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October 29

Malfunctioning Windows 10 on HP Pavilion

I recently bought a new HP Pavilion with Windows 10 OS. I understand now why MS has been pushing Windows 10 on the public free of charge. The OS is full of ads. Some of that I have been able to remove. Then there are two major problems. A message for updates comes in from MS and I click OK. The system begins to update, proverbial white dots begin to roll around a small circle, I can see the percentage of what has been done, until it reaches 70 or 71%. Then the progress stops and after 40 minutes or so the OS begins to undo the changes after saying that update could not have been completed. It's happened 5 times already, last time this morning. It take an hour for the whole procedure. It is irritating and offensive.

The second problem is that many windows I invoke (open) get frozen on the desktop, like IE, FireFox, etc. It drives me nuts. No button will respond to any click until I invoked the Task Manager with Ctrl+Alt+Del. Today is the first time when even this drastic intervention stopped working.

So, I have two serious problems. I wonder if anybody could comment on them.

I also have a Linux Ubuntu machine. It works like a clock. Very simple and effective. MS has packed this Pavilion with bizarre features nobody needs, like Cortana. I tried to uninstall Cortana but it is irreducible. How to remove Cortana? I am about to join Microsoft hate crowd. --AboutFace 22 (talk) 17:57, 29 October 2016 (UTC)

The first step should always be downloading and creating Windows 10 insallation media (See [ https://www.microsoft.com/en-us/software-download/windows10 ]), formatting the hard disk, and doing a new install without the stuff HP adds. Be prepared to spend some time waiting as windows installs updates.

Next you should remove the Microsoft-added bloatware. See [ http://www.makeuseof.com/tag/easily-remove-bloatware-windows-10/ ] or [ https://www.hackread.com/windows-10-is-spying-on-you/ ] to do this manually.

Finally, you might consider using Spybot Anti-Beacon. See [ https://www.safer-networking.org/spybot-anti-beacon/ ]

There are other tools that claim to do similar things:

[ https://www.whatswithtech.com/how-to-stop-windows-10-from-spying-on-you/ ]

[ http://neurogadget.net/2015/08/20/windows-10-privacy-tools/13016 ]

[ http://www.ghacks.net/2015/08/14/comparison-of-windows-10-privacy-tools/? ]

[ https://bgr.com/2015/08/14/windows-10-spying-prevention-privacy-tools/ ]

--Guy Macon (talk) 18:46, 29 October 2016 (UTC)

THANK YOU VERY MUCH. I APPRECIATE IT. --AboutFace 22 (talk) 00:49, 30 October 2016 (UTC)

"Hot air balloon problem"

Resolved

I just came up with this problem and am wondering about its computational difficulty (as in, its worst-case runtime).

A hot air balloon pilot wants to go from point $A\in \mathbb {Q} ^{n}$ to $B\in \mathbb {Q} ^{n}$. There is a function $\mathbf {v} (h):\mathbb {Q} \to \mathbb {Q} ^{n}$ which gives the wind vector at height h (here, "height" is not a position parameter, but instead, considered as just a parameter the pilot can increment/decrement - the height will not change if the pilot does not decide so). Because weights are a finite resource, the pilot can only move up a given number of units $K_{up}\in \mathbb {Q} ^{+}$ n times, and can only move down $K_{down}\in \mathbb {Q} ^{+}$ m times. Every minute, the pilot has three choices: go up by K_up, decrementing his n "up" movements, go down by K_down, decrementing his m "down" movements, or maintain the current height, which implies maintaining the existing velocity (for simplicity, assume v is time independent and depends only on height). Each minute, regardless of the decision, the velocity vector is added to the current position (so the vector is always in units of position per minute). Assume that height changes are fast enough such that one can ignore the effect of winds between the starting and ending heights for every height change. The decision problem then is, can the pilot make this journey? Formally, is there a sequence of height increments starting from height 0, as described above, such that the sum of all the resultant velocity vectors is the position vector from A to B?

This problem is clearly in NP because any solution could be easily verified by just walking the proposed solution sequence. But is it in P, or is it NP-intermediate or something?--Jasper Deng (talk) 19:06, 29 October 2016 (UTC)

If I understand this question correctly. Your asking about forecasting a flight path and thus being able to choose altitude/wind currents rather than using app like [1] to log your flight path/track flown. A bit like being able to fly an aerial route map.--Aspro (talk) 20:45, 29 October 2016 (UTC)

@Aspro: No, it should've been clear that this is a theoretical computer science problem I just came up with. It has many unrealistic assumptions that would preclude its use in practice.--Jasper Deng (talk) 22:12, 29 October 2016 (UTC)

