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  • 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 have your pie!

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As a followup to the main video about how quantum computers factor large numbers to

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