Here's how you can implement the expression

in $0,1,2,$ and $3$ address machines:

$0$ Address Machine $($Stack Machine$)$:

PUSH A

PUSH B

PUSH C

MUL

ADD

PUSH D

PUSH E

PUSH F

MUL

SUB

DIV

POP X

Explanation:

In this $0-$address stack machine code, the expression

is evaluated. It starts by pushing the values of A$,$ B$,$ and C onto a stack, then performs multiplication $($MUL$)$ to calculate

and addition $($ADD$)$ to compute

$.$ Next, it pushes the values of D$,$ E$,$ and F onto the stack and proceeds to multiply $($MUL$)$

$,$ followed by subtraction $($SUB$)$ to calculate

$.$

Finally, it performs division $($DIV$)$ to obtain the result of

and stores it in the variable X by popping it from the stack. This code demonstrates the operation of a $0-$address stack machine, where operands are pushed onto a stack, and operations are performed on the top elements of the stack.

Step $2$

$1$ Address Machine $($Accumulator Machine$)$:

LOAD A ; Load A into the accumulator

ADD B ; Add B to the accumulator

MUL C ; Multiply the accumulator by C

LOAD D ; Load D into the accumulator

SUB E ; Subtract E from the accumulator

MUL F ; Multiply the accumulator by F

SUB ; Subtract the top of the stack from the accumulator

STORE X ; Store the result in X

Explanation:

In this $2-$address machine code, the expression X $=($A $+$ B C$)-($D $-$ E F$)$ is computed step by step. It begins by loading the value of A into the accumulator, then adds B to the accumulator, followed by multiplying the accumulator by C$.$ Next, it loads the value of D into the accumulator, subtracts E from the accumulator, and multiplies the accumulator by F$.$

Finally, it subtracts the top value from the stack from the accumulator, which represents the result of the first part of the expression, and stores this result in variable X$.$ This code illustrates the operation of a $2-$address machine, where instructions typically involve two operands $$ one source and one destination.

Step $3$

$2$ Address Machine:

MOVE X$,$ A ; Load A into X

ADD X$,$ B ; Add B to X

MUL X$,$ C ; Multiply X by C

MOVE Y$,$ D ; Load D into Y

SUB Y$,$ E ; Subtract E from Y

MUL Y$,$ F ; Multiply Y by F

SUB X$,$ Y ; Subtract Y from X

Explanation:

In this $3-$address machine code, the expression X $=($A $+$ B C$)-($D $-$ E F$)$ is computed step by step. It begins by loading the value of A into the variable X$.$ Then, it adds the value of B to X$,$ followed by multiplying X by C$.$ Similarly, the value of D is loaded into the variable Y$,$ and then E is subtracted from Y$.$

Finally, it multiplies Y by F and subtracts Y from X to obtain the final result of the expression. This code illustrates the operation of a $3-$address machine, where instructions explicitly specify source and destination operands for each operation.

Step $4$

$3$ Address Machine:

LOAD X$,$ A ; Load A into X

LOAD Y$,$ B ; Load B into Y

LOAD Z$,$ C ; Load C into Z

MUL Y$,$ Z ; Multiply Y by Z and store in Y

ADD X$,$ Y ; Add X and Y$,$ store in X

LOAD U$,$ D ; Load D into U

LOAD V$,$ E ; Load E into V

MUL V$,$ F ; Multiply V by F and store in V

SUB U$,$ V ; Subtract V from U$,$ store in U

DIV X$,$ U ; Divide X by U and store in X

Explanation:

In this $3-$address machine code, the expression X $=($A B $+$ C$)/($D $-$ E F$)$ is computed step by step. It starts by loading the values of A$,$ B$,$ and C into variables X$,$ Y$,$ and Z$,$ respectively. Then, it multiplies Y and Z$,$ storing the result in Y$.$ Next, it adds X and Y$,$ storing the result back in X$.$ After that, it loads the values of D$,$ E$,$ and F into variables U$,$ V$,$ and W$,$ respectively. It multiplies V by W and stores the result in V$.$ Subsequently, it subtracts V from U$,$ storing the result in U$.$

Finally, it divides X by U$,$ obtaining the final result of the expression and storing it in X$.$ This code exemplifies the operation of a $3-$address machine, where each instruction specifies source and destination registers for the computation.

Answer

Each machine's instructions are tailored to its addressing format. In a $0-$address stack machine, you push operands onto the stack and perform operations on the stack's top elements. In a $1-$address machine, you use an accumulator to accumulate intermediate results. In $2-$address and $3-$address machines, you use explicit registers or memory locations for operations.

The example code assumes that you have suitable instructions for loading, storing, arithmetic operations, and moving data between registers or memory locations. The specific assembly instructions may vary depending on the architecture and assembly language of the target machine.