Solving Problems with Unknown Solution Length at Almost No Extra Cost

Benjamin Doerr; Carola Doerr; Timo Kötzing

doi:10.1007/s00453-018-0477-7

Algorithmica

February 2019, Volume 81, Issue 2, pp 703–748 | Cite as

Solving Problems with Unknown Solution Length at Almost No Extra Cost

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Benjamin Doerr
Carola Doerr
Timo Kötzing

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First Online: 03 July 2018

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Special Issue on Theory of Genetic and Evolutionary Computation

Abstract

Following up on previous work of Cathabard et al. (in: Proceedings of foundations of genetic algorithms (FOGA’11), ACM, 2011) we analyze variants of the (1 + 1) evolutionary algorithm (EA) for problems with unknown solution length. For their setting, in which the solution length is sampled from a geometric distribution, we provide mutation rates that yield for both benchmark functions OneMax and LeadingOnes an expected optimization time that is of the same order as that of the (1 + 1) EA knowing the solution length. More than this, we show that almost the same run times can be achieved even if no a priori information on the solution length is available. We also regard the situation in which neither the number nor the positions of the bits with an influence on the fitness function are known. Solving an open problem from Cathabard et al. we show that, for arbitrary $s\in {\mathbb {N}}$, such OneMax and LeadingOnes instances can be solved, simultaneously for all $n\in {\mathbb {N}}$, in expected time $O(n (\log (n))^2 \log \log (n) \ldots \log ^{(s-1)}(n) (\log ^{(s)}(n))^{1+\varepsilon })$ and $O(n^2 \log (n) \log \log (n) \ldots \log ^{(s-1)}(n) (\log ^{(s)}(n))^{1+\varepsilon })$, respectively; that is, in almost the same time as if n and the relevant bit positions were known. For the LeadingOnes case, we prove lower bounds of same asymptotic order of magnitude apart from the $(\log ^{(s)}(n))^{\varepsilon }$ factor. Aiming at closing this arbitrarily small remaining gap, we realize that there is no asymptotically best performance for this problem. For any algorithm solving, for all n, all instances of size n in expected time at most T(n), there is an algorithm doing the same in time $T'(n)$ with $T'=o(T)$. For OneMax we show results of similar flavor.

Keywords

Black-box optimization Evolutionary computation Runtime analysis Uncertainty Unknown solution length

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Notes

Acknowledgements

Parts of this work have been done while Timo Kötzing was visiting the École Polytechnique. This work was supported in part by the German Research Foundation under Grant FR 2988 (TOSU), by a public grant as part of the Investissement d’avenir project, reference ANR-11-LABX-0056-LMH, LabEx LMH, and from the Gaspard Monge Program for Optimization and Operations Research (PGMO) of the Jacques Hadamard Mathematical Foundation (FMJH).

Appendix A: Summable Sequences

In this section we recall definitions and results regarding summable sequences.

A sequence $\vec {p}=(p_i)_{i \in {\mathbb {N}}}$ is summable if the sequence $(S_n)_{n \in {\mathbb {N}}}$ of partial sums of absolute values $S_n:=\sum _{i=1}^n{|p_i|}$ converges.

The run time guarantees in Theorems 15, 16, 19, and 20 all decrease with increasing sequence $\vec {p}$. In the hope of finding best possible parameters for the $(1 + 1)~\text{ EA }_{\vec {p}}$ and the $(1 + 1)~\text{ EA }_{Q}$ it is therefore natural to ask for a largest summable sequence p. The following statement shows that such a sequence does not exist. More precisely, for every summable sequence $\vec {p}$, there is a summable sequence which decreases more slowly than $\vec {p}$. Such constructions are well known in the mathematics literature, cf. [2] for an easily-accessible example proving Theorem 41.

Theorem 41

(Folklore) For every summable sequence $\vec {p}:=(p_n)_{n \in {\mathbb {N}}}$ of positive terms $p_n>0$ there exists a summable sequence $\vec {q}:=(q_n)_{n \in {\mathbb {N}}}$ such that $\vec {q}=\omega (\vec {p})$.

Likewise, for every non-summable sequence $\vec {q}$ there exists a non-summable sequence $\vec {p}$ such that $\vec {p}=o(\vec {q})$.

It is interesting to note that, in both statements of Theorem 41, the sequence $\vec {q}$ can be chosen in a way that the expected run times of the $(1 + 1)~\text{ EA }_{\vec {p}}$ when fed with sequence $\vec {p}$ and $\vec {q}$, respectively, satisfy ${{\mathrm{E}}}[T_{\vec {q}}]=o({{\mathrm{E}}}[T_{\vec {p}}])$ (this is not automatically the case when $\vec {q}=\omega (\vec {p})$). But since this result also follows from Theorem 30, we do not prove these statements explicitly.

