The Advice Complexity of a Class of Hard Online Problems

Abstract

The advice complexity of an online problem is a measure of how much knowledge of the future an online algorithm needs in order to achieve a certain competitive ratio. Using advice complexity, we define the first online complexity class, AOC. The class includes independent set, vertex cover, dominating set, and several others as complete problems. AOC-complete problems are hard, since a single wrong answer by the online algorithm can have devastating consequences. For each of these problems, we show that $\log \left (1+(c-1)^{c-1}/c^{c}\right )n={\Theta } (n/c)$ bits of advice are necessary and sufficient (up to an additive term of $O(\log n)$) to achieve a competitive ratio of c. The results are obtained by introducing a new string guessing problem related to those of Emek et al. (Theor. Comput. Sci. 412(24), 2642–2656 2011) and Böckenhauer et al. (Theor. Comput. Sci. 554, 95–108 2014). It turns out that this gives a powerful but easy-to-use method for providing both upper and lower bounds on the advice complexity of an entire class of online problems, the AOC-complete problems. Previous results of Halldórsson et al. (Theor. Comput. Sci. 289(2), 953–962 2002) on online independent set, in a related model, imply that the advice complexity of the problem is Θ(n/c). Our results improve on this by providing an exact formula for the higher-order term. For online disjoint path allocation, Böckenhauer et al. (ISAAC 2009) gave a lower bound of Ω(n/c) and an upper bound of $O((n\log c)/c)$ on the advice complexity. We improve on the upper bound by a factor of $\log c$. For the remaining problems, no bounds on their advice complexity were previously known.

Appendix: A: Approximation of the Advice Complexity Bounds

In Theorems 3–8, bounds on the advice complexity of ASG were obtained. These bounds are tight up to an additive term of $O(\log n)$. However, within the proofs, they are all expressed in terms of the minimum size of a certain covering design or a quotient of binomial coefficients. In this appendix, we prove the closed formula estimates for the advice complexity stated in Theorems 3–8 and 11. Again, these estimates are tight up to an additive term of $O(\log n)$. The key to obtaining the estimates is the estimation of a binomial coefficient using the binary entropy function.

1.1 A.1 Approximating the Function B(n, c)

Lemma 15

For c>1, it holds that

$$\frac{1}{e\ln(2)}\frac{1}{c}\leq\log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)\leq \frac{1}{c}. $$

Proof

We prove the upper bound first. To this end, note that

$$\log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)\leq \frac{1}{c} \!\Leftrightarrow\! 1+\frac{(c-1)^{c-1}}{c^{c}}\leq 2^{1/c} \; \Leftrightarrow \; \left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{c}\!\leq\! 2. $$

Using calculus, one may verify that $\left (1+\frac {(c-1)^{c-1}}{c^{c}}\right )^{c}$ is decreasing in c for c>1. Thus, by continuity, it follows that

$$\begin{array}{@{}rcl@{}} \left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{c}&\leq&\lim_{c\rightarrow 1^{+}}\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{c} =\lim_{c\rightarrow 1^{+}}\left(1+\left(\frac{c-1}{c}\right)^{c-1} \, \frac{1}{c}\right)^{c}\\ &=&\lim_{c\rightarrow 1^{+}}\left(1+\frac{1}{c}\right)^{c} =2. \end{array} $$

For the lower bound, let $a=e\ln (2)$ and note that

$$\frac{1}{ac}\leq \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right) \; \Leftrightarrow \; 2\leq \left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{ac}. $$

Again, using calculus, one may verify that $\left (1+\frac {(c-1)^{c-1}}{c^{c}}\right )^{ac}$ is decreasing in c for c>1. It follows that

$$\begin{array}{@{}rcl@{}} \left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{ac}\!&\geq&\! \lim_{c\rightarrow\infty}\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)^{ac} =\lim_{c\rightarrow\infty}\left(1+\left(\frac{c-1}{c}\right)^{c-1} \, \frac{1}{c}\right)^{ac}\\ &=&\!\lim_{c\rightarrow\infty}\left(1+\frac{1}{e} \, \frac{1}{c}\right)^{ac} \,=\,\lim_{c\rightarrow\infty}\left(\!1\,+\,\frac{a/e}{ac}\right)^{ac} \,=\, e^{a/e}\,=\,e^{\ln (2)}=2. \end{array} $$

□

1.2 A.2 The Binary Entropy Function

In this section, we give some properties of the binary entropy function that will be used extensively in Appendix AA.4.

