752 lines
22 KiB
TeX
752 lines
22 KiB
TeX
\chapter{Complete search}
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\key{Complete search}
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is a general method that can be used
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to solve almost any algorithm problem.
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The idea is to generate all possible
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solutions to the problem using brute force,
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and then select the best solution or count the
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number of solutions, depending on the problem.
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Complete search is a good technique
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if there is enough time to go through all the solutions,
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because the search is usually easy to implement
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and it always gives the correct answer.
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If complete search is too slow,
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other techniques, such as greedy algorithms or
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dynamic programming, may be needed.
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\section{Generating subsets}
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\index{subset}
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We first consider the problem of generating
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all subsets of a set of $n$ elements.
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There are two common methods for this:
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we can either implement a recursive search
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or use bit operations of integers.
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\subsubsection{Method 1}
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An elegant way to go through all subsets
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of a set is to use recursion.
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The following function
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generates the subsets of the set
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$\{1,2,\ldots,n\}$.
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The function maintains a vector
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that will contain the elements of each subset.
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The search begins when the function is called
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with parameter 1.
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\begin{lstlisting}
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void gen(int k) {
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if (k == n+1) {
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// process subset v
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} else {
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gen(k+1);
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v.push_back(k);
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gen(k+1);
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v.pop_back();
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}
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}
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\end{lstlisting}
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The parameter $k$ is the next
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candidate to be included in the subset.
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The function considers two cases that both
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generate a recursive call:
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either $k$ is included or not included in the subset.
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Finally, when $k=n+1$, all elements have been processed
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and one subset has been generated.
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The following tree illustrates how the function is
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called when $n=3$.
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We can always choose either the left branch
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($k$ is not included in the subset) or the right branch
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($k$ is included in the subset).
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\begin{center}
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\begin{tikzpicture}[scale=.45]
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\begin{scope}
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\small
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\node at (0,0) {$\texttt{gen}(1)$};
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\node at (-8,-4) {$\texttt{gen}(2)$};
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\node at (8,-4) {$\texttt{gen}(2)$};
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\path[draw,thick,->] (0,0-0.5) -- (-8,-4+0.5);
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\path[draw,thick,->] (0,0-0.5) -- (8,-4+0.5);
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\node at (-12,-8) {$\texttt{gen}(3)$};
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\node at (-4,-8) {$\texttt{gen}(3)$};
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\node at (4,-8) {$\texttt{gen}(3)$};
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\node at (12,-8) {$\texttt{gen}(3)$};
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\path[draw,thick,->] (-8,-4-0.5) -- (-12,-8+0.5);
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\path[draw,thick,->] (-8,-4-0.5) -- (-4,-8+0.5);
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\path[draw,thick,->] (8,-4-0.5) -- (4,-8+0.5);
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\path[draw,thick,->] (8,-4-0.5) -- (12,-8+0.5);
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\node at (-14,-12) {$\texttt{gen}(4)$};
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\node at (-10,-12) {$\texttt{gen}(4)$};
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\node at (-6,-12) {$\texttt{gen}(4)$};
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\node at (-2,-12) {$\texttt{gen}(4)$};
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\node at (2,-12) {$\texttt{gen}(4)$};
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\node at (6,-12) {$\texttt{gen}(4)$};
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\node at (10,-12) {$\texttt{gen}(4)$};
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\node at (14,-12) {$\texttt{gen}(4)$};
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\node at (-14,-13.5) {$\emptyset$};
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\node at (-10,-13.5) {$\{3\}$};
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\node at (-6,-13.5) {$\{2\}$};
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\node at (-2,-13.5) {$\{2,3\}$};
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\node at (2,-13.5) {$\{1\}$};
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\node at (6,-13.5) {$\{1,3\}$};
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\node at (10,-13.5) {$\{1,2\}$};
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\node at (14,-13.5) {$\{1,2,3\}$};
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\path[draw,thick,->] (-12,-8-0.5) -- (-14,-12+0.5);
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\path[draw,thick,->] (-12,-8-0.5) -- (-10,-12+0.5);
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\path[draw,thick,->] (-4,-8-0.5) -- (-6,-12+0.5);
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\path[draw,thick,->] (-4,-8-0.5) -- (-2,-12+0.5);
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\path[draw,thick,->] (4,-8-0.5) -- (2,-12+0.5);
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\path[draw,thick,->] (4,-8-0.5) -- (6,-12+0.5);
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\path[draw,thick,->] (12,-8-0.5) -- (10,-12+0.5);
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\path[draw,thick,->] (12,-8-0.5) -- (14,-12+0.5);
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\end{scope}
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\end{tikzpicture}
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\end{center}
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\subsubsection{Method 2}
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Another way to generate subsets is to exploit
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the bit representation of integers.
