August 25

I have been searching for

Hi SF007. I think you meean Automatix (software), as "Automatix" redirects to a robot company.78.144.151.118 (talk) 10:26, 25 August 2008 (UTC)[reply]

Yes, yes, exactly, I already fixed the link, Thanks! SF007 (talk) 17:47, 25 August 2008 (UTC)[reply]

Unfortunately is has no seeds... :( SF007 (talk) 21:15, 25 August 2008 (UTC)[reply]

Automatix is no longer supported. Ultamatix has replaced it. --Russoc4 (talk) 18:22, 30 August 2008 (UTC)[reply]

I have following partitions on my HDD:

/devsda1 extended with /dev/sda5 and /dev/sda6 on it /dev/sda2 /dev/sda3 /dev/sda4

In sda6 and sda4 I have some Linux flavor installed. However, always when I try to boot from sda6, it boots from sda4.

Here is my menu.lst file:

 title Ubuntu1 (on /dev/hda4)
 root (hd0,3)
 kernel /casper/vmlinuz boot=casper root=/dev/ram ramdisk_size=1048576 rw acpi=off
 initrd /casper/initrd.gz

 title Ubuntu2 (on /dev/hda6)
 root (hd0,5)
 kernel /casper/vmlinuz boot=casper root=/dev/ram ramdisk_size=1048576 rw acpi=off
 initrd /casper/initrd.gz

What is wrong? —Preceding unsigned comment added by Mr.K. (talk • contribs) 11:25, 25 August 2008 (UTC)[reply]

Is it possible that your menu.lst is out of sync with what's actually written on your boot sector? --Sean 14:25, 25 August 2008 (UTC)[reply]

I recommended the super grub disk live cd in a previous thread, and I'll do it again. It might help you diagnose the problem. --NorwegianBlue ^talk 17:37, 25 August 2008 (UTC)[reply]

does anyone know...i just cant find it:( thanks--84.64.116.193 (talk) 12:49, 25 August 2008 (UTC)[reply]

You want to go to "Document Properties" and change the page size. —Bkell (talk) 13:50, 25 August 2008 (UTC)[reply]

You can't crop in the way you can in photoshop. You can just change the page size which means that elements outside of the page area won't be rendered or printed. But you can't crop a vector graphic in the same way that you can in photoshop with a raster graphic. It doesn't work that way. (Either in Inkscape or any other program, e.g. Illustrator) --98.217.8.46 (talk) 14:46, 25 August 2008 (UTC)[reply]

Isn't what you want the "Intersection" of a group of objects with another object? Just place a rectangle or any other shape on top of what you want to crop and perform a Path -> Intersection. --Juliano (T) 15:26, 25 August 2008 (UTC)[reply]

But how well this will work can depend on the complexity of the objects you are applying it to. I've had trouble getting things like that to work with very complicated (e.g. lots of nested groups) objects. --140.247.40.116 (talk) 20:55, 25 August 2008 (UTC)[reply]

Here's one possibility: Group the objects you want to crop, draw a rectangle over the objects, then select Object -> Clip -> Set. However, with this method the full objects are still stored. You can undo the clip by choosing Object -> Clip -> Release. --70.129.130.202 (talk) 01:23, 26 August 2008 (UTC)[reply]

Is it possible to print a document which is being scanned in at the same time like a photocopier using a scanner and printer? Clover345 (talk) 16:15, 25 August 2008 (UTC)[reply]

I don't think you can do that, at least not with most scanner/printer combinations. You can, however, make scanning and printing a little faster by scanning the next page while the current page is being printed. (You can start scanning the next page as soon as the previous page has finished scanning. You don't have to wait until it's come out of the printer.) --71.162.241.230 (talk) 16:53, 25 August 2008 (UTC)[reply]

This will depend on your hardware. My father's combined scanner-printer has a 'photocopy' button that works without any input from a computer. Algebraist 17:05, 25 August 2008 (UTC)[reply]

Does that scanner-printer start printing a page while the page is still being scanned? --71.162.241.230 (talk) 17:12, 25 August 2008 (UTC)[reply]

