Saturday, December 27, 2014

Hangman Challenge

My holiday gift to readers this year is a simple Hangman game:

___  ___  ___  ___

If you'd like to play, email me your first move. I'll reply and we'll begin a thread that way.

There's no gallows here, and there's no limit to the number of guesses you can take. Just keep guessing letters until you guess the word. Try to solve the puzzle using as few guesses as possible.

Later I'll post the mystery word, and I'll say a few words about the principles I used to select it.

Until then, enjoy the holidays! All my best for a healthy and happy 2015.

Saturday, December 20, 2014

Change for a Dollar

So it turns out that there are 293 ways to make change for a dollar using pennies, nickels, dimes, quarters, half-dollars, or dollar coins. If you limit yourself to the more common coins (pennies, nickels, dimes, or quarters), then there are 242 ways to make change for a dollar. I copied all of the combinations from Frank Morgan's post on this topic and transferred them to this online spreadsheet, so you can play with the data if you like.

Having all of the combinations handy on a computer makes possible some wacky puzzle ideas, like this one:

A customer went to the bank and gave the teller \$242 in dollar bills. The customer said, "Give me change for each one of these dollar bills, please—pennies, nickels, dimes, or quarters—and furthermore, I want no two of these dollar bills to be changed the same way." The teller obliged, and soon the customer had a large pile of coins in front of him. "On second thought," said the customer, concerned about the weight of the coins, "let's change as many of these pennies as we can for dollar bills." The teller did so. "And you know what?" the customer said. "Let's also make dollar bills out of as many of these nickels as we can." This was done. "OK," the customer said, "let's do the same for the dimes." When that was done, the customer said, "What the heck, let's change as many of these quarters as we can for dollar bills too." After this was done, the customer had \$241 in dollar bills, plus change for a dollar, and he walked happily out of the bank. How many pennies, nickels, dimes, and quarters was the customer finally left with?

I don't know if there is any realistic way to find the answer without using a computer. Anybody who's sufficiently spreadsheet-savvy should feel free to post it in the comments!

Wednesday, December 10, 2014

Lane Change (A Brachistochrone Problem)

Imagine driving your car along the freeway. Initially in the right lane, you decide to get into the left lane. Given that you could lose control of the car if its acceleration vector ever exceeds a given threshold, what is the shortest time in which you can effect the lane change?

(At the end of the maneuver, you must be moving forward at the speed you started with.)

The natural unit of time in this problem is \(\sqrt{\frac{2w}{a_0}}\), where \(w\) is the lane width and \(a_0\) is the acceleration threshold. Expressed in these units, I can make the lane change in \(\sqrt{2}\) ticks of the clock:

During the first half of the motion, the acceleration vector points due left; during the second half, the acceleration vector points due right. The forward velocity component thus remains constant over time.

Formula 1 drivers might maneuver in something like this fashion, but regular drivers probably make a tradeoff between the duration of the maneuver and the difficulty of executing it.

The following graph illustrates some of the difficulty. It shows the component of the acceleration vector along the instantaneous velocity vector (what I call \(a_\parallel\) in my textbook):

Perhaps you can see from the graph what an aggressive sequence this is!

Finally, here is the resulting speed curve:

Monday, December 8, 2014

Speaking of Andromeda...check out this cool image of it

This image shows what Andromeda would look like in the evening sky if it were brighter:

Surprising, no? Here is a larger version of the image. Astronomer Phil Plait has further details about the image at Slate.

Sunday, December 7, 2014

Left Turn at Andromeda, Cont'd

The title of my last post was a reference to the old Bugs Bunny cartoons in which Bugs says, "I knew I shoulda taken that left turn at Albuquerque." Here are some further thoughts on the two problems.

First, the problem of getting back to the origin with a velocity vector rotated 90 degrees. How long will it take, using a constant-thrust accelerator?

Measured in appropriate units of time, the 45-45-90 solution takes around 3.68 ticks of the clock:

Those slowdowns at the corners are painful to watch; one can do better. The best I'm able to do is \(\sqrt{10}\approx 3.16\) ticks of the clock:

Here is a side-by-side comparison:

As for the second problem, making the left turn without having to return to the origin first, the best I can do is \(\sqrt{4+2\sqrt{2}} \approx 2.61\) ticks of the clock:

If anyone can do either problem faster, let me know!

In the meantime, here's a supercut of Bugs Bunny: