Subset Sum Problem Revisited

See DPV Exercise 6.22
Problem: We are given a positive integer $t$ and a sequence $A = \langle a_1, a_2, \dots, a_n \rangle$ of (not necessarily distinct) $n$ positive integers. We want to find out whether some subsequence of $A$ sums to $t$.

Dynamic Programming Solution

Define $m(i, v)$, for all $0\le i \le n$ and $0\le v\le t$, to be \[ m(i, v) = \left\{ \begin{array}{ll} \mbox{true}, & \mbox{if some subsequence of $a_1$, $a_2$, $\ldots$, $a_i$ sums to $v$}\\ \mbox{false}, & \mbox{otherwise.} \end{array} \right. \]
We seek $m(n, t)$.

We have $m(0, v)$ true if and only if $v = 0$ because the empty sequence has exactly one subsequence—itself—and the empty sequence sums to 0.
We also have $m(i, 0)$ true for all $1\le i\le n$ because the empty sequence is a subsequence of any sequence, and it sums to 0.
Now let $i, v$ be positive integers. Suppose the sequence $\langle a_1, a_2, \dots, a_i\rangle$ contains a subsequence $\langle b_1, b_2, \dots, b_j\rangle$ that sums to $v$. These two mutually exclusive events exhaust all possibilites:
- Case 1: $a_i=b_j$. Then the sequence $\langle a_1, a_2, \dots, a_{i-1}\rangle$ contains the subsequence $\langle b_1, b_2, \dots, b_{j-1}\rangle$ that sums to $v-a_i$.
- Case 2: $a_i\ne b_j$. Then the sequence $\langle a_1, a_2, \dots, a_{i-1}\rangle$ contains the subsequence $\langle b_1, b_2, \dots, b_j\rangle$ that sums to $v$.

The above reasoning gives us the following recurrence. \[ m(i, v) = \left\{ \begin{array}{ll} \mbox{true}, & \mbox{if $v = 0$} \\ \mbox{false}, & \mbox{if $i=0$, $v>0$} \\ m(i-1,v), & \mbox{if $i>0$, $a_i>v>0$} \\ m(i-1,v-a_i) \vee m(i-1,v), & \mbox{if $i>0$, $v\ge a_i>0$} \end{array} \right. \]

[Step 1] Fill in a table of $m(\cdot, \cdot)$ values, We can fill the table row-by-row or column-by-column, with both the row and column indices increasing.
[Step 2] Return the value $m(n, t)$ as answer.

The cases $v=0$ and $i=0$ in the recurrence are the so-called artificial base cases. For some problems, they help simplify the recurrence. For this particular problem, they don’t do much.
Without using artificial base cases, we define $m(i, v)$, for all $1\le i \le n$ and $1\le v\le t$, and the recurrence is \[ m(i, v) = \left\{ \begin{array}{ll} \mbox{true}, & \mbox{if $i=1$, $a_i=v$} \\ \mbox{false}, & \mbox{if $i=1$, $a_i\ne v$} \\ m(i-1,v), & \mbox{if $i>1$, $a_i>v$} \\ m(i-1,v-a_i) \vee m(i-1,v) & \mbox{if $i>1$, $a_i\le v$} \end{array} \right. \]
If the set of numbers that add up to the target $t$ is desired, we can get them from the filled-out $M$ table directly without having to use an extra table.
Instead of asking whether some subsequence of $A$ summing to $t$ exists, we can ask how many subsequences of $A$ sum to $t$ instead.
Subset Sum is an NP-complete problem. (See DPV p.242.) Dynamic programming solves it in pseudo-polynomial time.
The 0-1 Knapsack Problem is similar to the Subset Sum Problem. (See DPV p.164–167.)