Older blog entries for fzort (starting at number 60)

28 Mar 2007 (updated 28 Mar 2007 at 10:47 UTC) »
Posts longos em português - eu não gosto

jarod: Advogato is an international forum. Please try to use English. It's a matter of common courtesy.

5 Mar 2007 (updated 5 Mar 2007 at 18:16 UTC) »
fejj: another trick is to use use bubble sort when the partition size gets below a certain threshold.

You might also like to check out the paper A Killer Adversary for Quicksort, in case you haven't already. :)

22 Feb 2007 (updated 22 Feb 2007 at 21:10 UTC) »
Bitwise 2007

So, Team Bandersnatch (bi and I) got in the top-50, which entitles us to t-shirts. I did one problem, he did three. :)

I'd post my solution to problem 6, like he did, but it looks fugly - it took a lot of hammering to make the square peg fit in the star-shaped hole.

Some of the problems were hardcore. None of the teams did problems 3 or 4.

Buzzword Assault

A company called D-Wave demo'ed an alleged superconducting 16-qubit quantum computer. Their PR department seems to claim it's able to solve NP-complete problems in polynomial time, which sounds like BS, but the thing looks pretty darn cool anyway. In both senses of the word (ha, ha).

2 Feb 2007 (updated 2 Feb 2007 at 09:09 UTC) »
Google Code Jam, Round 2

I got pwned. Big time.

That is all.

1 Feb 2007 (updated 1 Feb 2007 at 15:59 UTC) »
fejj: the standard way to do that is to replace the division with a multiplication by the reciprocal, multiplied by some power of 2. Then divide the result by that power of 2, which can be done with a right shift. In the case of division by 7, you'd end up with the increasingly more accurate expressions:

q = ((a<<3) + a) >> 6

q = ((a<<6) + (a<<3) + a) >> 9

q = ((a<<9) + (a<<6) + (a<<3) + a) >> 12

q = ((a<<12) + (a<<9) + (a<<6) + (a<<3) + a) >> 15

...

and so on (in binary, 1/7 is .001001001...). Optionally, the result can be fixed with something like

r = a - q*7; while (r >= 7) { q++; r -= 7; }

The multiplication by 7 can be replaced with shifts and adds, of course.

31 Jan 2007 (updated 31 Jan 2007 at 13:24 UTC) »
Google Code Jam, Round 1

My performance was dismal. I did the easy problem, then moved straight to the hard one. Then I threw away 20 minutes because I jumped straight to the code before understanding completely the problem statement and then had to start all over again, wasted another 20 minutes tracking down a stupid off-by-one bug, panicked for about 5 minutes, and finally ended up submitting a /wrong/ solution. Guess I need to work on the think-crush-under-pressure thing.

What makes it so frustrating is that the problem was not really that hard, and I was able to concoct a correct solution (here) without major difficulties afterwards, when the time pressure was gone.

Gah.

24 Jan 2007 (updated 24 Jan 2007 at 19:19 UTC) »
Google Code Jam

The qualification round was easier than I expected. It's probably just meant to filter out the curious. I was able to do both problems in about 30 minutes, and qualified to the next phase.

Tupper's Self-Referential Formula

... seems to be making the rounds on the internet. Two different people sent me the link. It's cute, but there's nothing magical about it, really - perhaps if some people look at it as a simple Perl script they'll realize that it's just a way to rasterize the big number, and the bitmap could be anything.

23 Jan 2007 (updated 23 Jan 2007 at 00:19 UTC) »

So, tomorrow is Google Code Jam Latin America. I'll give it a try, but I expect to get my arse kicked. Although I've been practicing a bit at SPOJ and acm.uva.es, I really suck at thinking under pressure. Ah well, let's see how it goes.

OpenSpecies:

1.


#define is_odd(x) (((x)&1) != 0)
#define is_even(x) (((x)&1) == 0)

2. if you're out to disprove the Collatz conjecture, use GMP or something.

30 Nov 2006 (updated 1 Dec 2006 at 00:13 UTC) »
mchirico, fejj: it's the Look and Say Sequence. See also: Conway's Cosmological Theorem.

It's all demonstrated in this wonderful IOCCC winner (hint file).

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