A polylog competitive algorithm
for the k-server problem
Nikhil Bansal (IBM)
Niv Buchbinder (Open Univ.)
Aleksander Madry (MIT)
Seffi Naor (Technion)
The k-server Problem
• k servers lie in an n-point metric space.
• Requests arrive at metric points.
• To serve request: Need to move some server there.
Goal: Minimize total distance traveled.
Objective: Competitive ratio.
1
2
Move Nearest Algorithm
3
The Paging/Caching Problem
Set of pages {1,2,…,n} , cache of size k<n.
Request sequence of pages 1, 6, 4, 1, 4, 7, 6, 1, 3, …
a) If requested page already in cache, no penalty.
b) Else, cache miss. Need to fetch page in cache
(possibly) evicting some other page.
Goal: Minimize the number of cache misses.
K-server on the uniform metric.
Server on location p = page p in cache
1
. . .
n
Previous Results: Paging
Paging (Deterministic) [Sleator Tarjan 85]:
• Any deterministic algorithm >= k-competitive.
• LRU is k-competitive (also other algorithms)
Paging (Randomized):
• Rand. Marking O(log k)
• Lower bound Hk
[Fiat, Karp, Luby, McGeoch, Sleator, Young 91].
[Fiat et al. 91],
tight results known.
K-server conjecture
[Manasse-McGeoch-Sleator ’88]:
There exists k competitive algorithm on any metric space.
Initially no f(k) guarantee.
Fiat-Rababi-Ravid’90: exp(k log k)
…
Koutsoupias-Papadimitriou’95: 2k-1
Chrobak-Larmore’91: k for trees.
Randomized k-server Conjecture
There is an O(log k) competitive algorithm for any metric.
Uniform Metric: log k
Polylog for very special cases (uniform-like)
Line: n2/3
exp(O(log n)1/2)
[Csaba-Lodha’06]
[Bansal-Buchbinder-Naor’10]
Depth 2-tree: No o(k) guarantee
Our Result
Thm: There is an O(log2 k log3 n) competitive*
algorithm for k-server on any metric with n points.
Key Idea: Multiplicative Updates
* Hiding some log log n terms
Our Approach
Hierarchically Separated Trees (HSTs) [Bartal 96].
Any Metric
O(log n)
Allocation Problem (uniform metrics): [Cote-Meyerson-Poplawski’08]
(decides how to distribute servers among children)
Allocation
instances
K-server on HST
Outline
•
•
•
•
•
•
Introduction
Allocation Problem
Fractional view of Randomized Algorithms
Fractional Caching Algorithm
What makes Allocation Problem harder?
The Fix
Allocation Problem
Uniform Metric
At each time t, request arrives at some location i
Request = (ht(0),…,ht(k))
[monotone: h(0) ¸ h(1) … ¸ h(k)]
Upon seeing request, can reallocate servers
Hit cost = ht(ki)
[ki : number of servers at i]
Total cost = Hit cost + Move cost
Eg: Paging = cost vectors (1,0,0,…,0)
*Total servers k(t) can also change (let’s ignore this)
Allocation to k-server
Thm [Cote-Poplawski-Meyerson]: An online algorithm for allocation
s.t. for any  > 0,
i) Hit Cost (Alg) · (1+) OPT
ii) Move Cost (Alg) · () OPT
gives ¼ O(d (1/d)) competitive k-server alg. on depth d HSTs
d = log (aspect ratio)
So,  = poly(1/) polylog(k,n) suffices
*HSTs need some well-separatedness
*Later, we do tricks to remove dependence on aspect ratio
We do not know how to obtain such an algorithm.
Outline
•
•
•
•
•
•
Introduction
Allocation Problem
Fractional view of Randomized Algorithms
Fractional Caching Algorithm
What makes Allocation Problem harder?
The Fix
Fractional View of Randomized
Algorithms
To specify a randomized algorithm:
i) Prob. distribution on states at time t.
ii) How it changes at time t+1.
Fractional view: Just specify some marginals.
Eg. Paging, actual algorithm = distribution over k-tuples
but, Fractional: p1,…,pn s.t. p1 + …+ pn = k
Cost: If p1,…,pn changes to q1,…,qn ,pay (1/2) i |pi – qi|
Not too weak: Fractional Paging -> Randomized Paging (2x loss)
Fractional Allocation Problem
xi,j : prob. of having j servers at location i (at time t)
j xi,j = 1
i j j xi,j · k
(prob. distribution)
(global server bound)
Cost: Hit cost with h(0),…,h(k) = j xi,j h(j)
Moving  mass from (i,j) to (i,j’) costs  |j’-j|
Surprisingly, fractional allocation does not give good
randomized alg. for allocation problem.
