753 lines
22 KiB
TeX
753 lines
22 KiB
TeX
\chapter{Complete search}
|
|
|
|
\key{Compelete search}
|
|
is a general method that can be used
|
|
for solving almost any algorithm problem.
|
|
The idea is to generate all possible
|
|
solutions for the problem using brute force,
|
|
and select the best solution or count the
|
|
number of solutions, depending on the problem.
|
|
|
|
Complete search is a good technique
|
|
if it is feasible to go through all the solutions,
|
|
because the search is usually easy to implement
|
|
and it always gives the correct answer.
|
|
If complete search is too slow,
|
|
greedy algorithms or dynamic programming,
|
|
presented in the next chapters,
|
|
may be used.
|
|
|
|
\section{Generating subsets}
|
|
|
|
\index{subset}
|
|
|
|
We first consider the case where
|
|
the possible solutions for the problem
|
|
are the subsets of a set of $n$ elements.
|
|
In this case, a complete search algorithm
|
|
has to generate
|
|
all $2^n$ subsets of the set.
|
|
|
|
\subsubsection{Method 1}
|
|
|
|
An elegant way to go through all subsets
|
|
of a set is to use recursion.
|
|
The following function \texttt{gen}
|
|
generates the subsets of the set
|
|
$\{1,2,\ldots,n\}$.
|
|
The function maintains a vector \texttt{v}
|
|
that will contain the elements in the subset.
|
|
The generation of the subsets
|
|
begins when the function
|
|
is called with parameter $1$.
|
|
|
|
\begin{lstlisting}
|
|
void gen(int k) {
|
|
if (k == n+1) {
|
|
// process subset v
|
|
} else {
|
|
gen(k+1);
|
|
v.push_back(k);
|
|
gen(k+1);
|
|
v.pop_back();
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The parameter $k$ is the number that is the next
|
|
candidate to be included in the subset.
|
|
The function branches to two cases:
|
|
either $k$ is included or it is not included in the subset.
|
|
Finally, when $k=n+1$, a decision has been made for
|
|
all the numbers and one subset has been generated.
|
|
|
|
For example, when $n=3$, the function calls
|
|
create a tree illustrated below.
|
|
At each call, the left branch doesn't include
|
|
the number and the right branch includes the number
|
|
in the subset.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.45]
|
|
\begin{scope}
|
|
\small
|
|
\node at (0,0) {$\texttt{gen}(1)$};
|
|
|
|
\node at (-8,-4) {$\texttt{gen}(2)$};
|
|
\node at (8,-4) {$\texttt{gen}(2)$};
|
|
|
|
\path[draw,thick,->] (0,0-0.5) -- (-8,-4+0.5);
|
|
\path[draw,thick,->] (0,0-0.5) -- (8,-4+0.5);
|
|
|
|
\node at (-12,-8) {$\texttt{gen}(3)$};
|
|
\node at (-4,-8) {$\texttt{gen}(3)$};
|
|
\node at (4,-8) {$\texttt{gen}(3)$};
|
|
\node at (12,-8) {$\texttt{gen}(3)$};
|
|
|
|
\path[draw,thick,->] (-8,-4-0.5) -- (-12,-8+0.5);
|
|
\path[draw,thick,->] (-8,-4-0.5) -- (-4,-8+0.5);
|
|
\path[draw,thick,->] (8,-4-0.5) -- (4,-8+0.5);
|
|
\path[draw,thick,->] (8,-4-0.5) -- (12,-8+0.5);
|
|
|
|
\node at (-14,-12) {$\texttt{gen}(4)$};
|
|
\node at (-10,-12) {$\texttt{gen}(4)$};
|
|
\node at (-6,-12) {$\texttt{gen}(4)$};
|
|
\node at (-2,-12) {$\texttt{gen}(4)$};
|
|
\node at (2,-12) {$\texttt{gen}(4)$};
|
|
\node at (6,-12) {$\texttt{gen}(4)$};
|
|
\node at (10,-12) {$\texttt{gen}(4)$};
|
|
\node at (14,-12) {$\texttt{gen}(4)$};
|
|
|
|
\node at (-14,-13.5) {$\emptyset$};
|
|
\node at (-10,-13.5) {$\{3\}$};
|
|
\node at (-6,-13.5) {$\{2\}$};
|
|
\node at (-2,-13.5) {$\{2,3\}$};
|
|
\node at (2,-13.5) {$\{1\}$};
|
|
\node at (6,-13.5) {$\{1,3\}$};
|
|
\node at (10,-13.5) {$\{1,2\}$};
|
|
\node at (14,-13.5) {$\{1,2,3\}$};
|
|
|
|
|
|
\path[draw,thick,->] (-12,-8-0.5) -- (-14,-12+0.5);
|
|
\path[draw,thick,->] (-12,-8-0.5) -- (-10,-12+0.5);
|
|
\path[draw,thick,->] (-4,-8-0.5) -- (-6,-12+0.5);
|
|
\path[draw,thick,->] (-4,-8-0.5) -- (-2,-12+0.5);
|
|
\path[draw,thick,->] (4,-8-0.5) -- (2,-12+0.5);
|
|
\path[draw,thick,->] (4,-8-0.5) -- (6,-12+0.5);
|
|
\path[draw,thick,->] (12,-8-0.5) -- (10,-12+0.5);
|
|
\path[draw,thick,->] (12,-8-0.5) -- (14,-12+0.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\subsubsection{Method 2}
|
|
|
|
Another way to generate the subsets is to exploit
|
|
the bit representation of integers.
