Work and World

How Shor’s Algorithm Factors 314191

As a followup to the main video about how
quantum computers factor large numbers to break encryption, I want to demonstrate how
Shor’s algorithm would factor a real live number! Like, maybe you were bequeathed a bank vault
full of pies, but the access code left to you was encrypted using the number 314191
and you can’t get to the pies until you know the factors. Luckily I happen to have a working quantum
computer. As a refresher, here’s a rough overview
how Shor’s algorithm factors large numbers quickly: for any crappy guess at a number
that shares factors with N, that guess to the power p over 2 plus or minus one is a
much much better guess, if we can find p. And we CAN find p almost immediately with
a single (if complex) quantum computation. So, first we make some random crappy guess,
like, I dunno, a hundred and one. Then we check to see if 101 shares a factor
with 314191 – it doesn’t. So our goal is to find the special power p
for which 101to the p over 2 plus or minus 1, is a better guess for a number that shares
factors with 314191. To do this, we need to find p so that 101
to the p is one more than a multiple of 314191. This is where we use my quantum computer which
can raise 101 to any power and calculate how much more that power is than a multiple of
314191. If we start with a superposition of all the
numbers up to 314191, then the quantum computation will give us the superposition of 101 plus
101 squared plus 101 cubed, and so on. and then the superposition of the remainders. So we measure just the state of the remainders,
and we’ll get one remainder as output – say, 74126. From which we know that the rest of the quantum
state is left in a superposition of the powers that resulted in the remainder of 74126, which
must all be “p” apart from each other, which I explained in the other video. Because we’re not actually dealing with
particularly big numbers, I’ve done the calculation and can tell you that this would
mean we had a superposition of 20 and 4367 and 8714 and so on, and the difference between
them is p. but in a real situation we of course wouldn’t know what the numbers in the superposition
are – we just know they’re separated with a period of p, or a frequency of 1 over p,
though we still don’t know what p is. The next step is to put the superposition
through a quantum Fourier transform, which would result in a superposition of 1 over
p plus 2 over p plus 3 over p and so on (this is a part I glossed over in the main video,
but for technical reasons the quantum Fourier transform doesn’t just output 1 over p – it
outputs a superposition of multiples of 1 over p). Again, because these are small numbers I can
tell you that we’d have a superposition of 1 over 4347 and 2 over 4347 and 3 over
4347 and so on, but in practice we wouldn’t actually know what they were. So, we measure the superposition, and we’d
randomly get one of the values as the output. Say, for example, 5 over 4347. And then we’d do the calculation again,
and get, say, 6 over 4347. And then 2 over 4347, and so on. Pretty soon we’d be able to tell that 1
over 4347 is the common factor of all of those, and so p is 4347. And you can check that 101 to the 4347 is
indeed exactly 1 more than a multiple of 314191 (though it’s a very very big multiple). So to get our better guess for a number that
shares a factor with 314191, let’s take 101 to the power of 4347 over 2 plus one-
oh, crap. 4347 is odd! So we can’t divide it by 2 and get a whole
number. So we have to start over. Well, let’s pick another random guess, say,
127. After going through the same process of creating
a simultaneous superposition of raising 127 to all possible powers and then doing a quantum
Fourier transform and so on, we’d end up finding that the value of p corresponding
to 127 is 17388. And so raising 127 to p over 2 gives 127 to
the 8694, plus or minus one, for our new and improved guess of a number that shares factors
with 314191. Using Euclid’s algorithm on 314191 with
127 to the 8694 + 1 gives a common factor of 829, and using it on 314191 with 127 to
the 8694 – 1 gives a common factor of 379. And 829 times 379 does indeed give us 314191!! So we can break the encryption and you can
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