Improve language
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Antti H S Laaksonen
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chapter07.tex
211
chapter07.tex
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@ -348,7 +348,6 @@ to form an empty sum.
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Otherwise we calculate the sum of all values
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of the form $\texttt{solve}(x-c)$ where $c$ is in \texttt{coins}.
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The following code constructs an array
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$\texttt{count}$ such that
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$\texttt{count}[x]$ equals
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@ -393,15 +392,13 @@ possibilities of dynamic programming.
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\index{longest increasing subsequence}
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Let us consider the following problem:
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Given an array that contains $n$
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numbers,
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our task is to find the
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Our first problem is to find the
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\key{longest increasing subsequence}
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of the array.
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This is a sequence of array elements
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in an array \texttt{t} of $n$ elements.
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This is a maximum-length
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sequence of array elements
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that goes from left to right,
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and each element of the sequence is larger
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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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@ -462,65 +459,81 @@ contains 4 elements:
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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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Let $\texttt{length}(k)$ denote
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the length of the
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longest increasing subsequence
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that ends at position $k$.
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Using this function, the answer to the problem
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is the largest of the values
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$f(0),f(1),\ldots,f(n-1)$.
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For example, in the above array
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the values of the function are as follows:
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Thus, if we calculate all values of
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$\texttt{length}(k)$ where $0 \le k \le n-1$,
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we will find out the length of the
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longest increasing subsequence.
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For example, the values of the function
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for the above array are as follows:
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\[
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\begin{array}{lcl}
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f(0) & = & 1 \\
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f(1) & = & 1 \\
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f(2) & = & 2 \\
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f(3) & = & 1 \\
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f(4) & = & 3 \\
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f(5) & = & 2 \\
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f(6) & = & 4 \\
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f(7) & = & 2 \\
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\texttt{length}(0) & = & 1 \\
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\texttt{length}(1) & = & 1 \\
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\texttt{length}(2) & = & 2 \\
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\texttt{length}(3) & = & 1 \\
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\texttt{length}(4) & = & 3 \\
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\texttt{length}(5) & = & 2 \\
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\texttt{length}(6) & = & 4 \\
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\texttt{length}(7) & = & 2 \\
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\end{array}
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\]
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When calculating the value of $f(k)$,
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there are two possibilities how the subsequence
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that ends at position $k$ is constructed:
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\begin{enumerate}
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\item The subsequence
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only contains the element at position $k$. In this case $f(k)=1$.
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\item The subsequence contains a subsequence
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that ends at position $i$ where $i<k$,
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followed by the element at position $k$.
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The element at position $i$ must be smaller
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than the element at position $k$.
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In this case $f(k)=f(i)+1$.
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\end{enumerate}
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For example, $\texttt{length}(6)=4$,
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because the longest increasing subsequence
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that ends at position 6 consists of 4 elements.
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For example, in the above example $f(6)=4$,
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because the subsequence $[2,5,7]$ of length 3
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ends at position 4, and by adding the element
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at position 6 to this subsequence,
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we get the optimal subsequence $[2,5,7,8]$ of length 4.
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To calculate a value of $\texttt{length}(k)$,
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we should find a position $i<k$
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for which $\texttt{t}[i]<\texttt{t}[k]$
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and $\texttt{length}(i)$ is as large as possible.
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Then we know that
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$\texttt{length}(k)=\texttt{length}(i)+1$,
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because this is an optimal way to add
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$\texttt{t}[k]$ to a subsequence.
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However, if there is no such position $i$,
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then $\texttt{length}(k)=1$,
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which means that the subsequence only contains
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$\texttt{t}[k]$.
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An easy way to calculate the
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value of $f(k)$ is to
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go through all previous values
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$f(0),f(1),\ldots,f(k-1)$ and select the best solution.
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The time complexity of such an algorithm is $O(n^2)$.
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Surprisingly, it is also possible to solve the
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problem in $O(n \log n)$ time. Can you see how?
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Since all values of the function can be calculated
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from its smaller values,
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we can use dynamic programming.
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In the following code, the values
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of the function will be stored in an array
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$\texttt{length}$.
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\begin{lstlisting}
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for (int k = 0; k < n; k++) {
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length[k] = 1;
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for (int i = 0; i < k; i++) {
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if (t[i] < t[k]) {
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length[k] = max(length[k],length[i]+1);
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}
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}
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}
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\end{lstlisting}
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This code works in $O(n^2)$ time,
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because it consists of two nested loops.
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However, it is also possible to implement
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the dynamic programming calculation
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more efficiently in $O(n \log n)$ time.
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Can you find a way to do this?
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\section{Paths in a grid}
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Our next problem is to find a path
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in an $n \times n$ grid
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from the upper-left corner to
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the lower-right corner such that
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the lower-right corner
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of an $n \times n$ grid, such that
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we only move down and right.
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Each square contains a number,
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Each square contains a value,
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and the path should be constructed so
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that the sum of the numbers along
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that the sum of the values along
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the path is as large as possible.
