[EM] Cloning: 2nd preference doing worse is implied by (P1)
Craig Carey
research at ijs.co.nz
Mon Feb 15 02:26:35 PST 1999
My last message did take the example far.
At 22:31 15.02.99 , Craig Carey wrote: ...
>At 18:56 15.12.99 , DEMOREP1 at aol.com wrote:
>
>>I repeat my observation about clones-
V1 =
>> 34 A>B>C
>> 33 B>C>A
>> 32 C>A>B
>>Adding D-- a 100 percent clone of B
V2 =
>> 34-x A>B>D>C
>> 33+2x B>D>C>A
>> 32-x C>A>B>D
>>A>B>D>C>A
>
>The winners of the 1st are {A} or {A,B}.
** If C loses the 1st it loses the 2nd
>** B should have the same in lose state in the 2nd as it has in the 1st,
> by SPC.
** Of the 2nd, if D wins then B wins (if B loses then D loses).
>It might be that D rather than A, is a winner of the 2nd.
----------------
The winners of the first are one of: {}, {A}, {A,B}, {A,C}, {A,B,C}.
Case W(V1)={A}: B & C & D lose the 2nd, so W(V2) = {A}
Case 0<x, W(V1)={A,B}: C loses V2 and B wins V2, so W(V2) = {A,B} or {B,D}
Case x<0, W(V1)={A,C}: B loses V2, hence D loses V2, so W(V2) = {A,C}
Case 3 = #W(V2): B wins V2. W(V2) = {A,B,C} or {A,B,D} or {B,C,D}
Result 1:
Using SPC B wins V2 iff B wins this 2 winner election:
V3=
34-x A>B
33+2x B
32-x C>A
Negate and solve 1 winner case:
-32+x C>A
-33-2x B
-34+x A>B Winner is ({B},{C}) iff (-33<-33-2x, -33-2x<-33)
So the 2 winners of V3 are: Winners = {A,C} if x<0, {A,B} if 0<x
A note on Demorep1 'clones':
If in this system (with only 2 papers shown), B loses and D wins:
c1 p,B,q,D,r
c2 s,B,t,D,u
then (P1) requires that B also loses this (and D's win-lose win-lose
state is so far unknown):
c1 p,D,q,B,r
c2 s,D,t,B,u
Now relabel and swap D with B, and hence D loses this.
c1 p,B,q,D,r
c2 s,B,t,D,u
That is is a contradiction and so the steps are a proof by contradiction
that Demorep1's clone rule is just a corollary of (P1). (I've assumed
I've understood the clone rule).
Craig Carey
Auckland, 15 December 1999
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