Cmpt 441-711: Homework 2 Question 6.22

Denote by Sk the set of floors that can be reached from floor 32 in k, but not in k − 1, steps. Then, obviously, S0 = 32. Now we give description of the algorithm for computing Sk, for any k > 0, assuming that all S0, S1, . . . , Sk−1 have been computed. From the definition of Sk it follows that any element x of Sk was reached from at least one of x − 11, where x − 11 ∈ Sk−1 and x + 6, where x + 6 ∈ Sk−1. Also, if some element y ∈ {1, . . . , 50} belongs to any of S0, S1, . . . , Sk−1, then y / ∈ Sk (as shorter path from 32 to y has already been found). Therefore, in order to compute Sk, we iterate through all y ∈ {1, . . . , 50} and for each of the y that does not belong to any of S0, S1, . . . , Sk−1 we check if any of y− 11 and y + 6 belongs to Sk−1. If it does, then we add y to Sk. In the case that some of Sj remains empty after this procedure then either the shortest paths to all of the floors have been found or there are some floors that can not be reached using the given operations. In this problem this can not happen as 11 and 6 are relatively prime (so path from i to j exists for any pair of floors), but for example in the case when we can only go 10 floors up and 5 floors down it is impossible to reach floor 33 starting from floor 32. It can be easily computed that S14 is the Sk containing 33, hence the answer to the first question is: we have to press an elevator’s button at least 14 times in order to get from floor 32 to floor 33. Using simple backtracking it can be found that, in total, we have to go 5 times up and 9 times down in order to reach floor 33. For the second part of the question, we first prove that in this case the answer to the first question (where 5 · 11 + 9 · 6 = 109 floors are visited) gives the answer to the second question as well. Assume on contrary that there exists another path where lower number of floors is visited. It implies that we have to take less than 9 button presses down or less than 5 button presses up (if none of these two is true than we end up with solution requiring at least 5 · 11 + 9 · 6 = 109 floors to be passed on our way). But, if we take less than 9 button presses down then, in total, we will pass by less than 9 · 6 = 54 floors down, implying that we will pass by less than 55 floors up. This further implies that we have less than 55 11 = 5 up button presses. This gives us solution where less than 9 + 5 = 14 button presses have been used, contradicting first part where we showed that 14 is the least number of button presses we need. The case when less than 5 button presses up are used is handled in analogous way. Therefore, our assumption is wrong and we conclude that the shortest time to go from 32 to 33 requires passing 109 floors.