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import select | |
import serial | |
serial_dev = '/dev/ttyS1' | |
select_timeout = 30 | |
port = serial.Serial(serial_dev, timeout=None) | |
port.flushInput() |
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#target photoshop | |
main(); | |
function main() { | |
if(!documents.length) return; | |
var doc = app.activeDocument; | |
try{ | |
var Path = activeDocument.path; |
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from __future__ import print_function | |
import random, itertools, sets | |
## The following represents thought process and has not been subjected to rigorus proof | |
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings. | |
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively. | |
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings) | |
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6). | |
# Therefore unique pairings for cycles through all 7 generates 6! = 720 (ie 7!/7) pairings. | |
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420 |
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from __future__ import print_function | |
import random, itertools, sets | |
## The following represents thought process and has not been subjected to rigorus proof | |
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings. | |
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively. | |
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings) | |
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6). | |
# Therefore unique pairings for cycles through all 7 generates 6! = 720 (ie 7!/7) pairings. | |
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420 |
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from __future__ import print_function | |
import random, itertools, sets | |
## The following represents thought process and has not been subjected to rigorus proof | |
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings. | |
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively. | |
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings) | |
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6). | |
# Therefore unique pairings for cycles through all 7 generates 6! = 720 (ie 7!/7) pairings. | |
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420 |
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from __future__ import print_function | |
import random, itertools, sets | |
## The following represents thought process and has not been subjected to rigorus proof | |
# Possible pairings can be represented as two varieties of cycles in the permutations of the residents due to the restriction on reflexive pairings. | |
# For each permutation there is a single cycle through all 7 name and a two disjoint cycles through 3 names and 4 names respectively. | |
# Total unique pairings are therefore less than 7!*2 = 10080 (less because cycles of a particular permutation generate equivalent pairings) | |
# There are 7! permutations. Cycles through all 7 that produce equivalent pairings can be elminiated by considering only permutations that begin with a particular digit (eg. 6). | |
# Therefore unique pairings for cycles through all 7 generates 6! = 720 (ie 7!/7) pairings. | |
# Considering 3,4 cycles inside of each permutation there should be (7!/(4!*3))(4!/4) = 420 |
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// Much code borrowed from the following | |
// https://github.com/practicalarduino/VirtualUsbKeyboard | |
// http://playground.arduino.cc/Learning/SoftwareDebounce | |
#include "UsbKeyboard.h" | |
#define BUTTON_PIN 12 | |
#define LED_PIN 13 | |
#define KEY_SPACE 0x2C |
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from VideoCapture import Device | |
import pygame | |
import datetime | |
import glob | |
cam = Device() | |
pic_dir = './' | |
pic_ext = '.jpg' | |
pics = glob.glob(pic_dir+'*'+pic_ext) |
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<html> | |
<head> | |
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script> | |
<script type="text/javascript"> | |
var drawOn = function (id) { | |
var lastPoint = null; | |
var canvas = document.getElementById(id); | |
var ctx = canvas.getContext('2d'); | |
ctx.lineWidth = 4; | |
var offset = jQuery('#'+id).offset(); |
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class word(object): | |
def __init__(self,w): | |
self.w = w | |
self.key = "%02d%s" % (len(w),w) | |
self.set = self._makeset(w) | |
def __str__(self): | |
return self.w | |
def __cmp__(self,other): | |
return cmp(other.key,self.key) | |
def __repr__(self): |