TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Subset sum polynomial-time many-one reduces to partition

Statement

SUBSET SUM polynomial-time many-one reduces to PARTITION.

Facts & Assumptions

Given: A SUBSET SUM instance $(a_{1}, \dots, a_{n}; k)$ and the total $A : = a_{1} + \dots + a_{n} .$

[L1]

A polynomial-time many-one reduction is a total polynomial-time function preserving membership in both directions, by Polynomial-time many-one reductions.

[F1]

SUBSET SUM asks for a subcollection summing to $k$ , and PARTITION asks for a split into two subcollections of equal total sum, by Subset sum and partition decision problems.

Proof

technique · direct

1.1

If $k > A$ , map the instance to the one-element list $(1)$ for PARTITION. This is a correct no-instance because no subcollection of $a_{1}, \dots, a_{n}$ can exceed the total $A$ , and the list $(1)$ cannot be partitioned into two equal sums. Thus it remains to treat the case $k \leq A$ . Define $b : = 2 A - k$ and $c : = A + k$ , and output the PARTITION instance with list $(a_{1}, \dots, a_{n}, b, c)$ .

F1givenconstruct

2.1

The new total sum is $A + b + c = A + (2 A - k) + (A + k) = 4 A$ , so any equal partition must split the numbers into two subcollections each summing to $2 A$ .

step 1.1algebra

3.1

If the SUBSET SUM instance is positive, choose $I \subseteq {1, \dots, n}$ with $\sum_{i \in I} a_{i} = k$ . Then $\sum_{i \in I} a_{i} + b = k + (2 A - k) = 2 A$ . The complementary numbers among $a_{1}, \dots, a_{n}$ therefore sum to $A - k$ , and $(A - k) + c = (A - k) + (A + k) = 2 A$ . So the augmented list is a yes-instance of PARTITION.

step 1.1step 2.1construct

3.2

Conversely, suppose the augmented list has a partition into two subcollections each summing to $2 A$ . The numbers $b$ and $c$ cannot lie on the same side because $b + c = 3 A > 2 A$ , and they cannot both be absent from one side because then that side would use only the original $a_{i}$ and so would sum at most $A < 2 A$ . Hence exactly one of $b, c$ lies on each side. The side containing $b$ must then contain original numbers summing to $2 A - b = 2 A - (2 A - k) = k$ . Those numbers witness a yes-instance of SUBSET SUM.

step 1.1step 2.1algebra

4.1

The construction adds only two binary integers computed from $A$ and $k$ , so it is polynomial-time. Steps 3.1 and 3.2 establish the required iff, and [L1] therefore gives a polynomial-time many-one reduction from SUBSET SUM to PARTITION.

L1step 1.1step 3.1step 3.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Fabrizio Iozzi and Luca Trevisan, Handout NP3 (standard reference, not scraped)