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A Quantitative Approach for Validating the Building-block Hypothesis

Chatchawit AporntewanDepartment of Mathematics

Faculty of Science, Chulalongkorn UniversityBangkok, 10330, ThailandChatchawit.A@chula.ac.th

Abstract- The building blocks are common structures ofhigh-quality solutions. Genetic algorithms often assumethe building-block hypothesis. It is hypothesized that thehigh-quality solutions are composed of building blocksand the solution quality can be improved by composingbuilding blocks. The studies of building blocks are lim-ited to some artificial optimization functions in which itis obvious that the building blocks exist. A large num-ber of successful applications has been reported with-out a strong evidence that proves the hypothesis. Thispaper proposes a quantitative approach for validatingthe building-block hypothesis. We define the quantity ofbuilding blocks and the degree of discontinuity by us-ing the chi-square matrix. We test the building-blockhypothesis with 15-bit onemax, 5 x 3-trap, parabola1 - (X2/1010), and two-dimensional Euclidian travelingsalesman problem (TSP). The building-block hypothe-sis holds for onemax, 5 x 3-trap, and parabola. In thecase of parabola, Gray coding gives a higher quantity ofbuilding blocks than that of binary coding. The hypoth-esis is accepted for random instances of TSP with a lowconfidence.

1 Introduction

The building blocks are common structures of high-qualitysolutions [6, 3]. Genetic algorithms often assume thebuilding-block hypothesis. It is hypothesized that the high-quality solutions are composed of building blocks and thesolution quality can be improved by composing buildingblocks. The studies of building blocks are limited to someartificial optimization functions in which it is obvious thatthe building blocks exist [7, 14]. A large number of suc-cessful applications has been reported without a strong ev-idence that proves the hypothesis. The genetic algorithmsenhanced with building-block identification give promis-ing solutions in real-world applications [11, 4]. However,it is not sufficient to prove the building-block hypothesis.Reaching optimal solution results from a number ofparame-ters that interact with each other, for example, genetic oper-ator, selection method, diversity control, and local search. Afair building-block measurement must decouple itself fromthe unnecessary parameters that involve finding optimal so-lutions.

Modern genetic algorithms are capable of identifyingand maintaining building blocks [5, 12, 4, 15]. Recent workturns to build a distribution of solutions [13]. The basic con-cept of the estimated distribution algonrithms (EDAs) startswith a uniform distribution of solutions. Next, a number of

Prabhas ChongstitvatanaDepartment of Computer Engineering

Faculty of Engineering, Chulalongkorn UniversityBangkok, 10330, ThailandPrabhas.C@chula.ac.th

solutions is drawn according to the distribution. Some goodsolutions (winners) are selected, and the distribution is ad-justed toward the winners (the winner-like solutions will bedrawn with a higher probability in the next iteration). Thesesteps are repeated until the optimal solution is found orreaching a termination condition. Building a distribution ofsolutions involves identifying dependency of solution bits.It can be done by a statistical method, for example, Bayesiannetwork [12]. EDAs focus on the accuracy of building dis-tribution (or identifying the dependency) which is essentialto improve solution quality. As mentioned earlier, execut-ing EDAs and obtaining the optimum are not sufficient toprove the building-block hypothesis. In contrast, testing thebuilding-block hypothesis requires measuring the quantityof dependency of solution bits. It will be shown that testingthe hypothesis is simple and fast. If the building-block hy-pothesis holds, one of the EDAs would be applied. Buildingan accurate distribution needs a great deal of computationaltime. It might be worth testing the building-block hypothe-sis before executing EDAs.

This paper proposes a quantitative approach for validat-ing the building-block hypothesis. We define the quantity ofbuilding blocks and the degree of discontinuity by using thechi-square matrix [1]. We test the building-block hypothe-sis with 15-bit onemax, 5x3-trap, parabola 1 - (x2/1010),and two-dimensional Euclidian traveling salesman problem(TSP). Onemax and trap functions are chosen for evaluationpurpose because the building-block information is knownbeforehand. The quantity of building blocks depends onhow the solutions are encoded. In some cases, the perfor-mance of genetic algorithms is enhanced by the use of Graycoding [10]. The parabola is tested with binary coding andGray coding. TSP represents real-world applications. Wechoose TSP because it is close to practical applications suchas vehicle routing, PCB design, X-ray crytallography, etc[8, 2]. The hypothesis is tested for three different codingsthat are commonly used in the TSP literature.

