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Transcript
FT
An Analysis of Evolutionary Algorithms for
Finding Approximation Solutions to Hard
Optimisation Problems
Jun He and Xin Yao
School of Computer Science, The University of Birmingham
Birmingham B15 2TT, England
[email protected]
DR
A
Abstract. In practice, evolutionary algorithms are often used to find
good feasible solutions to complex optimisation problems in a reasonable
running time, rather than the optimal solutions. In theory, an important
question we should answer is that: how good approximation solutions
can evolutionary algorithms produce in a polynomial time? This paper
makes an initial discussion on this question and connects evolutionary
algorithms with approximation algorithms together. It is shown that evolutionary algorithms can’t find a good approximation solution to two
families of hard problems.
1
Introduction
In many applications, evolutionary algorithms (EAs) are used to find a good
feasible solution for complex optimisation problems [1, 2]. There are some experiments that claim EAs can obtain higher quality solutions in a shorter running
time than existing algorithms. But in theory, we know little about this. We
should answer the question of how good approximation solutions EAs can produce in a polynomial time. In this paper, we aim to obtain some initial answers
to this question.
In combinatorial optimisation, there have already existed a theory on this
topic, i.e., approximation algorithms for N P -hard problems [3, 4]. Approximation algorithms have been developed in response to the impossibility of solving a
great variety of important problems. It aims to investigate the quality of solution
an algorithm can produce in a polynomial time for hard problems.
In this paper we investigate evolutionary algorithms under the framework of
approximation algorithms. The first thing that we will study is to identify what
kind of problems is hard to EAs and to describe their characteristics. Of course
N P -hard problems are naturally hard to EAs, but some problems in P class
are hard to EAs too. In this paper, we introduces a classification of EA-hard
problems, i.e., wide-gap far-distance and narrow-gap far-distance problems.
In the next step we analyse for what kind of EA-hard problems, EAs can’t
find a good approximation solution to them in a polynomial time. We present
2
two families of EA-hard problems, which are difficult for EAs to find a good
approximation solution. We describe the features of these families.
The paper is arranged as follows: Section 2 makes a link between evolutionary
algorithms and approximation algorithms; Section 3 introduces a classification
of EA-hard problems; Section 4 discusses the EA-hard problem which EAs can’t
find a good approximation solutions to them; Section 5 summarise the paper.
2
2.1
Approximation Algorithms and Evolutionary
Algorithms
Approximation Algorithms for NP-hard Problems
Approximation algorithms have developed in response to the impossibility of
solving a grate of important optimisation problems. If the optimal solution is
unattainable, then it is reasonable to sacrifice optimality and settle for a good
feasible solution that can be computed efficiently. In practise, we expect we can
find a good and satisfying, but maybe not the best solution in a polynomial time.
A survey about the past and recent achievements on this topic can be found in
[4]. In this section, we use some definitions and statements directly from [3].
Foremost among the concepts in approximation algorithms is that of a ǫapproximation algorithm. An approximation algorithm is always assumed to
be efficient or more precisely, polynomial. We also assume that approximation
algorithm delivers a feasible solution to some hard combinatorial optimisation
problem that has a set of instance {I}.
Given an optimisation (minimisation or maximisation) problem with a positive const function f , a polynomial algorithm, α, returns a feasible solution
fα (I) for each instance I of the problem; denote the optimal solution of I by
fˆ(I). Then α is said to be a ǫ-approximation algorithm for some ǫ > 0 if and
only if
| fα (I) − fˆ(I) |
≤ ǫ,
f (I)
for all instances I.
The above definition is based on the worst-case performance of algorithms,
which means that is suffices to have a single bad instance to render the value of
ǫ larger than it is for all other encountered instance. To describe this worst-case
analysis, we sometimes allow ǫ to be a function of the input.
An approximation algorithm is said to be good if it has an ǫ-approximation
for some constant ǫ > 0 for all instances.
2.2
Evolutionary Algorithms and Approximation Algorithms
From N P -hard theory [3], we know that we can’t find any efficient algorithm
for hard combinatorial optimisation problems at present and even forever. We
believe that EAs are not efficient algorithms too, although we don’t prove this
point. Instead of searching the exact solution to hard optimisation problems, we
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
3
expect EAs to find some good approximation solutions efficiently. But we had
to discuss whether EAs could do in this way.
