions are used for next level computation. Hence not only results but decisions are also generated.• Final collection of results: Bottom Up Approach.
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Longest Common Subsequence• Given two strings of characters, LCS is a method to
find a longest subsequence either continuous or non-continuous and is common in both the strings. The subsequence generation starts from comparison of first character of both the given strings.• The two strings may or may not be of equal length.• Let “A” and “B” be two strings of length “m” and “n”
respectively, and let “i” and “j” be two pointers to handle the two strings. • Let matrix c[m,n] be the storage matrix to store the results
of comparison related with two strings. The C[m,n] matrix will be handled using pointer “i” and “j”.
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If i=0 and j=0, then c[i,j] = 0If a[1..i] and b[1..j] is compared and a[i] is not
equal to b[j]. Then the comparison will be either In betweena[1..i-1] and b[1..j] or a[1..i] and b[1..j-1]
Thenc[i,j] = max[c[i-1,j], c[I,j-1]]If a[1..i] and b[1..j] is compared and a[i] is equal
to b[j].Then c[i,j] = c[i-1,j-1] + 1
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ExampleSTRING A = G D V E G T A B = G V C E K S T
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1 G2 D3 V4 E5 G6 T7 A
H = Halt
Case:String A = No CharacterString B = All CharactersResult = 0 and Halt State
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ExampleSTRING A = G D V E G T A B = G V C E K S T
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H1 G 0/H2 D 0/H3 V 0/H4 E 0/H5 G 0/H6 T 0/H7 A 0/H
H = Halt
Case:String A = All CharactersString B = No CharacterResult = 0 and Halt State
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ExampleSTRING A = G D V E G T A B = G V C E K S T
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1 G 0/H
1/D
H = HaltD = DiagonalS = Side
Case:String A = GString B = GResult = MatchUse diagonal value from the cell and add 1 to diagonal value0+1
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1 G 0/H
1/D 1/S
Case:String A = GString B = VResult = No MatchUse Side or Upper cell value and select maximum valueIf both values are same select UPPER cell value
1/S 1/S 1/S 1/S 1/S
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ExampleSTRING A = G D V E G T A B = G V C E K S T
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H 0/H 0/H
0/H 0/H 0/H 0/H 0/H
1 G 0/H 1/D 1/S 1/S 1/S 1/S 1/S 1/S
2 D 0/H 1/U
H = HaltD = DiagonalS = Side
Case:String A = DString B = GResult = No Match
Since the value in Upper cell is greater than Side cellUse UPPER Cell value
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ExampleSTRING A = G D V E G T A B = G V C E K S T
i/j 0 1 2 3 4 5 6 7G V C E K S T
0 0/H 0/H
0/H 0/H 0/H 0/H 0/H 0/H
1 G 0/H 1/D 1/S 1/S 1/S 1/S 1/S 1/S
2 D 0/H 1/U
H = HaltD = DiagonalS = Side
Case:String A = DString B = VResult = No Match
Since the value in Upper cell is same as Side cellUse UPPER Cell value
1/U1/U 1/U 1/U 1/U 1/U
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ExampleSTRING A = G D V E G T A B = G V C E K S Ti/j 0 1 2 3 4 5 6 7
G V C E K S T0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1 G 0/H 1/D 1/S 1/S 1/S 1/S 1/S 1/S2 D 0/H 1/U 1/U 1/U 1/U 1/U 1/U 1/U3 V 0/H 1/U 2/D 2/S 2/S 2/S 2/S 2/S4 E 0/H 1/U 2/U 2/U 3/D 3/S 3/S 3/S5 G 0/H 1/D 2/U 2/U 3/U 3/U 3/U 3/U6 T 0/H 1/U 2/U 2/U 3/U 3/U 3/U 4/D7 A 0/H 1/U 2/U 2/U 3/U 3/U 3/U 4/U
