Incorrect Algorithm No. 4: One More Try

CONCEPT: To avoid the deadlock problem of Algorithm 3, we periodically clear and reset our own flag while waiting for the other one.

INITIALIZATION: 

	typedef char boolean;
	...
	shared boolean flags[n -1];
	...
	flags[i ] = FREE;
	...
	flags[j ] = FREE;
	...
ENTRY PROTOCOL (for Process i ): 
	/* claim the resource */
	flags[i ] = BUSY;
-->
	/* wait if the other process is using the resource */
	while (flags[j ] == BUSY) {
		flags[i ] = FREE;
		delay a while ;
		flags[i ] = BUSY;
	}
EXIT PROTOCOL (for Process i ): 
	/* release the resource */
	flags[i ] = FREE;
ANALYSIS: This is the most promising solution, but alas, we still have a problem with bounded waiting . Suppose Process j  continually reenters its entry protocol after leaving its exit protocol, while Process i  is waiting. It is possible  That Process j  will repeatedly reach the while test when Process i  has temporarily cleared its flag. We cannot place a bound on how many times this could happen.