In order to determine the advice complexity ofASG, we will use some basic results from the theory of combinatorial designs. Letv≥k≥t be integers. A(v,k,t)-covering design is a family ofk-subsets (calledblocks) of av-set,X, such that anyt-subset ofX is contained in at least one block. Thesizeof a covering design, D, is the number of blocks inD. The covering number, C(v, k, t), is the smallest possible size of a (v, k, t)-covering design. The connection toASGis that for inputs tominASGwhere the number of 1s ist, an (n, ct, t)-covering design can be used to obtain a strictlyc-competitive algorithm. Many papers have been devoted to the study of covering numbers. See [13] for a survey. We make use of the following bounds on the size of a covering design:
ILemma 7(Erdős, Spencer [17]). For all natural numbersv≥k≥t,
v t
k t
≤C(v, k, t)≤
v t
k t
1 + ln k
t
(1)
We will state the obtained advice complexity bounds in terms of the following function:
B(n, c) = log
1 + (c−1)c−1 cc
n (2)
For c > 1, we show that B(n, c)±O(logn) bits of advice are necessary and sufficient to achieve a (strict) competitive ratio ofc, for any version ofASG. See Figure 1 for a graphical view. It can be shown thatn/(a·c) ≤B(n, c)≤n/c, where a=eln(2). In particular, if c=o(n/logn), we see thatO(logn) becomes a lower-order term. Thus, for this range ofc, we determine exactly the higher-order term in the advice complexity ofASG. Since this is the main focus of our paper, we will considerO(logn) a lower-order term. The case where c= Ω(n/logn) is treated in the full version of the paper [9].
J. Boyar, L. M. Favrholdt, C. Kudahl, and J. W. Mikkelsen 123
3.1 Advice Complexity of minASG
The following theorems show that covering numbers are closely related to the amount of advice needed for an algorithm,ALG, to achieve a strict competitive ratio ofc forminASG. ITheorem 8. For anyc >1, there exists a strictlyc-competitive algorithm forminASGu andminASGk readingb bits of advice, where
b= log
max
t:bctc<nC(n,bctc, t)
+O(logn) =B(n, c) +O(logn).
Proof (sketch). Letx=x1. . . xnbe an input string tominASGand sett=|x|1. The oracle encodesnandtin a self-delimiting way. Ift= 0 orct≥n, it is trivial to output a solution of the desired quality. In all other cases,ALGcomputes an optimal (n,bctc, t)-covering design by making a systematic enumeration of all possible solutions (using lexicographic order, say).
Note there must exist abctc-block Y from this covering design such that the characteristic vectory of Y satisfiesxvy. UsingdlogC(n,bctc, t)ebits of advice, the oracle can encode the index of Y. A lengthy calculation using (1) allows us to express this upper bound in
terms of the functionB(n, c) defined in (2). J
Note that an algorithm forminASGureadingb(n) bits of advice can only produce 2b(n) different outputs, one for each possible advice string. Consider the set of input stringsIn,t = {x∈ {0,1}n: |x|1=t}, and letYn,t be the corresponding set of output strings produced by the algorithm. If the algorithm is strictlyc-competitive, thenYn,t can be converted into an (n,bctc, t)-covering design. This gives a lower bound of log maxt:bctc<nC(n,bctc, t)
on the advice needed to achieve a strict competitive ratio ofc forminASGu.
Recall that forminASGk, the output produced by an algorithm can depend on both the advice read and the correct answer to requests in previous rounds. In particular, the proof of the lower bound forminASGusketched above breaks down forminASGk. However, by using a more complicated argument, Theorem 10 gives a lower bound ofB(n, c)−O(logn) on the advice needed to achieve a strict competitive ratio ofcforminASGk. Note that this matches the upper bound from Theorem 8 up to an additive term ofO(logn).
