Consider a processor that has 16 general purpose registers and it uses 2-byte instruction format for all its instructions. Variable-sized opcodes are permitted. There are three different types of instructions; M-type, R-type, and C-type. Each M-type instruction has 2 register operands and a 6-bit immediate operand. Each R-type instruction has 3 register operands. Each C-type instruction has a register operand and a 6-bit offset value. If there are 2 unique M-type opcodes and 7 unique R-type opcodes, which one of the following options gives the maximum number of unique opcodes possible for C-type instructions?

Let us analyze the instruction length and opcode requirements for each instruction format:

1. Instruction Size & Register Bits:

   • Total instruction size $= 2\text{ bytes} = 16\text{ bits}$.

   • Number of registers $= 16 \implies \log_2(16) = 4\text{ bits}$ are needed per register field.

2. M-type Instruction:

   • Operands: $2$ registers ($2 \times 4 = 8\text{ bits}$) $+ 6\text{-bit immediate} = 14\text{ bits}$.

   • Remaining bits for Opcode $= 16 - 14 = 2\text{ bits}$.

   • Total possible opcodes with $2\text{ bits} = 2^2 = 4$.

   • Given M-type opcodes used $= 2$.

   • Free (unused) combinations $= 4 - 2 = 2$.

3. R-type Instruction:

   • Operands: $3$ registers ($3 \times 4 = 12\text{ bits}$).

   • Remaining bits for Opcode $= 16 - 12 = 4\text{ bits}$.

   • Expanding the $2$ unused $2$-bit prefixes gives: $2 \times 2^{4-2} = 2 \times 4 = 8$ available combinations of $4$-bit opcodes.

   • Given R-type opcodes used $= 7$.

   • Free (unused) combinations $= 8 - 7 = 1$.

4. C-type Instruction:

   • Operands: $1$ register ($4\text{ bits}$) $+ 6\text{-bit offset} = 10\text{ bits}$.

   • Remaining bits for Opcode $= 16 - 10 = 6\text{ bits}$.

   • Expanding the $1$ unused $4$-bit prefix to a $6$-bit opcode gives:

     $$\text{Maximum C-type opcodes} = 1 \times 2^{6-4} = 1 \times 2^2 = 4$$

Thus, the correct option is (B).