This paper is organized as follows. Section 2 defines thequantity of building blocks and the chi-square matrix. Sec-tion 3 describes a methodology for validating the building-block hypothesis. Section 4 tests the building-block hypoth-esis with a number of functions. Section 5 concludes thepaper.

2 The Quantity of Building Blocks

Building blocks are inferred from a set of solutions. Anexample of four 15-bit solutions is shown below.

0-7803-9363-5/05/$20.00 ©2005 IEEE.

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V1V2V3 V4V5V6 V7V8V9 V1OV11V12 V13V14V15#1 000 1 1 1 000 1 11 1 11#2 1 1 1 000 1 1 1 000 000#3 1 1 1 000 1 1 1 000 1 1 1#4 1 1 1 000 000 1 1 1 000An inference might be that the above solutions are com-posed of aligned chunks of 000 and 111. The strongdependency between variables vi and vj+ and Vi+2 wherei = 1, 4, 7, 10, 13 can be detected by a statistical method.We do not give a precise definition of building blocks, butmeasuring the quantity ofbuilding blocks is possible.

Given a set of solutions, the quantity of building blocksranges between min and max. min means no buildingblocks. This implies that the solutions are random becauserandom bits do not have common structures. max indicatesthat the solutions are absolutely not random. For instance,all solution bits are zero. The quantity of building blocksinversely relates to randomness. We choose the chi-squarematrix for measuring randomness because computing thematrix is simple and fast [1].

Let M = (mij) be an exe symmetric matrix of num-bers. Let P be a population or a set of i-bit binary strings.The chi-square matrix is defined as follows.

rij = { ChiSquare(i, j) ; otherwsj; otherwise.

The ChiSquare(i,j) is defined as:

z (CourttY(i,j)n/4n

, xy E {00, 01,10, 11} (2)

xy

where the observed frequency Countxp (i, j) counts thenumber of solutions in which bit i is identical x and bit j isidentical to y. The expected frequencies of observing "00,""01," "10," "1 1" are n/4 where n is the number of solu-tions. If the solutions are random, the observed frequenciesare close to the expected frequencies and therefore the chi-square is low. The quantity of building blocks is defined asthe sum of all matrix elements. Since the matrix is sym-metric, we sum only a half of the matrix. The quantity ofbuilding blocks given a population, P, is defined as follows.

Qp = Emij, i < j (3)ii

The time complexity of computing the quantity of buildingblocks is o(e2n).

It is important to note that the chi-square matrix dealsonly with second-order interactions or marginal depen-dency. There exist also higher-order interactions that playa crucial role in optimization. However, our goal is not tobuild an accurate distribution of solutions. A sum of everypairwise couplings is sufficient for measuring the quantityof building blocks, because the sum counts the quantity ofhigher-order interactions. For examples:

V1 V2 V3 V4 Vl V2 V3 V4

#1 0 0 0 0 #1 0 0 0 0#2 0 0 1 1 #2 0 0 0 0#3 1 1 0 0 #3 1 1 1 1#4 1 1 1 1 #4 1 1 1 1

The set of four 4-bit solutions on the left represents thesecond-order interactions of v1 and V2, V3 and v4. Thereis no dependency between v1 and V3, v1 and V4, V2 andV3, v2 and V4 because the observed frequencies are iden-tical to the expected frequencies. The quantity of build-ing blocks is ChiSquare(1, 2) + ChiSquare(3, 4) = 4+ 4 = 8. The solutions on the right represent the higher-order interactions of v1, V2, V3, V4. The quantity ofbuilding blocks is ChiSquare(1, 2) + ChiSquare(1, 3) +ChiSquare(1, 4) + ChiSquare(2, 3) + ChiSquare(2, 4)+ChiSquare(3,4) =4+4+4+4+4+4=24. Theso-lutions on the right give a higher quantity of building blocksbecause of the high-order interactions.

3 A Methodology for Validating the Building-block Hypothesis

In the previous section, we have defined the quantity ofbuilding blocks that is a fundamental tool for validatingthe building-block hypothesis. The building-block hypoth-esis are separated in two hypotheses. The first hypothesisstates that the high-quality solutions are composed of build-ing blocks. The second hypothesis states that the solutionquality can be improved by composing building blocks. Webegin with a methodology for validating the first hypothe-sis. Then, a different methodology for validating the secondhypothesis will be presented.

