A geeky toy, for a geeky boy.

March 3rd, 2009 § 1 comment § permalink

I’m sure all of you know what the rubik’s cube is, if you don’t… I mean, really? man what was you thinking until now? Well If you REALLY don’t know it then you should at least google for it. Wikipedia will work as well.

rubik's cube

I always thought it was a logic puzzle, and after some unsuccessful tries I decided to cheat, googled for “rubik’s cube solution” and I found out a beautiful site, which explains everything you need to solve the cube in a very simple and fast way, it is of course the Lars Petrus method.

Once learnt a solution method there is no more “logic” in it. Where is the fun part? What’s the challenge?
I was terribly wrong. It is not a logic-puzzle, it is a memory-puzzle, and a hand-dexterity-puzzle.

You need to remember the moves of your solving methods, they are almost always the same (in the petrus method there are 4 basic moves), but the most challenging part is how fast can you solve the cube?

Hey “how-fast” sounds damn challenging, how can you measure that? You’ll need a stopwatch. If you are at least half geeky as I am you’d be already on google at the moment and starting to type something like “online stopwatch” or “online timer”.

There are tons of them, but all have a very annoying interface that make you use the mouse. Given that you are a very good speedcuber you don’t want to loose 0.3s to start to solve and an other 0.3s to stop the timer, do you?

Well I wouldn’t care as I need 1’30”+ to solve the cube, but I found fun to code a stopwatch designed to work with Miss Spacebar instead of your loveable mice.

nigma timer

the simple nigma timer is written in simple html/css/js and as a very simple interface, I hope you’ll enjoy it.

return ”

How to convert Microsoft Word .doc files to PDF from command line

January 14th, 2009 § 2 comments § permalink

I know lot of people need it, Google is full of requests by hundred, maybe thousands of users asking for a doc2pdf converter or this kind of thing. I need it too. It is useful to have all files in pdf format (and maybe all merged in one file only) and if you have a lot of files to convert by hand, believe me, you’re not going to have a nice day.

The easy way

It is pretty easy:

$ abiword --to=pdf filename.doc

I don’t think there is so much to explain here. It converts filename.doc to filename.pdf and saves it in the current directory. It was too easy. Why should you need an hard way? I don’t know, I’m sure I need one. Unfortunately abiword’s Microsoft doc file support is not so good, in fact it lacks of the math and image/clipart features. I’m not sure if this affects all versions of abiword but it is sure for the one that comes with ubuntu (actually it doesn’t come with it, you’ve to apt-get install it).

Anyway I really need to see plots and formulas. What you said? OpenOffice supports them. Check it out. Yes I know that, OpenOffice can read almost always plots and images in doc files. Bad luck seems to be here again, OpenOffice lacks of the same command line interface abiword has, so the only way is to open doc files one by one and click on the Export as PDF button. It is very frustrating. So, here is the hard way.

The hard way

Short version (for whom doesn’t like read me be but want to read so much): check the Python-UNO site.

Long version. You need to know what Python-UNO is

The Python-UNO bridge allows to

  • use the standard OpenOffice.org API from the well known python scripting language.
  • to develop UNO components in python, thus python UNO components may be run within the OpenOffice.org process and can be called from Java, C++ or the built in StarBasic scripting language.
  • create and invoke scripts with the office scripting framework (OOo 2.0 and later).

You can find the most current version of this document from http://udk.openoffice.org/python/python-bridge.html

Oh no! I’ll have to download this Python-UNO, read manuals to learn how to use those API and who knows if it’ll work…… No. Just don’t panic. I’m going to tell you something that will make this a not-so-hard way. The first thing is that if you have installed OpenOffice you’re at 50% of the work, in fact Pyhton-UNO comes with OpenOffice since version 1.1.

  • Pyhton-UNO comes with OpenOffice since version 1.1. You don’t have to download and install anything
  • Pyhton-UNO’s guys are so cool that in their code examples there is all of what we need.

From the examples page you can download the ooextract.py script. It has a very simple usage, we need to use it in this way:

$ openoffice -invisible "-accept=socket,host=localhost,port=2002;urp;"
$ python ooextract.py --pdf filename.doc

The result is almost the same of the one of the easy way but this will use OpenOffice for the conversion, so it will do it better. You also may like to write a little shell script to automate the conversion of a bunch of files, so there it is a very simple version:

#!/bin/bash

openoffice -invisible "-accept=socket,host=localhost,port=2002;urp;"
for i in *.doc; do
	python ooextract.py --pdf $i
done

Remember to kill OpenOffice when it ends :o) OpenOffice has now batteries included.

