Answer: 32 Instructions
Explanation:
Given,
16 Integer registers. So, we need $4$ bits to address any one of them.
64 Floating point registers. This requires $6$ bits to uniquely identify them.
Each instruction is $16$ bits long.
Type 1 Instructions:
4 instructions, each with 3 integer operands.
The 3 integers, each requires $4$ bits. So, $4*3$ bits for operands. We are left with $16 - 12 = 4$ bits.
With $4$ bits, $2^4 = 16$ opcodes are possible. Out of these we used $4$ opcodes. i.e $2$ bits. Let's say first two bits are fixed to $00$ and next two bits are used for $4$ different Type1 instructions.
$00\ 00\ ...$
$00\ 01\ ...$
$00\ 10\ ...$
$00\ 11\ ...$
Type 2 Instructions:
$8$ instructions, each with $2$ floating point register operands.
Here we need $6*2$ bits for operands, and remaining $16 - 12 = 4$ bits are left for opcodes.
So using these 4 bits, we need to get $8$ opcodes.
Here we can't use $00\ ...$ for any opcode since it will not distinguish Type 2 from Type 1. So, we are left with $12$ opcodes. And we are going to use $8$ out of these $12$ for type 2 instructions.
$01\ 00\ ...$
$01\ 01\ ...$
$01\ 10\ ...$
$01\ 11\ ...$
$10\ 00\ ...$
$10\ 01\ ...$
$10\ 10\ ...$
$10\ 11\ ...$
Type 3 Instructions:
$14$ instructions, with $1$ integer and $1$ floating type operand.
$4 + 6 = 10$ bits required for opcodes, remaining $16 - 10 = 6$ bits available for use in opcode.
The only valid combination left for this first $2$ bits is $11\ ...$. Rest have been used in Type1 and Type2 instructions.
So, we are left with $4$ bits for opcodes. With these $4$ bits we can have $2^4 = 16$ opcodes, out of which $14$ are required. So, we use all except last two opcodes:
$11\ 00\ 00\ ...$
$11\ ...$
$11\ 11\ 01\ ...$
These two opcodes are still left unassigned.
$11\ 11\ 10\ ...$
$11\ 11\ 11\ ...$
Type 4 Instructions:
$\mathbf{N}$ instructions, each with $1$ floating point operand.
We have $16 - 6 = 10$ bits for opcode. Out of $10$ bits, first $6$ bits can be either one of the two left opcodes(above). And any combination for remaining $4$ bits.
So we have $2 * 2^4$ opcodes. So, $\mathbf{N = 32}$.