DD-WRT isn't what it used to be anymore, the build environment is a mess, a lot of drivers are binary only and often you have to dig around in the forums to find which builds are stable. I had a lot of bad experiences with consumer grade wireless routers (Linksys WRT610N... $#!&) and building my own router was the best choice I ever made.
You can use OpenWRT on a broad range of devices and it has similar features as DD-WRT and also a very nice web-interface. It runs perfectly on low power embedded PC's such as PCEngines Alix (X86) or Ubiquity RouterStation (ARM). Nice enclosures and also complete pre-built systems can be found on eBay.
The best choice for wireless network cards would be Atheros based, using the ATH9K driver in Linux. Ubiquity (www.ubnt.com) makes some very decent high power versions (SR-71 series) and Wistron DNMA92 is perfect as a budget solution (can be found on the pcengines online store). The RouterStation Pro and some of the Alix boards allow you to connect multiple wireless network cards for Dual Band radios. I would strongly suggest to use 5 GHz in addition to 2.4 GHz for the devices that support it. The 2.4 GHz range is overcrowded.
16GB is nothing, just playing with large amounts of TNT in Minecraft requires this.
An average server has 32+ GB when hosting some decent VM's. Windows Server with Exchange will eat up 16GB in no time as well.
The odds will be quite big if the guy diving into your can of used cat litter is the same guy that has been been watching you with a telescope from across the street while you were recording those "private video's". Even bigger if you're a celebrity.
Or of course if you work for the CIA, FBI, NSA, etc... for some reason lost USB sticks with secret information always seem to get found somehow...
At least with your own computer you can delete the files, use encryption or simply throw the hard drive away in the dumpster. If you don't want your data being disclosed I wouldn't throw a harddrive in the dumpster. Who knows who'll put your "private video's" on youtube.
I understand that with Walborn's double-slit quantum eraser experiment that FTL communication is not possible, because to be able to see the interference pattern after the QWP's are inserted you'll need to measure the coincidence of reception of both the quantum entangled photons from both observers (which requires a normal "at speed-of-light" communication channel).
Walborn at al. only provided experimental proof for the quantum erasure effect (which was the scope of the experiment) and also showed that the order in which the photons are detected by both observers is not important.
But it is not made clear in the experiment described by Cramer how exactely FTL communication would (or would not) be possible, so what is the catch here?
We know that the two down-converted photons will be momentum-entangled, have a different polarisation and that both will always show the same behaviour (either wave-like or particle-like) because of the entanglement.
In the Walborn experiment it was not known which of these two photons went to which observer so a measurement from both observers was needed to derive which-path information. The interference pattern was destroyed by adding the QWP's (marking the path of the photons) and can be reconstructed with an appropriate measurement by the other observer (erasing the effect of the QWP's).
In the Cramer experiment the photons are split with a polarizing splitter to both observers. The presence of an interference pattern at the camera would be entirely dependent on a "choice" being made by the other observer controlling the optical switch for measuring which-path information or not. Cramer does not show that a coincidence measurement is done to construct the diffraction pattern like in the Walborn experiment!
It is not clear to me what results you would get with Cramer's experiment without a measurement of coincidence (requiring normal communication from both observers to a coincidence chamber). Would you be able to see any interference pattern at all?
I read the paper by Walborn et al. http://grad.physics.sunysb.edu/~amarch/Walborn.pdf and can only conclude that it is possible to transmit data between two obervers by carrying out the described "double-slit quantum eraser experiment" but nowhere I could find at what speed this communication would be.
Read the article and just think of this:
It is clear that the observer behind the double-slit window with quarter wave plates (observer #1) will detect an interference pattern (wave-like behaviour) as long as the entangled photon detected by observer #2 gets polarised and can therefore not give any information about which path the photon took through the double-slit window (by comparing polarization of both received signals).
In case the entangled photon detected by observer #2 remains UNpolarised the information could be used to derive which path the photon took through the double-slit window. Because of this the wave function will collapse and observer #1 will NOT see the interference pattern (which is particle-like behaviour).
By letting observer #1 switch the polariser in front of the detector "on" and "off" the interference pattern measured by observer #2 should then in turn respectively "appear" and "disappear" DISREGARDING the distance between observer #1 and #2.
BUT... does this indicate a faster than light communication between the two observers by means of just switching on/off a polariser on one side and measuring an interference pattern on the other?