In this paper, we assume the optimisation problem is Pseudo-Boolean minimum problem and the minimum value of the objective function fmin > 0. Let
x∗ is the approximation solution returned by an EA, then the performance ratio
of the EA is
| f (x∗ ) − f (xmin ) |
r=
.
fmin
Because EAs is a kind of stochastic algorithm, we use the expectation E [f (x∗ )]
instead of the result f (x∗ ) returned by a single running:
r=
| E [f (x∗ )] − f (xmin ) |
.
fmin
An EA is said to be a good approximation algorithm if it is an ǫ-approximation
for some constant ǫ. So when the upper bound of r(n) trends to +∞ while the
input size n increases, e.g., r(n) = log(n), the EA is a bad approximation algorithm.
3
Classification of EA-hard Problems
If a problem is easy to an EA, there is no need to investigate its approximation
solutions. So we should restrict our discussion on EA-hard problems. The first
question we should answer is what kind of problems is difficult to a given EA.
The study of this question leads to a classification of problems into the classes
of easy problems and hard problems for the EA.
3.1
EAs and Drift Analysis
Drift analysis is the mathematical tool used in this paper to investigate the
behaviour of EAs, more details can be found in [5–7].
In this paper EAs are considered for solving a Pseudo-Boolean minimisation
optimisation problem: Given an objective function f : S → R, where S is the
space {0, 1}n and R is the space of real numbers, the optimisation problem is to
find an xmin ∈ S such that
f (xmin ) = min{f (y), y ∈ S},
where such an xmin is called a global optimal solution.
Let x = {x1 , · · · , xN } be a population of N individuals, E be the population
space consisting of all populations, and ξt be the t-th generation of the population. Given an initial population ξ0 and let t = 0, EAs can be described by the
following three major steps.
Recombination: Individuals in population ξt are recombined. An offspring
(c)
population ξt is then obtained.
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
4
(c)
Mutation: Individuals in population ξt are mutated. An offspring population
(m)
ξt is then obtained.
Selection: Each individual in the original population ξt and mutated popula(m)
tion ξt is assigned a survival probability. Individuals are then selected to
survive into the next generation ξt+1 according to their survival probabilities.
The process of {ξt ; t = 0, 1, · · · } often can be modelled by a Markov chain,
In this paper we only discuss the EAs that can be modelled by a homogeneous
Markov chain. At generation t, let its transition probability from ξt = x to
ξt+1 = y to be:
P(ξt+1 = y | ξt = x) = P(x, y; t).
We can model EAs by a super-martingale too. This model was first used in
discussing the convergence of non-elitist selection strategies [8]. In the following
we introduce how to model an EA by a super-martingale.
Let Eopt be the subset in E, consisting of all populations which include an
optimal solution. We can define a nonnegative function d(x, Eopt ) to measure
the distance between a population x and the optimal set Eopt . In this paper, we
call d(x, Eopt ) a distance function, and without confusion denote it as d(x) in
short.
There are many ways to define a distance function d(x). A simple way is
d(x) = fmin − f (x).
Assume that, at generation t, population ξt = x, then the one-step mean
drift is defined by
E [d(ξt ) − d(ξt+1 ) | ξt = x]
X
P(x, y; t)d(y).
= d(x) −
(1)
y∈E
The drift can be decomposed into two parts, i.e., positive drift,
E + [d(ξt ) − d(ξt+1 ) | ξt = x]
X
= d(x) −
P (x, y)d(y),
{y∈E:d(y)<d(x)}
and negative drift,
E − [d(ξt ) − d(ξt+1 ) | ξt = x]
X
= d(x) −
P (x, y)d(y).
{y∈E:d(y)>d(x)}
For the process {d(ξt ); t = 0, 1, · · · }, if its one-step mean drift is always no
less than 0, i.e.
E [d(ξt ) − d(ξt+1 ) | ξt ] ≥ 0,
then {d(ξt )} is a super-martingale [9].
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
5
If {d(ξt )} is a super-martingale, the martingale theory [9] can be used to
study the convergence and first hitting times of stochastic sequence {d(ξt ); t =
0, 1, · · · }. Drift analysis is based on this model and some initial results have been
obtained in [5–7].
3.2
Classification of Hard problems
Given an EA, we divide optimisation problems into two convergence classes
based on the mean number of generations needed to solve the problems [10].
Denote the first hitting time of the population ξt to enter the optimal set Eopt
to be τ = min{t : ξt ∈ Eopt }.
Easy Class A: For the given EA, starting from any initial population x ∈ E,
the mean number of generations needed by the EA to solve the problem, i.e.,
E [τ | ξ0 = x], is polynomial in the input size n.
