Transcript Recursive & Dynamic Programming - monjur-ul

Recursive & Dynamic
Programming
LEVEL II, TERM II
CSE – 243
MD. MONJUR-UL-HASAN
LECTURER
DEPT OF CSE, CUET
EMAIL: [email protected]
Introduction to Recursion
2
 "Normally", we have methods that call other
methods.

For example, the main() method calls the square() method.
main()
square()
 Recursive Method:
 A recursive method is a method that calls itself.
compute()
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Why Recursive
3
 In computer science, some problems are more easily
solved by using recursive methods.
 In this course, will see many of examples of this.
 For example:



Traversing through directories of a file system.
Traversing through a tree of search results.
Some sorting algorithms recursively sort data
 For today, we will focus on the basic structure of
using recursive methods.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Situation
4
LIST
 Consider the following list of numbers:
24, 88, 40, 37
 Such a list can be defined as follows:
A LIST is a:
or a:
number
number
comma
LIST
 That is, a LIST is defined to be a single number, or a
number followed by a comma followed by a LIST
 The concept of a LIST is used to define itself
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Situation
5
 The recursive part of the LIST definition is used
several times, terminating with the non-recursive
part:
number comma LIST
24
,
88, 40, 37
number comma LIST
88
,
40, 37
number comma LIST
40
,
37
number
37
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Situation
6
GCD
 gcd(a,b)
=
=
a
gcd(b,a%b)
 gcd(4032, 1272) = gcd(1272, 216)
= gcd(216, 192)
= gcd(192, 24)
= gcd(24, 0)
= 24.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
; when b
; otherwise
Recursive Situation
7
Finding Power
The power function, p(x,n)=xn, can be defined recursively:
1
if n  0

p( x, n)  
x  p( x, n 1) else
 This leads to an power function that runs in O(n) time (for
we make n recursive calls).
 We can do better than this, however.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Situation
8
Recursive Squaring
 We can derive a more efficient linearly recursive algorithm
by using repeated squaring:
1
if x  0


p( x, n)  x  p( x, (n 1) / 2)2 if x  0 is odd
 p( x, n / 2)2
if x  0 is even

 For example,
24 = 2(4/2)2 = (24/2)2 = (22)2 = 42 = 16
25 = 21+(4/2)2 = 2(24/2)2 = 2(22)2 = 2(42) = 32
26 = 2(6/ 2)2 = (26/2)2 = (23)2 = 82 = 64
27 = 21+(6/2)2 = 2(26/2)2 = 2(23)2 = 2(82) = 128.
 Better then this solution is also available
 For Practice Try to solve (bn%P) in this same way for large number of n
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Situation
9
Reversing an Array
Algorithm ReverseArray(A, i, j):
Input: An array A and nonnegative integer indices i and j
Output: The reversal of the elements in A starting at
index i and ending at j
if i < j then
Swap A[i] and A[ j]
ReverseArray(A, i + 1, j - 1)
return
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Definition
10
 Base case: You must always have some base case
which can be solved without recursion
 Making Progress: For cases that are to be solved
recursively, the recursive call must always be a case
that makes progress toward the base case.
 Design Rule: Assume that the recursive calls work.
 Compound Interest Rule: Never duplicate work
by solving the same instance of a problem in separate
recursive calls.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Visualizing Recursion
11
 Recursion trace
 A box for each
recursive call
 An arrow from each
caller to callee
 An arrow from each
callee to caller
showing return value
Example recursion trace:
1
if n  0

f (n)  
else
n  f (n 1)
return 3*2 = 6
call
recursiveFactorial(3)
return 2*1 = 2
call
recursiveFactorial(2)
call
recursiveFactorial(1 )
call
recursiveFactorial(0)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
return 1*1 = 1
Return 1
final answer
Visualizing Recursion
12
fact(1)
fact(3)
main()
Time 2:
Push: fact(3)
fact(2)
fact(2)
fact(2)
fact(3)
fact(3)
fact(3)
main()
Time 3:
Push: fact(2)
Inside findFactorial(3):
if (number <= 1) return 1;
else return (3 * factorial
(2));
main()
Time 4:
Push: fact(1)
main()
Time 5:
Pop: fact(1)
returns 1.
Inside findFactorial(2):
if (number <= 1) return 1;
else return (2 * factorial
(1));
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
1
fact(3)
2
main()
main()
Time 6:
Pop: fact(2)
returns 2.
Inside findFactorial(1):
if (number <= 1) return 1;
else return (1 * factorial (0));
6
Time 7:
Pop: fact(3)
returns 6.
Several Types of Recursion
13
 Liner Recursion