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Each subset of a set of $n$ elements
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can be represented as a sequence of $n$ bits,
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which corresponds to an integer between $0 \ldots 2^n-1$.
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The ones in the bit sequence indicate
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which elements are included in the subset.
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The usual convention is that the $k$th element
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is included in the subset exactly when the $k$th last bit
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in the sequence is one.
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For example, the bit representation of 25
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is 11001, that corresponds to the subset $\{1,4,5\}$.
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The following code goes through all subsets
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of a set of $n$ elements
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\begin{lstlisting}
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for (int b = 0; b < (1<<n); b++) {
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// process subset b
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}
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\end{lstlisting}
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The following code shows how we can find
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the elements of a subset that corresponds to a bit sequence.
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When processing each subset,
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the code builds a vector that contains the
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elements in the subset.
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\begin{lstlisting}
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for (int b = 0; b < (1<<n); b++) {
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vector<int> v;
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for (int i = 0; i < n; i++) {
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if (b&(1<<i)) v.push_back(i+1);
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}
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}
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\end{lstlisting}
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\section{Generating permutations}
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\index{permutation}
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Next we will consider the problem of generating
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all permutations of a set of $n$ elements.
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Again, there are two approaches:
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we can either use recursion or go through the
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permutations iteratively.
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\subsubsection{Method 1}
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Like subsets, permutations can be generated
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using recursion.
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The following function goes
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through the permutations of the set $\{1,2,\ldots,n\}$.
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The function builds a vector that contains
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the elements in the permutation,
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and the search begins when the function is
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called without parameters.
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\begin{lstlisting}
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void gen() {
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if (v.size() == n) {
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// process permutation v
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} else {
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for (int i = 1; i <= n; i++) {
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if (p[i]) continue;
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p[i] = 1;
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v.push_back(i);
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gen();
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p[i] = 0;
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v.pop_back();
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}
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}
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}
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\end{lstlisting}
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Each function call adds a new element to
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the vector \texttt{v}.
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The array \texttt{p} indicates which
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elements are already included in the permutation:
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if $\texttt{p}[k]=0$, element $k$ is not included,
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and if $\texttt{p}[k]=1$, element $k$ is included.
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If the size of \texttt{v} equals the size of the set,
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a permutation has been generated.
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\subsubsection{Method 2}
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\index{next\_permutation@\texttt{next\_permutation}}
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Another method for generating permutations
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is to begin with the permutation
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$\{1,2,\ldots,n\}$ and repeatedly
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use a function that constructs the next permutation
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in increasing order.
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The C++ standard library contains the function
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\texttt{next\_permutation} that can be used for this:
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\begin{lstlisting}
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vector<int> v;
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for (int i = 1; i <= n; i++) {
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v.push_back(i);
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}
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do {
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// process permutation v
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} while (next_permutation(v.begin(),v.end()));
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\end{lstlisting}
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\section{Backtracking}
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\index{backtracking}
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A \key{backtracking} algorithm
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begins with an empty solution
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and extends the solution step by step.
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The search recursively
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goes through all different ways how
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a solution can be constructed.
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\index{queen problem}
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As an example, consider the \key{queen problem}
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where the task is to calculate the number
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of ways we can place $n$ queens to
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an $n \times n$ chessboard so that
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no two queens attack each other.
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For example, when $n=4$,
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there are two possible solutions to the problem:
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\begin{center}
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\begin{tikzpicture}[scale=.65]
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\begin{scope}
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\draw (0, 0) grid (4, 4);
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\node at (1.5,3.5) {\symqueen};
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\node at (3.5,2.5) {\symqueen};
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\node at (0.5,1.5) {\symqueen};
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\node at (2.5,0.5) {\symqueen};
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\draw (6, 0) grid (10, 4);
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\node at (6+2.5,3.5) {\symqueen};
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\node at (6+0.5,2.5) {\symqueen};
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\node at (6+3.5,1.5) {\symqueen};
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\node at (6+1.5,0.5) {\symqueen};
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\end{scope}
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\end{tikzpicture}
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\end{center}
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The problem can be solved using backtracking
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by placing queens to the board row by row.
