Facebook Puzzle Challenge - Liar Liar - Difficulty Easy

#!/usr/bin/python

import sys

import re



# Ensure we have the proper amount of arguments

if len(sys.argv) < 2:

  print "Please pass through a file to open"

  sys.exit(0)

  

# Let's open the file

filename = sys.argv[1]

try:

  file = open(filename,"r").read()

except:

  print "An error has occured when opening the file"



# Let's parse the data in to a dictionary

lines = file.split("\n")

people = {}

pointer = 1 # Where we currently are

while len(people) < int(lines[0].strip()):

  (person, amount) = re.sub(" \s+", " ", lines[pointer].strip()).split(" ")

  amount = int(amount)

  people[person] = []



  for x in range(1, amount + 1):

    pointer += 1

    people[person].append(lines[pointer].strip())

  pointer += 1  



# Sort out the liars from the ones telling truth

(truth, liars) = ([],[])



# Now let's go through and see if someone is telling the truth

def tellingTruth(name, trail=[]):

  # If we've already determined whether or not they're liars/telling truth

  if name in truth:

    return len(trail) % 2 != 0

  elif name in liars:

    return len(trail) % 2 == 0

  

  # Loop through

  if name not in trail:

    trail.append(name)

  for i in range(0, len(people[name])):

    if people[name][i] in trail: #we've already checked

      if len(trail) <= 2:

        return None

      else:

        return ( len(trail) - 1 ) % 2 == 0

    else:

      trail.append(people[name][i])

      return tellingTruth(people[name][i], trail)



# Let's go through and determine who's lying and who isnt

for i in range(0,len(people)):

  isTruth = tellingTruth(people.keys()[i], [])

  if isTruth == True:

    truth.append(people.keys()[i])

  elif isTruth == False:

    liars.append(people.keys()[i])



# Have a preliminary list, let's see if that solves

for person in truth:

  subjects = people[person]

  for subject in subjects:

    if subject not in liars:

      liars.append(subject)

  

for person in liars:

  subjects = people[person]

  for subject in subjects:

    if subject not in truth:

      truth.append(subject)

    

# Let's now output

if len(truth) > len(liars):

  print str(len(truth)) + " " + str(len(liars))

else:

  print str(len(liars)) + " " + str(len(truth))[/person][/person][/i][/i][/i][/i][/name][/i][/name][/i][/name][/name][/pointer][/person][/person][/pointer][/0][/1]

import java.util.*;

import java.io.*;



public class FacebookPuzzle20_1

{

	public static void main(String[] args)

	{

		try {

			String lines = "", str;

			int t = 0, l = 0;

			BufferedReader in = new BufferedReader(new FileReader("puzzle.txt"));



			while ((str = in.readLine()) != null)

	        	lines = lines + str + "\n";



			String[] lieChart = lines.split("\n");

			int mPos = lieChart[1].length();



			Map<String, Integer> m = new TreeMap<String, Integer>();

			for(int x = 0; x<lieChart.length; x++)

				if(x!=0) m.put(lieChart[x].split(" ")[0], (m.containsKey(lieChart[x].split(" ")[0]) ? (m.get(lieChart[x].split(" ")[0]) == 1 ? 0 : 1) : ((lieChart[x].length() == mPos) ? 1 : 0)));



			for(int y = 0; y<m.size(); y++) if(m.values().toArray()[y].toString().contains("1")){ t++; }else{ l++; } 

			System.out.println((t>=l) ? (t + " " + l) : (l + " " + t) );

		}catch(IOException e){}

	}

}[/y][/x][/0][/x][/0][/x][/0][/x][/1]

import java.util.*;

import java.io.*;



public class FacebookPuzzle20

{

	public static void main(String[] args)

	{

		try {

			String lines = "", str;

			int t = 0, l = 0;

			BufferedReader in = new BufferedReader(new FileReader("puzzle.txt"));



			while ((str = in.readLine()) != null)

	        	lines = lines + str + "\n";



			String[] lieChart = lines.split("\n");

			int mPos = lieChart[1].length();



			Map<String, Integer> m = new TreeMap<String, Integer>(); // Map of the liars and truth tellers, sorted by a 0 and 1, key being the name.

			for(int x = 0; x<lieChart.length; x++)

			{

				if(x!=0) // If it isnt the first line, this is where the person count exists.

				{

					String key = lieChart[x].split(" ")[0];

					int value = 0;

					boolean mContainsNameAlready = m.containsKey(key); // Name has already been in, if so flip it.



					if(mContainsNameAlready)

					{

						if(m.get(key) == 1) // If this person is a 1 already put 0, else, put 1

							m.put(key, 0);

						else

							m.put(key, 1);

					}

					else

					{

						if(lieChart[x].length() == mPos) // If telling a accusation (assume 1), else, being blamed (assume 0)

							m.put(key, 1);

						else

							m.put(key, 0);

					}

				}

			}



			// Organize our truth and liars integers.

			for(int y = 0; y<m.size(); y++) 

			{

				String valueOfMY = m.values().toArray()[y].toString();

				if(valueOfMY.contains("1")) // If 1 count t, else, count l

					t++; 

				else

					l++; 

			}

			

			// Determine if t or l is larger, print larger one first.

			if(t>=l)

				System.out.println(t + " " + l);

			else

				System.out.println(l + " " + t);



		}catch(IOException e){}

	}

}[/y][/x][/0][/x][/1]