Is this explanation confusing NP-hard and NP-complete?

My notes on P vs NP say the following:

Every problem x in the NP-hard class has the following properties:
– There is no known polynomial-time algorithm for x.
– The only known algorithms take exponential time.
– If you could solve x in polynomial-time, then you could solve them all in polynomial-time.

The last point sounds like it’s confusing NP-hard with NP-complete. I would appreciate it if someone would please clarify this.

reductions – Is this proof of the PARTITION problem being NP-hard correct?

This is the PARTITION problem:

Given a multiset S of positive integers, decide if it can be partitioned into two equal-sum subsets.

This is the SUBSET SUM problem:

Given a multiset S of integers and an integer T, decide if any subset of S sums to T.

The positive variant of SUBSET SUM is NP-complete.


SUBSET SUM can be reduced to PARTITION as follows:

Define S' = S + {c}, where c = 2T - sum(S). S' can be partitioned into two equal-sum subsets iff there is a subset in S summing to T.


Proof:

Partition S into A and B:

S = A + B

If there exists a subset in S summing to T, then S can be partitioned so either sum(A) = T or sum(B) = T. Suppose sum(A) = T. Let S' = A + B' where B' = B + {c}.

sum(B') = sum(S) - sum(A) + c
sum(B') = sum(S) - sum(A) + 2T - sum(S)
sum(B') = -sum(A) + 2T

By assumption, T = sum(A) therefore sum(B') = sum(A). This means S' can be partitioned into two equal-sum subsets.

If there does not exist a subset in S summing to T, then there cannot exist a c such that S' can be partitioned into two equal-sum subsets.

Suppose for contradiction c can exist.

Adding c to B creates S' = A + B'.

What value of T produces c such that sum(A) = sum(B') = sum(B) + c?

sum(A) = sum(B) + c
sum(A) = sum(B) + 2T - sum(S)
sum(A) = sum(B) + 2T - (sum(A) + sum(B))
sum(A) = sum(B) + 2T - sum(A) - sum(B)
2sum(A) = 2T
sum(A) = T

But A already sums to sum(A), contradicting the assumption that there is no subset of S summing to T.

Adding c to A is symmetric to adding c to B.


This is my attempt to prove PARTITION is NP-hard. Is this correct? Did I miss something?

Why this problem is NP-Hard?

I’m asking about the question described here:
Knapsack Problem with exact required item number constraint

Can’t we iterate over $binom{n}{L}$ options (which is polynomial), and for each option check if the constraints are met?

complexity theory – Is this variation of Max-Coverage NP-hard?

Setup

An instance of Max-Coverage is typically defined by a collection of $n$ sets $S = {s_1, s_2, dots, s_n}$, and a budget $k$, where the objective is to select a subset $Usubset S$ such that
$$big|Ubig| leq k,~text{ and }~big|cup_{sin U}sbig|~text{ is maximized}.$$

The variation I am interested in is as follows. Instead of being given a collection of sets, we are given a collection of $n$ pairs of sets, $S = big{p_1 ={s_{1, 0}, s_{1, 1}}, p_2={s_{2, 0}, s_{2, 1}}, dots, p_n={s_{n, 0}, s_{n, 1}}big}$. Further, instead of selecting $k$ sets, we now have to select one set from each pair of sets say $U = { s_{1, i_1}, s_{2, i_2}, dots s_{n, i_n}}$ s.t. $i_jin{0, 1}$ and
$$big|cup_{sin U}sbig|~text{ is maximized}.$$

Questions

1.] Is it NP-hard to optimally select one set from each pair such their union is maximized?

2.] Let $A$ be the universe of possible set elements and for each $i leq n$, we have $s_{i, 0} cup s_{i, 1} = A$ and $s_{i, 0} cap s_{i, 1} = emptyset$.

decision problem – Is it NP-hard to check whether for a $k$ there exist both a Cut and a Bisection of value $k$?

Input: An undirected, unweighted graph $G=(V,E)$.

A cut is defined as a partition $V=Adotcup B$.

A bisection is defined as a partition $V=Adotcup B$ with $|A|=|B|$ if $|V|$ is even (or $|A|= |B|+1$ if $|V|$ is odd).

We define the value of a cut/bisection $V=Adotcup B$ as $E(A,B)$, i.e. as the number of edges between the partitions.

Question:

Is it NP-hard to solve the following problem:
Given an integer $k$, do there exist both a Cut and a Bisection of value $k$?

The problem is in NP, because given a cut and a bisection, we can efficiently check whether both have value $k$.

I’m also wondering whether there are somewhat general techniques that help one decide the NP-hardness of a problem which is more or less two NP-hard problems slapped together by an equality.


Literature:

NP-completeness of Max-Cut: DOI:10.1016/0304-3975(76)90059-1

NP-completeness of Vertex Bisection: DOI=10.1.1.154.5438

optimization – Is the problem “find the sequence of $N$ numbers between 1 and $D$ with least cost”, NP-hard?