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ExampleSTRING A = G D V E G T A B = G V C E K S Ti/j 0 1 2 3 4 5 6 7
G V C E K S T0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1 G 0/H 1/D 1/S 1/S 1/S 1/S 1/S 1/S2 D 0/H 1/U 1/U 1/U 1/U 1/U 1/U 1/U3 V 0/H 1/U 2/D 2/S 2/S 2/S 2/S 2/S4 E 0/H 1/U 2/U 2/U 3/D 3/S 3/S 3/S5 G 0/H 1/D 2/U 2/U 3/U 3/U 3/U 3/U6 T 0/H 1/U 2/U 2/U 3/U 3/U 3/U 4/D7 A 0/H 1/U 2/U 2/U 3/U 3/U 3/U 4/U
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Example:2STRING A = X M J Y A U Z B = M Z J A W X U i/j 0 1X 2M 3J 4Y 5A 6U 7Z0 0/H 0/H 0/H 0/H 0/H 0/H 0/H 0/H1M 0/H 1/U 1/D 1/S 1/S 1/S 1/S 1/S2Z 0/H 1/U 1/U 1/U 1/U 1/U 1/U 2/D3J 0/H 1/U 1/U 2/D 2/S 2/S 2/S 2/S4A 0/H 1/U 1/U 2/U 2/U 3/D 3/S 3/S5W 0/H 1/U 1/U 2/U 2/U 3/U 3/U 3/U6X 0/H 1/D 1/U 2/U 2/U 3/U 3/U 3/U7U 0/H 1/U 1/U 2/U 2/U 3/U 4/D 4/S
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Algorithm:• Assumptions: Let the two strings be present in
array “a” size “m” and array “b” size “n” handled using pointers “i“ and “j”.
• The resultant array will be c[m,n] : one instance c[i,j]. The array “c” will be structure: c[i,j].val and c[i,j].dir = “u”, “s”, “d” and “h”
Algorithm lcs (a,b: c){ for i = 0 to m do for j = 0 to n do if a[i]=0 or b[j]=0{ c[i,j].val = 0; c[i,j].dir=‘h’}else
{ if (a[i] ≠ b[j]) c[i,j].val = max [c[i-1, j].val, c[i,j-1].val] if(c[i-1, j].val >= c[i,j-1].val) c[i,j].dir = ‘u’ else c[i,j].dir=‘s’ else c[i,j].val = c[i-1,j-1].val + 1 c[i,j].dir = ‘d’}
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Algorithm:• Printing LCS: Assumptions: Let “A”
be the given array. • Let “C” be the array containing
the details of LCS
Algorithm print_lcs (a,c,i,j){ if (i=0 or j=0) return; if(c[i,j] = ‘d’ {
print_lcs(a,c,i-1,j-1) print(a[i]) } else { if(c[i,j] = ‘u’ print_lcs(a,c,i-1,j) else print_lcs(a,c,i,j-1) }
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Exercises
• Question 1:• String 1 : E X P O N E N T I A L• String 2 : P O L Y N O M I A L
• String 1 : SUBSEQUENCE• String 2 : CONSEQUENCES
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Optimal Binary Search Tree (OBST)• Drawback of BST: The first value in the sequence is always used as
ROOT node.• If the data is present in sorted order, BST will be in the form of linked
list. This will increase search time and complexity equation will be not valid.• For example if dictionary word are to be stored in BST, it will be in the
form of linked list:Apple, Bat, Cat, Dog, Elephant, Fish, Gun, Horse, Ink, Jug, King, Lion
Since all the start characters are Greater than “A”, all the values will be on the Right side of the Apple.
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Example of word list
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Optimal Binary Search Tree• Need: Mechanism to decide out of the available values,
which value should be used as ROOT, so that the tree is balanced and COST of searching is minimum.• Requirement: For selection, the probability of data in the
domain text is required.• There are two values: Successful probability value and
Unsuccessful probability value.• Application: OBST is applied to generate the tree for the set
of values (keys) with defined successful and unsuccessful search probabilities.
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Example:•Creation of matrices:• “w” = probability matrix• “e” = evolution matrix• “r” = root matrix
Dimensions:n = Total number of keys given in the search spaceMatrix “w” = [1..n+1, 0..n]Matrix “e” = [1..n+1, 0..n]For example number of keys are K1, K2, K3, K4 then n=4 and matrix “w” and “e” will have dimensions as: [1..5, 0..4]
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Optimal Binary Search Tree
i 0 1 2 3 4pi 0.12 0.10 0.09 0.20qi 0.10 0.08 0.05 0.11 0.15
sum 0.10 0.20 0.15 0.20 0.35
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W 0 1 2 3 4
1 0.10 0.30 0.45 0.65 1.0
2 # 0.08 0.23 0.43 0.78
3 # # 0.05 0.25 0.60
4 # # # 0.11 0.46
5 # # # # 0.15
E 0 1 2 3 4
1 0.10 0.48/1 0.91/1 1.54/2 2.63/3
2 # 0.08 0.36/2 0.90/3 1.83/4
3 # # 0.05 0.41/3 1.16/4
4 # # # 0.11 0.72/4
5 # # # # 0.15
i 0 1 2 3 4
pi 0.12 0.10 0.09 0.20
qi 0.10 0.08 0.05 0.11 0.15
sum 0.10 0.20 0.15 0.20 0.35
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Example 2: OBSTi 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.25 0.10 0.15 0.25
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Solution: Example 2: AnimationW 0 1 2 3 4 5
1 0.10
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
e 0 1 2 3 4 5
1 0.10
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
i 0 1 2 3 4 5pi 0.05 0.20 0.05 0.10 0.15qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.25 0.10 0.15 0.25
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Empty matrix structure and formulaw 0 1 2 3 4 5
1
2 0.10
3 0.05 0.15
4 0.05 0.20
5 0.05 0.30
6 0.10
e 0 1 2 3 4 5
1 0.10
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
i 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15
qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.10 0.15 0.250.15
0.10
0.10+0.15=0.25
0.25
0.25
0.10+0.25=0.35
0.35
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Empty matrix structure and formulaw 0 1 2 3 4 5
1 0.25
2 0.10 0.35
3 0.05 0.15 0.30
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
e 0 1 2 3 4 5
1 0.10
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
i 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15
qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.15 0.25
0.10
0.25
0.25+0.25=0.50
0.50
0.10
0.35+0.10=0.45
0.45
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Empty matrix structure and formulaW 0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
e 0 1 2 3 4 5
1 0.10
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
i 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15
qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.25 0.10 0.15 0.25
0.10 0.50
0.45
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Empty matrix structure and formulae 0 1 2 2 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
e 0 1 2 3 4 5
1 0.10 0.45/1
2 0.10
3 0.05
4 0.05
5 0.05
6 0.10
i 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15
qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.25 0.10 0.15 0.25
0.10 0.50
0.45
e 0 1 2 3 4 5
1 0.10 0.45/1
2 0.10 0.50/2
3 0.05 0.25/3
4 0.05 0.30/4
5 0.05 0.45/5
6 0.10e[1,1] = e[1,0]+e[2,1]+w[1,1] 0.10+0.10+0.250.45e[2,2] = e[2,1]+2[3,2]+w[2,2] 0.10+0.05+0.350.50
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Empty matrix structure and formulae 0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
2 0.10 0.50/2
3 0.05 0.25/3
4 0.05 0.30/4
5 0.05 0.45/5
6 0.10
0.10 0.50
0.45
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
2 0.80/2
3 0.05
4 0.30/4
