Do the turn variable solution, array-of-flags solution satisfy the bounded waiting condition?

Comments

• KG AI MentorKG AI Mentor6 Jan 2026

  Quick AnswerYes, in the classic two-process implementations (Dekker/Peterson style) that use a turn variable and per-process flags, bounded waiting holds—the waiting time is bounded (effectively by 1 other entry). If you drop the turn or use a naive array-of-flags without a turning rule, bounded waiting may not be guaranteed. Key StepsUse the canonical two-process pattern: each process sets its flag and, when both want in, the turn variable decides who goes next. Turn-variable version: after setting its flag, a process sets turn to the other and waits while the other is waiting and turn == other. This bounds waiting to at most one other-entry. Array-of-flags variant (Peterson-like) relies on the same waiting condition: wait while flag[other] && turn == other; this preserves bounded waiting. If you implement a naive array-of-flags without a turning rule, starvation or unbounded waiting can occur; ensure a proper turn/update mechanism and visibility guarantees. Resources: Peterson’s algorithm (two-process mutual exclusion), Dekker’s algorithm, and textbooks like Silberschatz/Tanenbaum on operating systems. Lamport’s Bakery algorithm is a scalable next step for more than two processes. — KG AI Mentor 🤖