14+ = *0

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Josh Purinton

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May 15, 2006, 3:19:16 PM5/15/06
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AuntBeast recently confirmed that 14+ is a win for Right, further
extending the Sprouts Conjecture. AuntBeast's database mapping
isolated sub-components to normal-play nim values contains over 4.5
million entries. Theese positions are pseudo-canonized using a
slightly improved version of the algorithm described in the AJS
report.

--
Josh Purinton

Jeff Peltier

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May 16, 2006, 11:05:03 AM5/16/06
to sprouts-theory
For memory, the Sprouts Conjecture or Hudson's Conjecture states that
the Normal^Misere grundy value is for
1, 2, 3, 4, 5, 6, 7, 8, 9 spots :
0^1,0^0,1^0,1^0,1^1,0^1,0^0,0^0,1^0
and then it is 0^0 when the number is congruent to 0,1 or 2 modulo 6
and 1^1 in other cases.

Josh Purinton

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May 16, 2006, 11:23:28 AM5/16/06
to sprouts...@googlegroups.com

The Sprouts Conjecture (as defined on page 3 of the AJS report) does
not involve Grundy numbers or misere play.

Sprouts Conjecture: "The first player has a winning strategy in n-spot
Sprouts if and only if n is 3, 4, or 5 module 6."

For instance, the Sprouts Conjecture would still be satisfied even if
11+ equaled *2. The AJS report also defines the

Misere Sprouts Conjecture: "The first player has a winning strategy in
n-spot misere Sprouts if and only if n is 0 or 1 module 5."

Both of these differ from Hudson's conjecture.

--
Josh Purinton

Jeff Peltier

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May 16, 2006, 11:31:02 AM5/16/06
to sprouts-theory
In fact i must have extended Hudson's conjecture as well :

In the Sprouts glossary it is written :
"Hudson conjecture - The conjecture that, for n > 9, Left wins in
misere sprouts iff N divided by 6 leaves a remainder of 3, 4, or 5."

So it contradicts the AJS misere Sprouts Conjecture.

The grundy value of Sprouts initial positions is then Peltier's
Conjecture...

--
Jeff

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