Assuming there is a finite number of heights, then all the paths will be a linear combination of the f(v) values. Which ones of these can get you from A to B can then be solved as a simultaneous linear equation, that will give the times that you spend at each height. This will take polynomial time. Note that there may not be a single solution. So then you have to intersect your constraints on n and m. This will result in a more limited simplex that will contain all the answers - possible routes. If you are limited to doing this on minute boundaries, and not at arbitrary times, then you have more contraints, and will likely have no solution. But in anycase this should be solvable or proved you can't do it in polynomial time. Graeme Bartlett (talk) 01:03, 30 October 2016 (UTC)

The existence of such a linear combination is necessary, but not sufficient for a solution. The way the problem is constructed, one cannot "jump levels", i.e. I cannot go from height N to N + 2Kup without first passing through N + Kup or other intermediate levels (depending on K_down and K_up and whether they're equal). I'm not sure you could formulate these constraints as linear ones, especially when K_up is not the same as K_down.--Jasper Deng (talk) 01:27, 30 October 2016 (UTC)

I am assuming that you spend no time in your transitions of height. The possible levels should be n×m or less. Just being able to move one up or one down at a time is just yet another constraint. It would result in a smaller intersection of the simplexes with less than n×m dimensions. Read Linear programming, and Simplex algorithm. If it is an Integer programming problem, then it becomes much tougher. Graeme Bartlett (talk) 01:51, 30 October 2016 (UTC)

That assumption is correct. I have learned integer linear programming in the past but am unsure of its applicability here. To be clear, yes this is an instance of integer programming, because we are really looking at an integer lattice generated by the different values of v, since one may not take nonintegral linear combinations here. All known algorithms for a general ILP run in at least pseudopolynomial time (since ILP is NP-complete). An $O(mn)$ solution would still be only pseudopolynomial in m and n. Even if we did not restrict it to integers, I'm not sure of how one could encode the m and n constraints in only polynomially many inequalities.--Jasper Deng (talk) 04:12, 30 October 2016 (UTC)

Your problem-setup assumes perfect a priori knowledge of the field (the wind function at all positions), correct?

Your algorithmic running-time bound - and the result of your decision-problem - depends entirely on the properties of this field. If the field is entirely unconstrained, the decision is much harder - you might have to traverse every possible path, which is a graph search in dimensionality n × m × h. (This is not a realistic model of normal wind, but it's how you described the problem: wind can cross other flow lines and speeds can vary without constraint). If the field is uniform at all positions, the decision problem is easy, almost trivial. If there is some constraint, like a smoothness constraint, on the field, there might be a more efficient way to reduce the span of your search space.

I think I'd begin by looking at our article on graph connectivity to see what prior art exists. It's a safe bet to say that your problem is a special-case of some already-published theoretical work. We just need to figure out what obtuse mathematical keyword describes your graph and its connectivity property.

Nimur (talk) 13:53, 30 October 2016 (UTC)

The wind field has no smoothness or other conditions imposed on it, but yes, you know it as a given. I wanted to avoid thinking of it as a graph problem: a graph to represent it would ignore the vector arithmetic aspect of this problem, and IMO would not be easier than an integer programming formulation, as described above.--Jasper Deng (talk) 19:20, 30 October 2016 (UTC)

Because the question is "can the pilot make the journey?", I would place it in P, easily calculable using lazy evaluation. If, instead, you asked "what is the optimal journey?", I would not place it in P because it is unlikely that there is a linear solution that finds the optimal journey. Nearly all problems can be divided like this. The classical Traveling Salesman is hard because it asks for the minimal path. If, instead, you asked for any path, it would be an easy problem. 209.149.113.4 (talk) 11:33, 31 October 2016 (UTC)

@209.149.113.4: And your algorithm is? Lazy evaluation is not relevant here. And also, this asks for a solution within the aforementioned bounds on m and n height changes - therefore, one could easily restate the problem here as an optimization problem w.r.t. either of those variables. Also, a pure decision problem in NP is also generally not known to be in P - 3-SAT makes no mention of optimization at all and is NP-complete.--Jasper Deng (talk) 15:36, 31 October 2016 (UTC)

The discussion above seems to imply that there are nearly infinite paths to get to any position. Because this is limited to specific positions, I prefer to think in terms of graph theory and call each position a node. The reality is that there are very few paths to get to any specific node. You have a source (starting point) and sink (ending point). The starting point itself eliminates a large chunk of the possible paths. Think of a standard Cartesian graph. Assume the starting point is as 0,0 and all the nodes are in a large rectangle from 0,0 to 100,100. If the balloon can only rise or fall 1 node on the Y axis per turn, that means that the balloon can never reach 0,2 because that requires a rise of two nodes. Similarly, if the goal is to land at 100,0, being at 100,2 is failure because you can't drop two nodes. Being at 99,3 is failure. Being at 98,4 is failure. Of course, this all assumes that winds blow in a positive direction from X=0 to X=100. It should be apparent that a lazy evaluation from 100,0 would eliminate 1/4 of the nodes. Moving forward from 0,0 would eliminate 1/4 of the nodes as well. But, a lazy evaluation tends to be quicker at asking "can it be done." From 100,0, how can you get there? Only look at nodes that could possibly get you there. From those nodes, how can you get to them? Only look at nodes that get you to them. Keep backtracking. If you hit 0,0, it can be done. If you hit all dead-ends, it cannot. My gut feeling is that this could be converted into a static graph in linear time. Then, you have a source and sink on a static graph that just screams to use Dijkstra's algorithm. 209.149.113.4 (talk) 14:17, 1 November 2016 (UTC)

