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### কোর্স: কম্পিউটার বিজ্ঞান > অধ্যায় 2

পাঠ 7: দৈব অ্যালগোরিদম- দৈব অ্যালগোরিদম (সূচনা)
- শর্তাধীন সম্ভাব্যতা চিত্র ব্যাখ্যা করা হয়েছে
- মুদ্রাটি অনুমান কর
- দৈবভাবে মৌলিক সংখ্যা যাচাই (প্রস্তুতি)
- লেভেল 9: ভাগের প্রচেষ্টা বনাম দৈবভাবে ভাগ
- ফার্মার ছোট তত্ত্ব
- ফার্মার মৌলিক সংখ্যা যাচাই পরীক্ষা
- লেভেল 10: ফার্মার মৌলিক সংখ্যা যাচাই পরীক্ষা

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# ফার্মার মৌলিক সংখ্যা যাচাই পরীক্ষা

কীভাবে এবং কেন এটা কাজ করে সে সম্পর্কে দ্রুত একটি ধারণা। এটি তৈরি করেছে ব্রিট ক্রুজ

## আলোচনায় অংশ নিতে চাও?

কোন আলাপচারিতা নেই।

## ভিডিও ট্রান্সক্রিপ্ট

- [Instructor] Our goal is to define a series of instructions which can prove whether some input integer is
composite or else identify it as prime with some very
high degree of accuracy. Which we can define, and it must be efficient to do so. Meaning it should be fast
to calculate on any machine and even if the input size grows, which is our input integer N, it should still be fast. And so far we've learned
that at the lowest level this requires some known
pattern that all primes follow and very few composites follow. However, in the previous video
we did a visual demonstration of Fermat's little theorem and it provides us with
a very interesting rule. Given some prime number P
and some other integer A. Which is just less than P. We know now that P divides
A to the power of P minus A. We write this as A to the
power of P equals A mod P and that's the way you'll normally see it. Now because A and P share no common factor since P's a prime, we can use a cancellation law, and you'll sometimes see this written as A to the power of P
minus one is one mod P, and remember we can only do this because the greatest common divisor
of A and P equals one, and here I've just set up a demo so we can at first just
see this equation in action and get comfortable with it. Notice now if I input a prime for P, the output is always one
no matte what A I choose. If we input a composite number for P, we see that it usually isn't one, and anytime this equation output something that isn't one, we have a composite witness. This is now proof that the
P we chose cannot be prime, and that's the essence of this test, and before going any deeper, let's just step back and
construct the frame work for this test using this
pattern I just showed you. So our test starts. We are provided with some integer. Let's call it P, as input. Next we generate a random integer A. Which is less than P, and now we can as, is the
greatest common divisor of A and P one? If not, if the greatest
common divisor of A and P is greater than one, then
they share a common factor and we've proven that P is composite because a factor exists
and we can halt and exit and our algorithm will output composite. However, if yes, and we
can ask the key question, does A to the power of P
minus one mod P equal one? If not, we have a witness
that P is composite. We can halt and say we're
done, P's composite. Otherwise, if yes, if
our equation outputs one, then it should be prime, right? I coded up this series of instructions, and on the left hand side
we have Fermat's test, and on the right I just have
an existing trial division test and that's there because we
know that test is always correct and just at first glance
it seems like it's working. However, I found a problem. I hit the number 511,
and now the Fermat's test is saying it's prime, and
the trial division test is telling me that it's composite. 511 seems prime from the tests
perspective but it's not. Now let's just return back to our equation and see what happened. Well this is an example of
what we call a pseudoprime. It's a composite number
but there are certain A's we can choose that will output a one. That's incorrect again. So these A's which result
in a composite number outputting a one, we can call fools, because they're fooling us into
thinking the number's prime, but notice that if we
choose different A's, we seem to find many composite
witnesses instead of fools. So we could maybe just
step back and let's apply the same logic we used
in the divisibility test where we simply repeat
the experiment a few times and generate a new random A each time and hopefully we won't
pick a fool each time. Now it's been proven
that the number of fools must divide the total size
of the group we select from. Which means at most, half of the choices or half of the elements in
this pool could be fools. So since A is chosen randomly, the chance of finding a composite witness, which is what we want,
is at lest one half. After T iterations, the
probability that no witness will be found with a
composite number is at most one divided by two to the power of T. So after 20 trials the probability of mistakenly outputting a prime is just over one in a million. So that's the case where we just keep getting really unlucky
and generating random A's and every time we choose I fool. If you can let that sink in, that's really powerful to understand, and here now we can see our test in action with the increase number of trials. It seems to be working perfectly, and notice that in the worst case, which we know is when
we provide our algorithm with a prime, it's gonna do
the maximum amount of work. The Fermat test is much more
efficient than trial division. Especially because the number of steps doesn't scale with the input
and that's a key distinction. We set the number of trials and that's it. We never have to worry about our algorithm running millions and
millions of iterations as we did before. So I mean this is
quintessentially applied math. We are taking a pattern someone discovered and we're saving an immense
amount of computational cycles. However, there is one tiny
flaw for error in this system. Can you find it?