GCJ – Practice Contest – Old Magician

October 13th, 2008 § 0 comments § permalink

Wow, an other practice contest came out :^D here’s the first problem, very simple solution :^)

from __future__ import with_statement
 
def solve(w,b):
    return "BLACK" if b%2==1 else "WHITE"
 
def eatFile(fin, fout):
    with file(fin,'r') as f1:
        N = int(f1.readline().replace("\n",''))
        with file(fout,'w') as f2:
            for case in xrange(1,N+1):
                w,b = [int(i) for i in f1.readline().replace("\n",'').split(' ')]
                s = solve(w,b)
                #print "Case #%d: %s" % (case, s)
                f2.write("Case #%d: %s\n" % (case, s))
 
eatFile('A-large-practice.in','A-large-practice.out.txt')

GCJ 2008 – Round 1A: Minimum Scalar Product

August 5th, 2008 § 3 comments § permalink

This one was really really really too simple. The easiest problem of round 1. Unfortunately I didn’t took part at 1A, I did 1B and 1C and I did not pass :^( Anyway I’ll keep doing those problem just to have some fun :^)

def solve(a,b):
    a.sort()
    b.sort(reverse = True)
    return sum([i[0]*i[1] for i in zip(a,b)])

You have to pass a and b as lists, or if you prefer vectors… I think there is no more to comment here… I did it just thinking about it and I find the solution out in about 3 minutes, but I discovered that mathematicians already thought about that and gave it a name :^D

You can search on wikipedia for Rearrangement inequality. I mean, 4 lines of fully readable python… it was quite simple, wasn’t it?

return YAWN;

GCJ 2008: Fly Swatter

July 19th, 2008 § 8 comments § permalink

Consider this:
GCJ 2008 - Problem C - raquet

Believe it or not, but that’s a racquet! In fact the third problem of the Qualification Round of Google Code Jam 2008 is about balls and racquets (or rackets if you prefer)… well… actually it’s about probability :^D

You can find the whole text of the problem here, I’m reporting just the interesting part (IMHO):

Problem

What are your chances of hitting a fly with a tennis racquet?

To start with, ignore the racquet’s handle. Assume the racquet is a perfect ring, of outer radius R and thickness t (so the inner radius of the ring is R-t).

The ring is covered with horizontal and vertical strings. Each string is a cylinder of radius r. Each string is a chord of the ring (a straight line connecting two points of the circle). There is a gap of length g between neighbouring strings. The strings are symmetric with respect to the center of the racquet i.e. there is a pair of strings whose centers meet at the center of the ring.

The fly is a sphere of radius f. Assume that the racquet is moving in a straight line perpendicular to the plane of the ring. Assume also that the fly’s center is inside the outer radius of the racquet and is equally likely to be anywhere within that radius. Any overlap between the fly and the racquet (the ring or a string) counts as a hit.

The input file consists in N lines containing values of f, R, t, r and g separated by spaces.

I’m now trying to explain my approach. The probability we’re looking for it’s given by total_hitting_area/total_racquet_area, or 1 – total_not_hitting_area/total_racquet_area

total racquet area it’s simple:
GCJ 2008 - Problem C - big
I’ll call it big, and it’s given by pi*R^2

The problem is the other area. I’ll try to calculate the not hitting area. Calculus gives a solution to get area of any kind of shape: integrals, but I don’t think it’s necessary to use them in this case. First for all we need to simplify. I don’t care about the ball, I’ll care about the center of the ball. The center of the ball has to be a distance f (the ball radius) from the border of the racquet in order to donot hit it…

This is so the interesting area:
GCJ 2008 - Problem C - small
I’ll call it small, and has a radius of R-t-f so the area is given by pi*(R-t-f)^2

The probability of the center of the ball to be in the small area is given by small/big, and I call it q1

Now consider this:
GCJ 2008 - Problem C - square

The two red areas are identical, they are both composed by the same amount of green and white area, I can consider any of them as a square, its area is given by (g+2*r)^2

Note that small is composed by squares. Some of them are cut off, but that’s should not be a problem… it should

Now consider an hole:
GCJ 2008 - Problem C - gap
I’ll call the red area gap, it’s a little square with sides long g-2*f as I subtract 2 times the ball radius. In fact the center of the ball has to be at least at a distance of f from the sides of the white area in order to do not hit a string. So a gap is (g-2*f)^2

The probability of the center of the ball to pass through a gap contained in a square is given by gap/square and I’ll call it q2

In order to do not hit the racquet to events have to happen at the same time:

    1. The center of the ball has to be in the small area
    2. The center of the ball has to be in a gap area

These to events have probability q1 and q2 respectively.
The probability they are both verified is q = q1 * q2, and that’s the not successful case.

The successful case is p = 1 – q… the probability we were looking for…

It should work… it should… But it doesn’t.

