Exercises — Subtractors

$D=0\oplus 1=1$. $B_{out}=\bar{0}\cdot 1=1\cdot 1=1$. Why borrow: the top bit $A=0$ is smaller than the bottom bit $B=1$; $0-1$ cannot be done in this column, so we borrow $2$ from the next column ($0+2-1=1$) and mark $B_{out}=1$.

(b) $\bar{A}B$. Borrow needs top bit $=0$ (so $\bar{A}$) AND bottom bit $=1$ (so $B$). Option (a) is the half-adder carry; (c) borrows on the wrong row; (d) is never 1 with $B=1$.

True. Both are XOR of the inputs: adder $S=A\oplus B\oplus C_{in}$, subtractor $D=A\oplus B\oplus B_{in}$. Only the carry/borrow term differs (it flips a bar on $A$). See Adders.

$D=0\oplus 1\oplus 1=0$ (two 1's = even parity → 0). $B_{out}=\bar{0}\cdot 1+\bar{0}\cdot 1+1\cdot 1=1+1+1=1$. Sanity: arithmetic is $0-1-1=-2$, definitely needs a borrow, and the leftover in this column is $0$ ($-2+2=0$). Both match. ✓

$D=1\oplus 1\oplus 1=1$ (three 1's = odd parity → 1). $B_{out}=\bar{1}\cdot 1+\bar{1}\cdot 1+1\cdot 1=0+0+1=1$. Check: $1-1-1=-1$. Borrow $2$: $-1+2=1=D$, and we owe upstairs so $B_{out}=1$. ✓

Chain from LSB (bit0) upward; each column's $B_{out}$ becomes the next column's $B_{in}$.

• bit0: $A=0,B=1,B_{in}=0$. $D=0\oplus 1\oplus 0=1$, $B_{out}=\bar{0}\cdot 1+\bar{0}\cdot 0+1\cdot 0=1$.
• bit1: $A=1,B=1,B_{in}=1$. $D=1\oplus 1\oplus 1=1$, $B_{out}=\bar{1}\cdot 1+\bar{1}\cdot 1+1\cdot 1=1$.
• bit2: $A=1,B=0,B_{in}=1$. $D=1\oplus 0\oplus 1=0$, $B_{out}=\bar{1}\cdot 0+\bar{1}\cdot 1+0\cdot 1=0$.

Result $D_2{D}_1{D}_0=011=3$. Final borrow $=0$. Interpret: final borrow $0$ means no borrow escaped the top column → $A\ge B$, so the answer $3$ is a valid non-negative number. ✓ (See Ripple-Carry-Chains.)

• bit0: $A=1,B=0,B_{in}=0$. $D=1$, $B_{out}=\bar{1}\cdot 0+\bar{1}\cdot 0+0\cdot 0=0$.
• bit1: $A=1,B=1,B_{in}=0$. $D=0$, $B_{out}=\bar{1}\cdot 1+\bar{1}\cdot 0+1\cdot 0=0$.
• bit2: $A=0,B=1,B_{in}=0$. $D=1$, $B_{out}=\bar{0}\cdot 1+\bar{0}\cdot 0+1\cdot 0=1$.

Result bits $=101$, final borrow $=1$. Interpret: final borrow $1$ means the subtraction "underflowed" — $A<B$, so the true answer is negative. The $101$ is actually the wrapped magnitude; in two's complement the borrow-out being 1 flags that $3-6=-3$.

$D=1$ needs odd parity; $B_{out}=0$ means no borrow. Scan the 8 rows:

• $(1,0,0)$: $D=1,B_{out}=0$ ✓
• $(0,0,1)$: $D=1$ but $B_{out}=\bar{0}\cdot 0+\bar{0}\cdot 1+0=1$ ✗
• $(0,1,0)$: $B_{out}=1$ ✗
• $(1,1,1)$: $B_{out}=1$ ✗

Only $(A,B,B_{in})=(1,0,0)$. Meaning: $1-0-0=1$, a clean difference, no borrow needed.

