Any demoscener from Mensa?
category: residue [glöplog]
Ok, so, 2 chess players started a game, and the left the board after 4 moves (4 moves by white, 4 moves by black, you know white-black-white-black-white-black-white-black). The position was the one you see in the image. Knowing that they did only legal chess moves, what were the moves to get into that position?
Note: this is not a common chess problem. This is much more a logical puzzle using chess rules. It use to be a hard problem for chessmasters, but an medium-easy one for AI students. If you can, do it only thinking, paper and pen is for losers :D
I will not post more problems until tomorrow...
*There are 4 solutions to this problem
Maybe you could also post the typical set-up of the beginning of a chess game for the less chess-educated ones among us...
Adok, if a person doesn't know how to set-up the chess board, then maybe that person also doesn't know the rules of chess and then maybe is not going to even try to read this :D
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texel: that puzzle is so simple that even my sister who can't play chess gave me two solutions in few seconds :)
rmeht, tell her she left 2 solutions :D
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It's a trick question. Those pieces are yellow and blue, not black and white.
yeah but white always starts with a queen on a white square.
So the black and white pieces got lost somewhere and the horses went out to look for them?
The REALLY smart people know that Mensa is pointless.
No solutions to the problem yet?
How can this be a hard problem for chess masters?
The yellow horse head moves up to eat the leftmost blue horse head in four moves. Meanwhile the blue buttplug moves forward and the rightmost blue horse head moves in to eat the yellow horse head.
I guess the trick is to deduce that the blue horse heads have to switch places. You can move the buttplug and blue horse head in different order, and there are two possible routes for the yellow horse head.
Does this mean I qualify for teh mensa?
The yellow horse head moves up to eat the leftmost blue horse head in four moves. Meanwhile the blue buttplug moves forward and the rightmost blue horse head moves in to eat the yellow horse head.
I guess the trick is to deduce that the blue horse heads have to switch places. You can move the buttplug and blue horse head in different order, and there are two possible routes for the yellow horse head.
Does this mean I qualify for teh mensa?
Quote:
Does this mean I qualify for teh mensa?
It means you fail as a chess master.
At least one solution to the problem was really easy to figure out, didn't bother with the other three.
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Congratulations Doom, you win :)
Well, chessmasters use to think in useful moves, not in irrational moves (even being legal ones). I've tested it with about 10 people with 2000 to 2050 Europen ELO points, to 2 people with 2200 to 2300 and a grandmaster with 2450. I got no answer from them (I gave them about 30 minutes to think). As I wrote, for people used to AI, logical problems, tree searching with rules and that, it is much easier. And yes, the trick is to know right white/yellow horse eats left black/blue horse.
I have other problem for tomorrow... another good one. About difficulty of problems, I'm not going to post any TOO difficult one, only beatiful ones.
Well, chessmasters use to think in useful moves, not in irrational moves (even being legal ones). I've tested it with about 10 people with 2000 to 2050 Europen ELO points, to 2 people with 2200 to 2300 and a grandmaster with 2450. I got no answer from them (I gave them about 30 minutes to think). As I wrote, for people used to AI, logical problems, tree searching with rules and that, it is much easier. And yes, the trick is to know right white/yellow horse eats left black/blue horse.
I have other problem for tomorrow... another good one. About difficulty of problems, I'm not going to post any TOO difficult one, only beatiful ones.
I couldn't figure it out... does that make me a Chessmaster??
First, put 16 cards in a table upside-down like in (0). Then apply this algorithm:
for (n=1; n<=number_of_cards; n++)
{
for (i=n; i<=number_of_cards; i+=n)
{
flip_card_in_the_position(i);
}
}
In (1) and (2) you can see steps in the algorithm. Then you can see the result of complete execution.
As you can see, the only upside cards are 1, 4, 9 and 16. These are the first natural squares (1*1, 2*2, 3*3, 4*4).
If you get more cards, will the algorithm shows all and only natural squares? Demostrate it (I don't need a math prof, only a logical explanation in common language).
Note: This is a very famous problem, so maybe you have already solved it in discrete math class.
a card at position i is flipped "number of divisors of i" times.
squares have odd number of divisors. am i a chessmaster now?
squares have odd number of divisors. am i a chessmaster now?
anes, can you demostrate it?
"squares have odd number of divisors" < this is right, but why?
and, only squares have odd number of divisors? why?