I tried an other approach so. I calculate how many squares there are in the small area. That number is nos = small/square (where nos stands for number of squares)

We know there are as many gap as squares, that’s just because there is one gap inside each square.

In this way I can calculate the total gap area, that is just nos*gap and I call it totalgap

Do you remember the 1 – total_not_hitting_area/total_racquet_area… well the total_not_hitting_area is just our totalgap!

So we have q = totalgap/big… and p = 1 – q

I’ve found the same p as the one above… and it’s wrong.

I went so close the solution but there is something wrong, just to show you, these are the sample cases:
0.25 1.0 0.1 0.01 0.5
0.25 1.0 0.1 0.01 0.9
0.00001 10000 0.00001 0.00001 1000
0.4 10000 0.00001 0.00001 700
1 100 1 1 10

The Google’s results for these cases are:
Case #1: 1.000000
Case #2: 0.910015
Case #3: 0.000000
Case #4: 0.002371
Case #5: 0.573972

But mines are:
Case #1: 1.000000
Case #2: 0.920132
Case #3: 0.000000
Case #4: 0.002364
Case #5: 0.573156

They are so close.

You can download my solution file by clicking here.
That’s my implementation in python:

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def solve(f,R,t,r,g):
    if 2*f >= g: return 1.0
    t += f
    g -= 2*f
    r += f
 
 
    big = pi*R**2
    small = pi*(R-t)**2
    square = (2*r+g)**2
    gap = g**2
 
    # Way 1
 
    q1 = 1.0 * small/big
    q2 = 1.0 * gap/square
    q = q1*q2
 
 
    # Way 2
    '''
    nos = 1.0*small/square
    totalgap = gap*nos
    q1 = 1.0 * small/big
    q2 = 1.0 * totalgap/small
 
    q = q1*q2
    '''
 
 
    if 0 <= q:
        if q <= 1:
            p = 1-q
        else:
            p = 0.0
    else:
        p = 1.0
    #print 1-q
    return p
 
 
def sfs(s):
    'solve from string'
    s = s.replace('\n', '').split(' ')
    f,R,t,r,g = [float(i) for i in s]
    return solve(f,R,t,r,g)

I did one in C too, I was thinking python did something wrong with math…

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#include <stdio.h>
#define pi 3.1415926535897931
 
double solve(double f,double R,double t,double r,double g)
{
    double big,small,square,gap,q1,q2,q,p;
 
    if (2*f >=g )
    {
         return 1;
    }
 
    t = t + f;
    g = g - 2*f;
    r = r +f;
 
    big = pi*R*R;
    small = pi*(R-t)*(R-t);
    square = (2*r+g)*(2*r+g);
    gap = g*g;
 
    q1 = small/big;
    q2 = gap/square;
    q = q1*q2;
 
    if (0<=q)
    {
        if (q<=1)
        {
            p = 1-q;
        } else p = 0;
    } else p = 1;
 
    return p;
}
 
int main()
{
    printf("Case #1: %lf\n",solve(0.25, 1.0, 0.1, 0.01, 0.5));
    printf("Case #2: %lf\n",solve(0.25, 1.0, 0.1, 0.01, 0.9));
    printf("Case #3: %lf\n",solve(0.00001, 10000, 0.00001, 0.00001, 1000));
    printf("Case #4: %lf\n",solve(0.4, 10000, 0.00001, 0.00001, 700));
    printf("Case #5: %lf\n",solve(1, 100, 1, 1, 10));
}

There’s no need to say that the result was the very same.

I don’t know what’s wrong with my approach, all the others’ code I read used something involving integrals, sometime already solved and they just did the “definite” part.

Did you read someone’s code that is not using integrals but pure geometry and probability approach?
I think it’s possible, I think there is something very bastard I’m missing… don’t know what… do you?

return “grr”

GCJ 2008: Train Timetable

July 19th, 2008 § 0 comments § permalink

This one was the title of the second problem of Google Code Jam Qualification Round 2008.

This time my code is a bit messy, but I won’t adjust it for you :^D It’s the disadvantage of coding fast, at least for me. I often wrote something not necessary, someone call it gold planting, it’s just what I did in the maze problem with the draw method…

You can download my solution for the Train Timetable problem I wrote.

Here is the commend code.

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from __future__ import with_statement
from copy import copy

Don’t ask me why I imported copy, I think I did it while testing… maybe. I don’t remember it’s not useful
The first 3 are classes but I did not spent so much time projecting their structure. Tempo just means time in italian, I didn’t want to override the python name (I know that’s the name of a lib, it’s only an excuse, I’m italian, let me name things in italian sometime :^P)

tempo is not so useful in that form… anyway converts the format ‘hh:mm’ to an int representing minutes

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class tempo(object):
    def __init__(self, s):
        a = s.split(':')
        x = [int(i) for i in a]
        self.m = x[0]*60 + x[1]

Trip represents a trip, with starting and ending time.

