Brain Teasers

[Oorang]

8 is impossible and I'll prove it.
To make the conversation easier, let's number the tiles:
<table border="1"><tbody><tr><td bgcolor="cyan">1</td><td bgcolor="#33ff33">2</td><td bgcolor="cyan">3</td></tr><tr><td bgcolor="#33ff33">4</td><td bgcolor="red">5</td><td bgcolor="#33ff33">6</td></tr><tr><td bgcolor="#3333ff">7</td><td bgcolor="#33ff33">8</td><td bgcolor="#3333ff">9</td></tr></tbody></table>

• To solve this in 8 moves each piece would have to only move twice. All possible combinations of 0 or 1 moves from the starting position would leave any piece in an illegal finishing position.
  As there are 4 pieces any more than 2 more than two moves would cause you to run out moves before all pieces are in the proper finishing position, thus each piece must move twice.
• From each start position there are only two legal moves: 1 vertical, 2 horizontal; or 2 vertical and 1 horizontal (the green tiles).
• Of those 2 legal moves, for all starting positions, only 1 move puts you in a position to move to a "Finishing Tile" in the next move. That move is 1 vertical, 2 horizontal.
• So for a successful 8 move solution the first moves must be either: 1 to 6; 3 to 4; 7 to 6; or 9 to 4. There are no other options.
• From any green square each piece will now have only 1 move left, and only one square that will take them to a legal finishing position. This means each piece only has one possible finishing position for an 8 square victory.
• Now notice that 3 and 9 both both share the same one and only legal move (as do 1 and 7). Therefore if 3 takes it's one and only legal move. Then 9 has no possible legal first move and vice versa (the same for 1 & 7).
• In an 8 move scenario, every piece has only 1 legal move, and that legal move makes illegal another piece's only legal move. QED an 8 move solution is impossible.

Of course what this doesn't prove is that 16 is optimum I'll give you hint though... 5 is red for a reason

 

[Sandeep Warrier]

8 is impossible and I'll prove it.

I'm with Oorang on that one, it looks like 16 to me.

Until now I also can only do it in 16 moves.

Which is why I want Gettingbetter to post his solution

 

[Sandeep Warrier]

I'll give you hint though... 5 is red for a reason

Would that be that in a 3X3 grid there is no legal way a knight can land up on 5

And while we ponder on the most optimum solution....

In a 10X10 grid arrange 10 Queens so that all Queens are safe.

 

[crimson_b1ade]

New Brain Teaser:

Three New York supermodels are shown a selection of five scarfs at an Fifth Avenue department store. Three of the scarfs are deep blue and two, yellow. The three supermodels are placed in single file, facing forward, and then blindfolded. One scarf is draped on each, with two returned to the shelf. The blindfold is first removed from the supermodel in the back. She is asked if she can guess the color of her scarf by looking only at the two models in front of her. “No,” she says. The blindfold is removed from the supermodel in the middle, and she is asked the same question. (She can only look at the supermodel in front, not in back.) “I can’t,” she says. Immediately the supermodel in front, still blindfolded, blurts out, “I’m wearing a [blank] scarf. Can I keep it?”
What color scarf is she wearing? (Explain your answer)

 

[Cindy Ellis]

Blue...
The first model would know what scarf she was wearing if both models in front of her were wearing yellow. This leaves yellow-blue, blue-yellow, or blue-blue as options.
The second model could deduce that her scarf color was blue if the front model were wearing yellow, since yellow-blue is the only option that includes yellow for the first model.
So...the front model knows she can't be wearing yellow.

 

[crimson_b1ade]

We've got a winner!! YAY

Blue...
The first model would know what scarf she was wearing if both models in front of her were wearing yellow. This leaves yellow-blue, blue-yellow, or blue-blue as options.
The second model could deduce that her scarf color was blue if the front model were wearing yellow, since yellow-blue is the only option that includes yellow for the first model.
So...the front model knows she can't be wearing yellow.

 

[mikerickson]

..The second model could deduce that her scarf color was blue if the front model were wearing yellow, since yellow-blue is the only option that includes yellow for the first model...

How smart are the super-models?

I first heard this with condemned prisoners (lots of motivation), but scarves for models...I guess they'd be motivated.

OK, here's mine.

There are ten piles of bombs (15 bombs to a pile).
Nine of the piles contain conventional bombs, one has atomic bombs.
Some knuckle-head removed the signs so you don't know which pile has the atomic bombs.
However, you do know that a conventional bomb weighs 300 lb. and an atomic bomb weighs 400 lb.
You have a scale. (30,000 lb. load limit)
(The bombs are identical in size and appearance and the atom bombs don't leak radiation even if you had a Geiger counter.)

Using the scale only once, how do you determine which pile has the atomic bombs?

 

[crimson_b1ade]

you take a bomb from each of the piles until the total load on scale increases by 400lb.

....
There are ten piles of bombs (15 bombs to a pile).
Nine of the piles contain conventional bombs, one has atomic bombs.
Some knuckle-head removed the signs so you don't know which pile has the atomic bombs.
However, you do know that a conventional bomb weighs 300 lb. and an atomic bomb weighs 400 lb.
You have a scale. (30,000 lb. load limit)
(The bombs are identical in size and appearance and the atom bombs don't leak radiation even if you had a Geiger counter.)

Using the scale only once, how do you determine which pile has the atomic bombs?

 

[mikerickson]

One weighing only.
Load the scale, press a button, read the result.

No intermediate values.

 

[crimson_b1ade]

take one bomb from pile 1, two from pile 2, three bombs from pile 3 and so on. Take the difference in the total weight less 16,500 (300*1+300*2+300*3+300*4...so on until...300*10) divided by 100lbs (the difference between 300lbs and 400lbs) and the result is your pile with the atomic bombs.

...
OK, here's mine.

There are ten piles of bombs (15 bombs to a pile).
Nine of the piles contain conventional bombs, one has atomic bombs.
Some knuckle-head removed the signs so you don't know which pile has the atomic bombs.
However, you do know that a conventional bomb weighs 300 lb. and an atomic bomb weighs 400 lb.
You have a scale. (30,000 lb. load limit)
(The bombs are identical in size and appearance and the atom bombs don't leak radiation even if you had a Geiger counter.)

Using the scale only once, how do you determine which pile has the atomic bombs?