4. Consider two sequences X = ABCBDA and Y = CABBDA. Find LCS (Longest Common Subsequence) performing:4.a Compute c[] and b[] (like we did in Lec9.1), draw the table.4.b Print LCS with PrintLCSO method, describe the steps.4.c Convert PrintLCSO from recursive into an iterative algorithm. Write your Pseudocode with commentaries.4.d Modify the code (LCSLength() and PrintLCSO) adding a parameter I (length of a subsequence) s.t. it prints the first common subsequence of length / or prints "No common subsequence of length I" if there is no common subsequence of length 1. Computing the LengthLCSLength(X, Y, m, n)let b[1..m, 1..n] and c[0..m, 0..n] be new tables (2d arrays)for i = 1 tomc[i, 0] = 0 // first columnjo 1 2 3 4 5 6for j = 0 to ny, BDCABAx 0 0 0 00 0 0iВc[0, j] = 0 // first rowfor i = 1 tomfor j = 1 to nif x, == y;c[i, j] = c[i - 1, j - 1] + 1bli, j] = "R"else if c[i - 1, j] > c[i, j - 1]c[i, j] = c[i - 1, j]b[i, j] = “ ↑ "1 0 1 010 k 1+1 K 12в о+ 1+ 1↑ 1 R 2 + 2↑ 1↑ 1 k 2 +2 1 2 1 24B 01 ↑ 1↑ 21 2 K 3+ 3%3DD↑ 1 R 21 2 1 2 1 3 ↑ 3А↑ 1↑ 21 2 K 3↑ 3else c[i, j] = c[i, j - 1]b[i, j] = “+"7В↑ 2↑ 3return c and b4.4.A Printing the LCSPrintLCS(b,X, i, j)if i == 0 or j ==0return Noneif b[i, j] == "R"PrintLCS(b, X, i - 1, j - 1)jo 123 4 5 6x BD CABA3 AiВВprint x,X 0 0 00 0 0 0else if b[i, j] == "↑"PrintLCS(b, X, i - 1, j)1A 0 1010 k 1+1R 12B 0+ 1+ 1↑ 1 k 2 + 2else PrintLCS(b, X, i, j - 1)co 11↑ 1 k 2↑ 2+ 2 1 2Initial call: PrintLCS(b, X, m, n)4 B0k 1↑ 1 12 1 2 K 3 + 3Running time: 0(m+n).D0 11 2A 0 117 B011 2 1 2↑ 3 1 3Overall running time for LCS is↑ 2↑ 2 R 3 1 3 R 4Ө (mn) + Ө(m+n) —O(mn)↑ 2 1 2 1 3 R 4↑ 4

Answered: 4. Consider two sequences X = ABCBDA…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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4. Consider two sequences X = ABCBDA and Y = CABBDA. Find LCS (Longest Common Subsequence) performing:

4.a Compute c[] and b[] (like we did in Lec9.1), draw the table.

4.b Print LCS with PrintLCSO method, describe the steps.

4.c Convert PrintLCSO from recursive into an iterative algorithm. Write your Pseudocode with commentaries.

4.d Modify the code (LCSLength() and PrintLCSO) adding a parameter I (length of a subsequence) s.t. it prints the first common subsequence of length / or prints "No common subsequence of length I" if there is no common subsequence of length 1.

Computing the Length
LCSLength(X, Y, m, n)
let b[1..m, 1..n] and c[0..m, 0..n] be new tables (2d arrays)
for i = 1 tom
c[i, 0] = 0 // first column
jo 1 2 3 4 5 6
for j = 0 to n
y, BDCABA
x 0 0 0 00 0 0
i
В
c[0, j] = 0 // first row
for i = 1 tom
for j = 1 to n
if x, == y;
c[i, j] = c[i - 1, j - 1] + 1
bli, j] = "R"
else if c[i - 1, j] > c[i, j - 1]
c[i, j] = c[i - 1, j]
b[i, j] = “ ↑ "
1 0 1 0
10 k 1
+1 K 1
2
в о
+ 1
+ 1
↑ 1 R 2 + 2
↑ 1
↑ 1 k 2 +2 1 2 1 2
4
B 01 ↑ 1
↑ 2
1 2 K 3
+ 3
%3D
D
↑ 1 R 2
1 2 1 2 1 3 ↑ 3
А
↑ 1
↑ 2
1 2 K 3
↑ 3
else c[i, j] = c[i, j - 1]
b[i, j] = “+"
7
В
↑ 2
↑ 3
return c and b
4.
4.
A

Printing the LCS
PrintLCS(b,X, i, j)
if i == 0 or j ==0
return None
if b[i, j] == "R"
PrintLCS(b, X, i - 1, j - 1)
jo 1
2
3 4 5 6
x BD CABA
3 A
i
В
В
print x,
X 0 0 0
0 0 0 0
else if b[i, j] == "↑"
PrintLCS(b, X, i - 1, j)
1
A 0 10
10 k 1
+1R 1
2
B 0
+ 1
+ 1
↑ 1 k 2 + 2
else PrintLCS(b, X, i, j - 1)
co 11
↑ 1 k 2
↑ 2
+ 2 1 2
Initial call: PrintLCS(b, X, m, n)
4 B0k 1
↑ 1 12 1 2 K 3 + 3
Running time: 0(m+n).
D0 11 2
A 0 11
7 B01
1 2 1 2
↑ 3 1 3
Overall running time for LCS is
↑ 2
↑ 2 R 3 1 3 R 4
Ө (mn) + Ө(m+n) —
O(mn)
↑ 2 1 2 1 3 R 4
↑ 4

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