Cover the Grid

You must cover a 7x7 grid with triminos and tetrominos, without overlapping. What is the minimum number of pieces you can use in order to do that?

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Solution

Consider all cells in the grid which lie in an odd row and odd column - there are 16 of them. Since each of the two pieces can cover at most 1 of these cells, we neet at least 16 pieces. It is easy to give an example with 15 triminos and 1 tetromino.


Pinned Men

The following game is played under very specific rules - no pinned piece checks the opposite king. How can White mate Black in 2 moves?


Solution

First, White plays f3 and threatens mate with Qxe2. Indeed, blocking with the black rook on d4 will not help, because it will become pinned, which means that the rook on d6 will become unpinned, which will make the bishop on b6 pinned, and that will unpin the knight on c7, resulting in mate. Below are listed all variations of the game.

  1. ... Rd5 2. Qxe2#
  2. ... Bxa5 2. Kc8#
  3. ... Bxc7 2. Nxc7#
  4. ... Bxe8 2. Kxe8#
  5. ... Qxe7+ 2. Kxe7#
  6. ... Rd2 2. Bxd2#
  7. ... Rxd6+ 2. Qxd6#

Imprisoned Logicians

Two friends, logicians - Ein and Stein, get imprisoned in two distant cells in a castle. Both cells have just one door, and a window with 8 bars in the first cell, and 12 bars in the second cell. The first day both logicians get the same letter from the prison master:

"The total number of bars in the two prison cells in this castle is either 18 or 20. Starting tomorrow, every morning I will go first to Ein and then to Stein, and will ask how many bars does the other logician have. If one of you answers correctly, I will immediately let both of you leave the castle. If one of you answers incorrectly, I will execute both of you. Of course, you can always decide not to answer and just stay imprisoned.
I have sent a copy of this letter both to you and your friend. There is no point in trying to communicate with him - your cells are far away from each other and he won't hear you."

Will the logicians manage to escape the castle eventually? When will they do it?


Solution

On the second day, Stein will realize that the number of bars in Ein's cell is at most 18 (assuming the windows can have 0 bars on them), because otherwise he would have guessed correctly 20. On the third day Ein will realize that the number of bars in Stein's cell is between 2 and 18, because otherwise he would have guessed the exact number. Later Stein will realize that the number of bars in Ein's cell is between 2 and 16, because otherwise Ein would have guessed the exact number. On the fourth day Ein will realize that the number of bars in Stein's cell is between 4 and 16, then Stein will realize after that the number of bars in Ein's cell is between 4 and 14. On the fifth day Ein will realize that the number of bars in Stein's cell is between 6 and 14, and Stein will realize that the number of bars in Ein's cell is between 6 and 12. On the sixth day Ein will realize that the number of bars in Stein's cell is between 8 and 12, and Stein will realize that the number of bars in Ein's cell is between 8 and 10. Then Stein will conclude that it must be 8, and that the total number of bars in the castle is 20.


The Connect Game

Two friends are playing the following game:

They start with 10 dots and taking turns, connect any two of them which are not already connected. The player who first makes all dots connected to each other, loses.

Who will win the game?


Solution

The second player will win. First, he makes sure he gets 9 of the points connected to each other, leaving the tenth point isolated. After that, both players must keep connecting points among the connected 9, until exhaust all 36 edges between. Since 36 is an even number, the first player will be the one who will connect the tenth point to the first 9 and lose the game.


Seven Dwarfs

In one house deep in the forest seven dwarfs are living alone. The first dwarf is reading a book, the second dwarf is cooking, the third dwarf is playing chess, the fourth dwarf is tidying up the house, the fifth dwarf is washing the clothes, and the sixth dwarf is gardening. What is the seventh dwarf doing?


Solution

The seventh dwarf is also playing chess - 2 people are needed for this.


Wizards with Hats

There are 2 wizards and each of them has infinitely many hats on his head. Every hat has 50-50 chance to be white or black, and the wizards can see the hats of the other person, but not their own. Each wizard is asked to identify a black hat on his head without looking, and they win if both succeed to guess correctly. If the wizards are allowed to devise a strategy in advance, can they increase their chance of winning to more than 25%?


SOLUTION

Each wizard guesses the position of the lowest black hat on the head of the other wizard. Then the chance of winning becomes 1/4 + 1/16 + 1/64 + ... = 1/3. It can be shown that this is an optimal strategy as well.


Rubik's Chess

Last week we found out that Puzzle Pranks Co. have invented new type of puzzle - Rubik's Chess. The goal is simple - you get a scrambled cube with various chess pieces on its sides, and you must unscramble it so that on each side there is one mated King, assuming the kings can not capture the neighboring pieces (Queens, Rooks, Bishops, kNights).

Rubik's Chess Puzzle

We are usually good with this type of puzzles, but we spent our entire weekend trying to solve this one without any success. We even started wondering if it can be actually solved, so decided to share it with you and see if you can help us figure that out.

Below you can see the way the cube looks when seen through 8 different angles:

Remark: The orientations of the pieces are irrelevant of the final solution, i.e. they don't need to be consistent on each side.


SOLUTION

The Rubik's Chess puzzle can not be solved. Solution coming soon.


Superman

Manifold

In an age of video games and noisy high-tech toys, Brainwright's Manifold is a very rare find. I got truly delighted by the idea of folding little papers into origami, trying to achieve some easy to understand task. The goal is simple - you start with an 8x8 paper, which has 16 black, 16 white, and 32 empty squares printed on its front. You must make several folds, so that it ends up as a 4x4 paper with all black squares on one side and all white squares on the other.

For under $10 I was able to get a set of 100 puzzles, many of them quite challenging, which translates into several hours of gameplay. The difficulty of the puzzles gradually increases, which makes Manifold appropriate for all ages. Despite my highly positive impressions, I have to point out several (minor) flaws. First - the paper, despite being high quality and glossy, may not be the best choice for making origami with it. Second - there is too much empty space left at the edges of the papers, which may get in the way when one needs to make many folds in one puzzle. Finally - once solved, the papers naturally show the folds applied on them, so they may not be ideal for solving more than once. Despite these few gripes, Manifold receives my highest recommendations.

If you want to try some sample puzzles from the set before buying it, you can print them from this PAGE.

  • fun, simple gameplay
  • various difficulty levels
  • highly recommended