1005 lines
29 KiB
TeX
1005 lines
29 KiB
TeX
\chapter{Dynamic programming}
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\index{dynamic programming}
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\key{Dynamic programming}
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is a technique that combines the correctness
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of complete search and the efficiency
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of greedy algorithms.
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Dynamic programming can be used if the
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problem can be divided into subproblems
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that can be calculated independently.
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There are two uses for dynamic programming:
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\begin{itemize}
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\item
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\key{Findind an optimal solution}:
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We want to find a solution that is
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as large as possible or as small as possible.
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\item
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\key{Couting the number of solutions}:
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We want to calculate the total number of
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possible solutions.
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\end{itemize}
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We will first see how dynamic programming can
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be used for finding an optimal solution,
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and then we will use the same idea for
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counting the solutions.
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Understanding dynamic programming is a milestone
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in every competitive programmer's career.
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While the basic idea of the technique is simple,
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the challenge is how to apply it for different problems.
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This chapter introduces a set of classic problems
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that are a good starting point.
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\section{Coin problem}
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We first consider a problem that we
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have already seen:
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Given a set of coin values $\{c_1,c_2,\ldots,c_k\}$
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and a sum of money $x$, our task is to
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form the sum $x$ using as few coins as possible.
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In Chapter 6.1, we solved the problem using a
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greedy algorithm that always selects the largest
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possible coin for the sum.
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The greedy algorithm works, for example,
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when the coins are the euro coins,
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but in the general case the greedy algorithm
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doesn't necessarily produce an optimal solution.
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Now it's time to solve the problem efficiently
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using dynamic programming, so that the algorithms
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works for any coin set.
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The dynamic programming
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algorithm is based on a recursive function
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that goes through all possibilities how to
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select the coins, like a brute force algorithm.
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However, the dynamic programming
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algorithm is efficient because
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it uses memoization to
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calculate the answer for each subproblem only once.
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\subsubsection{Recursive formulation}
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The idea in dynamic programming is to
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formulate the problem recursively so
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that the answer for the problem can be
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calculated from the answers for the smaller
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subproblems.
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In this case, a natural problem is as follows:
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what is the smallest number of coins
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required for constructing sum $x$?
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Let $f(x)$ be a function that gives the answer
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for the problem, i.e., $f(x)$ is the smallest
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number of coins required for constructing sum $x$.
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The values of the function depend on the
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values of the coins.
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For example, if the values are $\{1,3,4\}$,
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the first values of the function are as follows:
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\[
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\begin{array}{lcl}
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f(0) & = & 0 \\
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f(1) & = & 1 \\
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f(2) & = & 2 \\
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f(3) & = & 1 \\
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f(4) & = & 1 \\
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f(5) & = & 2 \\
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f(6) & = & 2 \\
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f(7) & = & 2 \\
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f(8) & = & 2 \\
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f(9) & = & 3 \\
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f(10) & = & 3 \\
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\end{array}
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\]
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First, $f(0)=0$ because no coins are needed
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for sum $0$.
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Moreover, $f(3)=1$ because the sum $3$
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can be formed using coin 3,
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and $f(5)=2$ because the sum 5 can
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be formed using coins 1 and 4.
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The essential property in the function is
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that the value $f(x)$ can be calculated
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recursively from the smaller values of the function.
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For example, if the coin set is $\{1,3,4\}$,
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there are three ways to select the first coin
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in a solution: we can choose coin 1, 3 or 4.
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If coin 1 is chosen, the remaining task is to
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form the sum $x-1$.
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Similarly, if coin 3 or 4 is chosen,
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we should form the sum $x-3$ or $x-4$.
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Thus, the recursive formula is
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\[f(x) = \min(f(x-1),f(x-3),f(x-4))+1\]
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where the function $\min$ returns the smallest
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of its parameters.
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In the general case, for the coin set
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$\{c_1,c_2,\ldots,c_k\}$,
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the recursive formula is
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\[f(x) = \min(f(x-c_1),f(x-c_2),\ldots,f(x-c_k))+1.\]
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The base case for the function is
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\[f(0)=0,\]
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because no coins are needed for constructing
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the sum 0.
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In addition, it's a good idea to define
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\[f(x)=\infty,\hspace{8px}\textrm{jos $x<0$}.\]
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This means that an infinite number of coins
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is needed to create a negative sum of money.
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This prevents the situation that the recursive
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function would form a solution where the
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initial sum of money is negative.
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Now it's possible to implement the function in C++
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directly using the recursive definition:
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\begin{lstlisting}
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int f(int x) {
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if (x == 0) return 0;
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if (x < 0) return 1e9;
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int u = 1e9;
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for (int i = 1; i <= k; i++) {
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u = min(u, f(x-c[i])+1);
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}
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return u;
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}
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\end{lstlisting}
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The code assumes that the available coins are
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$\texttt{c}[1], \texttt{c}[2], \ldots, \texttt{c}[k]$,
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and the value $10^9$ means infinity.