|
|
Each subset of a set of $n$ elements
|
|
can be represented as a sequence of $n$ bits,
|
|
which corresponds to an integer between $0 \ldots 2^n-1$.
|
|
The ones in the bit representation indicate
|
|
which elements of the set are included in the subset.
|
|
|
|
The usual interpretation is that element $k$
|
|
is included in the subset if $k$th bit from the
|
|
end of the bit sequence is one.
|
|
For example, the bit representation of 25
|
|
is 11001 that corresponds to the subset $\{1,4,5\}$.
|
|
|
|
The following iterates through all subsets
|
|
of a set of $n$ elements
|
|
|
|
\begin{lstlisting}
|
|
for (int b = 0; b < (1<<n); b++) {
|
|
// process subset b
|
|
}
|
|
\end{lstlisting}
|
|
|
|
The following code converts each bit
|
|
representation to a vector \texttt{v}
|
|
that contains the elements in the subset.
|
|
This can be done by checking which bits
|
|
are one in the bit representation.
|
|
|
|
\begin{lstlisting}
|
|
for (int b = 0; b < (1<<n); b++) {
|
|
vector<int> v;
|
|
for (int i = 0; i < n; i++) {
|
|
if (b&(1<<i)) v.push_back(i+1);
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
|
|
\section{Generating permutations}
|
|
|
|
\index{permutation}
|
|
|
|
Another common situation is that the solutions
|
|
for the problem are permutations of a
|
|
set of $n$ elements.
|
|
In this case, a complete search algorithm has to
|
|
generate $n!$ possible permutations.
|
|
|
|
\subsubsection{Method 1}
|
|
|
|
Like subsets, permutations can be generated
|
|
using recursion.
|
|
The following function \texttt{gen} iterates
|
|
through the permutations of the set $\{1,2,\ldots,n\}$.
|
|
The function uses the vector \texttt{v}
|
|
for storing the permutations, and the generation
|
|
begins by calling the function without parameters.
|
|
|
|
\begin{lstlisting}
|
|
void haku() {
|
|
if (v.size() == n) {
|
|
// process permutation v
|
|
} else {
|
|
for (int i = 1; i <= n; i++) {
|
|
if (p[i]) continue;
|
|
p[i] = 1;
|
|
v.push_back(i);
|
|
haku();
|
|
p[i] = 0;
|
|
v.pop_back();
|
|
}
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
|
|
Each function call adds a new element to
|
|
the permutation in the vector \texttt{v}.
|
|
The array \texttt{p} indicates which
|
|
elements are already included in the permutation.
|
|
If $\texttt{p}[k]=0$, element $k$ is not included,
|
|
and if $\texttt{p}[k]=1$, element $k$ is included.
|
|
If the size of the vector equals the size of the set,
|
|
a permutation has been generated.
|
|
|
|
\subsubsection{Method 2}
|
|
|
|
\index{next\_permutation@\texttt{next\_permutation}}
|
|
|
|
Another method is to begin from permutation
|
|
$\{1,2,\ldots,n\}$ and at each step generate the
|
|
next permutation in increasing order.
|
|
The C++ standard library contains the function
|
|
\texttt{next\_permutation} that can be used for this.