Definition 11

The binary entropy function $H:[0,1]\rightarrow [0,1]$ is the function given by

$$H(p)=-p\log(p) - (1-p)\log(1-p),\;\text{for }0<p<1,$$

and H(0)=H(1)=0.

Lemma 16 (Lemma 9.2 in27)

For integers m,n such that 0≤m≤n,

$$\frac{2^{nH\left(m/n\right)}}{n+1}\leq \left(\begin{array}{c} n \\ m \end{array}\right)\leq 2^{nH\left(m/n\right)}. $$

Proposition 1

The binary entropy function H(p) has the following properties.

Proof

(H1): Follows from the definition.

(H2): For s>1,

$$\begin{array}{@{}rcl@{}} sH\left(\frac{1}{s}\right)&=&s\left(\log s + \frac{1-s}{s}\log (s-1)\right), \text{ by \textit{(H1)}}\\ &=&\log\left(\left(1+\frac{1}{s-1}\right)^{s-1}s\right) \leq \log (e\cdot s)= \log (e) + \log (s)\leq \log s +2. \end{array} $$

(H3): Note that H is smooth for 0<p<1. The derivative $H^{\prime }(p)$ can be calculated from the definition. The second-order derivative is

$$H^{\prime\prime}(p)=\frac{-1}{(1-p)p\ln (2)}\,, $$

which is strictly less than zero for all 0<p<1.

(H4): Fix t>0. The claim follows by showing that the partial derivative of $sH(\frac {t}{s})$ with respect to s is positive for all s>t.

$$\begin{array}{@{}rcl@{}} \frac{d}{ds}\left(sH\left(\frac{t}{s}\right)\right)&=&H\left(\frac{t}{s}\right)+sH^{\prime}\left(\frac{t}{s}\right)\left(-\frac{t}{s^{2}}\right) =H\left(\frac{t}{s}\right)-\frac{t}{s}H^{\prime}\left(\frac{t}{s}\right)\\ &=&-\frac{t}{s}\log\left(\frac{t}{s}\right)-\left(1-\frac{t}{s}\right)\log\left(1-\frac{t}{s}\right)\\ &&-\frac{t}{s}\log\left(\frac{s}{t}-1\right), \text{by Def.~11 and \textit{(H3)}}\\ &=&-\log \left(1-\frac{t}{s}\right)>0. \end{array} $$

(H5): H(p) is increasing for $0\leq p\leq \frac {1}{2}$ and decreasing for $\frac {1}{2}\leq p\leq 1$. If $\frac {1}{x}+\frac {1}{n}\leq \frac 12$, then the claim is trivially true (since then the difference is negative). Assume therefore that $\frac {1}{x}+\frac {1}{n}> \frac 12$. Under this assumption, $H(\frac {1}{x})$ increases and $H(\frac {1}{x}+\frac {1}{n})$ decreases as x tends to 2. Thus, $H(\frac {1}{x})-H(\frac {1}{x}+\frac {1}{n})$ increases as x tends to 2 and, hence,

$$ H\left(\frac{1}{x}\right)-H\left(\frac{1}{x}+\frac{1}{n}\right)\leq H\left(\frac{1}{2}\right)-H\left(\frac{1}{2}+\frac{1}{n}\right). $$

(3)