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More precisely, exactly one queen will
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be placed to each row so that no queen attacks
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any of the queens placed before.
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A solution has been found when all
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$n$ queens have been placed to the board.
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For example, when $n=4$,
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some partial solutions generated by
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the backtracking algorithm are as follows:
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\begin{center}
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\begin{tikzpicture}[scale=.55]
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\begin{scope}
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\draw (0, 0) grid (4, 4);
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\draw (-9, -6) grid (-5, -2);
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\draw (-3, -6) grid (1, -2);
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\draw (3, -6) grid (7, -2);
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\draw (9, -6) grid (13, -2);
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\node at (-9+0.5,-3+0.5) {\symqueen};
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\node at (-3+1+0.5,-3+0.5) {\symqueen};
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\node at (3+2+0.5,-3+0.5) {\symqueen};
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\node at (9+3+0.5,-3+0.5) {\symqueen};
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\draw (2,0) -- (-7,-2);
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\draw (2,0) -- (-1,-2);
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\draw (2,0) -- (5,-2);
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\draw (2,0) -- (11,-2);
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\draw (-11, -12) grid (-7, -8);
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\draw (-6, -12) grid (-2, -8);
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\draw (-1, -12) grid (3, -8);
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\draw (4, -12) grid (8, -8);
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\draw[white] (11, -12) grid (15, -8);
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\node at (-11+1+0.5,-9+0.5) {\symqueen};
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\node at (-6+1+0.5,-9+0.5) {\symqueen};
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\node at (-1+1+0.5,-9+0.5) {\symqueen};
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\node at (4+1+0.5,-9+0.5) {\symqueen};
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\node at (-11+0+0.5,-10+0.5) {\symqueen};
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\node at (-6+1+0.5,-10+0.5) {\symqueen};
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\node at (-1+2+0.5,-10+0.5) {\symqueen};
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\node at (4+3+0.5,-10+0.5) {\symqueen};
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\draw (-1,-6) -- (-9,-8);
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\draw (-1,-6) -- (-4,-8);
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\draw (-1,-6) -- (1,-8);
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\draw (-1,-6) -- (6,-8);
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\node at (-9,-13) {\ding{55}};
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\node at (-4,-13) {\ding{55}};
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\node at (1,-13) {\ding{55}};
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\node at (6,-13) {\ding{51}};
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\end{scope}
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\end{tikzpicture}
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\end{center}
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At the bottom level, the three first boards
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are not valid, because the queens attack each other.
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However, the fourth board is valid
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and it can be extended to a complete solution by
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placing two more queens to the board.
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\begin{samepage}
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The following code implements the search:
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\begin{lstlisting}
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void search(int y) {
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if (y == n) {
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c++;
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return;
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}
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for (int x = 0; x < n; x++) {
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if (r1[x] || r2[x+y] || r3[x-y+n-1]) continue;
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r1[x] = r2[x+y] = r3[x-y+n-1] = 1;
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search(y+1);
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r1[x] = r2[x+y] = r3[x-y+n-1] = 0;
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}
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}
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\end{lstlisting}
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\end{samepage}
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The search begins by calling \texttt{search(0)}.
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The size of the board is in the variable $n$,
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and the code calculates the number of solutions
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to the variable $c$.
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The code assumes that the rows and columns
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of the board are numbered from 0.
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The function places a queen to row $y$
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where $0 \le y < n$.
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Finally, if $y=n$, a solution has been found
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and the variable $c$ is increased by one.
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The array \texttt{r1} keeps track of the columns
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that already contain a queen,
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and the arrays \texttt{r2} and \texttt{r3}
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keep track of the diagonals.
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It is not allowed to add another queen to a
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column or diagonal that already contains a queen.
