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3 points by jmatt 5821 days ago | link | parent

This is readable and just fine if you are handing it in for a class.

The benefit of the pascal triangle is that through a simple number triangle you can find binomial coefficients. It has since been redefined as the binomial coefficient. But historically it was interesting because it could provide binomial coefficients through simpler math than n choose k.

If you have to calculate the whole triangle, then I would probably recursively define it and calculate it that way. (Pascal's formula shows that each subsequent row is obtained by adding the two entries diagonally above) Instead of constantly calculating it at each position with n choose k.

I guess you could use the binomial coefficient definition and memoize the choose and factorial. Memoization alone is what is saving you here. Luckily arc provides this functionality.

Anyways, this is a fun problem. Much easier to solve elegant and efficiently in arc, thanks to memoize. Try this in an imperative language and you'll definitely have to fall back on the recursive definition because all those factorials and chooses will kill the performance. There is one other benefit to using the recursive definition - you won't use any extra memory memoizing factorials. Instead you'll just have the chooses appearing in the tree.



1 point by lacker 5820 days ago | link

A bit nit picky, but memoization alone won't make the solution optimal, since you are dealing with very large numbers. n! has O(n log n) digits, so in the limit it will take O(n log n) time just to calculate n! from n * (n-1)!.

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1 point by jmatt 5820 days ago | link

I agree memoization isn't optimal.

But I don't know about your big-O estimate. Since the triangle is constructive and you are memoizing calculating n! from n * (n - 1)! will take O(1). It'll be the cached (n - 1)! value times n.

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1 point by lacker 5815 days ago | link

Doing a single multiplication is not O(1) when the numbers have more than a constant number of digits. I'm not precisely sure how long multiplying large numbers takes, but it's at least linear in the number of digits involved. And (n-1)! has O(n log n) digits.

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1 point by eds 5820 days ago | link

I don't follow how n! has O(n log n) digits. Mind explaining?

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2 points by lacker 5815 days ago | link

No prob. Sorry if I go on at too much length here ;-)

So a useful formula for counting digits is, a positive integer x has floor(log10(x)) + 1 digits. You can figure this out yourself by thinking, where does the digit-counting function bump up a notch? At 10, 100, 1000, etc. So asymptotically the number of digits in x is O(log x).

So n! has O(log n!) digits. The trickier part is figuring out or knowing that O(n log n) = O(log n!). Using log ab = log a + log b you can expand out the factorial:

  O(log n!) = O(log (n * (n-1) * ... * n * 1))
            = O(log n + log (n-1) + ... + log 2 + log 1)
            = O(n log n)
In case the last step isn't clear, you can do this splitting-in-half bounding trick. Since each element in the sum is less than log n you can bound from above with

  log n + log (n-1) + ... + log 2 + log 1 < n log n
And if you just take the larger half of the list you can bound from below with

  log n + log (n-1) + ... + log 2 + log 2 > log n + log (n-1) + ... + log (n/2)
                                          > (n/2) log (n/2)
which is itself O(n log n). So O(log n!) = O(n log n).

In general the rules of thumb you use to reduce O(log n!) are:

  1. complicated expressions inside factorials are ugly, you should simplify them
  2. O(sum of n things) is usually O(n * the biggest thing)
Make sense?

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1 point by kens 5820 days ago | link

You're multiplying O(n) numbers, each of which is O(log n) digits long, so the result is O(n log n) digits long.

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1 point by shader 5820 days ago | link

Usually, it seems to be either (n log n) or (n log n) - 1 digits.

And usually in this case I would leave of the O, as that usually refers to the performance of an algorithm in time or space. I suppose you could construe the number of digits to be "space" but multiplying O(n) numbers doesn't make that much sense.

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