From 15f11b09ec9eb131902d6a6ec183ed024e1aa5ba Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Sat, 4 Feb 2017 22:10:45 +0200 Subject: [PATCH] Corrections --- luku10.tex | 204 +++++++++++++++++++++++++++-------------------------- 1 file changed, 105 insertions(+), 99 deletions(-) diff --git a/luku10.tex b/luku10.tex index bc5fd01..d56302b 100644 --- a/luku10.tex +++ b/luku10.tex @@ -1,92 +1,94 @@ \chapter{Bit manipulation} -A computer internally manipulates data -as bits, i.e., as numbers 0 and 1. +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 for manipulating them. +can be used to manipulate them. It turns out that there are many uses for -bit operations in the implementation of algorithms. +bit operations in algorithm programming. \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. +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. -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 +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. -In this case, the bit representation of 43 +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}. -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. +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 $2^{31}-1$. -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, +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 representation of $-43$ +For example, the bit 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 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 $2^{32}-1$. -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 +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} -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, +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 code +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} -we increase $2^{31}-1$ by one to get $-2^{31}$. \section{Bit operations} @@ -97,9 +99,9 @@ we increase $2^{31}-1$ by one to get $-2^{31}$. \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 +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} @@ -114,16 +116,17 @@ 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 bit 1 in positions where at least one -of the numbers -$x$ and $y$ have bit 1. -For example, $22$ | $26$ = 30 because +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} @@ -139,10 +142,9 @@ For example, $22$ | $26$ = 30 because \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 +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} @@ -159,12 +161,12 @@ $\XOR$ & 11010 & (26) \\ The \key{not} operation \textasciitilde$x$ produces a number where all the bits of $x$ -have been inversed. +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 +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: @@ -180,60 +182,64 @@ $x$ & = & 29 & 00000000000000000000000000011101 \\ \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 $2^k$, +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 $2^k$ -rounding downwards. +rounded down to an integer. -\subsubsection{Bit manipulation} +\subsubsection{Applications} -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 +A number of the form $1 < < k$ has a one bit in position $k$, and all other bits are zero, -so we can manipulate single bits of numbers -using these numbers. +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 $x$: -The $k$th bit in $x$ is one if -$x$ \& $(1 < < k) = (1 < < k)$. -The formula $x$ | $(1 < < k)$ +\begin{lstlisting} +for (int i = 31; i >= 0; i--) { + if (x&(1<