@209.149.113.4: Except, that's just wrong. Please re-read the problem: the vectors' components are in general not of the same sign and "height" is a parameter separate from the position vector (so one does not talk of a "target height"). Also, in general, $K_{down}\neq K_{up}$. And in any case, you cannot solve this, nor can you reduce it, in linear time (or polynomial time at all, in fact), using your purported strategy, even in the special case you mentioned. This would at the minimum require $\Omega (|A-B|)$ time - pseudopolynomial in the storage size of A and B. You also ignore the constraints on the number of height changes allowed, which makes the allowable future travels from a given point dependent on previous movements.--Jasper Deng (talk) 14:30, 1 November 2016 (UTC)

I read that the pilot begins at height 0. Then, the pilot can increase or decrease height at will, once per minute. Therefore, height is not a continuous vector. It is a discrete set of values. I assume it cannot go negative. It obviously has a maximum value, which is the minimum of how many times it may increase height or decrease height (assuming the goal is to end at height 0). So, ignoring everything else in the entire problem, one part of it is to go up and down, beginning at height 0 and ending at height 0, correct? If not, you really need to restate the problem.

Next, you state that you want to know if the pilot can make the journey. What journey? A journey has a starting and ending point. You don't clearly state what the starting and ending point might be. I assume you have a starting point you will call point A. You then want to travel to point B. The goal is therefore to begin at height 0 at point A and end at height 0 and point B. If that is not the case, then you really need to restate the problem.

Assuming that the two above assumptions are correct, you have a pilot that begins as position A,0 and wants to go to position B,0. Why not restate that with numbers: 0,0 to 100,0. So, in this case, the journey is 100 units. We can consider the first number to be X on a Cartesian plane and the second number (height) to be Y. For every Y, you claim that there is a wind vector that increases/decreases the X velocity of the pilot. The pilot can increase/decrease Y or stay at the same Y per turn (minute). If that is not correct, you really need to restate the problem.

Assuming that everything is correct so far... This problem CAN be written as a graph theory problem. It CAN be written as a directed node problem. You keep claiming that it cannot be reduced to a simple graph without explaining why you believe that it cannot. I personally would treat it as a 3D graph. You begin at 0,0,0 (the third dimension is X velocity). What can you do? Go up. There is an issue here. Does the wind at height 1 affect velocity when you move into height 1 or does it affect velocity only if you remain at height 1? I will assume it affects you as soon as you enter the height. So, if wind velocity at height 1 was 3, moving up from 0,0,0 would move to 0,1,3. From there, you have 3 choices. You should see a decision tree (graph) forming. Just don't use depth-first search on an infinitely-sized decision tree. 209.149.113.4 (talk) 16:17, 1 November 2016 (UTC)

@209.149.113.4: No, please don't put words in my mouth. I never said it couldn't be reduced. I said it couldn't be reduced in polynomial time. I thought the problem's statement is clear: we have $A,B\in \mathbb {Q} ^{n}$, and height was explicitly stated to be a parameter independent of those. I also never restricted height to be nonnegative. The semantics of the wind vector were a bit vaguely stated, but can be formalized as follows: on taking the ith decision, you add the wind vector at the ending height to your total displacement. So if I start at the origin and increase my height, my displacement is now $\mathbf {v} (K_{up})$. You also keep ignoring the fact that in general, $K_{up}\neq K_{down}$ and are nonintegral (so at best, you're only treating a special case), and like I stated, due to the constraints m and n, your current displacement and height is insufficient data to determine what paths are allowable from a given point. And in any case, even your reduction is not polynomial time, and can't be. My problem also never constrained h = 0 at the end, only the start (in real life, one would parachute out!).

