L-5.3: 0/1 Knapsack Problem |Dynamic Programming |Recursive Equation |Recursion Tree Time Complexity
Gate Smashers video on recursive equation of 0-1 Knapsack
Importance of knowing the recursive equation for competitive exams, college, university exams, and interviews
Brute force method of considering or not considering each object
Total time complexity: 2^n
Dynamic programming approach to reduce time complexity
n: number of objects, m: total weight of knapsack
Pn: profit of nth object, Wn: weight of nth object
Two cases: consider nth object or not consider nth object
Remaining objects: n-1
Remaining weight: m-Wn
Profit: Pn
Remaining objects: n-1
Remaining weight: m
Profit: 0
Termination condition: if weight of an object is more than total knapsack weight, remove it
Termination condition: if no elements or remaining weight is 0, output is 0
Next, discuss example and recursion tree
Also discuss time complexity Chapter 1: Introduction to Recursive Equations and Recurrence Relations
Recursive equation and recurrence relation explained
Request for confirmation of understanding
Chapter 2: Explaining the Recursion Tree with an Example
Example of 0-1 knapsack problem with 4 objects and weight 4
Objects and their weights explained
Consideration and non-consideration of objects in the knapsack
Recursive calls and their effects on the number of objects and weight of the knapsack
Chapter 3: Termination Condition and Optimal Result
Termination condition when either the number of objects or the weight of the knapsack becomes 0
Recursive calls returning 0 at the termination condition
Adding the maximum result obtained from the recursive calls to get the optimal result
Chapter 4: Time Complexity Analysis
Brute force method and its time complexity of 2^n
Introduction to dynamic programming and its purpose of avoiding repeated calculations
Identification of repeated subproblems in the example
Calculation of the total number of unique and repeated problems
Reduction of time complexity from 2^n to n times m (or n times w) through dynamic programming
Data is stored in a table
The size of the table needs to be determined
The table is used to store the value of every subproblem
Storing the value of subproblems allows for direct use without re-evaluation
The size of the table is determined by (n + 1) times (m + 1)
The additional 1 is added because the table includes the 0th level
The size of the table is (n + 1) times (m + 1)
The reason for adding 1 is to account for the 0th level
The table starts from (n, m) and goes down to (0, 0)
An example table size is 5x5, which is 25 in total
The table can be made for any given values of n and m
In this case, a table of size 5x5 is used to store all the values
The time complexity can be expressed as O(n * m)
The time complexity is determined by the size of the table
In this case, the time complexity is O(5 * 5) or O(25)
The space required is determined by the size of the table
In normal cases, the space complexity is 2^n
However, with dynamic programming, the space complexity is reduced to (n + 1) * (m + 1)
Dynamic programming saves time
By storing values in a table, dynamic programming avoids re-evaluating subproblems
This leads to improved time efficiency
Thank you.
Gate Smashers video on recursive equation of 0-1 Knapsack
Importance of knowing the recursive equation for competitive exams, college, university exams, and interviews
Brute force method of considering or not considering each object
Total time complexity: 2^n
Dynamic programming approach to reduce time complexity
n: number of objects, m: total weight of knapsack
Pn: profit of nth object, Wn: weight of nth object
Two cases: consider nth object or not consider nth object
Remaining objects: n-1
Remaining weight: m-Wn
Profit: Pn
Remaining objects: n-1
Remaining weight: m
Profit: 0
Termination condition: if weight of an object is more than total knapsack weight, remove it
Termination condition: if no elements or remaining weight is 0, output is 0
Next, discuss example and recursion tree
Also discuss time complexity Chapter 1: Introduction to Recursive Equations and Recurrence Relations
Recursive equation and recurrence relation explained
Request for confirmation of understanding
Chapter 2: Explaining the Recursion Tree with an Example
Example of 0-1 knapsack problem with 4 objects and weight 4
Objects and their weights explained
Consideration and non-consideration of objects in the knapsack
Recursive calls and their effects on the number of objects and weight of the knapsack
Chapter 3: Termination Condition and Optimal Result
Termination condition when either the number of objects or the weight of the knapsack becomes 0
Recursive calls returning 0 at the termination condition
Adding the maximum result obtained from the recursive calls to get the optimal result
Chapter 4: Time Complexity Analysis
Brute force method and its time complexity of 2^n
Introduction to dynamic programming and its purpose of avoiding repeated calculations
Identification of repeated subproblems in the example
Calculation of the total number of unique and repeated problems
Reduction of time complexity from 2^n to n times m (or n times w) through dynamic programming
Data is stored in a table
The size of the table needs to be determined
The table is used to store the value of every subproblem
Storing the value of subproblems allows for direct use without re-evaluation
The size of the table is determined by (n + 1) times (m + 1)
The additional 1 is added because the table includes the 0th level
The size of the table is (n + 1) times (m + 1)
The reason for adding 1 is to account for the 0th level
The table starts from (n, m) and goes down to (0, 0)
An example table size is 5x5, which is 25 in total
The table can be made for any given values of n and m
In this case, a table of size 5x5 is used to store all the values
The time complexity can be expressed as O(n * m)
The time complexity is determined by the size of the table
In this case, the time complexity is O(5 * 5) or O(25)
The space required is determined by the size of the table
In normal cases, the space complexity is 2^n
However, with dynamic programming, the space complexity is reduced to (n + 1) * (m + 1)
Dynamic programming saves time
By storing values in a table, dynamic programming avoids re-evaluating subproblems
This leads to improved time efficiency
Thank you.