An answer:

Let's write out the microaddresses in binary:
00011
00110
01011
10011
10110
11011
Easiest observation is that the second-least-significant bit is always '1', so we can just use the other four bits for the opcode. Then the circuit is trivial: x3, x2, x1, and x0 go to µPC4, µPC3, µPC2, and µPC0 (note gap), respectively; whereas µPC1 is connected to a logic 1.

So the opcode assignment is:

Mnemonicµaddressopcode
ADD31
SUB62
XOR115
BR199
BEQ2210
BGT2713


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