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This function works but it is not efficient yet
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because it goes through a large number
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of ways to construct the sum.
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However, the function becomes efficient by
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using memoization.
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\subsubsection{Memoization}
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\index{memoization}
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Dynamic programming allows to calculate the
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value of a recursive function efficiently
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using \key{memoization}.
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This means that an auxiliary array is used
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for storing the values of the function
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for different parameters.
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For each parameter, the value of the function
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is calculated only once, and after this,
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it can be directly retrieved from the array.
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In this problem, we can use the array
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\begin{lstlisting}
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int d[N];
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\end{lstlisting}
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where $\texttt{d}[x]$ will contain
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the value $f(x)$.
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The constant $N$ should be chosen so
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that there is space for all needed
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values of the function.
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After this, the function can be efficiently
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implemented as follows:
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\begin{lstlisting}
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int f(int x) {
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if (x == 0) return 0;
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if (x < 0) return 1e9;
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if (d[x]) return d[x];
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int u = 1e9;
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for (int i = 1; i <= k; i++) {
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u = min(u, f(x-c[i])+1);
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}
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d[x] = u;
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return d[x];
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}
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\end{lstlisting}
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The function handles the base cases
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$x=0$ and $x<0$ as previously.
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Then the function checks if
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$f(x)$ has already been calculated
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and stored to $\texttt{d}[x]$.
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If $f(x)$ can be found in the array,
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the function directly returns it.
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Otherwise the function calculates the value
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recursively and stores it to $\texttt{d}[x]$.
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Using memoization the function works
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efficiently because it is needed to
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recursively calculate
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the answer for each $x$ only once.
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After a value $f(x)$ has been stored to the array,
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it can be directly retrieved whenever the
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function will be called again with parameter $x$.
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The time complexity of the resulting algorithm
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is $O(xk)$ when the sum is $x$ and the number of
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coins is $k$.
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In practice, the algorithm is usable if
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$x$ is so small that it is possible to allocate
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an array for all possible function parameters.
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Note that the array can also be constructed using
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a loop that calculates all the values
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instead of a recursive function:
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\begin{lstlisting}
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d[0] = 0;
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for (int i = 1; i <= x; i++) {
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int u = 1e9;
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for (int j = 1; j <= k; j++) {
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if (i-c[j] < 0) continue;
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u = min(u, d[i-c[j]]+1);
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}
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d[i] = u;
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}
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\end{lstlisting}
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This implementation is shorter and somewhat
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more efficient than recursion,
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and experienced competitive programmers
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often implement dynamic programming solutions
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using loops.
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Still, the underlying idea is the same as
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in the recursive function.
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\subsubsection{Constructing the solution}
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Sometimes it is not enough to find out the value
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of the optimal solution, but we should also give
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an example how such a solution can be constructed.
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In this problem, this means that the algorithm
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should show how to select the coins that produce
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the sum $x$ using as few coins as possible.
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We can construct the solution by adding another
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array to the code. The array indicates for
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each sum of money the first coin that should be
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chosen in an optimal solution.
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In the following code, the array \texttt{e}
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is used for this:
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\begin{lstlisting}
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d[0] = 0;
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for (int i = 1; i <= x; i++) {
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d[i] = 1e9;
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for (int j = 1; j <= k; j++) {
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if (i-c[j] < 0) continue;
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int u = d[i-c[j]]+1;
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if (u < d[i]) {
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d[i] = u;
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e[i] = c[j];
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}
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}
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}
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\end{lstlisting}
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After this, we can print the coins needed
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for the sum $x$ as follows:
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\begin{lstlisting}
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while (x > 0) {
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cout << e[x] << "\n";
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x -= e[x];
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}
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\end{lstlisting}
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\subsubsection{Counting the number of solutions}
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Let us now consider a variation of the problem
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that it's like the original problem but we should
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count the total number of solutions instead
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of finding the optimal solution.
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For example, if the coins are $\{1,3,4\}$ and
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the required sum is $5$,
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there are a total of 6 solutions:
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\begin{multicols}{2}
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\begin{itemize}
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\item $1+1+1+1+1$
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\item $1+1+3$
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\item $1+3+1$
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\item $3+1+1$
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\item $1+4$
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\item $4+1$
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\end{itemize}
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\end{multicols}
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The number of the solutions can be calculated
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using the same idea as finding the optimal solution.
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The difference is that when finding the optimal solution,
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we maximize or minimize something in the recursion,
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but now we will sum together all possible alternatives to
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construct a solution.
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In this case, we can define a function $f(x)$
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that returns the number of ways to construct
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the sum $x$ using the coins.
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For example, $f(5)=6$ when the coins are $\{1,3,4\}$.
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The function $f(x)$ can be recursively calculated
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using the formula
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\[ f(x) = f(x-c_1)+f(x-c_2)+\cdots+f(x-c_k)\]
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because to form the sum $x$ we should first
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choose some coin $c_i$ and after this form the sum $x-c_i$.
