The answer is the value of k0 is meaningless as long as we have a good random
distribution here when we raise a to the e power and the reason for that is
this value now v would be x1 + r to the e mod n that is because xb is x1.
The value for k0 is that minus x0 raised to the d power modn.
There’s nothing that gets rid of this random value and
there’s no reason this would be a meaningful value.
The value of k1 though is meaningful, and it's in fact the value of r, and the reason for that
now the value of k1 is equal to the value of v minus x1 so that removes the x1 from that value.
So that means the value of k is r to the e modn raised to the d power modn
and this is exactly the RSA decryption that will give us the value of the message
which in this case is equal to r.
That means Alice has now learned that random r is selected by Bob,
and if value is stored in one of these keys, she doesn't know which one,
the value of k0 is meaningless.
It’s a meaningless value that can’t be determined by Bob
because Bob doesn’t know the value of d.
That’s the important thing that we’ve done here.
Now, Alice has these two keys.
Depending on what value Bob picked, one of them is meaningful, has the value r, the other one is not.
