Fwd: Sprouts equivalences

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Yper Cube

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Jan 28, 2009, 2:05:34 AM1/28/09
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Jeff Peltier has provided counterexamples for the last 2 couples of equivalences in the Equivalences page (at wgosa).

Since the second one, (the }2.2.2.AB.} == }2.2.2.2.2.AB.} == ... ) is also dependent on the first one, (the }2.2.2A.} == }2.2.2.2.2A.} == ), my only choice is to retract both of these two couples.

After careful reexamination of my files, they seem not true :(

ypercube

---------- Forwarded message ----------
From: Jeff Peltier <jfpe...@gmail.com>
Date: Tue, Jan 27, 2009 at 6:06 PM
Subject: Re: Sprouts equivalences
To: Yper Cube <yper...@gmail.com>


... and as a bad news nver comes alone, for the same reason we have :

2.2.2A.}AB.}2B.}] is *4 while 2.2.2.2.2A.}AB.}2B.}] is *3
...

On Tue, Jan 27, 2009 at 4:52 PM, Jeff Peltier <jfpe...@gmail.com> wrote:
Yper,

Bad news, rereading your article, the one with AB.} seemed so incredible I tested it on the simplest partial game AB.} here is what Glop said:

(2.2.2.AB.}AB.}] 4) is a losing position and (2.2.2.2.2.AB.}AB.}] 3) is a losing position.

.....

--
Jean-François

Dan Hoey

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Jan 29, 2009, 10:49:03 PM1/29/09
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I am disappointed by the failure of this equivalence, too.

Let's consider two sequences of order-1 partial positions,
P_n = ^x:x.((2))*n and Q_n = ^x:x2.((2))*n. In traditional
Glop notation, that is P_0=}A.}, P_1=}A.2}, P_2=}A.2.2}, ...,
and Q_0=}A2.}, Q_1=}A2.2.}, Q_2=}A2.2.2.}, ....

As in my post on "A new equivalence", these partial positions X
are determined by E_x(X), I(X), and I_x(X), where E_x(X) is the
position that arises from using the pivot externally, I_x(X) is
the set of positions arising from using the pivot internally,
and I(X) is the set of partial positions that arising from
moving internally without using the pivot.

In tabular form, the results for the P_i are as follows.

E_x I_x I
P_0 *[0] {} {}
P_1 *[0] {*[0]} {}
P_2 *[1] {*[1]} {P_1}
P_3 *[0] {*[0]} {P_2}

Comparing P1 and P_3, we see the only difference is that P_2 is
in I(P_3}, but since P_1 is in I{P_2} the P_2 is reversible.
Therefore P_1 = P_3, and similarly P_n = P_(n+2) for n>0.
Furthermore,

E_x I_x I
P_2+*[1] *[0] {*[0]} {P_1+*[1], P_2}

which is equal to P_1, since both P_1+*[1] and P_2 reverse
to P_1. This means that P_n + *[1] = P_(n+1) for n>0.

Now consider the Q_i.

E_x I_x I
Q_0 *[0] {*[0]} {}
Q_1 *[1] {*[1]} {Q_0}
Q_2 *[0] {*[0],*[1]} {Q_1}
Q_3 *[1] {*[0],*[1],*[2]} {Q_2}
Q_4 *[0] {*[0],*[1],*[3]} {Q_3}
Q_5 *[1] {*[0],*[1],*[2]} {Q_4}

We would like to reverse Q_4 to Q_3 to make Q_5=Q_3.
Unfortunately, I(Q_3) has the option Q_2, which I(Q_5) does not
have. The same problem arises with any attempt to find
equivalence among the Q_i. With *[1],

E_x I_x I
Q_0+*[1] *[1] {*[1]} {Q_0}
Q_1+*[1] *[0] {*[0]} {Q_0+*[1], Q_1}
Q_2+*[1] *[1] {*[0],*[1]} {Q_1+*[1], Q_2}
Q_3+*[1] *[0] {*[0],*[1],*[3]} {Q_2+*[1], Q_3}
Q_4+*[1] *[1] {*[0],*[1],*[2]} {Q_3+*[1], Q_4}

we have that Q_0+*[1] = Q_1 (not surprising, since Q_0=P_1 and
Q_1=P_2), but no other equivalences appear.

I would be interested in seeing partial positions that can be
used to distinguish Q_n, Q_(n+1)+*[1], and Q_(n+2). I am
also interested in seeing whether a single partial position
can be used for arbitrarily large n, or whether more and more
complicated partial positions will be needed.

