[wolfgang]

12 Holes arrenged in a circle ,with numbers,1-12(like a Clock numbers).

The hole No. 12 is yours, and is empty.

There are Jewels in each hole equal to the number of the hole, i.e. in hole No. 1 is only one jewel, in No. 2 are 2jewels....and so on,till No. 11 which contains 11 jewels.

your aim is to collect as much jewels as you can in your hole.

1-choose any hole you wish,take the jewels from that hole then move clockwise among the holes puting one jewel in each hole you pass through, your hole must be also included.

2-when you reach with your last jewel to a hole having jewels,take them all( with your last jewel) and continue as in step 1.

3- if the last jewel in your hand ended in an empty hole...you lose the game!.

4-if you have enough jewels you should put a jewel in an empty hole when passing through it.

5-if you reach with your last jewel to your hole ( No. 12 ), repeat step 1.

Now enjoy collecting the jewels in your hole,and avoid falling in an empty hole.

And we`ll see who has the best strategy...

Have a nice time

Edited October 30, 2010 by wolfgang

[bonanova]

Nice puzzle.

similar to a closed awari [qv, via google].

It's the dual of Bulgarian Solitaire, described in

In Bulgarian Solitaire you create a pile by taking a card from existing piles.

In awari, you remove a pile and sow [as in planting seeds] one each into existing piles.

It will be interesting to see the winning strategy in your game.

[wolfgang]

Nice puzzle.

similar to a closed awari [qv, via google].

It's the dual of Bulgarian Solitaire, described in

In Bulgarian Solitaire you create a pile by taking a card from existing piles.

In awari, you remove a pile and sow [as in planting seeds] one each into existing piles.

It will be interesting to see the winning strategy in your game.

Thank you dear...anyhow...I tried with my own strategy....and I collected 53...

Edited October 30, 2010 by wolfgang

[Guest]

The way I see it, here's the result of what would happen with each hole you start from (note, this starts from hole 0 which is hole 12, the second one is hole 1):

Start at 1: 0,0,0,6,0,4,11,7,0,17,0,21

Start at +2: 0,1,0,5,0,9,6,7,0,17,10,11

Start at +3: 0,1,2,0,7,5,0,13,8,9,10,11

Start at +4: 0,1,2,3,0,9,6,7,0,17,10,11

Start at +5: 0,1,2,3,0,4,11,7,0,17,0,21

Start at +6: 0,1,2,3,4,5,0,13,8,9,10,11

Start at +7: 11,1,0,5,0,9,6,0,0,24,0,10

Start at +8: 0,1,2,3,0,9,6,7,0,17,10,11

Start at +9: 0,1,2,3,4,5,0,13,8,0,19,11

Start at +10: 0,1,2,3,0,9,6,7,0,17,0,21

Start at +11: 11,1,2,3,0,9,6,7,0,17,0,10

Unless I understood the riddle wrong...

[araver]

The way I see it, here's the result of what would happen with each hole you start from (note, this starts from hole 0 which is hole 12, the second one is hole 1):

Start at 1: 0,0,0,6,0,4,11,7,0,17,0,21

Start at +2: 0,1,0,5,0,9,6,7,0,17,10,11

Start at +3: 0,1,2,0,7,5,0,13,8,9,10,11

Start at +4: 0,1,2,3,0,9,6,7,0,17,10,11

Start at +5: 0,1,2,3,0,4,11,7,0,17,0,21

Start at +6: 0,1,2,3,4,5,0,13,8,9,10,11

Start at +7: 11,1,0,5,0,9,6,0,0,24,0,10

Start at +8: 0,1,2,3,0,9,6,7,0,17,10,11

Start at +9: 0,1,2,3,4,5,0,13,8,0,19,11

Start at +10: 0,1,2,3,0,9,6,7,0,17,0,21

Start at +11: 11,1,2,3,0,9,6,7,0,17,0,10

Unless I understood the riddle wrong...

Rule 5 -"If you reach with your last jewel to your hole ( No. 12 ), repeat step 1."

[araver]

65 so far.

The strategy involves many steps, see attached file.

66 is still theoretically possible but I haven't found one so far.

Edited October 30, 2010 by araver

[araver]

66 by starting with hole 7. 66Run.txt

I can see no analytical way to describe the solution though.

Edited October 30, 2010 by araver

[wolfgang]

Rule 5 -"If you reach with your last jewel to your hole ( No. 12 ), repeat step 1."

Thats right...thanks Araver !

[wolfgang]

65 so far.

The strategy involves many steps, see attached file.

66 is still theoretically possible but I haven't found one so far.

I'm afraid you didn`t understand the puzzle though....let me show you what did I mean..

1...2...3...4...5...6...7...8...9...10...11...0

if we take the jewel from hole No.1(only one)..we are going to put it in hole No.2(were only two jewels in),then we are going to take these 3 jewels( 2+1)from this hole leaving it empty,and put one jewel in hole 3,4 and5(here was our last jewel),making 6 jewels in it,and again we take these 6 jewels from hole 5( leaving it empty)and so on.by reaching the hole No. 12,we have Two possibilities :

1-the last jewel ended here....and this case we can choose any new hole.