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class Trip(object):
    def __init__(self,start,end):
        self.start = tempo(start)
        self.end = tempo(end)
        self.duration = self.end.m - self.start.m
        self.busy = False
    def __cmp__(self,x):
        return cmp(self.start.m,x.start.m)
    def use(self):
        self.busy = True
    def isfree(self):
        return self.busy == False
 
    is_free = property(isfree)

There is no need to put the duration attribute, but it did, thinking on it now it’s pretty useless… The interesting part is the flag busy, or is_free (just a joke with python property function…). A Trip is free if you did not use it yet. I added __cmp__ method because I’ll need to “sort” them, if trip X starts earlier than trip Y well…

X<y == True

It will be useful to understand what’s the first Trip to use…

A Train is identified by the time of the last thing it did (self.Time) and it has done something or not (self.init)

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class Train(object):
    def __init__(self,Time):
        self.Time = tempo(Time)
        self.init = False
    def TripTrough(self, trip, turnaround):
        'Returns True if the train can trip (and it does)'
        if (trip.is_free and self.Time.m+turnaround <= trip.start.m) or self.init ==False:
            self.init = True
            self.Time = trip.end
            trip.use()
            return True
        else:
            return False

The TripTrough method is very important (yes I know, it’s Through not Trough… my english sucks, sorry for that :^D )… Anyway… How does it work? It’s simple. It wants a trip and the turnaround time. the flag init it’s useful because if the trip starts at 0:03 and you have a turnaround of 5 you have to don’t care about the turnaround, you have no need to turn around… my check can’t work without an init flag… I don’t know if it’s clear..

Now something magical happens, I’m doing it bad I think, I’m playing with the “all name are references” of python… I changed the trip flag as the train used it, I could do it out of that method but maybe I was tired, bored, drunk, whatever… and I did it in this way :^D. I updated the time of the train too, obviously.

Table has the algorithm! Let me explain…

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class Table(object):
    def __init__(self,A,B,T): # trip lists... and turnaround
        self.A = A
        self.B = B
        self.T = T # turnaraund
    def solver(self, starting,ending,train):
        starting.sort()
        s = starting
        for i in s:
            if train.TripTrough(i,self.T):
                return self.solver(ending,starting,train)
        # else...
        return train
 
 
    def solve(self):
        tfa = []
        tfb = []
 
        check = True
        while check:
            freeA = [i for i in self.A if i.is_free]
            freeB = [i for i in self.B if i.is_free]
            #print freeA, freeB
            if len(freeB) > 0 and len(freeA)>0:
                mins = min(freeA),min(freeB)
            elif len(freeB) == len(freeA) == 0:
                break
            if len(freeA)>0 and (len(freeB) == 0 or mins[0]<mins[1]): #start from A
                tfa.append(self.solver(freeA,freeB,Train('00:00')))
            else:
                tfb.append(self.solver(freeB,freeA,Train('00:00')))
        return len(tfa),len(tfb)
  • A is the list of trips starting from station A going to B
  • B is the list of trips starting from station B going to A
  • T is the turnaround time of the case
  • solver it’s a recursive method.
    It takes 3 parameters:

    1. starting it’s the list of the trips starting from a station
    2. ending it’s the list of the trips starting from the other station
    3. train is the train that’s traveling forward and backward

    Well… I sort the trip list, in this way come first the one that will start first, then I begin to look for the first one that the train can travel through. If it can I can look (recursively part) for a return trip, just inverting starting with ending calling once again the solver method.

  • solve is the method you need to call to get the final result.
    • tfa is the list of trains starting from the A station
    • tfb is the list of trains starting from the B one
    • The part
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              check = True
              while check:

      it’s just useless… I could use while True… I won’t change check

    • freeA is the list of the trip starting in A that are not busy
    • freeB it’s the same as the above one but in B, obviously
    • if there is something in freeA and freeB I’d like to know what’s the trip starting earlier in freeA and the one in freeB
    • else if there is nothing in freeA and nothing in freeB there is no need to continue
    • if there is something in freeA and there is nothing in freeB or just the earliest trip in freeA starts before the earliest one in freeB I can set up a new train and make it travel starting from the earliest trip in freeA until it stops…
    • else I can set up a new train and make it travel starting from the earliest trip in freeB until it stops…
    • I added these trains to tfa and tfb
    • I return the number of items of tfa and tfb, as the problem asks..

You can download the solution here with the file processing too.

