Hi.

   My question is one of mathematics. Does a program exist which can solve for X in the following equation, where Y and Z are known?

Y^X=Z

   Thanks. Rocketshiporion Thursday 19-August-2010, 5:43am (GMT)

Hi there. If you're just doing a quick calculation or two, you can do it with a calculator using logarithms. If Y^X=Z, then X log Y = log Z, so X = (log Z / log Y). You could put this into the language of your choice if you wanted to automate it. Brammers (talk/c) 08:07, 19 August 2010 (UTC)[reply]

Another way to view this is (and get the same result) by the definition of log. ${\displaystyle \log {a}=b}$implies ${\displaystyle 10^{b}=a}$(if we are using log base 10, but any base will do for this definition). So, assume we are using log base Y. You have ${\displaystyle Y^{X}=Z}$, which implies ${\displaystyle \log _{Y}{Z}=x}$. The computer will not have log base Y, but to convert bases, you just divide by another base: ${\displaystyle \log _{Y}{Z}={{\log {Z}} \over {\log {Y}}}}$. And, from above, we know that ${\displaystyle {{\log {Z}} \over {\log {Y}}}=X}$. -- kainaw™ 15:03, 19 August 2010 (UTC)[reply]

The above assumes we're talking about real numbers. More generally, working with complex numbers, the solution may be multivalued -- see complex logarithm. 198.161.238.19 (talk) 14:50, 20 August 2010 (UTC)[reply]

Hi how to create a key stroke logger (software) using C++. Can you please give the details and the code?117.204.3.54 (talk) 13:10, 19 August 2010 (UTC)[reply]

Split off into new section Rojomoke (talk) 14:54, 19 August 2010 (UTC)[reply]

See Keystroke logging for details and google "open source c++ keylogger" for source. --Sean 15:19, 19 August 2010 (UTC)[reply]

There is an article on key logging at Keystroke logging. Your current request is overly broad...how do you plan to implement the key stroke logger: with an api hook, a blue pill hypervisor, or with a rootkit. The implementation is often system-dependent.Smallman12q (talk) 15:23, 19 August 2010 (UTC)[reply]

Without objecting to the question, can we stop this conversation somewhere short of giving explicit instructions on how to create one's own password-stealing trojan? If someone's going to try to hack into my machine, I prefer to know that they are smart enough to figure out how to do it on their own. --Ludwigs2 16:45, 19 August 2010 (UTC) [reply]

Or they were smart enough to avoid the hassle and simply slip a hardware keylogger on the keyboard input. -- kainaw™ 16:58, 19 August 2010 (UTC)[reply]

The good news, though, is that we can provide complete instructions and links to compiled programs, source-code, strategies, and everything: the information is all freely available to answer these sorts of responses and is easy to get. Fortunately, the people who want to use keyloggers invariably will find the instructions overly-complicated. Nimur (talk) 17:49, 19 August 2010 (UTC)[reply]

and let's keep it that way. I have a soft spot for intelligent criminals (call it anarcho-romanticism...), but I can't abide dumb ones. --Ludwigs2 18:15, 19 August 2010 (UTC)[reply]

Is there any known message whose message digest after applying one of the encryption algorithms like SHA# or MD# is exactly the same as the original message? 20.137.18.50 (talk) 15:08, 19 August 2010 (UTC)[reply]

It may be remotely possible...but its highly unlikely.Smallman12q (talk) 15:23, 19 August 2010 (UTC)[reply]

It's not unlikely for a fixed point to exist. The chance of a good hash function having a fixed point is 1 − (1 − 1/n)ⁿ where n is the number of outputs, such as 2¹²⁸ or 2¹⁶⁰. For any realistic hash size, this is almost exactly 1 − 1/e, or 63%. -- BenRG (talk) 19:17, 20 August 2010 (UTC)[reply]

SHA# and MD# are hash functions, not encryptions, so your question is unclear. They don't affect the original message. --Sean 15:25, 19 August 2010 (UTC)[reply]

Of course I meant that the output of the hash function would be equivalent to the input to it. 20.137.18.50 (talk) 15:30, 19 August 2010 (UTC)[reply]

