Unable to understand how to take 1-Address Instruction Opcode ? I am quiet Confused.

Question

A computer has 16-bit Instruction and 5-bit address if there are maximum 128 1-Address Instructions possible,then how many 2-Address Instructions used by Computer ?

Comments

@Ananya : Thanks, this link helps me to clear my doubt.

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@Anu

will ans not be $\left ( 2^{11}-128 \right )\times 2^{5}$

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answer will be (2¹⁶ - 128*2⁵ ) / 2¹⁰ = 60

Answer 1

This question uses expanding opcode technique.

Suppose we are provided with $x$ 2-address instructions, $y$ 1 address instructions and $z$ 0 address instructions

Then the formula is

Total 2-address instructions possible = $2^6$

$x< 2^{6}$

After utilizing x instructions, No of 1 address instructions remaining are

$(2^{6}-x)*2^{5}$

But we know we have $y$ 1 address instructions

So total 0 address instructions are

$((2^{6}-x)*2^{5})-y)*2^{5}$