Chapter 2.3, Problem 2.33PP

Program Plan Intro

Two’s complement negation:

The two’s-complement negation is represented by “${\text{-}}_{\text{w}}^{\text{t}}$”

For any number “x” such that ${\text{TMin}}_{\text{w}}\le \text{ x }\le {\text{TMax}}_{\text{w}}$, its “w-bit” two’s-complement negation is denoted by below formula 

${\text{-}}_{\text{w}}^{\text{t}}\text{x = }\left\{\begin{array}{l}{\text{TMin}}_{\text{w}}{\text{, x = TMin}}_{\text{w}}\\ \text{- x, }{\text{x > TMin}}_{\text{w}}\end{array}\right\}$

From the above equation, “TMin_w” is its own additive inverse, though any other value “x” has “-x” as its additive inverse.

• When “x” is “TMin_w”, then its two’s-complement negation value is “TMin_w”.
• When “x > TMin_w”, then its two’s-complement negation value is “–x”.

Two’s-Complement encodings:

• The two’s complement encoding is represented by the interpretation function of “B2T_w”.
  • The function “B2T_w” means “binary to two’s complement” of length “w”.
• The equation for “B2T_w” is

${\text{B2T}}_{\text{w}}\left(\overrightarrow{\text{x}}\right)\text{=B}{\text{-x}}_{\text{w-1}}{\text{2}}^{\text{w-1}}\text{+}{\displaystyle \sum _{\text{i}\text{=}\text{0}}^{\text{w}\text{-}\text{2}}{\text{x}}_{\text{i}}{\text{2}}^{\text{i}}}$

Example:

The example for two’s-complement negation is shown below:

Here assume bit length, “w” is “4” and “TMin₄” is “-8”.

Case 1: consider the value of “x” is “0”, then ${\text{-}}_{\text{4}}^{\text{t}}\text{0 = 0}$.

Case 2: consider the value of “x” is “6”

Step 1: compute the decimal value for given hex value

• The binary value for “0x6” is “0110”.
• Compute the ${\text{B2T}}_{\text{4}}\left(\text{0110}\right)$ by using the above equation of “B2T_w”

  $\begin{array}{c}{\text{B2T}}_{\text{4}}\left(\text{0110}\right)\text{=}{\text{-0×2}}^{\text{3}}\text{+}{\text{1×2}}^{\text{2}}\text{+}{\text{1×2}}^{\text{1}}\text{+}{\text{0×2}}^{\text{0}}\\ \text{=}\text{-0}\text{+}\text{4}\text{+}\text{2}\text{+}\text{0}\\ \text{=}\text{6}\end{array}$

${\text{B2T}}_{\text{4}}\left(\text{0110}\right)\text{=}\text{6}$

Therefore, decimal value for given hex value is “6”.

Step 2: compute the two’s-complement of decimal and hex value from step 1

Here, the decimal value of “x” from step 1 is “6” then check condition “x > TMin_w” that is “6 > -8”. So, compute the two’s-complement negation by using ${\text{-}}_{\text{w}}^{\text{t}}\text{x = - x}$

$\begin{array}{c}{\text{-}}_{\text{4}}^{\text{t}}\text{6 }\text{= -6}\\ \text{ = -6}\end{array}$

Therefore, the decimal value is “-6”. Here the value is in negative. So, need to compute two’s-complements.

Binary value for “6”: 0110

1’s complement: 1001

Add 1:       1

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2’s complement: 1010 (A)

Therefore, the corresponding hex value for “-6” is “A”.

Table for corresponding hexadecimal and binary value:

Hex digit	0	1	2	3	4	5	6	7
Decimal value	0	1	2	3	4	5	6	7
Hex digit	8	9	A	B	C	D	E	F
Decimal value	8	9	10	11	12	13	14	15