An Operating Systems Vade Mecum

Advance-claim algorithm 137Every time a resource is granted, the process that requested it must have had a claimhigher than the number of resources granted so far, so × will be placed in at least the S[0]column, if not in others. Therefore, the current value of S[0] is the number of availableresources, since we subtract one for each × in the column. If S[0] < 0, the state is notonly unsafe, it is also unrealizable, since more resources have been granted than exist.Next, let us see what happens if a process, say D, claims the entire set of resourcesand then acquires them all. We can see how S is calculated for r = 4:Index 0 1 2 3Original S 4 3 2 1D × × × × D × × × D × ×D ×Final S 0 0 0 0A process that acquires its entire claim makes a triangular ‘‘footprint’’ of × marks in ourpicture. In the case in which the initial claim is the entire set of resources, the footprint islarge enough to reduce every entry in S to 0. When a process has received its entireclaim, the footprint is complete and makes a triangle.Now let us say that one or more entries of S have become −1. For concreteness, letus say that S[3] is the leftmost column with an entry of −1. What is the largest possiblevalue for S[0], S[1], and S[2]? S[2] cannot be greater than 0 — that is, one larger thanS[3] — since every × in column 3 is in a line that also puts × in column 2, and column 2starts only 1 larger than column 3. S[2] cannot be negative, since column 3 was the firstwith a negative value. So S[2] must be 0. Similar reasoning shows that S[1] is either 0or 1, and S[0] is 0, 1, or 2.Since S[2] = 0 and S[3] = −1, there cannot be any line of exactly three × marks.Such a line would have × in column 2 but not in column 3 and therefore would havedecreased S[2] more than S[3], so S[2] would have to be less than 0. Therefore, no footprinthas a line with exactly three × marks. All footprints are either large and incomplete,with their last row having at least four × marks, or they are small, with their first row havingat most two × marks. All the small footprints can finish. They belong to processesthat form the start of a safe sequence. (There are at most two such processes, one with aclaim of 2 and only one line of footprint and one with a claim of 1 and one line of footprint.)After these processes have terminated, S[0] will be 2 and S[1] will be 1. Onlyprocesses with large incomplete footprints are left. Not a single one of them can finish,because each has at least three lines of footprint to complete, so S[0] must be at least 3.But S[0] is only 2. Therefore, there is no safe sequence. This argument does not dependon the fact that we chose column 3 as the first with a negative value.We have been assuming throughout that there is only one resource class. Thesituation becomes more complicated if there are several resource classes. It is notenough to find a safe sequence for each resource class independently; a sequence must befound that is safe with respect to each resource class.The advance-claim algorithm is conservative insofar as it cautiously avoids enteringan unsafe state even if this unsafe state has no deadlock. However, it achieves the