CSE 120: Homework #2 Solutions

Fall 2023

  1. The Intel x86 instruction set architecture provides an atomic instruction called XCHG for implementing synchronization primitives. Semantically, XCHG works as follows (although keep in mind it is executed atomically by the CPU):
    void XCHG (bool *X, bool *Y) {
      bool tmp = *X;
      *X = *Y;
      *Y = tmp;
    }
    

    Show how XCHG can be used instead of test-and-set to implement the acquire() and release() functions of the spinlock data structure described in the "Synchronization" lecture.

    struct lock {
        bool isAvailable = TRUE;
    }
    
    void acquire (struct lock* p_lock) {
        bool isAvailable = FALSE;
        while (!isAvailable) {
            XCHG (&(p_lock->isAvailable), &isAvailable);
        }
    }
    
    void release (struct lock* p_lock) {
        p_lock->isAvailable = TRUE;
    }
    

  2. One of the goals of this question is to give you practice with context switching and thread queue manipulation in Nachos, and the hope is that you will find it useful for working on project 1 (so there is value in doing this problem well before the due date).

    Consider the following test program for an implementation of KThread.join in Nachos. It begins when the main Nachos thread calls KThread.selfTest. You do not need to know the details of how join is implemented. All you need to know is that when a parent thread calls join on a child thread, the parent does one of two things: (1) if the child is still running, the parent blocks until the child finishes (at which point the parent is placed on the ready queue); (2) if the child has finished, the parent continues to execute without blocking. Assume join uses a wait queue of some kind in its implementation.

    private static class A implements Runnable {
        A () {}
        public void run () {
            KThread t2 = new KThread (new B()).setName ("B");
    	System.out.println ("foo");
    	t2.fork ();
    	System.out.println ("far");
    	t2.join ();
    	System.out.println ("fum");
        }
    }
        
    private static class B implements Runnable {
        B () {}
        public void run () {
            System.out.println ("fie");
        }
    }
    
    public static void selfTest() {
        KThread t1 = new KThread (new A()).setName ("A");
        System.out.println ("fee");
        t1.fork ();
        System.out.println ("foe");
        t1.join ();
        System.out.println ("fun");
    }
    

    Assume that the scheduler runs threads in FIFO order with non-preemptive scheduling (no preemptive time-slicing), and threads are placed on wait queues in FIFO order. Trace the execution of this program until it returns from selfTest and (a) write the sequence of context switches that occurred up this point, (b) write the output of the program, and (c) list the queues that the threads are on, and their relative order if more than one thread is on a queue.

        a. Context switches: main → A → B → A → main
        b. Output: fee foe foo far fie fum fun
        c. Thread queues when selfTest returns:
              currentThread: main
              readyQueue: (nothing, A & B have finished)
              join wait queue: (nothing, A & B have finished)

  3. A common pattern in parallel scientific programs is to have a set of threads do a computation in a sequence of phases. In each phase i, all threads must finish phase i before any thread starts computing phase i+1. One way to accomplish this is with barrier synchronization. At the end of each phase, each thread executes Barrier::Done(n), where n is the number of threads in the computation. A call to Barrier::Done blocks until all of the n threads have called Barrier::Done. Then, all threads proceed. You may assume that the process allocates a new Barrier for each iteration, and that all threads of the program will call Done with the same value.

    a. Write a monitor that implements Barrier using Mesa semantics.

    monitor Barrier {
      int called = 0;
      Condition barrier;
    
      Barrier () {
        barrier = new Condition();
      }
    
      void Done (int needed) {
        called++;
        if (called == needed) {
          called = 0;
          barrier.wakeAll();
        } else {
          barrier.sleep();
        }
      }
    }
    

    b. Implement Barrier using an explicit lock and condition variable. The lock and condition variable have the semantics described at the end of the "Semaphore and Monitor" lecture in the ping_pong example, and as implemented by you in Project 1.

    class Barrier {
      int called = 0;
      Lock lock;
      Condition barrier;
    
      Barrier () {
        lock = new Lock();
        barrier = new Condition(lock);
      }
    
      void Done (int needed) {
        lock.acquire();
    
        called++;
        if (called == needed) {
          called = 0;
          barrier.wakeAll();
        } else {
          barrier.sleep();
        }
    
        lock.release();
      }
    }
    

  4. Torrey Pines would like your help synchronizing surfers and the ocean. Using pseudocode, implement the class Surfing using locks and condition variables to synchronize multiple surfer threads with one ocean thread (do not manipulate interrupts). Your solution also cannot change the lock, condition variable, or thread classes, and do not use data structures other than locks and condition variables to store references to threads.

    You need only use pseudocode in your answers. Your pseudocode does not have to compile, and you can use whatever syntax you are most comfortable with. But it does have to look like code.

    The Surfing class can be in one of two states, either breaking or calm. Surfer threads invoke the paddle method to indicate the direction, left or right, they would like to surf the break. When calm, surfer threads block until the next wave arrives. When breaking, surfer threads block if the wave is not breaking in their direction, and otherwise return immediately.

    The ocean thread invokes the wave method indicating the direction the next wake is breaking (left, right, or both ways). It changes the state to breaking and wakes up all surfer threads waiting to catch waves in that direction, or wakes up all threads if the wave is breaking in both directions. It invokes the done method to indicate that the wave is finished breaking, changing the state back to calm. The ocean thread alternates invoking wave and done, and the state is initially calm.

    
    class Surfing {
      enum State { calm, breaking; }
      enum Direction { left, right, both; }
      State state;
      Direction breakdir;
      Lock lock;
      Condition left, right;
    
      Surfing () {
        lock = new Lock();
        left = new Condition(lock);
        right = new Condition(lock);
        cv = new Condition(lock);  // for alternate implementation with one cv
        state = calm; breakdir = both;
      }
    
      void paddle (Direction dir) {
        lock.acquire();
        if (state == breaking && (breakdir == dir || breakdir == BOTH)) {
           lock.release();
           return;
        }
        if (dir == left) {
          left.sleep();
        } else {
          right.sleep();
        }
        lock.release();
      }
    
      void wave (Direction dir) {
        lock.acquire();
        state = breaking;
        breakdir = dir;
        if (dir == left) {
          left.wakeAll();
        } else if (dir == right) {
          right.wakeAll();
        } else {
          left.wakeAll();
          right.wakeAll();
        }
        lock.release();
      }
    
      void done () {
        lock.acquire();
        state = calm;
        lock.release();
      }
    }
    

  5. Eleanor, Chidi, Tahani, and Jason are working on their term papers in CSE 120, which is a 10,000 word essay on My All-Time Favorite Race Conditions. To help them work on their papers, they have one dictionary, two copies of Roget's Thesaurus, and two coffee cups.

    Consider the following state:

    Is the system deadlocked in this state? Explain using a resouce allocation graph.

    The resource allocation graph, shown below, can be fully reduced. Chidi is not blocked. Erasing the edges incident on Chidi unblocks Tahani and Jason. Erasing their edges unblocks Eleanor.