Hard Class A: For the given EA, starting from some initial population x ∈ E,
the mean number of generations needed by the EA to solve the problem, i.e.,
E [τ | ξ0 = x], is exponential in the input size n.
The above classification is based on the worst-case analysis. A more practical
classification is based on the average-case analysis.
Easy Class B: For the given EA, starting from any initial population x in a
subset Ê of E where E \ Ê is exponentially small of the input size n, the
mean number of generations needed by the EA to solve the problem, i.e.,
E [τ | ξ0 = x], is polynomial in the input size n.
Hard Class B: For the given EA, starting from some population x in a subset
Ê of E where Ê is polynomial large of the input size n, the mean number
of generations needed by the EA to solve the problem, i.e., E [τ | ξ0 = x], is
exponential in the input size n.
In the following we focus on describing the characteristics of hard classes.
Before that, we give two lemmas.
Lemma 1. Assume the process {d(ξt ); t = 0, 1, 2, · · · } is a super-martingale and
the upper bound of the one-step mean drift is less than a positive cup , that is,
for any generation t ≥ 0 and d(ξt ) > 0,
E [d(ξt ) − d(ξt+1 ) | ξt ] ≤ cup ,
(2)
then the expectation of the first hitting time τ = min{t; ξt ∈ Eopt } satisfies
E [τ | ξ0 ] ≥
d(ξ0 )
.
cup
(3)
This lemma is a direct corollary of Theorem 4 in [5].
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
6
Lemma 2. Assume {ξt ; t = 0, 1, · · · } is a homogeneous absorbing Markov chain.
Denote m(x) = E [τ | ξ0 = x] to be the mean first hitting time starting from
population x. Let the distance function d(x) = m(x). Then d(x) satisfies
d(x) = P
0,
if x ∈ Eopt ,
(4)
d(x) − y∈E P(x, y)d(y) = 1, if x ∈
/ Eopt .
The above lemma follows directly from Theorem 10, Chapter II §3.1.1 in [11].
Now we can describe the features of hard problems. A sufficient and necessary
condition for a problem to be in the Hard Class A is given by Theorem 1.
Theorem 1. Given an EA and an optimisation problem, assume {ξt ; t = 0, 1, }
is a homogeneous Markov chain. The optimisation problem belongs to the Hard
Class A, if and only if there exists a distance function d(x) such that
1. d(x) satisfies: for some population x ∈ E : d(x) ≥ D(n), where D(n) is
exponential in the input size n,
2. The process {d(ξt ), t = 0, 1, · · · } is a super-martingale, and
3. The one-step mean drift satisfies: for any population ξt with d(ξt ) > 0,
E [d(ξt ) − d(ξt+1 ) | ξt ] ≤ cup ,
where cup > 0 is a positive constant or a polynomial of n.
Proof. (1) Sufficient condition.
From Lemma 1, we know that for every population x ∈
/ Eopt ,
m(x) = E [τ | ξ0 = x] ≥
d(x)
.
cup
Since for some population x : d(x) ≥ D(n), where D(n) is exponential in the
input size n, for the x, its mean first hitting time m(x) will also be exponential
in n.
(2) Necessary condition.
Let’s define the distance function to be d(x) = m(x). Since for some population x, m(x) is exponential in the input size n, the distance function satisfies:
for some population x : d(x) ≥ D(n), where D(n) is exponential in n.
From Lemma 2, we know that, for any x ∈
/ Eopt ,
X
d(x) −
P(x, y)d(y) = 1.
y∈E
Assume that at generation t, ξt = x ∈
/ Eopt , then we have
X
E [d(ξt ) − d(ξt+1 ) | ξt ] = d(x) −
P(x, y)d(y) = 1.
y∈e
Obviously the process {d(ξt )} is a super-martingale. Let cup = 1, then the
one-step mean drift satisfies: E [d(ξt ) − d(ξt+1 ) | ξt ] = cup .
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
7
According to the above theorem, we can give the Hard Class A an intuitive
explanation. Given an EA, an optimisation problem belongs to the Hard Class
A if there is a distance function and
Condition 1 in the theorem shows that under this distance, some populations
are far (exponential in the input size) away from the optimal set. That is,
the problem is a far-distance problem.
Condition 2 shows that the one-step mean drift towards the optimum is always
non-negative. This means the process {ξ; t = 0, 1, · · · } is a super-martingale.
Condition 3 shows that the one-step mean drift towards the optimal set is
limited and always less than a positive constant or a polynomial of n.