May be more than one recursive method available but for each
recursion it calls only one among them
 Tail Recursion

The recursive call will be the last statement of the method
 Chain Recursion

Function A will call Function B. function B will call Function C …… .
… last function will call function A again.
 Multiple Recursion

f(x) = a
=b
= a*f(x-1) + b*f(x-2)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
when x = 0
when x = 1
otherwise
Defining Arguments for Recursion
14
 In creating recursive methods, it is important to define the
methods in ways that facilitate recursion.
 This sometimes requires we define additional paramaters
that are passed to the method.
 For example, we defined the array reversal method as
ReverseArray(A, i, j), not ReverseArray(A).
To verify that a recursive definition works:


convince yourself that the base case(s) are handled correctly
ASSUME RECURSIVE CALLS WORK ON SMALLER PROBLEMS, then
convince yourself that the results from the recursive calls are combined to solve
the whole
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
When recursion?
15

when it is the most natural way of thinking about & implementing a
solution


when it is roughly equivalent in efficiency to an iterative solution



can solve problem by breaking into smaller instances, solve, combine
solutions
OR
when the problems to be solved are so small that efficiency doesn't
matter
think only one level deep
make sure the recursion handles the base case(s) correctly
 assume recursive calls work correctly on smaller problems
 make sure solutions to the recursive problems are combined correctly


avoid infinite recursion

make sure there is at least one base case & each recursive call gets
closer
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Alternative of Recursive
16
 any recursive method/class can be rewritten iteratively (i.e.,
using a loop)

but sometimes, a recursive definition is MUCH clearer
e.g., PermutationGenerator would be very difficult to conceptualize &
implement without recursion
 Use stack to model the recursive into iterative
method. [ use the data structure course concept]
 When overhead is low it is always better to use
recursive but in other case it is better to use iterative
solution.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursion pros and cons
17
 All recursive solutions can be implemented without




recursion.
Recursion is "expensive". The expense of recursion lies in
the fact that we have multiple activation frames and the fact
that there is overhead involved with calling a method.
If both of the above statements are true, why would we ever
use recursion?
In many cases, the extra "expense" of recursion is far
outweighed by a simpler, clearer algorithm which leads to
an implementation that is easier to code.
Ultimately, the recursion is eliminated when the compiler
creates assembly language (it does this by implementing
the stack).
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Think Your Own Recursion
18
 Try to solve the problem for base case
 Think in the middle of the problem with following
criteria:


Define that with same problem but with different domain
The change in domain should leads that to the base case.
 You need not to think all the problem together. Just
think one step from middle and the base case.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Generating Permutation Systematically
19
numerous applications require systematically generating permutations (orderings)
 e.g., word games, debugging concurrent systems, tournament pairings, . . .

want to be able to take some sequence of items (say a String of characters) and
generate every possible arrangement without duplicates

"123"  "123", "132", "213", "231", "312", "321"




"tape"  "tape", "taep", "tpae", "tpea", "teap", "tepa",
"atpe", "atep", "apte", "apet", "aetp", "aept",
"ptae", "ptea", "pate", "paet", "peta", "peat",
"etap", "etpa", "eatp", "eapt", "epta", "epat“
 how do we generate permutations systematically?
•
must maintain initial ordering
•
must get all permutations, with no duplicates
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Generating Permutation Systematically
20
 Generate all permutations that start with ‘t' , then 'a'
then ‘p‘ then ‘e’
 To generate permutations starting with ‘t', we need
to find all permutations of "ape"
 This is the same problem with simpler inputs.
 Use recursion
To get your permutation go:
http://home.att.net/~srschmitt/script_permutatio
ns.html
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
All Permutation Algorithm
21
consider P is a global array contain the string
exch (int i, int j)
{
int t = p[i]; p[i] = p[j]; p[j] = t; }
generate(int N)
{
int c;
if (N == 1) doit();
for (c = 1; c <= N; c++)
{ exch(c, N); generate(N-1); exch(c, N); }
}
 Invoke by calling
 generate(N);
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Challenge to Students (75 Marks)
22
Redesign the algorithm with no global variable
Redesign the algorithm that u develop in 1 to generate
permutation which contain k element from the n length
string
3. Redesign the algorithm of describe in 1 to generate all
combination from a string by taking k element each
time?
4. Design a correct algorithm to generate permutation
when the string contain repetitive character in the
given string
5. Design a correct Algorithm to generate Combination of
k element among n element where the string contain
repetitive character.
1.
2.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Marks
23



Fail to submit: -15 from your Lab work
Duplicate submission : add absolute 0 with your lab work
Correct solution will add marks as follows
Q1:
5
 Q2 & Q2: 10+10 =
20
 Q4
20
 Q5
30
 Total
75
This mark will add to you lab result.

Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive to Iterative
24
a when x  0

f ( x)  b when x  1
a * f ( x 1)  b * f ( x  2)

otherwise
 What is the problems with this recursion?
 Is there any direct solution available of this recursive
function?
 How can we test and find if any such available.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Guess-and-Test Method
25
 In the guess-and-test method, we guess a closed form
solution and then try to prove it is true by induction:
b
if n  2

T ( n)  
2T (n / 2)  bn log n if n  2
 Guess: T(n) < cn log n.
T (n)  2T (n / 2)  bn log n
 2(c(n / 2) log(n / 2))  bn log n
 cn(log n  log 2)  bn log n
 cn log n  cn  bn log n
 Wrong: we cannot make this last line be less than
cnlogn
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Guess-and-Test Method, Part 2
26
 Recall the recurrence equation:
b
if n  2

T ( n)  
2T (n / 2)  bn log n if n  2
 Guess #2: T(n) < cn log2 n.
T (n)  2T (n / 2)  bn log n
 2(c(n / 2) log 2 (n / 2))  bn log n
 cn(log n  log 2)2  bn log n
 cn log 2 n  2cn log n  cn  bn log n

if c > b.
 cn log 2 n
 So, T(n) is O(n log2 n).
 In general, to use this method, you need to have a good guess and you need
to be good at induction proofs.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Master Method
27
 Many divide-and-conquer recurrence equations
have the form:
c
if n  d

T ( n)  
aT (n / b)  f (n) if n  d
 The Master Theorem:
1. if f (n) is O(nlogb a ), then T (n) is (nlogb a ), where
 0
2. if f (n) is (nlogb a logk n), then T (n) is (nlogb a log k 1 n)
3. if f (n) is (nlogb a ), then T (n) is ( f (n)),
provided af (n / b)  f (n) for some  1.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Master Method, Example 1
28
 The form:
c
if n  d

T ( n)  
aT (n / b)  f (n) if n  d
 The Master Theorem:
1. if f (n) is O(nlogb a ), then T (n) is (nlogb a )
2. if f (n) is (nlogb a log k n), then T (n) is (nlogb a log k 1 n)
3. if f (n) is (nlogb a ), then T (n) is ( f (n)),
provided af (n / b)  f (n) for some  1.
 Example:
T (n)  4T (n / 2)  n
 Solution: logba=2, so case 1 says T(n) is O(n2).
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Master Method, Example 2
29
 The form:
c
if n  d

T ( n)  
aT (n / b)  f (n) if n  d
 The Master Theorem:
1. if f (n) is O(nlogb a ), then T (n) is (nlogb a )
2. if f (n) is (nlogb a log k n), then T (n) is (nlogb a log k 1 n)
3. if f (n) is (nlogb a ), then T (n) is ( f (n)),
provided af (n / b)  f (n) for some  1.
 Example:
T (n)  2T (n / 2)  n log n
 Solution: logba=1, so case 2 says T(n) is O(n log2 n).
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Master Method, Example 3
30
 The form:
c
if n  d

T ( n)  
aT (n / b)  f (n) if n  d
 The Master Theorem:
1. if f (n) is O(nlogb a ), then T (n) is (nlogb a )
2. if f (n) is (nlogb a log k n), then T (n) is (nlogb a log k 1 n)
3. if f (n) is (nlogb a ), then T (n) is ( f (n)),
provided af (n / b)  f (n) for some  1.
 Example:
T (n)  T (n / 3)  n log n
Solution: logba=0, so case 3 says T(n) is O(n log n).
As f(n) = nlogn is asymptotically larger then nlogba =n
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Divide & Conquer Algorithm
31
 Divide
problem into sub-problems
 Conquer by solving sub-problems recursively.
If the sub-problems are small enough, solve them
in brute force fashion
 Combine the solutions of sub-problems into a
solution of the original problem (tricky part)
Example
MergeSort
Finding the middle point in the
alignment matrix in linear space
Linear space
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
sequence alignment
Block Alignment
Four-Russians speedup
Constructing LCS in sub-quadratic time
Dynamic Programming
32
fibonacci(5)
fibonacci(4)
fibonacci(3)
+
fibonacci(2)
+
fibonacci(3)
fibonacci(2) + fibonacci(1)
fibonacci(2) + fibonacci(1)
 Dynamic Programming is an algorithm design technique
for optimization problems: often minimizing or
maximizing.
 Like divide and conquer, DP solves problems by combining
solutions to subproblems.
 Unlike divide and conquer, subproblems are not
independent.


