Name: Shor的算法如何影响314191（How Shor's Algorithm Factors 314191）
Uploaded: 2021-04-29T03:49:09.000Z
Duration: 5 min 52 s
Description: 【看影片學英語】數萬部 YouTube 影片，搭配英漢字典即點即查，輕鬆掌握單字發音與用法，長久累積看電影不必再看字幕。

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

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

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,

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,

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

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

So, we measure the superposition, and we'd randomly get one of the values as the output.

And then we'd do the calculation again, and get, say, 6 over 4347.

Pretty soon we'd be able to tell that 1 over 4347 is the common factor of all of those,

And you can check that 101 to the 4347 is indeed exactly 1 more than a multiple of 314191

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.

So we can't divide it by 2 and get a whole number.

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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