base address of a = 1000
“*” → has higher precedence the “+”
*(*(a + ** a + 2) + 3)

base address of a = 1000
“*” → has higher precedence the “+”
*(*(a + ** a + 2) + 3)

∴ No. of states in min. DFA = 6 states.
⇒ # of instruction possible = 24 = 16
∴ No. of free opcodes = 16 – x
Let x is number of R-type instruction exist
∴ No. of I-type instruction possible = (16 – x) * 22
8 = 64 – 4x
4x = 64 – 8 = 56
x = 14Saved in this browser. Nothing changes for anyone else.