Let F : {0, 1} -+ R be a fitness function. LetPi,.. ,Pm be m sets of n high-quality solutions. LetT1, . , Tm be m fitness thresholds such that T1 < ... <Tm. Each set Pi is made by randomly choosing n solutionsof which their fitness is greater than or equal to Ti. The setsof solutions Pi to Pm simulate a sequence of evolving pop-ulations generated by an optimization algorithm. We do notexecute an actual optimization algorithm because we wantto decouple our methodology from unnecessary parameterssuch as genetic operator, selection method, diversity con-trol, and local search.

The quantity of building blocks is computed for all PNto Pm. The result is shown in Figure 1. min and max arethe minimum and the maximum quantity of building blocks.min is the quantity of building blocks in a random popula-tion. max is the quantity of building blocks in a populationin which all bits are zero. We normalize min and max to0 and 1, respectively. The first hypothesis holds if and onlyif the quantity of building blocks does not fall within therejection area (being closer to min means more random-ness). Drawing the plot in Figure 1 requires some parame-ters. Here is a guideline for parameter settings.

Population size. If the population size is too small, thegap between min and max is narrowed. Subsequently, theresolution is not enough for distinguishing between a ran-dom population and a nonrandom population. The min andmax should be calculated to make sure that the chosen pop-ulation size gives a large resolution.

Fitness thresholds. T1 should not be too small becauseit makes P1 random (falling within the rejection area). Thefirst hypothesis involves high-quality solutions. T1 should

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be set at a fitness value that indicates "high quality". A ruleof thumb is to set Pi a little bit greater than the averagefitness of a random population. Tm can be at most the opti-mum. Increasing the number of fitness thresholds, m, likeszooming in the plot. Setting m to be too small makes loss ofvisual information because of zooming out. It is difficult tomake a good-looking plot at the first time without any priorknowledge about the fitness function. However, the sweetspot for setting m is large for trial-and-error method.

Rejection area. The first hypothesis is rejected if thequantity of building blocks is close to min (falling withinthe rejection area). It is difficult to draw a line that di-vides building blocks and randomness because the quan-tity of building blocks is defined over the whole population.However, we can test the randomness separately for everypairwise coupling. There are ( - 1)/2 pairwise couplings(the number of elements in the upper triangle of the chi-square matrix) where e is the number of solution bits. Thechi-square test is a test procedure for studying random data[9]. It is summarized as follows. We observe a solutionat bit i and bit j. The number of observations is set at afairly large number such that an expected frequency is fiveor more. An observation falls into four categories (threedegrees of freedom) that are "00," "01," "10," "1 1." Wecompute the chi-square value (an element of the chi-squarematrix). The selected percentage points of the chi-squaredistribution is shown below.

Chi-square value Percentage point11.345 99%7.815 95%6.251 90%

If the chi-square value is greater than 99% entry, the ob-served data is not sufficiently random. If the chi-squarevalue lies between the 95% and 99% entries, the observeddata is suspect. If the chi-square value lies between the 90%and 95% entries, the observed data might be almost suspect.Otherwise, the observed data passes the randomness test.We also reject the building-block hypothesis if more than ahalf of all pairwise couplings pass the randomness test.

The number of conducting tests. As testing the hy-pothesis is probabilistic, the hypothesis is accepted by a ma-jority vote or an average quantity of building blocks. Thenumber of conducting tests is up to your desired confidencein accepting the hypothesis.

Generating solutions, Pi, of which their fitness is greateror equal to a constant, Ti, is not obvious. Fortunately,we can choose a small problem size so that we can enu-merate all possible solutions (for example, choosing 8-cityTSP rather than 100-city TSP). Intuitively, if the hypothesisholds for 8-city TSP, the hypothesis would hold for 100-cityTSP. Enumerating all possible solutions may not be requiredif there is an effective method for generating a set of solu-tions that is subject to a fitness constraint.

