Per what is written in EWD310 "Hierarchical Ordering of Sequential Processes", it looks like number 5 has been chosen for educational purposes, in order to make it easier for students to understand algorithm designed to demonstrate solution of the problem.
This very paper further supports the idea that 5 is not really relevant to general problem, first by explicitly stating that "the problem could have been posed for 9 or 25 philosophers..." and next, by representing it in terms of two concurrently operating entities, "class A and class B, sharing the same resource..."
Solution used by Dijkstra introduces three "states of philosopher": thinking, eating, hungry. Code presented to solve the problem, operates these three states, along with an unrelated to it number of philosophers.
Would author has chosen number of philosophers 2, 3 or 4, this could cause confusion of the students reading the code, whether chosen number is related to the amount of states or something else. This can easily be tested by trying mentioned numbers in description quoted from EWD310 below: note for example how this would change
[0:1] and statements involving
As opposed to this, number 5 looks fairly innocent and doesn't invoke unneeded associations. One can say that it has been chosen to better illustrate that amount of philosophers is, well, arbitrary.
Mentioned algorithm is presented in EWD310 as follows:
...we associate with each philosopher a state variable, "C" say, where
C[i] = 0 means: philosopher
i is thinking
C[i] = 2 means: philosopher
i is eating.
we introduce for the last transition an intermediate state
C[i] = 1 means: philosopher
i is hungry
Now each philosopher will go cyclically through the states 0, 1, 2, 0 ...... The next question to ask is: when has the (dangerous) transition from 1 to 2 to take place for philosopher
In the universe we assume declared
semaphore mutex, initially = 1
integer array C[0:4], with initially all element = 0
semaphore array prisem[0:4] with initially all elements = 0
procedure test (integer value K);
if C[(K-1) mod 5] ≠ 2 and C[K]= 1
and C[(K+1) mod 5] ≠ 2 do
begin C[K]:= 2; V(prisem[K]) end;
(This procedure, which resolves unstability for
K when present, will only be called from within a critical section).
In this universe the life of philosopher
w can now be coded
cycle begin think;
C[w]:= 1; test (w);
C[w]:= 0; test [(w+l) mod 5];
test [(w-1) mod 5];
And this concludes the solution I was aiming at...