\chapter{Bit manipulation}

A computer internally manipulates data
as bits, i.e., as numbers 0 and 1.
In this chapter, we will learn how integers
are represented as bits, and how bit operations
can be used for manipulating them.
It turns out that there are many uses for
bit operations in the implementation of algorithms.

\section{Bit representation}

\index{bit representation}

The \key{bit representation} of a number
indicates which powers of two form the number.
For example, the bit representation of the number 43
is 101011 because
$43 = 2^5 + 2^3 + 2^1 + 2^0$ where
bits 0, 1, 3 and 5 from the right are ones,
and all other bits are zeros.

The length of a bit representation of a number
in a computer is static, and depends on the
data type chosen.
For example, the \texttt{int} type in C++ is
usually a 32-bit type, and an \texttt{int} number
consists of 32 bits.
In this case, the bit representation of 43
as an \texttt{int} number is as follows:

\[00000000000000000000000000101011\]

The bit representation of a number is either
\key{signed} or \key{unsigned}.
The first bit of a signed number is the sign
($+$ or $-$), and we can represent numbers
$-2^{n-1} \ldots 2^{n-1}-1$ using $n$ bits.
In an unsigned number, in turn,
all bits belong to the number and we
can represent numbers $0 \ldots 2^n-1$ using $n$ bits.

In an signed bit representation,
the first bit of a nonnegative number is 0,
and the first bit of a negative number is 1.
\key{Two's complement} is used which means that
the opposite number of a number can be calculated
by first inversing all the bits in the number,
and then increasing the number by one.

For example, the representation of $-43$
as an \texttt{int} number is as follows:

\[11111111111111111111111111010101\]

The connection between signed and unsigned numbers
is that the representations of a signed
number $-x$ and an unsigned number $2^n-x$
are equal.
Thus, the above representation corresponds to
the unsigned number $2^{32}-43$.

In C++, the numbers are signed as default,
but we can create unsigned numbers by
using the keyword \texttt{unsigned}.
For example, in the code
\begin{lstlisting}
int x = -43;
unsigned int y = x;
cout << x << "\n"; // -43
cout << y << "\n"; // 4294967253
\end{lstlisting}
the signed number
$x=-43$ becomes the unsigned number $y=2^{32}-43$.

If a number becomes too large or too small for the
bit representation chosen, it will overflow.
In practice, in a signed representation,
the next number after $2^{n-1}-1$ is $-2^{n-1}$,
and in an unsigned representation,
the next number after $2^{n-1}$ is $0$.
For example, in the code
\begin{lstlisting}
int x = 2147483647
cout << x << "\n"; // 2147483647
x++;
cout << x << "\n"; // -2147483648
\end{lstlisting}
we increase $2^{31}-1$ by one to get $-2^{31}$.

\section{Bit operations}

\newcommand\XOR{\mathbin{\char`\^}}

\subsubsection{And operation}

\index{and operation}

The \key{and} operation $x$ \& $y$ produces a number
that has bit 1 in positions where both the numbers
$x$ and $y$ have bit 1.
For example, $22$ \& $26$ = 18 because

\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
\& & 11010 & (26) \\
\hline
 = & 10010 & (18) \\
\end{tabular}
\end{center}

Using the and operation, we can check if a number
$x$ is even because
$x$ \& $1$ = 0 if $x$ is even, and
$x$ \& $1$ = 1 if $x$ is odd.

\subsubsection{Or operation}

\index{or operation}

The \key{or} operation $x$ | $y$ produces a number
that has bit 1 in positions where at least one
of the numbers
$x$ and $y$ have bit 1.
For example, $22$ | $26$ = 30 because

\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
| & 11010 & (26) \\
\hline
 = & 11110 & (30) \\
\end{tabular}
\end{center}

\subsubsection{Xor operation}

\index{xor operation}

The \key{xor} operation $x$ $\XOR$ $y$ produces a number
that has bit 1 in positions where exactly one
of the numbers
$x$ and $y$ have bit 1.
For example, $22$ $\XOR$ $26$ = 12 because

\begin{center}
\begin{tabular}{rrr}
& 10110 & (22)\\
$\XOR$ & 11010 & (26) \\
\hline
 = & 01100 & (12) \\
\end{tabular}
\end{center}

\subsubsection{Not operation}

\index{not operation}

The \key{not} operation \textasciitilde$x$
produces a number where all the bits of $x$
have been inversed.
The formula \textasciitilde$x = -x-1$ holds,
for example, \textasciitilde$29 = -30$.

