Wikipedia:Reference desk/Archives/Computing/2015 January 6
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January 6
editDifference between an LFSR and an NLFSR
editThe gist of feedback shift registers is that several units in an array of registers are used to generate the newest unit after the data has been shifted. But what is the difference between a linear feedback shift register and a nonlinear feedback shift register? Can I get examples of either one? — Melab±1 ☎ 01:32, 6 January 2015 (UTC)
- Take a long time and read/re-read linearity § boolean functions very slowly and carefully. It took me three or four read-throughs to decipher, and I'm already very familiar with this type of mathematics!
- Linearity is a specific, mathematical property. In the context of boolean algebra, the conventional definition is adapted to the boolean (binary) nature of the inputs and outputs, but it is intuitively the same.
- However, like many other mathematical definitions, we can't "simplify" the definition. The mathematical language in our definition is already exactly as succinct as possible to correctly describe what linearity means. So, just spend some time to wrap your head around the meaning of the definition, and intuitively understand why this definition implies certain useful properties and applications. A non-linear equation is harder to invert; if you want to use a shift register for cryptography, this is a useful property. Nimur (talk) 03:16, 6 January 2015 (UTC)
- Yeah, well examples in these articles always help. — Melab±1 ☎ 05:42, 6 January 2015 (UTC)
- An example of a LFSR is a (not very) pseudorandom bitstream produced by for , with representing logical exclusive or (addition in ) and being the seed. An example of a NLFSR is something like , with representing logical and (multiplication in ). The multiplication of previous bits makes it depend nonlinearly on those bits. Real-world examples are DVD-CSS (a stream cipher using LFSRs) and Trivium (cipher) (a stream cipher using NLFSRs). -- BenRG (talk) 07:12, 6 January 2015 (UTC)
- I do not understand that at all. I was hoping for examples in the form of shift registers themselves, like "XOR bits 5 and 7, shift to the right, and fill bit 0 with the previous calculation". — Melab±1 ☎ 16:46, 6 January 2015 (UTC)
- Does the sequence represent the register? Is the presence of solely XOR operations that make a feedback shift register linear? — Melab±1 ☎ 16:49, 6 January 2015 (UTC)
- No: formally, the bits represent an eigenvector of the linear function. The bits represent one instance of inputs into the function (i.e., any possible example of the values of register bits). So, a represents a characteristic of the function - i.e., the digital circuit; b represent the input to that circuit (i.e. the contents of the register) at any given time. Nimur (talk) 18:54, 6 January 2015 (UTC)
- Yes, what makes it linear is that there are only xor operations, and yes, in my notation represents the register. Initially its bits are . Then (in my simple LFSR) you xor the leftmost two bits together and shift that value in on the right, pushing the existing bits left (and the leftmost bit into the bit bucket). That gives you , where is the xor result. Then you do the same thing, getting , and so on. Using different indices for the different register values means that each has one unambiguous value, which is how mathematicians like to do things (and how some languages, like Haskell, do things). The destructive-assignment version would look something like while (true) { temp ← w + x; w ← x; x ← y; y ← z; z ← temp; }, where the bits are now called wxyz. Nimur is talking about something else is his reply above mine, probably the definition in the "Linearity" article. My a has nothing to do with that a and b -- BenRG (talk) 05:38, 7 January 2015 (UTC)
- NLFSRs are more random than LFSRs. Is it difficult to construct a secure NLFSR? How about an array of bytes that functions like a linear feedback register? — Melab±1 ☎ 00:59, 8 January 2015 (UTC)
- Yes, what makes it linear is that there are only xor operations, and yes, in my notation represents the register. Initially its bits are . Then (in my simple LFSR) you xor the leftmost two bits together and shift that value in on the right, pushing the existing bits left (and the leftmost bit into the bit bucket). That gives you , where is the xor result. Then you do the same thing, getting , and so on. Using different indices for the different register values means that each has one unambiguous value, which is how mathematicians like to do things (and how some languages, like Haskell, do things). The destructive-assignment version would look something like while (true) { temp ← w + x; w ← x; x ← y; y ← z; z ← temp; }, where the bits are now called wxyz. Nimur is talking about something else is his reply above mine, probably the definition in the "Linearity" article. My a has nothing to do with that a and b -- BenRG (talk) 05:38, 7 January 2015 (UTC)
What is the name of that dramatic effect?
editHi there,
I've seen many times,pictures with a dramatic effect.
Does anyone know its name?