The far-distance problems can be divided into two classes further, i.e., the
narrow-gap and wide-gap problems.
A narrow-gap far-distance problem is a far-distance problem such that: for
any point x ∈
/ Eopt , there exists a neighbouring point y with d(y) < d(x)
satisfying that the distance between them is polynomial in n, i.e., d(x)−d(y)
is polynomial in n.
A wide-gap far-distance problem is a far-distance problem such that: there exist
some points x ∈
/ Eopt and for any point y with d(y) < d(x), the distance
between them is exponential in n, i.e., d(x) − d(y) is exponential in n.
The Hard Class A consists of the above two types of problems. The difference
between them is the width of the gap between neighbouring points. In a narrowgap problem, all gaps are narrow (polynomial), but in a wide-gap problem, some
of the gaps are wide (exponential).
From Theorem 1, we also get a classification of EA-hard problems under the
meaning of the average-case analysis.
Theorem 2. Given an EA and an optimisation problem, assume {ξt ; t = 0, 1, }
is a homogeneous Markov chain. The optimisation problem belongs to the Hard
Class B in the meaning of the average-case analysis, if and only if there exists a
distance function d(x) such that
1. d(x) satisfies: for some populations x ∈ Ê : d(x) ≥ D(n), where D(n) is
exponential in the input size n, and Ê is polynomial large of the input size.
2. The process {d(ξt ), t = 0, 1, · · · } is a super-martingale, and
3. The one-step mean drift satisfies: for any population ξt with d(ξt ) > 0,
E [d(ξt ) − d(ξt+1 ) | ξt ] ≤ cup ,
where cup > 0 is a positive or a polynomial of n.
4
4.1
Analysis of EAs Finding Approximation Solutions to
Far-distance Problems
Analysis of Wide-gap Far-distance Problems
For a given EA, we discuss what kind of hard problems are difficult for the EA
to find a good approximation solution in a polynomial time. In this section,
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
8
we restrict our discussion in the worst-case analysis, i.e., Hard Class A. But a
similar discussion can be made to Hard Class B with minor changes.
Proposition 1. Given an EA and a wide-gap far-distance problem, that is,
there exists a distance function d(x) and
1. The maximum of the distance function is exponential, i.e. for some population
x ∈ E, where d(x) is exponential in the input size n;
2. The one-step mean drift towards the optimum is limited, i.e., for any population x with d(x) > 0,
X
0 ≤ d(x) −
d(y)P(x, y) ≤ cup ,
y∈E
where cup is a positive constant or a polynomial of n.
3. There exists a point x0 and for any point y with d(y) < d(x0 ), the distance
d(x) − d(y) between x0 and y is exponential.
If the cost function satisfies following conditions:
4. for any x with d(x) < d(x0 ),
lim
n→∞
f (x)
= +∞.
fmin
(5)
Then the EA is not a ǫ-approximation algorithm to the problem for any constant ǫ.
Proof. Assume the EA starts from any x with d(x) ≥ d(x0 ).
Let t is a polynomial time, then after t-generations, the mean drift is that:
E [d(ξ0 ) − d(ξt ) | ξ0 > 0] ≤ tcup ,
E [d(ξt ) | ξ0 > 0] ≥ d(ξ0 ) − tcup ,
which is still exponential.
This means that after a polynomial generation, d(ξt ) is still an exponential
of the input size (in mean), and ξt doesn’t get across the gap between (y, x0 ),
then d(ξt ) ≥ d(x0 ), so
f (ξt )
lim
= +∞.
n→∞ fmin
EA cannot find a good approximation solution for any constant ǫ.
Now we give an example to explain the above general statement in details.
The problem is defined by:

P
if Pi si ≤ n/8,
 1,
Pn
f (x) = 1 + n i=1 si , if
i si ≥ 2n/8,

infeasible,
else.
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
9
where x = (s1 · · · sn ) is a binary string.
This is a simple but typical wide-gap far-distance problem. It comes from
the ONE-MAX problem but with a wide gap between (n/8, 2n/8). Because the
gap grows with n, it is difficulty to get across it. Therefore one cannot get the
optimum efficiently, and moreover one cannot get close to the optimum efficiently.
A (1 + 1) EA is used to solve the problem. In the (1 + 1) EA, the individual
x = (s1 · · · sn ) is a binary string with length n.
Mutation At generation t, for the individual ξt , flip each of its bits to its
complement with probability 1/n. The mutated population is denoted as
(m)
ξt .