I don't know. Algebraist 17:13, 25 August 2008 (UTC)[reply]

My HP MPC does that, an old 1210 IIRC. --antilived^{T | C | G} 06:05, 26 August 2008 (UTC)[reply]

When you have a new computer, how do you take a snapshot of its initial software/hardware configurations so that in the future you can, if you want to, find out what changes later-installed software has made to the system? What are some good tools for this purpose? —Preceding unsigned comment added by 71.162.241.230 (talk) 16:47, 25 August 2008 (UTC)[reply]

On Windows (at least XP), Start -> Programs -> Accessories -> System Tools -> System Restore, and create a restore point. You'll need to keep track of the month/day/year of when you do it, since you restore by date. JeremyMcCracken (talk) (contribs) 03:05, 26 August 2008 (UTC)[reply]

Just a general note: So much stuff gets changed/edited/deleted in just the normal use of your computer such a comparison with comparison points more than a week apart would be hard to decipher. It IS possible, however. Let me see if I can dig some stuff up for your when I get home. --mboverload@ 03:08, 26 August 2008 (UTC)[reply]

Hi. I'm trying to do this problem using python, but I'm stuck at determining whether a number is a Lychrel number, since the function I'm defining to do it doesn't seem to be returning the right values. Here's the code:

#This function tests if n is a palindromic number, returning 1 if it is and 0 if it isn't.

   s = str(n)
   l = len(s)
   c = 0 #truth counter - is added to if the nth digit from the start is equal to the nth digit from the end. 
   if l <=1:
       return "Error"
   elif l % 2 == 0: #even number of digits case
       for i in range(0,l/2):
           if s[i] == s[l-(1+i)]:
               c = c+1
       if c == l/2: #compares c to what it should be if n is a palindrome
           return 1
       else:
           return 0
   else: #odd number of digits case
       for i in range(0,(l-1)/2):
           if s[i] == s[l-(1+i)]:
               c = c+1
       if c == (l-1)/2: #compares c to what it should be if n is a palindrome
           return 1
       else:
           return 0

   f = ""
   s = str(n)
   l = len(s)
   for i in range(0,l):
       f = f + s[(l-1)-i]
   return int(f)

   if c <= 5: #would be 50 for real testing, but this makes stepping through in the debugger quicker for known numbers.
       m = n + switch(n)
       if testpal(m):
           return 0
       else:
           testlych(m,c+1) #each call increases c by 1, so will be cut off by if statement above when needed
   else:
       return 1
       

print testlych(47,1) #should be 1, is 0

I think there might be something wrong with my understanding of the differences between 1/0/True/False/None, but I've written stuff before that uses the same approach, and seems to work. Thanks for any help you can give. 91.143.188.103 (talk) 17:09, 25 August 2008 (UTC)[reply]

It's not the best idea to return 1 when you mean True, even if a given language doesn't distinguish between the two, but strictly that shouldn't break things. But really I don't understand why testlych is recursive (wouldn't a loop do?). The following works, at least for the examples given on that article:

def reverse(s):
    result=''
    length = len(s)
    for x in range(0,length):
        result+=s[length-1-x]
    return result

def isPalindromicNumber(i):
    s = str(i)
    return s==reverse(s)

def testLychrel(i, maxIterations):
    iterations=0

    while iterations < maxIterations:
        if isPalindromicNumber(i):
            print 'resolves to palindrome %d after %d iterations' % (i,iterations)
            return
        
        i = i + int(reverse(str(i)))
        iterations += 1

    print 'failed'

-- Finlay McWalter | Talk 18:13, 25 August 2008 (UTC)[reply]

The original code was almost correct (although the rewrite is a lot cleaner). A call to testlych can go 3 different ways: it reaches the limit of 5 levels deep and returns 1, or it finds a palindrome and returns 0, or it recurses and returns... what?

testlych(m,c+1)

The value returned from the recursive call is not used. And what happens after the recursive call? The inner "else" branch is finished, and the outer "else" branch is not entered because the outer "if" was true. The next thing that happens is therefore falling of the end of the function, which is where your "None" came from. What it should do is return the value that the recursive call returned, like this:

return testlych(m,c+1)

Also, the "should be" comments by the test cases are wrong but that won't be too hard to fix. And yeah, the whole thing would have been more clear without recursion, but this looks like a beginner-level programming exercise. Let's not just tell the beginner "your approach sucks, start over" when they were just 1 keyword away from an working program, however convoluted it may be.