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For example, the rows and diagonals of
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the $4 \times 4$ board are numbered as follows:
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\begin{center}
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\begin{tikzpicture}[scale=.65]
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\begin{scope}
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\draw (0-6, 0) grid (4-6, 4);
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\node at (-6+0.5,3.5) {$0$};
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\node at (-6+1.5,3.5) {$1$};
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\node at (-6+2.5,3.5) {$2$};
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\node at (-6+3.5,3.5) {$3$};
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\node at (-6+0.5,2.5) {$0$};
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\node at (-6+1.5,2.5) {$1$};
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\node at (-6+2.5,2.5) {$2$};
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\node at (-6+3.5,2.5) {$3$};
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\node at (-6+0.5,1.5) {$0$};
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\node at (-6+1.5,1.5) {$1$};
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\node at (-6+2.5,1.5) {$2$};
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\node at (-6+3.5,1.5) {$3$};
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\node at (-6+0.5,0.5) {$0$};
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\node at (-6+1.5,0.5) {$1$};
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\node at (-6+2.5,0.5) {$2$};
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\node at (-6+3.5,0.5) {$3$};
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\draw (0, 0) grid (4, 4);
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\node at (0.5,3.5) {$0$};
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\node at (1.5,3.5) {$1$};
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\node at (2.5,3.5) {$2$};
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\node at (3.5,3.5) {$3$};
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\node at (0.5,2.5) {$1$};
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\node at (1.5,2.5) {$2$};
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\node at (2.5,2.5) {$3$};
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\node at (3.5,2.5) {$4$};
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\node at (0.5,1.5) {$2$};
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\node at (1.5,1.5) {$3$};
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\node at (2.5,1.5) {$4$};
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\node at (3.5,1.5) {$5$};
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\node at (0.5,0.5) {$3$};
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\node at (1.5,0.5) {$4$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$6$};
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\draw (6, 0) grid (10, 4);
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\node at (6.5,3.5) {$3$};
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\node at (7.5,3.5) {$4$};
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\node at (8.5,3.5) {$5$};
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\node at (9.5,3.5) {$6$};
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\node at (6.5,2.5) {$2$};
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\node at (7.5,2.5) {$3$};
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\node at (8.5,2.5) {$4$};
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\node at (9.5,2.5) {$5$};
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\node at (6.5,1.5) {$1$};
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\node at (7.5,1.5) {$2$};
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\node at (8.5,1.5) {$3$};
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\node at (9.5,1.5) {$4$};
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\node at (6.5,0.5) {$0$};
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\node at (7.5,0.5) {$1$};
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\node at (8.5,0.5) {$2$};
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\node at (9.5,0.5) {$3$};
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\node at (-4,-1) {\texttt{r1}};
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\node at (2,-1) {\texttt{r2}};
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\node at (8,-1) {\texttt{r3}};
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\end{scope}
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\end{tikzpicture}
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\end{center}
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The above backtracking
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algorithm tells us that
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there are 92 ways to place 8
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queens to the $8 \times 8$ chessboard.
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When $n$ increases, the search quickly becomes slow,
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because the number of the solutions increases
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exponentially.
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For example, calculating the ways to
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place 16 queens to the $16 \times 16$
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chessboard already takes about a minute
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on a modern computer
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(there are 14772512 solutions).
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\section{Pruning the search}
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We can often optimize backtracking
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by pruning the search tree.
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The idea is to add ''intelligence'' to the algorithm
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so that it will notice as soon as possible
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if a partial solution cannot be extended
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to a complete solution.
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Such optimizations can have a tremendous
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effect on the efficiency of the search.
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Let us consider a problem
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of calculating the number of paths
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in an $n \times n$ grid from the upper-left corner
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to the lower-right corner so that each square
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will be visited exactly once.
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For example, in a $7 \times 7$ grid,
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there are 111712 such paths.