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The base cases are $f(0)=1$ because there is exactly
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one way to form the sum 0 using an empty set of coins,
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and $f(x)=0$, when $x<0$, because it's not possible
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to form a negative sum of money.
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In the above example the function becomes
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\[ f(x) = f(x-1)+f(x-3)+f(x-4) \]
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and the first values of the function are:
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\[
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\begin{array}{lcl}
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f(0) & = & 1 \\
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f(1) & = & 1 \\
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f(2) & = & 1 \\
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f(3) & = & 2 \\
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f(4) & = & 4 \\
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f(5) & = & 6 \\
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f(6) & = & 9 \\
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f(7) & = & 15 \\
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f(8) & = & 25 \\
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f(9) & = & 40 \\
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\end{array}
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\]
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The following code calculates the value $f(x)$
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using dynamic programming by filling the array
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\texttt{d} for parameters $0 \ldots x$:
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\begin{lstlisting}
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d[0] = 1;
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for (int i = 1; i <= x; i++) {
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for (int j = 1; j <= k; j++) {
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if (i-c[j] < 0) continue;
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d[i] += d[i-c[j]];
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}
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}
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\end{lstlisting}
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Often the number of the solutions is so large
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that it is not required to calculate the exact number
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but it is enough to give the answer modulo $m$
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where, for example, $m=10^9+7$.
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This can be done by changing the code so that
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all calculations will be done in modulo $m$.
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In this case, it is enough to add the line
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\begin{lstlisting}
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d[i] %= m;
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\end{lstlisting}
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after the line
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\begin{lstlisting}
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d[i] += d[i-c[j]];
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\end{lstlisting}
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Now we have covered all basic
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techniques related to
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dynamic programming.
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Since dynamic programming can be used
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in many different situations,
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we will now go through a set of problems
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that show further examples how dynamic
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programming can be used.
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\section{Longest increasing subsequence}
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\index{longest increasing subsequence}
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Given an array that contains $n$
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numbers $x_1,x_2,\ldots,x_n$,
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our task is find the
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\key{longest increasing subsequence}
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in the array.
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This is a sequence of array elements
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that goes from the left to the right,
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and each element in the sequence is larger
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than the previous element.
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For example, in the array
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$6$};
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\node at (1.5,0.5) {$2$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$1$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$4$};
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\node at (6.5,0.5) {$8$};
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\node at (7.5,0.5) {$3$};
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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the longest increasing subsequence
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contains 4 elements:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\fill[color=lightgray] (1,0) rectangle (2,1);
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\fill[color=lightgray] (2,0) rectangle (3,1);
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\fill[color=lightgray] (4,0) rectangle (5,1);
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\fill[color=lightgray] (6,0) rectangle (7,1);
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\draw (0,0) grid (8,1);
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\node at (0.5,0.5) {$6$};
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\node at (1.5,0.5) {$2$};
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\node at (2.5,0.5) {$5$};
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\node at (3.5,0.5) {$1$};
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\node at (4.5,0.5) {$7$};
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\node at (5.5,0.5) {$4$};
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\node at (6.5,0.5) {$8$};
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\node at (7.5,0.5) {$3$};
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\draw[thick,->] (1.5,-0.25) .. controls (1.75,-1.00) and (2.25,-1.00) .. (2.4,-0.25);
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\draw[thick,->] (2.6,-0.25) .. controls (3.0,-1.00) and (4.0,-1.00) .. (4.4,-0.25);
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\draw[thick,->] (4.6,-0.25) .. controls (5.0,-1.00) and (6.0,-1.00) .. (6.5,-0.25);
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\footnotesize
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\node at (0.5,1.4) {$1$};
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\node at (1.5,1.4) {$2$};
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\node at (2.5,1.4) {$3$};
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\node at (3.5,1.4) {$4$};
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\node at (4.5,1.4) {$5$};
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\node at (5.5,1.4) {$6$};
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\node at (6.5,1.4) {$7$};
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\node at (7.5,1.4) {$8$};
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\end{tikzpicture}
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\end{center}
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Let $f(k)$ be the length of the
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longest increasing subsequence
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that ends to index $k$.
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Thus, the answer for the problem
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is the largest of values
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$f(1),f(2),\ldots,f(n)$.
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For example, in the above array
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the values for the function are as follows:
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\[
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\begin{array}{lcl}
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f(1) & = & 1 \\
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f(2) & = & 1 \\
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f(3) & = & 2 \\
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f(4) & = & 1 \\
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f(5) & = & 3 \\
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f(6) & = & 2 \\
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f(7) & = & 4 \\
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f(8) & = & 2 \\
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\end{array}
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\]
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When calculating the value $f(k)$,
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there are two possibilities how the subsequence
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that ends to index $k$ is constructed:
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\begin{enumerate}
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\item The subsequence
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only contains the element $x_k$, so $f(k)=1$.
