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Alex's solitaire game - care to try?


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8 replies to this topic

#1 bonanova

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Posted 17 October 2007 - 09:58 AM

Alex hadn't been seen for a week. And things just weren't
the same at Morty's. Jamie said it was cuz he'd lost his
last three bets to the geniuses over at Brainden. But Davey
opined Alex was not a sore loser. He'll be back, he said.
And it was no sooner said than Alex appeared at the door.
All smiles. And with a strange looking board under his arm.

Take a look, mates, he said, and laid it on the table:

Posted Image

What you see here is Alex's new version of the old solitaire
game that you play with pegs. There's one empty hole right
there in the center, and 36 other holes each with a peg.

You jump pegs over neighbors into empty holes, straight ahead,
along the lines ya see there, and remove the pegs that are
jumped. When ya can't make any more jumps, the game is over.
And if there's only one peg left, you've won.

Now who'll take my bet that none of ya is smart enough
to win? The last peg doesn't have to be in the center
hole, mind ya, but there can be only one peg standing
at the end.

Go ahead. Try a few games to get the idea. Then think it
over. If ya take the bet, you've got until Friday to write
down a winning game.

I'd advise against it, he winked. Cuz if you try and lose, then
I get free drinks till Christmas. Ya might be better off just
passing it on to the Brainden crowd. ;)

Any takers here?

Number the holes like this.

--------1 2 3 4
-------5 6 7 8 9
---10 11 12 13 14 15
16 17 18 [19] 20 21 22
---23 24 25 26 27 28
-----29 30 31 32 33
------34 35 36 37

and write moves like this:

[1] - 6 [12] 19 - [that means: move #1 - peg 6 jumps and removes peg 12 and lands in hole 19]
[2] - ....... etc. 35 total jumps and you win.

Sign up now if ya want to try, then post yer solution by Friday.

Edit:
Take a hint from Alex's wink:
Prove that the game can be won. Or not.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#2 KamZhiYhi

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Posted 17 October 2007 - 06:45 PM

i have tried this lots of times and would have to say that it cant be done or there is one specific way of going about it. i hate it when i sneezed cos i blew everything out of place ( i drew the board on a piece of paper and the pegs were paper too...)
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#3 bonanova

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Posted 17 October 2007 - 07:02 PM

i have tried this lots of times and would have to say that it cant be done or there is one specific way of going about it. i hate it when i sneezed cos i blew everything out of place ( i drew the board on a piece of paper and the pegs were paper too...)

Gezundheit!

I used a bigger board with pennies ...

There is a proof [that it can or can't be done] and it's cute.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#4 KamZhiYhi

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Posted 17 October 2007 - 07:13 PM

i always get left with corners and they are always on the opposite side of the board.
if the entire outline was a hexagon then it would be possible, i think i gotta try that next.

edit: didnt want to double post. i just re-read the speech thing and he says he would advise against it. so he is trying to put them off of it and he has spend way too much buying drinks for everyone else therefore i think that there is a soloution to the game.
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#5 bonanova

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Posted 17 October 2007 - 07:28 PM

The holes on the board can be colored red, green and blue
in such a way that each move involves a hole of each color.

Example: blue peg jumps a green peg and lands in a red hole.
Continue making all possible moves, and every hole will eventually be colored.

In terms of these colors, think about the starting position and
a winning end position. Can a series of moves go from the
starting position to a winning end position?

In terms of colors, what happens on each move?
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#6 Writersblock

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Posted 18 October 2007 - 04:38 AM

You will be able to lick your elbow before you solve this game. Here's why:

I don't know math beyond simple algebra, but let's look at this.

We have 4 numbered rings of 1, 6, 12, and 18.

Ring 1 is empty. Rings 2, 3, and 4 are full.

In any given move, you have to have 2 pegs that become 1 peg because of the elimination. Also, no peg can move without a corresponding peg to jump.

For any given move, you must either go A) from one ring to another or B) one spot to another within the same ring laterally, except in ring 4 where there are no lateral movements allowed. If there are 18 pegs in the outside ring, we HAVE to have 18 pegs for them to jump in rings 1, 2 & 3, or else they don't move. The moment we have one less peg in the middle then we have in ring 4, we have a problem because not all of the outside pegs can move and we are stuck. And that’s if you never move a peg from a lower ring to ring 4. No problem, because we have 12 and 6 in rings 2 & 3, right? That makes 18. We are fine you say.

But wait. Watch what happens when you start moving:

Start
Ring 1 (01) - 00
Ring 2 (06) - 06
Ring 3 (12) - 12
Ring 4 (18) - 18

Now, no ring can fill beyond its capacity (in parens). So let's jump from one ring to the other.

So we jump from ring 3 to ring 1 and eliminate the ring 2 peg.
Ring 1 (01) - 01
Ring 2 (06) - 05
Ring 3 (12) - 11
Ring 4 (18) - 18

We have a problem. We have 18 pegs in the outside row and only 17 in the center.

But, you say, you can even them up with another move:

Ring 1 (01) - 01, 01
Ring 2 (06) - 05, 06
Ring 3 (12) - 11, 10
Ring 4 (18) - 18, 17

Ok. But because of the configuration, there will always come a time when a lateral move within a ring is necessary. Also, there will always come a time where you have to move within rings 2 and 3 in order to get to other pegs in ring 4. This will put you further and further from a 1 to 1 ratio between ring 4 and the interior rings. The more laterals or inward moves you must make the more “stray” pegs you will have in ring 4 when you run out of interior pegs. You can sometimes make up ground by moving inward from ring 4, which at best puts you at an even state, but then you are back to square 1 and the next move puts you at a deficit again.

We cannot ever move all of the outside pegs regardless of formation. The only successful move is to put away the game. You are dead from the start.
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#7 KamZhiYhi

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Posted 18 October 2007 - 07:44 AM

well actually many people can lick their elbow.
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#8 Writersblock

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Posted 18 October 2007 - 07:45 AM

try
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#9 bonanova

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Posted 19 October 2007 - 06:59 AM

Well it's finally Friday, and no one has won the game, Alex confessed, and it's no wonder.
Writersblock even gave ya a reason why ... it just can't be done.
And here's another way to show it's impossible.

If ya color one third of the holes blue and one third green and one third red this way...

Posted Image

you'll see the first move jumps a green peg over a blue peg into
the center red hole. Or, blue jumps green into red. Either way,
starting with 12 of each color, ya now have 11 blue, 11 green
and 13 red. All odd numbers. And any move after that makes
an even number of each color. The colors stay all odd or all
even until there are no moves left.

But ya see, the winning layout has to have only one peg, say it's red.
That means there's an odd number [1] of red and an even number [0]
of blue and green. And that's just impossible.

I suppose it would have been sporting of me to give ya the colored
version of the board to start with, but ... heh heh ... i was running
out of drink money and ... well, you know...
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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