https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-xpa1/v/t1.0-9/s720x720/10898278_768211989920203_7270092323090769989_n.jpg?oh=9eb390a0e3db245027e48861f63341ae&oe=553DA691&__gda__=1429152119_8de2849e52b5c407b33fbec7508205d2
Exx8 (talk) 14:09, 6 January 2015 (UTC)
- High-dynamic-range imaging -- Finlay McWalterᚠTalk 14:11, 6 January 2015 (UTC)
- For additional clarification, as explained in that article, the actual images shown tend to be HDR images (perhaps constructed from multiple exposures) that have had Tone mapping to try and simulate display with a medium only capable of a much more limited dynamic range. Nil Einne (talk) 15:19, 6 January 2015 (UTC)
- Basically, neither a camera nor a computer screen can capture the very brightest light and the very dimmest at the same time. Parts of a natural image are always either over-exposed or under-exposed. Human eyes are MUCH better at that kind of thing - which is why conventional photography often fails to capture the beauty of a sunset or some other kind of dramatic natural lighting. By taking a bunch of photos of the exact same scene at different exposure settings, then combining the best parts of each of them, you can get closer to something that we see with our own eyes. SteveBaker (talk) 15:50, 7 January 2015 (UTC)
- ... more aptly, you can produce an image that is closer to what our visual perception makes our brain think we see with our own eyes. In actual fact, our eyes have a slightly wider color gamut and a somewhat wider dynamic range than most camera sensors, but we think our eyes have a near-infinite dynamic range because of the way our psychology distorts the actual signals produced by the biological photochemicals and the nerve impulses in our eyes. Our article on visual perception is a great place to start reading. Nimur (talk) 16:21, 7 January 2015 (UTC)
- That makes me think we actually never get to see what our eyes really see, because our brains have evolved to enhance and correct the image on the fly so well we can't even turn it off. JIP | Talk 19:20, 7 January 2015 (UTC)
- ... more aptly, you can produce an image that is closer to what our visual perception makes our brain think we see with our own eyes. In actual fact, our eyes have a slightly wider color gamut and a somewhat wider dynamic range than most camera sensors, but we think our eyes have a near-infinite dynamic range because of the way our psychology distorts the actual signals produced by the biological photochemicals and the nerve impulses in our eyes. Our article on visual perception is a great place to start reading. Nimur (talk) 16:21, 7 January 2015 (UTC)
- That's certainly true. We can't distinguish a spectrally pure shade of yellow light (eg from a Sodium lamp) from a mixture of pure red and pure green light that happen to produce the same visual effect. So for sure, we can't see more than a tiny, tiny amount of the information contained in the visible spectrum. That's a limitation of our eyes that our brains try to fudge over by making the best guess as to what a particular combination of red, green and blue "really" is. There are many other examples of that. There is simply not enough bandwidth available through the optic nerve to send a few million pixels of data every dozen or so milliseconds - so you know that a lot of that data has to be lost in the compression done on the data. That becomes evident when (for example) you convert an image from PNG to JPEG and it looks pretty much the same - despite the fact that maybe 90 to 95% of the data was thrown away. That only works because the JPEG encoding standard throws away stuff like high frequency color variations at near-constant brightness that our brains can't detect. SteveBaker (talk) 21:21, 7 January 2015 (UTC)
Help for VICE
editMy question about VICE above isn't getting any replies. It's possible no one here knows. So where could I find a forum I could ask for more help? JIP | Talk 20:08, 6 January 2015 (UTC)
- You can try sending a polite email to the developers' mailing list, which can be found on their webpage: vice-emu-mail@lists.sourceforge...
- On that page, they have additional instructions to help you file a bug-report, and how to ensure your mail gets through their filters.
- Nimur (talk) 21:26, 6 January 2015 (UTC)
- I'm not sure it's a bug. I might just not be understanding how VICE works. Can I still file a report? JIP | Talk 21:33, 6 January 2015 (UTC)
- Yes. If it's not a bug, the developers will be happy to promptly close it! You can also ask them via email first. Nimur (talk) 23:06, 6 January 2015 (UTC)
- Do I have to be on the VICE mailing list myself to receive replies or will they reply to my own e-mail address as well? JIP | Talk 20:35, 7 January 2015 (UTC)
- I'm not sure it's a bug. I might just not be understanding how VICE works. Can I still file a report? JIP | Talk 21:33, 6 January 2015 (UTC)
- You could try at StackExchange, or either of these emulator fora [1] [2]. There seems to be a lot of focus on NES era consoles and MAME, but you might still have better odds than here. SemanticMantis (talk) 22:41, 6 January 2015 (UTC)
Alarm clock in Windows 8
editDoes Windows (specifically, Windows 8) have any sort of alarm clock in it somewhere? Let's say that I am working on my computer. I want the computer to "beep" in an hour to remind me that I have to do something at that time. Is there anything in Windows that can do such a thing? If possible, I would prefer something that is already there, in the Windows operating system, as opposed to something that I have to download. Thanks. Joseph A. Spadaro (talk) 22:52, 6 January 2015 (UTC)
- If you want an alarm at a particular time of day, and don't mind it popping up a message box instead of ringing, you can use Task Scheduler, as described here for example. If you want a timer that goes off 60 minutes after you set it, the only way I know to do it is to calculate the time yourself and add a one-time task at that time. You can automate it with the command-line schtasks utility; the command looks like
schtasks /create /tn CommandLineMessage /f /sc once /st 23:45 /tr "msg * MESSAGE"
- where 23:45 is the time it should alert you in 24-hour syntax and MESSAGE is the alert message. -- BenRG (talk) 06:05, 7 January 2015 (UTC)
- Thanks. I will look into that. I'd really prefer an audio alarm, rather than a visual. But, I will see how this suggestion works. Thanks. Joseph A. Spadaro (talk) 23:32, 7 January 2015 (UTC)
- One way to make an audio alarm might be to use Task Scheduler to start Media Player (or your preferred player) with an appropriate sound file. 175.45.116.61 (talk) 23:48, 7 January 2015 (UTC)
ping 127.0.0.1 -n 3600 & C:\Windows\Media\Notify.wav
— Preceding unsigned comment added by 82.44.45.127 (talk • contribs)
- sleep 3600 would be better, but yeah, that's an easier solution if you don't mind leaving the console window open for the full hour. -- BenRG (talk) 00:22, 8 January 2015 (UTC)
Thanks. Joseph A. Spadaro (talk) 19:48, 10 January 2015 (UTC)