This algorithm solves the critical section problem for n processes in software. The basic idea is that of a bakery; customers take numbers, and whoever has the lowest number gets service next. Here, of course, "service" means entry to the critical section.
1 var choosing: shared array[0..n-1] of boolean; 2 number: shared array[0..n-1] of integer; ... 3 repeat 4 choosing[i] := true; 5 number[i] := max(number,number,...,number[n-1]) + 1; 6 choosing[i] := false; 7 for j := 0 to n-1 do begin 8 while choosing[j] do (* nothing *); 9 while number[j] <> 0 and 10 (number[j], j) < (number[i],i) do 11 (* nothing *); 12 end; 13 (* critical section *) 14 number[i] := 0; 15 (* remainder section *) 16 until false;
lines 1-2: Here, choosing[i] is true if Pi is choosing a number. The number that Pi will use to enter the critical section is in number[i]; it is 0 if Pi is not trying to enter its critical section.
lines 4-6: These three lines first indicate that the process is choosing a number (line 4), then try to assign a unique number to the process Pi (line 5); however, that does not always happen. Afterwards, Pi indicates it is done (line 6).
lines 7-12: Now we select which process goes into the critical section. Pi waits until it has the lowest number of all the processes waiting to enter the critical section. If two processes have the same number, the one with the smaller name - the value of the subscript - goes in; the notation "(a,b) < (c,d)" means true if a < c or if both a = c and b < d (lines 9-10). Note that if a process is not trying to enter the critical section, its number is 0. Also, if a process is choosing a number when Pi tries to look at it, Pi waits until it has done so before looking (line 8).
line 14: Now Pi is no longer interested in entering its critical section, so it sets number[i] to 0.
Why does Lamport's Bakery algorithm used a variable called choosing (see the algorithm, lines 1, 4, 6, and 8)? It is very instructive to see what happens if you leave it out. This gives an example of mutual exclusion being violated if you don't put choosing in. But first, the algorithm (with the lines involving choosing commented out) so you can see what the modification was:
1 var (* choosing: shared array [0..n-1] of boolean; *) 2 number: shared array [0..n-1] of integer; ... 3 repeat 4 (* choosing[i] := true; *) 5 number[i] := max(number,number,...,number[n-1]) + 1; 6 (* choosing[i] := false; *) 7 for j := 0 to n-1 do begin 8 (* while choosing[j] do ; *) 9 while number[j] <> 0 and 10 (number[j], j) < (number[i],i) do 11 (* nothing *); 12 end; 13 (* critical section *) 14 number[i] := 0; 15 (* remainder section *) 16 until false;
Suppose we have two processes just beginning; call them p0 and p1. Both reach line 5 at the same time. Now, we'll assume both read number and number before either addition takes place. Let p0 complete the line, assigning 1 to number, but p1 block before the assignment. Then p0 gets through the while loop at lines 9-11 and enters the critical section. It leaves, and attempts re-entry. It passes line 5, assigning 2 to number; it then goes into the critical section, as number is 0. Now p1 resumes execution. It completes the assignment in line 5; as it read number and number as 0, it assigns 1 to number. It then reaches the while loop at lines 9-11, and as number is 2, it will pass through the while loop and enter the critical section. Now both p0 and p1 are in the critical section.
The reason for choosing is to prevent the while loop in lines 9-11 from being entered when process j is setting its number[j]. Note that if the loop is entered and then process j reaches line 5, one of two situations arises. Either number[j] has the value 0 when the first test is executed, in which case process i moves on to the next process, or number[j] has a non-zero value, in which case at some point number[j] will be greater than number[i] (since process i finished executing statement 5 before process j began). Either way, process i will enter the critical section before process j, and when process j reaches the while loop, it will loop at least until process i leaves the critical section.
Department of Computer Science
University of California at Davis
Davis, CA 95616-8562