Consider sequences $p=(p_1,dots,p_N)$ (the order matters) of length $N$, where $p_iin{1,dots,D}$ for fixed $D$. Moreover, consider a cost function $c:{1,dots,D}^Ntomathbb{R}$ which comply $c(q)neq c(p), forall p,qin{1,dots,D}^N$. The problem is to find $pin{1,dots,D}^N$ with least cost $c(p)$. I want to show that this problem is NP-hard.

In my opinion, since all costs are different, there is no other way but to check all possible ($D^N$) combinations to find the optimum. Should this be enough to conclude that this problem is NP-hard? If so, what is the “formal argument” under this line of reasoning?

Alternatively, is there a standard reference which state that a problem like this is NP-hard? Which other NP-hard problem can be reduced to this one?

optimization – Is this variation of the traveling salesman problem NP-hard

Consider the following setting. You have $n$ cities, and there is a cost to travel from a city $i$ to a city $j$ given by $c_{ij}>0$ where $c_{ij}neq c_{ji}$. Moreover, if you are traveling to city $i$, you are allowed to stay exactly $d_i>0$ days. The problem is:

Find a travel schedule, this is: a number $m$ and a sequence of different cities $i_1,dots,i_min{1,dots,n}$, such that the cost $sum_{k=1}^{m-1} c_{i_k,i_{k+1}}$ is minimal and that the whole trip lasts at least $L$, this is $sum_{k=1}^m d_{i_k}geq L$.

There are many similarities between this problem and the traveling salesman one. The key differences are the non symmetric costs $c_{ij}neq c_{ji}$ and that instead of having to travel to all cities once, one just needs to travel to enough cities to make the trip last enough.

I understand that to show that this problem is NP-hard, I need to show that another NP-hard problem can be reduced to this one in polynomial time. I suspect that the original traveling salesman problem may work to do so through some clever trick I haven’t found yet. Moreover, since there is plenty of literature on the traveling salesman problem, I suspect a similar problem to this may already be shown to be NP-hard.

My question is: Do you see how to show this problem to be NP-hard? Or, do you recognize this problem to be something standard in some piece of literature you can point me out?

complexity theory – Is there any NP-hard problem which was proven to be solved in polynomial time or at least close to polynomial time?

I know this could be a strange question. But was there any algorithm ever found to compute an NP-problem, whether it be hard or complete, in polynomial time. I know this dabbles into the “does P=NP” problem, but I was wondering whether there were any particular instances when people were close to it. I’ve also heard about the factoring number problem being solved by a quantum computer by Shor’s algorithm. Would this count as an algorithm which solved an NP-problem in polynomial time please ? Please do ask if you want me to clarify any parts in my questions. Thanks in advance.

algorithms – show the problem of find two subsets such that the difference of them of two sets is smaller than a value, is NP-Hard

As input, given two finite sets of integers X = {x1,…,xm},Y = {y1,…,yn} ⊆ Z, and a non-negative integer v ≥ 0. The goal is to decide if there are non-empty subsets S ⊆ [m] and T ⊆ [n] such thatenter image description here

How to show this problem is NP-Complete? I’m quite confused.
What I got so far is to reduct from subset sum problems, since the form is set to less than v. So I need to have 2v+1 subset sum problems to verify

algorithms – Is the following problem NP-hard? (or have yous seen it before?)

Let’s formulate the decision problem form of this problem, which I’ll call Tree Scheduling (TS):

Given a number $k$, and a rooted tree with

  • tasks $t_1, dots, t_n$ for vertices, each having some integer duration $l_i$ and requiring some resource $r_i$ from a set $S$ of resources, and
  • arcs representing dependencies between tasks,

is there a schedule that satisfies all dependencies, avoids scheduling two jobs that use the same resource at overlapping times, and completes within $k$ time units?

This problem is NP-complete even with just two resources, by reduction from Partition (PART).

In PART, we are given a multiset of $m$ numbers $a_1, dots, a_m$ having $sum_i a_i = T$, and the task is to determine whether we can partition them into two multisets, each having sum $T/2$. Given an instance of PART, we construct an instance of TS with:

  1. A root task $t_r$ using resource 1, of duration 1;
  2. For each $1 le i le m$, a task $t_i$ using resource 1, having duration $l_i = a_i$ and on which $t_r$ depends;
  3. A task $t_{gap}$ using resource 2, of duration $T/2$ and on which $t_r$ depends;
  4. A task $t_{sep}$ using resource 1, of duration 1 and on which $t_{gap}$ depends;
  5. $k=T+2$.

The idea is that an optimal solution to an instance consisting of just $t_r$, $t_{gap}$ and $t_{sep}$ has a “gap” of length $T/2$ in the usage of resource 1 between running $t_{sep}$ and $t_r$. After adding the tasks $t_1, dots, t_m$, we can complete all tasks in $T+2$ time iff we are able to fill up this gap with exactly $T/2$ time units’ worth of tasks.