Permutation routing in n-cube

n-cube 1-cube 2-cube 3-cube 4-cube

Labeling 00 01 000 001 100 101 010 011 1 110 111 10 11 1000 1001 0000 0001 1100 1101 0100 0101 1010 1011 0010 0011 1110 1111 0110 0111

Definition of n-cube An n-cube: Number of nodes: N = 2n, each node has a distinct n-bit label. The node with label (i1, i2, …, in)  {0, 1}n and the node with label (j1, j2, …, jn)  {0, 1}n are connected if the “hamming distance” between (i1, i2, …, in) and (j1, j2, …, jn) is 1. Each node has degree n, so there are nN/2 edges in an n-cube.

Permutation Routing Each node i contains one packet m(i) that should be sent to the destination node d(i). Each destination node d(i) for each node i, for 0  i  N – 1, should form a permutation of {0, …, N – 1}, i.e., every node is the destination of exactly one packet

Example d(000)=010 d(001)=000 d(010)=110 d(011)=001 d(100)=111 101 010 011 110 111

The rule All interconnections between nodes occur in synchronous steps Each direction of each edge can carry at most one message in one step During a step, a node can send at most one packet to each of its neighbors That is, at most n outgoing messages for each node in each time step.

Two steps suffice d(000)=010 d(001)=000 d(010)=110 d(011)=001 100 101 010 011 110 111

Things can be complicated… 000 001 d(000)=111 d(001)=110 d(010)=101 d(011)=100 d(100)=011 d(101)=010 d(110)=001 d(111)=000 100 101 010 011 110 111

Need two time steps for this edge. Traffic control Need two time steps for this edge. 000 001 d(000)=111 d(001)=110 d(010)=101 d(011)=100 d(100)=011 d(101)=010 d(110)=001 d(111)=000 100 101 010 011 110 111

Oblivious Routing For any nodes i and j in {0, 1, 2, …, N–1}, the route between nodes i and j depends on the labels of i and j only. The “memoryless” feature is attractive for hardware design. For example, the bit-fixing algorithm is an oblivious routing algorithm.

A Question Let P and P’ be two oblivious paths in n-cube. If they overlap at some edge e and then separate at the endpoint x of e, can P and P’ rejoin again later at some node y? y x

Negative! The segments of P and P’ from x to y depend only on the labels of x and y. y x

Bit-Fixing Routing Algorithm: e a c h n d i , w s t m g f ( ) b y l v p u k 1 ; 2 : L j . I = O S ¯ ` ® + - T R

Example (0, 1, 0, 0, 1, 0)  (1, 1, 0, 0, 1, 0)  (1, 0, 0, 0, 1, 0)  (1, 0, 1, 0, 1, 0)  (1, 0, 1, 0, 1, 1).

The weakness Bit-fixing routing is a deterministic oblivious routing algorithm. Deterministic oblivious routing algorithms have a quite bad lower bound on the number of time steps required for permutation routing.

A SPAA’90 result [Kaklamanis, Krizanc, Tsantilas, SPAA’90] For any deterministic oblivious permutation routing algorithm on a network of N nodes each of degree n, there is an instance of permutation routing requiring ((N/n)0.5) time steps. Comment: The network does not have to be an n-cube.

Bit-fixing on n-cube By [Kaklamanis, Krizanc, Tsantilas, SPAA’90] The bit-fixing routing algorithm on an N-node n-cube has an instance of permutation routing requiring ((N/n)0.5)=((N/log N)0.5) time steps. We show a stronger lower bound There is an instance such that the bit-fixing algorithm requires (N0.5) time steps.

A bad instance for bit-fixing Consider the case that n is an even number, say, n = 2t for some positive integer t. For each node i in {0,1,2,…,N-1}, let d(i) be defined as follows. If (i1, i2, …, in) is the label of i, then let d(i) be the node with label (it+1,it+2,…,in, i1, i2, …, it).