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One of the paths is as follows:
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\begin{center}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(5.5,0.5) -- (5.5,3.5) -- (3.5,3.5) --
|
|
(3.5,5.5) -- (1.5,5.5) -- (1.5,6.5) --
|
|
(4.5,6.5) -- (4.5,4.5) -- (5.5,4.5) --
|
|
(5.5,6.5) -- (6.5,6.5) -- (6.5,0.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
We will concentrate on the $7 \times 7$ case,
|
|
because its level of difficulty is appropriate to our needs.
|
|
We begin with a straightforward backtracking algorithm,
|
|
and then optimize it step by step using observations
|
|
how the search can be pruned.
|
|
After each optimization, we measure the running time
|
|
of the algorithm and the number of recursive calls,
|
|
so that we will clearly see the effect of each
|
|
optimization on the efficiency of the search.
|
|
|
|
\subsubsection{Basic algorithm}
|
|
|
|
The first version of the algorithm does not contain
|
|
any optimizations. We simply use backtracking to generate
|
|
all possible paths from the upper-left corner to
|
|
the lower-right corner and count the number of such paths.
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 483 seconds
|
|
\item
|
|
recursive calls: 76 billions
|
|
\end{itemize}
|
|
|
|
\subsubsection{Optimization 1}
|
|
|
|
In any solution, we first move one step
|
|
down or right.
|
|
There are always two paths that
|
|
are symmetric
|
|
about the diagonal of the grid
|
|
after the first step.
|
|
For example, the following paths are symmetric:
|
|
|
|
\begin{center}
|
|
\begin{tabular}{ccc}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(5.5,0.5) -- (5.5,3.5) -- (3.5,3.5) --
|
|
(3.5,5.5) -- (1.5,5.5) -- (1.5,6.5) --
|
|
(4.5,6.5) -- (4.5,4.5) -- (5.5,4.5) --
|
|
(5.5,6.5) -- (6.5,6.5) -- (6.5,0.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
& \hspace{20px}
|
|
&
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}[yscale=1,xscale=-1,rotate=-90]
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(5.5,0.5) -- (5.5,3.5) -- (3.5,3.5) --
|
|
(3.5,5.5) -- (1.5,5.5) -- (1.5,6.5) --
|
|
(4.5,6.5) -- (4.5,4.5) -- (5.5,4.5) --
|
|
(5.5,6.5) -- (6.5,6.5) -- (6.5,0.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{tabular}
|
|
\end{center}
|
|
|
|
Hence, we can decide that we always first
|
|
move one step down,
|
|
and finally multiply the number of the solutions by two.
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 244 seconds
|
|
\item
|
|
recursive calls: 38 billions
|
|
\end{itemize}
|
|
|
|
\subsubsection{Optimization 2}
|
|
|
|
If the path reaches the lower-right square
|
|
before it has visited all other squares of the grid,
|
|
it is clear that
|
|
it will not be possible to complete the solution.
|
|
An example of this is the following path:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(6.5,0.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Using this observation, we can terminate the search
|
|
immediately if we reach the lower-right square too early.
|
|
\begin{itemize}
|
|
\item
|
|
running time: 119 seconds
|
|
\item
|
|
recursive calls: 20 billions
|
|
\end{itemize}
|
|
|
|
\subsubsection{Optimization 3}
|
|
|
|
If the path touches the wall so that there is
|
|
an unvisited square on both sides,
|
|
the grid splits into two parts.
|
|
For example, in the following path
|
|
both the left and right squares
|
|
are unvisited:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(5.5,0.5) -- (5.5,6.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Now it will not be possible to visit every square,
|
|
so we can terminate the search.
|
|
It turns out that this optimization is very useful:
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 1.8 seconds
|
|
\item
|
|
recursive calls: 221 millions
|
|
\end{itemize}
|
|
|
|
\subsubsection{Optimization 4}
|
|
|
|
The idea of the previous optimization
|
|
can be generalized:
|
|
the grid splits into two parts
|
|
if the top and bottom neighbors
|
|
of the current square are unvisited and
|
|
the left and right neighbors are
|
|
wall or visited (or vice versa).
|
|
|
|
For example, in the following path
|
|
the top and bottom neighbors are unvisited,
|
|
so the path cannot visit all squares
|
|
in the grid anymore:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (7, 7);
|
|
\draw[thick,->] (0.5,6.5) -- (0.5,4.5) -- (2.5,4.5) --
|
|
(2.5,3.5) -- (0.5,3.5) -- (0.5,0.5) --
|
|
(3.5,0.5) -- (3.5,1.5) -- (1.5,1.5) --
|
|
(1.5,2.5) -- (4.5,2.5) -- (4.5,0.5) --
|
|
(5.5,0.5) -- (5.5,4.5) -- (3.5,4.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
Thus, we can terminate the search in all such cases.
