CMPT 408: Theory of Computer Networks
Prof. Funda Ergun
Lecture 7 & 8: Routing in Hypercube
September 21, 2004
Scribe:
Hypercube
A hypercube of dimension d is constructed a follows:
 Take two hypercubes of dimension d-1
 Connect identically labeled nodes
 Add a prefix of 0 to all nodes of hypercube 1 of dimension d-1
 Add a prefix of 1 to all nodes of hypercube 2 of dimension d-1

Base case
o Hypercube of dimension 1
Properties
 An N-node hypercube is a graph where each node has degree log N
Proof: A hypercube of dimension (and degree) has N = 2d nodes
Base case: degree 1
Inductive hypothesis: Assume a hypercube of dimension d-1 has 2d-1 nodes.
Inductive step: To construct a d dimensional hypercube, we join together two
d-1 dimensional hypercubes.
The degree of each node becomes d.
The number of nodes is now 2*2d-1 = 2d

A dimension d hypercube has (d*N)/2 edges = (NlogN)/2 = 2*2d-1
Proof: N nodes each of degree d
Example: d = 3
Edges = 2*2d-1
= 3*22
= 12

Adjacent nodes differ by 1 bit only
000
101
100
110
001
011
110
111
Routing: Oblivious routing, routing at each node.
Oblivious: Algorithm (through path ) is determined at the source.
Example:
010011  110101
-Use exclusive OR and route in the direction desired.
010011
110011
110111
110101

Algorithm:
o Go from left to right on the source address.
o If the current bit is not equal to the corresponding bit int the destination,
change it.

Property:
o The length of a path from s (source) to t (destination) = # of bits that are
different in s, t ( = # of 1’s in s XOR t)
The diameter of the hypercube is d = log N of N nodes.
Proof: The addresses have d bits. If they’re all different in s, t
Hamilton Cycle Problem
 Go over each vertex without covering an edge or vertex twice.
 Comes back to the starting point.
 NP-hard: no one knows how to solve in 0(Nk) for any fixed k.
Gray Code: Each time you increment, you flip one bit.
Inductive Construction:
2 bit gray code:
00
01
11
10
3 bit gray code
010
011
001
000
Reversed
100
101
111
110
2 bit gray codes – reverse 1
First one gets prefix 0
Second one gets prefix 1
By induction, must prove that 000  100 and 010110
Theorem: If you follow the nodes corresponding to the gray code on the hypercube, you
get a Hamiltonian cycle.
Degree 3 Hypercube
000
011
0110
1000
0001
0101
0100
110
110
001
101
100
Degree 4 Hypercube
0000
0110
1110
Proof sketch: By induction, do identical Hamiltonian cycles on:
- the 2 dimensional d-1 hypercubes
- leave out the same edge
- then the endpoints differ only in their leading bits.
- By induction you can do this for any degree hypercube
Star Topology
Problem: Only one message can travel at a time
Solution: Fast nodes.
Tree Topology
Problem 1: Big diameter
Problem 2: Nodes need to share edges
All node pairs have a unique path between them.
Solution to the bandwidth problem: Fat Tree.
1011
1110
0111
111
1101
1100
0011
1001
1111
16 pairs
4 pairs
High bandwidth
Lower bandwidth
Low bandwidth
1 pair
Bandwidth increases exponentially higher up the tree.
Routing:
Permutation routing:
12345678
54387621
1
4
2
3
2,3
4
5
1
Issue: Max delivery time can be long if there is contention from a link.
Theorem: On the hypercube there exists a permutation that delivery time is ≥
└ N/d ┘ for any deterministic algorithm.
Randomized Routing:
For any s-t pair:
Pick an intermediate node v uniformly at random (can be done
with d independent coin tosses, 1 for each bit of v).
Phase 1: Route from s to v
Phase 2: Route form v to t.
Theorem: With high probability (1 – ε for small ε) the routing will take O(log N)
steps for any randomized algorithm.