Question

Consider $Z = X - Y$, where $X$, $Y$ and $Z$ are all in sign-magnitude form. $X$ and $Y$ are each represented in $n$ bits. To avoid overflow, the representation of $Z$ would require a minimum of:

• A. $n$ bits
• B. $n-1$ bits
• C. $n+1$ bits
• D. $n+2$ bits

Hint:

In sign-magnitude arithmetic, subtraction can double the maximum magnitude. So, always expect {one extra bit} beyond the original word length to avoid overflow.

Answer 1

The Correct Option is C

To solve this problem, we have to determine the minimum number of bits required to represent the result \(Z = X - Y\) without overflow, where \(X\) and \(Y\) are positive and negative numbers represented in sign-magnitude form using \(n\) bits.

In sign-magnitude representation, one bit is used for the sign (either positive or negative), and the remaining bits are used to represent the magnitude of the number.

1. Consider the bit configuration:
   • \(X\) can represent a range from \(-(2^{n-1} - 1)\) to \(2^{n-1} - 1\).
   • \(Y\) represents a similar range as \(X\), i.e., from \(-(2^{n-1} - 1)\) to \(2^{n-1} - 1\).
2. When subtracting \(Y\) (which can effectively be thought of as adding either a positive or negative number), \(Z = X - Y\) must account for the largest possible negative and positive results:
   • \(X - Y = 2^{n-1} - 1 - (-(2^{n-1} - 1))\)
   • This results in \(Z\) ranging from \(-2^{n-1}\) to \(2^{n-1} - 1\).
3. To represent \(Z\) correctly in sign-magnitude form without overflow, the magnitude must range up to \(2^{n-1}\). This requires an additional bit for the magnitude representation.
4. Thus, to cover numbers from \(-2^{n-1}\) to \(2^{n-1} - 1\), a minimum of \(n+1\) bits is necessary:
   • 1 bit for sign
   • n bits for magnitude

Therefore, the correct answer is that \(Z\) would require a minimum of n+1 bits in sign-magnitude form to avoid overflow.

Answer 2

Step 1: Understand sign-magnitude representation.
In sign-magnitude representation:
-- 1 bit is used for the sign
-- Remaining $(n-1)$ bits are used for the magnitude
So, the range of values representable by an $n$-bit sign-magnitude number is: \[ -(2^{n-1}-1) \;\text{to}\; +(2^{n-1}-1) \] Step 2: Analyze the subtraction $Z = X - Y$.
Subtraction can be rewritten as: \[ Z = X + (-Y) \] The worst-case magnitude of $Z$ occurs when:
-- $X$ is the largest positive number, and
-- $Y$ is the largest negative number.
That is: \[ X = +(2^{n-1}-1), \quad Y = -(2^{n-1}-1) \] Step 3: Compute the maximum possible value of $Z$.
\[ Z_{\max} = (2^{n-1}-1) - (-(2^{n-1}-1)) = 2 \times (2^{n-1}-1) \] \[ Z_{\max} = 2^n - 2 \] Step 4: Determine the number of bits required.
To represent a magnitude close to $2^n$, we need $n$ magnitude bits.
Including the sign bit, the total number of bits required is: \[ n + 1 \] Step 5: Conclusion.
To avoid overflow in sign-magnitude representation, $Z$ must be represented using \[ \boxed{n+1 \text{ bits}} \]