Consider a 16-bit hypothetical processor that supports 1-word-long instructions. The processor has 30 registers and 4 KB memory. If there are 11 two-address register-reference instructions and 10 one-address memory-reference instructions, how many zero-address instructions can be formulated?

For 4 KB memory, memory-address field size is log₂(4096) = 12 bits. For 30 registers, each register field requires 5 bits. In a fixed 16-bit instruction, remaining bits are opcode bits. Unused opcode combinations in one format can be expanded into another format.

Step 1: One-address format (1AF)

Total 1AF operations = 2⁴ = 16
Used 1AF operations = 10
Free 1AF opcodes = 16 − 10 = 6

Step 2: Two-address register format (2AF)

Extra opcode bits over 1AF = 6 − 4 = 2
Derived 2AF operations from free 1AF opcodes = 6 × 2² = 24
Used 2AF operations = 11
Free 2AF opcodes = 24 − 11 = 13

Step 3: Zero-address format (0AF)

Extra opcode expansion from 2AF to 0AF = 16 − 6 = 10
Total zero-address instructions = 13 × 2¹⁰ = 13312

Final Answer: 13312 (about 13K).