Complete the code for the provided Sequence class. Your code should provide the expected results when run through the provided SequenceTester class. Add code for these two methods in the Sequence class: public Sequence append(Sequence other) append creates a new sequence, appending this and the other sequence, without modifying either sequence. For example, if a is 1 4 9 16 and b is the sequence 9 7 4 9 11 then the call a. append(b) returns the sequence 1 4 9 16 9 7 4 9 11 without modifying a or b. public Sequence merge(Sequence other) merges two sequences, alternating elements from both sequences. If one sequence is shorter than the other, then alternate as long as you can and then append the remaining elements from the longer sequence. For example, if a is 1 4 9 16 and b is 9 7 4 9 11 then a. merge(b) returns the sequence without modifying a or b. 3. 1 9 4 7 9 4 16 9 11

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.\