function [teams1,nteams1,exitcode]=wctiebreaker(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 conference but different divisions, 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 the separation.
%
% The criteria are:
%
% 0. overall won-lost-tied percentage,
% 1. head-to-head sweep (applicable only if one team has defeated each of the others or if one team 
%    has lost to each of the others),
% 2. won-lost-tied percentage in the conference,
% 3. won-lost-tied percentage in games against common opponents (minimum of four),
% 4. strength of victory, and
% 5. strength of schedule.
%
% (Criteria 1 through 5 are called tiebreakers.) If none of these criteria differentiates among the 
% teams, then the function arbitrarily eliminates the last team.

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

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

% see if there is a head-to-head sweep
y=zeros(32,32);
for j=1:256
   if div(a(j,1))~=div(a(j,2))
      if a(j,3)==1
         y(a(j,2),a(j,1))= 1;
         y(a(j,1),a(j,2))=-1;
      elseif a(j,3)==0
         y(a(j,2),a(j,1))=-1;
         y(a(j,1),a(j,2))= 1;
      end
   end
end
% kth column of y now contains 1 for a win by team k, -1 for a loss by team k

Y=zeros(nteams0,nteams0);   % submatrix of y corresponding to the nteams0 teams
for i=1:nteams0
   for k=1:nteams0
      Y(k,i)=y(teams0(k),teams0(i));
   end
end

sweep=0;
for i=1:nteams0
   if sum(Y(:,i))==nteams0-1
      sweep=teams0(i);
      istar=i;
   end
end
if sweep>0             % team teams0(i) swept
   nteams1=1;
   teams1(1)=teams0(istar);
   exitcode=1;
   break
end

sweep=0;
for i=1:nteams0
   if sum(Y(:,i))==-(nteams0-1)
      sweep=teams0(i);
   end
end
if sweep>0             % team teams0(i) was swept
   nteams1=nteams0-1;
   r=0;
   for i=1:nteams0
      if teams0(i)~=sweep
         r=r+1;
         teams1(r)=teams0(i);
      end
   end
   exitcode=1;
   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=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 number of games against common opponents
commongames=zeros(nteams0,1);
for i=1:32
   if z(i)>0
      for k=1:nteams0
         commongames(k)=commongames(k)+y(i,teams0(k));
      end
   end
end

if min(commongames)>=4           % if at least 4 games against common opponents, do the calculations

   % 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

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=4;
   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=5;
   break
end

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

% if all else fails, just drop the last team on the list
nteams1=nteams0-1;
teams1=teams0;
exitcode=6;

end