The second hypothesis states that the solution quality canbe improved by composing building blocks. In other words,the common structures of Pk are similar to that of Pk+, be-cause the population Pk+, is produced by composing build-ing blocks from the previous population Pk. The secondhypothesis would be accepted if there exist a similarity or

Quantity of Building Blocks

max (1)1

min (0)

4C

Rejection Area.............Pannulatinn

1 2 3 m

Figure 1: Quantity of building blocks.

continuity between Pk and Pk+,. To validate the secondhypothesis, we define the degree of discontinuity betweentwo populations. The degree of discontinuity between Pkand Pk+1 is defined as:

Dpk,Pk±l =Zk(m -mtl)2, i < jii

(4)

where (mk) is the chi-square matrix of Pk and (mkjl) isthe chi-square matrix of Pk+,. The degree of discontinu-ity is plotted in Figure 2. min and max are the minimumand the maximum degree of discontinuity. min is alwayszero (Pk and Pk+1 are identical). max is the degree of dis-continuity between a random population and a populationin which all bits are zero. We normalize max to 1. Thesecond hypothesis holds if and only if the degree of discon-tinuity does not fall within the rejection area (being closeto max means a drastic change of common structures be-tween adjacent populations). The other parameter settingsare similar to that of the first hypothesis.

Degree of Discontinuity

ax (1) |.. ..........

min (0)

Rejection Area

P,P P,P PP.. P PPopulation

1 2 2 3 3 4 m-1 mPair

Figure 2: Degree of discontinuity.

4 Testing the Building-block Hypothesis

This section tests the building-block hypothesis with 15-bitonemax, 5 x 3-trap, parabola, and two-dimensional Euclid-ian TSP. The 15-bit onemax Fonemax : {, i}15{0, . . . 15} is defined as:

15

Fonemax = Ebii=l

(5)

where bi is the ith bit of the solution. The 5x3-bit trapF5X3 {0, 1}15 - {0 ... , 15 is defined as:

5

F5x3(Bl B2B3B4B5) =E F3(B%) (6)i=l

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'r-FIJUULIV

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where Bi E {0, 1}3. F3 denotes 3-bit trap F3: {0, }13 _+

{0, 1, 2, 3} that is defined as:

F3(bib2b3)={ -u ; ifu=3; otherwise, (7)

where u =E bi and bi E {0, 1}. The quantity of buildingblocks and the degree of discontinuity are shown in Figure3 (see also Table 2). The building-block hypothesis holdsfor 15-bit onemax and 5x3-trap because of high quantityof building blocks and low degree of discontinuity. In ad-dition, the number of pairwise couplings that pass the ran-domness test drops to zero. It is clear that the populationsare composed of building blocks and not random. It mayseem counter-intuitive that onemax (first-order interaction)gives a higher quantity of building blocks than that of 5 x 3-trap (third-order interaction). The high-quality solutions ofonemax are composed of "1" more than "0." The 5 x 3-trapsolutions are aligned chunks of "000" and "111." The one-max and 5 x 3-trap solutions of which the fitness is 12 areshown below.

15-bit onemax 5 x 3-trap#1 000111111111111 000000000111111#2 001011111111111 000000111000111#3 00110111l111111 000000111111000#4 001110111111111 000001111111111#5 001111011111111 000010111111111

By the chi-square definition, the onemax solutions are moredependent (less random). The quantity of building blocks is1143 and 720 for onemax and 5 x 3-trap, respectively. Theresult is counter-intuitive.

The parabola Fparabola {0, 1}32 -* E is defined as:

x2Fparabola(X) = 1 ° (8)

There are two alternatives for encoding solutions, the binarycoding and Gray coding. The 3-bit coding is shown in Ta-ble 1 (actual codings are 32 bits). The quantity of buildingblocks and the degree of discontinuity are shown in Figure4 (see also Table 3). It is obvious that Gray coding givesa higher quantity of building blocks. The high discontinu-ity of Gray coding is typical for the quantity of buildingblocks that increases sharply. The building-block hypothe-sis is rejected for binary coding because of the low quantityof building blocks and more than a half of pairwise cou-plings passing the randomness test. On the other hand, thehypothesis is accepted with a high confidence for Gray cod-ing.We randomize an instance of TSP (Figure 7). The num-

ber of cities is fixed at eight. Each city is randomly placedon a lOx 10 grid. The cost of traveling from city i to j isidentical to the distance between i and j. The fitness of asolution is 100 subtracted by tour length. There are threecodings that are commonly used. The first coding, each cityis tagged with a 3-digit binary number ("000" to "111").The first three bit of a binary solution is the starting city.The next city the salesman visits is the next three bits. Thesecond coding represents a tour by a matrix. If the salesmanvisits city i before j, the matrix element at row i and columnj is one. Otherwise, the matrix element is zero. The third

