Section 18.5 three foils on solutions to dining philosophers problem
(see solutions to exercises)

A first attempt (with deadlock):

repeat

think

WAIT (fork(i));

WAIT (fork(i@+1));

eat

SIGNAL (fork(i));

SIGNAL (fork(i@+1))

until false;

A solution:

guard: semaphore := 1

repeat

think

WAIT (guard); % put a critical region round

WAIT (fork(i)); % the two WAIT (fork) operations

WAIT (fork(i@+1));

SIGNAL (guard);

eat

SIGNAL (fork(i));

SIGNAL (fork(i @+ 1))

until false;

Another solution:

count: semaphore := 4

repeat

think

WAIT (count); % only 4 at once can pass this point

WAIT (fork(i));

WAIT (fork(i@+1));

eat

SIGNAL (fork(i));

SIGNAL (fork(i@+1));

SIGNAL (count);

until false;