Before proving Theorem 10, we remark that there is a close connection between results on the competitive ratio and the strict competitive ratio:
I Remark 9. Suppose that a minASG algorithm, ALG, is c-competitive. By definition, there exists a constant, α, such that ALG(σ) ≤ c·OPT(σ) +α. Then, one can construct a new algorithm, ALG0, which is strictly c-competitive and uses O(logn) additional advice bits as follows: Use O(logn) bits of advice to encode the length n of the input and use α· dlogne=O(logn) bits of advice to encode the index of (at most)αrounds in whichALG guesses 1 but where the correct answer is 0. Clearly, ALG0 can use this additional advice to achieve a strict competitive ratio ofc. This also means that a lower bound ofbon the number of advice bits required to bestrictlyc-competitive implies a lower bound ofb−O(logn) advice bits for beingc-competitive (where the constant hidden inO(logn) depends on the additive constant αof the c-competitive algorithm). We will use this observation in Theorem 10.
Note that the same technique can be used formaxASG.
ITheorem 10. For anyc >1, ac-competitive algorithm forminASGk orminASGumust read at leastb bits of advice, where
b≥log max
t:bctc<n n t
bctc t
!
−O(logn) =B(n, c)−O(logn) (3)
124 Advice Complexity for a Class of Online Problems
Proof (sketch). Fixn, cand letbbe the maximum number of advice bits read by the strictly c-competitive algorithm,ALG, over all inputs of lengthn. Suppose, by way of contradiction, that there exists somet such that bctc< nand b <log( nt
/ bctct
). Let In,t be the set of input strings of length nand Hamming weight t. We achieve the desired contradiction by showing that there is some input string inIn,t on which the algorithm incurs a cost of at leastbctc+ 1.
Note that|In,t|= nt
. By the pigeonhole principle, there must exist some advice string, ϕ, such that the set of input strings,In,tϕ ⊆In,t, of lengthnand Hamming weighttfor which ALGreads the adviceϕhas size
In,tϕ
> bctct .
We consider the computation ofALG, when reading the adviceϕ, as a game betweenALG and an adversary, ADV. From the advice, ALG learns that the input string belongs to In,tϕ . Because of the known history, at the beginning of round i, ALGalso knows the first i−1 bits of the input string. We say that a strings ∈In,tϕ is alive in round i if the first i−1 bits ofsare identical to those revealed in the firsti−1 rounds. If, in roundi, there exists some stringssuch thatsis still alive andsi= 1, thenALGmust answer 1. If not,ADVcould picks as the input string and hence ALGwould incur a cost of ∞. On the other hand, if no suchsexists in roundi, we may assume (without loss of generality) thatALGanswers 0.
Intuitively,ADVwants to maximize the number of rounds where ALGis forced to answer 1 but whereADVcan still give 0 as the correct answer.
Suppose that in some round, there aremstrings fromIn,tϕ which are alive. Lethbe the number of 1s that have yet to be revealed in each of these strings (this is well-defined since all strings fromIn,tϕ have the same number of 1s). We let L1(m, h) denote the minimum cost that the adversary can forceALGto incur in the remaining rounds when starting from this situation. The proof is finished by showing that for anym, h≥1,
L1(m, h)≥min
d:m≤ d
h
. (4)
Indeed, it follows from (4) thatL1
In,tϕ , t
≥ bctc+ 1. By induction onmandh, one can show that (4) is true by using the following adversary strategy: Letmthe number of strings alive in the current round,i. Furthermore, letm0be the number of themalive strings for which theith bit is 0, and letm1 =m−m0. If m0=m, thenADVhas to choose 0 as the correct bit in roundi. Ifm0< m, letd1be the smallest integer such thatm1≤ h−1d1
and letdbe the smallest integer such thatm≤ dh
. If d1 ≤d+ 1, the adversary chooses 1 as the correct bit in roundiand otherwise it chooses 0.
By Remark 9, the first inequality of (3) follows. The last equality follows from a simple but lengthy calculation showing that log
maxt:bctc<n n t
/ bctct
=B(n, c)−O(logn). J
3.2 Advice Complexity of maxASG
The advice complexity of maxASG is the same as that of minASG, up to an additive O(logn) term. This is not immediately obvious, but one can show that computing a c-competitive solution formaxASG, on input strings where the number of 0s isu, requires roughly the same amount of advice as computing ac-competitive solution forminASG, on input strings where the number of 1s isdu/ce.
ITheorem 11. FormaxASGuandmaxASGkand for any c >1,
b=B(n, c)±O(logn) (5)
bits of advice are necessary and sufficient to achieve a (strict) competitive ratio ofc.
J. Boyar, L. M. Favrholdt, C. Kudahl, and J. W. Mikkelsen 125