Exemplo n.º 1
0
    def scalar_one_elimination(self, **defaults_config):
        '''
        Equivalence method that derives a simplification in which
        a single scalar multiplier of 1 is eliminated.
        For example, letting v denote a vector element of some VecSpace,
        then ScalarMult(one, v).scalar_one_elimination() would return
            |- ScalarMult(one, v) = v
        Will need to know that v is in VecSpaces(K) and that the
        multiplicative identity 1 is in the field K. This might require
        assumptions or pre-proving of a VecSpace that contains the
        vector appearing in the ScalarMult expression.
        '''
        from proveit.numbers import one
        # from . import elim_one_left, elim_one_right
        from . import one_as_scalar_mult_id

        # The following seems silly -- if the scalar is not 1, just
        # return with no change instead?
        if self.scalar != one:
            raise ValueError(
                "For ScalarMult.scalar_one_elimination(), the scalar "
                "multiplier in {0} was expected to be 1 but instead "
                "was {1}.".format(self, self.scalar))

        # obtain the instance params:
        # _K is the field; _V is the vector space over field _K;
        # and _v is the vector in _V being scaled
        _K, _v, _V = one_as_scalar_mult_id.all_instance_params()

        # isolate the vector portion
        _v_sub = self.scaled

        # Find a containing vector space V in the theorem.
        # This may fail!
        _V_sub = list(VecSpaces.yield_known_vec_spaces(_v_sub))[0]

        # Find a vec space field, hopefully one that contains the mult
        # identity 1. This may fail!
        _K_sub = list(VecSpaces.yield_known_fields(_V_sub))[0]

        # If we made it this far, can probably instantiate the theorem
        # (although it could still fail if 1 is not in the field _K_sub)
        return one_as_scalar_mult_id.instantiate({
            _K: _K_sub,
            _v: _v_sub,
            _V: _V_sub
        })
Exemplo n.º 2
0
 def yield_known_hilbert_spaces(vec):
     '''
     Given a vector expression, vec, yield any Hilbert spaces
     known to contain vec.
     '''
     for vec_space in VecSpaces.yield_known_vec_spaces(vec, field=Complex):
         if vec_space in HilbertSpacesLiteral.known_spaces_memberships:
             # This vector space is already known to be an inner
             # product space.
             yield vec_space
         else:
             try:
                 deduce_as_hilbert_space(vec_space)
                 yield vec_space
             except NotImplementedError:
                 # Not known you to prove 'vec_space' is an inner
                 # product space.
                 pass