|
|
The following code generates the permutations
|
|
of the set $\{1,2,\ldots,n\}$ using the function:
|
|
|
|
\begin{lstlisting}
|
|
vector<int> v;
|
|
for (int i = 1; i <= n; i++) {
|
|
v.push_back(i);
|
|
}
|
|
do {
|
|
// process permutation v
|
|
} while (next_permutation(v.begin(),v.end()));
|
|
\end{lstlisting}
|
|
|
|
\section{Backtracking}
|
|
|
|
\index{backtracking}
|
|
|
|
A \key{backtracking} algorithm
|
|
begins from an empty solution
|
|
and extends the solution step by step.
|
|
At each step, the search branches
|
|
to all possible directions how the solution
|
|
can be extended.
|
|
After processing one branch, the search
|
|
continues to other possible directions.
|
|
|
|
\index{queen problem}
|
|
|
|
As an example, consider the \key{queen problem}
|
|
where our task is to calculate the number
|
|
of ways we can place $n$ queens to
|
|
an $n \times n$ chessboard so that
|
|
no two queens attack each other.
|
|
For example, when $n=4$,
|
|
there are two possible solutions for the problem:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (4, 4);
|
|
\node at (1.5,3.5) {$K$};
|
|
\node at (3.5,2.5) {$K$};
|
|
\node at (0.5,1.5) {$K$};
|
|
\node at (2.5,0.5) {$K$};
|
|
|
|
\draw (6, 0) grid (10, 4);
|
|
\node at (6+2.5,3.5) {$K$};
|
|
\node at (6+0.5,2.5) {$K$};
|
|
\node at (6+3.5,1.5) {$K$};
|
|
\node at (6+1.5,0.5) {$K$};
|
|
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The problem can be solved using backtracking
|
|
by placing queens to the board row by row.
|
|
More precisely, we should place exactly one queen
|
|
to each row so that no queen attacks
|
|
any of the queens placed before.
|
|
A solution is ready when we have placed all
|
|
$n$ queens to the board.
|
|
|
|
For example, when $n=4$, the tree produced by
|
|
the backtracking algorithm begins like this:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.55]
|
|
\begin{scope}
|
|
\draw (0, 0) grid (4, 4);
|
|
|
|
\draw (-9, -6) grid (-5, -2);
|
|
\draw (-3, -6) grid (1, -2);
|
|
\draw (3, -6) grid (7, -2);
|
|
\draw (9, -6) grid (13, -2);
|
|
|
|
\node at (-9+0.5,-3+0.5) {$K$};
|
|
\node at (-3+1+0.5,-3+0.5) {$K$};
|
|
\node at (3+2+0.5,-3+0.5) {$K$};
|
|
\node at (9+3+0.5,-3+0.5) {$K$};
|
|
|
|
\draw (2,0) -- (-7,-2);
|
|
\draw (2,0) -- (-1,-2);
|
|
\draw (2,0) -- (5,-2);
|
|
\draw (2,0) -- (11,-2);
|
|
|
|
\draw (-11, -12) grid (-7, -8);
|
|
\draw (-6, -12) grid (-2, -8);
|
|
\draw (-1, -12) grid (3, -8);
|
|
\draw (4, -12) grid (8, -8);
|
|
\draw[white] (11, -12) grid (15, -8);
|
|
\node at (-11+1+0.5,-9+0.5) {$K$};
|
|
\node at (-6+1+0.5,-9+0.5) {$K$};
|
|
\node at (-1+1+0.5,-9+0.5) {$K$};
|
|
\node at (4+1+0.5,-9+0.5) {$K$};
|
|
\node at (-11+0+0.5,-10+0.5) {$K$};
|
|
\node at (-6+1+0.5,-10+0.5) {$K$};
|
|
\node at (-1+2+0.5,-10+0.5) {$K$};
|
|
\node at (4+3+0.5,-10+0.5) {$K$};
|
|
|
|
\draw (-1,-6) -- (-9,-8);
|
|
\draw (-1,-6) -- (-4,-8);
|
|
\draw (-1,-6) -- (1,-8);
|
|
\draw (-1,-6) -- (6,-8);
|
|
|
|
\node at (-9,-13) {\ding{55}};
|
|
\node at (-4,-13) {\ding{55}};
|
|
\node at (1,-13) {\ding{55}};
|
|
\node at (6,-13) {\ding{51}};
|
|
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
At the bottom level, the three first subsolutions
|
|
are not valid because the queens attack each other.
|
|
However, the fourth subsolution is valid
|
|
and it can be extended to a full solution by
|
|
placing two more queens to the board.