Inserting into the definition of H gives

$$\begin{array}{@{}rcl@{}} H\left(\frac{1}{2}\right)\,-\,H\left(\frac{1}{2}\,+\,\frac{1}{n}\right)\!&=&\!1\,-\,\left(\,-\,\left(\frac{1}{2}\,+\,\frac{1}{n}\right)\log \left(\frac{1}{2}\,+\,\frac{1}{n}\right)\,-\,\left(\frac{1}{2}\,-\,\frac{1}{n}\right)\log \left(\frac{1}{2}\,-\,\frac{1}{n}\right)\!\right)\\ &=&\frac{1}{n} \log\left(\frac{\frac{1}{2}+\frac{1}{n}}{\frac{1}{2}-\frac{1}{n}}\right)\,+\,\frac{1}{2}\log\left(\!\left(\frac{1}{2}\,+\,\frac{1}{n}\right)\left(\frac{1}{2}\,-\,\frac{1}{n}\right)\!\right)\,+\,1\\ &=&\frac{1}{n} \log\left(\frac{n+2}{n-2}\right)+\frac{1}{2}\log\left(\frac{n^{2}-4}{4n^{2}}\right)+1 \end{array} $$

Since (n+2)/(n−2) is decreasing for n ≥ 3, it follows that $\log ((n+2)/$ $(n-2))\leq \log (5)$. Furthermore, $(n^{2}-4)/(4n^{2})\leq \frac {1}{4}$ for all n ≥ 3, and so $\frac {1}{2}\log \left ((n^{2}-4)/(4n^{2})\right )+1\leq 0$. We conclude that, for all n ≥ 3,

$$ H\left(\frac{1}{2}\right)-H\left(\frac{1}{2}+\frac{1}{n}\right)\leq \frac{\log(5)}{n}<\frac{3}{n}. $$

(4)

Combining (3) and (4) proves (H5). □

1.3 A.3 Binomial Coefficients

The following proposition is a collection of simple facts about the binomial coefficient that will be used in Appendices AA.4 and AA.5.

Proposition 2

Let $a,b,c\in \mathbb {N}$.

Proof

First, we prove (B1):

$$\left(\begin{array}{c} a \\ b \end{array}\right) = \frac{a!}{b!(a-b)!} = \frac{a}{a-b} \, \frac{(a-1)!}{b!(a-1-b)!} = \frac{a}{a-b} \left(\begin{array}{c} a-1 \\ b \end{array}\right)$$

(B2) follows directly from (B1).

To prove (B3), we calculate the two fractions separately:

$$\begin{array}{@{}rcl@{}} \frac{\left(\begin{array}{c} a \\ c \end{array}\right)}{\left(\begin{array}{c} b \\ c \end{array}\right)} & =& \frac{a!}{c!(a-c)!} \frac{c!(b-c)!}{b!} = \frac{a!}{(a-c)!} \frac{(b-c)!}{b!}\\ \frac{\left(\begin{array}{c} a \\ b \end{array}\right)}{\left(\begin{array}{c} a-c \\ a-b \end{array}\right)} & =& \frac{a!}{b!(a-b)!} \frac{(a-b)!(b-c)!}{(a-c)!} = \frac{a!}{b!} \frac{(b-c)!}{(a-c)!} = \frac{\left(\begin{array}{c} a \\ c \end{array}\right)}{\left(\begin{array}{c} b \\ c \end{array}\right)}. \end{array} $$

□

1.4 A.4 Approximating the Advice Complexity Bounds for minASG

The following lemma is used for proving Theorems 3–5.

Lemma 17

For c>1 and n ≥ 3,

$$\begin{array}{@{}rcl@{}} \log\left(\max_{t\colon \lfloor{ct}\rfloor< n} C(n,\lfloor{ct}\rfloor,t)\right)&\geq& \log\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right) \end{array} $$

(5)

$$\begin{array}{@{}rcl@{}} &\geq& \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n-2\log(n+1)-5 \end{array} $$

(6)

and

$$\begin{array}{@{}rcl@{}} \log\left(\max_{t\colon \lfloor{ct}\rfloor< n} C(n,\lfloor{ct}\rfloor,t)\right)&\leq& \log\left(\max_{t\colon \lfloor{ct}\rfloor < n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}n\right) \end{array} $$

(7)

$$\begin{array}{@{}rcl@{}} &\leq& \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n+3\log (n+1). \end{array} $$

(8)

