Subset sum

  • Input: A multi set $S$ of numbers and a natural number $t$
  • Question: Does $S$ contain a subset $A$ such that $\sum_{x \in A} x = t$? (e.g., $\{1,1,2,3,4,5\}$, by multiset it means duplicates are allowed)

Vertex cover

  • Input: An undirected graph $G$ and a number $k$.
  • Question: Does $G$ have a vertex cover of size $k$?

Show: vertex cover $\leq_P$ subset sum

My work so far: I need to somehow transform $(G, k) \to (S, t)$

enter image description here

Above is a example graph $G$ that I constructed. It's vertex covers are $\{v_1, v_3\}$ so the size of the vertex cover is $2$, thus $k = 2$.

This is basically $S$

\begin{array}{|c|c|c|c|c|} \hline & e_1 & e_2 & e_3 & e_4 \\ \hline v_1 & 1& 0& 0& 1\\ \hline v_2 & 1& 1& 0& 0\\ \hline v_3 & 0& 1& 1& 0\\ \hline v_4 & 0& 0& 1&1\\ \hline \end{array}

if $e_i$ and $v_j$ is incident I put $1$; otherwise $0$

Now what I need is a target $t$ to satisfy this.

The matrix above can be written in a set to be $S = \{1001, 1100, 0110, 0011\}$ if I were to add up all these values probably in base 2 since its binary, it would give me a value but I think I need a consistent value $t$, so this wont work. I'm also unsure how to integrate the $k = 2$ into this either.

Could someone help me construct this?


For simplicity write all the numbers in base $4$. Fix an arbitrary ordering of the edges of the graph $G$ and let $e_i$ be the $i$-th edge ($i= 1, \dots, |E(G)|$).

For each vertex $v \in V(G)$ define an integer $n_v$ of $|E(G)| + 1$ digits as follows:

  • the $i$-th least significant digit of $n_v$ is $1$ if $e_i$ is incident to $v$ and $0$ otherwise;
  • the most significant digit of $n_v$ is $1$.

For each edge $e_i \in E(G)$ let $m_{e_i} = 4^{i-1}$ (i.e., the $i$-th least significant digit of $m_e$ is $1$ and all the other digits are $0$).

The instance of subset sum is obtained by choosing $S = \{ n_v : v \in V(G) \} \cup \{m_e : e \in E(G) \}$ and $t = k 4^{|E(G)|} + 2\sum_{i=0}^{|E(G)|-1} 4^i$. In other words, the most significant digits of $t$ encode the number $k$ in base $4$, while the $|E(G)|$ least significant digits of $t$ are all equal to $2$.

If there is a vertex cover $V' \subseteq V(G)$ of $G$ of size $k$, let $E'$ be the set of edges $(u,v) \in E(G)$ such that exactly one of $u$ and $v$ is in $V'$. Then, $A = \{ n_v : v \in V' \} \cup \{ m_e : e \in E' \}$ is a subset of $S$ that sums to $t$. Indeed: $$ \sum_{x \in A} x = \sum_{v \in V'} n_v + \sum_{e \in E'} m_v = k 4^{|E(G)|} + 2 \sum_{e_i \in E \setminus E'} 4^{i-1} + 2 \sum_{e_i \in E'} 4^{i-1} = k 4^{|E(G)|} + 2 \sum_{e_i \in E} 4^{i-1} = t. $$

If there is a subset $A$ of $S$ that sums to $t$, then $A$ contains exactly $k$ numbers from $\{ n_v : v \in V(G) \}$ (as otherwise the most significant digits of $\sum_{x \in A} x$ would not match those of $t$). Moreover, $V' = \{v_i : n_i \in A \}$ is a vertex cover for $G$.

To see this, pick any $e_i = (u,v) \in E(G)$ and notice the $i$-th least significant digit of $t$ is $2$. Since, for $j=1,\dots, |E(G)|$, there are exactly $3$ numbers in $S$ whose $j$-th least significant digit is non-zero (namely $n_{u'}$, $n_{v'}$, and $m_{e_j}$, where $e_j = (u', v')$), we have that in the summation $\sum_{x \in A} x$ no carry occurs among the $|E(G)|$ least significant digits. We conclude that $A$ contains exactly two numbers from $\{n_u, n_v, m_e\}$, implying that $u \in V'$ or $v \in V'$ (possibly both).

In your example graph, for $k=3$ you get the following (bold rows denote the elements of $A$ when $V'=\{v_1, v_3, v_4\}$): $$ \begin{array}{|c|c|c|c|c|} \hline & & e_1 & e_2 & e_3 & e_4 \\ \hline \pmb{n_{v_1}} & \pmb{1} & \pmb{1}& \pmb{0}& \pmb{0}& \pmb{1}\\ \hline n_{v_2} & 1 & 1& 1& 0& 0\\ \hline \pmb{n_{v_3}} & \pmb{1} & \pmb{0} & \pmb{1} & \pmb{1} & \pmb{0}\\ \hline \pmb{n_{v_4}} & \pmb{1} & \pmb{0} & \pmb{0} & \pmb{1} & \pmb{1}\\ \hline \end{array} $$ $$ \begin{array}{|c|c|c|c|c|} \hline & & e_1 & e_2 & e_3 & e_4 \\ \hline \pmb{m_{e_1}} & \pmb{0} & \pmb{1} & \pmb{0} & \pmb{0} & \pmb{0} \\ \hline \pmb{m_{e_2}} & \pmb{0} & \pmb{0}& \pmb{1}& \pmb{0}& \pmb{0}\\ \hline m_{e_3} & 0 & 0& 0& 1& 0\\ \hline m_{e_4} & 0 & 0& 0& 0& 1\\ \hline \end{array} $$

$$ \begin{array}{|c|c|c|c|c|} \hline \phantom{m_{e_4}}& & e_1 & e_2 & e_3 & e_4 \\ \hline t & 3 & 2& 2& 2& 2\\ \hline \end{array} $$

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  • $\begingroup$ Nice idea to use base 4! $\endgroup$ – j_random_hacker Dec 8 '19 at 12:06

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