What you're asking in essence is whether there is a fixpoint for SHAx or MDx. It seems that no-one knows - that is, there is no analytic finding that says yay or nay, and brute force is a big job. This guy is apparently organising a brute force search for a fixedpoint in md5. While in theory such a finding might cast doubt on the security of a hash algorithm that had hitherto been thought secure, as long as they're fantastically rare (as, for example, are the weak keys of certain otherwise secure block ciphers) that probably wouldn't affect the practical security of a hash algorithm. I don't know of any evidence for either hash family that shows any algorithmic relationship between the tractable determination of collisions and the existence (or determination) of fixpoints. -- Finlay McWalter ☻ Talk 18:00, 19 August 2010 (UTC)[reply]

I wonder if there is an "officially" coined term for it in computing, rather than fixpoint. See some of the math behind it here.Smallman12q (talk) 18:07, 19 August 2010 (UTC)[reply]

Neither SHA-1 nor MD5 is perfect; that doesn't mean they necessarily have fixpoints. -- Finlay McWalter ☻ Talk 18:29, 19 August 2010 (UTC)[reply]

The term "fixed point" is reasonably common in computer science. It's an important concept; the Y combinator is used to find the fixed point of functions (in the lambda calculus, they're guaranteed to exist), and this can be used to make recursion happen. Paul (Stansifer) 19:11, 23 August 2010 (UTC)[reply]

The existence of a fixed point is not a weakness any more than the existence of collisions (which are a mathematical certainty, of course). Successfully finding a fixed point or a collision does demonstrate weakness. -- BenRG (talk) 19:17, 20 August 2010 (UTC)[reply]

When searching for a fixed point of a hash, how, precisely, do you describe the function? MD5 yields a 128 bit (16 byte) hash, but this is typically given as a 32 digit hexadecimal number, represented in ASCII with 32 bytes. Do you feed MD5 the 128 bit value or the 32 character string? The former seems mathematically "purer" in some sense, but the latter seems likely to be what most people would be interest in. -- 1.46.157.19 (talk) 12:23, 22 August 2010 (UTC)[reply]

Wouldn't they be equal? The 128-bit value when converted ASCII will yield the same result if the ASCII is converted to the 128-bit value.Smallman12q (talk) 14:54, 22 August 2010 (UTCle

No. The MD5sum of a 16 byte file consisting all of the value 0 will differ from the MD5sum of the 32 byte file consisting of all 0x30, the ASCII value of the symbol "0". A 128 bit file which yields a 128 bit MD5sum equal to itself is one sense of a fixed point, but so is a 32 byte file of ASCII characters whose MD5sum, in hex, represented in ASCII, is equal to itself. One such fixed point may exist while the other may not. -- 1.46.93.58 (talk) 00:37, 23 August 2010 (UTC)[reply]

MD5's output is defined to be a sequence of 16 bytes (RFC 1321, section 3.5). -- BenRG (talk) 23:04, 23 August 2010 (UTC)[reply]

I do think that a 16 byte fixed point would be of more interest to a mathematician doing anything other than a brute force search. The 32 byte "hex in ASCII" fixed point is almost more of a recreational mathematics question, as our choice or character representation is fairly arbitrary. You might as well be look for the 128 byte fixed point of the string which is the ASCII representation its hash in binary.

Still, the 32 byte fixed point is the fixed point of the function that returns a hex string, and it is much easier for the average user to verify, as they wouldn't need a hex editor to create the input file. "This guy"'s brute force search (mentioned above by Finlay McWalter) speaks of MD5 returning a "32-character, hexadecimal string" and states that "there could exist a string which, when hashed, returns itself" implying that he is interested in the 32 byte fixed point. Of the first dozen programs and scripts submitted to that site that I checked, only one searched for the 16 byte fixed point; the others searched for a 32 byte fixed point. -- ToE^T 12:36, 24 August 2010 (UTC)[reply]

Also of interest would be detection of a cycle of a hash function. They are guaranteed to exist, and repeated iterations of the hash from any starting point will eventually enter a cycle. Any fixed point is a cycle of period one. The hash is invertible on this restricted domain (domain restricted to its range) if and only if it is collision free over the restricted domain if and only if all cycles lack initial non-cyclic leads feeding into them. Barring unexpected properties of the hash, the birthday theorem should apply. So for MD5, on average it would take on the order of √(2¹²⁸) = 2⁶⁴ iterations of the hash to enter a cycle and complete one period, with the typical period being about half that. Note that 2⁶⁴ ≈ 18.4 billion billion and a billion seconds is over 31.5 years. I don't see how a search for the cycle resulting from a given starting point can be parallelized. -- ToE^T 18:25, 24 August 2010 (UTC)[reply]