Proof

We prove the upper and lower bounds separately.Upper bound: Fix n, c. By Lemma 1,

$$C(n,\lfloor{ct}\rfloor,t)\leq\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\left(1+\ln\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)\right). $$

Note that $1+\ln \left (\begin {array}{c} \lfloor {ct}\rfloor \\ t\end {array}\right )\leq n$since we consider only ⌊c t⌋<n. This proves (7). Now, taking the logarithm on both sides gives

$$\begin{array}{@{}rcl@{}} \log (C(n,\lfloor{ct}\rfloor,t)) &\leq&\! \log\!\left(\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\!\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\!\right)\,+\,\log n \!\leq\! \log\!\left(\!\frac{\left(\!\begin{array}{c} n \\ t\end{array}\!\right)}{\left(\!\begin{array}{c} \lceil{ct}\rceil\,-\,1 \\ t \end{array}\!\right)}\!\right)\,+\,\log n\\ &\leq&\! \log\left(\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\frac{\lceil{ct}\rceil-t}{\lceil{ct}\rceil}\left(\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\right)} \right)+\log n, \text{ by }(B1)\\ &\leq&\! \log\!\left(\!\frac{\left(\!\begin{array}{c} n \\ t\end{array}\!\right)}{\left(\!\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\!\right)}\!\right) \,+\, \log\left(\!\frac{\lceil{ct}\rceil}{\lceil{ct}\rceil\,-\,t}\!\right)\,+\,\log n\\ &\leq& \log\left(\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\right)} \right)+2\log n\,. \end{array} $$

(9)

Above, we have increased ⌊c t⌋ to ⌈c t⌉ in the binomial coefficient (at the price of an additive term of $\log n$). This is done since it will later be convenient to use that c t≤⌈c t⌉. Using Lemma 16, we get that

$$\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\right)} \leq \frac{2^{nH(t/n)}}{2^{\lceil{ct}\rceil H(t/\lceil{ct}\rceil)}}\left(\lceil{ct}\rceil+1\right), $$

and therefore

$$\begin{array}{@{}rcl@{}} \log\left(\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\right)} \right)\!&\leq&\! nH\left(\frac{t}{n}\right)\,-\,\lceil{ct}\rceil H\left(\!\frac{t}{\lceil{ct}\rceil}\!\right)+\log\left(\lceil{ct}\rceil +1\right)\\ &\leq& nH\left(\frac{t}{n}\right)-ctH\left(\frac{1}{c}\right)+\log(n+1), \text{ by \textit{(H4)}.} \end{array} $$

(10)

Define

$$M(n,t)=nH\left(\frac{t}{n}\right)-ctH\left(\frac{1}{c}\right). $$

Combining (9) and (10) shows that

$$ \log (C(n, \lfloor{ct}\rfloor, t))\leq M(n,t)+3\log (n+1)\,. $$

(11)

The function M is smooth. For any given input length n, we can determine the value of t maximizing M(n, t) using calculus. In order to simplify the notation for these calculations, define

$$x=\left(\frac{c}{c-1}\right)^{c}(c-1)+1, $$

and note that

$$\begin{array}{@{}rcl@{}} \log(x-1)&=&c\left(\log c +\frac{1-c}{c}\log (c-1)\right)\\ &=&cH\left(\frac{1}{c}\right), \text{ by \textit{(H1)}.} \end{array} $$

(12)

We want to determine those values of t for which $\frac {d}{dt}M(n,t)=0$:

$$\begin{array}{@{}rcl@{}} \frac{d}{dt}M(n,t)&=&\frac{d}{dt}\left(nH\left(\frac{t}{n}\right)-ctH\left(\frac{1}{c}\right)\right)=0\\ &\Leftrightarrow&\; nH^{\prime}\left(\frac{t}{n}\right) \cdot \frac{1}{n} -cH\left(\frac{1}{c}\right)=0\\ &\Leftrightarrow&\; \log\left(\frac{n}{t}-1\right)=cH\left(\frac{1}{c}\right), \text{ by \textit{(H3)}}\\ &\Leftrightarrow&\; \frac{n}{t}=2^{cH(1/c)}+1\\ &\Leftrightarrow&\; t=\frac{n}{2^{cH(1/c)}+1}\\ &\Leftrightarrow&\; t=\frac{n}{2^{\log(x-1)}+1}, \text{ by (12)}\\ &\Leftrightarrow&\; t=\frac{n}{x}. \end{array} $$

