%% Math 464 Linear Optimization (Spring 2023)
%% Lecture 10, 02/09/2023

%% bfs <==? vertex correspondences, and 
%% Degenerate bfs's

>> format rat
>> format compact

%% Restarting with the same LP

>> A = [1 1 -1 0 0; 3 1 0 -1 0; 3 2 0 0 1]
A =
       1              1             -1              0              0       
       3              1              0             -1              0       
       3              2              0              0              1       
>> Bind = [1 4 5];
>> B = A(:,Bind)
B =
       1              0              0       
       3             -1              0       
       3              0              1       

>> b = [2 4 10]';
>> xB = inv(B)*b
xB =
       2       
       2       
       4       
>> % This == A(2,0)


%% Let's find some basic solutions that are *NOT* feasible

>> Bind = [1 2 3];
>> B = A(:,Bind)
B =
       1              1             -1       
       3              1              0       
       3              2              0       
>> xB = inv(B)*b
xB =
      -2/3     
       6       
      10/3     
>> % This is H(-2/3,6)


>> Bind = [1 2 4];
>> B = A(:,Bind)
B =
       1              1              0       
       3              1             -1       
       3              2              0       
>> xB = inv(B)*b
xB =
       6       
      -4       
      10       
% This is (6,-4), the point of intersection of
% 3x1 + 2x2 = 10 and x1 + x2 = 2

% Finding the determinants of B in each case to
% fill the table of vertex-bfs correspondences

>> Bind = [1 4 5]; B = A(:,Bind); det(B)
ans =
      -1       
>> xB = inv(B)*b
xB =
       2       
       2       
       4       
>> % This == A(2,0)

>> Bind = [1 3 4]; B = A(:,Bind); det(B)
ans =
       3       
>> xB = inv(B)*b
xB =
      10/3     
       4/3     
       6       
>> % This == B(10/3,0)

>> Bind = [2 3 4]; B = A(:,Bind); det(B)
ans =
       2       
>> xB = inv(B)*b
xB =
       5       
       3       
       1       
>> % This == C(0,5)

>> Bind = [2 3 5]; B = A(:,Bind); det(B)
ans =
       1       
>> xB = inv(B)*b
xB =
       4       
       2       
       2       
>> % This == D(0,4)

>> Bind = [1 2  5]; B = A(:,Bind); det(B)
ans =
      -2       
>> xB = inv(B)*b
xB =
       1       
       1       
       5       
>> % This == E(1,1)

%% Degenerate bfs example: Adding x1 <= 10/3

>> oldA = A; % saving a copy of original A and b
>> oldb = b;

>> A = [ [A zeros(3,1)]; [1 0 0 0 0 1]]
A =
       1              1             -1              0              0              0       
       3              1              0             -1              0              0       
       3              2              0              0              1              0       
       1              0              0              0              0              1       

>> b = [b; 10/3]
b =
       2       
       4       
      10       
      10/3     

% We know Bind should include 1,3,4; we need one more column index
% from {2,5,6}. Any one of them would work here!

>> Bind = [1 2 3 4];
>> B = A(:,Bind)
B =
       1              1             -1              0       
       3              1              0             -1       
       3              2              0              0       
       1              0              0              0       
>> det(B)
ans =
      -2       
>> xB = inv(B)*b
xB =
      10/3     
       0       
       4/3     
       6       
>> % x1 = 10/3, x2=0, x3=4/3, x4=6; B(10/3,0)

>> Bind = [1 5 3 4];
>> B = A(:,Bind);det(B)
ans =
      -1       
>> xB = inv(B)*b
xB =
      10/3     
       0       
       4/3     
       6
>>       
>> % x1 = 10/3, x5=0, x3=4/3, x4=6; B(10/3,0)