REP-SET Example

\[\begin{align} C=\{ & \{1,11,14,18,20,23\}, \{5,12,15,18,21,24\}, \{2, 11, 16, 17\}, \\ & \{4, 5, 10,19,24,25\}, \{2,7,17,18,25\}, \{8,11,12,14,16\}, \\ & \{4,6,10,11,22,25\}, \{3,7,12,15,16,19\}, \{1, 6, 9, 12, 14, 19, 21, 24\} \} \end{align}\]

Minimal rep sets: \(\{11,24,7\}, \{11, 12, 25\}\), etc.

REP-SET (from MinVertexCover)

Suppose you have a polynomial-time solver for REP-SET, and you are given an instance of MinVertexCover to solve: a graph \(G\).

• Number the vertices \(1,2,\ldots, E\).
• Represent each edge by \(\{i,j\}\) if it connects vertex \(i\) to vertex \(j\). Let \(C\) be the set of these sets. \(O(E)\) or \(O(V^2)\), depending on graph structure.
  • If \(M\) is a vertex cover of \(G\), then each edge is touched by a vertex in \(M\), so \(M\) is a rep set for \(C\).
  • If \(M\) is a rep set for \(C\), then every set \(\{i,j\}\) contains at least one element of \(M\), so the vertices in \(M\) touch every edge, so \(M\) is a vertex cover of \(G\).
  • So \(M\) is a minimal vertex cover iff \(M\) is a minimal rep set.
• You just solved MinVertexCover in polynomial time!
  • Therefore, REP-SET is NP-hard.

MinVertexCover vs. VertexCover

MinVertexCover: Given a graph \(G\), what is size of the smallest subset of vertices you can select so that every edge is touched?

VertexCover: Given a graph \(G\) and an integer \(k\), is there a vertex cover in \(G\) consisting of \(k\) vertices?

• The complexity of these problems is polynomially equivalent.
  • Clearly, if you can do MinVertexCover, you can answer VertexCover.
  • Conversely, if you can do VertexCover, you can just repeat the process for \(k = 1,2,\ldots n\) to solve MinVertexCover.
    • This multiplies the running time by \(n\) (at most) so if you had a polynomial-time solver for VertexCover, you get a polynomial-time solver for MinVertexCover.

Other Decision Problem Versions

• MaxIndSet: Given an undirected graph \(G\), what is largest set of vertices in \(G\) with no edges between any of them?
  • IndSet: Given an undirected graph \(G\) and an integer \(k\), is there a set of \(k\) vertices in \(G\) with no edges between any of them?
• MaxClique: What is the size of the largest complete subgraph of a given graph?
  • Clique: Given a graph \(G\) and an integer \(k\), does \(G\) have a complete subgraph with \(k\) vertices?

SubsetSum (from VertexCover)

SubsetSum: Given a set \(X\) of positive integers and a target integer \(T\), is there some subset of \(X\) whose elements add up to \(T\)?

• Suppose we have a polynomial-time solver for SubsetSum.
  • Given a graph \(G\) and and integer \(k\):
    • Turn the graph into a set \(X\) of integers and choose a clever target value \(T\). (tricky part)
    • Show that \(G\) has a vertex cover of size \(k\) if and only if \(X\) has a subset that adds up to \(T\). (hard part)
    • Use our SubsetSum solver on this set to find a subset summing to \(k\).
  • We just solved VertexCover in polynomial time!
    • So SubsetSum is NP-hard.

Recall: Base 4 Integers

1. Write \(311_4\) in base 10.

2. Write your age in base 4.

3. Write \(\displaystyle{\sum_{i \in \{0,1,2,3,4\}} i \cdot 4^i}\) in base 4.

4. Write \(322222_4\) in base 10 using Sigma notation where convenient.

5. Compute the sum of \(111000_4\), \(101101_4\), \(100011_4\), \(110110_4\) in base 4.

Polynomial-time reduction

Given a graph \(G\) and and integer \(k\), make a set \(X\) of numbers:

• Represent each edge with the numbers \(4^0, 4^1, 4^2, \ldots, 4^{E-1}\).
  • \(00\cdots 01_4,\, 00\cdots 10_4,\, \ldots,\, 10\cdots 00_4\)
    
