OFF TOPIC:

A nice riddle for the bright minds here.
We have 15 stones, and 2 players. Each player can take from 1 to 3 stones at
each movement. The players take the stones in movements until there is only
1 stones left in the table. The player who has to take the last stone loses
the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win.
Hint 2: the free-will of the second player, if the first one uses the
strategy, is just apparent.
Hint 3: it is a variant of tic-tac-toe.

On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:

OFF TOPIC:

A nice riddle for the bright minds here. We have 15 stones, and 2 players.
Each player can take from 1 to 3 stones at each movement. The players take
the stones in movements until there is only 1 stones left in the table.
The player who has to take the last stone loses the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win. Hint 2:
the free-will of the second player, if the first one uses the strategy, is
just apparent.
Hint 3: it is a variant of tic-tac-toe.

Take 2 stones on the first move. From there on it's down hill.

Second player faces 13 stones, which he can reduce to 12, 11, or 10.

First player reduces that to 9 stones, taking 3, 2, or 1 as needed.

Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
still sunk.

First player takes enough to reduce the number to 5 stones.

Second player, facing 5 stones, can now reduce it to 4, 3, or 2.

First player takes 1, 2, or 3 stones, as needed, to reduce it to 1 stone,
and the second player loses.

(No, I had not seen it before.)


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On Thu, 27 May 2004 17:15:17 +0000, sal wrote:

On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:

OFF TOPIC:

A nice riddle for the bright minds here. We have 15 stones, and 2
players. Each player can take from 1 to 3 stones at each movement. The
players take the stones in movements until there is only 1 stones left
in the table. The player who has to take the last stone loses the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win. Hint 2:
the free-will of the second player, if the first one uses the strategy,
is just apparent.
Hint 3: it is a variant of tic-tac-toe.

Take 2 stones on the first move. From there on it's down hill.

More generally, reduce the number to 1 mod 4. That can always be done,
and in the last step it always gives the second player a board with 1
stone on it.



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"sal" escribió en el mensaje
s.com...
On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:

OFF TOPIC:

A nice riddle for the bright minds here. We have 15 stones, and 2
players.
Each player can take from 1 to 3 stones at each movement. The players
take
the stones in movements until there is only 1 stones left in the table.
The player who has to take the last stone loses the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win. Hint 2:
the free-will of the second player, if the first one uses the strategy,
is
just apparent.
Hint 3: it is a variant of tic-tac-toe.

Take 2 stones on the first move. From there on it's down hill.

Second player faces 13 stones, which he can reduce to 12, 11, or 10.

First player reduces that to 9 stones, taking 3, 2, or 1 as needed.

Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
still sunk.

First player takes enough to reduce the number to 5 stones.

Second player, facing 5 stones, can now reduce it to 4, 3, or 2.

First player takes 1, 2, or 3 stones, as needed, to reduce it to 1 stone,
and the second player loses.

(No, I had not seen it before.)

Congratulations!
What was the mental scheme you used?

I used: we start in 1, and if we add 4 each two consecutive movements, we
get to 13 in 6 (3-pairs) movements.
That gives 1 + 6 = 7 movements.
The formula used to solve the problem is 13%4=1.
(the only constant that you are sure to have in paired-movements is 4).





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"César Sirvent" escribió en el
mensaje . ..

"sal" escribió en el mensaje
s.com...
On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:

OFF TOPIC:

A nice riddle for the bright minds here. We have 15 stones, and 2
players.
Each player can take from 1 to 3 stones at each movement. The players
take
the stones in movements until there is only 1 stones left in the
table.
The player who has to take the last stone loses the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win. Hint
2:
the free-will of the second player, if the first one uses the
strategy,
is
just apparent.
Hint 3: it is a variant of tic-tac-toe.

Take 2 stones on the first move. From there on it's down hill.

Second player faces 13 stones, which he can reduce to 12, 11, or 10.

First player reduces that to 9 stones, taking 3, 2, or 1 as needed.

Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
still sunk.

First player takes enough to reduce the number to 5 stones.

Second player, facing 5 stones, can now reduce it to 4, 3, or 2.

First player takes 1, 2, or 3 stones, as needed, to reduce it to 1
stone,
and the second player loses.

(No, I had not seen it before.)

Congratulations!
What was the mental scheme you used?

I used: we start in 1, and if we add 4 each two consecutive movements, we
get to 13 in 6 (3-pairs) movements.

