Boolean Arithmetic

The Boolean Algebra has three basic arithmetic :

  1. AND :
    • Multiplication for SOP Logic Functions
    • \( F+G = !!(F+G) = !((!F)\cdot(!G)), \quad !x = NOT(x) \)
  2. OR :
    • Addition for SOP Logic Functions
    • \( F = !!(F) = !((!F)) \)
    • Let \( F = F_1 + F_2 \), then \( !F = !(F_1 + F_2) = ((!F_1)\cdot(!F_2)) \)
  3. NOT :
    • The complement set for a Logic Function
    • \( !F = !(F_1 + F_2) = ((!F_1)\cdot(!F_2)), F = F_1 + F_2 \)
The SOP means the Logic Function in the format of Sum Of Product.

In the computational Boolean Algebra, it is a symbolic computation. Therefore, we need an efficient and theoretical proven method to do this. In our Computational Boolean Algebra, we can always prove the correctness of the computation results.

Our Computational Boolean Algebra provides a fast computation for the operator NOT. Therefore

  1. OR : remained as summing up
  2. NOT : processed by fast searching algorithm.
  3. AND : processed by the computation of 2 NOTs and 1 OR.
Hence, the worst case of our computation is only the time comsumed by the computation of 2 NOTs.

Test Computation Speed Of Operator NOT

[ f ] = AndOr()
{
    1,-2,3,-4,-5,-6 ;
    -1,-2,3,4,-5,6 ;
    -1,2,3,-4,-5,6 ;
    1,-2,3,4,5,6 ;
    -1,-2,-3,4,-5,6 ;
    1,2,-3,4,5,6 ;
    1,2,-3,-4,-5,6 ;
    1,2,-3,-4,5,6 ;
    1,2,-3,4,5,6 ;
    -1,2,-3,-4,5,6 ;
}

[ g ] = Not(f);

Print("result:", g);

/*
The result should be :
//--------------------------------------------------//
/// Time for executing 'Not' : 312ms
"result:";
g = AndOr()
{
  1,2,3;
  2,-3,-6;
  1,-2,-3;
  -1,-2,-4;
  -1,2,3,4;
  -1,-2,4,5;
  -1,2,3,-4,5;
  -1,2,-3,4,6;
  1,-2,3,4,-5;
  1,-2,3,-4,5;
  -1,-2,4,-5,-6;
  -1,2,3,-4,-5,-6;
  1,2,-3,4,-5,6;
  -1,2,-3,-4,-5,6;
  1,-2,3,4,5,-6;
  1,-2,3,-4,-5,6;
}
*/



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