Table 1: Binary coding and Gray coding.Solution Binary coding Gray coding

-4 100 110-3 101 111-2 110 101-1 111 100+0 000 000+1 001 001+2 010 011+3 111 010

coding is very similar to the second. If the salesman travelsfrom city i to j, the matrix element at row i and column j isone. Otherwise, the matrix element is zero. The first codingresults in 24-bit solutions, but the second and the third cod-ing results in 56-bit solutions. We do not count the binarystrings that are invalid tours. The expected frequencies ofobserving a pairwise coupling being "00 ' "01," "'10," "1 ,'are not identical for TSP codings. Equation 2 assumes theidentical expected frequencies. In the case of TSP codings,the ChiSquare(i, j) is defined as follows.

(Ooo - Eoo)2 (Oo° - Eol)2 (O°o - Elo)2 (O11- El,)2Eoo Eol Elo El,

(9)The 0,Y and E.y are the observed frequencies and the ex-pected frequencies, respectively. The expected frequencies,E,y are computed by enumerating all possible solutions.The quantity of building blocks and the degree of discon-tinuity are shown separately in Figure 5 and 6 due to thedifferent lengths in codings (see also Table 4 and 5). Thebuilding-block hypothesis is rejected for the first and thesecond codings because the quantity of building blocks isobviously low. Plus more than a half of pairwise couplingspass the randomness test. The solutions encoded by the firstand the second codings are likely random bits rather thancomposing of some common structures. The third codinggives a higher quantity of building blocks that distinguishesthe third coding from the others. However, the quantityof building blocks is a little bit lower than that of 5x3-trap function. The 5x 3-trap is invented to have a sufficientamount of building blocks. In terms of quantity of build-ing blocks, the hypothesis would be accepted but more than40% of pairwise couplings pass the randomness test. Forthe third coding, we accept the hypothesis with a low confi-dence. The main point might not be accepting or rejectingthe hypothesis. Figure 5 and 6 suggest that the third codingis superior to the others. The third coding would be the bestchoice if we are going to solve TSP with an optimizer thatexploits common structures.

In practical optimization problems, there are manychoices for fitness function and coding. Designing fitnessfunction and coding is problematic because it is evaluatedby the end result (the quality of a best-so-far solution). Thesolution quality is affected by many parameters that inter-act with each other. Therefore it is difficult to find out aproper fitness function and a proper coding. Our method fortesting the building-block hypothesis is separated from the

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Quantity of Building Blocks1.01

0.5

nemax o+--5x3-trap +

Degree of DiscontinuityI A.e

0.5

P2 3 p pr Iuialuu

%emax -?x?-trap +

+ , +,,,,0.0 _ r-pui-un rair

Figure 3: Quantity of building blocks (left) and degree of discontinuity (right). The fitness functions are 15-bit onemax and5 x 3-trap. The population size is set at 100. Each point is averaged from 10 independent runs. The fitness thresholds are 8,9, 10, 11, 12 (see also Table 2).

Quantity of Building BlocksIt

Degree of Discontinuity4 n

Blnara codinanoGrag coaing-+

0.5

P2 P3 P4 pF PIIuaIEP 2 P2 1 P3 R P43'Population Pair

PI PS

Figure 4: Quantity of building blocks (left) and degree of discontinuity (right). The fitness function is parabola (1-/x21/01O)with binary and Gray codings. The population size is set at 500. Each point is averaged from 10 independent runs. Thefitness thresholds are 10-8, 10-6, 10-4, 10-2, 0 (see also Table 3).

Quantity of Building Blocks11 n%

Degree of Discontinuityn%

Coding 1 o

P1 P2 P3 P4

0.5

w6 propuiation

Coding 1 -

nr' U _ - ° -opuiaioln 5airIP 2 P4,P4 P34P4 P4P5 P56 P6,P7

Figure 5: Quantity of building blocks (left) and degree of discontinuity (right). The fitness function is 100 subtracted by a

tour length (coding 1). The population size is set at 500. Each point is averaged from 10 independent runs (an instance ofTSP is randomized every run). The fitness thresholds are 45, 48, 51, 54, 57, 60, 63 (see also Table 4).

