Explain 2-bit by 2-bit Array multiplier. Draw the flowchart for divide operation of two numbers in signed magnitude form.

Each partial product bit is one AND gate: $a_0b_0$, $a_1b_0$, $a_0b_1$ and $a_1b_1$. They are arranged like paper multiplication:

Column $2^0$ has one bit, so $p_0 = a_0b_0$. Column $2^1$ has two bits, which a half adder (HA1) adds: the sum is $p_1$ and the carry $c_1$ goes to the next column. Column $2^2$ has $a_1b_1$ and $c_1$, which a second half adder (HA2) adds: the sum is $p_2$ and the carry is $p_3$.

$$p_0 = a_0b_0 \qquad p_1 = a_1b_0 \oplus a_0b_1$$

$$p_2 = a_1b_1 \oplus (a_1b_0 \cdot a_0b_1) \qquad p_3 = a_1b_1 \cdot a_1b_0 \cdot a_0b_1$$

 a1b0 a0b1        a0b0 ----------------------> p0
   \   /
   [HA1] --sum-----------------------------> p1
     | c1
 a1b1 +--> [HA2] --sum----------------------> p2
              \--carry---------------------> p3

The circuit needs 4 AND gates and 2 half adders. For $A = 11$ and $B = 11$ (that is $3 \times 3$), the four AND outputs are all 1. Then $p_0 = 1$, $p_1 = 0$ with $c_1 = 1$, $p_2 = 1 \oplus 1 = 0$ and $p_3 = 1$. The product is $1001 = 9$.

Flowchart for division of signed magnitude numbers. The sign and the magnitude are handled separately. The dividend is in $A$ and $Q$ joined, the divisor is in $B$, and the sign bits are $A_s$, $B_s$ and $Q_s$.

 START
   |
 Qs = As xor Bs ; SC = n-1
   |
 EA = A + (not B) + 1       (try A - B)
   |
 E = 1 ?--yes--> DIVIDE OVERFLOW (stop)
   | no
 shift left AQ ; EA = A + (not B) + 1
   |
 E = 1 ? --yes--> Qn = 1 ; A = EA
   | no
 Qn = 0   (A stays, so no restore needed)
   |
 SC = SC - 1 ; SC = 0 ? --no--> back to shift
   | yes
 END : Q = quotient, A = remainder

The first subtraction checks that the quotient fits in $n$ bits. If $A \ge B$ already, the divide overflows. After that, each cycle shifts left, subtracts $B$ by adding its 2's complement, and checks the carry $E$. A carry of 1 means the subtraction succeeded, so the quotient bit is 1 and $A$ takes the new value. A carry of 0 means it failed, so the quotient bit is 0 and $A$ keeps its old value. After $n-1$ cycles $Q$ holds the quotient and $A$ the remainder. The quotient sign is $A_s \oplus B_s$, and the remainder takes the sign of the dividend.