Note that $\frac {d^{2}}{dt^{2}}M(n,t)=H^{\prime \prime }(\frac {t}{n})/n<0$ for all values of t, by (H3). Thus,

$$ M(n,t)\leq M\left(n,\frac{n}{x}\right), \text{ for all values of } t\,. $$

(13)

The value of $M(n, \frac {n}{x})$ can be calculated as follows:

$$\begin{array}{@{}rcl@{}} M\left(n, \frac{n}{x}\right)&=& n\,H\left(\frac{1}{x} \right)-c\,\frac{n}{x}\, H\left(\frac{1}{c}\right)\\ &=&n\left(\log(x)+\frac{1-x}{x}\log(x-1)-\frac{c}{x}H(1/c)\right), \text{ by \textit{(H1)}}\\ &=&\; n\left(\log(x)+\frac{1-x}{x}\log(x-1)-\frac{1}{x}\log(x-1) \right), \text{ by (12)}\\ &=&n\left(\log(x)-\log(x-1)\right) =n\log\left(\frac{x}{x-1}\right)\\ &=&n\log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right). \end{array} $$

(14)

Combining (11), (13), and (14), we conclude that

$$\log(C(n, \lfloor{ct}\rfloor, t))\leq n\log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)+3\log (n+1). $$

Lower Bound: By Lemma 1,

$$\log\left(\max_{t\colon \lfloor{ct}\rfloor< n} C(n,\lfloor{ct}\rfloor,t)\right) \geq \log\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right)\,.$$

This proves (5). In order to prove (6), first note that by Lemma 15,

$$\log\left(1+ \frac{(c-1)^{c-1}}{c^{c}} \right) n \leq \frac{n}{c}\,.$$

Thus, for $c \geq \frac {n}{2}$, the righthand side of (6) is negative, and hence, the inequality is trivially true.

Assume now that $c < \frac {n}{2}$. We will determine an integer value of t such that $\left (\begin {array}{c} n \\ t\end {array}\right )/\left (\begin {array}{c} \lfloor {ct}\rfloor \\ t\end {array}\right )$ becomes sufficiently large. First, we use Lemma 16:

$$\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\geq \frac{2^{n H(t/n)}}{(n+1)\cdot2^{\lfloor{ct}\rfloor H(t/\lfloor{ct}\rfloor)}} =\frac{2^{nH(t/n)-\lfloor{c t}\rfloor H(t/\lfloor{ct}\rfloor)}}{n+1} $$

It is possible that t=⌊c t⌋, but this is fine since H(1)=0. Using (H4), we see that

$$\lfloor{c t}\rfloor H\left(\frac{t}{\lfloor{ct}\rfloor}\right)\leq ct H\left(\frac{t}{ct}\right) = ct H\left(\frac{1}{c}\right). $$

Thus,

$$ \log\left(\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right)\!\geq\! nH\left(\frac{t}{n} \right)- c t H\left(\frac{1}{c} \right)-\log(n+1)\,=\, M(n,t) - \log (n+1). $$

(15)

Let $t^{\prime }=\frac {n}{x}$. We know that M(n, t) attains its maximum value when $t=t^{\prime }$. Since c>1, it is clear that x>c and hence $t^{\prime }<\frac {n}{c}$. It follows that $\lfloor {ct^{\prime }}\rfloor <n$. However, $t^{\prime }$ might not be an integer. In what follows, we will first argue that $\lfloor {c\lceil {t^{\prime }}\rceil }\rfloor <n$ and then that $M(n,\lceil {t^{\prime }}\rceil )$ is close to $M(n,t^{\prime })$. The desired lower bound will then follow by setting $t=\lceil {t^{\prime }}\rceil$.