$B=011\Rightarrow \bar{B}=100$. Carry-in $=1$ (this is the "+1"). Add $A=110$, $\bar{B}=100$, plus carry-in $1$: $110+100=1010,+1=1011.$ Keep the low 3 bits: $011=3$. The 4th bit (carry-out $=1$) is the $2^n$ we discard. So $6-3=3$. ✓ Why it works: $\bar{B}+1=2^n-B$ (two's complement), so $A+(2^n-B)=2^n+(A-B)$; the $2^n$ is exactly the discarded carry, leaving $A-B$. See Adder-Subtractor-Circuit.

$b_1+b_2=\bar{A}B+\overline{(A\oplus B)}{B}_{in}$. Expand $\overline{A\oplus B}=AB+\bar{A}\bar{B}$: $B_{out}=\bar{A}B+(AB+\bar{A}\bar{B})B_{in}=\bar{A}B+AB{B}_{in}+\bar{A}\bar{B}{B}_{in}.$ Both this and the standard SOP form $\bar{A}B+\bar{A}{B}_{in}+B{B}_{in}$ are true only on rows $\{(0,0,1),(0,1,0),(0,1,1),(1,1,1)\}$ — the four borrow rows of the truth table. Same function, verified. ✓

• $M=0$: $B_i\oplus 0=B_i$ (unchanged) and carry-in $=0$. Circuit computes $A+B$.
• $M=1$: $B_i\oplus 1={\bar{B}}_i$ (inverted) and carry-in $=1$. Circuit computes $A+\bar{B}+1=A-B$.

So the single control $M$ switches an adder into a subtractor by XOR-inverting $B$ and injecting the $+1$. This is the whole Adder-Subtractor-Circuit.

Substitute $A\to \bar{A}$: $\bar{A}B+(\bar{A}\oplus B)B_{in}$. Note $\bar{A}\oplus B=\overline{A\oplus B}$ (flipping one input of XOR flips the output). So we get $\bar{A}B+\overline{(A\oplus B)}{B}_{in}$ — exactly the full-subtractor borrow from L4.2. ✓ Takeaway: subtractor = adder skeleton with a bar on $A$ in the borrow-generate term. Same gates, one inverter difference.

• bit0: $A=1,B=1,B_{in}=0$: $D=0$, $B_{out}=\bar{1}\cdot 1+\bar{1}\cdot 0+1\cdot 0=0$.
• bit1: $A=0,B=0,B_{in}=0$: $D=0$, $B_{out}=0$.
• bit2: $A=0,B=1,B_{in}=0$: $D=1$, $B_{out}=\bar{0}\cdot 1=1$.
• bit3: $A=1,B=0,B_{in}=1$: $D=1\oplus 0\oplus 1=0$, $B_{out}=\bar{1}\cdot 0+\bar{1}\cdot 1+0\cdot 1=0$.

Result $=0100=4$, final borrow $=0$ → non-negative. $9-5=4$. ✓

• bit0: $A=0,B=1,B_{in}=0$: $D=1$, $B_{out}=1$.
• bit1: $A=1,B=0,B_{in}=1$: $D=1\oplus 0\oplus 1=0$, $B_{out}=\bar{1}\cdot 0+\bar{1}\cdot 1+0\cdot 1=0$.
• bit2: $A=0,B=1,B_{in}=0$: $D=1$, $B_{out}=1$.
• bit3: $A=0,B=0,B_{in}=1$: $D=0\oplus 0\oplus 1=1$, $B_{out}=\bar{0}\cdot 0+\bar{0}\cdot 1+0\cdot 1=1$.

Raw bits $=1101$, final borrow $=1$. Interpret: final borrow $1$ ⇒ underflow ⇒ answer negative. The bits $1101$ are the two's-complement of $-3$ (since $1101=-8+4+1=-3$). So $2-5=-3$. ✓ (See Twos-Complement-Representation.)

Why does subtractor borrow use $\bar{A}$ while adder carry uses $A$? ::: You borrow only when the top bit is $0$ (broke), so the generate term needs $\bar{A}$; addition carries when the top bit is $1$, so it needs $A$.