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\item We choose some index $i$ for which $i<k$
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and $x_i<x_k$.
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We extend the longest increasing subsequence
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that ends to index $i$ by adding the element $x_k$
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to it. In this case $f(k)=f(i)+1$.
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\end{enumerate}
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Consider calculating the value $f(7)$.
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The best solution is to extend the longest
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increasing subsequence that ends to index 5,
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i.e., the sequence $[2,5,7]$, by adding
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the element $x_7=8$.
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The result is
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$[2,5,7,8]$, and $f(7)=f(5)+1=4$.
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A straightforward way to calculate the
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value $f(k)$ is to
|
|
go through all indices
|
|
$i=1,2,\ldots,k-1$ that can contain the
|
|
previous element in the subsequence.
|
|
The time complexity of such an algorithm is $O(n^2)$.
|
|
Surprisingly, it is also possible to solve the
|
|
problem in $O(n \log n)$ time, but this is more difficult.
|
|
|
|
\section{Path in a grid}
|
|
|
|
Our next problem is to find a path
|
|
in an $n \times n$ grid
|
|
from the upper-left corner to
|
|
the lower-right corner.
|
|
Each square contains a number,
|
|
and the path should be constructed so
|
|
that the sum of numbers along
|
|
the path is as large as possible.
|
|
In addition, it is only allowed to move
|
|
downwards and to the right.
|
|
|
|
In the followig grid, the best path is
|
|
marked with gray background:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
\fill [color=lightgray] (0, 9) rectangle (1, 8);
|
|
\fill [color=lightgray] (0, 8) rectangle (1, 7);
|
|
\fill [color=lightgray] (1, 8) rectangle (2, 7);
|
|
\fill [color=lightgray] (1, 7) rectangle (2, 6);
|
|
\fill [color=lightgray] (2, 7) rectangle (3, 6);
|
|
\fill [color=lightgray] (3, 7) rectangle (4, 6);
|
|
\fill [color=lightgray] (4, 7) rectangle (5, 6);
|
|
\fill [color=lightgray] (4, 6) rectangle (5, 5);
|
|
\fill [color=lightgray] (4, 5) rectangle (5, 4);
|
|
\draw (0, 4) grid (5, 9);
|
|
\node at (0.5,8.5) {3};
|
|
\node at (1.5,8.5) {7};
|
|
\node at (2.5,8.5) {9};
|
|
\node at (3.5,8.5) {2};
|
|
\node at (4.5,8.5) {7};
|
|
\node at (0.5,7.5) {9};
|
|
\node at (1.5,7.5) {8};
|
|
\node at (2.5,7.5) {3};
|
|
\node at (3.5,7.5) {5};
|
|
\node at (4.5,7.5) {5};
|
|
\node at (0.5,6.5) {1};
|
|
\node at (1.5,6.5) {7};
|
|
\node at (2.5,6.5) {9};
|
|
\node at (3.5,6.5) {8};
|
|
\node at (4.5,6.5) {5};
|
|
\node at (0.5,5.5) {3};
|
|
\node at (1.5,5.5) {8};
|
|
\node at (2.5,5.5) {6};
|
|
\node at (3.5,5.5) {4};
|
|
\node at (4.5,5.5) {10};
|
|
\node at (0.5,4.5) {6};
|
|
\node at (1.5,4.5) {3};
|
|
\node at (2.5,4.5) {9};
|
|
\node at (3.5,4.5) {7};
|
|
\node at (4.5,4.5) {8};
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The sum of numbers is
|
|
$3+9+8+7+9+8+5+10+8=67$
|
|
that is the largest possible sum in a path
|
|
from the
|
|
upper-left corner to the lower-right corner.
|
|
|
|
A good approach for the problem is to
|
|
calculate for each square $(y,x)$
|
|
the largest possible sum in a path
|
|
from the upper-left corner to the square $(y,x)$.
|
|
We denote this sum $f(y,x)$,
|
|
so $f(n,n)$ is the largest sum in a path
|
|
from the upper-left corner to
|
|
the lower-right corner.
|
|
|
|
The recursive formula is based on the observation
|
|
that a path that ends to square$(y,x)$
|
|
can either come from square $(y,x-1)$
|
|
or from square $(y-1,x)$:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
\fill [color=lightgray] (3, 7) rectangle (4, 6);
|
|
\draw (0, 4) grid (5, 9);
|
|
|
|
\node at (2.5,6.5) {$\rightarrow$};
|
|
\node at (3.5,7.5) {$\downarrow$};
|
|
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Let $r(y,x)$ denote the number in square $(y,x)$.