Dan

Yper Cube

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Jan 30, 2009, 2:25:27 AM1/30/09
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Regarding the P_n and Q_n partial positions:

Lets say that R or R(A) is a partial position with A as parameter. Lets also say that for two partial positions P, R with A as aparameter (or P(A), R(A)) , we define P#R to be the position (not partial) that results after "gluing" these two partial positions together.

Now,
P_n = P_(n+2) for n>0
and
P_n + *1 = P_(n+1) for n>0

can be rewritten as:

For any R and any n>0, R#(P_n) +*1 =  R#(P_(n+1))

We obviously don't have the same results for the Q_n.
But we do have this:

For any R, if there is a k>2 such as, R#(Q_k) + *1 =  R#(Q_(k+1))
       then
            for any n>=k, R#(Q_n) + *1 =  R#(Q_(n+1))

But this k is dependent on R so the equivalences are not "global"

So, for example, while,
2.2.2A.}AB.}2B.}] is *4   and   2.2.2.2.2A.}AB.}2B.}] is *3
let R = }AB.}2B.}
Since, 2.2.2.2.2A.}AB.}2B.}] is *3
and 2.2.2.2.2.2A.}AB.}2B.}] is *2
we have k=4 and
2.2.2.2....2A.}AB.}2B.}] = *3  (even (>=4) number of .2. ) and
2.2.2.2.2....2A.}AB.}2B.}] = *2  (odd (>=5) number of .2. )

But this k (4) depends on }AB.}2B.} and cannot be expected to be the same for another partial position R(A).

Yper Cube

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Jan 30, 2009, 4:20:53 AM1/30/09
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So, Dan's question:

"I am also interested in seeing whether a single partial position can be used for arbitrarily large n ..."

can be rephrased as:

There exists an R such as
no k>=2 exists with
R#Q_k + *1  ==  R#Q_(k+1)

Dan Hoey

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Jan 30, 2009, 11:22:18 AM1/30/09
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Yper Cube wrote:

> We obviously don't have the same results for the Q_n.
> But we do have this:
>
> For any R, if there is a k>2 such as, R#(Q_k) + *1 = R#(Q_(k+1))
> then
> for any n>=k, R#(Q_n) + *1 = R#(Q_(n+1))
>
> But this k is dependent on R so the equivalences are not "global"

Thanks for making that explicit, Yper Cube. We can define lim_Q(R)=k
in this case, and lim_Q(R)=infinity if there is no such k. Now I can
see three possible situations for
limset(Q) = { lim_Q(x) : x a partial position}

1. limset(Q) contains infinity. That means there is some R such that
R#(Q_n) + *1 is never equal to R#(Q_(n+1)). This is the situation
that you explain in your next message.

2. limset(Q) is an infinite set, but does not contain infinity. That
would mean that there is a k as above for any R, but it may be
arbitrarily large.

3. limset(Q) has a finite maximum K. This would mean that for k > K,
Q_k + *1 = Q_(k+1) , even though we don't know how to prove it.

I really don't know which of these is most likely to be the case.

For any partial position S = }w.} where w is a string of symbols
defining S, we can define S_1=}2.w.}, S_2=}2.2.w.}, .... and
continue to define lim_S(R) and limset(S). The "lousy periodicity
theorem" I described earlier is dependent on case 2 or 3 holding
for every S . I have to withdraw that theorem; I'm not even
willing to offer it as a conjecture any more.

At least we now have a known unknown.

Dan

Dan Hoey

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Jan 30, 2009, 4:54:29 PM1/30/09
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I wrote:
> Yper Cube wrote:
>
>> We obviously don't have the same results for the Q_n.
>> But we do have this:
>>
>> For any R, if there is a k>2 such as, R#(Q_k) + *1 = R#(Q_(k+1))
>> then
>> for any n>=k, R#(Q_n) + *1 = R#(Q_(n+1))
>>
>> But this k is dependent on R so the equivalences are not "global"
>
> Thanks for making that explicit, Yper Cube.

I should say more--thanks for showing that this is the case. It is
a consequence of the particular sequence Q_k that if
R#Q_k + *1 = R#Q_(k+1) for k>1, then R#Q_(k+1) + *1 = R#Q_(k+2),
and therefore for all n>k.

> We can define lim_Q(R)=k in this case, and lim_Q(R)=infinity if
> there is no such k.

[...]


> limset(Q) = { lim_Q(x) : x a partial position}

[...]


> For any partial position S = }w.} where w is a string of symbols
> defining S, we can define S_1=}2.w.}, S_2=}2.2.w.}, .... and
> continue to define lim_S(R) and limset(S).

Here I must be a little more careful, and define lim_S(R) to be the
least k for which R#S_k + *1 = R#S_(k+1) = R#S_(k+2) + *1 , since
for general partial positions S we cannot conclude the second
equality from the first.

Dan

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