6.856 — Randomized Algorithms
Karger
Handout #5, February 16, 2011 — Homework 2 Solutions
Problem 1.
(a)
As in the class, we can construct an adversarial input as follows. When replying to al-
gorithm’s queries on leaves, we always reply such that for any tree, if not all leaves were
queried, then the value of the tree is not yet determined. In this case, for each internal node,
the algorithm is forced to evaluate each of its children.
For a node x, call val(x) the value of x.
We do this as follows. Suppose the algorithm queries leaf x, and let n1, n2, . . . , nh−1, nh=
xbe the path from root (n1) to xin the tree. Let nibe the lowest node such that there
exists a non-examined leaf in ni’s subtree (xbeing counted as “examined”). Then, if nihas
0 children with established value (ni+1 begin considered with non-established value), then
we choose val(x) = 0 (in which case, val(ni+1 ) = val(ni+2 ) = · · · =valnh= 0). If nihas
1 child with established value (ni+1 begin considered with non-established value), then we
choose val(x) = 1 (in which case, val(ni+1 ) = val(ni+2 ) = · · · =val(nh) = 1). Note that
the value of the first child of niis 0, and therefore the value of niremains un-established.
Also, note that by the definition of ni,nicannot have 2 children with established value (ni+1
being considered with non-established value).
Thus, for each node, the algorithm has to establish the value of all of its 3 children –
which means that the algorithm has to query all 3hleaves.
(b)
Consider a node xin the tree. Let its value be val(x). This means that there are two
children child1(x) and child2(x) such that val(x) = val(child1(x)) = val(child2(x)). If, for
each x, we know that these children child1(x) and child2(x), then we can only compute
the values of these children to compute the value of x. Thus, a non-deterministic algo-
rithm can just guess child1(x) and child2(x) and reccurse on those 2. If val(child1(x)) =
val(child2(x)) then we know for sure that val(x) = val(child1(x)) = val(child2(x)). If
val(child1(x)) 6=val(child2(x)) then we fail on that branch. At least one of the branches of
the non-deterministic algorithm will succeed and give the right answer.
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