Counting subsets
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Antti H S Laaksonen
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luku10.tex
123
luku10.tex
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@ -427,7 +427,7 @@ element can be any element.
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The dynamic programming values can be stored as follows:
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\begin{lstlisting}
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long long d[1<<n][n];
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int d[1<<n][n];
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\end{lstlisting}
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First, $f(\{k\},k)=1$ for all values of $k$:
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@ -459,93 +459,86 @@ Finally, the number of solutions can be
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calculated as follows:
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\begin{lstlisting}
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long long s = 0;
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int s = 0;
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for (int i = 0; i < n; i++) {
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s += d[(1<<n)-1][i];
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}
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\end{lstlisting}
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\subsubsection{Sums of subsets}
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\subsubsection{Counting subsets}
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As the last example, we consider the following problem:
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Each subset $x$
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of $\{0,1,\ldots,n-1\}$
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is assigned a value $c(x)$,
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and our task is to calculate for
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each subset $x$ the sum
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\[s(x)=\sum_{y \subset x} c(y).\]
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Using bit operations, the corresponding sum is
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\[s(x)=\sum_{y \& x = y} c(y).\]
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For example, the functions could be
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as follows when $n=3$:
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\begin{center}
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\begin{tabular}{rrr}
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$x$ & $c(x)$ & $s(x)$ \\
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\hline
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000 & 2 & 2 \\
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001 & 0 & 2 \\
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010 & 1 & 3 \\
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011 & 3 & 6 \\
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100 & 0 & 2 \\
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101 & 4 & 6 \\
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110 & 2 & 5 \\
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111 & 0 & 12 \\
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\end{tabular}
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\end{center}
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For example, $s(110)=c(000)+c(010)+c(100)+c(110)=5$.
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Our last problem in this chapter is as follows:
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We are given a collection $C$ of $m$ sets,
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and our task is to determine for each set
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the number of sets that are its subsets.
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For example, consider the following collection:
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\[C = \{\{0\}, \{0,2\}, \{1,4\}, \{0,1,4\}, \{1,4,5\}\}\]
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For any set $x$ in $C$,
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let $f(x)$ denote the number of sets (including $x$) in $C$
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that are subsets of $x$.
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For example, $f(\{0,1,4\})=3$, because the
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sets $\{0\}$, $\{1,4\}$ and $\{0,1,4\}$ are
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subsets of $\{0,1,4\}$.
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Using this notation, our task is to calculate the value of $f(x)$
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for every set $x$ in the collection.
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The problem can be solved in $O(2^n n)$ time
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by defining a function $f(x,k)$ that calculates
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the sum of $c(y)$ values such that $x$ can be
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turned into $y$ by changing zero or more one bits
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at positions $0,1,\ldots,k$ to zero bits.
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Using this function, the solution to the
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problem is $s(x)=f(x,n-1)$.
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We will assume that each set is
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a subset of $\{0,1,\ldots,n-1\}$.
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Thus, the collection can contain at most
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$2^n$ sets.
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A straightforward way to solve the problem
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is to go through all pairs of sets in the collection.
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However, a more efficient solution is possible
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using dynamic programming.
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Let $c(x,k)$ denote the number of sets in
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$C$ that equal to a set $x$
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if we are allowed to remove any subset of
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$\{0,1,\ldots,k\}$ from $x$.
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For example, in the above collection,
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$c(\{0,1,4\},1)=2$,
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where the corresponding sets are
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$\{1,4\}$ and $\{0,1,4\}$.
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It turns out that we can calculate all
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values of $c(x,k)$ in $O(2^n n)$ time.
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This solves our problem, because
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\[f(x)=c(x,n-1).\]
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The base cases for the function are:
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\begin{equation*}
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f(x,0) = \begin{cases}
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c(x) & \textrm{if bit 0 of $x$ is 0}\\
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c(x)+c(x \XOR 1) & \textrm{if bit 0 of $x$ is 1}\\
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c(x,-1) = \begin{cases}
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0 & \textrm{if $x$ does not appear in $C$}\\
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1 & \textrm{if $x$ appears in $C$}\\
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\end{cases}
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\end{equation*}
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For larger values of $k$, the following recursion holds:
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\begin{equation*}
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f(x,k) = \begin{cases}
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f(x,k-1) & \textrm{if bit $k$ of $x$ is 0}\\
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f(x,k-1)+f(x \XOR (1 < < k),k-1) & \textrm{if bit $k$ of $x$ is 1}\\
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c(x,k) = \begin{cases}
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c(x,k-1) & \textrm{if $k \notin x$}\\
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c(x,k-1)+c(x \setminus \{k\},k-1) & \textrm{if $k \in x$}\\
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\end{cases}
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\end{equation*}
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Thus, we can calculate the values of the function
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as follows using dynamic programming.
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The code assumes that the array \texttt{c}
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contains the values for $c$,
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and it constructs an array \texttt{s}
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that contains the values for $s$.
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We can conveniently implement the algorithm by representing
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the sets using bits.
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Assume that there is an array
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\begin{lstlisting}
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for (int x = 0; x < (1<<n); x++) {
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f[x][0] = c[x];
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if (x&1) f[x][0] += c[x^1];
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}
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for (int k = 1; k < n; k++) {
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for (int x = 0; x < (1<<n); x++) {
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f[x][k] = f[x][k-1];
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if (b&(1<<k)) f[x][k] += f[x^(1<<k)][k-1];
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}
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if (k == n-1) s[x] = f[x][k];
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}
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int d[1<<n];
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\end{lstlisting}
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that is initialized so that $d[x]=1$ if $x$ belongs to $C$
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and otherwise $d[x]=0$.
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We can now implement the algorithm as follows:
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Actually, a much shorter implementation is possible,
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because we can calculate the results directly
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into the array \texttt{s}:
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\begin{lstlisting}
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for (int x = 0; x < (1<<n); x++) s[x] = c[x];
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for (int k = 0; k < n; k++) {
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for (int x = 0; x < (1<<n); x++) {
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if (x&(1<<k)) s[x] += s[x^(1<<k)];
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for (int b = 0; b < (1<<n); b++) {
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if (b&(1<<k)) d[b] += d[b^(1<<k)];
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}
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}
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\end{lstlisting}
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The above code is based on the recursive definition
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of $c$. As a special trick, the function only uses
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the array $d$ to calculate all values of the function.
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Finally, for each set $x$ in $C$, $f(x)=d[x]$.
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