The result of the not operation at the bit level
depends on the length of the bit representation
because the operation changes all bits.
For example, if the numbers are 32-bit
\texttt{int} numbers, the result is as follows:

\begin{center}
\begin{tabular}{rrrr}
$x$ & = & 29 &   00000000000000000000000000011101 \\
\textasciitilde$x$ & = & $-30$ & 11111111111111111111111111100010 \\
\end{tabular}
\end{center}

\subsubsection{Bit shifts}

\index{bit shift}

The left bit shift $x < < k$ produces a number
where the bits of $x$ have been moved $k$ steps to
the left by adding $k$ zero bits to the number.
The right bit shift $x > > k$ produces a number
where the bits of $x$ have been moved $k$ steps
to the right by removing $k$ last bits from the number.

For example, $14 < < 2 = 56$
because $14$ equals 1110,
and it becomes $56$ that equals 111000.
Correspondingly, $49 > > 3 = 6$
because $49$ equals 110001,
and it becomes $6$ that equals 110.

Note that the left bit shift $x < < k$
corresponds to multiplying $x$ by $2^k$,
and the right bit shift $x > > k$
corresponds to dividing $x$ by $2^k$
rounding downwards.

\subsubsection{Bit manipulation}

The bits in a number are indexed from the right
to the left beginning from zero.
A number of the form $1 < < k$ contains a one bit
in position $k$, and all other bits are zero,
so we can manipulate single bits of numbers
using these numbers.

The $k$th bit in $x$ is one if
$x$ \& $(1 < < k) = (1 < < k)$.
The formula $x$ | $(1 < < k)$
sets the $k$th bit of $x$ to one,
the formula
$x$ \& \textasciitilde $(1 < < k)$
sets the $k$th bit of $x$ to zero,
and the formula
$x$ $\XOR$ $(1 < < k)$
inverses the $k$th bit of $x$.

The formula $x$ \& $(x-1)$ sets the last
one bit of $x$ to zero,
and the formula $x$ \& $-x$ sets all the
one bits to zero, except for the last one bit.
The formula $x$ | $(x-1)$, in turn,
inverses all the bits after the last one bit.

Also note that a positive number $x$ is
of the form $2^k$ if $x$ \& $(x-1) = 0$.

\subsubsection*{Additional functions}

The g++ compiler contains the following
functions for bit manipulation:

\begin{itemize}
\item
$\texttt{\_\_builtin\_clz}(x)$:
the number of zeros at the beginning of the number
\item
$\texttt{\_\_builtin\_ctz}(x)$:
the number of zeros at the end of the number
\item
$\texttt{\_\_builtin\_popcount}(x)$:
the number of ones in the number
\item
$\texttt{\_\_builtin\_parity}(x)$:
the parity (even or odd) of the number of ones
\end{itemize}
\begin{samepage}

The following code shows how to use the functions:
\begin{lstlisting}
int x = 5328; // 00000000000000000001010011010000
cout << __builtin_clz(x) << "\n"; // 19
cout << __builtin_ctz(x) << "\n"; // 4
cout << __builtin_popcount(x) << "\n"; // 5
cout << __builtin_parity(x) << "\n"; // 1
\end{lstlisting}
\end{samepage}

The functions support \texttt{int} numbers,
but there are also \texttt{long long} versions
of the functions
available with the prefix \texttt{ll}.

\section{Bit representation of sets}

Each subset of a set $\{0,1,2,\ldots,n-1\}$
corresponds to a $n$ bit number
where the one bits indicate which elements
are included in the subset.
For example, the bit representation for $\{1,3,4,8\}$
is 100011010 that equals $2^8+2^4+2^3+2^1=282$.

The bit representation of a set uses little memory
because only one bit is needed for the information
whether an element belongs to the set.
In addition, we can efficiently manipulate sets
that are stored as bits.

\subsubsection{Set operations}

In the following code, the variable $x$
contains a subset of $\{0,1,2,\ldots,31\}$.
The code adds elements 1, 3, 4 and 8
to the set and then prints the elements in the set.

\begin{lstlisting}
// x is an empty set
int x = 0;
// add numbers 1, 3, 4 and 8 to the set
x |= (1<<1);
x |= (1<<3);
x |= (1<<4);
x |= (1<<8);
// print the elements in the set
for (int i = 0; i < 32; i++) {
    if (x&(1<<i)) cout << i << " ";
}
cout << "\n";
\end{lstlisting}

The output of the code is as follows:
\begin{lstlisting}
1 3 4 8
\end{lstlisting}