(m)
(m)
Selection If f (ξt ) < f (ξt ), then let ξt+1 = ξt , otherwise, let ξt+1 = ξt .
Now we verify the problem satisfies the conditions in the above proposition.
Define a distance function as follows:
P
0,
if Pi si ≤ n/8,
d(x) =
(n/8)!, if
i si ≥ 2n/8,
It is obviously that the maximum of the distance function is exponential.
That is for some x, the distance function satisfies:
d(x) ≥ (n/8)!.
This means Condition 1 holds.
Next we need to verify Condition 2. Assume that at generation t, ξt = x
satisfies: d(x) ≥ 2n/8.
The drift E [d(ξt ) − d(ξt+1 ) | ξt ] can be decomposed into positive drift
E [(d(ξt ) − d(ξt+1 ))I{d(ξt ) > d(ξt+1 )} | ξt ],
and negative drift
E [(d(ξt ) − d(ξt+1 ))I{d(ξt ) < d(ξt+1 )} | ξt ].
Since this (1 + 1) EA use the elitist selection strategy, so there is no negative
drift.
We only need to consider the positive
drift. Assume x has k one-valued bits
P
and n − k zero-valued bits. Since i si ≥ 2n/8, so k ≥ 2n/8. A positive drift
will happen only if the following event happens: the number of one-valued bits
is reduced to no more than n/8, that mean at least n/8 bits should be flipped,
whose probability is
n/8
n
1
1
P(d(ξt+1 ) = 0 | ξt ) ≤
≤
.
n/8
n
(n/8)!
Then the positive drift satisfies:
E [(d(ξt ) − d(ξt+1 ))I{d(ξt ) > d(ξt+1 )} | ξt ]
= ((n/8)! − 0) P(d(ξt+1 ) = 0 | ξt )
≤ 1.
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
10
Then the drift condition 2 holds too.
Pn
Now let’s see condition 3. Let x0 such that i=1 si = 2n/8, then for any y
with d(y) < d(x0 ), the distance between y and x0 is (n/8)!. So there exists a
wide gap. This means Condition 3 holds.
The cost function satisfies: for any x with d(x) ≥ (n/8)!, it holds:
f (x)
= Ω(n).
fmin
This means Condition 4 holds.
For the given (1+1) EA, since the above function satisfies the four conditions,
so starting from any non-optimal, the EA cannot find a good approximation
solution.
4.2
Analysis of EAs finding approximation solutions to narrow-gap
long distance problems
Like the study of wide-gap problems, we also can give some narrow-gap fardistance problems, which EAs can’t find a good approximation solution too.
These hard problems can be described as follows:
Proposition 2. Given an EA and a narrow-gap far-distance optimisation problem, that is: there exists a distance function d(x) such that
1. The maximum of the distance is exponential, i.e. for some population x ∈ E :
d(x) is exponential in the input size n,
2. The one-step mean drift towards the optimum is limited, i.e., for any population ξt with d(ξt ) > 0,
0 ≤ E [d(ξt ) − d(ξt+1 ) | ξt ] ≤ cup ,
where cup > 0 is a constant or a polynomial of n.
3. The distance between any two neighbouring points is narrow, i.e., given any
x, there is another y with d(y) < d(x), d(x) − d(y) is polynomial of n.
If the cost function f (x) satisfies:
4. for any x with d(x) being an exponential of the input size n,
f (x)
= +∞,
n→+∞ fmin
lim
then the EA is not a ǫ-approximation algorithm to this problem.
Proof. Let’s the EA start at a population ξ0 where d(ξ0 ) is an exponential of n,
and t is a polynomial of n, then after t generations, the mean drift is that
E [d(x0 ) − d(ξt ) | ξ0 > 0] ≤ tcup ,
E [d(x0 ) − d(ξt ) | ξ0 > 0] ≥ d(x) − tcup ,
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
11
So d(ξt ) (in mean) is still exponential in n, the result at generation t satisfies:
lim
n→∞
f (ξt )
= +∞.
fmin
which means the EA is not an ǫ-algorithm.
In the following we use an example to explain the above general conditions.
We still consider the (1 + 1) EA given in the above subsection.
√
A typical example of narrow-gap far-distance problem is the long n − 1path problem given in [12]. The problem has the following feature: for any point,
there is a neighbouring point whose cost function is better than it. And the
neighbouring point is not far away, at most in a polynomial distance in n. So
the problem is a narrow-gap problem, which an EA could find a better solution
easily.
We start with defining the long k-path path which taken from [12].
