Hypercubes
Recall that the set of all n-bit strings is denoted by {0,1}n. The n-dimensional hypercube is a graph whose
vertex set is {0,1}n(i.e. there are exactly 2nvertices, each labeled with a distinct n-bit string), and with
an edge between vertices xand yiff xand ydiffer in exactly one bit position. i.e. if x=x1x2...xnand
y=y1y2...yn, then there is an edge between xand yiff there is an isuch that ∀j6=i,xj=yjand xi6=yi.
There is another equivalent recursive definition of the hypercube:
The n-dimensional hypercube consists oftwo copies of then−1-dimensional hypercube (the 0-subcube and
the 1-subcube), and with edges between corresponding vertices in the two subcubes. i.e. there is an edge
between vertex xin the 0-subcube (also denoted as vertex 0x) and vertex xin the 1-subcube.
Claim: The total number of edges in an n-dimensional hypercube is n2n−1.
Proof: Each vertex has nedges incident to it, since there are exactly nbit positions that can be toggled to
get an edge. Since each edge is counted twice, once from each endpoint, this yields a grand total of n2n/2.
Alternative Proof: By the second definition, it follows that E(n) = 2E(n−1) + 2n, and E(1) = 1. A
straightforward induction shows that E(n) = n2n−1.
We will prove that the n-dimensional hypercube is a very robust graph in the following sense: consider how
many edges must be cut to separate a subset Sof vertices from the remaining vertices V−S. Assume that S
is the smaller piece; i.e. |S| ≤ |V−S|.
Theorem: |ES,V−S| ≥ |S|.
Proof: By induction on n. Base case n=1 is trivial.
For the induction step, let S0be the vertices from the 0-subcube in S, and S1be the vertices in Sfrom the
1-subcube.
Case 1: If |S0| ≤ 2n−1/2 and |S1| ≤ 2n−1/2 then applying the induction hypothesis to each of the subcubes
shows that the number of edges between Sand V−Seven without taking into consideration edges that cross
between the 0-subcube and the 1-subcube, already exceed |S0|+|S1|=|S|.
Case 2: Suppose |S0|>2n−1/2. Then |S1| ≤ 2n−1/2. But now |ES,V−S| ≥ 2n−1≥ |S|. This is because by
the induction hypothesis, the number of edges in ES,V−Swithin the 0-subcube is at least 2n−1− |S0|, and
those within the 1-subcube is at least |S1|. But now there must be at least |S0| − |S1edges in ES,V−Sthat
cross between the two subcubes (since there are edges betwen every pair of corresponding vertices. This is
a grand total of 2n−1− |S0|+|S1|+|S0|− |S1|=2n−1.
CS 70, Fall 2004, Lecture 11 1
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