Using calculus, it can be verified that, for c>1, x/c is increasing in c. Hence,

$$\begin{array}{@{}rcl@{}} \frac{x}{c} & =& \left(\frac{c}{c-1}\right)^{c-1}+\frac{1}{c}\\ & \geq& \lim_{c\rightarrow 1^{+}} \left(\left(\frac{c}{c-1}\right)^{c-1}+\frac{1}{c} \right), \text{ for } c>1\\ & =& \lim_{c\rightarrow 1^{+}} \left(1+\frac{1}{c-1}\right)^{c-1}+ \lim_{c\rightarrow 1^{+}} \frac{1}{c} = \lim_{a\rightarrow 0^{+}} \left(1+\frac{1}{a}\right)^{a} +1 = 2\,. \end{array} $$

Thus, c≤x/2, and hence,

$$\lfloor{c\lceil{t^{\prime}}\rceil}\rfloor \leq c \left\lceil \frac{n}{x} \right\rceil < \frac{cn}{x}+c \leq \frac{n}{2}+c < n\,. $$

Note that $\frac {d}{dt}M(n,t)<0$ for $t>t^{\prime }$, so $M(n,\lceil {t^{\prime }}\rceil )\geq M(n,t^{\prime }+1)$. Combining this observation with (H2) and (H5), we get that

$$\begin{array}{@{}rcl@{}} M(n,\lceil{t^{\prime}}\rceil)&\geq& M(n, t^{\prime}+1) =nH\left(\frac{t^{\prime}+1}{n}\right)-c(t^{\prime}+1)H\left(\frac{1}{c}\right)\\ &=&nH\left(\frac{1}{x}+\frac{1}{n}\right)-c\frac{n}{x}H\left(\frac{1}{c}\right)-cH\left(\frac{1}{c}\right)\\ &\geq&\; nH\left(\frac{1}{x}+\frac{1}{n}\right)-c\frac{n}{x}H\left(\frac{1}{c}\right)-\log n-2, \text{ by \textit{(H2)}}\\ &\geq&\; nH\left(\frac{1}{x}\right)-c\frac{n}{x}H\left(\frac{1}{c}\right)-\log n-5, \text{ by \textit{(H5)}}\\ &=& M(n,t^{\prime})-\log n-5. \end{array} $$

By choosing $t=\lceil {t^{\prime }}\rceil$ in the $\max$, we conclude that

$$\begin{array}{@{}rcl@{}} \log\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right)&\geq& M(n,\lceil{t^{\prime}}\rceil)-\log (n+1), \text{ by~(15)}\\ &\geq& M(n,t^{\prime})-\log(n+1)-\log n-5\\ &\geq& n\log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)-2\log (n+1)-5, \text{ by~(14)}. \end{array} $$

□

The following lemma is used for proving Theorem 10.

Lemma 18

If c is an integer-valued function of n and c>1, it holds that

$$\log\left(\max_{t\colon ct< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} ct \\ t\end{array}\right)}\right)={\Omega}\left(\frac{n}{c}\right). $$

Proof

Assume that c is an integer-valued function of n, that c>1 and that c t<n. It follows that

$$\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} ct \\ t \end{array}\right)}=\frac{n!(ct-t)!}{(n-t)!(ct)!} \geq\frac{n (n-1){\cdots} (n-t+1)}{(ct)(ct-1)\cdots (ct-t+1)} $$

Let $t=\lfloor {\frac {n}{ec}}\rfloor$. Then

$$\begin{array}{@{}rcl@{}} \frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} ct \\ t \end{array}\right)}&=&\frac{n (n-1){\cdots} (n-t+1)}{(ct)(ct-1){\cdots} (ct-t+1)} \geq \frac{n (n-1){\cdots} (n-t+1)}{\frac{n}{e}(\frac{n}{e}-1){\cdots} (\frac{n}{e}-t+1)}\\ &=& \frac{n}{\frac{n}{e}}\frac{n-1}{\frac{n}{e}-1}\cdots\frac{n-t+1}{\frac{n}{e}-t+1} \geq e^{t}. \end{array} $$

Since

$$\log(e^{t})=t\log(e)\geq \left(\frac{n}{ec}-1\right)\log e = \frac{n}{e \, \ln(2) \, c}-\log(e)={\Omega}\left(\frac{n}{c}\right), $$

this proves the lemma by choosing $t=\lfloor {\frac {n}{ec}}\rfloor$. □

1.5 A.5 Approximating the Advice Complexity Bounds for maxASG

Lemma 20 of this section is used for Theorems 6–8. In proving Lemma 20, the following lemma will be useful.

