create small application for User Defined Voice Triggers

I am willing to discuss further about the project details and deliver the same to your specifications.

Hi I work towards providing reliable, relevant and robust IT solutions at most competitive prices to my customers. I ensure 100% customer satisfaction so lets start Thanks

Good Day How are you? Greetings From !! Samartha Solutions !! Thank you for this opportunity!! About Us: We have working with plenty of projects in WordPress, PHP, Joomla, Graphic Design, HTML, Website Design & Development, Content Writing, Android/IOS app and SEO. ***** WE ARE A GOOD OFFER FOR YOU, IF YOU WILL COME IN CONVERSATION ***** --We have a very strong and experienced team for web design and development. ***We have detail oriented and very sharp web developers*** ***We are Specialized in Website development/design from scratch, Web customization and Revamp *** --We believe in Client Satisfaction. We provide our clients with end to end solution within the prescribed deadline. Budget might never concern for us, We believe in long-term business relationship, So Let's start it with this important project. Eagerly waiting for your response. Thank you, Tushar Panchal Sr. Business Development Executive.

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void anneal(float x[], float y[], int iorder[], int ncity) This algorithm finds the shortest round-trip path to ncity cities whose coordinates are in the arrays x[1..ncity],y[1..ncity]. The array iorder[1..ncity] specifies the order in which the cities are visited. On input, the elements of iorder may be set to any permutation of the numbers 1 to ncity. This routine will return the best alternative path it can find. { int irbit1(unsigned long *iseed); int metrop(float de, float t); float ran3(long *idum); float revcst(float x[], float y[], int iorder[], int ncity, int n[]); void reverse(int iorder[], int ncity, int n[]); float trncst(float x[], float y[], int iorder[], int ncity, int n[]); void trnspt(int iorder[], int ncity, int n[]); int ans,nover,nlimit,i1,i2; int i,j,k,nsucc,nn,idec; static int n[7]; long idum; unsigned long iseed; float path,de,t; nover=100*ncity; Maximum number of paths tried at any temperature. nlimit=10*ncity; Maximum number of successful path changes before con- path=0.0; tinuing. t=0.5; for (i=1;i<ncity;i++) { Calculate initial path length. i1=iorder[i]; i2=iorder[i+1]; path += ALEN(x[i1],x[i2],y[i1],y[i2]); } i1=iorder[ncity]; Close the loop by tying path ends together. i2=iorder[1]; path += ALEN(x[i1],x[i2],y[i1],y[i2]); idum = -1; iseed=111; for (j=1;j<=100;j++) { Try up to 100 temperature steps. nsucc=0; for (k=1;k<=nover;k++) { do { n[1]=1+(int) (ncity*ran3(&idum)); Choose beginning of segment n[2]=1+(int) ((ncity-1)*ran3(&idum)); ..and end of segment. .. if (n[2] >= n[1]) ++n[2]; nn=1+((n[1]-n[2]+ncity-1) % ncity); nn is the number of cities } while (nn<3); not on the segment. idec=irbit1(&iseed); Decide whether to do a segment reversal or transport. if (idec == 0) { Do a transport. n[3]=n[2]+(int) (abs(nn-2)*ran3(&idum))+1; n[3]=1+((n[3]-1) % ncity); Transport to a location not on the path. de=trncst(x,y,iorder,ncity,n); Calculate cost. ans=metrop(de,t); Consult the oracle. if (ans) { ++nsucc; path += de; trnspt(iorder,ncity,n); Carry out the transport. } } else { Do a path reversal. de=revcst(x,y,iorder,ncity,n); Calculate cost. ans=metrop(de,t); if (ans) { ++nsucc; path += de; reverse(iorder,ncity,n); Carry out the reversal. } } if (nsucc >= nlimit) break; Finish early if we have enough suc- } cessful changes. printf("\n %s %10.6f %s %12.6f \n","T =",t, " Path Length =",path); printf("Successful Moves: %6d\n",nsucc); t *= TFACTR; Annealing schedule. if (nsucc == 0) return; If no success, we are done. } }