2-we have still more jewels in hand...so we put one jewel in it and continue to hole No. 1

Edited October 31, 2010 by wolfgang

[araver]

I'm afraid you didn`t understand the puzzle though....let me show you what did I mean..

1...2...3...4...5...6...7...8...9...10...11...0

if we take the jewel from hole No.1(only one)..we are going to put it in hole No.2(were only two jewels in),then we are going to take these 3 jewels( 2+1)from this hole leaving it empty,and put one jewel in hole 3,4 and5(here was our last jewel),making 6 jewels in it,and again we take these 6 jewels from hole 5( leaving it empty)and so on.by reaching the hole No. 12,we have Two possibilities :

1-the last jewel ended here....and this case we can choose any new hole.

2-we have still more jewels in hand...so we put one jewel in it and continue to hole No. 1

Yes, I am using these rules, I forgot to mention that I also positioned the holes as Anza Power did

0 (hole 12 which I call HOME for simplicity)...1...2...3...4...5...6...7...8...9...10...11

Since there in a circle, it makes no difference on the rules and it makes easier computations modulo 12.

So the first lines in my solution:

Starting with hole 7

Taking all jewels from hole 7 

Jewels taken 7

0 1 2 3 4 5 6 0 8 9 10 11 

Taking all jewels from hole 2 

Jewels taken 3

1 2 0 3 4 5 6 0 9 10 11 12 

should be read as (missing steps are now inserted):

Initial: 0 1 2 3 4 5 6 7 8 9 10 11

Starting with hole 7

Taking all jewels from hole 7 

Jewels taken 7

0 1 2 3 4 5 6 0 8 9 10 11

Putting one jewel in hole 8, still holding 6 in hand

0 1 2 3 4 5 6 0 9 9 10 11 

Putting one jewel in hole 9, still holding 5 in hand

0 1 2 3 4 5 6 0 9 10 10 11 

Putting one jewel in hole 10, still holding 4 in hand

0 1 2 3 4 5 6 0 9 10 11 11 

Putting one jewel in hole 11, still holding 3 in hand

0 1 2 3 4 5 6 0 9 10 11 12

Putting one jewel in hole 12(HOME), still holding 2 in hand

1 1 2 3 4 5 6 0 9 10 11 12  

Putting one jewel in hole 1, still holding 1 in hand

1 2 2 3 4 5 6 0 9 10 11 12 

Last jewel above hole 2 which is not empty so:

Taking all jewels from hole 2 

Jewels taken 3 (2 from hole and 1 in hand)

1 2 0 3 4 5 6 0 9 10 11 12 

I can provide a more verbatim solution (step-by-step) but it is much longer than the 439 lines I've already shown in 66Run.txt

Edited October 31, 2010 by araver

[superprismatic]

66 by starting with hole 7. 66Run.txt

I can see no analytical way to describe the solution though.

Please give a heuristic explanation of how you did this.

I'm pretty sure you wrote a program to find your solution(s),

so you must have thought up a strategy for it. I would

appreciate a quick outline of the strategy.

[araver]

Please give a heuristic explanation of how you did this.

I'm pretty sure you wrote a program to find your solution(s),

so you must have thought up a strategy for it. I would

appreciate a quick outline of the strategy.

there's nothing special with my strategy.

As I said, I have found no analytical insights on the game, except the fact that my intuition says it does not seem to be analytically "breakable".

So, yes, I wrote a computer program:

1) The game is pretty easy to simulate as long as you don't have a choice and then branch to all possible choices so you can simply try all the possibilities.

2) I can look ahead with only one modulo 12 computation to see if I land on an empty hole. However I did not find a simple way to look ahead more than 1 move at a time. This would have moderately improved the algorithm.

3) Retrograde analysis did not get me very far. It seems clear that the last positions (in reverse order) are:

66 0 0 0 0 0 0 0 0 0 0 0 

65 0 0 0 0 0 0 0 0 0 0 1 

64 0 0 0 0 0 0 0 0 0 2 0 

63 0 0 0 0 0 0 0 0 0 2 1 

but afterwards you don't have a single path to follow back.

I would have tried to encode possible end-game positions as a table so that I could detect earlier a winning sequence of choices. However my intuition is that the number of valid end-game positions is much larger than the number of end-game positions actually reachable from the proposed initial state.

So, I crossed my fingers and tried to brute force it Gave each initial choice a bound on the running time before terminating the entire tree. The hole-7 choice gave me the first solution. I have no guarantee that it's actually the quickest solution (least moves).

Edited October 31, 2010 by araver

[araver]

66RunExplained.txt

[wolfgang]

Yes, I am using these rules, I forgot to mention that I also positioned the holes as Anza Power did

0 (hole 12 which I call HOME for simplicity)...1...2...3...4...5...6...7...8...9...10...11

Since there in a circle, it makes no difference on the rules and it makes easier computations modulo 12.

