# Complexity Theory – Can a function that produces a set of edges with a specific "aggregate graph property" be multitasking?

Let $$G (V, E)$$ to be a graph.
Let $$S$$ either all of the combinations of $$| V |$$ edges.
Let $$A$$ And $$B$$ to be two subsets of $$S$$, or:

• each subset is a collection of all the elements of $$S$$ who share a common "global property"
• the overall properties of $$A$$ And $$B$$ are different
• $$| A | gt 0$$
• $$| B | gt 0$$
• $$| A cap B | geq 0$$
• $$A not subseteq B$$
• $$B not subseteq A$$

DEFINITION: Aggregate Property, $$P$$of $$Q in S$$
Given some $$Q in S$$, we must examine each element of $$Q$$ in order to check the property $$P$$.

• Example for $$Q in A$$:[sommedesarêtesdans[sumofedgesin[sommedesarêtesdans[sumofedgesin$$Q$$] $$leq 40$$
• Example for $$Q B$$:[bordsdans[edgesin[bordsdans[edgesin$$Q$$ Hamiltonian Cycle form in $$G$$]

Let $$alpha$$ to be an algorithm that returns an element $$A$$given $$G$$.
Let $$beta$$ to be an algorithm that returns an element $$B$$given $$G$$.

Each point represents an edge of the graph $$G$$:

As a reminder, the global properties of A and B are not the same, therefore:

Key idea 1: Any algorithm that starts at level 1 and ends at level 2 and produces a set of edges with a unique relationship, A, must be an algorithm for α and NOT β.

Let $$gamma$$ or any algorithm that returns a set of edges having both the properties A and B, and which does not provide one before the other (does not go to level 3, as described below).
By definition, $$gamma$$ is an algorithm for $$alpha$$.
By definition, $$gamma$$ is an algorithm for $$beta$$.
So $$alpha$$ and $$beta$$ can be equal.
Since $$alpha$$ and $$beta$$ can be equal, A and B are no longer distinct, because they can be insured using the same algorithm, $$gamma$$.
So, A and B are actually the same property.

Key idea 2If an algorithm, γ, which starts at level 1 and ends at level 2, produces a set of edges with 2 relations, M and N, then M and N must be identical.

In order to guarantee 2 distinct properties, $$gamma$$ must go to a 3rd level:

So now, $$gamma$$ has 2 options:

• Carry out $$alpha$$then run $$beta$$ (worst case = $$| A |$$, must search for the output of $$alpha$$ who has $$B$$-Property)
• Carry out $$beta$$then run $$alpha$$ (worst case = $$| B |$$, must search for the output of $$beta$$ who has $$A$$-Property)

Given the context above, is the time complexity of any algorithm $$gamma$$ (the one that produces a set of edges at the intersection of $$A$$ and $$B$$, so this product set will have 2 aggregate properties) delimited by the top $$| A |$$ or $$| B |$$?

If so, and we demonstrate a $$A$$ and $$B$$ with exponential sizes compared to the input graph, could we use this technique to show that $$gamma$$ Should it take an exponential time in the worst case?

This model does not apply to k-SAT because a solution to k-SAT must have only one aggregate property:

• truth values ​​make a true proposition

Whereas a solution for TSP must have 2 aggregate properties:

• sum of edges $$leq L$$
• the edges form the Hamiltonian cycle in $$G$$