|
|
The base cases for the recursive function
|
|
are as follows:
|
|
|
|
\[
|
|
\begin{array}{lcl}
|
|
f(1,1) & = & r(1,1) \\
|
|
f(1,x) & = & f(1,x-1)+r(1,x) \\
|
|
f(y,1) & = & f(y-1,1)+r(y,1)\\
|
|
\end{array}
|
|
\]
|
|
|
|
In the general case there are two
|
|
possible paths, and we should select the path
|
|
that produces the larger sum:
|
|
\[ f(y,x) = \max(f(y,x-1),f(y-1,x))+r(y,x)\]
|
|
|
|
The time complexity of the solution is $O(n^2)$,
|
|
because each value $f(y,x)$ can be calculated
|
|
in constant time using the values of the
|
|
adjacent squares.
|
|
|
|
\section{Knapsack}
|
|
|
|
\index{knapsack}
|
|
|
|
\key{Knapsack} is a classic problem where we
|
|
are given $n$ objects with weights
|
|
$p_1,p_2,\ldots,p_n$ and values
|
|
$a_1,a_2,\ldots,a_n$.
|
|
Our task is to choose a subset of the objects
|
|
such that the sum of the weights is at most $x$
|
|
and the sum of the values is as large as possible.
|
|
|
|
\begin{samepage}
|
|
For example, if the objects are
|
|
\begin{center}
|
|
\begin{tabular}{rrr}
|
|
object & weight & value \\
|
|
\hline
|
|
A & 5 & 1 \\
|
|
B & 6 & 3 \\
|
|
C & 8 & 5 \\
|
|
D & 5 & 3 \\
|
|
\end{tabular}
|
|
\end{center}
|
|
\end{samepage}
|
|
and the maximum total weight is 12,
|
|
the optimal solution is to select objects $B$ and $D$.
|
|
Their total weight $6+5=11$ doesn't exceed 12,
|
|
and their total value $3+3=6$ is as large as possible.
|
|
|
|
This task is possible to solve in two different ways
|
|
using dynamic programming.
|
|
We can either regard the problem as maximizing the
|
|
total value of the objects or
|
|
minimizing the total weight of the objects.
|
|
|
|
\subsubsection{Solution 1}
|
|
|
|
\textit{Maximization:} Let $f(k,u)$
|
|
denote the largest possible total value
|
|
when a subset of objects $1 \ldots k$ is selected
|
|
such that the total weight is $u$.
|
|
The solution for the problem is
|
|
the largest value
|
|
$f(n,u)$ where $0 \le u \le x$.
|
|
A recursive formula for calculating
|
|
the function is
|
|
\[f(k,u) = \max(f(k-1,u),f(k-1,u-p_k)+a_k)\]
|
|
because we can either include or not include
|
|
object $k$ in the solution.
|
|
The base cases are $f(0,0)=0$ and $f(0,u)=-\infty$
|
|
when $u \neq 0$. The time compexity of
|
|
the solution is $O(nx)$.
|
|
|
|
In the example case, the optimal solution is
|
|
$f(4,11)=6$ that can be constructed
|
|
using the following sequence:
|
|
\[f(4,11)=f(3,6)+3=f(2,6)+3=f(1,0)+3+3=f(0,0)+3+3=6.\]
|
|
|
|
\subsubsection{Solution 2}
|
|
|
|
\textit{Minimization:} Let $f(k,u)$
|
|
denote the smallest possible total weight
|
|
when a subset of objects
|
|
$1 \ldots k$ is selected such
|
|
that the total weight is $u$.
|
|
The solution for the problem is the
|
|
largest value $u$
|
|
for which $0 \le u \le s$ and $f(n,u) \le x$
|
|
where $s=\sum_{i=1}^n a_i$.
|
|
A recursive formula for calculating the function is
|
|
\[f(k,u) = \min(f(k-1,u),f(k-1,u-a_k)+p_k).\]
|
|
as in solution 1.
|
|
The base cases are $f(0,0)=0$ and $f(0,u)=\infty$
|
|
when $u \neq 0$.
|
|
The time complexity of the solution is $O(ns)$.
|
|
|
|
In the example case, the optimal solution is $f(4,6)=11$
|
|
that can be constructed using the following sequence:
|
|
\[f(4,6)=f(3,3)+5=f(2,3)+5=f(1,0)+6+5=f(0,0)+6+5=11.\]
|
|
|
|
~\\
|
|
It is interesting to note how the features of the input
|
|
affect on the efficiency of the solutions.
|
|
The efficiency of solution 1 depends on the weights
|
|
of the objects, while the efficiency of solution 2
|
|
depends on the values of the objects.
|
|
|
|
\section{Edit distance}
|
|
|
|
\index{edit distance}
|
|
\index{Levenshtein distance}
|
|
|
|
The \key{edit distance},
|
|
also known as the \key{Levenshtein distance},
|
|
indicates how similar two strings are.
|
|
It is the minimum number of editing operations
|
|
needed for transforming the first string
|
|
into the second string.