The algorithm it’s correct but the implementation sucks, I think I can code a better one with less line of code, I’ve just to spend some time on it… It was a time challenge and I was not thinking so much of optimization… I don’t know If I’ll post a revisited one… For now it’s all folks!

return ‘Bye’

GCJ 2008: Saving the Universe

July 18th, 2008 § 3 comments § permalink

That’s was the title of the first and very funny problem of Google Code Jam Qualification Round 2008

The urban legend goes that if you go to the Google homepage and search for “Google”, the universe will implode. We have a secret to share… It is true! Please don’t try it, or tell anyone. All right, maybe not. We are just kidding.

You can find (I think you need to register) the whole problem here.

I did it recursively and for the large input I had to setrecursionlimit up to 2000. I’ll show you why…

You can download my solution here.

That’s my implementation:

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import sys
sys.setrecursionlimit(20000)
 
def process(se, q, switch = 0):
    if len(q) == 0:
        return switch
    rank = {}
 
    notin = False
    for i in se:
        if q.count(i)>0:
            rank[i] = q.index(i)
        else:
            notin = i
 
    if notin == False:
        a,b = rank[se[0]],0
        for i in rank:
            if rank[i]>b:
                a = i
                b = rank[i]
        switch += 1
        return process(se,q[b:],switch)
 
    else:
        return switch

In words…

se is the list ( [‘Google’,’Yahoo’,…] ) of search engines
q is the list (like the one above) of queries

  • At the beginning I did no switches (switch = 0), but it’s a default value, when I’ll call it I’ll set it
  • If there is no query I don’t need to switch anymore, I can return the number of switches (switch)
  • rank is a dictionary. It’s just like ‘Search Engine': ‘position of the first query with the same name of this search engine’
  • if the name of a search engine does not appear in the query list I may do all searches with that search engines (the notin test)
  • else I look in the dictionary for what is the search engine that appear as later as possible in the query list and I use it, I add 1 to the switches counter, and return (recursively) the process giving the same se, the current switches number but a different query list!

The new query list is just the remaining queries. Example:

se = Yeehaw, NSM, Dont Ask, B9, Googol
q = Yeehaw,Yeehaw,Googol,B9,Googol,NSM,B9,NSM,Dont Ask,Googol

Call #1
We did no switch, so switch = 0
All search engines appears in the query list… Here is the list of each se with the index of the first query with the same name:

Yeehaw: 0, NSM: 5, Dont Ask: 8, B9: 3, Googol: 2

Dont Ask has an higher index so we can use that to search up to that index with it, and process the next call with the new query list:

se = Yeehaw, NSM, Dont Ask, B9, Googol
q = Dont Ask,Googol

Call #2
We did one switch, so switch = 1

Yeehaw, NSM and B9 do not appear in the query list, I can use any of them without do any other change, I return the current number of switches ;^) Quite simple ;^)

In the big file there was a query list of 999 items Like this:

se = [‘A’, ‘B’]
q = [‘A’, ‘B’,’A’, ‘B’,’A’, ‘B’,’A’, ‘B’,’A’, ‘B’,…]

ok? It did lots of recursion and went over the limit, I had to adjust that :^D

return ‘Bye’

GCJ: Always Turn Left

July 16th, 2008 § 1 comment § permalink

Hi there :^)

Let’s continue with the “Practice Problems” of GCJ, the next is a problem about perfect mazes.

This time the source code will be quite long, sorry :^D. I’m currently having some problems with wordpress and utf8 (if you know how to solve quickly please leave a comment), so the complete source code is available here.

We have to map the maze, square by square we have to know where you can move starting from that square. We use a code to do this.

Idea #1
Look at the map of “cases”. Choose one of them, any of them. To be short consider N as North, S as South and so on… Now consider (in example) E = 1 if you can move to East, else 0, and do it with the others too, then put the result in this order EWSN. It’s binary, yes. Got it? Try to convert it to hex and compare the result with the problem case-codes ;^)

I’ll import the alien-numbers to do the conversion (I know that’s simple and there are other ways, but you will see what I used to debug :^D)

I’m going to comment my implementation part by part. Let’s start.

