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July 9

I²C over COM port

hello, does someone by chance know of a program or a library that can bitbang the I²C protocol over an RS-232 port (or, rather, an FTDI style breakout board with TTL levels), for communicating with I²C devices, preferably under Linux? I'm aware of the electrical differences between I²C and TTL Asmrulz (talk) 04:29, 9 July 2016 (UTC)

Our I2C article has some suitable source code. Tevildo (talk) 08:39, 9 July 2016 (UTC)

Wattage

Is it possible to measure an L.E.D.'s actual watt using a typical Multimeter (supposing it can measure Current, apart from Voltage, both in AC and DC) ? 210.56.110.175 (talk) 05:42, 9 July 2016 (UTC)

Not with one measurement. You need to measure the current, disconnect the multimeter, reconnect the supply, then measure the voltage. Tevildo (talk) 09:29, 9 July 2016 (UTC)

Incidentally, if you have a simple arrangement with a resistor in series with the LED, you can measure the current by measuring the voltage across the resistor, and using the equation ${\displaystyle I={V \over R}}$. You still need to make a second measurement of the voltage across the LED, though. Tevildo (talk) 09:41, 9 July 2016 (UTC)

Many thanks Tevildo. If it's not taking too much advantage of your generosity, I'd beg you to be a bit more elaborate. For instance in first instruction you say "measure the current", now shall I measure it along with LED attached or LED removed from the circuit. And is it true that current is never measured like voltage (in parallel) but in series ? As for putting resistor in series with LED that ain't difficult, why not simply jump to this step ? Please tell how voltage across resistor and across LED ultimately give us wattage. Unless I am very much wrong I think current and wattage are separate things. 124.253.145.142 (talk) 17:00, 9 July 2016 (UTC)

No problem, it's what we're here for. I'm assuming your circuit is basically the same as the one to the right (from our LED article, with annotations). To find the power, we need to find the voltage across the LED and the current through it, and multiply them together.

Measuring the voltage across the LED is easy:

• Set your multimeter to "DC Volts".
• Connect the positive lead to point B on the diagram, and the negative lead to point C.
• The reading on the multimeter is the voltage across the LED. Let's call this ${\displaystyle V_{LED}}$. For example, it might be 3 volts.

There are two ways to measure the current. The first is to measure it directly with the multimeter:

• Disconnect the wire between the power source and the LED (between points A and B on the diagram).
• Set your multimeter to "DC Amps".
• Connect the multimeter positive lead to point A on the diagram (the supply positive output), and the negative lead to point B (the LED anode).
• The LED should now be on. The reading on the multimeter is the current in the LED. Let's call this ${\displaystyle I}$, as in the diagram.

The second way is to measure the voltage across the resistor. This assumes you know what its resistance is.

• The circuit should be connected as in the diagram (with the LED on).
• Set the multimeter to "DC Volts".
• Connect the positive lead to point C, and the negative lead to point D.
• The reading on the multimeter is the voltage across the resistor. Let's call this ${\displaystyle V_{R}}$.
• If the resistance of the resistor is ${\displaystyle R}$, the current in the LED is ${\displaystyle V_{R} \over R}$. For example, if ${\displaystyle R}$ is 100 ohms, and ${\displaystyle V_{R}}$ is 2 volts, the current (${\displaystyle I}$) is 20 mA (0.02 = 2 / 100).

Now we know ${\displaystyle V_{LED}}$ and ${\displaystyle I}$, the power of the LED (in watts) is ${\displaystyle W=I\times V_{LED}}$. Using the numbers from above, the power is 20 mA * 3 V = 60 mW. Hope this helps! Tevildo (talk) 18:08, 9 July 2016 (UTC)

I suggest you get a device like the Kill-A-Watt meter, which does the math for you and lists the wattage directly: . StuRat (talk) 17:08, 9 July 2016 (UTC)

This is for mains equipment, and the OP just wants to measure the power for a (DC) LED. Tevildo (talk) 18:08, 9 July 2016 (UTC)

To be more specific, if you try to measure the watts of an alternating current circuit by measuring volts and amperes separately and calculating the result, you end up with volt-amperes, not watts. In direct current circuits watts and volt-amperes are identical, so the technique works. The math is easy: volts times amperes equals watts, as Tevildo explained in detail above. --Guy Macon (talk) 22:09, 9 July 2016 (UTC)