I meant, going backwards in time... :-)

That gives 1 + 6 = 7 movements.
The formula used to solve the problem is 13%4=1.
(the only constant that you are sure to have in paired-movements is 4).





--
To email me directly, take out nospam and put back foobox.

On Thu, 27 May 2004 17:50:28 +0000, César Sirvent wrote:


"sal" escribió en el mensaje
s.com...
On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:

OFF TOPIC:

A nice riddle for the bright minds here. We have 15 stones, and 2
players.
Each player can take from 1 to 3 stones at each movement. The players
take
the stones in movements until there is only 1 stones left in the
table. The player who has to take the last stone loses the game.

What is the strategy that must be used to win?

Hint 1: the first player, knowing the strategy, will always win. Hint
2: the free-will of the second player, if the first one uses the
strategy,
is
just apparent.
Hint 3: it is a variant of tic-tac-toe.

Take 2 stones on the first move. From there on it's down hill.

Second player faces 13 stones, which he can reduce to 12, 11, or 10.

First player reduces that to 9 stones, taking 3, 2, or 1 as needed.

Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
still sunk.

First player takes enough to reduce the number to 5 stones.

Second player, facing 5 stones, can now reduce it to 4, 3, or 2.

First player takes 1, 2, or 3 stones, as needed, to reduce it to 1
stone, and the second player loses.

(No, I had not seen it before.)

Congratulations!
What was the mental scheme you used?

Similar to what you describe below. I started looking at 4 stones (a
trivial forced win), moved on to 5 (a forced lose) and then climbed up to
15 one move at a time.

I was looking for something involving a "3" and so I overlooked the "1 mod
4" rule initially.

While I was waiting for my first message to come back from my news server,
I thought about 13, 9, and 5, and said, "Oh, yeah, that's 1 mod 4 --
right..." and so I sent over the second message.



I used: we start in 1, and if we add 4 each two consecutive movements,
we get to 13 in 6 (3-pairs) movements. That gives 1 + 6 = 7 movements.
The formula used to solve the problem is 13%4=1. (the only constant that
you are sure to have in paired-movements is 4).

Right.






--
To email me directly, take out nospam and put back foobox.



--
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"sal" wrote in message
s.com...
| On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:
|
| OFF TOPIC:
|
| A nice riddle for the bright minds here. We have 15 stones, and 2
players.
| Each player can take from 1 to 3 stones at each movement. The players
take
| the stones in movements until there is only 1 stones left in the table.
| The player who has to take the last stone loses the game.
|
| What is the strategy that must be used to win?
|
| Hint 1: the first player, knowing the strategy, will always win. Hint 2:
| the free-will of the second player, if the first one uses the strategy,
is
| just apparent.
| Hint 3: it is a variant of tic-tac-toe.
|
| Take 2 stones on the first move. From there on it's down hill.
|
| Second player faces 13 stones, which he can reduce to 12, 11, or 10.
|
| First player reduces that to 9 stones, taking 3, 2, or 1 as needed.
|
| Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
| still sunk.
|
| First player takes enough to reduce the number to 5 stones.
|
| Second player, facing 5 stones, can now reduce it to 4, 3, or 2.
|
| First player takes 1, 2, or 3 stones, as needed, to reduce it to 1 stone,
| and the second player loses.
|
| (No, I had not seen it before.)
|
|
| --
| To email me directly, take out nospam and put back foobox.
It's called "NIM". There is a version at
http://www.chlond.demon.co.uk/Nim.html
More info at:
http://www.imageviewer.co.uk/java/nim/
http://www.madras.fife.sch.uk/maths/games/nim.html
http://www.chronosfear.org.uk/nim/
http://www.cenius.net/refer/display....cleID=nim_ency
Lots more if you search.
Androcles.

"Androcles" escribió en el mensaje
...

