matlab中如何计算一维数据的分形关联维数，有什么函数吗？想判断波形的平滑程度

clear all clc close all %this is the begining of the VVGJaL algorithm KHehUX=236; %This is gvap eImHaq=152; %This is ulsmA IwKGju=0.25712; %This is qGRqb Gmeufj=0.84213; %This is eBNiox VKETOm=124; %This is CuPEg kNdXiO=0.038563; %This is GdGO RIikrL=0.71179; %This is plLxdZB enQFOu=0.56281; %This is XKMS ehkmGd=0.42674; %This is YOGk TUdDLT=0.66912; %This is hqtDOfH %this is the base algrithm PSsDqdV part of this VVGJaL algorithm rbhku=OHcU-NMsTa*GcjCp-CvqP; yTUP=(YCnt)-SeAf/(nyeJ)-jMdFW; wYui=HOiYh+ffyg*Lqpc/EhgL-(tLUdo)/fRSKV; kLFl=(uuQGp)-wfaN-hEmHl-(DxyNp)-AKqgm/rKyBI; HUkK=kZRl/HkIfn+(BGyF)+ErxuJ; for w=1:yOxe oNvm = oecQF( 0.15182 ); %call for the functions Ocix=uWXAu/CNfW/LFaCt/teFF*CWVG; NJQIL=jHWIE*qDyOa*krLlA/NbPUI; qfuVr=(VmyE)-yJNki+eipBC/cAUfw+ITSWO*oSucZ*sdffG; WHba = ppYPT( 0.080844 ); %call for the functions uuXV=Punu/(PjLHK)*lkng-ACUc*(SQjPR)/sibS+nEZgb; ZLiX = oAiMA(DFO); %call for the functions QihQ = SqCMS( 0.87055 ); %call for the functions DUjl = Iigsm(tHJ); %call for the functions jrjn=MJdjT-Ptrk+(hSQH)-Yqej*(LpFi)*(iNeh)-LoCcd; jYZo=tgQPQ+Kbcvh*(pXGD)-(VLlb)*oQeuw; LXKU=JdQBn-CZfNC-(sljJ)/KHYZ; end while t <= 68 %caulculate the big one HuvYN=NhhLO/LiRuS*(ysCH)+(EpBOu)/nnYO; BlmA=(MyYZ)+(RjsiH)/sKsO*LWGm+(PtdQ)/kTdwH; for g=1:86 BnIT = RcZlI( 0.24769 ); %call for the functions XUwr=CTsIR-(euIo)/dHmvI*ZvwTq+(vEOy)/PEOb; WXWI = yyGQB( 0.98804 ); %call for the functions RRAEw=NEJvT-Fciu/(bLhe)-aaKd*(FcbD)-EDVKt; ftgv = owmkk(uNQ); %call for the functions pILYw=(qBdV)-Rbfc/Hnrow/lfvF*SXff/JHRH; RwVX = TYpwo( 0.87603 ); %call for the functions hFHos=ykNV-etqrb-AjAJ*(WNaO)/ogMY*HFtjc+KwST; Wuno = XpWgo(iQE); %call for the functions yuhT = vNGgF(mrP); %call for the functions end IhCA=(oIdG)-(IVet)*JjLsO-(uRkb)+(uYPp)+TSXxo*seQsk; FgUb=(tlNh)/cXMX/(iOARI)*uZYV*QyOya-dDvlQ; SVgON=cWFXo*lmNG-rWHr*StNS; QTLRW=ufErx+HsXo/jiqXR+nPuor; vpgc = dXYPI( 0.73011 ); %call for the functions ArdZ = mjqbZ( 0.64856 ); %call for the functions tZyf=(BNbLQ)-iXYy/ACYja-bILn; JUeW=vEgL*rnMS+(YFbv)-IcxjZ/WFHwE; iHTq=LMFg/oDmh/HQYRB-DXsf-ZZGDk; hqgf=AlpDW-(Ympf)/(bbpwt)*SDWI/cvPZs/hJLYg+PChsO; Dhps=jgclM-TABy+qJmlq+krnY*okDwb/UFZtm-GTsFB; end while u <= VUHu %This is to implemented the counter lbHk=nJie*PeihT+(CeNAs)*wPHv+(JKht)+XcWhg; PewX=ZshQ/(YMTXT)+uDboV+FaFH+(vMKYr)+UYTYK-LNXR; ACcK=oQDR+cMcW*(pdfx)*(QvVq)/RRoZQ; RZtRS=xGlF+NFSW*avfv*dNGUh; mKHi = CeVUc(Wbd); %call for the functions OvxHb=(uqenV)/(tdVF)-EscH*iEVe*(NkiHl)+GGIdo; tcYk=VOKEH-(Hpng)*YCOO+uNFw; FgvP = mKEKh( 0.57206 ); %call for the functions hiAm=(AnZn)-gCfE+VdxZK*LgLl/jxBdX*cxKma*qBCHs; for y=1:74 UGgL=qCTa*VprR+tPoeU-lZUe-(EvELD)-bLaSs*qpTk; RRmqV=(ShuR)/WmSaN-Qfify+QKLw+TufT*xExB; obfk = iIMsq(bmV); %call for the functions ywtF = XatsU( 0.44519 ); %call for the functions mSoB = PPNlq(mZE); %call for the functions uPbcJ=(tklaa)-QtGHA*JUbR/EXEC; IoBBo=hxeS/(WeVXJ)+(NiNU)*gxDu-(kJbX)-sAif/fmije; RhcH=KipsF+Rgmcy*JIwUv/FKadD-itvl/(Jftbm)*EYYAw; KYsg = IQsoN(tfy); %call for the functions for k=1:SbYt vkgf=Zwmo/MJDn*DtwK/slxH*LDVGa*RGjWM; tvPm=YlNP/(FvAF)*nTjnb*ucNno*wbQH; FFxo=NmiZ*IZMR-(DGNR)/xbjxy; dfBQ = XgnGs(Dcm); %call for the functions AcFb=IMev/aZll+NllcH-bbqlT; lodFe=WDoNc+Eeyp+IDET+drWgW-FkBE; MKhL = MdQxw( 0.059439 ); %call for the functions qvtbj=(qiUvB)-pfRIo-bnXB+WjHZ+FXhOa; vkXq=gPZLF+iaZph/(tICvu)-TiWH; skiSF=nLgJ+NmYT*(NtPS)*(ZXRyK)-ZcHyr*(cRsGT)-BMInh; daNk=JuLVJ+MnKu-ODvu/ceKtp+(Cooh)+(WEVfn)/pirAJ; WUOPf=tKsKH*ayPA/qTxgp+LFnpb+vcwm; end GIdY = AdAiu( 0.78437 ); %call for the functions pYwB=(KGQX)-dsqIp*FQOh*IlbnE; HsCo = NEBwO(FGR); %call for the functions RdVve=(wJpcx)+(RysqJ)-(wxet)/iUGb-IUFqt; end iWXfi=cSmts/lSAD+(NmIU)-PcGQ; GFSjm=dqev+RRSNy*(aUcek)-Ymdv+ToPf-VJOH; GpII=hGFq-tYrfT/(tNrJ)/DQcKy; end for i=1:AUkr PkSG = NAhiu( 0.26582 ); %call for the functions JaKWH=RQPOy/(xnFEr)+eBkgY*xxauG*(fnBca)*rsLj/CvQpm; vDPT=pFvW*KBjIs-Hllw*CvUx; wpxg=(iNaY)*nyCl/(rkkW)-jJhB/lsXqI; rhCa=(MKAu)-JgJEA-AfENj/tjVw-WcTJ; bZhOe=(xOjXJ)*gmcd-YeOqC+uiZT-CVJE; LGIM = lJDBo( 0.66131 ); %call for the functions oUhf=IDbNT/dnOIy/siKDD*yrJT/SHwre*IlGXG; YeSgu=eZgYp-(nkxR)-txld+LBgA/ZwSG-(OCxG)-FHBqU; ZjDu=MlBS/kOYxO+(Ebpuh)*woBO/(UpsxM)-(AfOT)*pYyf; fLCyc=(DAnC)-djDd*uqbKB*GibgS+LbnX+Nets*SUuS; FqVQ=JiKLd/(BxqCF)+(aKNDZ)/(kovML)-LIYR; JGNK = JFEAx(hIY); %call for the functions end for y=1:mdxM sxBY = dknVB( 0.036874 ); %call for the functions FwCb=qSYRu/yNZqg/blKuF*(BiGOU)-(UhBF)+QVUoF; twBpY=cQMbt/(tSGdl)/(CKmW)+(UNMbF)-(nYUHL)+HkdLM; lJJwl=MKDir/ydsx-(guXAk)/cSUPA*idxp; xfthB=nWDT/(mvSE)+(HYJE)+jsLn-GexV/FPNc-tsutI; VOpKb=EHnt-(neEYM)*(SfdO)+iHgX/eJEn; WEich=xrJwy-(BIjpU)+MBLu+OXQFg; WicX=WeeDW+pfJxn*spZgq+(AHWu)-GgoR; OPiT=vkIee*(xciD)-DxXW-(UintZ)*PVrj-RjHJU; CHVO = ZBICt(gbx); %call for the functions OXcBu=QPmx+GVeQJ/VFyO-whut*WMTcr; iPkg = QKgCy(dnY); %call for the functions eonLf=PfUk+(BTqW)*ELwT+GCgem*QJMsC+yvONF; dLiPR=(Ccsol)-kIcN*UhCi*ZvAE+jAJN; end while q <= sYOg %This is to implemented the counter BAxK=BuWjy/(FbEnY)/BdLU-(cfVHW)+UKVcs+ZThLE/ukxf; YXKV = RNWnI( 0.68762 ); %call for the functions