hillscottc

/* Keep stuff out of global by creating and calling an anon func.
*/
(function() {
  function square(x) { return x * x; }
  var hundred = 100;
  console.log(square(hundred));
})();
// → 10000

hillscottc

import unittest
from floydwarshall import floyd


class TestFloyd(unittest.TestCase):

    def test_floyd_1(self):
        inf = float('inf')
        g = {0: {0: 0,   1: 1,   2: 4},
             1: {0: inf, 1: 0,  2: 2},

hillscottc

# w = list of item weight or cost
# c = the cost matrix created by the dynamic programming solution
def getUsedItems(w,c):
    # item count
    i = len(c)-1
    currentW =  len(c[0])-1
    # set everything to not marked
    marked = []
    for i in range(i+1):
        marked.append(0)      

hillscottc

from __future__ import with_statement
import os
from fabric.api import local, settings, abort, sudo, env
from fabric.contrib.console import confirm

env.hosts = [
    # 'me@my-imac',
    'me@my-desk',
]

hillscottc

#!/usr/bin/env python
# -*- coding: utf-8

"""
Search longest common substrings using generalized suffix trees built with Ukkonen's algorithm

Author: Ilya Stepanov <code at ilyastepanov.com>

(c) 2013
"""

hillscottc

"""http://goo-apple.appspot.com/article/2e8d3c6a-2c38-48b9-96c6-240b4ded253a"""
class Node:
        def __init__(self, start, substr):
                self.start = start
                self.substr = substr
                self.branches = {}
              
def insert_into_tree(subroot, suffix, start):
        prefix_len = len(subroot.substr)
        new_suffix = str(suffix[prefix_len:])

hillscottc

"""http://stackoverflow.com/questions/2892931/longest-common-substring-from-more-than-two-strings-python"""

def long_substr(data):
    """Finds the longest common string in any arbitrary array of strings"""
    substr = ''
    if len(data) > 1 and len(data[0]) > 0:
        for i in range(len(data[0])):
            for j in range(len(data[0])-i+1):
                if j > len(substr) and all(data[0][i:i+j] in x for x in data):
                    substr = data[0][i:i+j]

hillscottc

def make_change(cents_needed, coin_vals):
    min_coins = [[0 for j in range(cents_needed + 1)]
                for i in range(len(coin_vals))]
    min_coins[0] = range(cents_needed + 1)

    for i in range(1, len(coin_vals)):
        for j in range(0, cents_needed + 1):
            if j < coin_vals[i]:
                min_coins[i][j] = min_coins[i-1][j]

hillscottc

import sys

def knapsack(items, maxweight):
    # Create an (N+1) by (W+1) 2-d list to contain the running values
    # which are to be filled by the dynamic programming routine.
    #
    # There are N+1 rows because we need to account for the possibility
    # of choosing from 0 up to and including N possible items.
    # There are W+1 columns because we need to account for possible
    # "running capacities" from 0 up to and including the maximum weight W.

hillscottc

"""
Dijkstra's algorithm for shortest paths.
http://code.activestate.com/recipes/577343-dijkstras-algorithm-for-shortest-paths/)

  - dijkstra(G, s) finds all shortest paths from s to each other vertex.
  - shortest_path(G, s, t) uses dijkstra to find the shortest path from s to t.

The input graph G is assumed to have the following representation:

  - A vertex can be any object that can be used as an index into a dictionary.