You hash a symmetric key and store the hash on your workstation. You call your friend and verbalize the symmetric key to him or her. Your friend then hashes the key using the exact same hash algorithm that you used (without random salting or anything clever). He or she sends you the hash which when you compare with your version of the hash, you find they are not the same hash. From this you conclude that: a. Your friend must have been on a different operating system which resulted in a slightly different implementation of the algorithm.
b. Your friend must have mis-typed the key when you verbalized it over the phone.
c. Your friend must have decrypted your hash to find the key before re-hashing.
d. Your friend must have double-hashed the key

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.\