We have introduced in equation (5) (Sect. 4.2) the sequences $(p_b^{s,\varepsilon }(n))_{n \in {\mathbb {N}}}$. We have also discussed that it was shown in [21, page 48] that for every $\varepsilon >0$ and every $s \ge 1$, the sequence $(p_2^{s,\varepsilon }(n))_{n \in {\mathbb {N}}}$ is summable. In the same work, Hardy also mentions that, for all $s \in {\mathbb {N}}$, the sequence $(p_2^{s,0}(n))_{n \in {\mathbb {N}}}$ is not summable; he attributes this result to De Morgan and Bertrand. From this result we easily get the following statement, which we need in the proof of our lower bounds in Sect. 6.2.3.

Lemma 42

Let $b>1$ and let $\vec {p}$ be a summable sequence of positive terms $0<p_n<1$. Then, for all $s \in {\mathbb {N}}$, $1/p_n = \omega (1/p_b^{s,0}(n))$; i.e., $1/p_n = \omega (n \log _b(n) \log ^{(2)}_b(n)\ldots \log ^{(s)}_b(n))$.

Proof

We show the statement for $b=2$. The general case follows from observing that, for all $s \in {\mathbb {N}}$ and all $b>1$, $1/p_b^{s,0}(n) = {\varTheta }(1/p_2^{s,0}(n))$.

We thus need to show that for all summable sequences $\vec {p}$ of positive terms $0<p_n<1$, all integers $s \in {\mathbb {N}}$, and all positive constants $C>0$ there exists an integer $n_0=n_0(\vec {p},s,C)$ such that, for all $n \ge n_0$, it holds that $1/p_n \ge C/p_2^{s,0}(n) = C n \log _2(n) \log ^{(2)}_2(n) \ldots \log _2^{(s)}(n)$.

For the sake of contradiction we assume that this is not the case. Then there exists a summable sequence $\vec {p}$ of positive terms $0<p_n<1$, an integer $s \in {\mathbb {N}}$, and a constant $C>0$ such that for all $n_0 \in {\mathbb {N}}$ there exists an integer $n \ge n_0$ such that $1/p_n \le C/p_2^{s,0}(n) = C n \log _2(n) \log ^{(2)}_2(n) \ldots \log _2^{(s)}(n)$. This implies that there exists a subsequence $\vec {q}$ of $\vec {p}$ such that $q_n \ge 1/\left( C n \log _2(n) \log ^{(2)}_2(n) \ldots \log _2^{(s)}(n)\right) $. However, the result of De Morgan and Bertrand states that this sequence is not summable. Hence, $\vec {p}$ cannot be summable. $\square $

Furthermore, we have mentioned in Sect. 4.2 that the summability of the sequence $(p_b^{\infty }(n))_{n \in {\mathbb {N}}}$ (cf. equation (7)) crucially depends on the base b of the logarithm. Indeed, using Cauchy’s condensation test, one can show the following.

Theorem 43

The sequence $(p_b^{\infty }(n))_{n \in {\mathbb {N}}}$ is summable if and only if $b<e:=\exp (1)$. More precisely,

1.

For all $b>1$ and all $s \in {\mathbb {N}}$ it holds that $\sum _{i=1}^{n}{p^{(s,0)}_b(i)} = {\varTheta }\left( \log ^{(s+1)}(n)\right) $.
2.

For all $1<b<e:=\exp (1)$, $\sum _{i=1}^{n}{p^\infty _b(i)} = {\varTheta }(1)$.
3.

$\sum _{i=1}^{n}{p^\infty _e(i)} = {\varTheta }(\log ^*(n))$.
4.

For all $b>e$, $\sum _{i=1}^{n}{p^\infty _b(i)} = \exp ({\varTheta }(\ln ^*(n)))$.

Proof

For this proof, we set

$$\begin{aligned} e(s) := \exp ^{(s)}(1) = \underbrace{\exp (\exp (\ldots (\exp }_{s \text { times}}(1)))). \end{aligned}$$

Let $s \in {\mathbb {N}}$ and $b>1$. We apply integration by substitution, in the formulation of Theorem 46 below and with $\varphi = \exp _b$, $s+1$ times to the function $x \mapsto p^{(s,0)}_b(x)$. Using $b^x p^{(s,0)}_b(b^x) = p^{(s-1,0)}_b(x)$ and that the derivative of $\exp _b$ is $\exp _b$ times $\ln (b)$ we obtain, for $n \ge e(s)$,

$$\begin{aligned} \int _{e(s)}^n p^{(s,0)}_b(x) \mathrm {d}x = \int _0^{\ln ^{(s+1)}(n)} (\ln (b))^{s+1} \; dx = {\varTheta }(\ln ^{(s+1)}(n)) = {\varTheta }(\log ^{(s+1)}(n)). \end{aligned}$$

