The method of solving is brute-force and the codes involved were optimized at my best.
All possible configurations of hours(n) spent over (N hours window) by each friend is given by either combR or comb.
Octave code for the subroutine : combR.m (depends on nested looping ann optimization)
Octave code for the subroutine : comb.m (So many for loops and functions were used finally the result will same as combR)
Surprisingly, both the methods were taking nearly equal time for smaller N values.
The final solution can be sought from friendsMeet(N) programme.
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Codes start here------>>>>
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combR.m code:
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function X=combR(N,n,Xpre)
if n>N
return ;
endif
if (n==1)
X=eye(N);
else
if nargin==3
X=Xpre;
else
X=combR(N,n-1);
endif
Y=[ ];
for col=1:size(X,2)
[rowidx]=find(X(:,col)==0);
for g=1:length(rowidx)
Y=[Y X(:,col)];
Y(rowidx(g),end)=1;
endfor
endfor
X=unique(Y',"rows")';
endif
endfunction
if n>N
return ;
endif
if (n==1)
X=eye(N);
else
if nargin==3
X=Xpre;
else
X=combR(N,n-1);
endif
Y=[ ];
for col=1:size(X,2)
[rowidx]=find(X(:,col)==0);
for g=1:length(rowidx)
Y=[Y X(:,col)];
Y(rowidx(g),end)=1;
endfor
endfor
X=unique(Y',"rows")';
endif
endfunction
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comb.m code:
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function [X]=comb(N, n)
%% Generates all possible combinations for N choose n
if n>N
return ;
endif
X=eye(N);
for B=1:n
for i=1:size(X,2)
for j=1:N
if(X(j,i)==1)
continue
endif
X=[X, X(:,i)];
X(j,end)=1;
endfor
endfor
[idx]=find(sum(X,1)==B);
X=X(:,idx);
X=unique(X',"rows")';
endfor
endfunction
%% Generates all possible combinations for N choose n
if n>N
return ;
endif
X=eye(N);
for B=1:n
for i=1:size(X,2)
for j=1:N
if(X(j,i)==1)
continue
endif
X=[X, X(:,i)];
X(j,end)=1;
endfor
endfor
[idx]=find(sum(X,1)==B);
X=X(:,idx);
X=unique(X',"rows")';
endfor
endfunction
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main function friendsMeet.m code:
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function H=friendsMeet(N)
%% Find number of hours each, two friends should wait
%% at a location so that they can meet for at least
%% for one hour
%% Input: (N) Available time window for both.
%% Each friend spends time of integer
%% hours starting from an integer clock hour.
%% The technique is Brute force or Trail-n-Error
if nargin==0
N=8;
endif
M=X=[];
for i=1:N
if i==1
Y=combR(N,i);
else
Y=combR(N,i,Y);
endif
X=[X, Y];
endfor
## j=0;
## for a=1:size(X,2)
## for b=1:size(X,2)
## M(++j)=X(:,a)'*X(:,b);
## Hc(j)=min(sum(X(:,a)),sum(X(:,b)));
## endfor
## endfor
M=(X'*X)(:);
Hc=full(sum(X));
Hc=min(Hc,Hc')(:); %%Symmetric matrix output
%hist(Hc(M>0))
H=mean(Hc(M>0));
V=std(Hc(M>0),1);
printf("For %f%% chances Spend atleast %d hours\n", 99.7, H+3*V);
printf("For %f%% chances Spend atleast %d hours\n", 95, H+2*V);
printf("For %f%% chances Spend atleast %d hours\n", 68, H+V);
endfunction
%% Find number of hours each, two friends should wait
%% at a location so that they can meet for at least
%% for one hour
%% Input: (N) Available time window for both.
%% Each friend spends time of integer
%% hours starting from an integer clock hour.
%% The technique is Brute force or Trail-n-Error
if nargin==0
N=8;
endif
M=X=[];
for i=1:N
if i==1
Y=combR(N,i);
else
Y=combR(N,i,Y);
endif
X=[X, Y];
endfor
## j=0;
## for a=1:size(X,2)
## for b=1:size(X,2)
## M(++j)=X(:,a)'*X(:,b);
## Hc(j)=min(sum(X(:,a)),sum(X(:,b)));
## endfor
## endfor
M=(X'*X)(:);
Hc=full(sum(X));
Hc=min(Hc,Hc')(:); %%Symmetric matrix output
%hist(Hc(M>0))
H=mean(Hc(M>0));
V=std(Hc(M>0),1);
printf("For %f%% chances Spend atleast %d hours\n", 99.7, H+3*V);
printf("For %f%% chances Spend atleast %d hours\n", 95, H+2*V);
printf("For %f%% chances Spend atleast %d hours\n", 68, H+V);
endfunction
