Luke Palmer wrote:
You know, \ and friends as xor is appealing to me.
Hmmmm. I quite like that too. :-)
Also, a question about superpositions: Is
$x = 1 | 2 | 3
equivalent to
$x = 1 | 2
$x |= 3
No. The precedence is wrong.
or
$x = (1 | 2) | 3
Yes.
or is there a difference at all?
No (if you mean between those two forms above).
Yes (if you mean between those two and C<all(1,2,3)>).
;-)
So the latter is either 3 or the
superposition 1 | 2.
Yes.
Does
$x == 1
still return true?
Yes.
I guess the root of my question is... do
superpositions search deep or shallow?
Think if it like this...
Suppose:
$x1 = any(1,2,3)
and
$x2 = 1|2|3;
$x1 has three superimposed states: C<1>, C<2>, and C<3>.
$x2 has two superimposed states: C<any(1,2)> and C<3>
If we write:
$x1 == 1
that means:
any(1,2,3) == 1
which means:
1==1 || 2==1 || 3==1
which is true.
If we write:
$x2==1
that means
any(any(1,2),3) == 1
which means:
any(1,2)==1 || 3==1
which means:
1==1 || 2==1 || 3==1
which is true.
So the effect is the same either way.
The only time you'd notice any difference between $x1 and $x2 is if you
asked for their eigenstates, in which case $x1 would give you
three states (C<1>, C<2>, and C<3>) and $x2 would give you two states
(C<any(1,2)> and C<3>).
Damian