In fact, I now can see that this problem is NP-complete by reduction from a variation of the subset sum problem. Suppose that we have a finite set of integers |S| indexed with natural numbers i from 1 to k = |S| (i.e. the ith element is denoted S_i; assume without loss of generality that this indexing represents a greatest-to-least sorting). Let A and B live in one-dimensional space. Set A = B = 0. Set m = n = 2k. Let $K_{up}=2,K_{down}=1$. Let v(h) be given as $S_{l}$ if $h=2l-1$ for integers l from 1 to k, S₁ at height 0 (ruling out the trivial solution), and zero otherwise. Then any admissible path consists of a nonempty multiset of integers in S summing to 0 - a variation of the subset sum problem that is NP-complete per this. Any subset of the values in S can be used: if I am at an odd-numbered height (with nonzero wind vector), I can skip the following entry of S by first going down by 1, and then going up by 2 twice. If I am at an even-numbered height and want to use the value of S at the next odd-numbered height, then I go up by 2, and go down by 1. At most, I need two up and two down operations per each of the k levels, hence the values of m and n.

Please, actually read what I wrote. Unless you are claiming P = NP, there is no way this problem is easy with a graph algorithm.--Jasper Deng (talk) 18:18, 1 November 2016 (UTC)

I believe this is where the problem comes in. Quoted from above: "...my displacement is now $\mathbf {v} (K_{up})$." I read that as claiming that the displacement jumps to whatever the vector magnitude might be at that height. But, within the real-world example of a balloon, wouldn't the displacement be changed by the vector instead of replaced by the vector? For example, if my displacement was currently 13 and I moved up to a height with a magnitude of 3, my displacement would be 16, not 3 - correct? If that is correct, which I believe it is, then this is very similar to a summing or knapsack problem. There is a catch though. Once I pick a height, I immediately eliminate many other options. That is based on my understanding that you can only move up or down a single "unit" of height. I cannot go from height 1 immediately to height 3. I have to pass through height 2 and be affected by that wind first. Then, it appears that the goal is to reach a specific value for displacement from the start, regardless of ending height. I know you keep pushing that height and displacement have no relationship to one another. I do not see how you can frame this as a balloon moving around and not apply both height and displacement to a single balloon. Are there two balloons? One has height and the other has displacement? I believe it is one balloon with height and displacement. But, it keeps getting clearer that what I am reading and what you are writing have practically nothing in common. 209.149.113.4 (talk) 19:36, 1 November 2016 (UTC)

@209.149.113.4: I also stated that height changes are atomic: You are affected only by the wind at the height you end up at after each view. "Displacement" is a vector quantity so I can't fathom why you are mentioning it as a scalar above. And like I said, this is a theoretical problem with little practical value; the "hot air balloon" construct is only an analogy. Also, your "unit" of moving up and down is once again generally not the same for those two operations.

When I say "height" is an independent parameter, I mean precisely that. It's every bit as independent of position as time is (don't even talk about relativity). Is this physical? No. But can I come up with this in theory? Definitely yes. If you can't understand what I think is a pretty clear statement of the problem, then I'm not sure what to say. Take it at face value - don't overthink it. The quoted text you referred to refers to the special case of starting from the origin. If I am working in two dimensions and I am already at (5, 24), and am making a decision while at height 9, then after choosing to move up, I update my displacement vector to (5, 24) + v(9 + K_up).

I must remind you that you originally claimed that this problem was in P. Good luck with that.--Jasper Deng (talk) 19:52, 1 November 2016 (UTC)

USB 3.0 Hub ports

Moved from Wikipedia:Reference desk/Miscellaneous § USB 3.0 Hub ports

 – 47.138.165.200 (talk) 22:26, 29 October 2016 (UTC)

1) What do call a hub box that consist of a Micro SD card slot, memory card slot along with few USB insertion ports? An example of a design or just the name of what it is called as a whole, is sufficient.

2) This point is irrelative to the above and is as follows:

My Laptop is from the year 2012 or so. I think the USB box port[s] attached is of v2.0. I wonder if v3.0 wired products will function as it should thereafter connecting to this v2.0 box, or, do I require a USB 3.0 box in order for a v3.0 wired product(s) to function efficiently? — Preceding unsigned comment added by 103.230.104.22 (talk) 19:08, 29 October 2016 (UTC)

If I understand your description correctly, it's generally called a "USB memory card reader" or something along those lines. USB is designed to be forwards- and backwards-compatible. Any USB 3.0+ device should work with a USB 2.0 port of the same connector type, but speeds will be limited to 2.0 speeds. --47.138.165.200 (talk) 22:37, 29 October 2016 (UTC)

I would call it a memory card reader with USB3.0 hub (or similar). That seems to be what other sites call it too [2] [3] [4]. (Last one is a 2.0 only.) Nil Einne (talk) 01:57, 30 October 2016 (UTC)

Any idea why/when the DC Power Plug should be used?; I'm happy with Nill's Point no: 1. However, I'm looking for more ports, 10+ or so. 103.230.105.20 (talk) 19:14, 31 October 2016 (UTC)

Again assuming I'm interpreting you correctly, the power plug is to provide power to the ports in the USB hub. Typically the hub can operate with or without this; you only need it if you plug in a USB device that draws a lot of power. --47.138.165.200 (talk) 03:08, 1 November 2016 (UTC)