Thus, we also get (using $p^\infty _b$ monotonically decreasing)

$$\begin{aligned} \sum _{i=2}^n{p^\infty _b(i)}\le & {} \int _{1}^n p^\infty _b(x) \mathrm {d}x \le \sum _{j=0}^{\ln ^*(n)} \int _{e(j)}^{e(j+1)} p^\infty _b(x) \mathrm {d}x\\= & {} \sum _{j=0}^{\ln ^*(n)} \int _{e(j)}^{e(j+1)} p^{(j,0)}_b(x) \mathrm {d}x = \sum _{j=0}^{\ln ^*(n)} \int _{0}^{1} (\ln (b))^{j+1} \mathrm {d}x\\= & {} \sum _{j=0}^{\ln ^*(n)} (\ln (b))^{j+1}. \end{aligned}$$

Clearly, for $b<e$, this sum converges for $n \rightarrow \infty $. Furthermore, for $b=e$, this partial sum equals $\ln ^*(n)$, while for $b > e$ it is $2^{O(\ln ^*(n))}$ as desired. A corresponding lower bound can be found analogously. $\square $

From Theorem 43 we also get the following result.

Corollary 44

Let $b\ge e$. For every summable sequence $\vec {p}$ it holds that $1/p_n = \omega (1/p_b^{\infty }(n)) = \omega (n \log _b(n) \log ^{(2)}_b(n) \ldots )$.

Corollary 44 can be proven in the same way as Lemma 42.

Appendix B: Useful Tools

We use this section to recall two useful theorems.

The following well-known lemma is applied a few times in our proofs.

Lemma 45

Let $\vec {a}=(a_n)_{n \in {\mathbb {N}}}$ and $\vec {b}=(b_n)_{n \in {\mathbb {N}}}$ be sequences of positive terms satisfying $\vec {a}= o(\vec {b})$. Set $A:=(A_n)_{n \in {\mathbb {N}}}$ with $A_n:=\sum _{i=1}^n{a_i}$ and $B:=(B_n)_{n \in {\mathbb {N}}}$ with $B_n:=\sum _{i=1}^n{b_i}$. Assume that $B=\omega (1)$. Then $A=o(B)$.

Proof

We need to show that for all $\varepsilon >0$ there exists an integer $n(\varepsilon ) \in {\mathbb {N}}$ such that $A_n/B_n \le \varepsilon $. Let $\varepsilon >0$. Since $\vec {a}= o(\vec {b})$ there exists an integer $n_{\varepsilon } \in {\mathbb {N}}$ such that $a_n/b_n < \varepsilon /2$ for all $n \ge n_{\varepsilon }$. Since $B=\omega (1)$ there exists $n_{\varepsilon ,B} \in {\mathbb {N}}$ such that $\sum _{i=1}^{n_{\varepsilon }-1}{a_i}/B_n \le \varepsilon /2$ for all $n \ge n_{\varepsilon ,B}$. We therefore obtain for all $n \ge \max \{n_{\varepsilon }, n_{\varepsilon ,B}\}$

$$\begin{aligned} \frac{A_n}{B_n}&= \frac{\sum _{i=1}^{n_{\varepsilon }-1}{a_i}}{B_n} + \frac{\sum _{i=n_{\varepsilon }}^{n}{a_i}}{B_n}\\&\le \frac{\varepsilon }{2} + \frac{\varepsilon \sum _{i=n_{\varepsilon }}^{n}{b_i}}{2B_n} \le \varepsilon . \end{aligned}$$

$\square $

When we apply integration by substitution, we refer to the following formulation.

Theorem 46

(Integration by Substitution) Let $m<n$ and let I be an Interval. Let $f: I \rightarrow {\mathbb {R}}$ be continuous and $\varphi : [m,n] \rightarrow I$. Then we have

$$\begin{aligned} \int _{\varphi (m)}^{\varphi (n)} f(x) \mathrm {d}x = \int _{m}^n f(\varphi (x))\varphi '(x) \mathrm {d}x. \end{aligned}$$

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Authors and Affiliations

Benjamin Doerr
- 1
Carola Doerr
- 2
Email author
Timo Kötzing
- 3

1.École Polytechnique, CNRS, LIX - UMR 7161PalaiseauFrance
2.Sorbonne Université, CNRS, LIP6ParisFrance
3.Hasso-Plattner-InstitutPotsdamGermany

Solving Problems with Unknown Solution Length at Almost No Extra Cost

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