The first known correct software solution to the critical-section problem for n processes with a lower bound on waiting of n ? 1 turns was presented by Eisenberg and McGuire.
The processes share the following variables: enum pstate {idle, want in, in cs}; pstate flag[n];
int turn;
All the elements of flag are initially idle; the initial value of turn is immaterial (between 0 and n-1). The structure of process Pi is shown in Figure 6.26.
Prove that the algorithm satisfies all three requirements for the critical-section problem.

4.11 painting a wall (1) prompt the user to input integers for a wall's height and width. calculate and output the wall's area (integer). note that in this case there is a new line after each prompt. (submit for 1 point). enter wall height (feet): 11 enter wall width (feet): 15 wall area: 165 square feet (2) extend to also calculate and output the amount of paint in gallons needed to paint the wall (floating point). assume a gallon of paint covers 350 square feet. store this value in a variable. output the amount of paint needed using the %f conversion specifier. (submit for 2 points, so 3 points total). enter wall height (feet): 11 enter wall width (feet): 15 wall area: 165 square feet paint needed: 0.471429 gallons (3) extend to also calculate and output the number of 1 gallon cans needed to paint the wall. hint: use a math function to round up to the nearest gallon. (submit for 2 points, so 5 points total). enter wall height (feet): 11 enter wall width (feet): 15 wall area: 165 square feet paint needed: 0.471429 gallons

Drag each label to the correct location on the imagelist the do’s and don’ts of safeguarding your password. keep yourself loggedin when you leave your computer. don’t write your password down and leave it whereothers can find it. share your password with your friends. each time you visit a website, retain the cookies on your computer. use a long password with mixed characters.

Write a method named addall that could be placed inside the hashintset class. this method accepts another hashintset as a parameter and adds all elements from that set into the current set, if they are not already present. for example, if a set s1 contains [1, 2, 3] and another set s2 contains [1, 7, 3, 9], the call of s1.addall(s2); would change s1 to store [1, 2, 3, 7, 9] in some order. you are allowed to call methods on your set and/or the other set. do not modify the set passed in. this method should run in o(n) time where n is the number of elements in the parameter set passed in.