Note: High power LEDs, also called compact light sources, were operated in pulse mode due more efficient power supply. The pulses may not get captured correctly by the multimeter. An oscilloscope lets you know, but note, the oscilloscope is connected to ground. Aware of shorcuts. --Hans Haase (有问题吗) 20:53, 12 July 2016 (UTC)

Thanks for reply, Hans. If you don't mind, I'd like to know what is meant by "connected to ground" ? And how does it disqualify an oscilloscope from correct and safe gauging. I think doesn't that simply mean that like many other electric appliances that run on A.C. mains, it has its outer metallic body attached to a 3rd wire that takes back any leaked current, through a cable common to whole house or complex, back to be sunk into the ground through a conductor buried considerably deep into it ? - O.P. — Preceding unsigned comment added by 150.129.198.38 (talk) 04:58, 13 July 2016 (UTC)

Unlike a portable multimeter that has two uncommitted probes (red +ve and black -ve), oscilloscope inputs are usually coaxial with the outer (shield or screen) permanently connected to the case and ground. If the LED circuit under investigation is powered by battery or some other floating power supply there is no problem. However if the LED circuit is part of mains-powered equipment whose circuit is connected to ground, the extra route to ground through the oscilloscope may cause a damaging short circuit. Some two-channel oscilloscopes offer a differential measurement mode that overcomes this limitation. It is also possible to "float" an oscilloscope without its ground connection but that strategy is risky even in the hands of experienced professionals. AllBestFaith (talk) 10:58, 13 July 2016 (UTC)

Chinese (and other exotic) Characters

Whenever a server sends traditional or simplified Chinese (same about some other languages also) as text, instead of occurring as they should, they rather appear to be rectangles confused from within, as shown in the picture here. Please tell me what should be done to make them occur naturally ? 124.253.145.142 (talk) 16:17, 9 July 2016 (UTC)

You will have to upgrade to an operating system with the Unicode fonts installed. You can install more fonts on your computer as well. Graeme Bartlett (talk) 22:35, 9 July 2016 (UTC)

Thanks, Graeme Bartlett...it works ! see → 活動写真

Airlines use of continuous stationary

Why do airlines still use dot matrix printers and continuous stationery? I hear them at gates, and I assume they're printing passenger lists. The only advantage I can think of is that if you're clutching a wodge of continuous paper, you know you've got it all. Hayttom (talk) 20:55, 9 July 2016 (UTC)

One advantage I can think of is that if the ink runs out, you still have the indentations in the paper you can read, especially if you run a pencil lead over it. But while we are on the topic, why do most retail establishments in the US still use thermal printers for receipts ? StuRat (talk) 23:23, 9 July 2016 (UTC)

Great follow-up question. I hope we get some answers.Hayttom (talk) 14:21, 10 July 2016 (UTC)

One obvious factor is cost. The best technology to replace these old technologies, as far as readability, is likely a black-and-white laser printer (color seems like an unnecessary expense). I imagine they cost more initially (but maybe not by much, if it only prints the width of a receipt or airline ticket). However, for these high volume operations, the more important issue is what it costs per print. Can anyone provide data on this, for dot-matrix, thermal, and black-and-white laser printers ? StuRat (talk) 15:13, 10 July 2016 (UTC)

Also, (nearly) all laser printers are page printers, i.e. they print full pages. Dot matrix printers are line printers. If you need a permanent record immediately, a page printer will waste a full page on each transaction, even if you only need a single line. --Stephan Schulz (talk) 19:03, 10 July 2016 (UTC)

But is their any inherent reason why laser printers can't be designed to print smaller areas (hopefully at reduced cost) ? StuRat (talk) 22:25, 10 July 2016 (UTC)

StuRat You mean like a laser receipt printer? Probably far more economical to use a thermal printer. (See my answer below↓ too) If someone wanted to throw enough money at it, it is probably possible to shrink a laser printer down like that. I imagine you could write to thermal paper with a laser, thus avoiding the use of toner? - 220 of 05:38, 11 July 2016 (UTC)

Those are most likely line printers, not dot-matrix. --Guy Macon (talk) 19:32, 10 July 2016 (UTC)

