Keytool error java io exception keystore password was incorrect.

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk,thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\

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Assume that two parallel arrays have been declared and initialized: healthoption an array of type char that contains letter codes for different healthcare options and annual cost an array of type int. the i-th element of annual cost indicates the annual cost of the i-th element of healthoption. in addition, there is an char variable, best2.write the code necessary to assign to best2 the health option with the lower annual cost, considering only the first two healthcare options. thus, if the values of healthoption are 'b', 'q', 'w', 'z' and the values of annualcost are 8430, 9400, 7050, 6400 your code would assign 'b' to best2 because 8430 is less than 9400 and is associated with 'b' in the parallel array. (we ignore 'w' and 'z' because we are considering only the first two options.)