Introduction

In this assignment, we will have further practice with condition variables and mutex locks. There are four problems that you need to solve.

Learning Objectives

At the end of this assignment, you should be able to:

Implement multi-threaded code using the pthreads library.
Solve basic concurrency problems using condition variables.

Getting the Source Code

We will do this lab in the main branch of your labs repository. To make sure you are on the right branch, check it out using:

  $ git branch

The branch you are currently on will be highlighted for you (with a * next to its name).

If you are working on the main or master branch, then follow these instructions:

  $ git fetch upstream
  $ git pull upstream main

At this stage, you should have the latest copy of the code, and you are good to get started. The starter code is contained under the threads_basics/ directory.

If you are currently on a different branch (say you are still on clab_solution from a previous lab), then we need to switch to main or master (depending on your default’s name).

First, add, commit, and push your changes to the clab_solution to make sure you do not lose any progress you did on the last lab. To check the status of your current branch, you can use:

  $ git status

This will show you all the files you have modified and have not yet committed and pushed. Make sure you add those files, then commit your changes, and push them.

If git push complains about not knowing where to push, you’d want to push the current branch you are on. So for example, if I am working on clab_solution, then I’d want to do git push origin clab_solution.

Now, you are ready to swap back into main (or master).

$ git checkout main

Then, grab the latest changes using:

$ git fetch upstream
$ git pull upstream main

At this stage, you should have the latest copy of the code, and you are good to get started. The starter code is contained under the condvar_challenge/ directory.

Band (25 points)

This problem refers to the code in the file band.c. Imagine a group of friends are getting together to play music. They are, however, arriving at different times; arriving can happen at any time (e.g., when some other friends are playing). They share a single stage on which at most one band can play at a time.

There are three different kinds of friends: drummers, singers, and guitarists, and the follow the following set of rules:

It takes one of each kind to make a band.
Only 1 band can be playing at once.
Once conditions (1) and (2) are met, the players can start playing and stop playing in any order.
All 3 players must stop playing before a new set of 3 can start playing.

Example output

drummer arrived
drummer arrived
guitarist arrived
guitarist arrived
singer arrived
drummer playing
guitarist playing
singer playing
singer arrived
singer arrived
drummer arrived
guitarist arrived
drummer finished playing
guitarist finished playing
singer finished playing
singer playing
guitarist playing
drummer playing
singer finished playing
guitarist finished playing
drummer finished playing
guitarist playing
drummer playing
singer playing
guitarist finished playing
drummer finished playing
singer finished playing
Everything finished.

Little Red Hen (25 points)

This problem refers to the code in littleredhen.c. This system has four threads: (1) the duck, (2) the cat, (3) the dog, and (4) little red hen. The duck, the cat, and the dog eat bread, while the little red hen is the one that makes the bread.

The little red hen makes seven loaves per batch, but she only has the patience for six batches. The little red hen will only make a batch if there are no loaves left.

The other three animals (the duck, the cat, and the dog) each want to eat 14 loaves of bread, but only one of them can be in the kitchen at a time (to avoid fights over who gets what bread).

When the duck, cat, or dog notice that there are no loaves of bread available, they complain to the little red hen and wait (in the kitchen) for the next batch to be ready.

Use condition variables and mutex locks to enforce the above constraints. Note that the global numLoaves variable should be left as is (i.e., you cannot make it a semaphore or a condition variable).

Sample output

Below is a sample output from running a trace that satisfies the above constraints.

Sleepy Cat          : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Lazy Dog            : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Sleepy Cat          : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Sleepy Cat          : Hey, Little Red Hen, make some more bread!
Little Red Hen      : A fresh batch of bread is ready.
Little Red Hen      : I'm fed up with feeding you lazy animals! No more bread!
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Noisy Yellow Duck   : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : Mmm, this loaf is delicious.
Sleepy Cat          : Mmm, this loaf is delicious.
Noisy Yellow Duck   : I've had my fill of bread. Thanks, Little Red Hen!
Lazy Dog            : Mmm, this loaf is delicious.
Lazy Dog            : I've had my fill of bread. Thanks, Little Red Hen!
Sleepy Cat          : I've had my fill of bread. Thanks, Little Red Hen!
Everything finished.

Priority (25 points)

This problem refers to the code in priority.c. In this system, there are threads numbered 1 through 6 and a critical section. The thread numbers are priorities, where thread 6 is the one with the highest priority, thread 5 is next, etc.

If the critical section is empty, any thread can enter. While a thread is in the critical section, no other thread can enter, regardless of its priority. When a thread leaves the critical section and there are threads waiting, the highest priority waiting thread is allowed to enter the critical section.

Make sure in your code that a newly arriving thread cannot jump into the critical section as the current thread finishes, bypassing priority rules.

Write a solution that satisfies the above constraints using mutex lock and condition variables.

Three Jobs (25 points)

This problem refers to the code in threeJobs.c. In this system, imagine there is a shared memory space called house.

There are 3 different kinds of operations on house: carpenters, painters, and decorators. For any particular kind of operation, there can be an unlimited number of threads doing the same operation at once (e.g., unlimited carpenter threads, etc.). However, only one kind of operation can be done at a time (so even a single carpenter should block all painters and vice versa).

Use mutex locks and condition variables to enforce this constraint. You don’t have to worry about starvation (e.g. that constantly arriving decorators might prevent carpenters from ever running) - though maybe it would be fun to consider how you would solve in that case.

This is similar to the readers/writers problem BTW.

Sample output

Below is a sample of the output generated by a trace of the program that satisfies the above constraint.

starting carpentry
starting carpentry
starting carpentry
finished carpentry
finished carpentry
finished carpentry
starting painting
starting painting
starting painting
finished painting
finished painting
finished painting
starting decorating
starting decorating
starting decorating
finished decorating
finished decorating
finished decorating
Everything finished.