cphb/luku10.tex

\chapter{Bit manipulation}

All data in a program is internally stored 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 to manipulate them.
It turns out that there are many uses for
bit operations in algorithm programming.

\section{Bit representation}

\index{bit representation}

Every nonnegative integer can be represented as a sum
\[c_k 2^k + \ldots + c_2 2^2 + c_1 2^1 + c_0 2^0,\]
where each coefficient $c_i$ is either 0 or 1,
and the bit representation of such a number is
$c_k \cdots c_2 c_1 c_0$.
For example, the number 43 corresponds to the sum
\[1 \cdot 2^5 + 0 \cdot 2^4 + 1 \cdot 2^3 + 0 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0,\]
so the bit representation of the number is 101011.

In programming, the length of the bit representation
depends on the data type chosen.
For example, in C++ the type \texttt{int} is
usually a 32-bit type and an \texttt{int} number
consists of 32 bits.
Thus, 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}.
Usually a signed representation is used,
which means that both negative and positive
numbers can be represented.
A signed number of $n$ bits can contain any
integer between $2^{n-1}$ and $2^{n-1}-1$.
For example, the \texttt{int} type in C++ is
a signed type, and it can contain any
integer between $2^{31}$ and $2^{31}-1$.

The first bit in a signed representation
is the sign of the number (0 for nonnegative numbers
and 1 for negative numbers), and
the remaining $n-1$ bits contain the value of the number.
\key{Two's complement} is used, which means that the
opposite number of a number is calculated by first
inverting all the bits in the number,
and then increasing the number by one.

For example, the bit representation of $-43$
as an \texttt{int} number is as follows:
\[11111111111111111111111111010101\]

In a signed representation, only nonnegative
numbers can be used, but the upper bound of the numbers is larger.
A signed number of $n$ bits can contain any
integer between $0$ and $2^n-1$.
For example, the \texttt{unsigned int} type in C++
can contain any integer between $0$ and $2^{32}-1$.

There is a connection between signed and unsigned
representations:
a number $-x$ in a signed representation
equals the number $2^n-x$ in an unsigned representation.
For example, the following code shows that
the signed number $x=-43$ equals the unsigned
number $y=2^{32}-43$:
\begin{lstlisting}
int x = -43;
unsigned int y = x;
cout << x << "\n"; // -43
cout << y << "\n"; // 4294967253
\end{lstlisting}

If a number is larger than the upper bound
of the bit representation, the number will overflow.
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 following code,
the next number after $2^{31}-1$ is $-2^{31}$:
\begin{lstlisting}
int x = 2147483647
cout << x << "\n"; // 2147483647
x++;
cout << x << "\n"; // -2147483648
\end{lstlisting}

\section{Bit operations}

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

\subsubsection{And operation}

\index{and operation}

The \key{and} operation $x$ \& $y$ produces a number
that has one bits in positions where both
$x$ and $y$ have one bits.
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.
More generally, $x$ is divisible by $2^k$
exactly when $x$ \& $(2^k-1)$ = 0.

\subsubsection{Or operation}

\index{or operation}

The \key{or} operation $x$ | $y$ produces a number
that has one bits in positions where at least one
of $x$ and $y$ have one bits.
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 one bits in positions where exactly one
of $x$ and $y$ have one bits.
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 inverted.
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 $x < < k$ appends $k$
zeros to the end of the number,
and the right bit shift $x > > k$
removes the $k$ last bits from the number.
For example, $14 < < 2 = 56$,
because $14$ equals 1110
and $56$ equals 111000.
Similarily, $49 > > 3 = 6$,
because $49$ equals 110001
and $6$ equals 110.

Note that $x < < k$
corresponds to multiplying $x$ by $2^k$,
and $x > > k$
corresponds to dividing $x$ by $2^k$
rounded down to an integer.

\subsubsection{Applications}

A number of the form $1 < < k$ has a one bit
in position $k$, and all other bits are zero,
so we can use such numbers to access single bits of numbers.
For example, the $k$th bit of a number is one
exactly when $x$ \& $(1 < < k)$ is not zero.
The following code prints the bit representation
of an \texttt{int} number $x$:

\begin{lstlisting}
for (int i = 31; i >= 0; i--) {
    if (x&(1<<i)) cout << "1";
    else cout << "0";
}
\end{lstlisting}

It is also possible to modify single bits
of numbers using a similar idea.
For example, the expression $x$ | $(1 < < k)$
sets the $k$th bit of $x$ to one,
the expression
$x$ \& \textasciitilde $(1 < < k)$
sets the $k$th bit of $x$ to zero,
and the expression
$x$ $\XOR$ $(1 < < k)$
inverts 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 $x$ | $(x-1)$
inverts 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 provides the following
functions for counting bits:

\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 functions can be used as follows:
\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 can be used with \texttt{int} numbers,
but there are also \texttt{long long} versions
of the functions
available with the prefix \texttt{ll}.