|
|
|
|
\begin{samepage}
|
|
The following code implements the search:
|
|
\begin{lstlisting}
|
|
void search(int y) {
|
|
if (y == n) {
|
|
c++;
|
|
return;
|
|
}
|
|
for (int x = 0; x < n; x++) {
|
|
if (r1[x] || r2[x+y] || r3[x-y+n-1]) continue;
|
|
r1[x] = r2[x+y] = r3[x-y+n-1] = 1;
|
|
search(y+1);
|
|
r1[x] = r2[x+y] = r3[x-y+n-1] = 0;
|
|
}
|
|
}
|
|
\end{lstlisting}
|
|
\end{samepage}
|
|
The search begins by calling \texttt{search(0)}.
|
|
The size of the board is in the variable $n$,
|
|
and the code calculates the number of solutions
|
|
to the variable $c$.
|
|
|
|
The code assumes that the rows and columns
|
|
of the board are numbered from 0.
|
|
The function places a queen to row $y$
|
|
when $0 \le y < n$.
|
|
Finally, if $y=n$, one solution has been found
|
|
and the variable $c$ is increased by one.
|
|
|
|
The array \texttt{r1} keeps track of the columns
|
|
that already contain a queen.
|
|
Similarly, the arrays \texttt{r2} and \texttt{r3}
|
|
keep track of the diagonals.
|
|
It is not allowed to add another queen to a
|
|
column or to a diagonal.
|
|
For example, the rows and the diagonals of
|
|
the $4 \times 4$ board are numbered as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
\draw (0-6, 0) grid (4-6, 4);
|
|
\node at (-6+0.5,3.5) {$0$};
|
|
\node at (-6+1.5,3.5) {$1$};
|
|
\node at (-6+2.5,3.5) {$2$};
|
|
\node at (-6+3.5,3.5) {$3$};
|
|
\node at (-6+0.5,2.5) {$0$};
|
|
\node at (-6+1.5,2.5) {$1$};
|
|
\node at (-6+2.5,2.5) {$2$};
|
|
\node at (-6+3.5,2.5) {$3$};
|
|
\node at (-6+0.5,1.5) {$0$};
|
|
\node at (-6+1.5,1.5) {$1$};
|
|
\node at (-6+2.5,1.5) {$2$};
|
|
\node at (-6+3.5,1.5) {$3$};
|
|
\node at (-6+0.5,0.5) {$0$};
|
|
\node at (-6+1.5,0.5) {$1$};
|
|
\node at (-6+2.5,0.5) {$2$};
|
|
\node at (-6+3.5,0.5) {$3$};
|
|
|
|
\draw (0, 0) grid (4, 4);
|
|
\node at (0.5,3.5) {$0$};
|
|
\node at (1.5,3.5) {$1$};
|
|
\node at (2.5,3.5) {$2$};
|
|
\node at (3.5,3.5) {$3$};
|
|
\node at (0.5,2.5) {$1$};
|
|
\node at (1.5,2.5) {$2$};
|
|
\node at (2.5,2.5) {$3$};
|
|
\node at (3.5,2.5) {$4$};
|
|
\node at (0.5,1.5) {$2$};
|
|
\node at (1.5,1.5) {$3$};
|
|
\node at (2.5,1.5) {$4$};
|
|
\node at (3.5,1.5) {$5$};
|
|
\node at (0.5,0.5) {$3$};
|
|
\node at (1.5,0.5) {$4$};
|
|
\node at (2.5,0.5) {$5$};
|
|
\node at (3.5,0.5) {$6$};
|
|
|
|
\draw (6, 0) grid (10, 4);
|
|
\node at (6.5,3.5) {$3$};
|
|
\node at (7.5,3.5) {$4$};
|
|
\node at (8.5,3.5) {$5$};
|
|
\node at (9.5,3.5) {$6$};
|
|
\node at (6.5,2.5) {$2$};
|
|
\node at (7.5,2.5) {$3$};
|
|
\node at (8.5,2.5) {$4$};
|
|
\node at (9.5,2.5) {$5$};
|
|
\node at (6.5,1.5) {$1$};
|
|
\node at (7.5,1.5) {$2$};
|
|
\node at (8.5,1.5) {$3$};
|
|
\node at (9.5,1.5) {$4$};
|
|
\node at (6.5,0.5) {$0$};
|
|
\node at (7.5,0.5) {$1$};
|
|
\node at (8.5,0.5) {$2$};
|
|
\node at (9.5,0.5) {$3$};
|
|
|
|
\node at (-4,-1) {\texttt{r1}};
|
|
\node at (2,-1) {\texttt{r2}};
|
|
\node at (8,-1) {\texttt{r3}};
|
|
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Using the presented backtracking
|
|
algorithm, we can calculate that,
|
|
for example, there are 92 ways to place 8
|
|
queens to an $8 \times 8$ chessboard.