.iso files don't get burned

I have a Windows 7 laptop. When I click on an .iso files the system shows it's readiness to burn it to blank disk. I put in a blank DVD and trigger process. But after all the show that disk is being written to, it turns out that disk is blank (though a closer visual inspection of DVD shows that slight darkening of tracks that for a moment makes one believe that data has been written to it.) But put it back in computer and it's empty. Why so happens ? Some overlook on my part in contemplating the process ?210.56.108.118 (talk) 15:48, 30 October 2016 (UTC)

Did you write click on the iso, select burn, and then select the destination drive? Graeme Bartlett (talk) 03:46, 31 October 2016 (UTC)

Another point to check - does the ISO file match the medium? You're not trying to burn a CD ISO on a DVD, or a DVD-DL ISO on a single-layer DVD? Tevildo (talk) 16:46, 1 November 2016 (UTC)

October 30

Bit and Qubit Register

Two part question:

1. I'm not really sure what a bit register is or what is so special about it. It just seems like a bunch of bits in a row! Could someone explain somewhat intuitively what it is and why it is important?
2. Going on with the bit register, I'm reading about quantum computing and a "qubit register" was one thing that came up. What is this/why is it so special? Again, this just seems like a bunch of qubits in a row. I do know that it can be written like |A>|B>|C>...|Z> or |ABC...Z> where A, B, C,...,Z are qubits, but I don't really understand intuitively what is going on here.

Any help would be appreciated, thanks! Hnerd (talk) 19:58, 30 October 2016 (UTC)

You can look at our articles Processor register and Quantum register, though they may not be helpful. Hardware register and flip-flop (electronics) may be more useful. You are correct in that they are a bunch of bits stored in a row. They usually have facilities to do things with the bits, such as shift register or other bitwise operations. They will have ways to load, and store values, ways to test what is there, and likely ways to clear them to all zeros. Graeme Bartlett (talk) 03:40, 31 October 2016 (UTC)

This article shows an animation of a small 4-bit register. AllBestFaith (talk) 17:53, 1 November 2016 (UTC)

Waiting for USB-C

I'm thinking about buying a high-end Lenovo X1 Yoga laptop (the model with the OLED screen and 16 GB RAM (which unfortunately only comes with the i7-6600U which I wouldn't choose if I could get more than 8 GB on this machine with an i5, but the RAM is soldered and configurations are limited)), and was wondering if I should really be waiting until the next gen comes out with a USB-C port. Can you guys point me to 1) a site that predicts when the next X-1 Yoga will be available based on Lenovo's release cycle and 2) a description of what I might be missing in a year or two by having a laptop with three USB 3.0 ports but no USB C. It seems that the big USB-C advantage is having a single cable type and single port type for multiple purposes. Is this all just a matter of convenience, or is there something that I will likely be wishing I could do a year or two from from now, "If only I had USB-C."? -- 66.203.139.164 (talk) 22:26, 30 October 2016 (UTC)

Yes, it's mostly convenience, as well as future-proofing. USB Type-C is an attempt at making "one cable to rule them all" for modern external communications buses. You likely won't miss much unless you're devoted to getting all the latest bleeding edge gizmos. At most you'll need a hub or adapter, if you get something that only has a USB Type-C connector and want to connect it to your laptop. --47.138.165.200 (talk) 03:23, 31 October 2016 (UTC)

See also this and this. --76.71.5.45 (talk) 05:34, 31 October 2016 (UTC)

While what you right is mostly correct, there are some things which will be different. For starters, if the device is USB type C, it may also be USB 3.1 gen 2 so will be capable of double the data rate, although I suspect it'll be quite difficult to find a device where you can notice this.

Additionally, many fancier Intel based laptops with type C also have Thunderbolt with it. Does the device you mentioned have Thunderbolt at all? You mention high-end so I would expect it to. Are you likely to want to use it? Thunderbolt (interface)#Thunderbolt 3 also has several advantages over earlier versions. Outside of certain niche products, Thunderbolt doesn't seem to have been that successful but it's possible this will change with USB type C. Notably, there's currently a big move towards support external GPUs which so far haven't been that successful.