It seems to me that the "communication" between the two entangled photons is instantaneous as if there was no physical distance between them. But to be able to construct a graph with the interference pattern (or lack thereof) you will need information from BOTH observers. The only way to get information from observer #1 and #2 to construct the interference graph is by using a normal/classical communication channel which is still limited by the speed of light.
Please correct me if I'm wrong!
MythTV can handle HDTV/encrypted content in Europe
on
MythTV Vs. TiVo, Round 2
·
· Score: 2, Informative
This comparison would be completely different in Europe! In europe nearly every digital TV channel broadcasted over Cable, OTA or Sattelite is encrypted with one of many encryption standards (Conax, Irdeto, SECA, etc.).
Instead of a "CableCARD", which is used for viewing encrypted content in the US, a "Conditional Access Module" (CAM) is used in Europe, Africa and most Asian countries for all digital broadcast methods (DVB-C, -S and -T). Most TV companies supply set-top boxes with a built-in decoder and a smart-card, but the smart-card can also be used in other receivers or in a PC when you have the right CAM.
There are a lot of TV cards that can use CAM's and are very well supported by MythTV, for instance: http://http//knc1.com/gb.htm/.
Receiving HDTV or Encrypted content with MythTV is no problem in Europe at least.
The TiVo doesn't seem to exist in Europe, so I wouldn't be able to compare it to TiVo myself, since I never saw one. A very popular digital TV receiver / DVR in Europe is the Dreambox: http://www.dream-multimedia-tv.de/.
The Dreambox is an open platform, is linux-powered and doesn't have any "problems" with DRM or whatsoever. The only limitation the Dreambox and other set-top boxes have is a lack of raw computing power and that's why I prefer to have all my home entertainment on a HTPC.
And that's where the Windows (MCE) vs. Linux discussion comes back!
Slashdot Mirror
Thanks for the tip, I found them here:
https://thepiratebay.org/torre...
https://thepiratebay.org/torre...
it roughly translates to "Little Bing"
Slashdot needs several things...
Unicode support being not the least of them.
maybe you should check out Beta... I heard a lot of good things about it :)
Interesting, it's a main ingredient of Nivea creme.
Indians... I kid i kid. And investors from 'the city'. But if it's truly so great then where are all these new booming tech startups.
Never attribute to malice that which is adequately explained by stupidity. Or incompetence in this case.
Who says we have to limit the integer size to 32 bits? Unix time FTW!
A certain planet from a galaxy far far away
DD-WRT isn't what it used to be anymore, the build environment is a mess, a lot of drivers are binary only and often you have to dig around in the forums to find which builds are stable. I had a lot of bad experiences with consumer grade wireless routers (Linksys WRT610N... $#!&) and building my own router was the best choice I ever made. You can use OpenWRT on a broad range of devices and it has similar features as DD-WRT and also a very nice web-interface. It runs perfectly on low power embedded PC's such as PCEngines Alix (X86) or Ubiquity RouterStation (ARM). Nice enclosures and also complete pre-built systems can be found on eBay. The best choice for wireless network cards would be Atheros based, using the ATH9K driver in Linux. Ubiquity (www.ubnt.com) makes some very decent high power versions (SR-71 series) and Wistron DNMA92 is perfect as a budget solution (can be found on the pcengines online store). The RouterStation Pro and some of the Alix boards allow you to connect multiple wireless network cards for Dual Band radios. I would strongly suggest to use 5 GHz in addition to 2.4 GHz for the devices that support it. The 2.4 GHz range is overcrowded.
Sucks if your motherboard doesn't support 16GB RAM or if you have a laptop.
16GB is nothing, just playing with large amounts of TNT in Minecraft requires this. An average server has 32+ GB when hosting some decent VM's. Windows Server with Exchange will eat up 16GB in no time as well.
I would recommend this one with a NEC chipset (2-port version): http://www.dealextreme.com/p/2-port-usb-3-0-superspeed-pci-e-controller-card-35681 There's also a 4-port version with a VIA chipset, but I haven't tried it yet: http://www.dealextreme.com/p/high-speed-usb-3-0-4-port-pci-e-express-card-5gbps-100865
They all got blended:
http://www.youtube.com/watch?v=qg1ckCkm8YI
The odds will be quite big if the guy diving into your can of used cat litter is the same guy that has been been watching you with a telescope from across the street while you were recording those "private video's". Even bigger if you're a celebrity. Or of course if you work for the CIA, FBI, NSA, etc... for some reason lost USB sticks with secret information always seem to get found somehow...