As to the use of thermal printers, they're very quick, but a key advantage may be that there's only one consumable (a thermal roll) rather than paper/ribbon or paper/ink or paper/toner - relatively speaking, this makes them a very much more easy to maintain than a printer which requires both. --Tagishsimon (talk) 22:34, 10 July 2016 (UTC)

re StuRats' query about Thermal Printers, I agree the big advantage of thermal printers may be speed. Where I live now they use a small thermal printer on a 'kiosk' at the library to print list of book etc borrowed. I have worked a lot (in the past) with inkjet log printers, and I was surprised that in this application were practically instant, i.e you tap the touchscreen asking for a 'receipt' and bang its there. Barely get the chance to lift my finger off the screen. Agree with Tagishsimon, in other words. I also think it's likely they can economically be made very small, likely more so than a laser. Thermals seem to be practically universal for cash register receipts.220 of 05:38, 11 July 2016 (UTC)

Re the speed of thermal receipt printers, source here says up to 250 mm per second, so 10 inches/sec, 600 in/minute or ≈50 feet,(15 metres) per minute. That is pretty fast! 220 of 05:49, 11 July 2016 (UTC)

Airlines were among the earliest earliest adopters of computer technology. It's entirely possible that some of what we see today just comes from the old adage of "if it ain't broke, don't fix it". If it works and fits their requirements, (including costs of consumables and break down frequency), sometimes it would cost a lot more to "re-engineer" something to work with a newer "peripheral" so you make do with the old one, until it becomes unviable. We have these old Dialogic Inc. fax cards called "brooktrout", they're old and obsolete but to replace the service will cost a fortune, so we just keep using them until they fail and then replace them with a different service. They might last 10 years or they might last 2 months but there's no point paying to change them over while they're still working. I'm not saying THAT is what's happening with line printers in airports, I'm just giving one example of why seemingly old obsolete computer hardware doesn't get replaced. Vespine (talk) 03:45, 11 July 2016 (UTC)

See If It Ain't Broke, Don't Fix It: Ancient Computers in Use Today. ---Guy Macon (talk) 04:32, 11 July 2016 (UTC)

July 11

Algorithm for Union of Sets

Let us denote ${\displaystyle I=\{1,\dots ,n\}}$, and denote ${\displaystyle P_{k}(A)=\{B|B\subseteq A,|B|\leq k\}}$.

I am looking for an efficient algorithm to transform ${\displaystyle \{A_{i}\}\in P_{n}(P_{n}(I))}$ into ${\displaystyle \{\cup _{a\in A_{i}}{a}\}}$.

Thanks in advance! 31.154.81.65 (talk) 10:25, 11 July 2016 (UTC)

Expressed with words, ${\displaystyle P_{k}(A)}$ is a set of k-element subsets of some ${\displaystyle A}$, right?

Then ${\displaystyle P_{n}(I)}$ is a set of n-element subsets of ${\displaystyle I}$. However, ${\displaystyle I=\{1,\dots ,n\}}$ is itself an n-element set, so it has just one n-element subset: ${\displaystyle I\subseteq I}$, hence ${\displaystyle P_{n}(I)=\{I\}}$.

Consequently ${\displaystyle P_{n}(P_{n}(I))}$ is a set of n-element subsets of a singleton ${\displaystyle P_{n}(I)=\{I\}}$, so it is ${\displaystyle \{\{I\}\}}$ if ${\displaystyle n=1}$, or an empty set otherwise.

Alas I can't get the ${\displaystyle \{A_{i}\}\in \ldots }$ part. Do you mean 'a singleton belonging to'...? If so, ${\displaystyle \{A_{i}\}\in P_{n}(P_{n}(I))}$ could only be ${\displaystyle \{A_{i}\}=\{I\}=\{\{1\}\}}$, so ${\displaystyle A_{i}=\{1\}}$ for ${\displaystyle n=1}$ (and does not exist at all for ${\displaystyle n\neq 1}$) and I can't see the reason for the ${\displaystyle i}$ index at ${\displaystyle A_{i}}$... --CiaPan (talk) 12:07, 11 July 2016 (UTC)

I'm sorry for the typo... I fixed my mistake (replacing equality sign with inequality sign). 31.154.81.65 (talk) —Preceding undated comment added 13:36, 11 July 2016 (UTC)

So, P_n(I) is simply the power set of I (where I = {1, 2, ... n}), and could be denoted P(I). Then P_n(P_n(I)), or more simply P_n(P(I)), is a truncated power set of P(I), containing only those sets of sets which have n or fewer elements from P(I). And your {A_i} is an element of that truncated power set of the power set?