Some (many?) laptops using type C also seem to support 1.5A, or even 3A for charging devices. Type A adapters are not supposed to enable these power delivery modes, it's incredibly non-compliant and generally results in scathing reviews from Benson Leung [5]. You could use an external power adapter (something you plug into the wall) or power pack (something with a battery) with a type C port to charge your devices, but that's more than just a simple adapter and so may be annoying to take around with you and use. While there are various means to deliver more than 500 or 900mA over type A, this relies on a variety of different signalling methods to indicate support. Laptop support for these isn't always good. Worse, device support for these can be spotty. Perhaps not surprising, Apple's method which allows up to 2.4A is best supported. But it seems this is actually being abandoned by some devices which use type C. E.g. from what I've heard, the Nexus 5x does not support these methods so is limited to 500 or 900mA over type A unless you use a non compliant adapter. Note that even if the laptop supports outputting more than 500 or 900mA over its type A port, it's possible it expects the signalling methods to tell it and won't be happy by a non-compliant passive adapter which simply tells the device it can draw more. In other words, without type C, newer phones and tablets may charge slowly off your laptop but could potentially charge a fair amount faster with type C. </p

It's possible charging advantages could be in reverse. Some laptops, especially those with Thunderbolt 3, seem to support charging over the type C port, even if they have their own charging port. They may need some form of USB higher voltage power delivery for this to work. And from what I've read implementations are a bit hit and miss and the moment, with limited info on what's required, so some devices will be able to charge some laptops but not others. Some even seem to use non standard methods although I think a lot of these were designed before USB power delivery over type C was finalised. Still if it does work, it may mean you can use a power pack you have for your phone, or an adapter to charge your laptop in a cinch.

There are a bunch of other scenarios which could arise e.g. relating to DisplayPort and HDMI. A minor point but outside of charging, the simplest and most common adapters for type C are small type C plug to microUSB socket. These are easy to carry around and work well because microUSB (to type A) cables are fairly common or you can carry your own. You can also carry a type C to type A cable. What is a lot less common is type C to type A USB3.0 cable or adapter. You could get one and carry it around but you may want to carry a type C to microUSB adapter with you at all times (to make it easier to attach to chargers and other stuff with microUSB) and maybe even a microUSB cable (especially if you have other devices you take with you) and maybe even a type C cable (if there's anything type C you could attach it to. If you don't carry a USB3.0 cable or adapter, this means if your phone or tablet does have USB3.1 gen 1 (i.e. USB3.0) or USB3.1 gen 2 speed support, you are potentially not going to be using it with your laptop when you're carrying it around. Or to put it a different way, especially once you have a phone or tablet or whatever with type C and assuming you ever want to attach it to your laptop, not having type C on your laptop is likely to mean a lot more adapters than having a laptop with type C but no type C or tablet.

Nil Einne (talk) 07:58, 1 November 2016 (UTC)

Thank you. -- 66.203.139.164 (talk) 16:11, 2 November 2016 (UTC)

November 1

Bypass computer's soundcard while using headphone amp?

So I just bought a USB headphone amp, the Fiio A3. Perhaps I naively assumed I could link it to the computer via USB and it would somehow bypass the computer's onboard sound card and just forward it to the amp, but that appears to not be the case? Looks like the only way is to run an 8th inch cable out of the computer's headphone jack into the aux-in of the amp? That's dumb... Is there a way to forward the audio output via USB? Thanks! NIRVANA2764 (talk) 16:03, 31 October 2016 (UTC)

I can't find a definitive manual for this, but everything I'm seeing on line suggests it's just a headphone amp with a battery, and that the microUSB port is solely for charging the battery. If you wanted to bypass the onboard sound card, you would have needed to buy a USB external soundcard. -- Finlay McWalter··–·Talk 16:27, 31 October 2016 (UTC)

Do you think the external soundcard bit would make a world of difference? NIRVANA2764 (talk) 16:29, 31 October 2016 (UTC)

For the average person (youtube, spotify, games), for output, probably no. If you wanted to drive some fancy multi-channel home theatre thing, maybe, I don't know. If you were involved in amateur or professional musical or audio production or editing (e.g. music recording, webcasting) where you needed to hook up decent microphones or musical instruments (often with an XLR or instrument-cable connection) then you want a USB audio interface like a Focusright or a Presonus, and not a "soundcard" per se. -- Finlay McWalter··–·Talk 16:38, 31 October 2016 (UTC)

Cool, thanks. I'm just gonna be using it with my laptop playing mp3s through some Shure SRH440s or Grado SR60s, just wanted something to make the sound louder and beefier but now I'm wondering if I should return this and pay a bit more for something with a DAC... NIRVANA2764 (talk) 16:43, 31 October 2016 (UTC)

Most evidence I've seen based on actual scientific inspired testing rather than flawed (i.e. non blind) "audiophile" testing or anecdotal reports suggests nowadays the DAC in most computers is good enough and the audiopath is likewise clean enough, so a good enough amp/driver is all you're likely to need (if that). Whether your device qualifies I have no idea. In other words, while it may seem nice to keep the signal digital as long as possible, ultimately it's probably better to ensure the device you have is sufficient rather than worrying about whether it has digital or analog input. The Hydrogen Audio forums [6] is a good place to get advice on consumer products largely free from the nonsense that seems to pervade a lot of audiophile groups. While it mostly focuses on the digital side of things, there is a place for discussions of hardware. Note that while USB sound cards are common, you could also use a device with HDMI input or TOSLINK/coax S/PDIF assuming your computer has them (I'm assuming it has at least one). Nil Einne (talk) 01:57, 1 November 2016 (UTC)

