Number System Practice Questions

The bit pattern consists of a sign bit $S=1$, exponent $E=10000011_2 = 131$, and mantissa $F=1010...0_2$.
The true exponent is calculated as $e = E - 127 = 131 - 127 = 4$.
The significand, including the implicit leading 1, is $1.F = 1.101_2 = 1 + 2^{-1} + 2^{-3} = 1 + 0.5 + 0.125 = 1.625$.
The decimal value is given by $(-1)^S \times (1.F) \times 2^e = -1 \times 1.625 \times 2^4$.
Calculating this gives $-1 \times 1.625 \times 16 = -26$.

First, convert $(7601)_8$ to decimal form:
$N^2 = 7 \cdot 8^3 + 6 \cdot 8^2 + 0 \cdot 8^1 + 1 \cdot 8^0 = 3584 + 384 + 1 = 3969$
Taking the square root gives $N = \sqrt{3969} = 63$.
Next, convert option B $(143)_6$ to decimal to verify:
$(143)_6 = 1 \cdot 6^2 + 4 \cdot 6^1 + 3 \cdot 6^0 = 36 + 24 + 3 = 63$
Since $N = 63 = (143)_6$, this is the correct option.

Expanding both sides of the equation $(12x)_3 = (123)_x$ yields:
$1(3^2) + 2(3^1) + x(3^0) = 1(x^2) + 2(x^1) + 3(x^0)$
$9 + 6 + x = x^2 + 2x + 3 \implies x^2 + x - 12 = 0$
Factoring the quadratic equation results in $(x+4)(x-3) = 0$, giving $x = -4$ or $x = 3$.
For base $3$, the digit $x$ must be less than $3$ ($x < 3$), and for base $x$, the digit $3$ must be less than $x$ ($x > 3$).
Neither $x = -4$ nor $x = 3$ satisfies these constraints, so no valid solution exists.

Gray code is defined by the property that adjacent values differ by only a single bit. In contrast, standard binary coding requires multiple bits to change simultaneously during transitions, such as $011_2 \to 100_2$, which increases the probability of transient errors. By restricting transitions to a single bit change, Gray code minimizes the likelihood of decoding errors in digital systems and sensors. Due to this characteristic of reducing switching errors, it is appropriately termed the "minimum error code".

Given $a = 0.52 \times 10^{-1}$ and $b = 0.28 \times 10^{13}$.
To compute $b + a$, we must align $a$ to the exponent of $b$, shifting $a$ by $13 - (-1) = 14$ places.
Because the computer mantissa is limited to 8 digits, $a$ becomes $0.00000000 \times 10^{13}$ when aligned with $b$ due to precision loss.
Therefore, $b + a = 0.28000000 \times 10^{13} + 0.00000000 \times 10^{13} = 0.28000000 \times 10^{13} = b$.
Consequently, $(b + a) - b = b - b = 0$.

Booth's algorithm performs an addition or subtraction when there is a bit transition $(q_i, q_{i-1})$ in the multiplier, where $q_{-1} = 0$.
The multiplier $0111\ 0111\ 1011\ 1101$ has the LSB-to-MSB sequence: $1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0$.
Transitions $(q_i \neq q_{i-1})$ occur at the following pairs:
$(1,0), (0,1), (1,0), (0,1), (1,0), (0,1), (1,0), (0,1)$.
There are exactly $8$ such transitions, each triggering an addition or subtraction operation.

In the IEEE 754 single-precision standard, a 32-bit floating-point number consists of a sign bit ($S$), an 8-bit exponent field ($E$), and a 23-bit fraction ($M$).
The hexadecimal value $0x00000000$ translates to $32$ binary zeros, resulting in $S=0$, $E=0$, and $M=0$.
In this standard, the combination where the exponent $E=0$ and the fraction $M=0$ is reserved to represent the special value zero ($\pm 0$).
Because the sign bit $S=0$, this specific representation corresponds to $+0$.
Thus, $0x00000000$ is categorized as the special value $+0$.

To convert the decimal fraction $0.75$ to the octal system, multiply the fraction by the target base, $8$, and record the integer parts.

1. Multiply the fraction by $8$: $0.75 \times 8 = 6.0$.
2. The integer part is $6$, and the remaining fractional part is $0$.
3. Since the fractional part is $0$, the conversion process terminates.
4. Thus, the decimal $0.75$ is equivalent to $0.6$ in the octal system.

