Select all of the following statements that are true 0 After the execution of the banker's algorithm, if Finish[i] == false for all processes i, then the system is in a safe state. The banker's algorithm can only be applied when there are single units of each resource type only. Unsafe system states can lead to deadlocks. Deadlock avoidance aims to ensure that a system of processes will never enter an unsafe state. Claim edges in resource allocation graphs indicate resource requests that a process may send out in the future. In order for the banker's algorithm to be applicable, when a process starts up, it must state in advance the maximum number of resources it may request in the

Select all of the following statements that are true 0 After the execution of the banker's algorithm, if Finish[i] == false for all processes i, then the system is in a safe state. The banker's algorithm can only be applied when there are single units of each resource type only. Unsafe system states can lead to deadlocks. Deadlock avoidance aims to ensure that a system of processes will never enter an unsafe state. Claim edges in resource allocation graphs indicate resource requests that a process may send out in the future. In order for the banker's algorithm to be applicable, when a process starts up, it must state in advance the maximum number of resources it may request in the

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Modify the reader-writer lock example to help waiting writers. When the number of current readers exceeds R, then block readers until the lock becomes free. Modify the appropriate functions. Do NOT worry about ANY errors. lock L; condition CV; // 0 unheld; -1 writer has it;> 0 readers have it int num readers = 0; int R; } acquire_rlock () { lock (&L); while (num_readers < 0) wait (&CV, &L); num_readers++; unlock (&L); release () { lock (&L); num_readers unlock (&L); if (num_readers num readers = 0; } else num_readers--; if (num_readers acquire_wlock () { // this does not change! lock (&L); while (num readers != 0) wait (&CV, &L); -1; } unlock (&L); == == -1) 0) { broadcast (&CV, &L);
The following answers are claimed to be properties of loop invariants. However, a number of them are false. Which ones are they? Select one or more O a. Loop invariants help us ensure progress from one iteration to the next. b. Loop invariants describe changes in the output data structure as a result of the assumptions made by the inductive hypothesis. C. cop invariants describe changes in the output data structure as a result of the instructions executed following the inductive step. Loop Invariarts descibe changes in the output data structure as a result of the instructions executed following the Inductive step. d. Loop invariants guarantee termination of the algorithm.
Correct answer will be upvoted else Multiple Downvoted. Computer science. You are given a multiset of abilities of two. All the more unequivocally, for every I from 0 to n selective you have cnti components equivalent to 2i. In one activity, you can pick any one component 2l>1 and partition it into two components 2l−1. You ought to perform q inquiries. Each inquiry has one of two sorts: "1 pos val" — appoint cntpos:=val; "2 x k" — ascertain the base number of tasks you really want to make basically k components with esteem lower or equivalent to 2x. Note that all inquiries of the subsequent kind don't change the multiset; that is, you simply work out the base number of activities, you don't perform them. Input The principal line contains two integers n and q (1≤n≤30; 1≤q≤2⋅105) — the size of cluster cnt and the number of inquiries. The subsequent line contains n integers cnt0,cnt1,… ,cntn−1 (0≤cnti≤106). Next q lines contain inquiries: one for each line.…
If only the premises below have been provided, and paying very careful attention to the order in which each step of an argument is presented, which of the following arguments is valid? Vr D(r) 3r R(r) Select one: D(k) R(k) D(k) A R(k) 3r (D(r) A R(x)) R(j) D(k) R(j) A D(k) Vr (R(x) ^ D(x)) R(k) D(k) R(k) A D(k) 3r (R(x)A D(x)) R(k) D(k) R(k) A D(k) Vr (R(x) ^ D(x)) D(k) R(j) D(j) A RG) Vr (D(x} A R(x)) Rj) D(k) RG) A D(k) 3r (R(x} A D(x)) D(k) R(k) D(k) A R(k) Vr (D(r} A R(x)) D(k) Rj) D(k) A R(j) 3r (D(r) A R(x)) O none of these options
Using c++ Apply both breadth-first search and best-first search to a modified version of MC problem. In the modified MC, a state can contain any number of M’s and any number of C’s on either side of the river. Assume the goal is always to move all the persons on the left side to the right side. The Initial state should be a parameter given to the program at beginning of execution. As in the original problem, boat capacity =2, the boat cannot move by itself, and on either side C’s should not outnumber M’s. For best-first search, you need to come up with an appropriate heuristic. In addition to solving the problem, your grade will also be based on th effectiveness of the heuristic. As an example, the program should execute as follows. Initial state… Enter number of M’s on left side of the river: 3 Enter number of C’s on left side of the river: 1 Enter number of M’s on right side of the river: 0 Enter number of C’s on right side of the river: 0 Enter location of the boat: L The output…
QUESTION 3 Examine the following loop segment Prove that it is correct by proving the invariant Q Q:j = x - i // returns the value of (x-y) given non- negative integers x, y local variables: integers i,j i=0 j= x while iy do j=j-1 i= l+1 end while //j now has the value of (x - y) return j
x = 1; y = 2; z = 1; n = 5; while (z
C
Maintaining the Loop Invariant (LI & not exit & codeloop → LI): Suppose that LI (the statement of the loop invariant before the iteration) is true,the exit condition exit is not, and we have executed another iteration of thealgorithm.
OZ PROGRAMMING LANGUAGE Exercise 1. (Efficient Recurrence Relations Calculation) At slide 54 of Lecture 10, we have seen aconcurrent implementation of classical Fibonacci recurrence. This is: fun {Fib X} if X==0 then 0 elseif X==1 then 1 else thread {Fib X-1} end + {Fib X-2} end end By calling Fib for actual parameter value 6, we get the following execution containing several calls ofthe same actual parameters.For example, F3, that stands for {Fib 3}, is calculated independently three times (although it providesthe same value every time). Write an efficient Oz implementation that is doing a function call for a givenactual parameter only once.Consider a more general recurrence relation, e.g.:F0, F1, ..., Fm-1 are known as initial values.Fn = g(Fn-1, ..., Fn-m), for any n ≥ m.For example, Fibonacci recurrence has m=2, g(x, y) = x+y, F0=F1=1
Prove the following implication. Submit your proof in the file uncurry.log, following the same instructions as in Task 2 P (Q→ R) = (PAQ) → R (Functional programmers may recognize this as uncurrying.)
Find the error in the following argument: Theorem. All computer programs contain the same number of bugs. Proof. If we show that any set of n programs contains the same number of bugs, then we have proved the theorem. We argue by induction on n. If n=1 then in any set of 1 program, all the programs contain the same number of bugs, so the statement is true, in this case. Now suppose that for every set of programs containing less then n programs, all the programs have the same number of bugs. Let DD be a set of n programs p1,p2,…,pn. Now the set D1={p2,…,pn} contains n−1 programs and hence they all contain the same number of bugs, by the induction hypothesis. Similarly the set D2={p1,…,pn−1} contains n−1 programs and hence they all contain the same number of bugs. In particular p1 and pn contains the same number of bugs as the other programs in the set. Hence all the programs contain the same number of bugs.