If you really want to go the DAC route, I noticed some discussion here [7] although it's mostly concentrated on speakers. I also noticed the FiiO D3 (D03K) seems to get good comments on those forums e.g. [8] [9], and it seems to be quite cheap although I don't know if the amp is good enough for headphones (which again is the main factor since it's quite difficult to design a DAC which isn't good enough to be come irrelevant nowadays). For pure amps the open source Objective2 headphone amp and CMoy seem to be interesting choices [10] although if you're not making them yourself, I'm not sure if they're necessarily and cheaper or better than more commercial products if you concentrate on the actual performance of these products rather than the hype and don't get suckered in my products which aren't actually any better than cheaper alternatives. Nil Einne (talk) 02:29, 1 November 2016 (UTC)

Text limit in 4000 bytes

So my friend asks me to fill out her university application form for a university in South Korea, and one of the requirements in the application is to fill some sort of personal statement with a maximum size of 4000 bytes.Now I find it really weird, because usually people define the limit as in word limit in essay, or character limit as in SMS or Twitter. But bytes? Is this having something to do with UTF-8 and Korean language? How do you even count it? Fëanor Engineering (talk) 07:53, 1 November 2016 (UTC)

SMS are actually limited to 140 bytes each. The number of characters depends on the code set used. Nil Einne (talk) 08:03, 1 November 2016 (UTC)

Should also mention that Concatenated SMS also requires a few bytes for the header to allow recomposition hence have fewer characters per SMS than a single SMS. Nil Einne (talk) 13:00, 1 November 2016 (UTC)

In UTF-8 there would be 2 bytes per character. So that would be about 2000 Korean characters. Graeme Bartlett (talk) 08:31, 1 November 2016 (UTC)

Actually a 2-byte UTF-8 sequence can only encode Unicode characters between U+0080 and U+07FF. All Hangul glyphs have values larger than U+07FF, so it takes three bytes to encode one Hangul glyph. So 4000 bytes would only allow 1333 Hangul glyphs encoded in UTF-8. Encoding in UTF-16 or UCS-2 would allow 2000 Hangul glyphs, if such an encoding is allowed by the university. CodeTalker (talk) 13:56, 1 November 2016 (UTC)

Is there a chance you could contact the university admissions office and ask what coding they use, or ask for additional technical details regarding the size requirement? And if you're concerned about it affecting her application, just contact them in your own name and don't mention whom you're helping. I doubt that it's something that is of relevance to the application itself (it's not like the word limit that you mention, which might be imposed so that the admissions folks don't have to read ten-page essays), so they probably wouldn't mind explaining how they calculate the total size. Might as well throw in something extra, like "I'd like to know beforehand, so I don't write up something that's a little too long and have to weaken it by cutting parts after I've finished". Nyttend (talk) 18:05, 3 November 2016 (UTC)

LG-639 Widgets

A webpage listing all the widgets available along with a guidance on how to install each widgets please? 103.230.106.4 (talk) 18:33, 1 November 2016 (UTC)

November 2

human

[Question moved to WP:RD/S by Fortuna Imperatrix Mundi.] Tevildo (talk) 21:05, 2 November 2016 (UTC)

Угьш ъькяея шгся сяякеБ redux

On 24 October I left a hasty note asking "Угьш ъькяея шгся сяякеБ", which was reverted as "gibberish". I repeat the question - in a more readable form: "What makes this happen?" "This" as astute readers will have realised, is a keyboard suddenly switching to Cyrillic. Platform is one of the the library's Windows 7 machines.

Supplementary, of course, is "How do I get it back to normal if it happens again?"

All the best: Rich Farmbrough, 23:43, 2 November 2016 (UTC).

This is the relevant page from Microsoft. This is a more advanced troubleshooting procedure if the obvious steps don't work. See also Language bar. Tevildo (talk) 00:04, 3 November 2016 (UTC)

• I'm curious about how you typed that. When I switch to Russian (on MacOS) and try to type "What makes this happen?", I get "Црфе ьфлуы ершы рфззут?". How many wildly different Cyrillic layouts exist? —Tamfang (talk) 20:21, 3 November 2016 (UTC)

Tamfang I too am curious. I did not do any of the steps Tevildo mentions (and no admin access I think), useful though they may be in future to fix such an issue. At most I may have pressed CtrlAlt\ or some such. Code pages such as Windows-1251 and Macintosh Cyrillic encoding might be relevant. All the best: Rich Farmbrough, 20:52, 3 November 2016 (UTC).

