Use this to show that a matrix with the consecutive l's property is TU.

3. A 0-1 matrix B has the consecutive 1 's property if for any column j, bij = b;.'j = 1 with i < i' implies b1j = 1 for i < l < i'. A more general sufficient condition for total unimodularity is: Matrix A is TU if . (i) aij E {+1,-1,0} for all i,j. (ii) For any subset M ofthe rows, there exists a partition (M1,M2) of M such that each column j satisfies Use this to show that a matrix with the consecutive l's property is TU.

aii - aiji | < 1 ie M1 ie M2