Subproblems may share subsubproblems,
However, solution to one subproblem may not affect the solutions to
other subproblems of the same problem. (More on this later.)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Dynamic Programming: How to think?
33
 DP reduces computation by
 Solving subproblems in a bottom-up fashion.
 Storing solution to a subproblem the first time it is solved.
 Looking up the solution when subproblem is encountered again.
 Key: determine structure of optimal solutions
Elements of Dynamic Programming
 Optimal substructure
 Overlapping subproblems
 Memorization
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Optimizing Substructure
34
 Show that a solution to a problem consists of making a




choice, which leaves one or more subproblems to solve.
Suppose that you are given this last choice that leads to an
optimal solution.
Given this choice, determine which subproblems arise and
how to characterize the resulting space of subproblems.
Show that the solutions to the subproblems used within the
optimal solution must themselves be optimal. Usually use
cut-and-paste.
Need to ensure that a wide enough range of choices and
subproblems are considered.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Optimal SubStructure
35
 Optimal substructure varies across problem domains:
 1. How many subproblems are used in an optimal solution.
 2. How many choices in determining which subproblem(s) to use.
 Informally, running time depends on (# of subproblems
overall)  (# of choices).
 How many subproblems and choices do the examples
considered contain?
 Dynamic programming uses optimal substructure bottom
up.


First find optimal solutions to subproblems.
Then choose which to use in optimal solution to the problem.
 Pieces of larger problem have a sequential
dependency
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Overlapping Subproblems
36
 The space of subproblems must be “small”.
 The total number of distinct subproblems is a
polynomial in the input size.
A recursive algorithm is exponential because it solves the
same problems repeatedly.
 If divide-and-conquer is applicable, then each problem
solved will be brand new.

Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
DP Approach
37
 Forward Approach
 Backward Approach
Note that if the recurrence relations are formulated using
the forward approach then the relations are solved
backwards . i.e., beginning with the last decision
 On the other hand if the relations are formulated using
the backward approach, they are solved forwards.

You can also represent the problem by a multistage
graph
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
A Simple DP Approach
38
Fibonacci Numbers
Int fib(int n)
{
static int knownFib[MAXFIB];
int x;
if(knownFib[n]==0)
{
if(n==1 || n==2)
knownFib[n] = 1;
else knownFib[n] = fib(n-1) + fib(n-2);
}
}
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Matrix-chain multiplication
39
 What will be the number of scalar multiplication in a matrix
multiplication of a 3X5 and 5X 10 matrix?
 What will be new matrix’s direction?
 What is the number of scalar multiplication of the following
matrices
A1 X A2 X A3 X A4
Where, the damnations are 3x5, 5x4, 4x2 and 2x5
We have the following choice:
((A1  A2)  A3)  A4, # of scalar multiplications:
3 * 5 * 4 + 3 * 4 * 2 + 3 * 2 * 5 = 114
(A1  (A2  A3))  A4, # of scalar multiplications:
3 * 5 * 2 + 5 * 4 * 2 + 3 * 2 * 5 = 100
(A1  A2)  (A3  A4), # of scalar multiplications:
3 * 5 * 4 + 3 * 4 * 5 + 4 * 2 * 5 = 160
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Function
40
 Let m(i, j) denote the minimum cost for computing
Ai  Ai+1  …  Aj
0
if i  j
m(i, j)  minm(i, k)  m(k  1, j)  p p p 
if i  j
i 1 k i
 ik j
 Computation sequence :
m(1,4)
m(1,3)
m(1,2)
m(2,4)
m(2,3)
m(3,4)
 Time complexity : O(n3)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
8 -40
41
Since subproblems
overlap, we don’t use
recursion.
Instead, we construct optimal
subproblems “bottom-up.”
Ni,i’s are easy, so start with
them. Then do length 2,3,…
subproblems,and so on.
Running time: O(n3)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Matrix Chan Multiplication
42
 Algorithm matrixChain(S):
 Input: sequence S of n matrices to be multiplied
 Output: number of operations in an optimal
paranethization of S
1. for i ← 1 to n-1 do
Ni,i ← 0
1.
2. for b ← 1 to n-1 do
for i ← 0 to n-b-1 do
1.
1.
2.
3.
1.
2.
j ← i+b
Ni,j ← +infinity
for k ← i to j-1 do
Ni,j ← min{Ni,j , Ni,k +Nk+1,j +di dk+1 dj+1}
Si,j ← k //for which Ni,j Minimum
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Matrix Chan Multiplication
43
 Print the Optimal Structure
OPTIMAL-MAT(s,i,j)
{
if(i==j)
then print “A”I
else print “(“
OPTIMAL-MAT(s,i,s[i,j)
OPTIMAL-MAT(s,s[i,j]+1, j)
print “)”;
}
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Longest Common Subsequence (LCS)
44
 Problem: Given 2 sequences, X = x1,...,xm and
Y = y1,...,yn, find a common subsequence whose
length is maximum.
springtime
ncaa tournament
basketball
printing
north carolina
krzyzewski
Subsequence need not be consecutive, but must be in order.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
LCS (Native Algorithm)
45
 For every subsequence of X, check whether it’s a
subsequence of Y .
 Time: Θ(n2m).


