Paper and Puzzle a Day 6

It’s been a while since I posted a paper and puzzle since I last made a blog post on a paper and puzzle. Other things have been occupying my time, but hopefully, I’ll be going back to my usual schedule with the Paper and Puzzle a Day posts.

So, first, a paper. I’ve been looking into some computational geometry problems and one particularly interesting paper that I’ve been reading is that of what are called “hinged dissections.” The paper’s title is appropriately titled, “Hinged Dissections Exist.” A hinged dissection is one where all pieces of a dissected shape are connected with a series of “hinges.” For example, the following figure shows a hinged dissection of a triangle into a square discovered by Dudeney in 1902.

Dudeney’s hinged dissection of triangle into square.

The paper shows that any finite collection of polygons of equal area has a hinged dissection such that it can folded into any polygon that is in the collection. It’s quite an intricate paper, and I’m still reading some of the details but it’s quite interesting so far. I may post more in future Paper and Puzzle a Day posts about this paper.

Now, for the puzzle. Since we are talking about computational geometry, I will use a computational geometry puzzle for today.

A tangram consists of the following pieces that form a square.

Even though all of the following figures are made from the same seven pieces, why does the equivalent black and white pieces seem to differ by a piece?

TangraMagic figures

Puzzle courtesy of http://www.archimedes-lab.org/workshoptangram4.html.

Now, the solution to Day 5’s puzzle:

The trick to this puzzle is not to be bogged down with probability calculations. There is a way to ensure that at least 5 prisoners are released regardless of what the distribution of hats is (in which case, the fact that the hats are distributed randomly is irrelevant). Each prisoner decides beforehand what her partner is. Then, one partner decides to say the color hat that is opposite the color of her partner’s hat and the other says the same color as that on the partner’s head. When the hats are distributed, the partners find each other and carry out the strategy. If the hats on both partners’ heads are the same color, then the partner who says the same color will answer correctly. If the hats are different colors, then the partner who says the different color answers correctly. Therefore, one partner in each pair will always answer correctly.

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