(Sidebar: doesn't python have a built-in string reverse function? It's weird seeing 2 people in a row reinvent that wheel. perl's got one...) --tcsetattr (talk / contribs) 07:40, 26 August 2008 (UTC)[reply]

I didn't say anything that remotely resembles "your approach sucks", I said "I don't understand why"; my code is essentially the OP's code tidied up a bit. Python does have a string reverse feature, where s[::-1] is the reverse of string s (cf Python extended slice notation), but I used essentially the author's approach so he'd know what was going on. -- Finlay McWalter | Talk 09:30, 26 August 2008 (UTC)[reply]

Thanks, I rewrote it using a loop and that worked, but now I get why what I had wasn't working. 91.143.188.103 (talk) 16:18, 26 August 2008 (UTC)[reply]

Why the heck are you guys messing around with strings? In C++:

 bool isPalindromic ( int n )
 {
   int rev_n = 0 ;
   while ( n > 0 )
   {
     rev_n = rev_n * 10 + n % 10 ;
     n /= 10 ;
   }
   return n == rev_n ;
 }

Way faster - using much less memory and no external libraries.

SteveBaker (talk) 00:28, 27 August 2008 (UTC)[reply]

By the time you get to that return, won't n be 0, so will always return false? That's what happened when I tried translating it into python. Copying over the value of n to use in the calculations, and using the original n for the comparison, I get

def isPalindromic(n):
 	rev_n = 0
 	new_n = n
 	while new_n > 0:
 		rev_n = rev_n * 10 + new_n % 10
 		new_n = new_n/10
 	return n == rev_n

which works. (as an aside, this is the first time I've come across when python's floor division is actually helpful, rather than an encumbrance). 91.143.188.103 (talk) 14:25, 27 August 2008 (UTC)[reply]

A shopping website says windows Vista home basic OEM DVD (for sale). What does the OEM here mean? Is it the normal windows vista? Can I buy and will I be able to install it ? —Preceding unsigned comment added by 59.92.108.49 (talk) 17:23, 25 August 2008 (UTC)[reply]

OEM means original equipment manufacturer. It ls the normal Windows Vista. It has one limitation, however: once you install it on your computer and activate it, it will be locked to that computer, and you will not be not permitted to install it on a future computer that you buy, even if you uninstall it on the first computer. --NorwegianBlue ^talk 17:43, 25 August 2008 (UTC)[reply]

Plus, support provided by Microsoft for the product is more limited. I think you get one free support incident, which expires after 1 year, or something like that. After that, it's $$$. -- Consumed Crustacean (talk) 01:30, 26 August 2008 (UTC)[reply]

I have a really old computer that was given to me long ago, running

A piece of diagnostics software like

PC100, but of course it doesn't have to be. Another way to go about it would be to look up serial numbers stamped on the parts and simply feeding them into Google; getting a positive identification that way is pretty likely. Good luck, you may well have your work cut out for you, though. -- Captain Disdain (talk) 19:03, 25 August 2008 (UTC)[reply

]

Before you rush out and buy a memory stick - you might want to check places like eBay and Craigslist for used computers. For something that old, it might actually be cheaper to buy an ancient used computer than to buy that memory stick. As components like that become obsolete, their new prices go up - not down! SteveBaker (talk) 18:21, 26 August 2008 (UTC)[reply]

What is a good free replacement for the wireless control in windows? Something I can download. Thank you. 79.75.211.94 (talk) 18:39, 25 August 2008 (UTC)[reply]