Using the bit representation of a set,
we can efficiently implement set operations
using bit operations:
\begin{itemize}
\item $a$ \& $b$ is the intersection $a \cap b$ of $a$ and $b$
(this contains the elements that are in both the sets)
\item $a$ | $b$ is the union $a \cup b$ of $a$ and $b$
(this contains the elements that are at least
in one of the sets)
\item $a$ \& (\textasciitilde$b$) is the difference
$a \setminus b$ of $a$ and $b$
(this contains the elements that are in $a$
but not in $b$)
\end{itemize}

The following code constructs the union
of $\{1,3,4,8\}$ and $\{3,6,8,9\}$:

\begin{lstlisting}
// set {1,3,4,8}
int x = (1<<1)+(1<<3)+(1<<4)+(1<<8);
// set {3,6,8,9}
int y = (1<<3)+(1<<6)+(1<<8)+(1<<9);
// union of the sets
int z = x|y;
// print the elements in the union
for (int i = 0; i < 32; i++) {
    if (z&(1<<i)) cout << i << " ";
}
cout << "\n";
\end{lstlisting}

The output of the code is as follows:
\begin{lstlisting}
1 3 4 6 8 9
\end{lstlisting}

\subsubsection{Iterating through subsets}

The following code iterates through
the subsets of $\{0,1,\ldots,n-1\}$:

\begin{lstlisting}
for (int b = 0; b < (1<<n); b++) {
    // process subset b
}
\end{lstlisting}
The following code goes through
subsets with exactly $k$ elements:
\begin{lstlisting}
for (int b = 0; b < (1<<n); b++) {
    if (__builtin_popcount(b) == k) {
        // process subset b
    }
}
\end{lstlisting}
The following code goes through the subsets
of a set $x$:
\begin{lstlisting}
int b = 0;
do {
    // process subset b
} while (b=(b-x)&x);
\end{lstlisting}
% Esimerkiksi jos $x$ esittää joukkoa $\{2,5,7\}$,
% niin koodi käy läpi osajoukot
% $\emptyset$, $\{2\}$, $\{5\}$, $\{7\}$,
% $\{2,5\}$, $\{2,7\}$, $\{5,7\}$ ja $\{2,5,7\}$.

\section{Dynaaminen ohjelmointi}

\subsubsection{Permutaatioista osajoukoiksi}

Dynaamisen ohjelmoinnin avulla on usein mahdollista
muuttaa permutaatioiden läpikäynti osajoukkojen läpikäynniksi.
Tällöin dynaamisen ohjelmoinnin tilana on
joukon osajoukko sekä mahdollisesti muuta tietoa.

Tekniikan hyötynä on,
että $n$-alkioisen joukon permutaatioiden määrä $n!$
on selvästi suurempi kuin osajoukkojen määrä $2^n$.
Esimerkiksi jos $n=20$, niin $n!=2432902008176640000$,
kun taas $2^n=1048576$.
Niinpä tietyillä $n$:n arvoilla permutaatioita ei ehdi
käydä läpi mutta osajoukot ehtii käydä läpi.

Lasketaan esimerkkinä, monessako
joukon $\{0,1,\ldots,n-1\}$
permutaatiossa ei ole 
missään kohdassa kahta peräkkäistä lukua.
Esimerkiksi tapauksessa $n=4$ ratkaisuja on kaksi:
\begin{itemize}
\item $(1,3,0,2)$
\item $(2,0,3,1)$
\end{itemize}

Merkitään $f(x,k)$:llä,
monellako tavalla osajoukon
$x$ luvut voi järjestää niin,
että viimeinen luku on $k$ ja missään kohdassa
ei ole kahta peräkkäistä lukua.
Esimerkiksi $f(\{0,1,3\},1)=1$,
koska voidaan muodostaa permutaatio $(0,3,1)$,
ja $f(\{0,1,3\},3)=0$, koska 0 ja 1 eivät
voi olla peräkkäin alussa.