Suppose someone is running two instances of the same program, for example, Firefox. Both instances send requests for information to the same IP address (for example, Google), but the instances are requesting different data. A packet arrives at the requesting computer. How does the operating system know which of the two instances the packet should be routed to? Jc3s5h (talk) 16:58, 19 August 2010 (UTC)[reply]

By creating an internal representation for each unique connection. Specific operating systems have different strategies: most Linux-like systems use connect() (on the client) and accept() (on the server) to re-assign the communication to any free port. In other words, both sides are initialized using the "known" port, but then are usually shuffled to a random, "temporary-use" port-number unique to the particular client/server socket-pair. After the initial connection is negotiated (e.g. on Port 80 for HTTP), the socket-pair gets reassigned to a different, free port. Correction - the server always uses the known "destination" port; only the client uses a randomly assigned "source" port; and the two ports (source and destination) form a uniquely-identifiable socket-pair between client and server. You can see this list of re-numerated ports if you use the netstat command. Nimur (talk) 17:15, 19 August 2010 (UTC)[reply]

Thanks, that is a clear answer. Jc3s5h (talk) 17:35, 19 August 2010 (UTC)[reply]

Just to make it clear, I don't think there's any shuffling. The source port is some randomly assigned port above port 1024, the destination port (the web server) is port 80. There's no port reassignment. The source port is (almost always) different than the destination port. On *nix machines (and maybe windows too, I'm not sure) non-root processes cannot open sockets below port 1025. Shadowjams (talk) 06:05, 20 August 2010 (UTC)[reply]

Ah, you're right. The server-side (i.e., whomever started with listen()) may choose to re-assign the port, but it is usually not necessary. There is enough state, provided by the IP-address and client-side random port, to allow multiple communications streams between the same host and same server that are initialized on the same server-port-number (each socket-pair between host and client is uniquely identified by the client's random port-number). Server-side shuffling is therefore optional, and uncommon. (FTP uses this technique, with a "side-channel" socket that I think can be randomly assigned on both ends, with specific ports negotiated through the first socket). Nimur (talk) 17:07, 20 August 2010 (UTC)[reply]

I want to download service pack SP6a for NT4, but the download page from microsoft.com says "The page cannot be displayed because an internal server error has occurred" and has been saying that for months. Where can I download a proper version of SP6a, because the last one I tried was some kind of virus. 82.44.54.4 (talk) 17:07, 19 August 2010 (UTC)[reply]

I'm afraid you're not going to be able to get it from Microsoft as it's not been supported since 2004, although I did check Technet and MSDN just in case, but no. Although I have no way of verifying it's authenticity this page is by someone in the same boat as you who's uploaded a local copy to their server. The comments on the page (assuming genuine) seem to suggest it's okay. Normally I'd say you should always get it from the original source, but since this is going to be impossible I think that's probably the best you can do. Still, use at your own risk etc!  ZX81  ^talk 17:31, 19 August 2010 (UTC)[reply]

Thanks, that's awesome! There appear to be four different downloads. Which one is the best, "High Encryption", "Standard Encryption", "56-bit Alpha", "128-bit Alpha"? 82.44.54.4 (talk) 18:59, 19 August 2010 (UTC)[reply]

You probably don't want Alpha (you'd know if you did!) and most likely you want the first "High Encryption" link (128-bit). The Standard version is 56-bit because at the time there was legal problems with encryption higher than that and exporting to other countries.  ZX81  ^talk 19:09, 19 August 2010 (UTC)[reply]

It is simply bad practice to encourage the use of 3rd-party distributions of Windows patches - they are probably violating distribution licenses and are difficult to vet for authenticity. NT Server hasn't been developed since 1999 - so you should have started considering an upgrade in 1999 - and it has been officially EOL'ed for more than five years - this means you should have definitely stopped using it in 2004. (The grace-period of reasonable usage is long since expired!) There are dozens of cheap and low-cost alternatives from Microsoft to help you upgrade to a more modern version; and there are lots of free software alternatives if you have a zero-dollar budget; but we should not advise you to seek out unofficially-distributed upgrades for (obsolete, proprietary) software . Nimur (talk) 17:41, 19 August 2010 (UTC)[reply]