Importing

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from __future__ import with_statement
from copy import deepcopy
import aliensys

with statement with input and output files (I did know show that on the previous post)

deepcopy is very useful. Remember that in python names are just a reference. I had serious problem debugging my code due to this!

aliensys… well you can now imagine why I’m using this… but that’s not the only reason :^D

class Mode

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class mode(object):
    def code(self):
        HEX = '0123456789abcdef'
        BIN = '01'
        number = ''
        for i in (self.e,self.w,self.s,self.n):
            number += str(1*i)
        return solve(number + ' ' + BIN + ' ' + HEX)
 
    def __init__(self, e = False, w = False, s = False, n = False):
        self.e = e
        self.w = w
        self.s = s
        self.n = n
    def change(self, k, val):
        if k == 'e':
            self.e = val
        elif k == 'w':
            self.w = val
        elif k == 'n':
            self.n = val
        elif k == 's':
            self.s = val
 
    def draw(self):
        MAZE = u'.???????????????'
        HEX =   '0123456789abcdef'
        return solve(self.code()+' '+HEX+' '+MAZE)
    def __str__(self):
        try:
            return self.code()
        except AttributeError:
            return 'tmpNone'
    def __repr__(self):
        try:
            return str(self.code())
        except AttributeError:
            return 'tmpNone'
    def draw(self):
        MAZE = u'.???????????????'
        HEX =   '0123456789abcdef'
        return solve(self.code()+' '+HEX+' '+MAZE)

__init__ I just get EWSN and save them.
code It returns the case-code using EWSN

Now come one of the interesting parts :^)
Note that I pasted it as is, sorry for the question marks, it’s a wordpress problem with utf8, anyway it should be clear in my perfect maze solution file. Check it.

Debugging this was not simple. I used the __str__ and __repr__ because the IDLE debugger uses them, but it was not enough. I looked for utf8 symbols that do the trick, I can’t show those in this post (damned wordpress) but some aliens could use them as numeral system… and I’ve a converter to read that :^D

I can draw a “mode” of a square of the maze…

class Square

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class sqr(object):
    def __init__(self, x,y,m):
        self.x = x
        self.y = y
        self.m = m
 
    def move(self, d, update=True):
        if d == 'e':
            self.x+=1
            if update: self.m.w=True
        elif d == 'w':
            self.x-=1
            if update: self.m.e=True
        elif d == 'n':
            self.y-=1
            if update: self.m.s=True
        elif d == 's':
            self.y+=1
            if update: self.m.n=True
 
    def __eq__(self, x):
        return ((self.x == x.x) and (self.y == x.y))
    def __cmp__(self, x):
        if ((self.x == x.x) and (self.y == x.y)):
            return 0
        elif self.y==x.y:
            return cmp(self.x,x.x)
        else:
            return cmp(self.y,x.y)

We have to know the position of a square (x and y) and its mode.
We can move a square (to E,W,N or S), and from that position it could (if want) come back to the old position (the update part)

__cmp__ and __eq__ are used in the Maze class because I need to sort squares. Given a list of squares comes first the square that has the minor y and minor x, then comes the one that has the same y but greater x else a greater y and the minor x… and so on… I think it will be clearer in the next class…

ls2str

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def ls2str(ls):
    r = ''
    for i in ls:
        r+=str(i)
    return r

That’s a simple function that convert a list to a string… No. str(ls) is just like print ls and that’s not what we will need ;^)

class Maze
Now the big part :^D I’ll slice it in more parts because is very long (120 lines) and not in the same order of the real file. Anyway I’m trying to save line numbers so that you can check ;^)

__init__

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    def __init__(self, i2o, o2i):
        self.i2o = list(i2o) # in 2 out
        self.o2i = list(o2i) # out 2 in
        self.head = 's'
        self.sqrs = []
        self.init=False

The problem gives us two path, as string, the first is from the entrance to the exit (i2o) and the second is from the exit to the entrance (o2i). Our head is poting to S, we know no squares now, because we haven’t started yet ;^)

have_sqr

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    def have_sqr(self, s):
        check = False
        for i in self.sqrs:
            if s == i:
                check = True
                break
        return check

It’s simple, it there is already a square with the same x and y returns True, else False… it will be useful :^D

turn

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    def turn(self, direction):
        if direction == 'R':
            for i in zip(list('ewsn'), list('snwe')):
                if i[0] == self.head:
                    self.head = i[1]
                    return True
        elif direction == 'L':
            for i in zip(list('ewsn'), list('nsew')):
                if i[0] == self.head:
                    self.head = i[1]
                    return True
        else:
            return False

When you are walking trough the maze you don’t change square if you turn. Returns True if you turned (and change the direction, read with attention those lines), else returns False

walk
That’s one of the longest parts. We want to walk trough the maze using a path….

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    def walk(self, path, reverse=False):
        pre = False
        for i in path:
            if pre:
                pass
            elif reverse:
                pre = deepcopy(self.sqrs[-1])
            else:
                pre = sqr(0,-1,mode())
 
            if self.turn(i) == False:
                if self.have_sqr(pre) == False:
                    self.sqrs.append(pre)
                index = self.sqrs.index(pre)
                self.sqrs[index].m.change(self.head, True)
                new = deepcopy(pre)
                new.m = mode() # resetting modes
                new.move(self.head)
                pre = deepcopy(new) # remember to update pre
        if self.have_sqr(new) == False: # the last ;)
            self.sqrs.append(deepcopy(new))
            if reverse == False: # Changing direction...
                self.last = self.sqrs[-1]
                for dirs in zip(list('ewns'), list('wesn')):
                    if self.head == dirs[0]:
                        self.head = dirs[1]
                        break

It’s simple. Don’t panic :^)
We start from a square, the path tells what to do. While it says to turn, we turn, if it says to go we move to a new square… are you sure is it new? Check it, if not add it, else update the modes.