  • \(O(E^2)\) time, because each expression has length \(O(E)\).
• Represent each vertex with the number \(\displaystyle{a_v = 4^E + \sum_{i\in \Delta(v)} 4^i}\).
  • \(1\delta_{E-1}\delta_{E-2}\cdots\delta_1{\delta_0}\) in base 4, where the \(\delta\)’s indicate which edges touch the vertex.
  • \(O(VE^2)\) time (conservatively)

This is a polynomial-time reduction (in \(V\) and \(E\), so also in \(n\), the input size).

Target value \(T\)

Set \(T = \displaystyle{k\cdot 4^E + \sum_{i=0}^{E-1} 2 \cdot 4^i}\).

• \(k222\cdots 22_4\), where the leading \(k\) may involve several digits.
  • Mixing base ten and base 4:

\[ T = k\cdot 4^E + 222\cdots 22_4 \]

There are two directions to prove:

• If \(G\) has a vertex cover of size \(k\), then \(X\) has a subset that sums to \(T\).
• If \(X\) has a subset that sums to \(T\), then \(G\) has a vertex cover of size \(k\).

Vertex cover \(\Rightarrow\) Subset sum

Suppose \(G\) has a vertex cover of size \(k\). Make a subset of \(X\):

• Include all the numbers for all the vertices in the cover.
• For each edge that touches only one vertex in the cover, include it’s number.

• Each place value \(E-1, \ldots, 1, 0\) will now have the digit \(2\), because each edge touches 1 or 2 vertices.
• Each vertex contributes \(4^E\) also, i.e., a \(1\) in place value \(E\). So the total sum will be \(T = k\cdot 4^E + 222\cdots 22_4\).

Subset sum \(\Rightarrow\) Vertex cover

Suppose some subset of \(X\) sums to \(T\).

• The subset must include exactly \(k\) vertex numbers, because there can’t be any carries in the lower \(E\) digits.
  • So the vertex numbers represent \(k\) vertices.
• Each edge number can only contribute \(1\) to any place value.
  • So each place value must have at least one \(1\) contributed by a vertex number (maybe two).
  • So each edge touches at least one vertex (maybe two).
• Therefore the represented vertices form a vertex cover of \(G\).

Recall: Dynamic Programming SubsetSum

• We have a set \(X\) of positive integers (so no duplicates).
• Given a target value \(T\), is there some subset of \(X\) that adds up to \(T\)?

Subset Sum: Recursive Specification

• For any \(x\in X\), the answer is TRUE for target \(T\) if
  • The answer is TRUE for \(X \setminus \{x\}\) and target \(T\), or
  • The answer is TRUE for \(X \setminus \{x\}\) and target \(T-x\).
• In other words, the cases are:
  • without: There is a subset not containing \(x\) that works.
  • with: There is a subset containing \(x\) that works.
• Base Cases:
  • If \(T = 0\), the answer is TRUE. (\(\emptyset\) works!)
  • If \(T < 0\), the answer is FALSE. (everything’s positive)
  • If \(X = \emptyset\), the answer if FALSE (unless \(T=0\), of course.)

Memoized Dynamic Programming Version

// X[1..n] is an array of positive integers (n is constant)
// SS[1..(n+1), 0..T] is a boolean array

FastSubsetSum(T):
    Initialize SS[*, 0] to TRUE
    Initialize SS[n + 1, *] to FALSE for * > 0
    Initialize SS[* , *] to FALSE for * < 0    // hmmm...
    for i <- n downto 1
        for t <- 1 to T
            with <- SS[i + 1, t - X[i]]
            wout <- SS[i + 1, t]
            SS[i, t] <- (with OR wout)
    return S[1, T]

• Space and Time are both \(O(nT)\).
• However, \(O(nT)\) is bad when \(T\) is large.
  • If \(m\) is the number of bits in \(T\), then \(T \in O(2^m)\), so the space and time become \(O(n\cdot 2^m)\).
  • So in terms of the size of the input, this algorithm is still exponential.