It's even worse practice to run the system unpatched. Besides, migration costs for that server is likely to be large. 121.72.203.118 (talk) 12:53, 20 August 2010 (UTC)[reply]

But ah, he is running it unpatched, as there have no doubt been exploits that affect NT4 SP6a that have not been addressed due to the aforementioned end of life taking place many many years ago. Using a patch from an unknown source to hold up such an old system is like playing Russian Roulette with two bullets instead of just one. --144.191.148.3 (talk) 14:21, 20 August 2010 (UTC)[reply]

It seems strange that Microsoft would remove the links to NT4 service packs when they still keep the old Windows 95, Windows 3.1 and DOS help pages and update packages available to view and download, with the "This article was written about products for which Microsoft no longer offers support. Therefore, this article is offered "as is" and will no longer be updated" disclaimer on them. Is there some specific reason why they removed NT4 downloads? 82.44.54.4 (talk) 18:59, 19 August 2010 (UTC)[reply]

They may not have deliberately removed it; it may simply be that it's no-one's job to maintain it any more, and some unrelated website change broke it and no-one's taking responsibility for either fixing it or removing the link entirely. -- Finlay McWalter ☻ Talk 19:31, 19 August 2010 (UTC)[reply]

If you have a chance to obtain SP6 (without the a), the patch to upgrade from SP6 to SP6a is still available here: http://download.microsoft.com/download/winntsp/patch/6.0a/nt4/en-us/q246009i.exe (Intel) / http://download.microsoft.com/download/winntsp/patch/6.0a/alpha/en-us/q246009a.exe (Alpha). -- 78.43.71.155 (talk) 17:11, 20 August 2010 (UTC)[reply]

Here's the most official download site I could find: ftp://ftp.lrz-muenchen.de/pub/comp/platforms/pc/winnt/sp6a/usa/ - obviously, it's not microsoft.com, but it's the official site of the computer centre for Munich's universities and for the Bavarian Academy of Sciences and Humanities (http://www.lrz.de/english/) so it's unlikely to be a hacked version. And as you can see from the file path, it's the US version of SP6a, not the German one. -- 78.43.71.155 (talk) 17:23, 20 August 2010 (UTC)[reply]

What kinds of files are viruses typically embedded in? Would a .pdf or an excel file be likely to infect a computer? What kinds of files are safe to back-up after a virus has infected a system? 138.192.58.227 (talk) 17:18, 19 August 2010 (UTC)[reply]

I think we can probably come up with a pretty good taxonomy of standard files viruses hide in, a pared down version of the information at Computer_viruses#Vectors_and_hosts. The obvious start:

• Executables. On Windows that means EXE and COM files in particular. No big surprise there.
• Scripts. Obvious Windows candidates are BAT and VBS (because most people won't have a system set up to run other scripting languages out of the box, I don't think).
• Files that can contain scripts inside of them. This is the big one, because all of the MS Office products can do this. So we're talking DOC, DOT, XLS, PPT, and MDB files at the minimum. Probably more.

Now the tricky bit is that last category, because what files can and can't have scripts embedded in them (in a way that they will actually execute, which is crucial), and whether their security can be compromised in weird ways, is a very big question. PDF viruses do exist, though I don't think they're common. Excel files, definitely, yes. As for backing something up after infection... it's a tricky proposition if you're going by just file name alone. --Mr.98 (talk) 17:50, 19 August 2010 (UTC)[reply]

Are there instances of viruses infecting your MSOffice files? Or are these only from downloading files that already contain the virus? 138.192.58.227 (talk) 18:56, 19 August 2010 (UTC)[reply]

Apparently they can be spread by worms. But there are millions of viruses — presumably there would be some that would mix-and-match their vectors. Again, approaching it from the standpoint of "what can be backed up" is a little tricky. --Mr.98 (talk) 19:22, 19 August 2010 (UTC)[reply]

This link explains how to disable macros in Microsoft Office. I believe that as long as Microsoft Office is set to ignore macros, Office files that have been infected do not pose a risk to you (now if you send them to someone else, you may infect them). Other Wikipedians will correct me if I'm mistaken.--el Aprel (^facta-_facienda) 03:23, 21 August 2010 (UTC)[reply]

That doesn't stop the ever-popular buffer overflow which doesn't function through macros, but rather through a handily crafted document.Smallman12q (talk) 23:12, 21 August 2010 (UTC)[reply]