If B is below A and you can go from A to B you can also go from B to A, so A has S = 1 and B has N = 1 ;^)

walkall

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    def walkall(self):
        if self.init:
            pass
        else:
            self.walk(self.i2o)
            self.walk(self.o2i, True)
            self.init = True

To have a clear and correct map we need to walk using i2o and o2i too, this means to walk trough all paths…

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    def getlist(self):
        self.walkall()
        m = deepcopy(self.sqrs)
        for i in m:
            if i == self.last:
                index = m.index(i)
                del m[index]
        xs = [i.x for i in m]
        xmin = min(xs)
        xm = max(xs)
        ys = [i.y for i in m]
        ym = max(ys)
 
        m.sort()
        r = []
        a = []
 
        for y in xrange(0,ym+1):
            a = []
            for x in xrange(xmin,xm+1):
                c = sqr(x,y,mode())
                if 1 == 0:
                    pass
                else:
                    if c in m:
                        i = m.index(c)
                        a.append(m[i].m)
                        del m[i]
                    elif xmin<x<xm:
                        a.append('0')
            r.append(ls2str(a))
 
        return r

The requested output is a list of line, each one contains the modes of the squares of that maze’s associated line. We need to order self.sqrs that’s why redefined the __cmp__ method for a square.

We’ve to delete the starting and the ending point from the map too, this method is too long for me, I think there is a better way to do this… but was the first way that turned in my mind…

draw

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    def draw(self):
        MAZE = u'.???????????????'
        HEX =   '0123456789abcdef'
        maxs = []
        p=[]
        for i in self.getlist():
            l = solve(i+' '+HEX+' '+MAZE)
            maxs.append(len(l))
            p.append(l)
        m = max(maxs)
        for i in p:
            print i.center(m)

My draw idea was cool enough that I forgot I’ve already implemented in the mode class… i did it again! o_O Don’t ask me why, maybe I was drunk when I wrote that :^O

Anyway… we can draw a maze now :^) really :^D

File eater
Final part, we have an input file, and we’ve to produce an output one…
eatFile will do the trick

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def eatFile(path_i, path_o, show=True, draw=False):
    with file(path_i, 'r') as f_in:
        lines = f_in.readlines()
        n = int(lines[0].replace('\n', ''))
        del lines[0]
        with file(path_o, 'w') as f_out:
            for i in xrange(0,n):
                s = lines[i].replace('\n', '').split(' ')
                m = maze(s[0],s[1])
                if show:
                    print 'Case #%d:' % (i+1)
                if draw:
                    m.draw()
                    print '\n\n'
 
                if draw:
                    f_out.write('Case #%d:\n' % (i+1))
                for j in m.getlist():
                    f_out.write(j+'\n')
                del m

Working with text files is painless with python. Using the with statement I don’t have to open and close files, it’s cool :^D

The mechanism is simple:

  • Get the number of lines
  • For each line of the input file:
    1. Create a maze
    2. Print and draw if you have to
    3. write to the file output
  • Finish :^D

and now the operative part….

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# Solving the question... (with draws :D)
import time
start = time.time()
before = time.time()
path1 = 'B2-small.in'
path2 = 'B2-small.out.txt'
eatFile(path1,path2,True,True)
print 'Small in',time.time() - before,'secs'
 
 
before = time.time()
path1 = 'B2-large.in'
path2 = 'B2-large.out.txt'
eatFile(path1,path2,True,True)
print 'Large in',time.time() - before,'secs'
print 'Both in',time.time() - start,'secs'

lines 243 and 250 are enough but I wanted to measure the time it tooks to solve the problem :^D
On my pc the final lines were:
Small in 9.17199993134 secs
Large in 247.765999794 secs
Both in 256.983999968 secs

But I had too many thing opened while testing it that’s not a valid test, anyway try it on your computer and tell me your time ;^)

Conclusions
There few notes I’d like to leave.
I spent time, like 2 hours, to solve this. And it is not optimized (I wrote a change method for mode but I don’t use it in sqr when I should, but there are lots of these little things). I was coding thinking of the future but in the future i forgot of the past and… it’s, maybe, TOO object oriented… I’m too slow for this competition I guess… I hope someone will find this useful, my code need to be shorter!