\section{Representing 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 of $\{1,3,4,8\}$
is 100011010 that equals $2^8+2^4+2^3+2^1=282$.

The benefit in using a bit representation is
that the information whether an element belongs
to the set requires only one bit of memory.
In addition, we can implement set operations
efficiently as bit operations.

\subsubsection{Set operations}

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

\begin{lstlisting}
// x is an empty set
int x = 0;
// add elements 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}

Set operations can be implemented as follows:
\begin{itemize}
\item $a$ \& $b$ is the intersection $a \cap b$ of $a$ and $b$
\item $a$ | $b$ is the union $a \cup b$ of $a$ and $b$
\item \textasciitilde$a$ is the complement $\bar a$ of $a$
\item $a$ \& (\textasciitilde$b$) is the difference
$a \setminus b$ of $a$ and $b$
\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}

\subsubsection{Iterating through subsets}

The following code goes 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
the 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}

\section{Dynamic programming}

\subsubsection{From permutations to subsets}

Using dynamic programming, it is often possible
to turn an iteration over permutations into
an iteration over subsets so that
the dynamic programming state
contains a subset of a set and possibly
some additional information.

The benefit in this is that
$n!$, the number of permutations of an $n$ element set,
is much larger than $2^n$, the number of subsets.
For example, if $n=20$, then
$n!=2432902008176640000$ and $2^n=1048576$.
Hence, for certain values of $n$,
we can efficiently go through subsets but not through permutations.

As an example, consider the problem of
calculating the number of
permutations of a set $\{0,1,\ldots,n-1\}$,
where the difference between any two consecutive
elements is larger than one.
For example, when $n=4$, there are two such permutations:
$(1,3,0,2)$ and $(2,0,3,1)$.

Let $f(x,k)$ denote the number of valid permutations
of $x$ where the last element is $k$ and
the difference between any two successive
elements is larger than one.
For example, $f(\{0,1,3\},1)=1$,
because there is a permutation $(0,3,1)$,
and $f(\{0,1,3\},3)=0$, because 0 and 1
cannot be next to each other.

Using $f$, the solution to the problem equals

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

\noindent
The dynamic programming states can be stored as follows:

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

\noindent
First, $f(\{k\},k)=1$ for all values of $k$:

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

\noindent
Then, the other values can be calculated
as follows:

\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
The variable $b$ goes through all subsets, and each
permutation is of the form $(\ldots,j,i)$
where the difference between $i$ and $j$ is
larger than one and $i$ and $j$ belong to $b$.

Finally, the number of solutions can be
calculated as follows:

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

\subsubsection{Sums of subsets}

Finally, we consider the following problem:
Every subset $x$
of $\{0,1,\ldots,n-1\}$
is assigned a value $c(x)$,
and our task is to calculate for
each subset $x$ the sum
\[s(x)=\sum_{y \subset x} c(y).\]
Using bit operations, the corresponding sum is
\[s(x)=\sum_{y \& x = y} c(y).\]
The following table shows an example of
the functions when $n=3$:
\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}
For example, $s(110)=c(000)+c(010)+c(100)+c(110)=5$. 

The problem can be solved in $O(2^n n)$ time
by defining a function $f(x,k)$ that calculates
the sum of $c(y)$ values such that $x$ can be
turned into $y$ by changing zero or more one bits
at positions $0,1,\ldots,k$ to zero bits.
Using this function, the solution to the
problem is $s(x)=f(x,n-1)$.

The base cases for the function are:
\begin{equation*}
    f(x,0) = \begin{cases}
               c(x)          & \textrm{if bit 0 of $x$ is 0}\\
               c(x)+c(x \XOR 1) & \textrm{if bit 0 of $x$ is 1}\\
           \end{cases}
\end{equation*}
For larger values of $k$, the following recursion holds:
\begin{equation*}
    f(x,k) = \begin{cases}
               f(x,k-1)          & \textrm{if bit $k$ of $x$ is 0}\\
               f(x,k-1)+f(x \XOR (1 < < k),k-1)    & \textrm{if bit $k$ of $x$ is 1}\\
           \end{cases}
\end{equation*}

Thus, we can calculate the values of the function
as follows using dynamic programming.
The code assumes that the array \texttt{c}
contains the values for $c$,
and it constructs an array \texttt{s}
that contains the values for $s$.
\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}

Actually, a much shorter implementation is possible,
because we can calculate the results directly
into the array \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}