hw1Sol.pdf - CSE 5500 HW1 Solutions Answer 1 We assume that...

CSE 5500: HW1 Solutions
Answer 1
We assume that the graph is given to us as an adjacency list. (Can’t use adjacency matrix since just
reading the matrix would require O ( | V
2
| ) time.) We perform a breadth-first (or depth-first) traversal of
the graph, mark nodes as “visited” when they are encountered, and also keep track of the search path by
including, at each node, back-pointers to the previous node. If the search encounters a node that has been
previously “visited” then a cycle exists, otherwise no cycle exists. If a previously visited node is encountered,
we backtrack using the back-pointers to output the nodes of the cycle. This is a simple problem and further
details and proof of correctness are omitted (left to the reader).
The BFS (or DFS) itself takes O ( | V | + | E | ) time and the backtracking step, if executed, requires another
O ( | V | ) time, yielding an O ( | V | + | E | )-time algorithm for the problem.
Answer 2
Part (a): Here, we have k = 2 jars available for testing. Consider the following strategy. We divide the
ladder with n rungs into different levels, each containing ⌈
√
n ⌉ rungs (except for the top-most level which
may contain fewer than ⌈
√
n ⌉ rungs). This would result in at most ⌊
√
n ⌋ levels. E.g., if n = 10, then we
would have 3-levels with 4, 4, and 2 rungs, respectively, from bottom to top. We now drop the first jar from
the bottommost rung of each level, going from the bottom to top, until the jar breaks. We now have three
possible cases: (i) The jar breaks at level x where x = 1. (ii) The jar breaks at level x for x > 1. And (iii)
The jar does not break on any of the tested rungs. For case (i), we can conclude that the jar breaks on the
first rung of the ladder. For case (ii), we test the second jar from each rung of level x − 1 from bottom to
top and find the first rung where the second jar breaks. For case (iii), the first jar did not yet break, so we
continue testing that jar from each rung of the topmost level, from bottom to top, and find the first rung
where it breaks. Thus, in either case, we break no more than 2 jars to find the highest safe rung.
We claim that the total number of drop required in this strategy is O (
√
n ). This is easy to verify using
the three possible cases above, as follows. In case 1, we needed only 1 drop. In case 2, we needed x drops for
the first jar and at most ⌈
√
n ⌉ drops for the second jar. Since x can be no greater than the total number of
levels, we have x ≤
√
n . Thus, the total number of drops in this case is O (
√
n . In case 3, we have up to
√
n
drops of the first jar to get to the topmost level and an additional, at most, ⌈
√
n ⌉ drops to find the highest
safe rung in the topmost level. Thus, the total number of drops is O (
√
n ), and lim
n →∞
√
n
n
= 0, as required.