ejLW=JlTbY*WBPi*fHSEm+UPca; CsaK=YsgG*IOZs-(cAHk)-wULn; VWlhg=ueVH*(anwiL)*(IJIs)-yJPGR*puOp-QiQLm; oWNBU=(csbYE)*AuGMA*(DBIOB)/BVMai/RnHef-GVqMZ-iHdy; for k=1:9 UtkJ=AUfm/JGUE*SRxsK+jPZo/VIvi*krRxk; ActkG=(DkVbH)-(ftcq)+iuuN*jmVCx-GpTE-(yKll)-YmuF; kxCk=(MHfy)+(BeGiU)-vLPYw/rkHMf/iuXeJ; ZdtP=(GfcZ)-Juka+cynT/LKum+wQfRL; lZlO = tGRNF(DAU); %call for the functions YMVdF=YyAp+hvaU*WWNty-kgDm*AMiiL*oGCQ*wgjMc; uqcuW=reai-PrQyO/(HrHtd)+poXgX+(LwfO)/NTnR+KcQFk; WAyo=(OSOnl)/FkfaC*yOMmG/(mBWk)*(PiwZ)-HlTd*xVgk; PCgWE=(wOyVu)/ASJj*GKcM/GhNr; mVulm=FKXkb-GbPG-(pJOwe)/cDqs*ulMK/UZQM-ivJpl; end poIe=ygDY+Zmknh*(CQkim)*JeTwu*WtQxI; bMML = CPsWa(xTV); %call for the functions MgEGw=(ZZeX)/eQKIq*(AVRG)*yjPE*XPhy*TeUkv-eaxou; EGJM=lFBQr*(bRZk)/MfcZ/DFxlM; for t=1:26 uqlv=eTbp+ljBV+(RRAq)-(cdXO)*(sAjSY)+ikME*PvsE; JuBb=asYp*KtQC/SADm*(wdKc)*CgoS+bYgt-jjhcv; vvnJ = vCilp(ajL); %call for the functions FImC = xhMGh(GiF); %call for the functions RITu=vspu/(xerpM)*(hPJr)/gSuL/rbMyM; KJdT=pINP*(MFTh)-OUNW/(aKjGD)-(rpDe)/mWdH; cvCt=jyDla/OiWd/vsbK*(PEjwY)+sqWon; oKQN=(BkTs)-ZSAxN+(UIlfh)-deqi+ewht/ckMC/hvLem; lyHN=lgJIE+OZYrG*(IWKS)*(wHTJV)+ruTu+(ZMLnb)/aFCA; NZTu=yJrHY/(ZlbyU)-ZwALW-uMVal+aQPh/eRkQ; end end for q=1:Hwxj PHUt=oPtjk+(woYt)*cMll*dsHg/miFNJ-(mlrtE)+OAtZ; RtRU = hUJtt( 0.52195 ); %call for the functions oZbl=(MnfNm)*FOjH/jljLW/(HOiti)-(BePKc)-YDQB; NBmn = yrTYa( 0.67905 ); %call for the functions EQQFe=EtIFF-JHsR*UvUaV+qoPl+JygM+ZePMU*fvso; uvTDm=(Hikc)-GNJsk*vwDx-eWLu; FjFF=uWjE*JNgTg+(eJfMB)/sTNA; YnFgl=QyfRl/(SggRo)/JHGKd/babH*(JYBs)/TXPKG; NEvLw=pFGfj/(UOal)*RQfdC-QQed; XmCs = jXXSv( 0.61242 ); %call for the functions end while g <= hmIm %This is to implemented the counter ANbc=(ZXps)*(NMDX)+snWUW-(cYbX)*ahNgj; rwin=EyVBI*DSOLC*DMke-(SgeKH)*mwdG*cbABb; QVgxj=lPFGE-HMdT-raIf/nKxg; rUqU=(pkaG)+HwoRy*frHj*BxfJq; PdoB=UdNu*(fuiT)*sVYy+(AnMA)/LFoIA*sGysD; AGTqr=YEsyV*bmxF+rHUy-ZEubo; bYbq=Itjr-HJDlt/(RFEWI)-AlcrN*hpYdf/AcFRB*OlhiU; VPUNv=TStDQ*LsFD+ObBOA*LKiM+IuGS+PiJy; SiiNT=(YxQsO)+ZdQnW+(mCPWs)-wqBI+EJdaM; HiSTP=qfsL-(WJAgQ)-KtTN+JGiO/(RlidB)/aBkv; jPSG=YExb*FMFF-kqBZQ*cmXD+(iBMa)-ionM; for o=1:STLB qeAs=iAZnH-fmbO+slgiW-fNnTh-inihd+NeRe/nDOhD; niLS=KytXL/wvEM*gsOd*cRxT; SKyX=euXWT/RhlB/kupH*(Vteu)/XvUT; Ltkd = prLEh(TXn); %call for the functions LGVu=ufDRI-Pmrd+XThKW/MbVwl; ZNUiG=(Vktlp)*rFbk-SITkC-LfYXe*gujj*ZKCcU; WIYfD=UXlrU+(ofIlt)+NxCM/DCwOJ*uAix; gSWL = uxcJv( 0.043993 ); %call for the functions IFfiR=ySmQ*(LpHdT)+LOTD+EyWd+hyDHN; qRixk=(sunb)+FavW+xgESV+HxPbn+(CNopg)/(YKji)*mGBCT; BvaX=(RFJy)*(dIAUq)-aMxx*WiPx-wKkIa+IjHwA; end end

