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united-coders.com authors christian harms nico heid datenschutz impressum raspberry pi supply switch, start and shut down your pi like your pc posted on august 24, 2015 by nico heid posted in uncategorized as you might have guessed, there is always at least one raspberry pi doing something at my place. two are already in datacenters providing services. one over at edis.at in austria and one at raspberrycolocation.com (pcextreme b.v. in the netherlands). i did a post on that a while back: all you need to know about raspberry pi colocation offers . the one at home currently has a hifiberry digi+ on its gpio and is connected to the stereo. it functions as my standalone music player. it runs musicbox ( or volumio ) which is capable of connecting to various music sources, including my spotify account. it also has to capability of being an airplay and dnla receiver, in case i want to stream audio from my pc or android devices. as this pi has only one purpose, it doesn’t need to be powered on when nobody is at home. powering up via usb connection or power plug and shutting it down via ssh is a bit tedious. additionally, the pi draws some power when shut down while still being connected to the the power supply. this is where pi supply switch comes to your rescue. it provides an easy method of switching the pi on and off via buttons. once it is switched off, the power supply will cut the power after a short time. this will also work when powering down via ssh. the power supply is connected to the 3v of the pi, if the pi is shut down, the 3v are not provided any more and the circuit cuts the power to the pi. for the shutdown via button to work, you need to listen on the interrupt for the push of the button: # import the modules to send commands to the system and access gpio pins from subprocess import call import rpi.gpio as gpio # define a function to keep script running def loop(): raw_input() # define a function to run when an interrupt is called def shutdown(pin): call('halt', shell=false) gpio.setmode(gpio.board) # set pin numbering to board numbering gpio.setup(7, gpio.in) # set up pin 7 as an input gpio.add_event_detect(7, gpio.rising, callback=shutdown, bouncetime=200) # set up an interrupt to look for button presses loop() # run the loop function to keep script running a more in depth article, schematics and code can be found on raspberry-pi-geek.com now if you’re like me, you probably forget to shut down the machine when you go to bed, as it just idles away near the stereo. so add an additional line in the crontab to shut it down at night, when no other machine is on, when nobody played music in a while or the stereo is off ( your stereo has ip networking right?). linux raspberry pi infinite house of pancakes – code jam 2015 posted on july 13, 2015 by nico heid posted in uncategorized by michael stern the infinite house of pancakes seems to be rather easy to solve at first glance. i did pick this puzzle right after solving standing ovation, thinking it should be only slightly harder, given it has only 9pts compared to the 7pts from standing ovation . i was wrong. my idea on how to solve this was simple, too simple. i did put all plate sizes in an array, located the highest plate(s) and took half of the pancakes off. i compared the number of steps needed before and after, and if it was better or equal after, i repeated the process until it did not get better. voila. no, “incorrect”. and it kept saying incorrect to me until i had to depart, leaving the puzzle behind but still taking it along in the head, trying to figure out what was wrong. let’s check with an example. 9 3 1 -> 5 4 3 1 -> 3 2 4 3 1 -> 3 2 2 2 3 1 -> 2 1 2 2 2 2 1 1 9 turns on the starting number of steps if we just let it be. i use 1 move to split the 9 into 5 and 4 and get to 6 steps (1 move + 5 eats). better, so we keept that. use another move (2 now) to split the 5 which gets to 6 (2+4) which is equal, lets still keep that. now split the 4 (3rd move) giving us 6 again (3+3). oh well, now split the 3s (needs 2 moves) giving us 7 (5+2) which is worse. so our answer is 6 steps. i was missing the cruical point: it might be better to split a stack in multiple stacks, not just take half of it off . splitting the 9 into three 3s and not into 5 and 4 is the move that makes the difference. 9 3 1 -> 3 3 3 3 1 which is 2 moves and 3 eats. giving you a 5 (2+3), which is better than the 6 we got with our previous algorithm. even though the puzzle crossed my mind a couple times after the qualification round, i did have to wait and check the solutions page to get that idea, which makes me feel a bit ashamed. i do only find comfort in the fact, that during the round less than half of the contestants solved this correctly. after being given that hint, the implementation was done rather quickly with less code than i did write for the first algorithm. try ever plate size from one to max and see how many steps it would take. -> on github public void solve() { private int checkplates(list plates) { int max = integer.max_value; int fullestplate = plates.stream().max(integer::compare).get(); for (int i = 1; i <= fullestplate; i++) { int moves = 0; for (int p : plates) { if (p > i) { moves += p / i; if (p % i == 0) { moves--; // if it divides evenly we actually took all pancakes off the plate, which is one move more than we need } } } if (moves + i < max) { max = moves + i; } } return max; } the code jam solutions page is even more elaborate with some more optimizations, so head over there and check it out. code jam code puzzle google code jam 2014 – magic trick posted on april 18, 2014 by nico heid posted in uncategorized image by steven depolo even though i could not participate in this years google code jam qualification round, i want to provide you my solution for magic trick. i actually failed to implement it correctly on the first try. my idea was that after the volunteer chose the row in the first grid, all the cards of this row needed to be in different rows on the second grid. which is probably how you’d set up the trick too, if you’d perform it. the thing is, you can still be lucky by placing only the correct card in the row the volunteer actually picks into it’s own new row. the other three can be either in the same row, or wherever. it doesn’t matter. let’s assume we start with this as our first grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 if you think of your pattern as a matrix it becomes clear, that by laying down the transponse (read the columns and write them as rows) of your matrix, you can always solve what the volunteer picked. 1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16 this would be the logical thing to do, but you can also just depend on your good luck and try something like this: 1 2 3 5 4 5 6 7 9 10 11 12 13 14 15 16 if the user actually picked the 4, it is still solveable. but if he picked 1, 2 or 3 you’d have no chance to determin which card it was because you failed to spread them into different rows. and this was the mistake, i forgot that the magician can be lucky even if he did not lay out the cards as transponse of his first grid. so here’s my solution ( or go to github ) public void solve() { int cases = in.nextint(); for (int j = 1; j <= cases; j++) { out.print("case #" + j + ": "); // first card int rowwithcard = in.nextint(); in.nextline(); list row1 = getrow(rowwithcard); //second card rowwithcard = in.nextint(); in.nextline(); list row2 = getrow(rowwithcard); if (!iswizardwithoutfailure(row1, row2)) { out.println("bad magician!"); continue; } string result = findcard(row1, row2); if (result == null) { out.println("volunteer cheated!"); continue; } else { out.println(result); } } } // get the card from the chosen row, every other row is not interesting private list getrow(int chosenrow) { list cardrow = new arraylist(); for (int row = 1; row <= 4; row++) { if (chosenrow == row) { cardrow.addall(arrays.aslist(in.nextline().split(" "))); } els

