/* by bell G6 challenge from Oma Falk Write a program that takes 8 hints and finds the 3 members of group G1.. G6. The hints give a set of numbers and the groups which are completely in it. Run https://code.sololearn.com/cjxEQ9Gvd4y1/?ref=app Each run generates a random different puzzle. */ #include #include #include #include #include #include // std::shuffle using namespace std; // Function to adjust probabilities for each digit to be present 2 times void adjustDigit(vector &p){ double sum,msum=0.,t; int ic,ig; for(int i=0; i<9; i++) { sum=0.; ic=0; ig=0; for(int j=0; j<6; j++) { t=p[i+j*9]; if(t>0.95)ig++; if(t<0.95&&t>0.001){ ic++; sum+=t;} } if(sum>msum)msum=sum; sum=((sum>0.1)?(2.-ig)/sum : 1.0); cout<0.0003)p[i+j*9]*=sum; //avoid rounding errors } } cout<<"highest sum was :"< &p,vector cl,vector o8){ string nu="012345678",s; bool finite=true; double sum,t; int ic,ig,il; for(int i=0; i<6; i++){ ic=0; ig=0; sum=0.; for(int j=0; j<9; j++) { t=p[j+i*9]; if(t>0.001)ic+=1; //are there only three possible? if(t>0.95)ig+=1; //are there three correct? if(t<0.95&&t>0.001)sum+=t; } cout<0.005)p[j+i*9]=1.; } } //now constrain absence in groups else if(ig==2){ //eliminate incompatibilities with clues for(int j=0; j<8; j++) { il=0; if(cl[2*j]!=i&&cl[2*j+1]!=i){ s=o8[j]; for(unsigned k=0;k<9;k++){ if(p[k+i*9]>0.95&&s.find(nu[k])!=string::npos)il++; } if(il==2){ for(unsigned k=0;k<9;k++){ t=k+i*9; if(p[t]<0.9&&p[t]>0.15&&s.find(nu[k])!=string::npos){ p[t]=0.0001; cout<<"G"< o6={"178","085","461","237","204","536"}; //not using, just for info vector o8={"0832761","5083241","5082764","5032761","0832641","5027641","5832741","5832764"}; //digits in hints vector os8; //will contain the sorted hints vector clu={0,3,1,4,1,4,3,5,2,4,2,4,0,3,3,5}; //groups in hints cout<<"1/3 is the base probability for each digit in each G clue because each digit will appear twice considering all six clues"< prob(54,1./3.); m=0; cout<0.95)cout<

ProbabilityHintsGroupsOmaFalk