Consider we have n pointers that need to be swizzled, and swizzling one point will take time t on average. Suppose that if we swizzle all pointers automatically, we can perform the swizzling in half the time it would take to swizzle each separately. If the probability that a pointer in main memory will be followed at least once is p, for what values of p is it more efficient to swizzle automatically than on demand?

Use a physical stopwatch to record the length of time it takes to run the program. calculate the difference obtained by calls to the method system.currenttimemillis() just before the start of the algorithm and just after the end of the algorithm. calculate the difference obtained by calls to the method system.currenttimemillis() at the start of the program and at the end of the program so that the elapsed time includes the display of the result. use the value returned by the method system.currenttimemillis() just after the end of the algorithm as the elapsed time.