This relates to a question I asked earlier. Most popup ads that get past the popup blocker have a URL at the top which cannot be changed, and most of them are small and rectangular. Then there are ads which have a changeable URL along with the back button, forward button, etc. that one would expect to see on a regular web page.Vchimpanzee · talk · contributions · 17:56, 19 August 2010 (UTC)[reply]

When popping up a window with Javascript, you can disable components of the browser window, such as the address bar, buttons, status bar, etc... Most browsers allow you to disable the disabling of window components. -- kainaw™ 19:44, 19 August 2010 (UTC)[reply]

What I'm saying is that there is no "address bar" as such in the window, or buttons, in most pop-up ads. I'm trying to figure out the correct terminology. I probably should have said pop-up window in my earlier question because the problem wasn't happening with a regular ad.Vchimpanzee · talk · contributions · 19:49, 19 August 2010 (UTC)[reply]

Kainaw is correct that it has to do with the specifics of the Javascript that launch the window, and how the browser interprets it. See, e.g. this page that describes the settings coders can use to customize the behavior of the popup window. Different browsers interpret these a bit differently — I don't think Firefox will let you totally hide the address of a page, for example, because of the security risk of not knowing what site you are connected to. Instead, it interprets that parameter as "show but don't let them modify the address," if I recall correctly. --Mr.98 (talk) 20:15, 19 August 2010 (UTC)[reply]

Right. I'm looking for the term for an ad with "show but don't let them modify the address". I used the term "address bar" to mean you could modify. There was one method that lets there be a toolbar, the other type. That would have been a simpler way to say it than "back button, forward button, etc." so that first version would be "without a toolbar". It still doesn't say what the different versions are called.Vchimpanzee · talk · contributions · 20:36, 19 August 2010 (UTC)[reply]

Firefox calls this the "menubar" and the bar with the "forward/back" buttons the "navigation bar". Firefox also has a Bookmarks toolbar, a Status Bar, and other addons can add their own "bar". Javascript can enable or disable any of these, if you permit it. If you want to preclude Javascript from affecting these, you can use Firefox's about:config interface to control Javascript access (described in detail at Developer.Mozilla.org). There is also a "standard" menu that lets you configure some, but not all, options for JavaScript permissions. Nimur (talk) 23:06, 19 August 2010 (UTC)[reply]

If you're just looking for a term, I doubt there is one for that specific thing. It's just a popup window, and if you want to say it lacks a back button, well, you just say that. --Mr.98 (talk) 16:19, 20 August 2010 (UTC)[reply]

Hi. What kind of knowledge should a Computer Tech Associate have, what are the major responsibilities of this position, and what qualifications (if any) are expected? Thanks. ~AH1^(TCU) 18:45, 19 August 2010 (UTC)[reply]

It depends entirely on the job. "Computer Tech Associate" could be anything from a salesman at a computer store (who only needs to know how to get people to fork over cash) to a networking assistant (who only needs to know how to keep the cables untangled) to a computer repair assistant (who only needs to know how to fdisk-format-reinstall) to a computer electronics tech (who only needs to know how to use a grounding strap when swapping out components). You need to ask whomever is hiring a Computer Tech Associate as it is a very general term, like medical assistant or construction assistant. The specific job creates the qualifications and expectations. -- kainaw™ 19:43, 19 August 2010 (UTC)[reply]

Hi. Say you're in the UK, and you dial a UK mobile number, and that phone happens to be in another country. What, in simple terms, are the steps that enable the network to locate that phone, given that (I assume) the number you dialled is not guaranteed to be globally unique.

The phone number is globally unique. All numbers issued in country XYZ are unique within XYZ and there is an implied country code prefixed to it. So UK phone 07777-777777 has the unique global number +44-7777-777777. Whilst some other countries "7777-777777" will become (say) +123-7777-777777. I don't KNOW how it works, but I do know that turning your phone on in another country "registers" it with that foreign network. I imagine that that netwrok lets the "UK" know that the relevant phone number in now in the particular country. -- SGBailey (talk) 22:38, 19 August 2010 (UTC)[reply]

