A new upper bound on the work function algorithm for the k-server problem

Abstract

The k-server problem was introduced by Manasse et al. (in: Proceedings of the 20th annual ACM symposium on theory of computing, Chicago, Illinois, USA, pp 322–333, 1988), and is one of the most famous and well-studied online problems. Koutsoupias and Papadimitriou (J ACM 42(5):971–983, 1995) showed that the work function algorithm (WFA) has a competitive ratio of at most $2k-1$ for the k-server problem. In this paper, by proposing a potential function that is different from the one in Koutsoupias and Papadimitriou (1995), we show that the WFA has a competitive ratio of at most $n-1$, where n is the number of points in the metric space. When $n<2k$, this ratio is less than $2k-1$.

Appendix A: Proof of Lemma 2.1

The following is the proof of Lemma 2.1 given in Koutsoupias and Papadimitriou (1995).

Definition A.1

A function w is called quasiconvex if for all configurations A, B, there exists a bijection $h: A \rightarrow B$ such that for all bipartitions of A into X, Y:

$$\begin{aligned} w(A)+w(B)\ge w(X\cup h(Y))+w(h(X)\cup Y). \end{aligned}$$

(1)

Notice that the union ($\cup$) in the definition denotes the union of multisets. Before showing that all work functions are quasiconvex, the following lemma provides a stronger form of the quasiconvexity condition by restricting the set of possible bijections.

Lemma A.1

If there exists a bijection h that satisfies the conditions in Definition A.1, then there exists a bijection $h'$ that satisfies the same conditions and $h'(x)=x$ for all $x\in A\cap B$.

Proof

Let h be a bijection from A to B that satisfies the conditions of the definition above and maps the maximum number of elements in $A\cap B$ to themselves. Assume that, for some $a\in A \cap B$, we have $h(a)\ne a$ . Define a bijection $h'$ that agrees with h everywhere except that

$$\begin{aligned} h'(a)=a~~~~~~ \text{ and }~~~~~~ h'(h^{-1}(a))=h(a) \end{aligned}$$

($h'$ interchanges the values of h on a and $h^{-1}(a))$.

Consider now a bipartition of A into X and Y and assume (without loss of generality) that $h^{-1}(a)\in X$. If $a\in X$, then $h(X)= h'(X)$ and $h(Y)=h'(Y)$ and (1) holds for $h'$. Otherwise, when $a\notin X$, we derive the quasiconvexity condition for X and Y from the quasiconvexity condition for $X'=X+a$ and $Y'=Y-a$ as follows:

Since, $h(Y')=h'(Y')$ and $h(X')=h'(X')$, we have that

$$\begin{aligned} X'\cup h(Y')=X'\cup h'(Y')=(X+a)\cup h'(Y-a)=X\cup h'(Y), \end{aligned}$$

and similarly, $h(X')\cup Y'=h'(X)\cup Y$. From these, we get

$$\begin{aligned} \begin{array}{cl} w(A)+w(B) &{}\ge w(X'\cup h(Y'))+w(h(X')\cup Y')\\ {} &{} =w(X\cup h'(Y))+w(h'(X)\cup Y). \end{array} \end{aligned}$$

Therefore, $h'$ satisfies the quasiconvexity condition. Because $h'$ maps at least one more element in $A\cap B$ to itself than h, it contradicts the assumption that h maps the maximum number of elements in $A\cap B$ to themselves.

We conclude that $h(a)=a$ for all $a\in A\cap B$, and the lemma holds. $\square$

Lemma A.2

(Quasiconvexity lemma) All work functions are quasiconvex.

Proof

The proof is by induction on the number of requests.

Recall that the initial work function $w_e(X)$ of a configuration X is equal to $D(A_0,X)$, where $A_0$ is the initial configuration. So we have

$$\begin{aligned} w(A)+w(B)=D(A_0,A)+D(A_0,B). \end{aligned}$$

Fix two matchings $M(A_0,A)$ and $M(A_0,B)$ that realize the minima of $D(A_0,A)$ and $D(A_0,B)$. Each point $x_j$ in $A_0$ is matched to some point $a_j$ in A and $b_j$ in B. Consider the bijection $h: A\rightarrow B$ that maps each $a_j$ to $b_j$. For any bipartition of A into X and Y, $w(X+h(Y))+w((h(X) + Y)$ is equal to the sum of two minima matchings between $A_0$, $X+ h(Y)$ and $A_0$, $h(X)+Y$. Since we can rearrange the matchings $M(A_0,A)$ and $M(A_0,B)$ to obtain two matchings (not necessarily minima) between $A_0$, $X+ h(Y)$ and $A_0$, $h(X)+Y$, it follows that $w(A)+w(B)\ge w(X+h(Y))+w(h(X)+Y)$.

For the induction step, assume that w is quasiconvex. We want to show that the resulting $w'$ after request r is also quasiconvex.

Fix two configurations A and B. Using Fact 1 to express $w'$ in terms of w, we get that $w'(A)=w(A-a+r)+d(r,a)$ for some $a\in A$; similarly $w'(B)=w(B-b+r)+d(r,b)$ for some $b\in B$. The induction hypothesis is that w is quasiconvex, so there exists a bijection h from $A-a+r$ to $B-b+r$ that satisfies the quasiconvexity condition. Furthermore, Lemma A.1 allows us to assume that $h(r)=r$.

