# Dynamic Programming

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Question 1 |

Which of the following standard algorithms is not Dynamic Programming based.

Bellman–Ford Algorithm for single source shortest path | |

Floyd Warshall Algorithm for all pairs shortest paths | |

0-1 Knapsack problem | |

Prim's Minimum Spanning Tree |

**Dynamic Programming**

**Discuss it**

Question 1 Explanation:

Prim's Minimum Spanning Tree is a Greedy Algorithm. All other are dynamic programming based.

Question 2 |

We use dynamic programming approach when

We need an optimal solution | |

The solution has optimal substructure | |

The given problem can be reduced to the 3-SAT problem | |

It's faster than Greedy |

**Dynamic Programming**

**Discuss it**

Question 2 Explanation:

http://www.geeksforgeeks.org/dynamic-programming-set-2-optimal-substructure-property/
Option (D) is incorrect because Greedy algorithms are generally faster than Dynamic programming. See http://www.geeksforgeeks.org/greedy-algorithms-set-1-activity-selection-problem/

Question 3 |

An algorithm to find the length of the longest monotonically increasing sequence of numbers in an array A[0 :n-1] is given below.
Let Li denote the length of the longest monotonically increasing sequence starting at index i in the array.

Which of the following statements is TRUE?

Which of the following statements is TRUE?

The algorithm uses dynamic programming paradigm | |

The algorithm has a linear complexity and uses branch and bound paradigm | |

The algorithm has a non-linear polynomial complexity and uses branch and bound paradigm | |

The algorithm uses divide and conquer paradigm. |

**Dynamic Programming**

**Discuss it**

Question 3 Explanation:

Question 4 |

Kadane algorithm is used to find:

Maximum sum subsequence in an array | |

Maximum sum subarray in an array | |

Maximum product subsequence in an array | |

Maximum product subarray in an array |

**Dynamic Programming**

**Discuss it**

Question 4 Explanation:

Kadane algorithm is used to find the maximum sum subarray in an array. It runs in O(n) time complexity. See details of the algorithm here.

Question 5 |

Four matrices M1, M2, M3 and M4 of dimensions pxq, qxr, rxs and sxt respectively can be multiplied is several ways with different number of total scalar multiplications. For example, when multiplied as ((M1 X M2) X (M3 X M4)), the total number of multiplications is pqr + rst + prt. When multiplied as (((M1 X M2) X M3) X M4), the total number of scalar multiplications is pqr + prs + pst.
If p = 10, q = 100, r = 20, s = 5 and t = 80, then the number of scalar multiplications needed is

248000 | |

44000 | |

19000 | |

25000 |

**Dynamic Programming**

**Discuss it**

Question 5 Explanation:

It is basically matrix chain multiplication problem. We get minimum number of multiplications using ((M1 X (M2 X M3)) X M4).
Total number of multiplications = 100x20x5 (for M2 x M3) + 10x100x5 + 10x5x80 = 19000.

Question 6 |

The subset-sum problem is defined as follows. Given a set of n positive integers, S = {a1 ,a2 ,a3 ,…,an} and positive integer W, is there a subset of S whose elements sum to W? A dynamic program for solving this problem uses a 2-dimensional Boolean array X, with n rows and W+1 columns. X[i, j],1 <= i <= n, 0 <= j <= W, is TRUE if and only if there is a subset of {a1 ,a2 ,...,ai} whose elements sum to j. Which of the following is valid for 2 <= i <= n and ai <= j <= W?

X[i, j] = X[i - 1, j] ∨ X[i, j -ai] | |

X[i, j] = X[i - 1, j] ∨ X[i - 1, j - ai] | |

X[i, j] = X[i - 1, j] ∧ X[i, j - ai] | |

X[i, j] = X[i - 1, j] ∧ X[i -1, j - ai] |

**Dynamic Programming**

**Discuss it**

Question 6 Explanation:

X[I, j] (2 <= i <= n and ai <= j <= W), is true if any of the following is true
1) Sum of weights excluding ai is equal to j, i.e., if X[i-1, j] is true.
2) Sum of weights including ai is equal to j, i.e., if X[i-1, j-ai] is true so that we get (j – ai) + ai as j
See http://www.geeksforgeeks.org/dynamic-programming-subset-sum-problem/ for details.