The number is globally unique; the full number is +44 7xxxxxxxxx, and it's the same number regardless of where the phone roams. Calling a UK mobile that's roaming in Darkest Peru entails calling the same number as you'd call if it was in the next room. The second part of your question is routing. Phones (strictly SIMs) are identified by their IMSI codes. Say you turn on a UK Vodafone mobile in Darkest Peru. It searches for all the available towers, hoping one is Vodafone UK. It doesn't find one, so it connects to anything, say a Claro Peru one. This handshake exchanges the SIM's MCC and MNC codes, 234 and 15 respectively, which identify it as a Vodafone UK. The Claro central office connects to the Vodafone UK central office and verifies that Vodafone will accept payment on this SIM (big telco providers have direct relationships with one another, smaller ones may use a big one as an intermediary). Vodafone confirms this, and remembers that the mobile is in Claro Peru's territory. So if someone in the UK phones that 07xxxxxxxxx number, Vodafone knows to route the call to Claro. Things are simpler if the phone, in Peru, makes a call - Claro routes the call as it would one of its own, but it sends the billing transaction to Vodafone. -- Finlay McWalter ☻ Talk 22:49, 19 August 2010 (UTC)[reply]

One thing that doesn't seem to be implemented by the GSM handshake is a a way to prioritise non-home network selection. So if that phone, turned on in Peru, can't see a Vodafone connection, but seeing towers run by Claro, Avanza, and Movistar, it will pick whichever it likes (often it gives the user the choice). If Vodafone has a deal with Avanza that makes the calls cheaper, that doesn't affect the handshake. In this circumstance I've received a text message from my home network saying "you could get cheaper calls if you connected to Avanza instead", but it's down to me to do it manually. -- Finlay McWalter ☻ Talk 22:56, 19 August 2010 (UTC)[reply]

To clarify one thing that may not have been obvious from FW's answer. If someone calls you while roaming, the call will always go thorough your home operator (at least AFAIK), even if it's from someone sitting right next to you. This isn't really that surprising, when someone calls you, your number is a UK one, specifically a UK Vodafone one so the network of the person calling you connect it to Vodafone UK, who then have to send it back to you since they know you're there. This is I think one of the key reasons why receiving a call is usually rather expensive and in fact receiving a call from someone in the UK is often the same price, perhaps even cheaper then receiving a call from someone where you're actually located. (Calling someone locally is often far cheaper, also for the caller of course.)

In theory with smart IP networks, it wouldn't be that hard (relatively speaking) for the initial connection to go thorough Vodafone UK, but then Vodafone UK tells the network of the calling party to route it to the network the phone is currently connected to (although there's need to be some way to handle voicemail and stuff like that) but I don't know if this is implemented if at all (it may be for networks run by the same parent company particularly if in a similar geographical location, I know for Vodafone NZ you could use your phone in Australia and only pay normal Vodafone NZ local rates IIRC, you didn't even need to register for roaming which you normally had/have to do).

BTW, while FW is right I think about the lack of prioritisation, that's I think only a problem if your network actually has a deal with the other networks. If there are 3 networks where you're located, but your network only has a deal with one, the other 2 networks which just tell you to bugger off when you try to connect to them (well it will probably say you can connect to me but only for emergency calls). (Or if it does connect, you'll see the emergency calls only and hopefully try a different one.) So at worst it may slow down getting a network.

Nil Einne (talk) 00:14, 20 August 2010 (UTC)[reply]

The mobile operators mostly have little financial incentive to implement smart routing (in practice I think all the CO equipment is quite capable of it, if configured to do so) - they like all that lovely roaming money they get, from temporary customers to don't know, and mostly can't shop around. The EU directive capping inter-member roaming charges (here) may incentivize them to do it (although they're still getting 37p a minute, so it's not that much of an incentive). It's quite possible that the EU will eventually want to abolish inter-member roaming fees altogether (it's really quite a marked obstacle to a single market) and you can bet the operators will route traffic very efficiently in that event. -- Finlay McWalter ☻ Talk 00:38, 20 August 2010 (UTC)[reply]

It's worth remembering of course it's not just the mobile operators that need to cooperate for this but other telcos as well. If someone in Peru calls you from a landline in Peru on your Vodafone UK mobile, then the Peruvian telco needs to have some sort of smart routing system so that's capable of learning it should route to the Peruvian mobile operator who's network your using rather then to the UK. Of course in general the mobile companies and the other telcos are related. Nil Einne (talk) 07:10, 22 August 2010 (UTC)[reply]

