"switching box" with microcontroller and relays
Control 220V AC line using MCU
"C" sizeof with a type or variable Motivated by :   Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of units of some commutative ring? Which finite abelian groups are the group of units of some finite commutative ring?   It  that even for fields this is not simple to answer.  I expect the general question (Which groups are the group of units of some ring?, with no finiteness hypotheses) is even harder. I have a HP Pavilion dm4-2070us which has an Intel Corporation Centrino Wireless-N 1000 adapter.   The problem is that my computer doesn't show the wireless networks after I upgraded my system to Ubuntu 11.10. Following is my output for sudo lshw -class network:  *-network DISABLED          description: Wireless interface    product: Centrino Wireless-N 1000    vendor: Intel Corporation    physical id: 0    bus info: pci@0000:01:00.0    logical name: wlan0    version: 00    serial: 8c:a9:82:81:5a:30    width: 64 bits    clock: 33MHz    capabilities: pm msi pciexpress bus_master cap_list ethernet physical wireless    configuration: broadcast=yes driver=iwlagn driverversion=3.0.0-12-generic firmware=39.31.5.1 build 35138 latency=0 link=no multicast=yes wireless=IEEE 802.11bgn    resources: irq:42 memory:c2500000-c2501fff  *-network    description: Ethernet interface    product: AR8151 v2.0 Gigabit Ethernet    vendor: Atheros Communications    physical id: 0    bus info: pci@0000:08:00.0    logical name: eth0    version: c0    serial: 2c:41:38:07:f3:e3    size: 100Mbit/s    capacity: 1Gbit/s    width: 64 bits    clock: 33MHz    capabilities: pm msi pciexpress vpd bus_master cap_list ethernet physical tp 10bt 10bt-fd 100bt 100bt-fd 1000bt-fd autonegotiation    configuration: autonegotiation=on broadcast=yes driver=atl1c driverversion=1.0.1.0-NAPI duplex=full firmware=N/A ip=192.168.1.7 latency=0 link=yes multicast=yes port=twisted pair speed=100Mbit/s    resources: irq:44 memory:c1400000-c143ffff ioport:2000(size=128)   The output for rfkill list all:  0: hp-wifi: Wireless LAN     Soft blocked: no     Hard blocked: no 1: hp-bluetooth: Bluetooth     Soft blocked: no     Hard blocked: no 2: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no 3: hp-wwan: Wireless WAN     Soft blocked: no     Hard blocked: no   I have also searched about the status of the WiFi network being shown as disabled. Somehow everything pointed to the fact that my wireless device is switched off and it can be switched on from Windows only. However, I don't have windows partition and it has just Ubuntu 11.10 so I don't see any point in switching the device on in Windows. Accessing raster attribute table in QGIS? I have some Python code that makes Raster Attribute Tables for GeoTiffs using GDAL.    In ArcGIS I can see these attribute tables fine and they behave how I like (as classifications) but in QGIS the whole concept of an attribute table seems to be missing.    Is there any way to access a RAT in QGIS? How can I define a mathematical function as a LaTeX macro? How do I rename a directory via the command line? I have got the directory /home/user/oldname and I want to rename it to /home/user/newname. How can I do this in a terminal? Automatically Locate Included Images Enable disk encryption after installation Solutions of $x^2 \equiv 1 \pmod N$ I am stuck with this exercise:     Show that $\forall s &gt;0 \quad \exists N &gt;0$ such that $$x^2 \equiv 1\pmod N$$ has more than $s$ solutions.   I think I cannot use the theory if quadratic residues, because the modulus is not prime, so maybe I must prove it with square roots or roots of unity, but I don't know how to approach it at all.  Thanks for any help. Is there an MD5 Fixed Point where md5(x) == x? Is there a fixed point in the MD5 transformation, i.e. does there exist x such that md5(x) == x? How to prevent iMessage syncing without access to device I foolishly logged into someone's iPad with my Apple ID. This person is now receiving my full communications on iMessage. I know I can remove my account from that iPad but I don't have access to it. How can I stop the device from getting my iMessages without physical access? How are Clash of Clans multiplayer battles determined Are battles determined by level or trophies? Just curious. Present tense for future events Why does it sound perfectly natural to say Our flight leaves tomorrow at 6pm but weird to say It rains tomorrow at 6pm? What kind of scenario, if any, could make the rain sentence sound natural? Is there a function to round a float in C or do I need to write my own?       float conver = 45.592346543;   I would like to round the actual value to one decimal place, conver = 45.6. Can't install/uninstall packages in Debian I'm trying to install/uninstall packages in Debian but getting errors.    I can't remove or add any package, always getting the error.   I tried:  apt-get --remove purge package apg-get remove package dpkg -r package More and more frequently in the review queues, I am spending time either fixing a post or considering whether an edit is valid, etc, only to make changes and have them denied/not count because someone else has blindly hit the approve button in the meantime.  My proposal is for the system to "lock" a post in the review queue once it has been presented to someone for review.  Do not let anyone else (in the review queue) do anything with it until it is released by the last user it was presented to (or a certain amount of time has elapsed).    By "lock" I mean that it won't show up in anyone elses list until released.  I did NOT mean that it would be in multiple queues and be unactionable in some of them.  This would fit well as the queues only present one at a time anyway.  This will allow people trying to make good decisions the time to do it and prevent someone else from making that effort a waste of time.  As support, look to some of the published reasoning behind the auto-question ban.  A large portion of it is that bad questions waste the time of the community.  Taking the time to ensure you are making a good decision in the /review queue only to have it not count because someone was badge-whoring and clicking the approve button as fast as they could also wastes the users time... arguably the more (for lack of a better way to put it) valuable user who is willing to invest some time and thought into making the site better. Filtered Questions Asked By Accepted Answer/ No Accepted Answer I think it's helpful if we can go to a page and view all our own questions that   Does not have an accepted answer, and/or Does not have an answer yet.   The reason I asked for them is because, I sometimes need to review through all the questions, and pick those that don't have proper answers so that I can perform further operations on them ( such as offering a bounty, or accept an existing answer). How to tell Google to crawl distinct part of the webpage? What approach should I take to make the website Google-friendly, .  The problem is that when a specific content is being requested (eg. ) user is taken to the page using scrollbar, though the surrounding content is still being loaded. How to tell Google that for this URL  only the #36 entry should be crawled.  Is there a tag or other way to select the surrounding content as irrelevant for the URL? This is the proof, which I mostly understand except for one bit:  You have $h_1 \in H_1$ and $h_2 \in H_2$.   We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore  $$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $$  and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$  $$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$$ $$ h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$$ $$ h_2^{-1}h_1h_2 = h_1 \{1_G\}$$   Multiply both sides on the left by $h_2$ giving us  $$ h_1h_2 = h_2 h_1 $$  The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$." I can't write Chinese characters in edited posts, but OP could when posting