Example, n = 4. d(0000)=0000 d(0001)=0100 d(0010)=1000 d(0011)=1100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

Traffic jam at an edge e Consider the directed edge e from node (000…01 000…00) to node (000…00 000…00) The message sent from each node with label (???…?1 000…00) has to pass edge e before reaching its destination (000…00 ???…?1). There are 2n/2-1 = (N0.5) such messages.

Breaking the lower bound by randomization… Q: Any proposals for how to randomize a permutation routing in an n-cube?

Valiant and Brebner’s algorithm [STOC’81]

Valiant-Brebner Phase 0: Comment: For each i = 0 to N – 1, choose an intermediate node t(i) uniformly at random from {0, 1, …, N – 1}. Comment: Note that t is not necessarily a permutation on {0, 1, …, N – 1}.

Valiant-Brebner Phase 1: Phase 2: Comments: Route each message m(i) from node i to its temporary destination t(i) using bit-fixing algorithm. Phase 2: Route each message m(i) from node t(i) to its real destination d(i) using bit-fixing algorithm. Comments: No randomization in these two steps. Phase 2 starts only after Phase 1 is completely finished.

The waiting list Each node keeps a waiting list for each of its n neighbors. The waiting list operates in a First-In-First-Out manner. For multiple messages joining a particular waiting list at the same time, the tie can be broken arbitrarily.

How good is Valiant-Brebner? Theorem With probability at least 1-O(N -1) the number of time steps required by Valiant-Brebner’s randomized oblivious routing algorithm is only O(log N). Comments: O(log N) is much better than the lower bound for deterministic oblivious algorithms. The bound is the same even for the special case that d(i) = i for each index i = 0, 1, …, N – 1.

Prof. Herman Chernoff’s bound, Annal of Mathematical Statistics 1952 Math tool Prof. Herman Chernoff’s bound, Annal of Mathematical Statistics 1952

Bernoulli trial L e t X b a i n r y d o m v l f u 1 ( s p c , ) w h - ) w h - ¡ . E [ ] = ¹ P V ¾ 2

Bernoulli trials L e t X = P w h r a c 1 · s d p b y o m v l f u . E [ ] : V ( ¡ )

Poisson trials L e t X = P w h r a c 1 · s d p b y o m v l f u . E [ ] : V ( ¡ )

Chernoff bound T h e o r m L t X = + ¢ , w ; : a i d p ( P s ) l u c 1 + ¢ n , w ; : a i d p ( P s ) l u c < [ ] f 2 . y v b > ¹ · µ ¶ E ¡

Analyzing Valiant-Brebner

If a packet is routed to a randomly chosen node t= (t1, If a packet is routed to a randomly chosen node t= (t1,...,tn), we can think each ti is 0 (or 1) with probability ½. For a given packet M, let T(M) be the number of steps for M to finish Phase 1. For a given edge e, let X(e) denote the total number of packets traversing edge e during Phase 1. In each step of Phase 1, M is either traversing an edge or waiting in a queue when some other packet traverses an edge on M’s route.

Lemma : Let e1,...,em be the m <= n edges traversed by a packet M in Phase 1. Then T(M)  X(e1) + X(e2) +...+ X(em). Call a path P=(e1,...,em) of m  n edges that follows the bit-fixing algorithm a possible packet path, with ei=(vi-1, vi). There are at most 22n such paths. Why? For any possible packet path P =(e1,...,em), define T(P) = X(e1) + X(e2) +...+ X(em). Note that the probability that Phase 1 takes more than K steps is bounded by the probability that for some possible packet path P, T(P) >= K.

Since T(P) = X(e1) + X(e2) + Since T(P) = X(e1) + X(e2) +...+ X(em), it is natural to use Chernoff bound, but unfortunately X(ei) are not independent. To overcome this, we first use Chernoff bound to prove that with high probability (w.h.p.) no more than 6n different packets cross any edge of P. Then condition on this event to derive a high-probability bound on the total number of transitions these packets make through edges of P, again by Chernoff bound.

