by Xav de Matos, Jan 18, 2011 5:00pm PST

As I sit here and type this the only thing I can think about are the piles of previews and interviews I have to write out for the rest of this week. When I say "rest of this week," what I really mean is tonight because I don't have any other time to do it. Thankfully, I'm taking a breather beginning Friday and returning on Monday night.

I haven't traveled for non-work related reasons in a while, so it will be nice to not have to worry about my schedule, or laptop, or voice recorder. Or taking piles of notes.

Speaking of notes, remember to jot down this note for yourself: "Keep refreshing Shacknews tomorrow morning because Nintendo is set to reveal its launch details for the 3DS in North America tomorrow."

Here's the video game news of the day from the Shack to tide you over while you wait until the morning:

Here are some random musings from around the web:

• I had a programming test yesterday. On of the questions was:

Given an array of numbers
[4, 6, 8, 3]

Work out how many steps it takes to make them all the same number if you can only increment them by +1/-1 per step.

I worked out the average of the array then took the difference between the average and the number furtherest away from it. Was this correct?

Thread Truncated. Click to see all 31 replies.

• My explanation/proof for this is:

If you take the biggest and smallest number, picking any number in between them will give a combined distance equal to the difference of the numbers. Picking a number outside of them will give a combined distance larger than this. Therefore we know the target number has to be somewhere between them (ignoring the other two numbers right now).

Now we take the two remaining numbers, and use the same reasoning to show that their minimum distance is also produced when the target number is between them. This number also happens to fall in the range which produces the minimum distance for the first two numbers.

This reasoning could be extrapolated for any array size.

So I would sort the array, and then sum the differences between the k'th and (n-k)'th index for k=0 to the length of the array.