$(9+8+1\mathrm{.}5+2\mathrm{.}5+30=51$ points $)40$min. $1\mathrm{.}2\mathrm{.}4$ State diagram: Complete the $7$ missing state transition conditions for $7$ transitition arrows.

Complete RTL for the three states. In state C$\_$I$\_$J $($compare I and J$),$ you would like to

decrement

$($I$/$J$/$I or J$/$I and J$/$I

unconditionally and $J$ conditionally $/$ neither I nor $J).$ Crucial point: You want to arrive in

INS$\_$AI$\_$BJ with the right combination of I and $J$ whether you arrive from C$\_$I$\_$J or INS$\_$AI $.$

If $B$ is an $\frac{?}{\text{b}\text{a}\text{r}}(8)-$bit number $(B[7$:$0]),$ the first pair to be compared in INS$\_$AI$\_$BJ is

$($A $[7],$ B $[7]/$ A $[6],$ B $[6]/$ neither$).$ And if $B$ is a $16-$bit number $),$ the first pair to be

compared in INS$\_$AI$\_$BJ is

neithe

A and $B$ are negative numbers represented in $2\text{'}$s complement notation. $A$ is $16-$bit in size and $B$

is $8-$bit in size.

For $A$ to be equal to $B,$ all the $8$A bits $(A[14$:$7])$ shall be

$($all zeros$/$all ones$)$

$($and$/$or$)$

the rest of the $7$A bits $($A$[6$:$0])$ shall be equal to the corresponding $7B$ bits $($B$[6$:$0]).$

For $A$ to be less than $B($like in $-3$ is less than $-2),$ it is enough if any of the $8$A bits $(A[14$:$7])$ is

a

$($zero $/$ one$).$ On the other hand, if all those $8$A bits $(A[14$:$7])$ are

$($all zeros$/$all ones$),$ then, for $A$ to be less than $B,$ we need the $7-$bit $A(A[6$:$0])$ to be

$($lower $/$ higher$)$ than the corresponding $7-$bit $B(B[6$:$0]).$ Here we compare the two $\%-$bit numbers

treating them as

$($signed $/$ unsigned$)7-$bit numbers.

$1\mathrm{.}2$ Mealy machine design: Browse through the state diagram on the next page first. Here, we perform

serial inspection of bits of $A($or bits of A and $B)$ to compare them. $A$ is a $16-$bit number, but

for this part of the question, B can be any where between $8-$bit to $16-$bit number. Here, we are

allowed to inspect at a time $($in a clock$)$ one bit of A $($A $[$I$])$ and $($simultaneously$,$ if needed$)$ one

bit of $).$ The I and the $J$ are indices into the A and $B$ respectively. I is initialized to $15.$

$J$ is initiated to Jini $($Jinitial$)$ which can be anywhere between $7$ through $15($corresponding

to the $B$ sizes of $8-$bit to $16-$bit$).$ You will be needing to compare I and $J$ to see when they are

equal.

Note: Your TA says that $,$after START is given, you should not take more than $16$ clocks. After

all, $A$ is $16$ bits and $B$ is at most $16$ bits. So decrement I and$/$or $J$ as soon as possible!

Suppose B is an $8-$bit number.

State an example of A and $B($in binary$)$ such that the conclusion is drawn in the least number of clocks.

$A=$

;$B=$

How many ciocks are spent in INs$\_$A$1\_$Bu state for the above numbers?

State an example of A and $B($in binary$)$ such that the conclusion is drawn in the most number of clocks.

A $-.$

How many clocks are spent in INS$\_$AI$\_$BJ state for the above numbers?

Pts$)1\mathrm{.}2\mathrm{.}2$ Since A and $B$ are negative, there is no point looking at $A[15]$ and $B[$Jini$].$ True $/$ False

pts $1\mathrm{.}2\mathrm{.}3$ Since $A\text{'}$s size is fixed at $16-$bit, if $A$ is equal to $B($equal in value, but not necessarily in size$),$

it takes the same number of clocks to compare A and $B,$ irrespective of the size of $B.$ True $/$ False