URL analysis for united-coders.com

https://www.united-coders.com/nico-heid/raspberry-pi-colocation/
https://www.united-coders.com/christian-harms/google-code-jam-2013-the-lawnmower/
https://www.united-coders.com/datenschutz/
https://www.united-coders.com/nico-heid/raspberry-pi-supply-switch/
https://www.united-coders.com/tag/code-jam/
https://www.united-coders.com/page/2/
https://www.united-coders.com/page/10/
https://www.united-coders.com/tag/raspberry-pi/
https://www.united-coders.com/nico-heid/scalable-wordpress-hosting/#more-689
https://www.united-coders.com/christian-harms/code-puzzles-and-permutations/
https://www.united-coders.com/tag/project-euler/
https://www.united-coders.com/tag/android/
https://www.united-coders.com/nico-heid/automated-encrypted-windows-server-backup/
https://www.united-coders.com/christian-harms/google-code-jam-2013-tic-tac-toe-tomek-solution/
https://www.united-coders.com/tag/java/

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Domain Name: UNITED-CODERS.COM
Registry Domain ID: 1549020498_DOMAIN_COM-VRSN
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Registrar URL: http://www.godaddy.com
Updated Date: 2017-03-23T10:02:37Z
Creation Date: 2009-03-24T13:08:08Z
Registry Expiry Date: 2018-03-24T13:08:08Z
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Name Server: NS29.DOMAINCONTROL.COM
Name Server: NS30.DOMAINCONTROL.COM
DNSSEC: unsigned
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  ARGS domain =united-coders.com

  PORT 43

  TYPE domain

DOMAIN

  NAME united-coders.com

  CHANGED 2017-03-23

  CREATED 2009-03-24

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