In other words, you have a list of at most n lists of integers, the integers themselves ranging from 1 to n, the lists of integers presumably unsorted, with the additional constraints that the individual lists of integers have no duplication and that no two lists of integers from your list of lists would be the same even if sorted, and you want an efficient way to merge this list of lists into a single list containing, without duplication, exactly those integers which appeared in any list of your list of lists. Presumably you hope that these constraints will allow for a more efficient algorithm. Note that our article merge algorithm discusses the merging of sorted lists; I don't know what we have on the merging of unsorted lists without duplication. -- ToE 14:51, 11 July 2016 (UTC)

Okay, I'll read the article about merge algorithm carefully. Thank you for the idea to read this! 31.154.81.65 (talk) 05:55, 12 July 2016 (UTC)

If you can tell us your actual task, it would help us know if we are up against an XY problem. The multiple roles of n -- it is both the maximum value of the integers in your sets and is the maximum size of the elements which are in the truncated power set of the power set of I -- strike me as unlikely to arise naturally.

Your task, as I understand it, has the (perhaps slight) advantages over simply merging unsorted lists with possible duplication in that you know that the individual sets of integers themselves contain no duplication. Note that some languages offer set abstract data types which are often implemented "using more efficient data structures, particularly various flavors of trees, tries, or hash tables." Our article mentions the expected union and add operations, but I see that Python's built-in set type also offers an update operation, which "update the set, adding elements from all others." (In essence, a ∪= augmented assignment.) Even if you wish to roll your own solution, the library implementations may offer suggestions for optimization. -- ToE 20:31, 12 July 2016 (UTC)

• Assuming ToE did represent correctly the problem, a solution that is optimal in time (but disastrous in space) would be to read all the lists once to find the maximum value ${\displaystyle n_{max}\leq n}$, then create a boolean array of ${\displaystyle n_{max}}$ elements set to false, then read all the list and flip the bits in the array when the value is found. If you know ${\displaystyle n}$ you do not even need to do the first pass, and there is no way to do it more efficiently (you will eventually have to read all the elements of all the lists, and practically, the bit-flipping is easy to vectorize in many programming languages such as Python). Returning the sorted union is trivial then.

The downside of course is that if there is reason to suspect that ${\displaystyle n_{max}}$ is much larger than the size of the final list, the array will be much bigger than it would need to be by other methods. Tigraan 09:01, 12 July 2016 (UTC)

uninstall Yahoo! powered

Yesterday I installed an update of Freemake Video Downloader. It seemed to install a browser hijacker. It installed "Yahoo! Powered" and I can't uninstall it. Is that a browser hijacker? Bubba73 16:44, 11 July 2016 (UTC)

Well, I ran Revo Uninstaller and that seemed to get it. Bubba73 19:22, 11 July 2016 (UTC)

That would be Search.yahoo.com browser hijacker -- There are instructions for manual removal which are somewhat complicated (depending on OS & browser). This seems to be the most recent and comprehensive:. --2606:A000:4C0C:E200:A073:98E5:BA6B:E905 (talk) 23:35, 12 July 2016 (UTC)

July 12

Would the graphics card R9 390X or gtx1070 fit on the motherboard GA-H81M-S1?

Would the graphics card R9 390X or gtx1070 fit on the motherboard GA-H81M-S1?201.79.70.197 (talk) 12:19, 12 July 2016 (UTC)

It should since those you mentioned are standard PCIe video cards anyway, but what matters more is if they fit inside the case you're using. Blake Gripling (talk) 12:57, 12 July 2016 (UTC)

And you want to make sure it works with the power supply. The GTX1070 is rated around 150W, so it should work easily with any power supply that is at least 500W. The R9 390X is rated around 250W. So, you will want 600+W. Because the price isn't much different between 500 and 600W power supplies, I usually get more than I need. If you are using a small power supply (like a 350W that comes free with many cases), you will have issues with using either card. 209.149.113.4 (talk) 14:10, 12 July 2016 (UTC)

Oh Yes, I know about the power supply thing, and I will buy one if needed. Anyway my current Graphics card is gt 730. Thanks for the help 201.79.78.164 (talk) 16:53, 12 July 2016 (UTC)

While the case is far more likely to be a problem, I wouldn't assume they definitely fit all motherboards. Some motherboards may place components (particularly heatsinks) which will block excessively long cards, either for the main slot or the second slot if it's a double width card (as many are). Admittedly this isn't very common nowadays for a single card, still you should always check. I don't know what the PCIe standards for physical clearance are but I'm pretty sure they did not and possibly still don't require the sort of physical clearance some cards need. So a motherboard could be fully compliant but still not be able to fit the card. Nil Einne (talk) 06:02, 13 July 2016 (UTC)

Movie maker sought

What is highly praised?