Excess-3 code represents a decimal digit $N$ as $N+3$.
Adding two Excess-3 encoded numbers yields $(N_1+3) + (N_2+3) = (N_1+N_2) + 6$.
This result is in Excess-6 format, which is $3$ greater than the target Excess-3 format $(N_1+N_2+3)$.
When no carry is generated (indicating the sum is less than 9), we must subtract $3$ ($0011_2$) from the result to convert the Excess-6 representation back to the correct Excess-3 format.
Thus, subtracting $0011$ is necessary to obtain the correct sum.

The given expression can be rewritten using powers of $2$:
$3 \times 2^9 + 7 \times 2^6 + 5 \times 2^3 + 3 = (2+1) \times 2^9 + (4+2+1) \times 2^6 + (4+1) \times 2^3 + (2+1)$.
Expanding these terms, we obtain:
$(2^{10} + 2^9) + (2^8 + 2^7 + 2^6) + (2^5 + 2^3) + (2^1 + 2^0)$.
This simplifies to the sum $2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^3 + 2^1 + 2^0$.
Each term corresponds to a digit $1$ in the binary representation, positioned at the exponents $10, 9, 8, 7, 6, 5, 3, 1,$ and $0$.
Counting these positions results in exactly $9$ ones.

Using the identity $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$ for $x > 0$, we simplify the first term:
$\cot^{-1}\left\{\frac{\beta(\alpha-\beta) + 1 + \beta^2}{\alpha-\beta}\right\} = \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) = \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right) = \tan^{-1}\alpha - \tan^{-1}\beta$
Similarly, for the second term:
$\cot^{-1}\left\{\frac{\gamma(\beta-\gamma) + 1 + \gamma^2}{\beta-\gamma}\right\} = \cot^{-1}\left(\frac{1+\beta\gamma}{\beta-\gamma}\right) = \tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right) = \tan^{-1}\beta - \tan^{-1}\gamma$
For the third term, since $\gamma-\alpha < 0$, we use the identity $\cot^{-1}(x) = \pi + \tan^{-1}(1/x)$ for $x < 0$:
$\cot^{-1}\left\{\frac{1+\alpha\gamma}{\gamma-\alpha}\right\} = \pi + \tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right) = \pi + \tan^{-1}\gamma - \tan^{-1}\alpha$
Summing the expressions:
$(\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + (\pi + \tan^{-1}\gamma - \tan^{-1}\alpha) = \pi$

An 8-bit exponent in two's complement has a range of $[-2^7, 2^7 - 1]$, which is $[-128, 127]$.
A 16-bit mantissa in two's complement has a range of $[-2^{15}, 2^{15} - 1]$, where the smallest positive value is $1$ and the largest positive value is $2^{15} - 1$.
The smallest positive value is calculated using the smallest positive mantissa and the minimum exponent: $1 \times 10^{-128}$.
The largest positive value is calculated using the largest positive mantissa and the maximum exponent: $(2^{15} - 1) \times 10^{127}$.
Comparing this result with the options, it matches Option D.

A system with $n$ ternary logic elements (flits) can represent $3^n$ unique values because each element has 3 possible states. To represent at least $256$ values, we must solve for the smallest integer $n$ such that $3^n \ge 256$. Evaluating powers of 3, we find:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$ (which is less than $256$)
$3^6 = 729$ (which is greater than $256$)
Therefore, $6$ flits are required to represent at least $256$ values.

To convert the binary number $01111100110111100011_2$ to hexadecimal, partition the bits into groups of four starting from the right:
$0111 \mid 1100 \mid 1101 \mid 1110 \mid 0011$
Convert each 4-bit group to its corresponding hexadecimal digit:
$0111_2 = 7_{16}$, $1100_2 = C_{16}$, $1101_2 = D_{16}$, $1110_2 = E_{16}$, and $0011_2 = 3_{16}$.
Concatenating these values yields the final result of $7CDE3_{16}$.

In Booth's bit-pair recoding, the multiplier coefficient $c_i$ for a bit group $(x_{i+1}, x_i, x_{i-1})$ is determined by the expression $c_i = -2x_{i+1} + x_i + x_{i-1}$. This encoding rule effectively determines the scalar applied to the multiplicand $Y$. For Row 5, with the bit pattern $100$, the coefficient is calculated as $c_i = -2(1) + 0 + 0 = -2$, resulting in the partial product $-2Y$. For Row 8, with the bit pattern $111$, the coefficient is calculated as $c_i = -2(1) + 1 + 1 = 0$, resulting in a partial product of $0$. These calculations confirm that the missing values are $-2Y$ and $0$.

