knapsack-output.txt

#1) Optimal greedy solution for fractional knapsack

1. items are sorted by value-to-weight ratios as below
2. pick items until the current item's weight exceeds weight_capacity
3. for the last item, only pick portion (weight_capacity - weight_current) / weight(item)

https://en.wikipedia.org/wiki/Continuous_knapsack_problem

*** KNAPSACK CONTENTS (REUSED FOR THE OTHER VARIANTS) ***
weight_capacity = 166
(val  wt  val/wt)
(20   10   2.00)
(30   30   1.00)
(25   30   0.83)
(15   20   0.75)
(10   15   0.67)
(20   35   0.57)
(5    10   0.50)
(20   50   0.40)
(15   45   0.33)
(10   35   0.29)

*** GREEDY-KNAPSACK EXECUTION ***
added item with (value, weight) = (20, 10)
added item with (value, weight) = (30, 30)
added item with (value, weight) = (25, 30)
added item with (value, weight) = (15, 20)
added item with (value, weight) = (10, 15)
added item with (value, weight) = (20, 35)
added item with (value, weight) = (5, 10)
added PORTION 0.32 of item with (value, weight) = (20, 50), contributing only (6.4, 16.00)
current weight of knapsack is 166 (full)
current value of knapsack is 131.4

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#2) recursive top-down discrete 0-1 knapsack with no optimization

https://en.wikipedia.org/wiki/Knapsack_problem#Definition

final items in knapsack = [[20, 50], [25, 30], [30, 30], [15, 20], [10, 15], [5, 10], [20, 10]]
number of items taken = 7
value = 125
weight = 165
number of recursive calls = 962

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#3) dynamic programming solution to discrete 0-1 knapsack

https://en.wikipedia.org/wiki/Knapsack_problem#0/1_knapsack_problem

max value = 125
number of iterations = (weight_capacity+1) * len(items) = 1670
the number of columns in the table is weight_capacity+1, since it goes from 0 through weight_capacity
this also allows for items with weights in the range [0, weight_capacity]

tracing solution from table
took item with (value, weight) = (20, 10)
took item with (value, weight) = (5, 10)
took item with (value, weight) = (10, 15)
took item with (value, weight) = (15, 20)
took item with (value, weight) = (30, 30)
took item with (value, weight) = (25, 30)
took item with (value, weight) = (20, 50)
value = 125
weight = 165

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#4) recursive top-down 0-1 discrete knapsack with memoization

number of recursive calls = 155
final items in knapsack = [[20, 50], [25, 30], [30, 30], [15, 20], [10, 15], [5, 10], [20, 10]]
value = 125
weight = 165