Hotashi - Jack’s Blog

There’s a post today over at coding horror that proposed a very simple probability problem. After more than 400 comments (and counting) I’m very surprised that so many people have the wrong answer. The question is this:

Let’s say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?

Actually, I’m not sure if by “odds” Jeff meant this, or probability — as is the case in most everyday use. But assuming that he meant probability, then the solution is as follows.

Let A be the event of having a girl.
Let B be the event of having a boy.

Assume that P(A) = P(B) = 0.5

We know that the person has a girl, thus we are trying to find P(B|A) (probability of B given that A has occurred).

Now we note that A and B are independent events, thus P(B|A) = P(B) = 0.5 (conditional probability).

And there you go, the probability of the person having a boy given that person has a girl is 0.5 (or 50%).

Happy New Year’s folks!

I’ve decided to try out the Publish/Subscribe pattern for some of the new things we’re creating at work. So far I think it’s working out really well. The neatest thing is that a Pub/Sub system can be created in JavaScript in very little amount of code. And with the help of jQuery, you can shorten the amount of code even more.

Here is the entire system in jQuery.

jQuery.subscribe = function(signal, scope, fnName) {
    var curryArgs = Array.prototype.slice.call(arguments, 3);
    $(window).bind(signal, function () {
        var normalizedArgs = Array.prototype.slice.call(arguments, 1);
        scope[fnName].apply(signal, curryArgs.concat(normalizedArgs));
    });
    return jQuery;
};

jQuery.publish = function(signal) {
    $(window).trigger(signal, Array.prototype.slice.call(arguments, 1));
    return jQuery;
};

Notice that the subscriber function allows for partial applicationof the handler function. This can be useful in some cases.

Thought I’d post up my solution to Problem 28 on Project Euler.

After the answer was submitted I read through some of the other code the people have posted. It seems that there are a lot more programmers than mathematicians because almost all of the solutions were brute-force algorithms… and quite a few were written in assembly. *shrug*

Anyway, here is the problem.

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of both diagonals is 101.

What is the sum of both diagonals in a 1001 by 1001 spiral formed in the same way?

The first you to notice is that you don’t have to actually construct the spiral in order to solve this problem. It can be shown quite simply that in an nxn spiral, the values of the four corners are as follows:

top-right: n^2
top-left: n^2 - n + 1
bottom-left: n^2 - 2n + 2
bottom-right: n^2 - 3n + 3

Adding all that up yields 4n^2 -6n + 6.

At this point you can easily loop through from 3 to 1001 (in increments of 2) and add up the results. Just make sure you add a 1 to the results for the base case. :)

Of course, if you’re like me you wouldn’t be satisfied with that result. All we’re dealing with are some simple summations, so the result for an nxn spiral should be easily expressed by a formula. In fact, here’s the formula.

Sum(i = 1:500)[4(2*i + 1)^2 - 6(2i + 1) +6]

The beginning reads “sum from i=1 to 500″… I’m not too sure how to type out sigma notations properly.

Converting the summation into closed form gives the formula we’re looking for. I’m leaving out the steps, but it’s fairly simple anyway.

(16n^3)/3 + (10n^2)/2 + 26n/3

And here is the complete solution in Python.

from math import pow
def prob28(n): return (16*pow(n, 3) + 26*n)/3 + 10*pow(n, 2)
print prob28(500) + 1

I’m sure everyone’s heard of the show America’s Best Dance Crew. Last year’s winner, the Jabbawokeez, are now famous across all of North America. They are super talented, no question about that, but sadly the average American doesn’t even know the state of world B-boying. I think most people in North America think B-boy crews in the US = best in the world. That’s totally wrong though… in fact crews in the US is not even close to being in the top. Talk to anyone know who’s what they’re talking about and they’ll say that currently South Koreans are the best b-boys in the world. Followed closely by France and Japan. In fact, in recent years, South Korean crews have been winning a lot of the annual Battle of the Year competitions, whereas US has not even reached the finals.

Here’s the crew that won last year’s BOTY, Extreme Crew. Enjoy!

Google just finished digitizing newspapers that date back to early 1900s, for their news archives search.

For example, here’s an article from the Pittsburgh Post-Gazette for the moon landing. Pretty cool eh?

So Globe & Mail’s archives search goes back to 2000, while Google’s goes back to early 1900s. And I think NY Times recently went back as far as 1800s? — or maybe early 1900s too. Globe and Fail!