The Gray code $h_3h_2h_1h_0$ corresponds to binary $n$, and we determine the Gray code $g_3g_2g_1g_0$ for $n+1$. For each $n \in \{0, \dots, 15\}$, we map the binary $n$ to Gray code $h = G(n)$ and $n+1$ to Gray code $g = G((n+1) \pmod{16})$. For example, when $n=3$, $h = G(3) = 0010_2 = 2_{10}$, and $n+1=4$, so $g = G(4) = 0110_2$, where $g_2 = 1$. Evaluating $g_2$ for all $n$, we find $g_2=1$ exactly when $h \in \{2, 4, 5, 6, 7, 12, 13, 15\}$. This corresponds to the function $g_2(h_3h_2h_1h_0) = \sum(2, 4, 5, 6, 7, 12, 13, 15)$.

To find the normalized floating-point representation, we process the number $0.239 \times 2^{13}$.

1. Normalization: Convert the decimal $0.239$ to binary. Approximating to the required precision gives $0.239 \approx 0.00111101_2 = 1.11101000_2 \times 2^{-3}$. Thus, the input is $1.11101000_2 \times 2^{-3} \times 2^{13} = 1.11101000_2 \times 2^{10}$.
2. Exponent: The exponent is $10$. In Excess-64 format, we add the bias: $10 + 64 = 74$. Converting to binary, $74_{10} = 1001010_2$.
3. Mantissa: The normalized form uses an implicit leading $1$. Extracting the fractional part $11101000$ yields the $8$-bit mantissa.
4. Assembly: Concatenating the sign ($0$ for positive), the $7$-bit exponent ($1001010$), and the $8$-bit mantissa ($11101000$) gives $0100101011101000_2$.
5. Hexadecimal: Grouping bits into $4$s ($0100$, $1010$, $1110$, $1000$) results in $4, A, E, 8$ in hexadecimal.

The 8-bit two's complement representation of $-57$ is $11000111_2$. Booth's algorithm utilizes pairs of bits $(b_i, b_{i-1})$ with an implicit $b_{-1} = 0$, where the encoding rule is $01 \to 1$, $10 \to -1$, and $00/11 \to 0$. Evaluating $11000111_2$ with $b_{-1}=0$:

1. $(b_0, b_{-1}) = (1, 0) \to -1$
2. $(b_1, b_0) = (1, 1) \to 0$
3. $(b_2, b_1) = (1, 1) \to 0$
4. $(b_3, b_2) = (0, 1) \to 1$
5. $(b_4, b_3) = (0, 0) \to 0$, $(b_5, b_4) = (0, 0) \to 0$, $(b_6, b_5) = (1, 0) \to -1$, and $(b_7, b_6) = (1, 1) \to 0$.
   The resulting recoded sequence from most significant to least significant is $0, -1, 0, 0, 1, 0, 0, -1$, which matches option A.

First, convert both octal numbers to decimal:
$(34.4)_{8} = 3 \times 8^1 + 4 \times 8^0 + 4 \times 8^{-1} = 24 + 4 + 0.5 = 28.5_{10}$
$(23.4)_{8} = 2 \times 8^1 + 3 \times 8^0 + 4 \times 8^{-1} = 16 + 3 + 0.5 = 19.5_{10}$
Next, multiply the decimal values: $28.5 \times 19.5 = 555.75_{10}$.
Convert $555$ to octal: $555 \div 8 = 69$ with remainder $3$, $69 \div 8 = 8$ with remainder $5$, $8 \div 8 = 1$ with remainder $0$, and $1 \div 8 = 0$ with remainder $1$, giving $(1053)_{8}$.
Convert the fraction $0.75$: $0.75 \times 8 = 6.0$, giving $.6_{8}$.
Thus, $(34.4)_{8} \times (23.4)_{8} = (1053.6)_{8}$.

To convert $(657)_8$ to hexadecimal, first represent each octal digit as a 3-bit binary number: $6_8 = 110_2$, $5_8 = 101_2$, and $7_8 = 111_2$.
Concatenating these gives the binary sequence $(110101111)_2$.
Next, partition the binary bits into groups of four starting from the right: $(0001 \ 1010 \ 1111)_2$.
Converting each 4-bit group to hexadecimal yields $(0001)_2 = 1_{16}$, $(1010)_2 = A_{16}$, and $(1111)_2 = F_{16}$.
Thus, $(657)_8$ is equivalent to $(1AF)_{16}$.