Thanks for reading

return ‘Bye’

GCJ: Alien Numbers

July 10th, 2008 § 3 comments § permalink

In these days I started to study python. That’s a very cool language and I have lots of things to learn, I need exercise. What is better than google code jam to practise?

I will share with you my solution for the first of the “Practice Problems”, it’s about aliens’ numeral system :^) and I’ve, obviously, written my solution in python ;^)

I’m posting it to discuss them with you, maybe I’ll do the same when I’ll have the time to do the others. It doesn’t want to be a spoil so if you want to solve this alone just go and code, when you’ve finished you can come back and comment my code and tell me if you coded a better one :^)

Well, the text of the problem is:

Problem

The decimal numeral system is composed of ten digits, which we represent as “0123456789” (the digits in a system are written from lowest to highest). Imagine you have discovered an alien numeral system composed of some number of digits, which may or may not be the same as those used in decimal. For example, if the alien numeral system were represented as “oF8″, then the numbers one through ten would be (F, 8, Fo, FF, F8, 8o, 8F, 88, Foo, FoF). We would like to be able to work with numbers in arbitrary alien systems. More generally, we want to be able to convert an arbitrary number that’s written in one alien system into a second alien system.

Input

The first line of input gives the number of cases, N. N test cases follow. Each case is a line formatted as

alien_number source_language target_language

There are some other things, if you want you can read the whole text here.

Ok? That’s should be not so difficult. I’m human, coder and engineering student, I hear aliens speaking every day, it’s ok. The problem is to speak with them :^)

We’re going to:

  • Get the number we want convert from system1 to system2
  • Convert to decimal that number from system1, we’re human and used to see decimal number :^)
  • We know what number it is, we can now convert the decimal number to system2

Well try to think on how we convert from binary to decimal and viceversa, or from hex to decimal and viceversa…

Here is my implementation:

class aliensys(object):
    def __init__(self, stringa):
        self.symbols = list(stringa)
        self.N = len(self.symbols)
    def a2d(self, n):
	"Convert n from alien to decimal"
        s = 0
        n = list(n)
        n.reverse()
        for i in xrange(0, len(n)):
            s += self.symbols.index(n[i])*self.N**i
        return s
    def d2a(self, n):
	"Convert n from decimal to alien"
        s = ''
        while n>=1:
            r = n%self.N
            n/=self.N
            s = self.symbols[r]+s
        return s
    def convert(self, n, target): # target must be an aliensys
        to = self.a2d(n)
        return target.d2a(to)
 
 
def solve(string):
    s = string.replace('\n', '')
    s = s.replace('\r', '')
    n, src, tgt = s.split(' ') # number, source, target
    src = aliensys(src)
    tgt = aliensys(tgt)
    return src.convert(n,tgt)

And it’s so simple to use:

>>> solve('CODE O!CDE? A?JM!.')
'JAM!'

ps: did you see that? how WP-syntax is highlighting “string” and “self” on my script? They’re not keyword in pyhton, str is, it shouldn’t do that… I’ve to fix this :\

yield ‘Bye’

WordPress code highlighter, the choosen one

July 9th, 2008 § 0 comments § permalink

I’m going to write about programming in next posts, that’s means I need a way to show some code here and there. My posts need a code highlighter to be clear :^)

I googled for a while to find a decent code highlighter. I found several code highlighters, some do it on the fly in JavaScript, some others do it server side, and some one let Google do it :^D. I’ve found a comparison too.

Well, I installed all of these, tried them and finally WP-syntax wins! Why?

  • It’s server-side, if you do not trust me or my site who cares? you will see highlighted code anyway ;^)
  • Supports a painless copy & paste,  when you copy from most of the other highlighters and paste in a simple text editor the code comes with more \n and # than what the author wrote… and maybe with line numbers too!
  • Supports line numbers, and the coolest part is that you can start from the line you want, it could be very useful!
  • And last but not least: I like it :^D, i’ll maybe fix the css to fit with the theme when I’ll end it (that’s cool but I want one written by me :^)

The only very bad thing is that the switching from HTML visual and WYSIWYG visual is very painful, I hope they’ll improve that!

Anyway here is the proof :^)

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def fooer(count=200,start=2): # a decent foo should have at least 2 'o'
    for i in xrange(start, start+count):
        yield 'f' + 'o'*i
 
def allfoo():
    ppl = ('I','You','He','She','It','We','You','They')
    ippl = iter(ppl)
    for s in fooer(len(ppl)):
        i = ippl.next()
        if i in ('He','She','It'):
            s += 'es'
        print i,s
 
allfoo()

That’s, actually, the proof of how you can code very useless things but they are highlighted pretty well anyway :^D

return ‘Bye’