Lemma 19

For all n,c, it holds that

$$ \max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{u/c}\rceil \\ n-u \end{array}\right)}\leq n \left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right). $$

On the other hand, it also holds that

$$ \max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{u/c}\rceil \\ n-u \end{array}\right)}\geq \frac{1}{n}\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right). $$

Proof

Let

$$f_{n,c}(t)=\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)} \: \text{ and } \: g_{n,c}(u)=\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{u/c}\rceil \\ n-u \end{array}\right)}\,.$$

In order to prove the upper bound, we show that f _{n, c} (⌊u/c⌋) ≥ g _{n, c} (u)/n, for any integer u, 0<u<n. Note that ⌊u/c⌋<u, since c>1.

$$\begin{array}{@{}rcl@{}} f_{n,c}(\lfloor{u / c}\rfloor)&=&\frac{\left(\begin{array}{c} n \\ \lfloor{u / c}\rfloor \end{array}\right)}{\left(\begin{array}{c} \lfloor{c\lfloor{u/c}\rfloor}\rfloor \\ \lfloor{u / c}\rfloor\end{array}\right)}\\ &\geq&\; \frac{\left(\begin{array}{c} n \\ \lfloor{u / c}\rfloor \end{array}\right)}{\left(\begin{array}{c} u \\ \lfloor{u / c}\rfloor\end{array}\right)}, \text{ by } \mathit{(B2)}\\ &=&\;\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lfloor{u /c}\rfloor \\ n-u \end{array}\right)}, \text{ by } \mathit{(B3)}\\ &\geq&\;\frac{u-\lfloor{u/c}\rfloor}{n-\lfloor{u/c}\rfloor} \, \frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{u /c}\rceil \\ n-u \end{array}\right)}, \text{ by } \mathit{(B1)}\\ &\geq&\;\frac{u-\lfloor{u/c}\rfloor}{n-\lfloor{u/c}\rfloor} \: g_{n,c}(u)\\ &\geq&\frac{1}{n} \, g_{n,c}(u), \text{ since } u-\lfloor{u/c}\rfloor \geq 1. \end{array} $$

By (B1), the second last inequality is actually an equality, unless u/c is an integer.

In order to prove the lower bound, we will show that g _{n, c} (⌈c t⌉) ≥ f _{n, c} (t)/n, for any integer t with ⌊c t⌋<n. Note that t<⌈c t⌉, since c>1.

$$\begin{array}{@{}rcl@{}} g_{n,c}(\lceil{ct}\rceil)&=&\frac{\left(\begin{array}{c} n \\ \lceil{ct}\rceil \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\lceil{ct}\rceil/c}\rceil \\ n-\lceil{ct}\rceil \end{array}\right)}\\ &\geq&\; \frac{\left(\begin{array}{c} n \\ \lceil{ct}\rceil \end{array}\right)}{\left(\begin{array}{c} n-t \\ n-\lceil{ct}\rceil \end{array}\right)}, \text{ by }\mathit{(B2)}\\ \end{array} $$