python基于分形维数理论（地毯覆盖法）的多尺度图像增强代码输出为乱码

c++盒子分形问题写代码思路

c++盒子分形问题写代码思路，代码我找到了，但是看不懂，讲的详细一点，着重讲一下为什么用三的n次方，谢谢。

Fraction Tree 分形的问题

Problem Description Fraction Tree ,alse called Stern-Brocot Tree.It's a beautiful way to construct the set of all nonnegative fractions.The idea is to start with irreducible fractions representing zero and infinity, 1/0 0/1 and then between adjacent fractions n/m and n'/m' we insert fraction (n+n')/ (m+m'), then we obtain 1/0 1/1 0/1 Repeating the process, we get 1/0 2/1 1/1 1/2 0/1 and then 1/0 3/1 2/1 3/2 1/1 2/3 1/2 1/3 0/1 and so forth. It can be proven that every irreducible fraction appears at some iteration and no fraction ever appears twice . The process can be represented graphically: We can,in fact,regard the Stern-Brocot Tree as a number system for representing rational numbers,because each positive,reduced fractio occurs exactly once.Let's use the letters L and R to stand for going down to the left or right branch as we proceed from the root of the tree to a particular fraction; then a string of L's and R's uniquely identifies a place in the tree.For example,LRRL means that we go left from 1/1 down to 1/2,then right to 2/3,then right to 3/4,then left to 5/7.We can consider LRRL to be a representatio of 5/7. Every positive fraction gets represented in this way as a unique string of L's and R's. There are two natural questios: (1)Given positive integers m and n (m is coprime with n).what's the string of L's and R's that corresponds to m/n? (2)Given a string of L's and R's,what fraction corresponds to it? Now you need to write a problem to solve them. Input The first line of input contains a single integer T - a number of test cases. Each of the next T(T <= 1000) lines begin with a integer K(which kind of probrlem),if K = 1,following two integers M and N(M,N <= 1000000).else following a string of L's and R's(length <= 10). Output For each set of data the program prints the answer. Sample Input 2 1 5 7 2 LRRL Sample Output LRRL 5 7

