wall05 by TRUONG Quoc Tuan

Juan Sebastian

jsebastian2020 — 2021-11-09 07:32:40 UTC

a. get all possible combination of items
calculate the total weight for all these combination
filter combination that exceed the weight limit
calculate the total value for all the filtered combination
sort the items by value and get the highest valued combination

b. divide each of the value by the weight to get value per kg per item.
while the total weight is below the weight limit, insert the item starting from the highest value per kg until the total weight is equal to weight limit. If weight limit will b exceeded with inserting the current item, try to insert the next highest value per kg item.

c. yes. such as when the weight limit is 8kg. optimal is 5kg+2kg = $14, but greedy algorithm gives 2kg +3kg = $11

Zhi Ying

zhiyingho20201 — 2021-11-09 08:37:29 UTC

a) Try all combinations of items that are under or equal to 15kg and return the combination of items with the optimal sum of value.

b) Find the highest value to weight ratio and return the combination of items that is closest to 15kg

Jonathan

2021-11-09 08:41:36 UTC

a.
Initialise local max that stores combination of items with max value under weight limit
1. Calculate the total value and weight of all possible combinations of items, if weight <= limit and value > local max so far, replace local max
2. Return local max

b.
1. Calculate the value per kg of each item
2. While limit > 0 and items remaining
limit -= item weight
remove item from item list

Zuyao

zuyaoliu2020 — 2021-11-09 09:06:55 UTC

a) Try every single combination and calculate the weight and value. Filter out those that are under 15kg and compare to pick the one with the highest value

b) Find the items with the greatest value as compared to weight and put them in till the threshold of 15kg is hit

c) 7kg

greedy algo --> 2kg +3kg = $11
highest value --> 2kg + 5kg = $14

Quinn

quinncheong2019 — 2021-11-09 09:15:01 UTC

a) Try out all possible combinations that the bag can hold and return the combination with the highest sum

b) rank-order the ratio of kg to value. Then proceed to take as much of each kg-value pair. If not, then proceed to take the next kg-value pair.

c) 7kg because the greedy will take E-D instead of B-E