Fix a possible packet path P=(e1,...,em) with m edges and ei=(vi-1, vi). Call a packet active at a node vi-1 on P if it reaches vi-1 and has the possibility of crossing edge ei to vi. Call a packet active if it is active at some vertex of P. Suppose vi-1 and vi differ in the j-th bit then for packet to be active at vi-1 its j-th bit cannot have been fixed by the bit-fixing algorithm when reaching vi-1 .

Let Hk be a 0-1 random variable such that Hk=1 if the packet starting at node k is active, 0 otherwise. Hk are independent. Why? Let H= H0 +...+ HN-1. What is E[H]? Consider all the active packets at vi-1 = (b1 ,...,bj-1, aj, aj+1,..., an) and assume vi = (b1 ,...,bj-1, bj, aj+1,..., an) .

Before fixing the j-th bit, what are the starting addresses of packets that can reach vi-1 = (b1 ,...,bj-1, aj, aj+1,..., an)? (* ,.....,*, aj, aj+1,..., an). How many such packets? Each of such packet can reach vi-1 only its random destination is of the form: (b1 ,...,bj-1, *, *,..., *). There are 2n-j+1 such random destinations.

There are at most 2j-1 packets with starting addresses of the form: (* ,.....,*, aj, aj+1,..., an) that can reach vi-1 = (b1 ,...,bj-1, aj, aj+1,..., an). Each of such packet can reach vi-1 with random destination of the form: (b1 ,...,bj-1, *, *,..., *). Thus, for such a packet, the probability of visiting vi-1 is 2-(j-1). Why? Hence, the average active packet number per vertex on P is 1. Why?

Since there are m vertices on P, E[H]  m  n. By Chernoff bound, we have Pr( H  6n  6E[H] )  2-6n. Use Pr(A)= Pr(A|B)Pr(B)+Pr(A|  B)Pr(B)  Pr(B) + Pr(A|  B). Pr( T(P) 30n)  Pr(H 6n) + Pr( T(P) 30n | H < 6n)  2-6n + Pr( T(P) 30n | H < 6n).

Pr( T(P) 30n)  2-6n + Pr( T(P) 30n | H < 6n). Claim: Pr( T(P) 30n | H < 6n)  2-3n-1 . With the claim we have: Pr( T(P) 30n)  2-3n Observe that if a packet leaves the path P, it cannot return to the path in Phase 1. Why?

Consider an active packet on the path P at vertex vi-1 . Suppose the vertices of ei differ in the j-th bit. What is the probability that the packet crosses ei = (vi-1, vi)? The probability of crossing ei is at most ½. vi-1 vi

vi-1 vi View an active packet at a node vi -1 on the path P that might cross edge ei as a trial. The trial is successful if the packet leaves the path, but a failure if the packet stays on the path. The successful probability is at least ½. Observe that if there are at most 6n active packets, then there are at most 6n successes.

Note that each trial is independent. Claim : vi-1 vi Note that each trial is independent. Claim : Pr( the active packets cross edges of P more than 30n times)  Pr( flip fair coin 36n times with successes fewer than 6n times). with at most 6n active packets.

Let Z be the number of heads (successes) in 36n fair coin flips. Pr( T(P)  30n | H  6n)  Pr( Z  6n)  exp(-18n (2/3)2/2), by Chernoff bound = e-4n  2-3n-1 Pr( T(P) 30n)  Pr(H 6n) + Pr( T(P) 30n | H < 6n)  2-3n.

There are at most 22n possible packet paths P in hypercube. Pr(  P such that T(P) 30n)  22n 2-3n = 2-n = O(N-1). This completes the analysis of Phase 1. Phase 2 can be viewed as running Phase 1 backwards. Hence no packet spends more than 30n steps in Phase 2 with probability 1- O(N-1).

Finally, the total number of packet traversals across the edges of any packet path during Phase 1 and Phase 2 is bounded by 60n steps with probability 1- O(N-1).

Conclusion Deterministic oblivious routing algorithms may have poor performance for some specific cases Valiant-Brebner’s randomized routing algorithm in n-cube has asymptotically optimal performance with with high probability.