|
|
When $n$ increases, the search quickly becomes slow
|
|
because the number of the solutions increases
|
|
exponentially.
|
|
For example, calculating the ways to
|
|
place 16 queens to the $16 \times 16$
|
|
chessboard already takes about a minute
|
|
(there are 14772512 solutions).
|
|
|
|
\section{Pruning the search}
|
|
|
|
A backtracking algorithm can often be optimized
|
|
by pruning the search tree.
|
|
The idea is to add ''intelligence'' to the algorithm
|
|
so that it will notice as soon as possible
|
|
if is not possible to extend a subsolution into
|
|
a full solution.
|
|
This kind of optimization can have a tremendous
|
|
effect on the efficiency of the search.
|
|
|
|
Let us consider a problem where
|
|
our task is to calculate the number of paths
|
|
in an $n \times n$ grid from the upper-left corner
|
|
to the lower-right corner so that each square
|
|
will be visited exactly once.
|
|
For example, in the $7 \times 7$ grid,
|
|
there are 111712 possible paths from the
|
|
lower-right corner to the upper-right corner.
|
|
One of the paths is as follows:
|
|
|
|
\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 it is computationally suitable difficult.
|
|
We begin with a straightforward backtracking algorithm,
|
|
and then optimize it step by step using observations
|
|
how the search tree 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 doesn't contain
|
|
any optimizations. We simply use backtracking to generate
|
|
all possible paths from the upper-left corner to
|
|
the lower-right corner.
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 483 seconds
|
|
\item
|
|
recursive calls: 76 billions
|
|
\end{itemize}
|
|
|
|
\subsubsection{Optimization 1}
|
|
|
|
The first step in a solution is either
|
|
downward or to the 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}
|
|
|
|
Thus, we can decide that the first step
|
|
in the solution is always downward,
|
|
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 case:
|
|
|
|
\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 branch
|
|
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 at both sides,
|
|
the grid splits into two parts.
|
|
For example, in the following case
|
|
both the left and the 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 branch.
|
|
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 case
|
|
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}
|
|
The search becomes even faster when we terminate
|
|
the search branch in all such cases:
|
|
|
|
\begin{itemize}
|
|
\item
|
|
running time: 0.6 seconds
|
|
\item
|
|
recursive calls: 69 millions
|
|
\end{itemize}
|
|
|
|
~\\
|
|
Now it's a good moment to stop optimization
|
|
and remember our starting point.
|
|
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 after the optimizations.
|
|
|
|
This is a usual phenomenon in backtracking
|
|
because the search tree is usually large
|
|
and even simple optimizations can prune
|
|
a lot of branches in the tree.
|
|
Especially useful are optimizations that
|
|
occur at the top of the search tree because
|
|
they can prune the search very efficiently.
|
|
|
|
\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 equally large parts.
|
|
A separate search is performed
|
|
for each of the parts,
|
|
and finally the results of the searches are combined.
|
|
|
|
The meet in the middle technique can be used
|
|
if there is an efficient way to combine the
|
|
results of the searches.
|
|
In this case, 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
|
|
an integer $x$.
|
|
Our task is to find out if it is possible
|
|
to choose some numbers from the list so that
|
|
the sum of the numbers 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 the list remains the same but $x=10$,
|
|
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 time complexity of this solution is $O(2^n)$
|
|
because there are $2^n$ possible subsets.
|
|
However, using the meet in the middle technique,
|
|
we can create 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 given as input
|
|
to two lists $A$ and $B$ that each contain
|
|
about half of the numbers.
|
|
The first search generates all subsets
|
|
of the numbers in the list $A$ and stores their sums
|
|
to list $S_A$.
|
|
Correspondingly, the second search creates
|
|
the list $S_B$ from the list $B$.
|
|
After this, it suffices to check if it is possible
|
|
to choose one number from $S_A$ and another
|
|
number 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, assume 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 the lists
|
|
$S_A=[0,2,4,6]$ and $S_B=[0,5,9,14]$.
|
|
The sum $x=15$ is possible to form
|
|
because we can choose the number $6$ from $S_A$
|
|
and the number $9$ from $S_B$.
|
|
This choice 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 created
|
|
using the numbers in $S_A$ and $S_B$. |