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Knapsack Variants and State Design

What This Concept Is​

In 0/1 knapsack, each indexed item is available once. Positive integer weights and nonnegative integer capacity make a finite capacity table possible; values may be signed. The lab maximizes value at at most capacity, with the empty selection allowed. Duplicate records still represent different items.

Let D(i,c) be the best value using the first i items and capacity c. D(0,c)=0. For item i-1 with weight w and value v, either skip it or, if w<=c, take it and combine with D(i-1,c-w). Both branches use the previous item layer. The answer is D(n,W), with reconstruction by comparing adjacent layers.

Why It Matters Here​

Small wording changes alter the state and initialization. "Fill exactly W", "at least one item", "unlimited copies", "number of ways", and "maximize value" are separate contracts. A table initialized to zero everywhere silently claims all capacities are attainable, which is wrong for exact-use feasibility.

Concrete Examples​

For [(2,3),(3,4),(4,5)], W=5, prefix rows at capacities 0..5 are:

no items: 0 0 0 0 0 0
item 0:   0 0 3 3 3 3
item 1:   0 0 3 4 4 7
item 2:   0 0 3 4 5 7

Value 7 is attained by indices 0 and 1. The final row is nondecreasing in capacity because unused capacity is allowed. That property need not hold for exact fill.

Loop direction. With one (2,3) item and W=4, descending capacity updates give 3; ascending updates give 6 by reusing the current item. The latter is valid for an unbounded variant, not for the 0/1 contract. Full layers avoid this ambiguity and preserve simple reconstruction.

Exact fill. Set D(0,0)=0, other base capacities to unreachable. Combine only reachable predecessors. For a sole weight-2 item, exact capacity 3 is impossible even though at-most capacity 3 admits the item. Do not use an arbitrary finite negative sentinel when valid values can be arbitrarily negative.

Bounded copies. Binary grouping converts a count k into groups of sizes 1,2,4,... and a remainder whose total is k. Each group becomes one 0/1 item. For k=6, groups 1,2,3 represent every count 0..6. This preserves achievable quantities for optimization, but multiple group subsets can encode the same quantity; naive counting after grouping can overcount original objects.

Common Confusion / Misconceptions​

O(nW) arithmetic work is pseudo-polynomial: W is a numeric value encoded in only O(log(W+1)) bits. It is not a polynomial-time solution in general input bit length. The full witness-producing table uses O((n+1)(W+1)) cells. Rolling value-only storage does not automatically supply safe parents. Zero-weight items are excluded by the lab contract; unlimited positive-value zero-weight items would make an unbounded variant's objective unbounded.

How To Use It​

List item multiplicity, weight domain, capacity meaning and empty-choice policy. Prove the take/skip partition by item identity. Estimate both the capacity table and output before allocating. Validate small cases by subset enumeration and check weight, distinct IDs and value independently. For huge W, consider an alternative state, a Pareto frontier or meet-in-the-middle only with its own analysis.

Check Yourself​

Which layer must a 0/1 transition read? Why can negative-value items be skipped in this contract? Why is binary grouping dangerous for counting distinct selections?

Mini Drill or Application​

Items (2,3),(3,4), W=4: compare ratio-first greedy, 0/1 optimum and unbounded optimum. Feedback: greedy takes the first for 3; 0/1 takes the second for 4; unbounded takes two first items for 6. Three algorithms are answering two contracts.

Read This Only If Stuck​

CLRS: 0/1 versus fractional knapsack and DP design. Use Skiena's index for the knapsack problem discussion. Derive the actual lab contract yourself.