5 0.05 0.45/5
6 0.10
e[1,2] = Two options for root between values 1 and 2 are r=1 and r=2e[1,2] will be minimum of e[1,0]+e[2,2]+w[2,2] 0.10+0.50+0.50 =1.10 [r=1]e[1,1]+e[3,2]+w[2,2] 0.45+0.05+0.50 =1.00 [r=2]
0.10
0.25/3
First0.80
0.50
0.05
0.45
Second1.00
Final0.80/2
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Empty matrix structure and formulae 0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
0.10 0.50
0.45
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
2 0.80/2
3 0.05 0.25/3 0.60/4
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
0.10 0.50
0.45
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Empty matrix structure and formulae 0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
0.10 0.50
0.45
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
1.30/2
2 0.80/2
3 0.05 0.25/3 0.60/4
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
0.10 0.50
0.45
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
1.30/2
2 0.10 0.50 0.80/2 1.30/2
3 0.05 0.25/3 0.60/4
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
1.30/2
2 0.10 0.50 0.80/2 1.30/2
3 0.05 0.25/3 0.60/4 1.25/4,5
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
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Empty matrix structure and formulae 0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
0.10 0.50
0.450.45
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
1.30/2
1.80/2
2 0.10 0.50 0.80/2 1.30/2
3 0.05 0.25/3 0.60/4 1.25/4,5
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
e 0 1 2 3 4 5
1 0.10 0.45/1 1.00/2
1.30/2
1.80/2
2 0.10 0.50 0.80/2 1.30/2
2.25/5
3 0.05 0.25/3 0.60/4 1.25/4,5
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
![Page 34: Dynamic Programming chandakmb@gmail.comchandakmb@gmail.com, hodcs@rknec.eduhodcs@rknec.edu Tutorial &Practice on Longest Common Sub-sequence](https://reader037.vdocument.in/reader037/viewer/2022103006/56649eac5503460f94bb2f19/html5/thumbnails/34.jpg)
Empty matrix structure and formula0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
0.10 0.50
0.450.45
0 1 2 3 4 5
1 0.10 0.45/1 1.00/2 1.30/2 1.80/2 2.70/2
2 0.10 0.50 0.80/2 1.30/2 2.25/5
3 0.05 0.25/3 0.60/4 1.25/4,5
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
k2
K3,K4,K5K1
Refer the cell35 from theMatrix andFind root InformationTake next rootas 4 or 5Consider : 4
Cost of TreeAnd main root
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Empty matrix structure and formula0 1 2 3 4 5
1 0.25 0.60 0.75 1.00
2 0.10 0.35 0.60 0.85
3 0.05 0.15 0.30 0.55
4 0.05 0.20 0.45
5 0.05 0.30
6 0.10
0.10 0.50
0.450.45
0 1 2 3 4 5
1 0.10 0.45/1 1.00/2 1.30/2 1.80/2 2.70/2
2 0.10 0.50 0.80/2 1.30/2 2.25/5
3 0.05 0.25/3 0.60/4 1.25/4,5
4 0.05 0.30/4 0.85/5
5 0.05 0.45/5
6 0.10
k2
K3,K4,K5K1
k3k1
k3 k5
![Page 36: Dynamic Programming chandakmb@gmail.comchandakmb@gmail.com, hodcs@rknec.eduhodcs@rknec.edu Tutorial &Practice on Longest Common Sub-sequence](https://reader037.vdocument.in/reader037/viewer/2022103006/56649eac5503460f94bb2f19/html5/thumbnails/36.jpg)
Empty matrix structure and formula
0 1 2 3 4 51 0.10 0.45/
11.00/2
1.30/2
1.80/2
2.70/2
2 0.10 0.50 0.80/2
1.30/2
2.25/5
3 0.05 0.25/3
0.60/4
1.25/4,5
4 0.05 0.30/4
0.85/5
5 0.05 0.45/5
6 0.10
k2
K3,K4,K5K1
k3k1
k3 k5
i 0 1 2 3 4 5
pi 0.05 0.20 0.05 0.10 0.15
qi 0.10 0.10 0.05 0.05 0.05 0.10
sum 0.10 0.15 0.25 0.10 0.15 0.25
do d1
d2 d3 d4 d5
Verification: 1x0.20 = 0.202x(0.05+0.05) = 0.203x(0.10+0.10) + 3x(0.05+0.15)=1.204x(0.05+0.05+0.05+0.10) = 1.00TOTAL = 2.70
L1
L2
L3
L4
pi
qi