I hope the homework went (is going) well!

Tomorrow (today) we’ll be talking about totally different. We’re gonna dive into some randomized algorithms with the Schwartz-Zippel Lemma.

Try to look it up on the internet before class so you’re not totally lost but here’s some basics:

We have a polynomial a_0 + a_1 x + … + a_d x^d that we’d like to check if it’s the zero-polynomial. That’s easy if we have it in coefficient form (just check if they’re all zero!). But what I give you something like (1 – x)(x + 2x^2 -19x^2)…(1 – x^3)? Well you’d have to multiply it all out (which takes exponential time) and check the coefficients.

😦

Schwartz-Zippel thought: what if you check that polynomial of degree d at some random point? If the value is not 0, then the polynomial can’t be the zero-polynomial. If it was zero, then there’s a good chance the polynomial is zero everywhere. So just repeat this test a bunch of times and if you always get zero, then its probably zero.

It gives us some cool applications so look forward to that!

What a great lecture today!

We first looked at the algebraic or deterministic way of determining if a polynomial is identically 0. You just have to check d+1 points for a polynomial of degree d. That’s great! So why bother with the randomized approach?

If you have a polynomial of degree d in two variables, then you have to check (d+1)(d+1) = d^2 + 2d + 1 points. Now it looks like our linear time algorithm is going to pay off.

We’ll look at how the probability bound changes when we go from one variable to several variables on Friday.

At the end, we looked at a very interesting application of this. We have two sets A and B. We want to check if they’re equal. Well, just sort ’em and check each element one by one. That takes O(n lg n) time, but what if n is huge??

Let A = [a_0, …, a_(n-1)] and B = [b_0, …, b_(n-1)]. Let P_A = (x – a_0)(…)(x – a_(n-1)) and P_B = (x – b_0)(…)(x – b_(n-1)). Now we can check if A = B by asking if (P_A – P_B) == 0. Using our probabilistic algorithm, we can pull this off in linear time and get the correct answer with very very very high probability.

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