• Thanks for all the answers! When I said "not guaranteed to be globally unique" I meant "not guaranteed to be globally unique unless a country code (or some other relevant code) is appended". So, really is it just that if you don't append a country code then you're assumed to be calling a mobile issued in the same country that you're calling from, and the appropriate country code is transparently appended by the network to uniquify the number? —Preceding unsigned comment added by 86.184.25.4 (talk) 00:53, 20 August 2010 (UTC)[reply]

I think that, if you fail to add a country code, then the assumed country is the one that the phone is registered in. i.e. if you're in Peru and call 07777 777 777 on a UK Vodafone phone, then I believe you get a UK number. Open to correction, though.--Phil Holmes (talk) 08:02, 20 August 2010 (UTC)[reply]

I wonder if this depends on the network you are using. Note the idea of transparently appending the country code is probably unnecessary anyway. If the local operator routes it locally, the country code is irrelevant. If it sends it to Vodafone UK, once it reaches Vodafone UK the country code is likely irrelevant. In any case, I would suspect most operators won't recommend it in case it doesn't do what you expect. Personally I never enter a stored number in my mobile without the country code. It seems pointless. Nil Einne (talk) 07:17, 22 August 2010 (UTC)[reply]

For months now my laptop fan has ceased to work, i.e. there is no exhaust air coming out even when the laptop is at 80 C. I have been using a hack of an external fan + cooling pad, which sometimes keeps the internal temperatures at 40-50C. The CPU is often cooler than the motherboard air which I take to be a sign that the big laptop fan has failed and not the CPU heatsink fan.

Now, I haven't taken it apart to look at lint clogging, etc. I have already sprayed the outside of my laptop (and its vents) with compressed air. Is it likely that my fan has ceased to work completely, or will spraying my fan with compressed air free it up again? Also I can't find any laptop fans for my model in stock. How interchangeable are laptop fans? I kinda want to upgrade to an extra loud, extra powerful one anyway. John Riemann Soong (talk) 23:33, 19 August 2010 (UTC)[reply]

The fans are very often part of a fan unit, which includes a duct or heat-pipe, and you'd end up having to replace the whole thing. As such units are contorted to the weird spaces available, they're generally model-specific. I've never known dust etc. to foul a fan so completely that it doesn't work, just that it becomes noisy and inefficient. Note that your BIOS may have a screen that shows the fan's speed (or try Speedfan). -- Finlay McWalter ☻ Talk 23:43, 19 August 2010 (UTC)[reply]

On the Mac, iStat Pro can show fan speed, amongst other things such as temperatures. Chevymontecarlo 07:04, 20 August 2010 (UTC)[reply]

First a background: I've not been working in the IT/programming industry for several years, although computer science was my major. I am currently in the process of looking for work here, but the economy stinks so companies don't want to hire people with very little experience. The ones that do mostly require some sort of certification, e.g., A+ certification. My questions:

1. How can I go about getting one of these certifications?
2. What's the best way to go about learning popular languages and frameworks like .NET and ASP (everyone seems to be looking to hire for these)?

I am willing to put up some money for the training, but I'd sure like to not have to spend more than 1 or 2 grand total. I live in the US. Magog the Ogre (talk) 23:59, 19 August 2010 (UTC)[reply]

Microsoft has some pretty good online training material for things like ASP.NET (some of which is free). Their info for training and certification is here (comparable things are run by Oracle, Redhat, Cisco etc.). Exams that go toward certification are generally administered by specialist testing companies (MS seems to use Prometric a lot), where you go to their testing centre (which is "local"). These tests aren't terribly cheap (MS ones in the US administered by Prometric seemed to cost $125 each); how many you need to take depends on the certification, but it seems to take several. That's the only cost you need to do; for many things they also sell self-study guides and of course (if a company with deep pockets is around to pay for it) optional training. -- Finlay McWalter ☻ Talk 00:55, 20 August 2010 (UTC)[reply]

And if you're wondering, the test is almost always entirely computer based. All the testing company does is verify your identity (so you haven't sent someone knowledgeable in to do the exam for you) and they watch that you don't cheat. For some certifications (like Cisco networking things) you really have to buy a simulator (as you obviously can't be experimenting on a room full of Cisco switches). -- Finlay McWalter ☻ Talk 00:59, 20 August 2010 (UTC)[reply]