Fix indices and grammar

This commit is contained in:
Antti H S Laaksonen 2017-04-18 20:22:56 +03:00
parent f551b3b422
commit a8e456ce15
1 changed files with 15 additions and 13 deletions

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@ -122,7 +122,7 @@ Thus, the recursive formula is
where the function $\min$ gives the smallest
of its parameters.
In the general case, for the coin set
$\{c_0,c_1,\ldots,c_{k-1}\}$,
$\{c_1,c_2,\ldots,c_k\}$,
the recursive formula is
\[f(x) = \min(f(x-c_1),f(x-c_2),\ldots,f(x-c_k))+1.\]
The base case for the function is
@ -392,6 +392,7 @@ possibilities of dynamic programming.
\index{longest increasing subsequence}
Let us consider the following problem:
Given an array that contains $n$
numbers,
our task is to find the
@ -399,7 +400,7 @@ our task is to find the
of the array.
This is a sequence of array elements
that goes from left to right,
and each element in the sequence is larger
and each element of the sequence is larger
than the previous element.
For example, in the array
@ -487,23 +488,24 @@ that ends at position $k$ is constructed:
\begin{enumerate}
\item The subsequence
only contains the element at position $k$. In this case $f(k)=1$.
\item The subsequence is constructed
by adding the element at position $k$ to
a subsequence that ends at position $i$
where $i<k$ and the element at position $i$
is smaller than the element at position $k$. In this case $f(k)=f(i)+1$.
\item The subsequence contains a subsequence
that ends at position $i$ where $i<k$,
followed by the element at position $k$.
The element at position $i$ must be smaller
than the element at position $k$.
In this case $f(k)=f(i)+1$.
\end{enumerate}
For example, in the above example $f(7)=4$,
For example, in the above example $f(6)=4$,
because the subsequence $[2,5,7]$ of length 3
ends at position 5, and by adding the element
at position 7 to this subsequence,
ends at position 4, and by adding the element
at position 6 to this subsequence,
we get the optimal subsequence $[2,5,7,8]$ of length 4.
An easy way to calculate the
value of $f(k)$ is to
go through all previous values
$f(1),f(2),\ldots,f(k-1)$ and select the best solution.
$f(0),f(1),\ldots,f(k-1)$ and select the best solution.
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. Can you see how?
@ -517,7 +519,7 @@ the lower-right corner such that
we only move down and right.
Each square contains a number,
and the path should be constructed so
that the sum of numbers along
that the sum of the numbers along
the path is as large as possible.
The following picture shows an optimal
@ -563,7 +565,7 @@ path in a grid:
\end{scope}
\end{tikzpicture}
\end{center}
The sum of numbers on the path is 67,
The sum of the numbers on the path is 67,
and this is the largest possible sum on a path
from the
upper-left corner to the lower-right corner.