[ExI] The list is in the timeline

[spike]

From: extropy-chat [mailto:extropy-chat-bounces at lists.extropy.org] On Behalf Of Adrian Tymes
Sent: Wednesday, December 09, 2015 5:14 PM
To: ExI chat list <extropy-chat at lists.extropy.org>
Subject: Re: [ExI] The list is in the timeline

 

On Wed, Dec 9, 2015 at 6:32 AM, spike <spike66 at att.net <mailto:spike66 at att.net> > wrote:

Your comment has me pondering evil twins and triplets in general.  The evil
twin thing is fairly simple, the terms unambiguous.  But once we introduce
the complexity of triplets…  just feels like you should have one on
either end of the malicious/benevolence spectrum with one exactly at zero.

 

>…Evil in different ways.  Imagine a set of triplets accelerated away from one another by a shockwave, but with local scenery strongly suggesting each one is stationary.  Each triplet would think themselves "not moving" and the other two as "in motion"….  Each of these would regard the other two, from their own perspective, as evil.  Adrian

 

Sure.  The possibilities are intriguing with three triplets.  The good/evil axis is not so straightforward or objective as something like wealth.  We can imagine a kind of rock/paper/scissors paradox in a three-way comparison of evil vs righteous.  We can imagine competitions, where all manner of irony and paradox arises.  

Everyone is the hero of her own autobiography.  Every member of every triple would always see herself as the righteous one.  So we could query each and have her identify which of her siblings is the evil and which is the neutral.  That way each sibling would get one vote as being the good one (made by themselves) but one sibling might get two evils and the other get two neutrals.  In that case the third would by necessity have one good one neutral and one evil vote.  So the double neutral wins the title with a net one point.  The one good, one neutral one evil vote getter scores a zero, so they are the neutral, and the remaining hapless triplet ends with a score of negative 1, and is identified as the evil triplet.

But wait, there’s more.  If an evil person identifies you as evil, then that counts as a vote for good, ja?  You multiply their assessment of your benevolence score by their benevolence score to get yours.

Case 1, Alice, Betty and Carla try to determine who is the evil triplet, who is the neutral and who is the righteous.  Each identifies herself as good, by the reasoning presented earlier (who among us considers himself evil?)  So imagine the symmetrical case first, where each triplet identifies herself as righteous and each gets on evil vote and one neutral, so their scores come out zero, multiply thru and all three get zero.  None of their opinions of each other or themselves is worth a damn:

	
Alice

Betty

Carla


Alice

1

0

-1


Betty

-1

1

0


Carla

0

-1

1

	
0

0

0

 

Case 2, unsymmetrical.  Alice gets one righteous and two neutrals, Betty gets one righteous, one neutral and one evil (for a zero) and Carla gets two evils, one of which is cast by a positive 1, so it stays A=1, B=0 and C=-1.  This comes about if Betty switches her vote and the other two hold:

	
Alice

Betty

Carla


Alice

1

0

-1


Betty

0

1

-1


Carla

0

-1

1

	
1

0

-1

 

How evil is that!  Carla is unanimously identified, even auto-identified as evil!  

But wait, being identified as evil by an evil person is the same as a righteous (my enemy’s enemy is my friend.)  So Alice got one vote from Alice, 1 times 1 is one so she stays at 1 and Betty gets one vote from the negative 1 scoring evil Carla, so multiplying Carla’s negative 1 by the negative 1 she gave Betty makes Betty’s new score 1 from herself and (-1)(-1) = 2. 

	
Alice

Betty

Carla


Alice

1

0

-1


Betty

0

1

-1


Carla

0

-1

1

	
1

0

-1

				

multiply-->

1

2

-3

 

Case 3, Carla, realizing that that she is stuck with evil, but not feeling particularly evil, decides on an alternate strategy.  She votes herself evil and Alice righteous, identifying Betty as the neutral.  Well now, this is an interesting turn, for Alice ends up with 2 points, Betty with 1 and Carla with -3.

	
Alice

Betty

Carla


Alice

1

0

-1


Betty

0

1

-1


Carla

1

0

-1

	
1

1

-3

				

multiply-->

-2

2

0

 

So after the vote, it come out to A=-2, B=2, C=0.

But wait, we need to vote again, because now Alice’s opinion is doubled and reversed, Betty’s is doubled and Carla’s opinion is not worth a damn.

Alice catches on to this evil scheme by which she was identified as evil, and votes herself the evil twin.

Weirdness ensues.

spike

 

 

 

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