Quantity of Building Blocks

0.5

nUn4 -

Degree of Discontinuity

0.5

Ponulation.. ............

n ° * Population Pair-P1,P2 P2 ,P3 P3 ,P4 P4 ,P5 Ps ,P6 P6 P7

Figure 6: Quantity of building blocks (left) and degree of discontinuity (right). The fitness function is 100 subtracted bya tour length (codings 2 and 3). The population size is set at 500. Each point is averaged from 10 independent runs (aninstance of TSP is randomized every run). The fitness thresholds are 45, 48, 51, 54, 57, 60, 63 (see also Table 5).

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0.5

Biaarv coina oGryung +-

0.5

1 U

Prniiiatinsn Psuir__TA r ----.. I,-"..I- Oannslatinnu.v _

I 0 1.0U

nn, nsilatinnUVP- n n

I sU I.U I_

O * - f_, ,l_*;_an_A

i n 1 .u

.............. ...........

.,..p1II-

,p P3 4 P4 'pi

U.U_

0.0 - ----------- ---

pi P2 P3 P4

2r- - T-r - --1 -

r' T;_;5- -1 - -L I J _L J _ _

L - T - -I r'L T

r r T

_+ _ _-I- -s_- +_ _-I _

L _1_J _ 1__LV J. ___

iX

r T r T INI0

_ + _ I, _ + _ v __ _

L - 1 - J _ _I_ - L _ 1 - J _I.I_-

4_-T - -1

_ + _

I

L7- I--- T- -

_ +I_

_ I1 _

- T--1

_ +- --

;3 1 _

Coding 1:

000 101 010 001 110 100 111 011

Coding 2:0 1 2 3 4 5 6 7

0x 1 111111loxx o 1 1 o2 0 1 X 1 1 0 1 13 0 0 0x 0 0 0 o4 0 0o 1 X 0 015 o 1 1 1 1 Xx6 0 0o 1 1 0 X 17 0 0o 1 0 0 0 X

Coding 3:0 1 2 3 4 5 6 7

0 x 0 0 0 0 1 0 01 0 x 0 0 0 0 1 02 0 1 X 0 0 0 0 03 1 0 0x 0 0 0 o4 0 0 0o X 0 0 15 0 o 1 0 0 x 0 0

6 0 0 0o 1 0 X 07 0 0o 1 0 0 0 X

Figure 7: Three codings for TSP. A tour (left) is encoded to a binary string and matrices (right). The tour length is 28.58and the fitness is 100 - 28.58 = 71.42.

Table 2: Averaged number of pairwise couplings that pass the randomness test (max = 105). The fitness functions are15-bit onemax and 5 x 3-trap. This table corresponds to the data in Figure 3.

Table 3: Averagedparabola (1 x2/I

Fitness Averaged number of pairwise couplings that pass the randomness testfunction P1 P2 P3 P4 P5Onemax 26% 3% 0% 0% 0%5 x 3-trap 72% 45% 21% 3% 0%

I number of pairwise couplings that pass the randomness test (max = 496). The fi10) with binary and Gray codings. This table corresponds to the data in Figure 4.

Coding Averaged number of pairwise couplings that pass the randomness testPi P2 P3 P4 P5

Binary 88% 86% 76% 77% 67%Gray 80% 58% 41% 36% 22%

Table 4: Averaged number of pairwise couplings that pass the randomness test (max = 276). The fitness functions is 100subtracted by tour length (coding 1). This table corresponds to the data in Figure 5.

Coding Averaged number of pairwise couplings that pass the randomness testP1 P2 P3 P4 P5 P6 P7

1 91% 90% 84% 80% 70% 67% 59%

Table 5: Averaged number of pairwise couplings that pass the randomness test (max = 1540). The fitness functions is 100subtracted by tour length (codings 2 and 3). This table corresponds to the data in Figure 6.

Coding Averaged number of pairwise couplings that pass the randomness testP1 P2 P3 P4 P5 P6 P7

2 93% 92% 92% 92% 89% 86% 83%3 88% 86% 81% 73% 65% 53% 40%

optimization. A large number of fitness functions and cod-

ings can be tested in a short time by setting a small problemsize.

5 Conclusion

Two important measurements are defined, the quantity ofbuilding blocks and the degree of discontinuity. We showthat the building-block hypothesis can be tested for a smallproblem size. The building-block hypothesis holds for 15-bit onemax, 5 x 3-trap, and parabola (Gray coding). For TSP

with the third coding, the hypothesis is accepted with a lowconfidence. Future work will be to explore the quantityof building blocks in a wide range of real-world applica-tions. The final outcome is a number of problem domainsfor which the building-block identification and compositionare applicable.

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