4.5 0/1 Knapsack - Two Methods - Dynamic Programming
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- čas přidán 19. 02. 2018
- 0/1 Knapsack Problem
Dynamic Programming
Two Methods to solve the problem
Tabulation Method
Sets Method
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Remember to always write the weight in ascending order on the left of the table.
bro did you watch the video.
@@jay-rathod-01 He is not wrong
i dont think it is necessary as by the end all of the objects are considered and you would have filled the cells with maximum profit
Just for future readers: the order of the items does not matter; no need to sort them.
```
import random
m = 8
p = 1, 2, 5, 6
w = 2, 3, 4, 5
items = list(zip(p, w))
random.shuffle(items)
results = [0] * (m + 1)
for item_profit, item_weight in items:
for target_weight in range(m, item_weight - 1, -1):
results[target_weight] = max(results[target_weight],
results[target_weight - item_weight] + item_profit)
print(results)
```
@@theundescribable1643aq1
Truly sir, watch many 0/1 knapsack video, got too puzzeled, how to solve this tabulation, readings article for the same, but gets on the sport clearation here, thank u so much sir for sharing ur vast knowledge with the students like us
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Words are not enough to Thank you Sir!
After going through a lot of videos, finally i found something which has cleared all my doubts regarding this problem. Thankyou 🙏🙏
SET method for solving 0/1 Knapsack starts at 19:26. And for Tabular method watch the lecture from the beginning. Thanks
🤣 thanks
Hey can you tell me that while we eliminate an ordered pair using dominance rule...so for applying that rule our ordered pairs should be arranfed in ascending order of profits right??
yes profits should be in ascending order @@aarushitiwari4787
@@aarushitiwari4787 arranging the weights in ascending is the key here...because then we always check which prev weight to include or not
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Sir, I have never studied before the set method of 0/1 knapsack, after watching this video, i got to know about the set method which is very easy method, Thank you so much Sir
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Your method is the easiest, which explains the two method. It makes the algorithm easy to understand. Thanks a lot sir.
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best video in knapsack ever! and I have watched a bunch to understand how it works! Thank you so much!!!
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Same bro
Best approach to solve any dp:
1. Learn to develop recursive logic from scratch. (TC is worst, you might end up with TLE)
2. Memoize the same recursive logic. (optimize TC but end up using extra space. SC is probably worse now)
3. Learn the art of transforming recursive logic to tabular. (you end up saving call stacks)
4. Optimize the space complexity as well. (now you have optimized space as well)
PS: thank you sir, if you weren't there, probably many wouldn't have tried in the first place.
Love the explanation! Can you elaborate what is meant by TC, TLE and SC?
@@alexwhitewood6480 TC : Time complexity
TLE: Time limit Exceeded
SC: Space complexity
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That was really good. Much better than most of the other explanations out there. Just need to find a way to code it now.
This might help somebody:
In the last row of tabulation method, the formula basically says we are checking the max of:
1) the best profit we have till now for a bag with weight w,
versus
2) if we put current item, the profit we could have + the max profit in remaining bag capacity we calculated. So, for e.g. consider 7kg bag. 5kg we put for last row item. Remaining 2. What's the best profit we can have with that 2kg.
Hey, we do that formula just at last row right..??
@@samhita2917 Yes.
Breakdown of the formula:
⭐V: DP table which stores values of subproblems
⭐V[i, w]: Maximum profit by considering the first 'i' elements in a bag of weight 'w'
⭐V[i-1, w]: Case when the current object is not included(~0) and the bag with current weight 'w' must be filled with the maximum profit possible (stored at 'i-1')
⭐V[i-1, w-w[i]] + p[i]: Case when the current object is included(~1) and the remaining part of the bag 'w-w[i]' must be filled with the maximum profit possible (stored at 'i-1')
(~ Hence the name 0/1 knapsack problem)
My brain got damaged
Thnx
why 0/1 knapsack needs a 2d array to memoize whereas house robber needs 1d array ? both are similar problems
@@jaideepsingh7955 In house robber we calculate Profit/weight where as in 0/1 Knapsack we can't divide the objects thus we store the profit value..
Brain left
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Right bro
In w - w[i] expression, the first w refers to the capacity and the second w refers to the weights list. It would be better to rename one of them. for example declaring the capacity as c.
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This was probably the best explaination video on 0-1 knapsack problem on you tube (even better than the one uploaded by geeks for geeks ). Thank you so much for this .
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