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Some fixes

This commit is contained in:
Antti H S Laaksonen 2017-05-25 23:23:54 +03:00
parent 43312caa92
commit d92fc7ea20
1 changed files with 16 additions and 19 deletions
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35
chapter05.tex
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@ -56,7 +56,7 @@ void search(int k) {
When the function \texttt{search}
is called with parameter $k$,
it decides whether to include
it decides whether to include the
element $k$ in the subset or not,
and in both cases,
then calls itself with parameter $k+1$
@ -137,7 +137,7 @@ the last bit corresponds to element 0,
the second last bit corresponds to element 1,
and so on.
For example, the bit representation of 25
is 11001, that corresponds to the subset $\{0,3,4\}$.
is 11001, which corresponds to the subset $\{0,3,4\}$.
The following code goes through the subsets
of a set of $n$ elements
@ -183,7 +183,7 @@ using recursion.
The following function \texttt{search} goes
through the permutations of the set $\{0,1,\ldots,n-1\}$.
The function builds a vector \texttt{permutation}
that describes the permutation,
that contains the permutation,
and the search begins when the function is
called without parameters.
@ -349,10 +349,10 @@ void search(int y) {
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;
if (column[x] || diag1[x+y] || diag2[x-y+n-1]) continue;
column[x] = diag1[x+y] = diag2[x-y+n-1] = 1;
search(y+1);
r1[x] = r2[x+y] = r3[x-y+n-1] = 0;
column[x] = diag1[x+y] = diag2[x-y+n-1] = 0;
}
}
\end{lstlisting}
@ -368,13 +368,13 @@ When the function \texttt{search} is
called with parameter $y$,
it places a queen to row $y$
and then calls itself with parameter $y+1$.
However, if $y=n$, a solution has been found
Then, if $y=n$, a solution has been found
and the variable \texttt{count} is increased by one.
The array \texttt{r1} keeps track of the columns
that already contain a queen,
and the arrays \texttt{r2} and \texttt{r3}
keep track of the diagonals.
The array \texttt{column} keeps track of columns
that contain a queen,
and the arrays \texttt{diag1} and \texttt{diag2}
keep track of diagonals.
It is not allowed to add another queen to a
column or diagonal that already contains a queen.
For example, the columns and diagonals of
@ -437,9 +437,9 @@ the $4 \times 4$ board are numbered as follows:
\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}};
\node at (-4,-1) {\texttt{column}};
\node at (2,-1) {\texttt{diag1}};
\node at (8,-1) {\texttt{diag2}};
\end{scope}
\end{tikzpicture}
@ -619,9 +619,6 @@ the path can turn either left or right:
(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);
\node at (4.5,6.5) {$a$};
\node at (6.5,6.5) {$b$};
\end{scope}
\end{tikzpicture}
\end{center}
@ -716,10 +713,10 @@ to choose some numbers from the list so that
their sum is $x$.
For example, given the list $[2,4,5,9]$ and $x=15$,
we can choose the numbers $[2,4,9]$ to get $2+4+9=15$.
However, if $x=10$,
However, if $x=10$ for the same list,
it is not possible to form the sum.
An easy solution to the problem is to
A simple algorithm to the problem is to
go through all subsets of the elements and
check if the sum of any of the subsets is $x$.
The running time of such an algorithm is $O(2^n)$,