Analysis of Algorithm Chapter 7 : Dynamic Programming

Analysis of Algorithm Chapter 7 : Dynamic Programming

University

9 Qs

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Analysis of Algorithm Chapter 7 : Dynamic Programming

Analysis of Algorithm Chapter 7 : Dynamic Programming

Assessment

Quiz

Mathematics

University

Practice Problem

Hard

Created by

วัชรศักดิ์ ศิริเสรีวรรณ

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9 questions

Show all answers

1.

MULTIPLE SELECT QUESTION

45 sec • 2 pts

Which are two key properties of the problem that could lead to the solution using dynamic programming?

Overlapping subproblem

Optimal substructure

Non-overlapping subproblem

Greedy subproblem

Polynomial time complexity

2.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

Which are the reasons that cause the time from the recursive control structure becomes too high ?

There are too many function calls on smaller size input

The large table are required to store data

Using the recursive stack for collecting function call value

The n at base condition is relatively low compared to the input n.

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is a key point of dynamic programming?

The storage of results in the smaller case to use for solving the bigger one.

The recursive calls of function

The division of the problem into sub-problems for easier solving

Including multiple ways of solutions to solve the problem based on inputs

4.

MULTIPLE SELECT QUESTION

45 sec • 2 pts

Which are types of dynamic programming based on storing

Tabulation

Dynamic memory

Top up

Memoization

Static memory

5.

FILL IN THE BLANK QUESTION

1 min • 1 pt

Media Image

From the tabulation derived from the dynamic programming approach of Longest Increasing Subsequence algorithm, what is the value of A + B

Note that if two elements have the equal LIS value, choose the one with the less value.

6.

MULTIPLE SELECT QUESTION

45 sec • 1 pt

Media Image

Based on this resulting tabulation, which elements are NOT in the longest increasing subsequence?

47

46

18

38

25

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Given N be the number of items, C be the capacity of the knapsack, W be the list of item weights and V be the list of item values, which is the dimension of the tabulation storing the result of this 0/1 knapsack problem.

N x (max(W)+1)

N x (C+1)

C x (len(W)*len(V))

len(W) x len(V)

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