I, too, wondered; the second word has six letters versus five in "makes", and the "happen" word has the second letter doubled, as if it were "haapen". Of course, if you made a couple of typos, that would resolve both issues. Nyttend (talk) 22:53, 3 November 2016 (UTC)

Угьш ъькяея шгся сяякеБ
What makses this issken

Assuming that the unique letters and some critical ones (H, S, etc.) are correct, I get this transcription. An odd set of typos, but when the computer isn't putting out what you meant to put in, it's not surprising that you'd not catch some mistakes. Nyttend (talk) 22:58, 3 November 2016 (UTC)

November 3

Adobe products for students

Which non-free adobe softwares are meant for students. If I am looking for time pass, not professional or business works. 1.39.39.187 (talk) 14:06, 3 November 2016 (UTC)

Many schools use site licenses with Adobe to provide students access to non-free products. SemanticMantis (talk) 17:06, 3 November 2016 (UTC)

Transparent cell copying in Excel

In Windows Paint, it's possible to copy/paste blank space: if you check "Transparent Selection" in the "Select" dropdown, all pixels that are precisely the same color as the secondary color (i.e. the one that's created when you use erase) will get ignored when you copy/paste part of the screen. If your background color is white, you can copy

 1 1

into

1 1

and get

1111

That's all very convenient, but I'm trying to work with Excel. In this program, it seems that empty cells are always treated as "actively" empty, i.e. comparable to being treated as white pixels in Paint instead of being totally ignored, as in my example. Is there a possibility of copy/pasting "blank" or "transparent" cells in this manner, so that when I paste an empty cell onto a cell with contents, the target cell remains unchanged? Basically, I'm in a situation with the following cells:

1 1 1 1 1 1
 1 1 1 1 1 1

I'd like to end up with a single row of 111111111111 without retyping everything or copy/pasting individual cells (this is a highly simplified version of my real situation), but I've not been able to figure it out with Google or by talking with my library reference desk. It's always been easy to do this with vertical stuff, e.g. taking

1
 1
1
 1

and putting it into a single column (just move it to Notepad; empty cells become tabs, and use find/replace to delete the tabs), but Notepad doesn't have a way to get rid of line breaks as it does column breaks. Nyttend (talk) 18:14, 3 November 2016 (UTC)

• Try this. Select the column or row you want to paste into. Right click and select "Paste special". Select the checkbox that says "Skip Blanks". Select "OK." This seems to work in Excel 2013. --Jayron32 18:26, 3 November 2016 (UTC)

• Works, all right. It totally avoids the risk of transcription errors while saving a ton of time. Thanks a lot! Nyttend (talk) 18:36, 3 November 2016 (UTC)

A more spreadsheety solution would be to create a new row, enter "=Concatenate(A1,B1)" in C1 and drag across. If they are really "1", then "=A1+B1" or just pressing the Sigma button will do.

If you want to get rid of line-breaks in text, consider using Notepad++ which has regex search and replace.

All the best: Rich Farmbrough, 20:58, 3 November 2016 (UTC).

Thanks for those, too. The situation actually wasn't just ones, and I've encountered other situations in which I didn't want an empty cell to get pasted on top of a cell with contents, so this route the thing I needed best in the long run. Nyttend (talk) 22:51, 3 November 2016 (UTC)

November 4

Is there any current manufacturer of Canadian multilingual keyboards?

I would like to know whether keyboards compliant with CAN/CSA Z243.200.92 and approved by the Office québécois de la langue française still exist. I wasn’t able to find a recently produced keyboard in e-commerce websites (eBay, Amazon) which can be shipped to Europe. I would like to test it. Thanks in advance.--Carnby (talk) 16:18, 4 November 2016 (UTC)

Getting the most out of a laptop battery, when all you want is a typewriter

How can I obtain the longest running time when I only want to type a text on my laptop? Is any OS (maybe some Linux dist) optimized for this? I don't need any fancy graphics, indeed, I don't need any at all. If necessary I can work in a command-line type of environment, no matter how ugly it looks like. If emacs is available, that would make me happy, but it's not a must.--Hofhof (talk) 18:11, 4 November 2016 (UTC)

Use a Linux distro and boot to command-line, I guess. emacs is available on every general-purpose distro, and it works just fine in a terminal (unsurprisingly, as emacs is older than modern personal computers). I'm not sure you'll see much of a difference with modern hardware though. But I could be wrong, and obviously you can experiment and compare battery life with a full GUI environment versus CLI. If you really want to maximize battery life, you'll likely want to do some fiddling with system settings, including possibly the kernel. PowerTOP will likely be useful for figuring out what's using power, and you can find tons more advice with Web searches for things like "linux laptop power saving". (If you settle on a distro, try adding your distro name to the search as well.) --47.138.165.200 (talk) 23:10, 4 November 2016 (UTC)