|
Size: 2740
Comment:
|
Size: 96
Comment:
|
| Deletions are marked like this. | Additions are marked like this. |
| Line 4: | Line 4: |
{{{#!latex Given an $n-$bit binary string, $I$, the leftmost bit indicates the sign of an integer in $1$s complement representation. In this left most position a $1$ indicates a negative value while a $0$ indicates a positive value. The representation for positive integers corresponds to unsigned representation where the leftmost bit must contain a $0$. Negative integers are formed by reversing all bits to form the bitwise complement of the corresponding positive integer. If we represent $I$ by the $n-$bit binary sequence, $b_{n} \ldots b_1 $ then $-I$ in one's complement is given by $\overline{b_n } \ldots \overline{b_1}$ where $\overline{b_i}=1-b_i$ for all $i$.\bigskip \noindent\textbf{Let's see what that looks like in Math speak}\bigskip Let $I$ be a negative one's complement integer. The value of $I$ is obtained by forming its one's complement: \begin{equation} -I = \sum_{i=0}^{n-1}(1-b_i)\cdot2^i = \sum_{i=0}^{n-1}2^i - \sum_{i=0}^{n-1} b_i \cdot 2^i. \end{equation} Thus, \begin{equation} I = \sum_{i=0}^{n-1}b_i \cdot 2^i - (2^n - 1). \end{equation} Negative one's complement integers are formed by subtracting a bias of $2^n - 1$ from the positive integers. Taking into account the sign bit $bn$, the value for a positive or negative (n+1) bit one's complement integer is: \begin{equation} I = \sum_{i=0}^{n-1}b_i \cdot 2^i - b_n (2^n - 1). \end{equation} Recalling that the left most bit only represents the sign, the range of values for an $n-$bit one's complement integer is $-(2^{n-1}-1)$ to $2^{n-1}-1$.\bigskip \noindent\textbf{Examples:}\bigskip Since the complement of $0$ is $2^{n+1}-1$, there are different representations for $+0$ and $-0$ in one's complement. Examples of $8$-bit one's complement numbers: \[ \begin{array}{cr} Binary & Decimal \\ 00000000 & 0 \\ 11111111 & -0 \\ 00000011 & 3 \\ 11111100 & -3 \\ \end{array} \] The range of $8-$bit one's complement integers is $-127$ to $+127$. Addition of signed numbers in one's complement is performed using binary addition with end-around carry. If there is a carry out of the most significant bit of the sum, this bit must be added to the least significant bit of the sum. To add decimal 17 to decimal -8 in 8-bit one's complement:\bigskip \begin{center} \begin{tabular}{rrrrrr} & & $0001$ & $0001$ & & $(17)$ \\ $+$ & & $1111$ & $0111$ & & $(-8)$ \\ \cline{1-4}\cline{6-6} & $1$ & $0000$ & $1000$ & & \\ & & \multicolumn{1}{l}{$\hookrightarrow $} & $+1$ & & \\ \cline{3-4} & & $0000$ & $1001$ & $=$ & $(9)$% \end{tabular} \end{center} }}} |