So the first lines in my solution:

Starting with hole 7

Taking all jewels from hole 7 

Jewels taken 7

0 1 2 3 4 5 6 0 8 9 10 11 

Taking all jewels from hole 2 

Jewels taken 3

1 2 0 3 4 5 6 0 9 10 11 12 

should be read as (missing steps are now inserted):

Initial: 0 1 2 3 4 5 6 7 8 9 10 11

Starting with hole 7

Taking all jewels from hole 7 

Jewels taken 7

0 1 2 3 4 5 6 0 8 9 10 11

Putting one jewel in hole 8, still holding 6 in hand

0 1 2 3 4 5 6 0 9 9 10 11 

Putting one jewel in hole 9, still holding 5 in hand

0 1 2 3 4 5 6 0 9 10 10 11 

Putting one jewel in hole 10, still holding 4 in hand

0 1 2 3 4 5 6 0 9 10 11 11 

Putting one jewel in hole 11, still holding 3 in hand

0 1 2 3 4 5 6 0 9 10 11 12

Putting one jewel in hole 12(HOME), still holding 2 in hand

1 1 2 3 4 5 6 0 9 10 11 12  

Putting one jewel in hole 1, still holding 1 in hand

1 2 2 3 4 5 6 0 9 10 11 12 

Last jewel above hole 2 which is not empty so:

Taking all jewels from hole 2 

Jewels taken 3 (2 from hole and 1 in hand)

1 2 0 3 4 5 6 0 9 10 11 12 

I can provide a more verbatim solution (step-by-step) but it is much longer than the 439 lines I've already shown in 66Run.txt

I don`t want the details...I know they are very long steps...but u can give us the hole numbers only,

so u begin with hole number 7,and ended at last in hole number 12 wining two jewels...then which holes are next?

[araver]

I don`t want the details...I know they are very long steps...but u can give us the hole numbers only,

so u begin with hole number 7,and ended at last in hole number 12 wining two jewels...then which holes are next?

7,4,1,5,1,4,8,1,1,4,7,4,2,9,10,11,6,11,8,2,11,10,11,9,11,2,11,4,11,6,11,9,11,5,11,6,11,8,11,10,11.

41 total choices.

[superprismatic]

there's nothing special with my strategy.

As I said, I have found no analytical insights on the game, except the fact that my intuition says it does not seem to be analytically "breakable".

So, yes, I wrote a computer program:

1) The game is pretty easy to simulate as long as you don't have a choice and then branch to all possible choices so you can simply try all the possibilities.

2) I can look ahead with only one modulo 12 computation to see if I land on an empty hole. However I did not find a simple way to look ahead more than 1 move at a time. This would have moderately improved the algorithm.

3) Retrograde analysis did not get me very far. It seems clear that the last positions (in reverse order) are:

65 0 0 0 0 0 0 0 0 0 0 1
64 0 0 0 0 0 0 0 0 0 2 0
63 0 0 0 0 0 0 0 0 0 2 1

66 0 0 0 0 0 0 0 0 0 0 0 

but afterwards you don't have a single path to follow back.

I would have tried to encode possible end-game positions as a table so that I could detect earlier a winning sequence of choices. However my intuition is that the number of valid end-game positions is much larger than the number of end-game positions actually reachable from the proposed initial state.

So, I crossed my fingers and tried to brute force it Gave each initial choice a bound on the running time before terminating the entire tree. The hole-7 choice gave me the first solution. I have no guarantee that it's actually the quickest solution (least moves).

Thanks. I also wrote a quick subroutine to carry out the steps between choices.

I was in a bit of a hurry, so I just made random choices thinking that the search

tree would bush too much. I didn't realize that the number of empty holes grows quickly.

Even with my random choices, I got to 55 in a few seconds! Thanks again for the

nice explanation.

[araver]

I don`t want the details...I know they are very long steps...but u can give us the hole numbers only,

so u begin with hole number 7,and ended at last in hole number 12 wining two jewels...then which holes are next?

smaller winning sequences (in terms of how many choices you make):

2, 1, 2, 2, 11, 9, 8, 5, 9, 3, 11, 9, 11, 11, 1, 11, 10, 11, 3, 11, 7, 11, 10, 11, 5, 11, 9, 11, 7, 11, 9, 11,Choices: 32

2, 1, 2, 3, 7, 11, 6, 1, 6, 11, 4, 11, 1, 3, 9, 8, 4, 11, 9, 11, 1, 11, 3, 11, 5, 11, 10, 11, 7, 11, 9, 11,Choices: 32

2, 1, 3, 5, 11, 11, 8, 3, 10, 1, 4, 1, 9, 10, 9, 10, 8, 11, 9, 11, 1, 11, 3, 11, 5, 11, 10, 11, 7, 11, 9, 11,Choices: 32

2, 1, 5, 2, 3, 7, 3, 6, 6, 5, 11, 6, 8, 10, 11, 7, 5, 11, 8, 11, 2, 11, 9, 11, 4, 11, 6, 11, 8, 11, 10, 11,Choices: 32

There might be easier winning strategies, still searching

Edited November 1, 2010 by araver

[wolfgang]

7,4,1,5,1,4,8,1,1,4,7,4,2,9,10,11,6,11,8,2,11,10,11,9,11,2,11,4,11,6,11,9,11,5,11,6,11,8,11,10,11.

41 total choices.

woooow...Fantastic!