Question 7 |

In the above question, which entry of the array X, if TRUE, implies that there is a subset whose elements sum to W?

X[1, W] | |

X[n ,0] | |

X[n, W] | |

X[n -1, n] |

**Dynamic Programming**

**Discuss it**

Question 7 Explanation:

If we get the entry X[n, W] as true then there is a subset of {a1, a2, .. an} that has sum as W.
Reference: http://en.wikipedia.org/wiki/Subset_sum_problem

Question 8 |

A sub-sequence of a given sequence is just the given sequence with some elements (possibly none or all) left out. We are given two sequences X[m] and Y[n] of lengths m and n respectively, with indexes of X and Y starting from 0.
We wish to find the length of the longest common sub-sequence(LCS) of X[m] and Y[n] as l(m,n), where an incomplete recursive definition for the function l(i,j) to compute the length of The LCS of X[m] and Y[n] is given below:

l(i,j) = 0, if either i=0 or j=0 = expr1, if i,j > 0 and X[i-1]=Y[j-1] = expr2, if i,j > 0 and X[i-1]!=Y[j-1]

expr1 ≡ l(i-1, j) + 1 | |

expr1 ≡ l(i, j-1) | |

expr2 ≡ max(l(i-1, j), l(i, j-1)) | |

expr2 ≡ max(l(i-1,j-1),l(i,j)) |

**Dynamic Programming**

**GATE-CS-2009**

**Discuss it**

Question 8 Explanation:

In Longest common subsequence problem, there are two cases for X[0..i] and Y[0..j]

1) The last characters of two strings match. The length of lcs is length of lcs of X[0..i-1] and Y[0..j-1] 2) The last characters don't match. The length of lcs is max of following two lcs values a) LCS of X[0..i-1] and Y[0..j] b) LCS of X[0..i] and Y[0..j-1]

Question 9 |

Consider two strings A = "qpqrr" and B = "pqprqrp". Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let y be the number of such longest common subsequences between A and B. Then x + 10y = ___.

33 | |

23 | |

43 | |

34 |

**Dynamic Programming**

**GATE-CS-2014-(Set-2)**

**Discuss it**

Question 9 Explanation:

//The LCS is of length 4. There are 3 LCS of length 4 "qprr", "pqrr" and qpqr
A subsequence is a sequence that can be derived from another sequence by selecting zero or more elements from it, without changing the order of the remaining elements. Subsequence need not be contiguous. Since the length of given strings A = “qpqrr” and B = “pqprqrp” are very small, we don’t need to build a 5x7 matrix and solve it using dynamic programming. Rather we can solve it manually just by brute force. We will first check whether there exist a subsequence of length 5 since min_length(A,B) = 5.
Since there is no subsequence , we will now check for length 4. “qprr”, “pqrr” and “qpqr” are common in both strings.
X = 4 and Y = 3
X + 10Y = 34
This solution is contributed by

**Pranjul Ahuja**Question 10 |

Let A1, A2, A3, and A4 be four matrices of dimensions 10 x 5, 5 x 20, 20 x 10, and 10 x 5, respectively. The minimum number of scalar multiplications required to find the product A1A2A3A4 using the basic matrix multiplication method is

1500 | |

2000 | |

500 | |

100 |

**Dynamic Programming**

**GATE-CS-2016 (Set 2)**

**Discuss it**

Question 10 Explanation:

We have many ways to do matrix chain multiplication because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result of the matrix chain multiplication obtained will remain the same. Here we have four matrices A1, A2, A3, and A4, we would have:
((A1A2)A3)A4 = ((A1(A2A3))A4) = (A1A2)(A3A4) = A1((A2A3)A4) = A1(A2(A3A4)).
However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. Here, A1 is a 10 × 5 matrix, A2 is a 5 x 20 matrix, and A3 is a 20 x 10 matrix, and A4 is 10 x 5.
If we multiply two matrices A and B of order l x m and m x n respectively,then the number of scalar multiplications in the multiplication of A and B will be lxmxn.
Then,
The number of scalar multiplications required in the following sequence of matrices will be :
A1((A2A3)A4) = (5 x 20 x 10) + (5 x 10 x 5) + (10 x 5 x 5) = 1000 + 250 + 250 = 1500.
All other parenthesized options will require number of multiplications more than 1500.

There are 14 questions to complete.