A processor has 64 registers and uses 16-bit instruction format. It has two types of instructions: I-type and R-type. Each I-type instruction contains an opcode, a register name, and a 4-bit immediate value. Each R-type instruction contains an opcode and two register names. If there are 8 distinct I-type opcodes, then the maximum number of distinct R-type opcodes is _________.

### 1. Register Address Bits: Number of registers = $64 = 2^6$, so 6 bits are required to specify a register.

### 2. Instruction Formats (16 bits total):

• I-type instruction format:

  • Register: 6 bits

  • Immediate value: 4 bits

  • Opcode field size for I-type: $16 - (6 + 4) = 6$ bits

  • Total available bit combinations for I-type opcode field: $2^6 = 64$

  • Given number of distinct I-type opcodes: $8$

  • Unused patterns in the I-type opcode field: $64 - 8 = 56$

• R-type instruction format:

  • Register 1: 6 bits

  • Register 2: 6 bits

  • Opcode field size for R-type: $16 - (6 + 6) = 4$ bits

### 3. Expanding Opcode Analysis: Notice that the I-type opcode field is 6 bits, while the R-type opcode field is 4 bits. Since the 4-bit prefix of the instruction determines whether it can be expanded into R-type instructions:

• Each unused 6-bit I-type opcode pattern can be used for R-type only if we look at prefix overlaps, or more precisely: Let $k$ be the number of 4-bit patterns allocated to R-type instructions. Then $16 - k$ patterns of length 4 are available to expand into 6-bit I-type opcodes. Each such 4-bit pattern yields $2^{6-4} = 2^2 = 4$ distinct 6-bit opcodes. Therefore, the total number of I-type opcodes possible is:

  $$(16 - k) \times 4$$

Given that there are 8 distinct I-type opcodes:

Thus, the maximum number of distinct 4-bit R-type opcodes is 14.