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’

Google Code Jam 2008 – Qualification Round

July 18th, 2008 § 0 comments § permalink

I’ll be short.

Qualification Round has just ended.

It was very cool, there were 3 problems, I wasn’t able to solve the last, I’ll discuss with you them all in these days, now let me go to sleep, it’s 2 a.m. in Italy :^D

return ‘yawn’

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’

Coding

July 8th, 2008 § 0 comments § permalink

I like to develop software, application in general but I prefer web applications.

Being a software developer could be the most boring and maybe painful thing to do you you don’t like to. You’ve to face everyday to problems that you don’t have and… you’ve to solve them, if you don’t get it like a challenge you get stressed from this.

I like it, really. If you do not like problem solving you may think I’m kinda mad or something, well… maybe I am :^D (but I’m not the only one)

I’ll try to run this blog and share with readers my opinions, that’s the first (and I hope the less meaningful) post and reading this in the future maybe will make me laugh, but who cares…

return ‘Bye’