If it was possible to download the entire terrain map of google earth, how much data would it consist of? —Preceding unsigned comment added by 86.162.38.227 (talk) 18:44, 25 August 2008 (UTC)[reply]

I can't find anything about the terrain (elevation) data, but in late 2006 the raw imagery took up 70.5

terabytes. It's likely somewhere in the order of hundreds of terabytes now. — Matt Eason ^{(Talk • Contribs)} 19:18, 25 August 2008 (UTC)[reply

]

Does Google even offer it for download at once? Kushal (talk) 20:42, 25 August 2008 (UTC)[reply]

The elevation data probably doesn't take up more than a few tens of gigabytes. Heightfields don't take up much space, and irregular meshes take up even less. --Carnildo (talk) 23:47, 25 August 2008 (UTC)[reply]

What i meant was how much data would the entire satellite map be? —Preceding unsigned comment added by 86.136.40.38 (talk) 17:04, 26 August 2008 (UTC)[reply]

Well, their maximum resolution is tremendously variable - in places, they have aircraft-based photography down to as little as 10 cm per pixel - in others it's ancient "Landsat 7" satellite data that is at 15 meters per pixel. They don't hold photos of oceans at significant resolution so we can neglect that and concentrate only on the land area. The earth has about 150 million square kilometers of land area. So, we can come up with a range of values.

• The most it could be would be if the whole thing was held at maximum quality (10cm per pixel) with 3 bytes (red+green+blue) per pixel and no 'lossy' data compression: then you'd need 3 x 10,000 x 10,000 bytes per square kilometer (300 Mbytes) for a total of 150Mx300M bytes. That's 45,000 terabytes. However, those images would certainly be highly compressed - you can easily get 50:1 image compression if you aren't too fussy about quality - but you're still down to 1,000 terabytes. That's not an unreasonable amount for a large company like Google - you could store it on a computer cluster costing maybe a couple of million dollars...but it's WAY too much to have at home!

• At the other end of the scale, much of the wild/undeveloped places of the world has only 15 meter resolution satellite data. Then you only need 3 x 66 x 66 = 13,300 bytes per square kilometer, 150M x 13K = 1,950 Gigabytes without compression - 40 Gigabytes with aggressive compression. That's certainly do-able. Most computers have drives of at least 100Gbytes these days...it would fit handily on a single dual-layer Blu-Ray disk. So it's a fair chunk - but definitely do-able.

I believe that Google's elevation data came originally from the NASA Space Shuttle mission that used radar techniques to measure the elevation of the entire earth to about 5m precision. You can actually legally download and use that from the NASA website. Height data takes a couple of bytes per "pixel" - but it's much more compressible than photographic data because there terrain typically slopes fairly gently and there are large flat areas in any typical chunk of terrain. Storing it as a pre-triangulated "mesh" is an effective means of compressing the data - and it makes it easier to draw. I did actually store this stuff on my PC once - but I don't recall how big it all was.

So for the color data - the answer is somewhere between "impossible" and "just about do-able" - depending on what percentage of the earth's surface is stored at high resolution. Sadly, I have no clue what that percentage is...but I think it's safe to say that it's going to be too much to store on a "normal" home computer. For the height data, it's definitely in the "do-able" range.

SteveBaker (talk) 18:13, 26 August 2008 (UTC)[reply]

Can iPods support videos from Youtube? --88.104.205.21 (talk) 19:20, 25 August 2008 (UTC)[reply]

Yes, please download the flv file with some kind of flv download helper (available as addons on the ever more popular

Mozilla Firefox), and then use iSquint

to convert it into an mp4 file. iSquint can add the product to iTunes for you and then you know how to copy it to your iPod. (Help is available on helping move files from iTunes to the iPod as well, just let us know you need help.)

