We assign to N the value N 1, so that the changes the value N refers to.

What is the smallest possible value for n?

By choosing a suitable, suitable large value of n.

The value for N is specified using ticks N within the declare blocks' s directive section.

We also have to change the value of n to keep reducing n as we go.

Removing the minimum value of a heap is log n.

We've now shown that for each value of n, T(n) is always upper bounded by 4n.

Now N refers to a value 4, but that doesn't change the value A refers to.

And how many of those will n have the same value?

The result is multiplied by the current value of n à   n is decreased by one.

We put in the value of n, and we call it again.

I'll just make a table mapping N to the value of Fibonacci of N. 1, 1, 2, 3, 5, 8.

The name N now refers to the same value as A referred to, but we can't mutate that value.

Is it n, n 1 over n, n over n 1, n divided by n 1, n 1 divided by n, or is it n squared 1 over n 1?

We're going to start with n as the value that was passed in.

This loop runs over all pairs of nodes n², but it does this from each value from 0 to n 1.

So whatever was the value of n, which is the input, the result of inc is the value, plus one.

So, here's how delta debugging works first, we split the input into n subsets where initially n has a value of 2.

Θ(n²), O(n²), and O(n³).

We want to get the value of n squared, and that's what we return.

Question 2 In how many of those frames will n have the same value?

Otherwise, we'll figure out the value of the chart by calling ourselves recursively on n 1 and n 2 and adding them together.

Okay. So basically (log n)⁷ beats 9n(log n)² and n² ³ beats (log n)⁷ and 9n(log n)² beats n² ³.

Then you're going to get a low p value just because you had big n.

Now, if K is n 2 that's n times n 2 where n².

So I subscript n by n is the n by n identity matrix.

So altogether we're talking about n n² or n³.

n times n minus one times n minus two.

If we divide through by n, we get 2 n n and divided by n is actually n.

Russian(n, n) for lots of different values of n.

Log n is piddly as n gets big compared to n, and n is actually pretty piddly as n gets big compared to n².

n

N

N

N

N

N

N

Keep shoot 'n. Keep shoot 'n.

Here we have n n 1.

Could there be n 1 ways, n 1 squared, n 1 factorial, or 2 n 1?

Now, she hides everything except the value of n and e, because n and e make up her public key, think of it as an opened lock.

A way to define factorial isà   for any value n, we can compute the factorial, which is the number of ways of arranging n items.

So, it's really n times n plus m which is (n² nm).

With the very long list that were wanting to tap values of, you might as well just sort the whole thing. n, well what happens with n? n where look we're comparing n log n to n( n) which is n³ ². n³ ² is asymptotically larger than n log n so we're still better off just sorting the whole list.