IGNOU MMTE002 Solved Assignment 2024  M.Sc. MACS
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IGNOU MMTE002 Assignment Question Paper 2024
mmte002solvedassignment2024b772e899b2ce40f786862afea878ebb1
 a) The maximum subsequence sum problem is defined as follows: If
${a}_{1},{a}_{2},\dots ,{a}_{n}$ ${a}_{1},{a}_{2},\dots ,{a}_{n}$ a_(1),a_(2),dots,a_(n) a_1, a_2, \ldots, a_n are in${a}_{1},{a}_{2},\dots ,{a}_{n}$ $\mathbf{Z}$ $\mathbf{Z}$ Z \mathbf{Z} , find the maximum value$\mathbf{Z}$ $\sum _{k=i}^{j}{a}_{i}$ $\sum _{k=i}^{j}\u200a{a}_{i}$ sum_(k=i)^(j)a_(i) \sum_{k=i}^j a_i , for all$\sum _{k=i}^{j}{a}_{i}$ $\mathrm{i},\mathrm{j},1\le i\le j\le n$ $\mathrm{i},\mathrm{j},1\le i\le j\le n$ i,j,1 <= i <= j <= n \mathrm{i}, \mathrm{j}, 1 \leq i \leq j \leq n . We assume that the answer is 0 if all the$\mathrm{i},\mathrm{j},1\le i\le j\le n$ ${a}_{i}$ ${a}_{i}$ a_(i) a_i are negative or if the sum is empty. The following algorithm finds a solution to the problem. Here, we assume that${a}_{i}$ ${a}_{i}\text{}\mathrm{s}$ ${a}_{i}\text{}\mathrm{s}$ a_(i)s a_i \mathrm{~s} are stored in the array${a}_{i}\text{}\mathrm{s}$ $\mathrm{A}$ $\mathrm{A}$ A \mathrm{A} .$\mathrm{A}$
b) Analyse the algorithm and find an upper bound for the run time of the above algorithm.
2) a) With the help of an example, explain the following:
i) Algorithm.
ii) Input and output for an algorithm.
iii) Running time of an algorithm.
b) Using Fig. 7.1 in page 147 of the book as the model, illustrate the operation of PARTITION on the array
3)
a) For the set of keys
b) Using Fig. 6.3 in page 134 of the book as a model, illustrate the operation of BUILDMAXHEAP on the array
 a) Show the results of inserting the keys
b) Suppose the ConnectedComponents is run on the undirected graph
5) a) Show how mergesort sorts the array
b) For the following set of points, describe how the CLOSESTPAIR algorithm finds a closest pair of points:
6) a) In the Coin changing problem, we have to give change for
b) Determine an LCS of
7) a) Show the
source vertex.
 a) Show the comparisons the naive string matcher makes for the pattern 𝑃 = 0100 with 01100010010100100.
c) Compute the values
9) a) Find all the solutions of the equation
b) Let
c) Compute the DFT of the vector
MMTE002 Sample Solution 2024
mmte002solvedassignment2024ss8e24e61006c94b4384f6a5bf6ef5ab5c
 a) The maximum subsequence sum problem is defined as follows: If
${a}_{1},{a}_{2},\dots ,{a}_{n}$ ${a}_{1},{a}_{2},\dots ,{a}_{n}$ a_(1),a_(2),dots,a_(n) a_1, a_2, \ldots, a_n are in${a}_{1},{a}_{2},\dots ,{a}_{n}$ $\mathbf{Z}$ $\mathbf{Z}$ Z \mathbf{Z} , find the maximum value$\mathbf{Z}$ $\sum _{k=i}^{j}{a}_{i}$ $\sum _{k=i}^{j}\u200a{a}_{i}$ sum_(k=i)^(j)a_(i) \sum_{k=i}^j a_i , for all$\sum _{k=i}^{j}{a}_{i}$ $\mathrm{i},\mathrm{j},1\le i\le j\le n$ $\mathrm{i},\mathrm{j},1\le i\le j\le n$ i,j,1 <= i <= j <= n \mathrm{i}, \mathrm{j}, 1 \leq i \leq j \leq n . We assume that the answer is 0 if all the$\mathrm{i},\mathrm{j},1\le i\le j\le n$ ${a}_{i}$ ${a}_{i}$ a_(i) a_i are negative or if the sum is empty. The following algorithm finds a solution to the problem. Here, we assume that${a}_{i}$ ${a}_{i}\text{}\mathrm{s}$ ${a}_{i}\text{}\mathrm{s}$ a_(i)s a_i \mathrm{~s} are stored in the array${a}_{i}\text{}\mathrm{s}$ $\mathrm{A}$ $\mathrm{A}$ A \mathrm{A} .$\mathrm{A}$
for
loop in lines 28 is:MaxSum
is the maximum subsequence sum of the subarray MaxSum
correctly represents the maximum sum of any subsequence that ends at or before the element 
Initialization (Beginning of the loop,
$i=1$ $i=1$ i=1 i = 1 ): Before the first iteration,$i=1$ MaxSum
is set to 0, which is correct since we have not encountered any elements of the array and the maximum subsequence sum of an empty array is 0. 
Maintenance (During the loop): At each iteration, we add
$A[i]$ $A[i]$ A[i] A[i] to$A[i]$ Sum
. IfSum
becomes greater thanMaxSum
, we updateMaxSum
toSum
because we have found a larger subsequence sum ending at$i$ $i$ i i . If$i$ Sum
is negative, we set it to 0 since a maximum subsequence sum cannot be negative (as per the problem statement, we take the maximum sum to be 0 in this case). This maintains the loop invariant becauseMaxSum
continues to represent the maximum sum of any subsequence that ends at or before$i$ $i$ i i .$i$ 
Termination (End of the loop): When the loop terminates (after the final iteration when
$i=n$ $i=n$ i=n i = n ), the loop invariant tells us that$i=n$ MaxSum
holds the maximum subsequence sum of the entire array$A[1..n]$ $A[1..n]$ A[1..n] A[1..n] , because every element has been considered, and$A[1..n]$ MaxSum
has been updated accordingly throughout the loop.
By using the loop invariant, we have shown that at the end of each iteration of the loop, the
MaxSum
variable holds the maximum subsequence sum for the part of the array processed so far. After the loop terminates, MaxSum
will therefore hold the maximum subsequence sum for the entire array. This proves that the algorithm correctly finds the maximum subsequence sum for the array 
Initialization (Line 1): Initializing
Sum
andMaxSum
takes constant time, so this part is$O(1)$ $O(1)$ O(1) O(1) .$O(1)$ 
For Loop (Lines 28):
 The loop iterates
$n$ $n$ n n times, where$n$ $n$ $n$ n n is the length of the array$n$ $A$ $A$ A A .$A$  Inside the loop, updating
Sum
(Line 4) and the comparison (Line 5) are constanttime operations, so they each take$O(1)$ $O(1)$ O(1) O(1) time per iteration.$O(1)$  Updating
MaxSum
(Line 6) and resettingSum
(Line 8) are also constanttime operations, executed at most once per iteration, so they also take$O(1)$ $O(1)$ O(1) O(1) time per iteration.$O(1)$
 The loop iterates
 Overall Runtime: Combining the initialization and the loop, the overall runtime of the algorithm is
$O(1)+O(n)=O(n)$ $O(1)+O(n)=O(n)$ O(1)+O(n)=O(n) O(1) + O(n) = O(n) .$O(1)+O(n)=O(n)$
The upper bound for the runtime of the algorithm is
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