presented by nick farnan 11 11 08
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New Protocols for Asymmetric Communication Channels

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Presented by Nick Farnan 11-11-08. New Protocols for Asymmetric Communication Channels. Asymmetric Communications. Setup: Client needs to send server an n -length bit string x

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asymmetric communications
Asymmetric Communications
  • Setup:
    • Client needs to send server an n-length bit string x
    • The communication channel to send from server to client has much greater bandwidth than the one from client to server
    • The server knows D, the distribution of probabilities of receiving any give n-length bit string from the client (Note that the client has no knowledge of D)‏
asymmetric communications1
Asymmetric Communications
  • In his “A Mathematical Theory of Communication”, Shannon provided a limit to the amount of lossless compression of any communication with Shannon entropy
  • Entropy is defined to be a measure of uncertainty in a random variable (in this case a bit string)‏
  • A repeating string (i.e. “00000...”) has minimal entropy as it is completely predictable, while a random bit string (could be any bit string of length n with equal probability) has maximum entropy
asymmetric communications2
Asymmetric Communications
  • From this, we can draw the conclusion that the absolute minimum expected number of bits that the client will have to send to the server to describe x to the server is the entropy of x, H(x)‏
asymmetric communications3
Asymmetric Communications
  • Consider a list of all n length bit string ordered from least to greatest
  • Let F(x) be the sum of D(y) for all y < x
  • Let F\'(x) = F(x) + 1/2(D(x))‏
  • Given a real number v in the range (0, 1], a bit string y is considered to be a v-split string if it is the largest n bit string such that F\'(y) <= v
bit efficient split
  • By using split strings, the server can do a sort of binary search through the space of n bit strings to determine the string x that the client wishes to send.
bit efficient split2
  • Fact: F\'(x) is the midpoint of the interval (F(x), F(x) + D(x))‏
  • After m rounds, given that D(x) >= 2^(1-m), and that F\'(x) is contained within (um, vm), at the end of the current round, y == x == z, and hence the string has been discovered!
  • Hence, if it takes at most m+1 rounds to discover the string being sent, the client m+1 bits indicating which side of the split string the x falls on
bit efficient split3
  • Now, as:m+1 = ceil(lg(1/D(x)))+1 <= lg(1/D(x))+2
  • The expected number of rounds can be bounded by the sum over all x of:D(x) * (lg(1/D(x)) + 2=sum(D(x) * lg(1/D(x))) + 2=H(D) + 2
  • A constant factor from optimal!
the hotlinking problem
The Hotlinking Problem
  • Proposed for the Web to make particular pages on a site more accessible
  • Given a rooted tree where each leaf has a probability attached to it (let D be the distribution of these probabilities), the Hotlinking problem moves subtrees up the tree closer to to the root in order to minimize the cost of the tree as a whole.
the hotlinking problem1
The Hotlinking Problem
  • Definitions:
    • c(u) is the sum of all of the probabilities of the leaves of the subtree rooted at node u, this is the cost of u
    • c(u,v) where u is the parent of v, is c(v), this is the cost of edge u-v
    • d(u, T) is the length of the path from the root of T to u
    • c(T) is the sum of the costs of all the edges in tree T
    • L(T) is the set of leaves of T
the hotlinking problem2
The Hotlinking Problem
  • Adoption
    • A node u can adopt one of its decedents (w) by removing w from it\'s parents list of children and adding it instead to u\'s list of children
the hotlinking problem3
The Hotlinking Problem
  • A k-hotlink-assignment of T is obtained by performing up to k adoptions (where k is some integer) at the root of T, creating T1, and the further find a k-hotlink-assignment for each subtree of T1 that is a child of the root.
  • This problem has been show to be generalized as: at each node u, adopt a its descendants until all descendants of u are either leaves or have a cost <= (2/(k+2)) * c(u)‏
the hotlinking problem4
The Hotlinking Problem
  • The resulting tree will have a cost of at most: (H/(lg(k + 2) – 1)) + 1, where H is the entropy of D, the distribution of the leaves
hotlinking in asymmetric communications
Hotlinking in Asymmetric Communications
  • Let T be a complete binary tree with 2nleaves (where n is the length of the string being sent from client to server) and a label on each edge such that all edges leading to left children have the label 0 and all the right are labeled 1
  • Each vertex is then labeled with its path from the root (i.e. going left, right then left from the root would lead to a node labelled “010”)‏
hotlinking in asymmetric communications1
Hotlinking in Asymmetric Communications
  • Both client and server maintain r1 and r2 where r1~r2 is the string r that the client wishes to send.
  • Initially, r1 is set to empty and the server does not know r2
  • Find T\', the k-hotlink-assignment of T
  • For each round of the protocol, the server considers all of the children (w) of node u in T whose label is r1, and send to the client all l where l is label(w) - r1
hotlinking in asymmetric communications2
Hotlinking in Asymmetric Communications
  • The client then replies with the index in the last server message of the longest l that is a prefix of r2
  • The client then removes l from the beginning of r2, and both client and server add l to r1
  • If |r1| == n, the server outputs r1 and the communication terminates
hotlinking in asymmetric communications3
Hotlinking in Asymmetric Communications
  • The client can then be expected to send c(T\') * ceil(lg(k + 2))‏
  • Given that c(T\') = (H/(lg(k + 2) – 1)) + 1, the client can then be expected to send H*(ceil(lg(k+2))/(lg(k+2)-1)) + ceil(lg(k+2)) bits
  • setting e = (ceil(lg(k+2))/(lg(k+2)-1)) – 1, and ce = ceil(lg(k+2)), then, a protocol where client is expected to send (1+e)H + cebits is achieved