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function [P,S] = permutation(V,X,NFFT,FS,win,OVERLAP,MT,thres) 
%permutation function 
%解决频域不确定性问题 
NF = NFFT/2+1; 
D  = ceil(MT*FS/(1000*(NFFT-OVERLAP))); 
MA = ones(D,1)/D; 
S  = ones(1,NF);  
P  = zeros(1,NF); 
TH = min(thres,1); 
 
B1 = specgram(X(:,1),NFFT,FS,win,OVERLAP); 
B2 = specgram(X(:,2),NFFT,FS,win,OVERLAP); 
L  = size(B1,2); 
 
for f=1:NF, 
   
  Z = inv(V(2*f-1:2*f,:))*[B1(f,:);B2(f,:)]; 
 
  tmp1 = V(2*f-1:2*f,:)*[Z(1,:);zeros(1,L)];  
  tmp2 = V(2*f-1:2*f,:)*[zeros(1,L);Z(2,:)]; 
   
  C11(f,:) = tmp1(1,:); C12(f,:) = tmp1(2,:); 
  C21(f,:) = tmp2(1,:); C22(f,:) = tmp2(2,:); 
   
  t1 = conv(MA,(ones(1,2)*abs(tmp1))); 
  t2 = conv(MA,(ones(1,2)*abs(tmp2))); 
 
  t1=t1/norm(t1); t2=t2/norm(t2); 
   
  tmp(f)=t1*t2'; 
  if tmp(f)>TH, 
    S(f)=0; 
    C11(f,:) = B1(f,:); C21(f,:) = zeros(1,L);  
    C12(f,:) = B2(f,:); C22(f,:) = zeros(1,L); 
  end 
end 
 
[tmp,l]=sort(tmp); 
f=l(1); k=f; 
y1 = zeros(1,L); y2 = y1; 
 
for f=l(2:NF), 
   
  t1=conv(MA,(ones(1,2)*abs([C11(f,:);C12(f,:)]))); 
  t2=conv(MA,(ones(1,2)*abs([C21(f,:);C22(f,:)]))); 
 
  if norm(t1)~=0, 
    t1=t1/norm(t1); 
  end 
  if norm(t2)~=0, 
    t2=t2/norm(t2); 
  end 
   
  y1=ones(1,3)*[y1;abs(C11(k,:));abs(C12(k,:))]; 
  y2=ones(1,3)*[y2;abs(C21(k,:));abs(C22(k,:))]; 
   
  ny1 = conv(MA,y1);ny1 = ny1/norm(ny1); 
  ny2 = conv(MA,y2);ny2 = ny2/norm(ny2); 
   
  if (abs(t1*ny1')+abs(t2*ny2') < abs(t1*ny2')+abs(t2*ny1')), 
    P(f)=1; 
  end 
  k=f; 
end