$$\begin{array}{@{}rcl@{}} &=&\frac{\left(\begin{array}{c} n \\ n-\lceil{ct}\rceil \end{array}\right)}{\left(\begin{array}{c} n-t \\ n-\lceil{ct}\rceil \end{array}\right)}\\ &=&\;\frac{\left(\begin{array}{c} n \\ n-t \end{array}\right)}{\left(\begin{array}{c} \lceil{ct}\rceil \\ t \end{array}\right)}, \text{ by }\mathit{(B3)}\\ &=&\frac{\left(\begin{array}{c} n \\ n-t \end{array}\right)}{\frac{\lceil{ct}\rceil}{\lceil{ct}\rceil-t}\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}, \text{ by }\mathit{(B1)}\\ &=&\frac{\lceil{ct}\rceil-t}{\lceil{ct}\rceil} \, f_{n,c}(t)\\ &\geq&\frac{1}{n} \, f_{n,c}(t), \text{ since } \lceil{ct}\rceil-t \geq 1 \text{ and } \lceil{ct}\rceil \leq n. \end{array} $$

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Lemma 20

Let c>1 and n ≥ 3. It holds that

$$\begin{array}{@{}rcl@{}} \log\left(\max_{u\colon 0<u< n} C(n,n-\left\lceil\frac{u}{c}\right\rceil,n-u)\right)\!&\geq&\! \log\left(\max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\frac{u}{c}}\rceil \\ n-u \end{array}\right)}\right)\\ &\geq&\! \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n-3\log n-6. \end{array} $$

Furthermore,

$$\begin{array}{@{}rcl@{}} \log\left(\max_{u\colon 0<u< n} C(n,n-\left\lceil\frac{u}{c}\right\rceil,n-u)\right)\!&\leq&\! \log\left(\!\max_{u \colon 0<u<n}\frac{\left(\!\begin{array}{c} n \\ u \end{array}\!\right)}{\left(\begin{array}{c} n\,-\,\lceil{\frac{u}{c}}\rceil \\ n\,-\,u \end{array}\right)} \, n\!\right)\\ &\leq& \log\left(1\,+\,\frac{(c\,-\,1)^{c\,-\,1}}{c^{c}}\right)n\,+\,4\log(n\,+\,1). \end{array} $$

Proof

We prove the lower bound first.

$$\begin{array}{@{}rcl@{}} &&\log\left(\max_{u\colon 0<u< n} C(n,n-\left\lceil\frac{u}{c}\right\rceil,n-u)\right)\\ &&\qquad\geq\log\left(\max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\frac{u}{c}}\rceil \\ n-u \end{array}\right)}\right), \text{ by Lemma ~1}\\ &&\qquad\geq \log\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)}\right)- \log n, \text{ by Lemma~19}\\ &&\qquad\geq \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n-2\log(n+1)-5 - \log n, \text{ by (6)}\\ &&\qquad\geq \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n-3\log n-6, \text{ since } n \geq 3. \end{array} $$

We now prove the upper bound.

$$\begin{array}{@{}rcl@{}} && \log\left(\max_{u\colon 0<u< n} C(n,n-\left\lceil\frac{u}{c}\right\rceil,n-u)\right)\\ &&\qquad\leq \log\left(\max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\frac{u}{c}}\rceil \\ n-u \end{array}\right)}\left(1\,+\,\ln\left(\begin{array}{c} n-\lceil{u/c}\rceil \\ n-u \end{array}\!\right)\!\right)\!\right)\!,\\ &&\qquad\qquad\;\;\text{by Lemma~1}\\ &&\qquad\leq \log\left(\max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\frac{u}{c}}\rceil \\ n-u \end{array}\right)} \, n\right)\\ &&\qquad= \log\left(\max_{u \colon 0<u<n}\frac{\left(\begin{array}{c} n \\ u \end{array}\right)}{\left(\begin{array}{c} n-\lceil{\frac{u}{c}}\rceil \\ n-u \end{array}\right)}\right) + \log n\\ &&\qquad\leq \log\left(\max_{t\colon \lfloor{ct}\rfloor< n}\frac{\left(\begin{array}{c} n \\ t\end{array}\right)}{\left(\begin{array}{c} \lfloor{ct}\rfloor \\ t\end{array}\right)} \, n\right) + \log n, \text{ by Lemma~19}\\ &&\qquad\leq \log\left(1+\frac{(c-1)^{c-1}}{c^{c}}\right)n+4\log (n+1), \text{ by (8).} \end{array} $$

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