the MIPS ALU operations below:

MIPS ALU. Computes a result as follows:

c

$-----------$

$C=A+B$

$C=BBC=B$

$C=BSaC=A-B$

$C=hat(C)BC=(A|B)$

$C=(A+B)?1$:$0C=(A>0)?1$:$0$

$C=(A=0)?1$:$0$

Determine the mapping of Opcode bits to ALU OP bits.

We will be implementing the Opcodes in white. Notice that some of the opcodes will use the ALU,

and some will not. Of the ones that will use the ALU, we need to map four of the bits to the ALU

OP input so it will provide the correct output when the Opcode is found in our instruction.

Your job is to figure out a particular mapping that will work, using the bits we have. Notice the

overlaps in the operations that have the SPECIAL opcode and those that don't: this will be the key

to figuring out how to go about producing the correct translation.

For this project, you only need to figure out a mapping for the functions that the ALU supports $($in

other words, don't worry about mapping the Branch, Jump, Move, Load, Store, or Set instructions.$)$

Question: Using the last $4$ digits of your student ID to construct the test patterns for input data A

and B$.$ Assume your student ID is: $x_6{x}_5{x}_4{x}_3{x}_2{x}_1{x}_0,$ treat each digit as Hexadecimal number,

then:

$A=0x(x_3(A)x_{2}B{x}_{1}Cx{x}_{0}D)$;

$B=0x(x_{0}C{l}_1{x}_1,D_{2}E{x}_3(F)).$

For example, if your student ID is $1234567,$ then take last $4$ digits $("4567"),$ your data input to the

ALU will be $("0x"$ means $16-$based hexadecimal format$)$:

$6$C $7$D