|
|
The allowed editing operations are as follows:
|
|
\begin{itemize}
|
|
\item insert a character (e.g. \texttt{ABC} $\rightarrow$ \texttt{ABCA})
|
|
\item remove a character (e.g. \texttt{ABC} $\rightarrow$ \texttt{AC})
|
|
\item change a character (e.g. \texttt{ABC} $\rightarrow$ \texttt{ADC})
|
|
\end{itemize}
|
|
|
|
For example, the edit distance between
|
|
\texttt{LOVE} and \texttt{MOVIE} is 2
|
|
because we can first perform operation
|
|
\texttt{LOVE} $\rightarrow$ \texttt{MOVE}
|
|
(change) and then operation
|
|
\texttt{MOVE} $\rightarrow$ \texttt{MOVIE}
|
|
(insertion).
|
|
This is the smallest possible number of operations
|
|
because it is clear that one operation is not enough.
|
|
|
|
Suppose we are given strings
|
|
\texttt{x} of $n$ characters and
|
|
\texttt{y} of $m$ characters,
|
|
and we want to calculate the edit distance
|
|
between them.
|
|
This can be efficiently done using
|
|
dynamic programming in $O(nm)$ time.
|
|
Let $f(a,b)$ denote the edit distance
|
|
between the first $a$ characters of \texttt{x}
|
|
and the first $b$ characters of \texttt{y}.
|
|
Using this function, the edit distance between
|
|
\texttt{x} and \texttt{y} is $f(n,m)$,
|
|
and the function also determines
|
|
the editing operations needed.
|
|
|
|
The base cases for the function are
|
|
\[
|
|
\begin{array}{lcl}
|
|
f(0,b) & = & b \\
|
|
f(a,0) & = & a \\
|
|
\end{array}
|
|
\]
|
|
and in the general case the formula is
|
|
\[ f(a,b) = \min(f(a,b-1)+1,f(a-1,b)+1,f(a-1,b-1)+c),\]
|
|
where $c=0$ if the $a$th character of \texttt{x}
|
|
equals the $b$th character of \texttt{y},
|
|
and otherwise $c=1$.
|
|
The formula covers all ways to shorten the strings:
|
|
\begin{itemize}
|
|
\item $f(a,b-1)$ means that a character is inserted to \texttt{x}
|
|
\item $f(a-1,b)$ means that a chacater is removed from \texttt{x}
|
|
\item $f(a-1,b-1)$ means that \texttt{x} and \texttt{y} contain
|
|
the same character ($c=0$),
|
|
or a character in \texttt{x} is transformed into
|
|
a character in \texttt{y} ($c=1$)
|
|
\end{itemize}
|
|
The following table shows the values of $f$
|
|
in the example case:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
%\fill [color=lightgray] (5, -3) rectangle (6, -4);
|
|
\draw (1, -1) grid (7, -6);
|
|
|
|
\node at (0.5,-2.5) {\texttt{L}};
|
|
\node at (0.5,-3.5) {\texttt{O}};
|
|
\node at (0.5,-4.5) {\texttt{V}};
|
|
\node at (0.5,-5.5) {\texttt{E}};
|
|
|
|
\node at (2.5,-0.5) {\texttt{M}};
|
|
\node at (3.5,-0.5) {\texttt{O}};
|
|
\node at (4.5,-0.5) {\texttt{V}};
|
|
\node at (5.5,-0.5) {\texttt{I}};
|
|
\node at (6.5,-0.5) {\texttt{E}};
|
|
|
|
\node at (1.5,-1.5) {$0$};
|
|
\node at (1.5,-2.5) {$1$};
|
|
\node at (1.5,-3.5) {$2$};
|
|
\node at (1.5,-4.5) {$3$};
|
|
\node at (1.5,-5.5) {$4$};
|
|
\node at (2.5,-1.5) {$1$};
|
|
\node at (2.5,-2.5) {$1$};
|
|
\node at (2.5,-3.5) {$2$};
|
|
\node at (2.5,-4.5) {$3$};
|
|
\node at (2.5,-5.5) {$4$};
|
|
\node at (3.5,-1.5) {$2$};
|
|
\node at (3.5,-2.5) {$2$};
|
|
\node at (3.5,-3.5) {$1$};
|
|
\node at (3.5,-4.5) {$2$};
|
|
\node at (3.5,-5.5) {$3$};
|
|
\node at (4.5,-1.5) {$3$};
|
|
\node at (4.5,-2.5) {$3$};
|
|
\node at (4.5,-3.5) {$2$};
|
|
\node at (4.5,-4.5) {$1$};
|
|
\node at (4.5,-5.5) {$2$};
|
|
\node at (5.5,-1.5) {$4$};
|
|
\node at (5.5,-2.5) {$4$};
|
|
\node at (5.5,-3.5) {$3$};
|
|
\node at (5.5,-4.5) {$2$};
|
|
\node at (5.5,-5.5) {$2$};
|
|
\node at (6.5,-1.5) {$5$};
|
|
\node at (6.5,-2.5) {$5$};
|
|
\node at (6.5,-3.5) {$4$};
|
|
\node at (6.5,-4.5) {$3$};
|
|
\node at (6.5,-5.5) {$2$};
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The lower-right corner of the table
|
|
indicates that the edit distance between
|
|
\texttt{LOVE} and \texttt{MOVIE} is 2.