Consider now a bijection $h': A\rightarrow B$ that agrees with h everywhere, except that $h'(a)=b$. We show that $h'$ satisfies the requirements of the quasiconvexity condition of $w'$. Consider a bipartition of A into X and Y and without loss of generality assume that $a\in X$. We have:

$$\begin{aligned} \begin{array}{cl} w'(A)+w'(B) &{}=w(A-a+r)+w(B-b+r)+d(r,a)+d(r,b)\\ &{}=w((X-a+r)\cup Y)+w(B-b+r)+d(r,a)+d(r,b)\\ &{}\ge w((X-a+r)\cup h(Y))+w(h(X-a+r)\cup Y)\\ &{}\quad +d(r,a)+d(r,b)\\ &{}= w((X-a+r)\cup h'(Y))+w((h'(X)-b+r)\cup Y)\\ &{}\quad +d(r,a)+d(r,b)\\ &{}\ge w'(X\cup h'(Y))+w'(h'(X)\cup Y), \end{array} \end{aligned}$$

where the first inequality is based on the quasiconvexity of w and the second one on Fact 1. So, $w'$ is quasiconvex and the lemma follows. $\square$

Now we use the quasiconvexity condition to prove the next two lemmata. In fact, the following weaker condition derived from quasiconvexity is used:

$$\begin{aligned} \forall a\in A~ \exists b\in B: ~~w(A)+w(B)\ge w(A-a+b)+w(B-b+a). \end{aligned}$$

Lemma A.3

Let w be a work function. Consider a new request at r and the resulting work function $w'$. If A is a minimizer of r with respect to w, then A is also a minimizer of r with respect to $w'$.

Proof

It suffices to show that for all configurations B:

$$\begin{aligned} w'(B)-\sum \limits _{b\in B} d(r,b)\ge w'(A)-\sum \limits _{a\in A} d(r,a), \end{aligned}$$

or equivalently:

$$\begin{aligned} w'(B)-\sum \limits _{b\in B} d(r,b)+w(A)\ge w'(A)-\sum \limits _{a\in A} d(r,a)+w(A). \end{aligned}$$

In order to show this we need the following: From Fact 1, we get that there exists $b'\in B$ such that

$$\begin{aligned} w'(B)=w(B-b'+r)+d(r,b'). \end{aligned}$$

Using quasiconvexity, we get that there exists $a'\in A$ such that

$$\begin{aligned} w(B-b'+r)+w(A)\ge w(B-b'+a')+w(A-a'+r). \end{aligned}$$

Finally, since A is a minimizer of r, we have that

$$\begin{aligned} w(B-b'+a')-\sum \limits _{b\in B-b'+a'} d(r,b)\ge w(A)-\sum \limits _{a\in A}d(r,a). \end{aligned}$$

Putting all these together:

$$\begin{aligned}\begin{array}{cl} &{}w'(B)+w(A)-\sum \limits _{b\in B} d(r,b) \\ &{}\quad =w(B-b'+r)+d(r,b')+w(A)-\sum \limits _{b\in B} d(r,b) \\ &{}\quad =w(B-b'+r)+w(A)-\sum \limits _{b\in B-b'+r} d(r,b) \\ &{}\quad \ge w(B-b'+a')+w(A-a'+r)-\sum \limits _{b\in B-b'+r} d(r,b) \\ &{}\quad =w(B-b'+a')+w(A-a'+r)+d(r,a')-\sum \limits _{b\in B-b'+a'} d(r,b) \\ &{}\quad \ge w(A)+w(A-a'+r)+d(r,a')-\sum \limits _{a\in A} d(r,a) \\ &{}\quad \ge w(A)+w'(A)-\sum \limits _{a\in A} d(r,a), \end{array} \end{aligned}$$

where the last inequality is based on Fact 1. The lemma follows. $\square$

The following lemma has the same premises with Lemma A.3, but a different conclusion:

Lemma A.4

Let w be a work function. Consider a new request at r and the resulting work function $w'$. If A is a minimizer of r with respect to w, then the extended cost occurs at A, that is

$$\begin{aligned} w'(A)-w(A)=\max \limits _{X}\{w'(X)-w(X)\}. \end{aligned}$$

Proof

The proof is rather similar to the proof of Lemma A.3. Notice first that it suffices to show that for all configurations B:

$$\begin{aligned} w'(A)+w(B)\ge w'(B)+w(A). \end{aligned}$$

By Fact 1, we get that there exists $a'\in A$ such that

$$\begin{aligned} w'(A)=w(A-a'+r)+d(r,a'). \end{aligned}$$

Using quasiconvexity, we also get that there exists $b'\in B$ such that

$$\begin{aligned} w(A-a'+r)+w(B)\ge w(A-a'+b')+w(B-b'+r). \end{aligned}$$

Finally, since A is a minimizer of r with respect to w:

$$\begin{aligned} w(A-a'+b')-\sum \limits _{a\in A-a'+b'} d(r,a)\ge w(A)-\sum \limits _{a\in A} d(r,a), \end{aligned}$$

which is equivalent to

$$\begin{aligned} w(A-a'+b')+d(r,a')\ge w(A)+d(r,b'). \end{aligned}$$

Combining all these we get:

$$\begin{aligned} \begin{array}{cl} w'(A)+w(B) &{}=w(A-a'+r)+d(r,a')+w(B) \\ {} &{}\ge w(A-a'+b')+d(r,a')+w(B-b'+r) \\ {} &{}\ge w(A)+d(r,b')+w(B-b'+r) \\ {} &{}\ge w(A)+w'(B). \end{array} \end{aligned}$$

Again, the last inequality is based on Fact 1. $\square$

Lemmata A.3 and A.4 can be combined into Lemma 2.1, which characterizes where the extended cost occurs. The proof of Lemma 2.1 is established.