Funktion $f$ avulla ratkaisu tehtävään
on summa

\[ \sum_{i=0}^{n-1} f(\{0,1,\ldots,n-1\},i). \]

\noindent
Dynaamisen ohjelmoinnin tilat voi
tallentaa seuraavasti:

\begin{lstlisting}
long long d[1<<n][n];
\end{lstlisting}

\noindent
Perustapauksena $f(\{k\},k)=1$ kaikilla $k$:n arvoilla:

\begin{lstlisting}
for (int i = 0; i < n; i++) d[1<<i][i] = 1;
\end{lstlisting}

\noindent
Tämän jälkeen muut funktion arvot
saa laskettua seuraavasti:

\begin{lstlisting}
for (int b = 0; b < (1<<n); b++) {
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (abs(i-j) > 1 && (b&(1<<i)) && (b&(1<<j))) {
                d[b][i] += d[b^(1<<i)][j];
            }
        }
    }
}
\end{lstlisting}

\noindent
Muuttujassa $b$ on osajoukon bittiesitys,
ja osajoukon luvuista muodostettu
permutaatio on muotoa $(\ldots,j,i)$.
Vaatimukset ovat, että lukujen $i$ ja $j$
etäisyyden tulee olla yli 1
ja lukujen tulee olla osajoukossa $b$.

Lopuksi ratkaisujen määrän saa laskettua näin
muuttujaan $s$:

\begin{lstlisting}
long long s = 0;
for (int i = 0; i < n; i++) {
    s += d[(1<<n)-1][i];
}
\end{lstlisting}

\subsubsection{Osajoukkojen summat}

Oletetaan sitten, että jokaista
joukon $\{0,1,\ldots,n-1\}$
osajoukkoa $x$ vastaa arvo $c(x)$ ja
tehtävänä on laskea kullekin
osajoukolle $x$ summa
\[s(x)=\sum_{y \subset x} c(y) \]
eli bittimuodossa ilmaistuna
\[s(x)=\sum_{y \& x = y} c(y). \]
Seuraavassa on esimerkki funktioiden arvoista,
kun $n=3$:
\begin{center}
\begin{tabular}{rrr}
$x$ & $c(x)$ & $s(x)$ \\
\hline
000 & 2 & 2 \\
001 & 0 & 2 \\
010 & 1 & 3 \\
011 & 3 & 6 \\
100 & 0 & 2 \\
101 & 4 & 6 \\
110 & 2 & 5 \\
111 & 0 & 12 \\
\end{tabular}
\end{center}
Esimerkiksi $s(110)=c(000)+c(010)+c(100)+c(110)=5$. 

Tehtävä on mahdollista ratkaista ajassa $O(2^n n)$
laskemalla arvoja funktiolle $f(x,k)$:
mikä on lukujen $c(y)$ summa, missä $x$:stä saa $y$:n
muuttamalla millä tahansa tavalla bittien $0,1,\ldots,k$
joukossa ykkösbittejä nollabiteiksi.
Tämän funktion avulla ilmaistuna $s(x)=f(x,n-1)$.

Funktion pohjatapaukset ovat:
\begin{equation*}
    f(x,0) = \begin{cases}
               c(x)          & \textrm{jos $x$:n bitti 0 on 0}\\
               c(x)+c(x \XOR 1) & \textrm{jos $x$:n bitti 0 on 1}\\
           \end{cases}
\end{equation*}
Suuremmille $k$:n arvoille pätee seuraava rekursio:
\begin{equation*}
    f(x,k) = \begin{cases}
               f(x,k-1)          & \textrm{jos $x$:n bitti $k$ on 0}\\
               f(x,k-1)+f(x \XOR (1 < < k),k-1)    & \textrm{jos $x$:n bitti $k$ on 1}\\
           \end{cases}
\end{equation*}

Niinpä funktion arvot voi laskea seuraavasti
dynaamisella ohjelmoinnilla.
Koodi olettaa, että taulukko \texttt{c} sisältää
funktion $c$ arvot ja muodostaa taulukon \texttt{s},
jossa on funktion $s$ arvot.
\begin{lstlisting}
for (int x = 0; x < (1<<n); x++) {
    f[x][0] = c[x];
    if (x&1) f[x][0] += c[x^1];
}
for (int k = 1; k < n; k++) {
    for (int x = 0; x < (1<<n); x++) {
        f[x][k] = f[x][k-1];
        if (b&(1<<k)) f[x][k] += f[x^(1<<k)][k-1];
    }
    if (k == n-1) s[x] = f[x][k];
}
\end{lstlisting}

Itse asiassa saman laskennan voi toteuttaa lyhyemmin
seuraavasti niin, että tulokset lasketaan
suoraan taulukkoon \texttt{s}:
\begin{lstlisting}
for (int x = 0; x < (1<<n); x++) s[x] = c[x];
for (int k = 0; k < n; k++) {
    for (int x = 0; x < (1<<n); x++) {
        if (x&(1<<k)) s[x] += s[x^(1<<k)];
    }
}
\end{lstlisting}

