mgritter

from scipy.misc import imread
import numpy as np

# These are flower images from the PROJCAM garden bundle.
path = "C:\Users\Mark\Documents\ProcJam\PROCJAM2016-Tess2D\PROCJAM2016-Tess2D\Garden\\"

names = [ "flower{:02d}.png".format( i ) for i in xrange( 0, 22 ) ]
original = [ imread( path + name ) for name in names ]

def flatten( image ):

mgritter

from scipy.misc import imread
import numpy as np

# These are flower images from the PROJCAM garden bundle.
path = "C:\Users\Mark\Documents\ProcJam\PROCJAM2016-Tess2D\PROCJAM2016-Tess2D\Garden\\"

names = [ "flower{:02d}.png".format( i ) for i in xrange( 0, 22 ) ]
original = [ imread( path + name ) for name in names ]

def flatten( image ):

mgritter

0x0748bd33b4f11064f70cb9e2d56534cdb7dfa40c

mgritter

import heapq
import math
from fractions import gcd

D = 33
n = 10000

mx = int( math.sqrt( n ) )
my = int( math.sqrt( n / D ) )

mgritter

const INPUT : &str = "R4, R4, L1, R3, L5, R2, R5, R1, L4, R3, L5, R2, L3, L4, L3, R1, R5, R1, L3, L1, R3, L1, R2, R2, L2, R5, L3, L4, R4, R4, R2, L4, L1, R5, L1, L4, R4, L1, R1, L2, R5, L2, L3, R2, R1, L194, R2, L4, R49, R1, R3, L5, L4, L1, R4, R2, R1, L5, R3, L5, L4, R4, R4, L2, L3, R78, L5, R4, R191, R4, R3, R1, L2, R1, R3, L1, R3, R4, R2, L2, R1, R4, L5, R2, L2, L4, L2, R1, R2, L3, R5, R2, L3, L3, R3, L1, L1, R5, L4, L4, L2, R5, R1, R4, L3, L5, L4, R5, L4, R5, R4, L3, L2, L5, R4, R3, L3, R1, L5, R5, R1, L3, R2, L5, R5, L3, R1, R4, L5, R4, R2, R3, L4, L5, R3, R4, L5, L5, R4, L4, L4, R1, R5, R3, L1, L4, L3, L4, R1, L5, L1, R2, R2, R4, R4, L5, R4, R1, L1, L1, L3, L5, L2, R4, L3, L5, L4, L1, R3";

#[derive(Debug)]
enum Heading {
    North,
    South,
    East,
    West
}

mgritter

# Problem:
# How many possible mazes are there on an NxN grid? Where a maze is a
# collection of links between grid nodes such that each node has 1-4 links
# and has a path to every other node.
#   -- @relsqui, https://twitter.com/relsqui/status/970555868390465536
#
# This counting approach implements a method similar to that described in
# "Spanning Trees in Grid Graphs", Paul Raff, https://arxiv.org/pdf/0809.2551.pdf
#
# The key idea is to create a transition matrix counting the number of

mgritter

words = set()

with open( "sowpods.txt", "r" ) as wordFile:
    for x in wordFile:
        words.add( x.rstrip() )

maxLen = max( len(x) for x in words )

alphabet = [ 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
             'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',

mgritter

var shuffleGrammar = {
    "origin": "#buildstack##element##element##element##element##element##element##element##element##element#",
    "element" : "#stack#[stack:POP]",
    "e1" : "[stack:1][e1:]",
    "e2" : "[stack:2][e2:]",
    "e3" : "[stack:3][e3:]",
    "e4" : "[stack:4][e4:]",
    "e5" : "[stack:5][e5:]",
    "e6" : "[stack:6][e6:]",
    "e7" : "[stack:7][e7:]",

mgritter

import heapq
from fractions import gcd
import math
import sys
import time

def initialValue( maxA, b, power ):
    y = b ** power
    # Better too low than too high and miss one
    a = min( int( math.ceil( math.pow( y, 1.0 / 3.0 ) ) ),

mgritter

#!/usr/bin/python3

from steem import Steem
from pprint import pprint
import sys

startTime = "2018-05-01T00:00:00"
endTime = "2018-05-31T23:59:59"

if len( sys.argv ) > 1: