It is well known the Subset-Sum problem is NP-complete. This can be shown using a reduction from the 3SAT problem. I am wondering: is there any other NP-Complete problem that could be reduced to the Subset-Sum problem?

# Tag: NPcomplete

## Proving a problem as NP-complete

According to this article, A problem X can be proved to be NP-complete if an already existing NP-complete problem (say Y) can be polynomial-time reduced to current problem X. The problem also needs to be NP. Now my question is “Do we also need to prove that problem X can be reduced to at least one NP problem?”. According to the NP-complete problem definition, each and every NP problem must be reducible to NP-complete problem. As problem X is NP, are we not supposed to prove that this NP problem can be reduced to other NP-complete problems? Why does this reduction have to be only one way to prove a problem is NP-complete?

## Prove/Disprove: Every two non-trivial NP-complete problems are decreasing reducible?

We say that two languages $ L_1,L_2$ are decreasing reducible if there exists a **polynomial time** reduction $ f:\Sigma^*\to\Sigma^* $ and there exists $ n\in\mathbb{N}$ such that for every $ x\in\Sigma^*$ satisfying $ |x|\ge n \implies |f(x)|\lt |x|$ .

Assuming $ P\ne NP$

Prove\Disprove: Every two NP-complete languages $ L_1,L_2$ are decreasing reducible.

I’d appreciate a hint or direction

## Reduction from np-complete problem to unknown complexity problem and viceversa

Suppose I have two problems. B NP-complete, A of unknown complexity.

Question:

if I show that B <= A I can state that A is also NP-complete because the two required conditions are satisfied: (i) A is in NP (ii) i reduced a NP -complete problem to A.

if I show that A <= B I can say that B is at least hard as A, so A is at least NP-complete but can be harder.

Are these statements correct?

## Prove NP-completeness for union of NP-complete language and language in P

Given disjoint languages $ X$ and $ Y$ , where $ X$ is NP-complete and $ Y\in P$ , how do I prove that $ X\cup Y$ is NP-complete?

My idea is to prove that $ (X\cup Y)\in NP$ and then prove that $ X\cup Y$ is NP-hard:

I can prove $ (X\cup Y)\in NP$ by saying that since both are in NP, they each have polytime verifiers. Thus we can have a polytime verifier that on any input will use these verifiers and accept if any accept, reject otherwise.

I am stuck on the part of proving NP-hardness. The idea of arbitrary languages is throwing me off; I am trying to reduce a NP-complete problem (SAT for example) to $ X\cup Y$ and using the fact that $ X$ has some method to reduce to it already but I am lost as for what to do with $ Y$ . I am thinking that given an input for SAT, I need to somehow change the input so that I can relate acceptance in $ X\cup Y$ to acceptance in SAT; same for rejection.

Any guidance would be appreciated; thanks!

## Subset with modified condition, is it still NP-complete?

So i know the conditions required for a problem to be NP-Complete is that it has to lie within NP and has to be NP-hard. The given problem I have is subset sum. However, the conditions have been change to sum ≤ M and sum ≥ M from sum = M. My initial reaction is that the two problems are no longer NP-complete since they can both be solved within polynomial time.

- Check each element and see if there exists at least one smaller than M.
- Add all positive integers and see if the sum of all elements is larger than M.

Since it isn’t NP Hard, it cannot therefore be NP-complete.

Am I thinking/approaching this correctly?

## Is it NP-complete to test if a graph contains $t$ $k$-cliques?

Let $ (G,t,k)$ – a graph with $ t$ cliques with $ k$ vertices (there are $ t$ cliques of size $ k$ in graph $ G$ ), for $ t,k > 100$ . How to prove that $ (G,t,k)$ is NP-complete?

It is obvious that it is in NP. I have tried to prove that $ k$ -CLIQUE language $ (G,k)$ is reducible to a $ (G,t,k)$ language. But I can’t get the idea, how to get $ t$ of $ k$ -CLIQUES.

## Is Graph 2-Coloring NP-Complete?

I know there is a polynomial time algorithm for 2-coloring. But should the answer be “No” or should it be “Maybe”? Since 2-coloring is in NP and we dont know if there is a reduction of any NP-complete to 2-color.

## Is this possible when it comes to the relations of P, NP, NP-Hard and NP-Complete?

I saw an image that describes the relations of P, NP, NP-Hard and NP-Complete which look like this :

https://en.wikipedia.org/wiki/NP-hardness#/media/File:P_np_np-complete_np-hard.svg

I wonder if the following is possible ? Which means, P = NP, but not all of them are in NP-Hard :

Edit : I want to add this : I’m not here to say if the original image is wrong or right, I’m just here to ask a question if my image contains a possible situation. In other words, is it correct to assume that all 3 images are possible ?

## Prove that “Finishing the degree in three years” problem is NP-Complete

I was asked in an interview the following question: We’ll define the “Finishing the degree in three years” problem in the following manner: Given a list of courses C={c1, c2,…, cn}, where every course ci∈C is given a list Ri⊆C of the precondition courses for this course, and also we are given the maximal amount of courses in each year m1, m2, m3, and we need to decide if it is possible to finish the degree in three years. Meaning, can we divide the courses for three years Y1, Y2, Y3 ⊆ C, such that the following conditions are satisfied: 1) All courses are studied within 3 years: Y1∪Y2∪Y3=C. 2) Each courses is studied only after all of its precondition courses have been studied. meaning: for any ci∈Y1: Ri⊆Ø, for any ci∈Y2: Ri⊆Y1, for any ci∈Y3: Ri⊆Y1∪Y2. 3) The number of courses studied in each year is not higher than maximum allowed. Meaning: |Yi|<=mi for each i∈{1,2,3}. Prove that the Finishing the degree in three years problem is NP-Complete.

I was told that the reduction should be from the clique problem but it has been a week since I’ve started thinking about it and I couldn’t prove it. If someone can please post a thorough solution I would appreciate it a lot. Thanks for your help!