Let n ≥ 1 hold. For all k > 1 where (n − 1)/k being an integer, the long
k-path of dimension n is a sequence of bit strings from {0, 1}. The long k-path of
dimension 1 is defined as P1k := (0, 1). The long k-path of dimension n is defined
using the long k-path of dimension n − k as basis as follows. Let the long k-path
k
of dimension n−k be given by Pn−k
= (v1 , · · · , vl ). Then we define the sequences
of bit strings S0 , Bn , and S1 from {0, 1}n, where S0 := (0k v1 , 0k v2 , · · · , 0k vl ),
S1 := (1k vl , 1k vl−1 , · · · , 1k vl ), and Bn := (0k−1 1vl , 0k−2 11vl , · · · , 01k−1 vl ). The
points in Bn build a bridge between the points in S0 and S1 , that differ in the k
leading bits. Therefore, the points in Bn are called bridge point. The resulting
long k-path Pnk is constructed by appending S0 , Bn , abd S1 , so Pnk is a sequence
of | Pnk |=| S0 | + | S0 | + | Bn | + | S1 | points. We call | Pnk | the length of Pnk .
The i-th point on the path Pnk is denoted as pi ; pi+j is called the j-th successor
of pi .
The length of the long path Pnk is that [12]
Lemma 3. The long k-path of dimension n has length | Pnk |= (k + 1)2(n−1)k −
(k + 1).
A important property of long k-paths is the simple rule that holds for the
Hamming distance between each point and its successors [12]:
Lemma 4. Let n and k be given such that the long k-path Pnk is well defined.
For all i with 0 < i < k, the following holds. If x ∈ Pnk has at least i different
successors on the path, the i-th successor of x has Hamming distance i of x and
all other points on the path that are successors of x have Hamming distance
different from i.
We define the cost function f (x) as follows:
f (x) =
i;
if x is the pi point on Pnk ,
k
1+ | Pn |; if x is not a point on Pnk ,
DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT DRAFT
12
In the following we will verify the problem satisfies the above four conditions
of Proposition 2. We define a distance function as follows:
d(x) = f (x) − 1,
It is obvious that the maximum of the distance function is exponential, i.e.
for some point x is exponential in the input size n. Then Condition 1 holds.
Now thing we should verify is the drift condition 2.
According to the definition of objective function for the k-path and the elitist
selection strategy, we know there is no negative drift. So we only need to estimate
the positive drift.
Assume at step t, ξt = pi is still not an optimum
point, then from Lemma 4,
√
we know that if it is able to move forward j < n − 1 points, the probability is
j
1
.
P(ξt+1 = i − j | ξt ) ≤ O(1)
n
√
and if it is able to move forward j ≥ n − 1 points, the probability is
P(ξt+1
√n−1
1
= i − j | ξt ) ≤ O(1)
.
n
Then the drift satisfies that
E [d(ξt ) − d(ξt+1 ) | ξt ]
X
=
(di − di−j )P(d(ξt ) − d(ξt+1 ))
j
≤ O(1)
X
j
√
j< n−1
+O(1)
X
√
j≥ n−1
j
1
n
√n−1
1
j
n
From Lemma 3, we know
E [d(ξt ) − d(ξt+1 ) | ξt ]
√
≤ O(1) + O(1) | Pn
≤ O(1) + O(1)O(n4
= O(1).
n−1
√
√n−1
1
|2
n
√
n−1 − n−1
n
)
Condition 3 holds because the distance between any two neighbouring points
is narrow, i.e., given any point x, there is another y with d(y) < d(x), d(x)− d(y)
is polynomial.
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13
And the fitness function satisfies Condition 4: for any x with d(x) being an
exponential, it holds:
f (x)
lim
= +∞.
n→+∞ fmin
Now we have shown the problem satisfies the four conditions. Then the EA
can not find a good approximation solution to it.
5
Conclusion
In this paper we have linked evolutionary algorithms with approximation algorithms together and discussed the question of how good approximation solutions
EAs can find in a polynomial time. It is shown that for some EA-hard problem,
e.g., two families of far-distance problems, EAs cannot find a good approximation
solution to them.
This is just a starting study based on Pseudo-Boolean optimisation problems.
Although many experimental results claim that EAs can find some high quality
solution for complex solutions, we still know little about them in theory. In the
future, we will check these statements by approximation algorithms and study
how well EAs find a good approximation solution.
Acknowledgments: The work is partially supported by EPSRC (GR/R52541/01)
and State Laboratory of Software Engineering at Wuhan University.
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