5000 Karma, help cant figure this simple game out

[BrothaBill]

It cant be that hard, first one to post up a viable solution gets the karma

http://www.ebaumsworld.com/pearl.shtml

You pick pearls out of a any row you want, object to leave the guy with one pearl and not you with one left to pick. I cant figure it out, seems it would be easy

 

[blut wump]

From what you describe it sounds like Nim, which has a solution based on binary pairing.

 

[blut wump]

Yes, it's Nim and gives the same start position each time with you to start. It can be won without fail.

Delete the entire second row of four pearls
It deletes one from the third row
Delete one from the top row
It deletes one from the bottom row
Delete three from the third row
It deletes one from the top row
Delete two from the bottom row
It deletes one from the top row
Delete either remaining pearl
It deletes the last one and loses

The 'trick' is to form pairs of powers of two starting with the highest available. Having paired-up all the pearls possible, you delete the spare ones to attain a winning position. There is some adjustment to this strategy right at the end depending on whether the victory condition is to leave or take the final pearl.

 

[H_T_]

it's not your fault, BB. you are so brilliant, you are actually a simpleton. does that make sense? like rainman.

 

[BrothaBill]

Yes, it's Nim and gives the same start position each time with you to start. It can be won without fail.

Delete the entire second row of four pearls
It deletes one from the third row
Delete one from the top row
It deletes one from the bottom row
Delete three from the third row
It deletes one from the top row
Delete two from the bottom row
It deletes one from the top row
Delete either remaining pearl
It deletes the last one and loses

The 'trick' is to form pairs of powers of two starting with the highest available. Having paired-up all the pearls possible, you delete the spare ones to attain a winning position. There is some adjustment to this strategy right at the end depending on whether the victory condition is to leave or take the final pearl.

I tried what you said exactly and it didnt work

 

[blut wump]

I just played it again and it varied its response this time. Apply the 'trick' and you will always win.

Removing the row of four is a winning first move. Right at the end of the game you will be aiming to break away from the 'trick' to leave three single pearls.

 

[blut wump]

Take a look here:
http://world.std.com/~reinhold/math/nim.html

 

[BrothaBill]

I just played it again and it varied its response this time. Apply the 'trick' and you will always win.

Removing the row of four is a winning first move. Right at the end of the game you will be aiming to break away from the 'trick' to leave three single pearls.

Im not following, its responses are completely different if I remove the four pears than you say.
Which row of pearls, the one with four, I counted that as the third row the first time. Then 'it' made the moves you suggested so its a bit confusing as to how you are stating the solution

 

[exotic_island_girl]

Nope that didn't work ...

 

[blut wump]

I was counting the top row as the one nearest the dude's wrist. The row of four as then being row 2 etc.

I'm always getting the start position of

I I I
I I I I <----- Remove this row
I I I I I
I I I I I I

Forming binary pairs:
Fours in rows 3 & 4
Twos in rows 1 & 4
Ones in rows 1 & 3
Row two is wholly spare

If you post back its response I'll give a next move with the pairs analysis.

[I I]
I I I I
[I I I I]
[I I I I] [I I]

 

[BrothaBill]

I was counting the top row as the one nearest the dude's wrist. The row of four as then being row 2 etc.

I'm always getting the start position of

I I I
I I I I <----- Remove this row
I I I I I
I I I I I I

Forming binary pairs:
Fours in rows 3 & 4
Twos in rows 1 & 4
Ones in rows 1 & 3
Row two is wholly spare

If you post back its response I'll give a next move with the pairs analysis.

I will a bit later, watching football right now. K to you for the effort. I can figure it out from here
Thanks

 

[gonelifting]

Yes, it's Nim and gives the same start position each time with you to start. It can be won without fail.

Delete the entire second row of four pearls
It deletes one from the third row
Delete one from the top row
It deletes one from the bottom row
Delete three from the third row
It deletes one from the top row
Delete two from the bottom row
It deletes one from the top row
Delete either remaining pearl
It deletes the last one and loses

The 'trick' is to form pairs of powers of two starting with the highest available. Having paired-up all the pearls possible, you delete the spare ones to attain a winning position. There is some adjustment to this strategy right at the end depending on whether the victory condition is to leave or take the final pearl.

Yeah, that worked. I tried it once and it did'nt. When I tried the second time, it worked. I hate that guy's laugh.

k to bw