There are two $4-$bit unsigned values $A(A_3{A}_2{A}_1{A}_0)$ and $B(B_3{B}_2{B}_1{B}_0)$ provided as input with a

maximum value of $7.$ So $A_3$ and $B_3$ will be $0.$

The objective of this homework is to determine the equations for two output bits ZF and SF$.$

The ZF bit should be $1$ if $A==B.$ The $SF$ bit should be $1$ if $B>A.$

The homework can be broken down into the following steps:

Calculate the equations for determining the $1\text{'}$s complement of $B.$ The inputs of this step

are $B_3{B}_2{B}_1{B}_0$ and the outputs of this step are $O_3{O}_2{O}_1{O}_0.$ Hint: $O_0$ only depends on $B_0,O_1$

only depends on $B_1,$ etc. It is not necessary to do truth table.

Calculate the equations for determining $-$B$.$ This is the $2\text{'}$s complement of $B.$ The inputs of

this step are $O_3{O}_2{O}_1{O}_0$ and the outputs of this step of $N_3{N}_2{N}_1{N}_0.$ The $2\text{'}$s complement of a

number is found by adding $1$ to the $1\text{'}$s complement of the number. So we are adding

$O_3{O}_2{O}_1{O}_0$ and $0001.$ Refer to the below table from the first lecture about adding $2$ numbers.

For the first bit, the input $A$ in the table is $O_0$ and the input $B$ in the table is $1.$ The sum is $N_0.$

$N_0$ is $1$ if $O_0$ is $0.$ The carry$-$out $C_1$ is $1$ if $O_0$ is $1.$ For the second bit, the input $A$ in the table

is $O_1$ and the input $B$ in the table is $C_1.$ The sum is $N_1*N_1$ is $1$ if $O_1{C}_1^{\text{'}}+O_1^{\text{'}}{C}_1.C_2$ is $1$ if

$O_1{C}_1.$ Similarly, calculate the equations for $N_2,C_3,N_3.$ The equations can be left as function

of the inputs of this step $(O_3{O}_2{O}_1{O}_0)$; it is not necessary to calculate them as function of $B$

$(B_3{B}_2{B}_1{B}_0).$