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The following picture shows an optimal
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@ -566,19 +579,29 @@ path in a grid:
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\end{scope}
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\end{tikzpicture}
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\end{center}
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The sum of the numbers on the path is 67,
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The sum of the values on the path is 67,
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and this is the largest possible sum on a path
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from the
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upper-left corner to the lower-right corner.
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We can approach the problem by
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calculating for each square $(y,x)$
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the maximum sum on a path
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from the upper-left corner to square $(y,x)$.
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Let $f(y,x)$ denote this sum,
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so $f(n,n)$ is the maximum sum on a path
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Assume that the rows and columns of the
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grid are numbered from 1 to $n$,
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and $\texttt{value}[y][x]$ equals the value
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of square $(y,x)$.
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Let $\texttt{sum}(y,x)$ denote the maximum
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sum on a path from the upper-left corner
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to square $(y,x)$.
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Now $\texttt{sum}(n,n)$ tells us
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the maximum sum
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from the upper-left corner to
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the lower-right corner.
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For example, in the above grid,
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$\texttt{sum}(5,5)=67$.
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We can recursively calculate the sums
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as follows:
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\[ \texttt{sum}(y,x) = \max(\texttt{sum}(y,x-1),\texttt{sum}(y-1,x))+\texttt{value}[y][x]\]
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The recursive formula is based on the observation
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that a path that ends at square $(y,x)$
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@ -597,27 +620,31 @@ or square $(y-1,x)$:
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\end{tikzpicture}
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\end{center}
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Let $r(y,x)$ denote the number in square $(y,x)$.
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The base cases for the recursive function
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are as follows:
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Thus, we select the direction that maximizes
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the sum.
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We assume that $\texttt{sum}(y,x)=0$
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if $y=0$ or $x=0$ (because no such paths exist),
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so the recursive formula also works when $y=1$ or $x=1$.
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\[
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\begin{array}{lcl}
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f(1,1) & = & r(1,1) \\
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f(1,x) & = & f(1,x-1)+r(1,x) \\
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f(y,1) & = & f(y-1,1)+r(y,1)\\
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\end{array}
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\]
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Since the function \texttt{sum} has two parameters,
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the dynamic programming array also has two dimensions.
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For example, we can use an array
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\begin{lstlisting}
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int sum[N][N];
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\end{lstlisting}
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and calculate the sums as follows:
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\begin{lstlisting}
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for (int y = 1; y <= n; y++) {
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for (int x = 1; x <= n; x++) {
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sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x];
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}
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}
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\end{lstlisting}
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In the general case there are two
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possible paths, and we should select the path
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that produces the larger sum:
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\[ f(y,x) = \max(f(y,x-1),f(y-1,x))+r(y,x)\]
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The time complexity of the solution is $O(n^2)$,
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because each value $f(y,x)$ can be calculated
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in constant time using the values of the
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adjacent squares.
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Note that it is not needed to separately handle the
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cases where $y=1$ or $x=1$, because we use a
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one-indexed array whose values are initially zeros.
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The time complexity of the algorithm is $O(n^2)$.
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\section{Knapsack}
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@ -625,10 +652,10 @@ adjacent squares.
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\key{Knapsack} is a classic problem where we
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are given $n$ objects with weights
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$p_1,p_2,\ldots,p_n$ and values
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$a_1,a_2,\ldots,a_n$.
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$w_1,w_2,\ldots,w_n$ and values
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$v_1,v_2,\ldots,v_n$.
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Our task is to choose a subset of the objects
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such that the sum of the weights is at most $x$
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such that the sum of the weights is at most $W$
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and the sum of the values is as large as possible.
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\begin{samepage}
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@ -644,26 +671,26 @@ D & 5 & 3 \\
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\end{tabular}
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\end{center}
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\end{samepage}
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and the maximum allowed total weight is 12,
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and $W=12$,
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an optimal solution is to select objects $B$ and $D$.
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Their total weight $6+5=11$ does not exceed 12,
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and their total value $3+3=6$ is the largest possible.
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In this case, the sum of weights is
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$6+5=11 \le 12$, and the sum of values is $3+3=6$.
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This task can be solved in two different ways
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This problem can be solved in two different ways
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using dynamic programming.
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We can either regard the problem as maximizing the
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total value of the objects or
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minimizing the total weight of the objects.
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as minimizing the total weight of the objects.
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\subsubsection{Solution 1}
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\textit{Maximization:} Let $f(k,u)$
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denote the largest possible total value
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when a subset of objects $1 \ldots k$ is selected
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such that the total weight is $u$.
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The solution to the problem is
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the largest value
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$f(n,u)$ where $0 \le u \le x$.
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\textit{Maximization:} Let $\texttt{value}(k,u)$
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denote the maximum value
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of a subset of objects $1 \ldots k$
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whose weight is $u$.
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Thus, the solution to the problem is
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\[\max_{0 \le u \le W} \texttt{value}(n,u).\]
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A recursive formula for calculating
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the function is
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\[f(k,u) = \max(f(k-1,u),f(k-1,u-p_k)+a_k),\]
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