Kushal (talk) 20:41, 25 August 2008 (UTC)[reply]

Greasemonkey for Firefox and and one of the best Youtube scripts ever go to here to find downloads. [2]. Rob^{Not an admin}  06:35, 28 August 2008 (UTC)[reply]

I have a Lexmark 2480 series printer (looks like this). Its serial number is 05141445687 2007/04. The CD says "Installation software and user's guides for use with Windows and Macintosh" but I AU CNET says the printer does not work with Mac. Please help. Kushal (talk) 20:33, 25 August 2008 (UTC)[reply]

Which version of the Mac OS? And have you tried just, you know, adding the printer in OS X? Leopard includes several Lexmark printer drivers... -- Consumed Crustacean (talk) 01:15, 26 August 2008 (UTC)[reply]

I am on Tiger 10.4.11 :( Any hope? Kushal (talk) 02:40, 26 August 2008 (UTC)[reply]

Is it the same as a Z2400? According to [3], that seems to be bundled with Tiger. Again, try just adding the printer through OSX's printer utility, or http://127.0.0.1:631. If it's not bundled, the drivers should be downloadable at [4] (or some page linked off it), if they're available at all. If they're not, and they're not on the CD, then you're screwed. -- Consumed Crustacean (talk) 02:48, 26 August 2008 (UTC)[reply]

I'd trust the manual more than a website. One thing- it's not an old enough printer to have drivers for OS 9 is it? JeremyMcCracken (talk) (contribs) 03:08, 26 August 2008 (UTC)[reply]

If it's the X2480 (Again, there are some naming confusion here. Is it X or Z or some other letter? On the Lexmark website, the only product that matches 2480 is X2480.), OpenPrinting lists no open-source drivers for X2480, which likely means that OS X doesn't come with drivers for it. Existing threads on the OpenPrinting forum seeking drivers for similar X2450, X2470 turned up absolutely nothing. --71.141.120.68 (talk) 03:55, 26 August 2008 (UTC)[reply]

This is what the printer looks exactly like. I am not sure what model the printer is. The CD that comes with the printer says it is a 2480 series. Could someone help me, please? School is starting and having a printer, scanner, copier would be awesome. I don't need to do a lot of printing in my room but having a printer/scanner in my room would be very convenient. Kushal (talk) 16:11, 26 August 2008 (UTC)[reply]

I'm getting the feeling I'm typing to a brickwall here.

1. Try the CD
2. Try simply installing the printer to see if the driver is bundled with OS X

If those fail, you're screwed. Simple as that. On the Lexmark site, I see that exact printer listed as "X2480", and they do not have OS X drivers available for download. They have Mac knowledge base articles linked from it, which might mean that the driver is included in OS X, but it might mean absolutely nothing. -- Consumed Crustacean (talk) 19:50, 26 August 2008 (UTC)[reply]

Thanks for your input. I know chances are slim but is there a slim possibility that I be able to use the printer for scanning, at least? Kushal (talk) 03:37, 27 August 2008 (UTC)[reply]

No response? I am pretty sure an industrious programmer working at home would have figured out how if Lexmark just freed the software at their end. Big bummer. Boo, Lexmark! Kushal (talk) 19:29, 27 August 2008 (UTC)[reply]

Is there a way to encapsulate some text in a SPAN in a way to keep it from breaking?

That is, I know non-breaking spaces can be used to keep spaces from breaking. But I want to keep the contents of an arbitrary SPAN from breaking.

In this particular case, I have something like this:

<img src="someimage.png">.¹

I know an inline IMG is not exactly best practice but for this particular application it's sort of the best thing I've found to work (using pure-CSS solutions fails more than it works in this particular case when it comes to cross-browser compatibility).

What properties should I give to the SPAN to keep the browser from breaking the line at the period and entity? One obvious way is to encapsulate the entire thing in a TABLE element but that's about the worst idea I can think of from a standards point of view. --140.247.40.116 (talk) 20:49, 25 August 2008 (UTC)[reply]

Setting white-space to nowrap should do the trick — Matt Eason ^{(Talk &#149; Contribs)} 20:57, 25 August 2008 (UTC)[reply]

Brilliant, thank you! --140.247.40.116 (talk) 21:21, 25 August 2008 (UTC)[reply]

Hello Wikipedia,

I would really like your thoughts on online storage companies. Essentially, rather than spend my money on a big clunky laptop with lots of storage space, i'd love to take advantage of having a good internet connection and buy a light, portable laptop and store all my work (Adobe illustrations etc) online. (i do have a normal external hard drive but really hate using it -its just too 'fiddly' plugging it in and out.. i'm quite badly organised.)

