Wikipedia:Reference desk/Archives/Computing/2007 December 25

= December 25 =

.ivr format
What is .ivr format? I have a video that was sent to me in email, that plays with Realplayer with no problem, but I don't know what the .ivr format is. Corvus cornix talk  00:37, 25 December 2007 (UTC)
 * Internet Video Recording (IVR) is a proprietary video format used by version 11 and later of RealPlayer when streams are saved on disk. › mys id (☎∆) 08:41, 25 December 2007 (UTC)
 * Thanks.  Corvus cornix  talk  18:27, 25 December 2007 (UTC)

Nokia 2610 USB port?
at the bottom of a nokia 2610, between the charger plug and the headphone plug is a port which looks like a USB port. Does anyone have any idea what that is? the phone has a TOTAL memory of 2 MB so the port can not have much use as a data port, can it? Kushalt 03:01, 25 December 2007 (UTC)

Graphics Card
I have purchased a game that requires "Nvidia Geforce 6600 or better, or ATI Radeon 9800Pro or better" I have tried searching around to find out what better is, and cant really come to a conclusion. Is there some sort of list or ranking of the cards that defines what, in this context, the better cards are. I currently have a Radeon X600, and it isn't new and wasn't state of the art when it was, so I'm assuming that it wont work, especially since it has trouble with a lower spec game. Though this seems partly because the game thinks it doesnt support Direct X 9.0c. So can someone offer me guidance to supported cards and how good they are? ΦΙΛ Κ 15:23, 25 December 2007 (UTC)
 * I use gpureview to compare the cards. Sorry, looks like you're out of luck :( Also, how good am I answering questions on christmas morning!! -- ⁪ffroth 15:58, 25 December 2007 (UTC)


 * So you think this card would do the trick? And if I did purchase that card, how long would it be before I'd be in a similar situation where retail RPG games would be demanding more than this could give again? ΦΙΛ Κ  17:11, 25 December 2007 (UTC)
 * Not exactly high-end but it's still a very solid card. You should never count on being able to play future games (it'll be low range in less than a year and worthless in 2-3) but it will be able to play basically all games out now if you have a decent cpu. -- ⁪ffroth 17:23, 25 December 2007 (UTC)
 * I dont want to overspend, as I only play games for a bit of fun, nothing remotely serious, but I'd like a card that is still mid range within a year, is that achievable? Or am I being overly hopeful. 172.200.130.39 (talk) 20:15, 25 December 2007 (UTC)

Optimising Euler Project 5 in Ruby
Below is Euler Project Problem 5, implemented in Ruby. However, the algorithm takes ages I stopped it on about 100,000,000 and I was wondering if you could provide some suggestions for speeding it up? Also if anyone wants to tell me if I'm close or if there is an error and I've stepped over my bounds, please do. The program works fine for faster cases of range_low and range_high. 217.44.16.75 (talk) 15:35, 25 December 2007 (UTC)


 * 1) Find the smallest number divisible by every number between range_low and range_high.

satisfy = false range_low, range_high = 2, 20 target = range_high

while (satisfy == false)

for i in (range_low..range_high)

if (target % i == 0) puts target.to_s + " mod " + i.to_s + " OK" satisfy = true if i >= range_high else break end

end

target += range_high unless satisfy == true

end

puts target

gets


 * The result grows rather rapidly as the range gets longer: in particular, if the range is 1 (or 2) to N, then the answer will be the primorial of N . To avoid integer overflow errors, you should probably be using BigNums.  Also, I don't think   is quite correct; if the upper end of the range isn't a prime number, the answer doesn't necessarily have to be a multiple of it (just consider the range from 3 to 4).


 * In any case, your algorithm is an extremely inefficient one, since it ends up trying a lot of wrong answers. It would be more practical to write a routine to compute the least common multiple of two numbers (hint: use Euclid's algorithm) and apply it repeatedly.  Note that you'll still need to use BigNums to represent the answer if the range isn't very short.  —Ilmari Karonen (talk) 18:15, 25 December 2007 (UTC)


 * I think you have misunderstood the problem statement. It seems that we are looking for the LCM of all numbers in the range. It doesn't have anything to do with primorials, and by definition it is a multiple of any number in the range, the high end included.


 * You're right, I'm not sure what I was thinking when I wrote the parts which I've now struck out. *slaps forehead*  Probably something like "lowest number that shares a prime factor with each number in the range".  —Ilmari Karonen (talk) 21:22, 26 December 2007 (UTC)


 * I agree that this is extremely inefficient, and that the sane method to do this is using the formula $$\textrm{lcm}(a,b)=\frac{ab}{\textrm{gcd}(a,b)}$$ repeatedly, each time adding the next number in the range, and using Euclid's algorithm for finding the gcd. -- Meni Rosenfeld (talk) 13:08, 26 December 2007 (UTC)

USB Hard drive structure
Yesterday my cousin opened his USB hard-disk(these HDDs are also known as portable hard drives) and found that USB hard-disk is just a laptop hard-disk with a USB interface attached to it. Are all USB hard-drives are made up with same structure? —Preceding unsigned comment added by Muhammad Hamza (talk • contribs) 17:48, 25 December 2007 (UTC)


 * I wouldn't claim that all USB harddisks are built like this, but those I've seen either contain a 2.5" laptop harddisk or a regular 3.5" harddisk, with a simple USB/IDE or, these days, an USB/SATA converter. I assume it's cheaper to use standard harddisks than building special ones with USB interfaces directly on the circuit board. --Dapeteばか 18:09, 25 December 2007 (UTC)


 * You can even buy an empty chassis 2.5 or 3.5" that includes the USB interface and plug in your own HD off the appropriate size. It's a good option if you already have a suitable drive knocking around or if you want to replace the drive that's in your machine. A 3.5" drive might be designed to take less trauma than a 2.5" drive though. --Seans Potato Business 22:41, 25 December 2007 (UTC)