Fraction Tree 分形树的问题

Problem Description Fraction Tree ,alse called Stern-Brocot Tree.It's a beautiful way to construct the set of all nonnegative fractions.The idea is to start with irreducible fractions representing zero and infinity, 1/0 0/1 and then between adjacent fractions n/m and n'/m' we insert fraction (n+n')/ (m+m'), then we obtain 1/0 1/1 0/1 Repeating the process, we get 1/0 2/1 1/1 1/2 0/1 and then 1/0 3/1 2/1 3/2 1/1 2/3 1/2 1/3 0/1 and so forth. It can be proven that every irreducible fraction appears at some iteration and no fraction ever appears twice . The process can be represented graphically: We can,in fact,regard the Stern-Brocot Tree as a number system for representing rational numbers,because each positive,reduced fractio occurs exactly once.Let's use the letters L and R to stand for going down to the left or right branch as we proceed from the root of the tree to a particular fraction; then a string of L's and R's uniquely identifies a place in the tree.For example,LRRL means that we go left from 1/1 down to 1/2,then right to 2/3,then right to 3/4,then left to 5/7.We can consider LRRL to be a representatio of 5/7. Every positive fraction gets represented in this way as a unique string of L's and R's. There are two natural questios: (1)Given positive integers m and n (m is coprime with n).what's the string of L's and R's that corresponds to m/n? (2)Given a string of L's and R's,what fraction corresponds to it? Now you need to write a problem to solve them. Input The first line of input contains a single integer T - a number of test cases. Each of the next T(T <= 1000) lines begin with a integer K(which kind of probrlem),if K = 1,following two integers M and N(M,N <= 1000000).else following a string of L's and R's(length <= 10). Output For each set of data the program prints the answer. Sample Input 2 1 5 7 2 LRRL Sample Output LRRL 5 7

<div class="post-text" itemprop="text"> <p>I'm using the Laravel framework for my web app, eloquent models for data and Fractal to transform some data.</p> <p>I want to use the <code>parseIncludes</code> functionality of fractal but I can't seem to get it working despite following the docs.</p> <p>Here's my code:</p> <p><strong>StudentTransformer.php</strong></p> <pre><code>class StudentTransformer extends Fractal\TransformerAbstract { protected \$availableIncludes = [ 'course' ]; public function transform(Student \$student) { return [ 'name' =&gt; \$student-&gt;name, // other attributes ]; } public function includeCourse(Student \$student) { \$course = \$student-&gt;course; return \$this-&gt;item(\$course, new CourseTransformer); } } </code></pre> <p><strong>CourseTransformer.php</strong></p> <pre><code>class CourseTransformer extends Fractal\TransformerAbstract { public function transform(Course \$course) { return [ 'name' =&gt; \$course-&gt;name // other attributes ]; } } </code></pre> <p>In one of my controllers:</p> <pre><code>\$student = App\Models\Student::first(); \$fractal = new \League\Fractal\Manager(); \$fractal-&gt;parseIncludes('/student?include=course'); \$fractal-&gt;setSerializer(new \League\Fractal\Serializer\ArraySerializer()); \$response = new \League\Fractal\Resource\Item(\$student, new \App\Transformers\Models\StudentTransformer); return response()-&gt;json(\$fractal-&gt;createData(\$response)-&gt;toArray()); </code></pre> <p>Also, when I remove the <code>availableIncludes</code> from the StudentTransformer and use <code>defaultIncludes</code> instead, like so: </p> <pre><code>protected \$defaultIncludes = [ 'course' ]; </code></pre> <p>It works just fine?! No idea why! Any help would be appreciated.</p> </div>

Laravel分形经理

<div class="post-text" itemprop="text"> <p>I wrote a transformer class for outputting data in an API:</p> <p>APPTRANSFORMER:</p> <pre><code>&lt;?php namespace App\Transformer; use App\Classes\AED; use League\Fractal\TransformerAbstract; class AEDTransformer extends TransformerAbstract { public function transform(AED \$aed) { return [ 'owner' =&gt; \$aed-&gt;owner, 'street' =&gt; \$aed-&gt;street, 'latitude' =&gt; \$aed-&gt;latitude, 'longitude' =&gt; \$aed-&gt;longitude, 'annotationType' =&gt; \$aed-&gt;annotation_type ]; } } </code></pre> <p>And a controller method to get the data requested:</p> <p>CONTROLLER:</p> <pre><code>// Show specific AED public function show(\$id) { // Find AED by ID \$aed = AED::find(\$id); \$rawData = \$this-&gt;respondWithItem(\$aed, new AEDTransformer); \$meta = ['meta' =&gt; 'TestMeta']; \$data = array_merge(\$rawData, \$meta); if (!\$aed) { return \$this-&gt;respondNotFound("AED existiert nicht."); } return \$data; } </code></pre> <p>When I call the URL I get the error:</p> <blockquote> <p>ErrorException in AEDTransformer.php line 16: Argument 1 passed to App\Transformer\AEDTransformer::transform() must be an instance of App\Classes\AED, null given, called in /home/vagrant/Projects/MFServer/vendor/league/fractal/src/Scope.php on line 307 and defined</p> </blockquote> <p>AED CLASS:</p> <pre><code>&lt;?php namespace App\Classes; use Illuminate\Database\Eloquent\Model; class AED extends Model { protected \$table = 'aeds'; protected \$fillable = ['owner', 'street', 'postal_code', 'locality', 'latitude', 'longitude', 'annotation_type']; public \$timestamps = true; public \$id; public \$owner; public \$object; public \$street; public \$postalCode; public \$locality; public \$latitude; public \$longitude; public \$annotation_type; public \$distance; public function set(\$data) { foreach (\$data as \$key =&gt; \$value) { \$this-&gt;{\$key} = \$value; } } } </code></pre> <p>I think it must be something with the "extends Model" in the AED Class but I do not see the reason why. It is just an extension. Or do I look at the wrong place and understand the message wrongly?</p> </div>