|
|
After this optimization, the search will be
|
|
very efficient:
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 0.6 seconds
|
|
\item
|
|
recursive calls: 69 millions
|
|
\end{itemize}
|
|
|
|
~\\
|
|
Now it is a good moment to stop optimizing
|
|
the algorithm and see what we have achieved.
|
|
The running time of the original algorithm
|
|
was 483 seconds,
|
|
and now after the optimizations,
|
|
the running time is only 0.6 seconds.
|
|
Thus, the algorithm became nearly 1000 times
|
|
faster thanks to the optimizations.
|
|
|
|
This is a usual phenomenon in backtracking,
|
|
because the search tree is usually large
|
|
and even simple observations can effectively
|
|
prune the search.
|
|
Especially useful are optimizations that
|
|
occur during the first steps of the algorithm,
|
|
i.e., at the top of the search tree.
|
|
|
|
\section{Meet in the middle}
|
|
|
|
\index{meet in the middle}
|
|
|
|
\key{Meet in the middle} is a technique
|
|
where the search space is divided into
|
|
two parts of about equal size.
|
|
A separate search is performed
|
|
for both of the parts,
|
|
and finally the results of the searches are combined.
|
|
|
|
The technique can be used
|
|
if there is an efficient way to combine the
|
|
results of the searches.
|
|
In such a situation, the two searches may require less
|
|
time than one large search.
|
|
Typically, we can turn a factor of $2^n$
|
|
into a factor of $2^{n/2}$ using the meet in the
|
|
middle technique.
|
|
|
|
As an example, consider a problem where
|
|
we are given a list of $n$ numbers and
|
|
a number $x$.
|
|
Our task is to find out if it is possible
|
|
to choose some numbers from the list so that
|
|
their sum is $x$.
|
|
For example, given the list $[2,4,5,9]$ and $x=15$,
|
|
we can choose the numbers $[2,4,9]$ to get $2+4+9=15$.
|
|
However, if $x=10$,
|
|
it is not possible to form the sum.
|
|
|
|
A standard solution for the problem is to
|
|
go through all subsets of the elements and
|
|
check if the sum of any of the subsets is $x$.
|
|
The running time of such a solution is $O(2^n)$,
|
|
because there are $2^n$ subsets.
|
|
However, using the meet in the middle technique,
|
|
we can achieve a more efficient $O(2^{n/2})$ time solution.
|
|
Note that $O(2^n)$ and $O(2^{n/2})$ are different
|
|
complexities because $2^{n/2}$ equals $\sqrt{2^n}$.
|
|
|
|
The idea is to divide the list into
|
|
two lists $A$ and $B$ such that both
|
|
lists contain about half of the numbers.
|
|
The first search generates all subsets
|
|
of the numbers in $A$ and stores their sums
|
|
to a list $S_A$.
|
|
Correspondingly, the second search creates
|
|
a list $S_B$ from $B$.
|
|
After this, it suffices to check if it is possible
|
|
to choose one element from $S_A$ and another
|
|
element from $S_B$ so that their sum is $x$.
|
|
This is possible exactly when there is a way to
|
|
form the sum $x$ using the numbers in the original list.
|
|
|
|
For example, suppose that the list is $[2,4,5,9]$ and $x=15$.
|
|
First, we divide the list into $A=[2,4]$ and $B=[5,9]$.
|
|
After this, we create lists
|
|
$S_A=[0,2,4,6]$ and $S_B=[0,5,9,14]$.
|
|
In this case, the sum $x=15$ is possible to form,
|
|
because we can choose the number $6$ from $S_A$
|
|
and the number $9$ from $S_B$,
|
|
which corresponds to the solution $[2,4,9]$.
|
|
|
|
The time complexity of the algorithm is $O(2^{n/2})$,
|
|
because both lists $A$ and $B$ contain $n/2$ numbers
|
|
and it takes $O(2^{n/2})$ time to calculate the sums of
|
|
their subsets to lists $S_A$ and $S_B$.
|
|
After this, it is possible to check in
|
|
$O(2^{n/2})$ time if the sum $x$ can be formed
|
|
using the numbers in $S_A$ and $S_B$. |