1. Using a N-bit adder, design the following N-bit comparators. A comparator's output is 1 if the condition is met, 0 otherwise. a. greater than (A > B, where A and B are N-bit signed integers).
b. smaller or equal (A <= B)
Sketch and submit your schematics for these two circuits. Schematics can be hand-drawing and then scanned into PDF for submission, or using Digital.
2. Using (System) Verilog design the 32-bit ALU as discussed in class but with an additional output 'Zero' that is 1 when all bits of the output Y is O. Write a testbech to test your design.
3. Design and test a 4-bit rotator that has two inputs: A and rotamt, and two outputs Yleft and Yright using (System) Verilog. The rotation amount (number of bits to be rotated) is given in rotamt and the output of left and right rotation will be in Yleft and Yright, respectively.

E. i hope it's you done your thing

So is there any ways to explain this better?

192.168.112.5  hope this : d

Public static int mystery(int a[], int x) {     int c = 0;     for(int i = 0; i < a.length; i++) {        if (a[i] == x) c++;     } return c; }the method counts how many times an integer shows up in an integer array.the c variable is declared in the method to keep track of how many times the certain value has shown up.the for-loop loops through the entire array.the line    if (a[i] == x) c++;   checks if each element of the array is equal to the inputted integer to check. for each number in the array that is equal to the inputted integer x, the c variable will be incremented by 1. otherwise it will do nothing.then c is returned at the end of the loop.for example, if we have, in the main method with the mystery method coded in the class:     int arraytest[] = {2, 23, 23, 23, 3};     int testme = 23;     int count = mystery(arraytest, testme);     system.out.println(count); then the print out will be    3 because there are three 23s in that array.