2m subsequences of X to check.
Each subsequence takes Θ(n) time to check:
scan Y for first letter, for second, and so on.
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
LCS: Optimal Substructure
46
Theorem
Let Z = z1, . . . , zk be any LCS of X and Y .
1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.
2. If xm  yn, then either zk  xm and Z is an LCS of Xm-1 and Y .
3.
or zk  yn and Z is an LCS of X and Yn-1.
Notation:
prefix Xi = x1,...,xi is the first i letters of X.
This says what any longest common subsequence
must look like;
do you believe it?
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Recursive Solution
47
if  emptyor  empty,
0

c[ ,  ]  c[ prefix , prefix ]  1
if end( )  end( ),
max(c[ prefix ,  ], c[ , prefix ]) if end( )  end( ).

c[springtime, printing]
c[springtim, printing]
c[springtime, printin]
[springti, printing] [springtim, printin] [springtim, printin] [springtime,
printi]
[springt, printing] [springti, printin] [springtim, printi] [springtime, print]
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
Memorization
48
if  emptyor  empty,
0

c[ ,  ]  c[ prefix , prefix ]  1
if end( )  end( ),
max(c[ prefix ,  ], c[ , prefix ]) if end( )  end( ).

p
•Keep track of c[,] in a
table of nm entries:
s
•top/down
p
•bottom/up
r
i
n
g
t
i
m
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
e
r
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t
i
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LCS: Length Finding Algorithm
49
LCS-LENGTH (X, Y)
1. m ← length[X]
2. n ← length[Y]
3. for i ← 1 to m
4.
do c[i, 0] ← 0
5. for j ← 0 to n
6.
do c[0, j ] ← 0
7. for i ← 1 to m
8.
do for j ← 1 to n
9.
do if xi = yj
10.
then c[i, j ] ← c[i1, j1] + 1
11.
b[i, j ] ← “ ”
12.
else if c[i1, j ] ≥ c[i, j1]
13.
then c[i, j ] ← c[i 1, j ]
14.
b[i, j ] ← “↑”
15.
else c[i, j ] ← c[i, j1]
16.
b[i, j ] ← “←”
17. return c and b
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
50
p
r
i
n
t
i
n
g
0
0
0
0
0
0
0
0
0
s 0
0
0
0
0
0
0
0
0
s
p 0
1
1
1
1
1
1
1
1
P
r 0
1
2
2
2
2
2
2
2
r
i 0
1
2
3
3
3
3
3
3
i
n 0
1
2
3
4
4
4
4
4
n
g 0
1
2
3
4
4
4
4
5
g
t 0
1
2
3
4
5
5
5
5
t
i 0
1
2
3
4
5
6
6
6
i
m 0
1
2
3
4
5
6
6
6
m
e 0
1
2
3
4
5
6
6
6
e
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET
p
r
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LCS: Constructing
51
PRINT-LCS (b, X, i, j)
1. if i = 0 or j = 0
2.
then return
3. if b[i, j ] = “ ”
4.
then PRINT-LCS(b, X, i1, j1)
5.
print xi
6.
elseif b[i, j ] = “↑”
7.
then PRINT-LCS(b, X, i1, j)
8. else PRINT-LCS(b, X, i, j1)
Md. Monjur-ul-hasan. Lecturer, Dept. of CSE, CUET