|
|
The table also shows how to construct
|
|
the shortest sequence of editing operations.
|
|
In this case the path is as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\begin{scope}
|
|
\draw (1, -1) grid (7, -6);
|
|
|
|
\node at (0.5,-2.5) {\texttt{L}};
|
|
\node at (0.5,-3.5) {\texttt{O}};
|
|
\node at (0.5,-4.5) {\texttt{V}};
|
|
\node at (0.5,-5.5) {\texttt{E}};
|
|
|
|
\node at (2.5,-0.5) {\texttt{M}};
|
|
\node at (3.5,-0.5) {\texttt{O}};
|
|
\node at (4.5,-0.5) {\texttt{V}};
|
|
\node at (5.5,-0.5) {\texttt{I}};
|
|
\node at (6.5,-0.5) {\texttt{E}};
|
|
|
|
\node at (1.5,-1.5) {$0$};
|
|
\node at (1.5,-2.5) {$1$};
|
|
\node at (1.5,-3.5) {$2$};
|
|
\node at (1.5,-4.5) {$3$};
|
|
\node at (1.5,-5.5) {$4$};
|
|
\node at (2.5,-1.5) {$1$};
|
|
\node at (2.5,-2.5) {$1$};
|
|
\node at (2.5,-3.5) {$2$};
|
|
\node at (2.5,-4.5) {$3$};
|
|
\node at (2.5,-5.5) {$4$};
|
|
\node at (3.5,-1.5) {$2$};
|
|
\node at (3.5,-2.5) {$2$};
|
|
\node at (3.5,-3.5) {$1$};
|
|
\node at (3.5,-4.5) {$2$};
|
|
\node at (3.5,-5.5) {$3$};
|
|
\node at (4.5,-1.5) {$3$};
|
|
\node at (4.5,-2.5) {$3$};
|
|
\node at (4.5,-3.5) {$2$};
|
|
\node at (4.5,-4.5) {$1$};
|
|
\node at (4.5,-5.5) {$2$};
|
|
\node at (5.5,-1.5) {$4$};
|
|
\node at (5.5,-2.5) {$4$};
|
|
\node at (5.5,-3.5) {$3$};
|
|
\node at (5.5,-4.5) {$2$};
|
|
\node at (5.5,-5.5) {$2$};
|
|
\node at (6.5,-1.5) {$5$};
|
|
\node at (6.5,-2.5) {$5$};
|
|
\node at (6.5,-3.5) {$4$};
|
|
\node at (6.5,-4.5) {$3$};
|
|
\node at (6.5,-5.5) {$2$};
|
|
|
|
\path[draw=red,thick,-,line width=2pt] (6.5,-5.5) -- (5.5,-4.5);
|
|
\path[draw=red,thick,-,line width=2pt] (5.5,-4.5) -- (4.5,-4.5);
|
|
\path[draw=red,thick,->,line width=2pt] (4.5,-4.5) -- (1.5,-1.5);
|
|
\end{scope}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The last characters of \texttt{LOVE} and \texttt{MOVIE}
|
|
are equal, so the edit distance between them
|
|
equals the edit distance between \texttt{LOV} and \texttt{MOVI}.
|
|
We can use one editing operation to remove the
|
|
character \texttt{I} from \texttt{MOVI}.
|
|
Thus, the edit distance is one larger than
|
|
the edit distance between \texttt{LOV} and \texttt{MOV}, etc.
|
|
|
|
\section{Counting tilings}
|
|
|
|
Sometimes the dynamic programming state
|
|
is more complex than a fixed combination of numbers.
|
|
As an example,
|
|
we consider a problem where our task
|
|
is to calculate the number of different ways to
|
|
fill an $n \times m$ grid using
|
|
$1 \times 2$ and $2 \times 1$ size tiles.