A gap example
Allocation Problem on 2 points
Left
Right
Requests alternate on locations.
Left: (1,1,…,1,0)
Right: (1,0,…,0,0)
Any integral solution must pay (T) in T steps.
Claim: Fractional Algorithm pays only T/(2k) .
XL,0 = 1/k xL,k = 1-1/k
XR,1 = 1
No move cost. Hit cost of 1/k on left requests.
Fractional Algorithm Suffices
Thm (Analog of Cote et al): Suffices to have fractional
allocation algorithm with (1+,()) guarantee.
Gives a fractional k-server algorithm on HST
Thm (Rounding): Fractional k-server alg. on HSTs ->
Randomized Alg. with O(1) loss.
Thm (Frac. Allocation): Design a fractional allocation
algorithm with () = O(log (k/)).
Outline
•
•
•
•
•
•
Introduction
Allocation Problem
Fractional view of Randomized Algorithms
Fractional Caching Algorithm
What makes Allocation Problem harder?
The Fix
Fractional Paging Algorithm
Current state, p1,…,pn s.t. i pi =k.
Say request at 1 arrives.
Algorithm: Need to bring 1-p1 mass into p1.
Rule: For each page i decrease pi / 1–pi + 
0
1
0
1
1-p1
p1
1-p2
p2
Pg 1
Pg 2
0
1
0
1
1-pn
…
pn
Pg n
Intuition: If pi close to 1, be more conservative in evicting.
Multiplicative Update: d(1-p) / (1-p)
( = 1/k)
Potential Fn. Analysis
Will Show: On(t) + (t) - (t-1) · O(log k) Off(t)
 = 1/k
Contribution of page i to : 0 if pi =0, log(k+1) if pi=1.
 =0 if online and offline state coincide.
Suppose, page 1 is requested.
Analyze in two steps: First offline moves, then online
If offline moves a server  increases by · log(k+1)
The claim holds easily.
Analysis
Will show: On(t) + (t) - (t-1) · O(log k) Off(t)
Suppose  mass enters p1.
pi decreases by dpi =  (1-pi+) / N
Offline Online
On = {i : pi > 0}
Key Obs: For each page i in On\Off
 decreases by dpi ¢ d / dpi = dpi (1/(1-pi+)) = /N
At least |On|-k such pages, so potential drop >=  (|On|-k)/N
Claim: Potential drop ¸ .
Pf:
N = i 2 On (1-pi+) = |On|-k + |On| ¼ |On|-k.
What makes Allocation Harder?
Paging: If location 1 requested,
Know that OPT also has server on 1.
Allocation Problem: Not so clear.
Say, already 10.5 servers at location 1.
Should we add even more?
May be OPT just has 1 server there.
An Instructive Example
Case 1: K servers, and allocation problem on 2k locations.
Cost Vectors = (1,0,…,0)
At Locations: 1, 2, …, 2k 1, 2,…, 2k, …
1
2
...
2k
Case 2: 2 k servers in total.
1/2 server on locations 1,…,k-1
and k+1/2 on location k.
Vector (1,0,…,0) on location 1,2,…,k-1 and vector (1,1,…,1,0,…,0) on k.
k k+1
Right thing: Release the servers on location k.
Much better solution: 1 server on each location.
The Fix
Suppose cost vector j = (,,…,,0,…,0) at location 1.
(i.e. cost  if · j servers, 0 otherwise)
Hit cost Y= (x1,0+ …+ x1,j)
Increase servers by ¼ Y
(location 1)
0
1
2
j
j+1
k
Recall j xij = 1, 8i
Fix number: For each location i (including 1), rebalance prob.
mass by multiplicative update.
……
Each xi,j (except last j)
increases / xi,j
Proof Idea
Location i
OPT
ON
Eg: Location i contributes 3 log (1+k) to .
Key observation: For every cut · j,
xi,· j increases / xi,· j.
Lemma: For any offline state,  pays for online movement.
Pf: Each cut for j > k*i, decreases potential by /|ON-k|.
There are |ON-k| such cuts.
Concluding Remarks
Removing dependence on aspect ratio.
HST -> Weighted HST with O(log n) depth.
Extend Allocation to weighted star.
Main question: Can we remove dependence on n.
1. Metric -> HST
2. But even on HST (lose depth of HST)
Thank you