function [teams1,nteams1,exitcode]=divtiebreaker(teams0,nteams0,a,conf,div,wltpct,cwltpct,dwltpct,sov,sos)

% Copyright 2006 by William S. Krasker

% Given a set of nteams0 teams from the same division, whose indices are in the array teams0, this
% function runs through a sequence of criteria until one is found for which not all nteams0 teams
% are equally good. The function then returns the number nteams1 of teams that are tied for the
% best according to that criterion, the array teams1 containing their indices, and the exitcode
% showing the criterion that achieved separation.
%
% The criteria are:
%
% 0. overall won-lost-tied percentage,
% 1. won-lost-tied percentage in games among the teams,
% 2. won-lost-tied percentage in the division,
% 3. won-lost-tied percentage versus common opponents,
% 4. won-lost-tied percentage in the conference,
% 5. strength of victory, and
% 6. strength of schedule.
%
% (Criteria 1 through 6 are called tiebreakers.) If none of these criteria differentiates among the 
% teams, then the function arbitrarily eliminates the last team. (Notice that if nteams0=2, then 
% the second criterion reduces to head-to-head.)

teams1=zeros(4,1);

for ii=1:1
   
%--------------------------------------------------
   
% find best win percentage
v=zeros(4,1);
x=0;
for i=1:nteams0
   x=max(x,wltpct(teams0(i)));
end

% find teams with best win percentage
k=0;
for i=1:nteams0
   if wltpct(teams0(i))==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on win percentage
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=0;
   break
end

%--------------------------------------------------

% compute win percentages in games among the teams
w=zeros(4,1);
for j=1:256  
   for i=1:nteams0
      for k=1:nteams0        
         if (a(j,1)==teams0(i)) & (a(j,2)==teams0(k))
            if a(j,3)==1
               w(i)=w(i)+1;
            elseif a(j,3)==0
               w(k)=w(k)+1;
            else
               w(i)=w(i)+0.5;
               w(k)=w(k)+0.5;
            end
         end        
      end
   end   
end
w=w/( 2*sum(w(1:nteams0))/nteams0 );

% find best win percentage in games among the teams
x=max(w(1:nteams0));

% find teams with best win percentage in games among the teams
k=0;
v=zeros(4,1);
for i=1:nteams0
   if w(i)==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on win percentage in games among the teams
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=1;
   break
end

%--------------------------------------------------

% find best win percentage in the division
v=zeros(4,1);
x=0;
for i=1:nteams0
   x=max(x,dwltpct(teams0(i)));
end

% find teams with best win percentage in the division
k=0;
for i=1:nteams0
   if dwltpct(teams0(i))==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on win percentage in the division
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=2;
   break
end

%--------------------------------------------------

% find common opponents
y=zeros(32,32);
for j=1:256
   y(a(j,1),a(j,2))=y(a(j,1),a(j,2))+1;
   y(a(j,2),a(j,1))=y(a(j,2),a(j,1))+1;
end
z=ones(32,1);
p=zeros(32,1);
for i=1:nteams0
   z=z.*y(:,teams0(i));         % positive elements of z correspond to common opponents
   p(teams0(i))=1;              % positive elements of p correspond to the teams under consideration
end

% find win percentage against common opponents
w=zeros(32,1);
num=zeros(32,1);
for j=1:256
   if (p(a(j,1))>0) & (z(a(j,2))>0)
      num(a(j,1))=num(a(j,1))+1;
      if a(j,3)==1
         w(a(j,1))=w(a(j,1))+1;
      elseif a(j,3)==-1
         w(a(j,1))=w(a(j,1))+0.5;
      end
   elseif (z(a(j,1))>0) & (p(a(j,2))>0)
      num(a(j,2))=num(a(j,2))+1;
      if a(j,3)==0
         w(a(j,2))=w(a(j,2))+1;
      elseif a(j,3)==-1
         w(a(j,2))=w(a(j,2))+0.5;
      end
   end
end
q=zeros(4,1);
for i=1:nteams0
   q(i)=w(teams0(i))/num(teams0(i));
end

% find best win percentage against common opponents
x=max(q(1:nteams0));

% find teams with best win percentage against common opponents
k=0;
v=zeros(4,1);
for i=1:nteams0
   if q(i)==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on win percentage against common opponents
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=3;
   break
end

%--------------------------------------------------

% find best win percentage in the conference
v=zeros(4,1);
x=0;
for i=1:nteams0
   x=max(x,cwltpct(teams0(i)));
end

% find teams with best win percentage in the conference
k=0;
for i=1:nteams0
   if cwltpct(teams0(i))==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on win percentage in the conference
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=4;
   break
end

%--------------------------------------------------

% find best strength of victory
v=zeros(4,1);
x=0;
for i=1:nteams0
   x=max(x,sov(teams0(i)));
end

% find teams with best strength of victory
k=0;
for i=1:nteams0
   if sov(teams0(i))==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on strength of victory
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=5;
   break
end

%--------------------------------------------------

% find best strength of schedule
v=zeros(4,1);
x=0;
for i=1:nteams0
   x=max(x,sos(teams0(i)));
end

% find teams with best strength of schedule
k=0;
for i=1:nteams0
   if sos(teams0(i))==x
      k=k+1;
      v(k)=teams0(i);
   end
end

if k < nteams0    % tie broken based on strength of schedule
   nteams1=k;
   teams1(1:k)=v(1:k);
   exitcode=6;
   break
end

%--------------------------------------------------

% if all else fails, just drop the last team on the list
nteams1=nteams0-1;
teams1=teams0;
exitcode=7;

end