|
|
For example, one valid solution
|
|
for the $4 \times 7$ grid is
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=.65]
|
|
\draw (0,0) grid (7,4);
|
|
\draw[fill=gray] (0+0.2,0+0.2) rectangle (2-0.2,1-0.2);
|
|
\draw[fill=gray] (2+0.2,0+0.2) rectangle (4-0.2,1-0.2);
|
|
\draw[fill=gray] (4+0.2,0+0.2) rectangle (6-0.2,1-0.2);
|
|
\draw[fill=gray] (0+0.2,1+0.2) rectangle (2-0.2,2-0.2);
|
|
\draw[fill=gray] (2+0.2,1+0.2) rectangle (4-0.2,2-0.2);
|
|
\draw[fill=gray] (1+0.2,2+0.2) rectangle (3-0.2,3-0.2);
|
|
\draw[fill=gray] (1+0.2,3+0.2) rectangle (3-0.2,4-0.2);
|
|
\draw[fill=gray] (4+0.2,3+0.2) rectangle (6-0.2,4-0.2);
|
|
|
|
\draw[fill=gray] (0+0.2,2+0.2) rectangle (1-0.2,4-0.2);
|
|
\draw[fill=gray] (3+0.2,2+0.2) rectangle (4-0.2,4-0.2);
|
|
\draw[fill=gray] (6+0.2,2+0.2) rectangle (7-0.2,4-0.2);
|
|
\draw[fill=gray] (4+0.2,1+0.2) rectangle (5-0.2,3-0.2);
|
|
\draw[fill=gray] (5+0.2,1+0.2) rectangle (6-0.2,3-0.2);
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\draw[fill=gray] (6+0.2,0+0.2) rectangle (7-0.2,2-0.2);
|
|
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|
\end{tikzpicture}
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\end{center}
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and the total number of solutions is 781.
|
|
|
|
The problem can be solved using dynamic programming
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|
by going through the grid row by row.
|
|
Each row in a solution can be represented as a
|
|
string that contains $m$ characters from the set
|
|
$\{\sqcap, \sqcup, \sqsubset, \sqsupset \}$.
|
|
For example, the above solution consists of four rows
|
|
that correspond to the following strings:
|
|
\begin{itemize}
|
|
\item
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|
$\sqcap \sqsubset \sqsupset \sqcap \sqsubset \sqsupset \sqcap$
|
|
\item
|
|
$\sqcup \sqsubset \sqsupset \sqcup \sqcap \sqcap \sqcup$
|
|
\item
|
|
$\sqsubset \sqsupset \sqsubset \sqsupset \sqcup \sqcup \sqcap$
|
|
\item
|
|
$\sqsubset \sqsupset \sqsubset \sqsupset \sqsubset \sqsupset \sqcup$
|
|
\end{itemize}
|
|
|
|
Let $f(k,x)$ denote the number of ways to
|
|
construct a solution for the rows $1 \ldots k$
|
|
in the grid so that string $x$ corresponds to row $k$.
|
|
It is possible to use dynamic programing here
|
|
because the state of a row is constrained
|
|
only be the state of the previous row.
|
|
|
|
A solution is valid if row $1$ doesn't contain
|
|
the character $\sqcup$,
|
|
row $n$ doesn't contain the character $\sqcap$,
|
|
and all successive rows are \emph{compatible}.
|
|
For example, the rows
|
|
$\sqcup \sqsubset \sqsupset \sqcup \sqcap \sqcap \sqcup$ and
|
|
$\sqsubset \sqsupset \sqsubset \sqsupset \sqcup \sqcup \sqcap$
|
|
are compatible, while the rows
|
|
$\sqcap \sqsubset \sqsupset \sqcap \sqsubset \sqsupset \sqcap$ and
|
|
$\sqsubset \sqsupset \sqsubset \sqsupset \sqsubset \sqsupset \sqcup$
|
|
are not compatible.
|
|
|
|
Since a row consists of $m$ characters and there are
|
|
four choices for each character, the number of different
|
|
rows is at most $4^m$.
|
|
Thus, the time complexity of the solution is
|
|
$O(n 4^{2m})$ because we can check the
|
|
$O(4^m)$ possible states for each row,
|
|
and for each state, there are $O(4^m)$
|
|
possible states for the previous row.
|
|
In practice, it's a good idea to rotate the grid
|
|
so that the shorter side has length $m$
|
|
because the factor $4^{2m}$ dominates the time complexity.
|
|
|
|
It is possible to make the solution more efficient
|
|
by using a better representation for the rows as strings.
|
|
It turns out that it is sufficient to know the
|
|
columns of the previous row that contain the first square
|
|
of a vertical tile.
|
|
Thus, we can represent a row using only characters
|
|
$\sqcap$ and $\Box$ where $\Box$ is a combination
|
|
of characters
|
|
$\sqcup$, $\sqsubset$ and $\sqsupset$.
|
|
In this case, there are only
|
|
$2^m$ distinct rows and the time complexity becomes
|
|
$O(n 2^{2m})$.
|
|
|
|
As a final note, there is also a surprising direct formula
|
|
for calculating the number of tilings:
|
|
\[ \prod_{a=1}^{\lceil n/2 \rceil} \prod_{b=1}^{\lceil m/2 \rceil} 4 \cdot (\cos^2 \frac{\pi a}{n + 1} + \cos^2 \frac{\pi b}{m+1}).\]
|
|
This formula is very efficient because it calculates
|
|
the number of tilings on $O(nm)$ time,
|
|
but since the answer is a product of real numbers,
|
|
a practical problem in using the formula is
|
|
how to store the intermediate results accurately.
|
|
|
|
|