Who knows who'll put your "private video's" on youtube.
I understand that with Walborn's double-slit quantum eraser experiment that FTL communication is not possible, because to be able to see the interference pattern after the QWP's are inserted you'll need to measure the coincidence of reception of both the quantum entangled photons from both observers (which requires a normal "at speed-of-light" communication channel).
Walborn at al. only provided experimental proof for the quantum erasure effect (which was the scope of the experiment) and also showed that the order in which the photons are detected by both observers is not important.
But it is not made clear in the experiment described by Cramer how exactely FTL communication would (or would not) be possible, so what is the catch here?
We know that the two down-converted photons will be momentum-entangled, have a different polarisation and that both will always show the same behaviour (either wave-like or particle-like) because of the entanglement.
In the Walborn experiment it was not known which of these two photons went to which observer so a measurement from both observers was needed to derive which-path information. The interference pattern was destroyed by adding the QWP's (marking the path of the photons) and can be reconstructed with an appropriate measurement by the other observer (erasing the effect of the QWP's).
In the Cramer experiment the photons are split with a polarizing splitter to both observers. The presence of an interference pattern at the camera would be entirely dependent on a "choice" being made by the other observer controlling the optical switch for measuring which-path information or not. Cramer does not show that a coincidence measurement is done to construct the diffraction pattern like in the Walborn experiment!
It is not clear to me what results you would get with Cramer's experiment without a measurement of coincidence (requiring normal communication from both observers to a coincidence chamber). Would you be able to see any interference pattern at all?
I read the paper by Walborn et al. http://grad.physics.sunysb.edu/~amarch/Walborn.pdf and can only conclude that it is possible to transmit data between two obervers by carrying out the described "double-slit quantum eraser experiment" but nowhere I could find at what speed this communication would be.
Read the article and just think of this:
It is clear that the observer behind the double-slit window with quarter wave plates (observer #1) will detect an interference pattern (wave-like behaviour) as long as the entangled photon detected by observer #2 gets polarised and can therefore not give any information about which path the photon took through the double-slit window (by comparing polarization of both received signals).
In case the entangled photon detected by observer #2 remains UNpolarised the information could be used to derive which path the photon took through the double-slit window. Because of this the wave function will collapse and observer #1 will NOT see the interference pattern (which is particle-like behaviour).
By letting observer #1 switch the polariser in front of the detector "on" and "off" the interference pattern measured by observer #2 should then in turn respectively "appear" and "disappear" DISREGARDING the distance between observer #1 and #2.
BUT... does this indicate a faster than light communication between the two observers by means of just switching on/off a polariser on one side and measuring an interference pattern on the other?
It seems to me that the "communication" between the two entangled photons is instantaneous as if there was no physical distance between them. But to be able to construct a graph with the interference pattern (or lack thereof) you will need information from BOTH observers. The only way to get information from observer #1 and #2 to construct the interference graph is by using a normal/classical communication channel which is still limited by the speed of light.
Please correct me if I'm wrong!
This comparison would be completely different in Europe! In europe nearly every digital TV channel broadcasted over Cable, OTA or Sattelite is encrypted with one of many encryption standards (Conax, Irdeto, SECA, etc.).
Instead of a "CableCARD", which is used for viewing encrypted content in the US, a "Conditional Access Module" (CAM) is used in Europe, Africa and most Asian countries for all digital broadcast methods (DVB-C, -S and -T). Most TV companies supply set-top boxes with a built-in decoder and a smart-card, but the smart-card can also be used in other receivers or in a PC when you have the right CAM.
There are a lot of TV cards that can use CAM's and are very well supported by MythTV, for instance: http://http//knc1.com/gb.htm/.
Receiving HDTV or Encrypted content with MythTV is no problem in Europe at least.
The TiVo doesn't seem to exist in Europe, so I wouldn't be able to compare it to TiVo myself, since I never saw one. A very popular digital TV receiver / DVR in Europe is the Dreambox: http://www.dream-multimedia-tv.de/.
The Dreambox is an open platform, is linux-powered and doesn't have any "problems" with DRM or whatsoever. The only limitation the Dreambox and other set-top boxes have is a lack of raw computing power and that's why I prefer to have all my home entertainment on a HTPC.
And that's where the Windows (MCE) vs. Linux discussion comes back!