Mystery Company Recruiting Talent With a Puzzle

An anonymous reader writes "Google has previously used coding competitions to locate top talent. In a new twist on the idea, an anonymous tech company is posting a help-wanted ad that challenges developers to find out who the company is. A little digging and text mashing reveals a website containing a Web 2.0 puzzle that makes notpron look like child's play. So, fellow developers, who is this company, and, well, what is the significance of the date '01-18-08?'" Update: 12/12 20:20 GMT by KD : Replaced link to a removed Craigslist ad with a mirror.

Comments

One word...

[DigiWood]

Cloverfield. 01-18-08 is the release date.

Save some time

[madsheep]

If anyone wants to save some time (like 30-60 seconds) with Base64 to Ascii:

eyAnOicgPT4gJycsICcgJyA9PiAnLScsICdzXG4nID0+ICdzLmNvbVxuJyB9 converts to { ':' => '', ' ' => '-', 's\n' => 's.com\n' }

Here's the contact info (spoiler warning)

[mukund]

Just base64 decode the string that appears to be made of random chars. You get:

{ ':' => '', ' ' => '-', 's\n' => 's.com\n' }

Apply that to the subject in the contact details. You get:

http://wanted-master-software-developers.com/

That was pretty easy. The test then seems to move to web programming and I'm not interested.

Re:Base64

[jcaldwel]

Its a find-and-replace that turns the title:
Wanted: Master Software Developers

Into:
http://wanted-master-software-developers.com/

... and the test continues...

Source code

[Raynor]

Am I the only one who grabbed their /js/ and peeked at the code...

"// Note: It is not necessary to reverse-engineer this file in order to complete the contest"

I did no testing of any sort... inside framework.pack.js it says

p.setAttribute("title","list, uniquify, relativity");
p.appendChild(document.createTextNode("Ford's, success, has, the, country, almost, financially, industrially, mechanically, exhibits, in, higher, than, persons, have, thought, possible, contradictory, requirements, of, efficiency, increase, great, workers, cost, consumer, And, cost, cost, consumer, And, cost, cost, consumer, And, workers, workers, workers, workers, to, repeated, great, increase, quality, increase, great, great, increase, quality, efficiency, efficiency, which, are, of, contradictory, contradictory, requirements, of, possible, have, have, thought, possible, have, have, persons, than, than, most, persons, persons, than, most, exhibits, exhibits, exhibits, exhibits, financially, financially, financially, financially, almost, the, the, country, almost, Ford's, Ford's, success, has")); ... That was easy.

Re:Source code

[marcansoft]

Crap, I'm a moron. I had a bug in the regular expression, and I managed to work around it with the other code, causing the problems. This caused the code to insert a 0 between everything, yielding the alternating +/- numbers, and the "o" oddity was caused by two separators together in the source text.

Turns out all you need is one differentiation and a non-retarded regular expression that doesn't insert empty words between each pair of non-word characters.

I've also made the variable names resemble less those of the problem (read: not single-character madness). And added some comments.

#!/usr/bin/python

import re

data = "Ford's, success, has, the, country, almost, financially, industrially, mechanically, exhibits, in, higher, than, persons, have, thought, possible, contradictory, requirements, of, efficiency, increase, great, workers, cost, consumer, And, cost, cost, consumer, And, cost, cost, consumer, And, workers, workers, workers, workers, to, repeated, great, increase, quality, increase, great, great, increase, quality, efficiency, efficiency, which, are, of, contradictory, contradictory, requirements, of, possible, have, have, thought, possible, have, have, persons, than, than, most, persons, persons, than, most, exhibits, exhibits, exhibits, exhibits, financially, financially, financially, financially, almost, the, the, country, almost, Ford's, Ford's, success, has"

text = "Ford's success has startled the country, almost the world, financially, industrially, mechanically. It exhibits in higher degree than most persons would have thought possible the seemingly contradictory requirements of true efficiency, which are: constant increase of quality, great increase of pay to the workers, repeated reduction in cost to the consumer. And with these appears, as at once cause and effect, an absolutely incredible enlargement of output reaching something like one hundredfold in less than ten years, and an enormous profit to the manufacturer"

def do_list(string):
    return re.split(r"[ ,.:]+",string.lower()) # split by any combination of space, comma, period, colon.

def do_uniquify(lst):
    out_l = []
    for i in lst:
        if i not in out_l: # ignore dupes
            out_l.append(i)
    return out_l

def do_relativity(textlist,datalist):
    last_pos = 0 # keep track of last position
    out_l = []
    for word in datalist:
        index = textlist.index(word) # find index in source text
        out_l.append(index-last_pos) # differentiate index
        last_pos = index
    return out_l

textlist = do_list(text)
datalist = do_list(data)
uniquelist = do_uniquify(textlist)
relative_numbers = do_relativity(uniquelist,datalist)
# stringize all numbers, join with commas.
print ",".join(map(str,relative_numbers))

Re:Difficult test? Hardly.

[Furry Ice]

Am I the only one who enjoyed the challenge of solving the problem the way it was intended? Someone correctly guessed that this is like Tetris, where true is a block and false is empty space. However, it's unlike Tetris in some key ways. If you try to solve it, you'll see how as you hit test cases that your code fails on. Here's my function, which passes all tests. I had to try three different algorithms because new information about the behavior of the blocks necessitated starting from scratch with more complexity twice.

f = function(d) {
    var height = d.length;
    var width = d[0].length;
    var find_base = function(t, i) {
        for (j = 0; j < width; j++) {
            if (d[i][j]) {
                if (d[i+1][j]) {
                    t[j] = true;
                }
                if (j > 0 && j < (width - 1)) {
                    if (d[i+1][j-1] && d[i+1][j+1]) {
                        t[j] = true;
                    }
                }
            }
        }
    };
    var add_sticky = function(t, i) {
        while (true) {
            var stop = true;
            for (j = 0; j < width; j++) {
                if (d[i][j] && !t[j]) {
                    if (j > 0 && t[j-1]) {
                        t[j] = true;
                        stop = false;
                    }
                    if (j < (width - 1) && t[j+1]) {
                        t[j] = true;
                        stop = false;
                    }
                }
            }
            if (stop) {
                break;
            }
        }
    };
    var i, j;
    var t = new Array(width);
    for (i = height - 2; i >= 0; i--) {
        for (j = 0; j < width; j++) {
            t[j] = false;
        }
        find_base(t, i);
        add_sticky(t, i);
        for (j = 0; j < width; j++) {
            if (d[i][j] && !t[j]) {
                d[i][j] = false;
                d[i+1][j] = true;
            }
        }
    }
};