Given algorithm (pseudo-code): 1.Print "Enter any number" 2.Input n1 3.n1cube = n1 * n1 * n1 4.print n1cube (a) What is the output of this algorithm if the input is 3 (n1=3 in line 2)? (b) What does this algorithm do? (c) Modify this algorithm so that it outputs (prints) the result of multiplication of two number n1 and n2 (both n1 and n2 are input values by user).

The following two transactions result in a deadlock. Show through examples how this deadlock can be dealt with or avoided using conservative and basic two-phase locking algorithms. T'1 read_lock (Y); read_item (Y); write_lock (X); (waits for X) T'2 read_lock (X); read_item (X); write_lock (Y); (waits for Y)

If f(n) is the number of times that y=y+A[i]; is executed in the code segment below, which of the following is the correct big-Oh notation for f? y=0;for(j=0;j<n;j++) { for (i=0;i<n;i++) { y=y+A[i]; }} Group of answer choices O(1) O(n2) O(nlog2n) O(n)

Assume n is the number of inputs and that the log function is base 2. Treat this piece of code as a pseudocode. Give the big-Oh characterization, in terms of n for the following algorithm: for i in range(n*logn): for j in range(i): for k range(1000): print('I did it!')

JAVA Make a function that will get the value of ln(n!). use Math.log() function in replace to ln()

Please do it in python

Bottom: Given b, set the low-order b bits of x to 1; the others to 0. For example, if b is 3, x should be 7. Pay special attention to the edge cases: if b is 32 x should be −1; if b is 0 x should be 0. Do not use - in your solution. Write code in Java. Provided input(s): b Permitted: 40 operations (may use !, ~, +, <<, >>, &, ^, |) Hint: The obvious solution ~(0xFFFFFFFF << b) won’t work. Bit shifts always do a modulo on their right-hand operand, so a << b does the same thing as a << (b % (8*sizeof(a)). Thus, << 32 and << 0 do the same thing.