Requirement is, something that can make me select points (from and to) of a video clip, more than one time in one go… – An easy to use tool is sought. A portable software is desirable, however…

Apostle (talk) 18:11, 12 July 2016 (UTC)

Arranging video clips in a desired sequence can be done using a program called a Non-linear editing system. Misplaced Pages has a List of video editing software and a Comparison of video editing software. AllBestFaith (talk) 23:41, 12 July 2016 (UTC)

What product is currently out there that people are admiring the most and are behaving like this about in order to acquire in this day and age?; relevant to my desire?... -- Apostle (talk) 04:31, 14 July 2016 (UTC)

Password Protector; Lockdown protection sought

Peeps, what’s highly praised? Btw, I possess Folderlock (currently installed; this provided Bitlocker – I believe bitlocker is unbreakable, and good, unless you guys re-correct me), Keepass (Portable acquired, and wired), AxCrypt (needs internet connection)... -- Apostle (talk) 18:17, 12 July 2016 (UTC)

Apostle, it is not clear what your question is. Are you asking for "consumer feedback" about some encryption products? If so, Misplaced Pages's reference desk is probably not the place to use; it is intended for factual questions (eg "which company sells product X") rather than opinion. Tigraan 11:33, 13 July 2016 (UTC)

If you read Misplaced Pages's article on BitLocker you will see that it is good but not unbreakable.--Shantavira| 14:38, 13 July 2016 (UTC)

What product is currently out there that people are admiring the most and are behaving like this about in order to acquire in this day and age? A name, with/without a reference would be of help -- Apostle (talk) 04:31, 14 July 2016 (UTC)

July 13

Windows 9x

Theoretically if Microsoft had continued to develop and invest in the DOS-based Win9x line instead of switching to NT, could they have produced a 64-bit OS running on-top of DOS that took full advantage of modern hardware capabilities? — Preceding unsigned comment added by 89.185.7.252 (talk) 14:39, 13 July 2016 (UTC)

There is no reason that DOS couldn't be altered and compiled to take advantage of a 64-bit CPU. Keep in mind that jumping to 64-bits isn't always an advantage in every single possible application. So, playing an old DOS game in 64-bit DOS won't make it better. 209.149.113.4 (talk) 16:06, 13 July 2016 (UTC)

What happens exactly when you clone your SIM card

If you clone your SIM card and use both the new and old card at the same time, how will the phone company react? How will they discover it? Is there a central server with a list of all numbers and where they are? What if one card accesses the network through roaming? Llaanngg (talk) 16:21, 13 July 2016 (UTC)

Yes, they do keep a database, See Network switching subsystem#Home location register (HLR), International mobile subscriber identity, and Mobility management. There are special clone SIMs that disables the first SIM when switched on, but as far as I know the carrier has to turn on support for such devices.

Also see:

• https://security.stackexchange.com/questions/64294/how-does-a-sim-card-prevent-cloning
• Radio fingerprinting (warning: poor quality article; needs to be expanded and references added)

--Guy Macon (talk) 18:13, 13 July 2016 (UTC)

Spy potential of Pokemon Go

I've been seeing a wave of newsvertisements for Pokemon Go that is unrivalled since, well, Facebook Live's snuff video PR campaign, including inducing players to jump a fence to find a dead body, and it's left me thinking about how useful this app may or may not be to spy agencies, police, or even ordinary private dicks.

• Who is in charge of placing the Pokemon monsters or other targets? Is there a straightforward way for a third party to set up an attractive feature on some property they want inspected?

• How do players find the targets they are looking for? Is there a general 'directory' that leads them to the land they are supposed to reconnoiter?

• Is the game known to upload the video the phone is taking in order to put the cartoon figure on the screen, or is that done locally?

• Does the game have a way of collecting data about things in the area, e.g. other cell-phones, Wi-Fi identifying numbers, etc.?

• If a player trespasses in order to happen across evidence, is there any possible legal argument that he is a "police agent" if it could be shown that police put the target on his map, or does his personal decision to trespass to play the game put an end to any attempt to claim a warrantless search?

• I've read about players being lured to police stations to hunt Pokemon, with cops complaining about it; is there any way to lure a particular player to a police station without luring all of them, say if that particular one had outstanding warrants?

• Are players being lured near crack houses and other highly hazardous features that someone might want to have periodic video surveillance of? I didn't see this one but it seems like an understandable "accident" to have happen.

Wnt (talk) 23:00, 13 July 2016 (UTC)

July 14

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