In prior classes, you learned that signed values are represented in $2\text{'}$s complement form. So subtraction such as A$-$B is simply A $+2\text{'}$s complement of B$.$ Also, in hardware subtraction of any two values is done as if they are signed values. You may declare $P$ and $Q$ as unsigned char $($or int$)$ in your C code, but $P-Q$ is done as $P+2\text{'}$s complement of $Q.$

eg: $3+5=8($are they signed or unsigned$)?$ answer is the same, still $8.$

eg: $3-5=-2$ even if $3$ and $5$ are looked at as unsigned. We cannot say $3-5$ is $2$ because $3$ and $5$ are unsigned. ALU hardware makes the answer $11111$dots$11110$ which is $-2.$

Q$1)$ An arithmetic and logic unit $($ALU$)$ performs integer and logic operations. Implement $($draw a minimized logic circuit$)$ the logic of a simple ALU that performs $2-$bit addition and subtraction. Assume the operands are $A_1{A}_0$ and $B_1{B}_0,$ and the output is $C_1{C}_0.$ The control signal $Op$ is only one bit, which selects addition or subtraction. Two ways to solve this. $($a$)$ Use $2$ bit adder $($half adder $+$ full adder$)$ with inputs to it being modified based on Op signal, or $($b$)$ start from a truth table. Then you may have to develop a truth table and use K maps for $5$ variables leading to a large circuit. See next page for help. Recall that subtraction is done with adding after making the subtrahend negative $(2\text{'}$s complement$)$ regardless of whether arguments are signed or unsigned.