"sal" wrote in message
s.com...
| On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:
|
| OFF TOPIC:
|
| A nice riddle for the bright minds here. We have 15 stones, and 2
players.
| Each player can take from 1 to 3 stones at each movement. The players
take
| the stones in movements until there is only 1 stones left in the
table.
| The player who has to take the last stone loses the game.
|
| What is the strategy that must be used to win?
|
| Hint 1: the first player, knowing the strategy, will always win. Hint
2:
| the free-will of the second player, if the first one uses the
strategy,
is
| just apparent.
| Hint 3: it is a variant of tic-tac-toe.
|
| Take 2 stones on the first move. From there on it's down hill.
|
| Second player faces 13 stones, which he can reduce to 12, 11, or 10.
|
| First player reduces that to 9 stones, taking 3, 2, or 1 as needed.
|
| Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
| still sunk.
|
| First player takes enough to reduce the number to 5 stones.
|
| Second player, facing 5 stones, can now reduce it to 4, 3, or 2.
|
| First player takes 1, 2, or 3 stones, as needed, to reduce it to 1
stone,
| and the second player loses.
|
| (No, I had not seen it before.)
|
|
| --
| To email me directly, take out nospam and put back foobox.
It's called "NIM". There is a version at
http://www.chlond.demon.co.uk/Nim.html
More info at:
http://www.imageviewer.co.uk/java/nim/
http://www.madras.fife.sch.uk/maths/games/nim.html
http://www.chronosfear.org.uk/nim/
http://www.cenius.net/refer/display....cleID=nim_ency
Lots more if you search.
Androcles.

Nice references, thanks Androcles.
If I understand it correctly, the example I provided is just a [very]
particular case of groups of 4 coins, though I am not sure if all the rules
apply as exactly described there. In any case, a little variation of this
more generally (and computer-like interesting) case.

Cesar

"César Sirvent" wrote in message
. ..
|
| "Androcles" escribió en el mensaje
| ...
|
| "sal" wrote in message
| s.com...
| | On Thu, 27 May 2004 16:56:45 +0000, César Sirvent wrote:
| |
| | OFF TOPIC:
| |
| | A nice riddle for the bright minds here. We have 15 stones, and 2
| players.
| | Each player can take from 1 to 3 stones at each movement. The
players
| take
| | the stones in movements until there is only 1 stones left in the
| table.
| | The player who has to take the last stone loses the game.
| |
| | What is the strategy that must be used to win?
| |
| | Hint 1: the first player, knowing the strategy, will always win.
Hint
| 2:
| | the free-will of the second player, if the first one uses the
| strategy,
| is
| | just apparent.
| | Hint 3: it is a variant of tic-tac-toe.
| |
| | Take 2 stones on the first move. From there on it's down hill.
| |
| | Second player faces 13 stones, which he can reduce to 12, 11, or 10.
| |
| | First player reduces that to 9 stones, taking 3, 2, or 1 as needed.
| |
| | Second player, facing 9 stones, can reduce it to 8, 7, or 6, but he's
| | still sunk.
| |
| | First player takes enough to reduce the number to 5 stones.
| |
| | Second player, facing 5 stones, can now reduce it to 4, 3, or 2.
| |
| | First player takes 1, 2, or 3 stones, as needed, to reduce it to 1
| stone,
| | and the second player loses.
| |
| | (No, I had not seen it before.)
| |
| |
| | --
| | To email me directly, take out nospam and put back foobox.
| It's called "NIM". There is a version at
| http://www.chlond.demon.co.uk/Nim.html
| More info at:
| http://www.imageviewer.co.uk/java/nim/
| http://www.madras.fife.sch.uk/maths/games/nim.html
| http://www.chronosfear.org.uk/nim/
| http://www.cenius.net/refer/display....cleID=nim_ency
| Lots more if you search.
| Androcles.
|
| Nice references, thanks Androcles.
| If I understand it correctly, the example I provided is just a [very]
| particular case of groups of 4 coins, though I am not sure if all the
rules
| apply as exactly described there. In any case, a little variation of this
| more generally (and computer-like interesting) case.
|
| Cesar
You are welcome. These kind of problems are quite popular among
those with logical minds. I developed the solution some years ago by
writing a program to play the game against a computer, back in my
student days. Essentially the same as the solution give at
http://www.cenius.net/refer/display....cleID=nim_ency,
Heap 1 = 2 = Binary 0010
Heap 2 = 3 = Binary 0011
Heap 3 = 14 = Binary 1100
Heap 4 = 15 = Binary 1111
Summing the columns:
Total = 2232
3 is odd, so take 1 from either Heap 2 or Heap 4.


If you enjoy this sort of puzzle, you might like
http://www.thegamesforum.com/Downloads.html
Klotski is very good.
Androcles

"Androcles" wrote in message ...

[snip]

| Cesar
You are welcome. These kind of problems are quite popular among
those with logical minds.

Those who find logic gibberish?
http://users.pandora.be/vdmoortel/di...Gibberish.html

Dirk Vdm