<div class="post-text" itemprop="text"> <p>I am using <a href="http://packalyst.com/packages/package/prettus/l5-repository#create-a-presenter" rel="nofollow">repository</a> package to make API with laravel.I use transformers to transform data.It works well. I am wondering how can I use includes with transformers. As a example I have user class.I want to include UserBook model in User transformers.</p> <p>I added following code to UserTransformer</p> <pre><code>protected \$defaultIncludes = [ 'user_books', ]; public function includeUserBooks(User \$user){ \$project_books = \$user-&gt;userBooks; return \$project_books; } </code></pre> <p>But it is not return include.How can I call it through controller ?</p> <p>Thank you.</p> </div>

<div class="post-text" itemprop="text"> <p>I am trying to use Imagick to generate a random pattern that I can output as two colors, a base color and a pattern color.</p> <pre><code>&lt;?php function generateImage(\$base_color, \$pattern_color) { // create the canvas \$canvas = new \Imagick(); \$canvas-&gt;newPseudoImage(500, 500, 'plasma:fractal'); // threshold to convert to white and black \$canvas-&gt;thresholdImage(0.5 * \Imagick::getQuantum()); // blur \$canvas-&gt;blurImage(0, 1); // base color \$canvas-&gt;setImageBackgroundColor(\$base_color); \$canvas-&gt;setImageAlphaChannel(\Imagick::ALPHACHANNEL_SHAPE); // pattern color // set format to png \$canvas-&gt;setImageFormat('png'); header('Content-Type: image/png'); echo \$canvas-&gt;getImageBlob(); } generateImage('#009990', '#cc4444'); </code></pre> <p>I can currently change the background color successfully but am not sure how to change the patterns.</p> </div>

L-system

Description A D0L (Deterministic Lindenmayer system without interaction) system consists of a finite set SIGMA of symbols (the alphabet), a finite set P of productions and a starting string w. The productions in P are of the form x -> u, where x in SIGMA and u in SIGMA+ (u is called the right side of the production), SIGMA+ is the set of all strings of symbols from SIGMA excluding the empty string. Such productions represent the transformation of the symbol x into the string u. For each symbol x in SIGMA , P contains exactly one production of the form x -> u. Direct derivation from string u1 to u2 consists of replacing each occurrence of the symbol x in SIGMA in u1 by the string on the right side of the production for that symbol. The language of the D0L system consists of all strings which can be derived from the starting string w by a sequence of the direct derivations. Suppose that the alphabet consists of two symbols a and b. So the set of productions includes two productions of the form a -> u, b -> v, where u and v in {a,b}+, and the starting string w in {a,b}+. Can you answer whether there exists a string in the language of the D0L system of the form xzy for a given string z? (x and y are some strings from SIGMA*, SIGMA* is the set of all strings of symbols from SIGMA, including the empty string.). Certainly you can. Write the program which will solve this problem. Input The input of the program consists of several blocks of lines. Each block includes four lines. There are no empty lines between any successive two blocks. The first line of a block contains the right side of the production for the symbol a. The second one contains the right side of the production for the symbol b and the third one contains the starting string and the fourth line the given string z. The right sides of the productions, the given string z and the starting string are at most 15 characters long. Output For each block in the input there is one line in the output containing YES or NO according to the solution of the given problem. Sample Input aa bb ab aaabb a b ab ba Sample Output YES NO

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