My two main concerns are: 1) Is there a reliable 'brand' that won't lose my data or make it difficult to get at.

2) do they generally offer some guarantee of privacy? i don't mind a computer looking at my stuff (and then targeting adverts accordingly -e.g. gmail) but i do mind a human doing it.

Is my web 2.0 dream a possibility? if any knows any good companies then that would be great.

Thanks,

82.22.4.63 (talk) 21:13, 25 August 2008 (UTC) p.s. i'm in the UK if that makes any difference at all[reply]

1) Go for bigger brands. With the size of their infrastructure, they are more likely to be reliable.

I suggest you avoid

Web2

. It is not yet widespread, and I have found that the tried and tested stuff just works, simply.

Have you considered the size of Adobe Illustrator files? They might be large.

What platform have you got? If it's

Mac OSX, avoid MobileMe

for the meantime. They have had lots of problems.

Are you going to use a backup utility, like

Time Machine (software)? Possibly rsync?My name is anetta (talk) 22:02, 25 August 2008 (UTC)[reply

]

Amazon S3?My name is anetta (talk) 22:12, 25 August 2008 (UTC)[reply]

How fast is a "good internet connection"? For comparison, a USB hard drive is equivalent to about a 480 megabit per second connection, while an internal hard drive is about 1500 megabits per second. --Carnildo (talk) 23:56, 25 August 2008 (UTC)[reply]

Personally, I don't like the idea of relying exclusively on someone else to protect data that I value. Two things that make backing up to the external drive easier: (1) Reorganizing my data so that for the most part I select only a few top-level folders (e.g., one for work stuff, one for photos), and (2) having backup software that makes it easy to regularly do full backups ("copy everything in these folders") and, on a daily basis, do incremental ones ("copy everything that's new or changed"). The typical daily backup takes less than 4 minutes; the full backup, around 30.

This doesn't relate directly to your question, but for convenience I use a small docking station for my laptop. All the peripherals (printer, external hard drive, wireless connection for the mouse) are plugged into the docking station; reconnecting my laptop means only two cables (power cord and USB to docking station). I keep a spare power cord, spare mouse, and 2-gigabyte USB drive in my laptop bag. No crawling under the desk; much less fiddly. — OtherDave (talk) 09:14, 26 August 2008 (UTC)[reply]

For a simple backup of a single file I was working on, I used to "Save as" every time I opened it, and rename it as a sensible name and then followed by a number, which increased by one each time I saved it. Suppose I began with John, I would "Save as" "John2", then "John3" and so on, which was a defence against a file becoming corrupt. Of course, space wasn't at a premium, and you need to remember the number you saved as.My name is anetta (talk) 09:21, 26 August 2008 (UTC)[reply]

I downloaded X11 from the Apple website, but every time I try to install it I get an error message saying that 'a newer version is already installed'. I have looked for this newer version on my iBook, but the Mac can't find it (it only finds the one I downloaded). I am sure I didn't install it from my Tiger DVD (which I have left at home - I am away for a few days), so the only version I should have should be the one I just downloaded. Can anyone explain what is happening here and is there any possible fix for this? Regards.--

If you do have X11, it'll be in Applications > Utilities > X11.app. Check to make sure you don't have it in there. --98.217.8.46 (talk) 14:26, 26 August 2008 (UTC)[reply]

I recall that you couldn't download

Mac OS X v10.4 (Tiger) or later from the Apple website (the download is only for 10.3?). X11.app is included as an optional install on the original install DVD. If you don't have the DVD, you can also try to download it from this unofficial site (I am not endorsing this site). --Spoon! (talk) 18:18, 26 August 2008 (UTC)[reply

]