Re: Automatic LIIA Independent of Locking Order
Toby Pereira via Election-Methods <[email protected]> Fri, 1 May 2026 21:41:25 +0000 (UTC)
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Gustav and Kristofer - I might have not thought that through properly. I w=
as only considering removing candidates from one end of the pecking and not=
the other.
I also agree it can be of theoretical interest and that there may be even s=
ome cases where it is of practical use. But when I see it listed alongside =
other criteria as if it's an equal, I wonder how it made the list.
But regardless of practicalities, and just from a theoretical "good candida=
te" standpoint, if you have an A>B>C>A cycle and A is the winner, I don't h=
ave any intuition that tells me B should automatically be considered better=
than C, which LIIA would suggest is the case. (Obviously B beats C pairwis=
e but we have a cycle.)
Toby
On Friday, 1 May 2026 at 19:09:49 BST, Kristofer Munsterhjelm <km-elmet=
@munsterhjelm.no> wrote: =20
=20
On 2026-05-01 18:28, Toby Pereira via Election-Methods wrote:
> I'm not sure if this relates to your question at all, but any method can=
=20
> easily be converted to an LIIA-passing method, without changing the=20
> winner. Instead of using the method's "natural" finishing order, declare=
=20
> just the winner initially and then for 2nd place, remove the winner from=
=20
> the process and find the new winner and declare them to be 2nd, and so on=
.
Is that true? Consider the definition as stated on Electowiki:
>> LIIA requires that both of the following conditions always hold:
>> If the option that finished in last place is deleted from all the
>> votes, then the order of finish of the remaining options must not
>> change. (The winner must not change.)
>> If the winning option is deleted from all the votes, the order of
>> finish of the remaining options must not change. (The option that
>> finished in second place must become the winner.)
Suppose that you construct a method like the above, where the winner is=20
the winner of the original method, then the second-place candidate is=20
the winner with the original winner removed, etc. It is then not at all=20
clear that removing the loser (the candidate ranked last) will preserve=20
the ranking of the other candidates.
Another reason that this seems wrong is, a method that satisfies=20
majority and LIIA must also satisfy Condorcet, and then Smith, and then=20
ISDA.
It must satisfy the property that if A is ranked immediately ahead of B,=20
then A beats B pairwise; suppose otherwise, then eliminate all=20
candidates ranked below B. This shouldn't change anything. Then=20
eliminate all candidates ranked above A. This shouldn't change anything,=20
either. Then majority requires that A win.
From this, it satisfies Condorcet because suppose A is the CW but not=20
ranked first, then the candidate ranked above A must beat A pairwise,=20
which is a contradiction.
Similar reasoning leads to Smith (since if not Smith, then someone in=20
the Smith set is ranked below someone not in it) and ISDA (because you=20
can eliminate everybody outside the Smith set as we've established the=20
Smith set must be ranked before any non-Smith candidate).
But the given construction would let you make a "LIIA" method that ranks=20
any candidate first, even a Condorcet loser. Which doesn't seem right.
> Going off on a tangent, I've always felt that LIIA has somehow found its=
=20
> way into the "standard list" of election method criteria without any=20
> proper scrutiny of its utility. It's not clear what purpose it serves.=20
> It sounds good because it has "IIA" in it, but it doesn't really have=20
> much, if anything, to do with the IIA criterion. It's certainly not a=20
> stepping stone towards it.
>=20
> I think when I mentioned this before, Kristofer said that if the winner=
=20
> drops out for some reason, then you can just elect 2nd place as the=20
> order wouldn't change if you ran the election again without the original=
=20
> winner. But the flipside of this is that after the election, 2nd place=20
> might be found to be ineligible for some reason, and there would be some=
=20
> elections where an LIIA-failing method would save you from the=20
> embarrassment of the original 3rd placed candidate becoming the new winne=
r.
It might also have some uses in Condorcet STV methods. Suppose a class=20
of STV-like methods is constructed like this:
=C2=A0=C2=A0=C2=A0 1. If we've filled every seat, exit.
=C2=A0=C2=A0=C2=A0 2. If anybody has more than a Droop quota of the first p=
references:
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 2.1. Elect and eliminate that candida=
te.
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 2.2. Redistribute surpluses based on =
first preferences.
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 2.3. Go to 1.
=C2=A0=C2=A0=C2=A0 3. Otherwise:
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 3.1. Determine a winning order by som=
e base method X
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 3.2. Eliminate the loser according to=
X
=C2=A0=C2=A0=C2=A0 =C2=A0=C2=A0=C2=A0 3.3. Go to 1.
There are two cases where small initial differences may be amplified to=20
cause widely diverging outcomes: election (where electing A instead of B=20
may change who gets elected next) and elimination (similar to IRV's=20
chaos). If you use a LIIA method, then the second source vanishes,=20
because eliminating the loser of X doesn't change the order of victory=20
of the other candidates.
This might lead to a more orderly method, possibly fewer monotonicity=20
violations, etc. I don't know this for sure: it's just an intuitive=20
argument.
-km
=20
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<html><head></head><body><div class=3D"ydp936f3a49yahoo-style-wrap" style=
=3D"font-family:Helvetica Neue, Helvetica, Arial, sans-serif;font-size:16px=
;"><div></div>
<div>Gustav and Kristofer - I might have not thought that through p=
roperly. I was only considering removing candidates from one end of the pec=
king and not the other.</div><div><br></div><div>I also agree it can be of =
theoretical interest and that there may be even some cases where it is of p=
ractical use. But when I see it listed alongside other criteria as if it's =
an equal, I wonder how it made the list.</div><div><br></div><div>But regar=
dless of practicalities, and just from a theoretical "good candidate" stand=
point, if you have an A>B>C>A cycle and A is the winner, I don't h=
ave any intuition that tells me B should automatically be considered better=
than C, which LIIA would suggest is the case. (Obviously B beats C pairwis=
e but we have a cycle.)</div><div><br></div><div>Toby</div><div><br></div>
=20
<div id=3D"ydp936f3a49yahoo_quoted_8117967279" class=3D"ydp936f3a49=
yahoo_quoted">
<div style=3D"font-family:'Helvetica Neue', Helvetica, Arial, s=
ans-serif;font-size:13px;color:#26282a;">
=20
<div>
On Friday, 1 May 2026 at 19:09:49 BST, Kristofer Mu=
nsterhjelm <[email protected]> wrote:
</div>
<div><br></div>
<div><br></div>
=20
=20
<div><div dir=3D"ltr">On 2026-05-01 18:28, Toby Pereira via=
Election-Methods wrote:<br></div><div dir=3D"ltr">> I'm not sure if thi=
s relates to your question at all, but any method can <br></div><div dir=3D=
"ltr">> easily be converted to an LIIA-passing method, without changing =
the <br></div><div dir=3D"ltr">> winner. Instead of using the method's "=
natural" finishing order, declare <br></div><div dir=3D"ltr">> just the =
winner initially and then for 2nd place, remove the winner from <br></div><=
div dir=3D"ltr">> the process and find the new winner and declare them t=
o be 2nd, and so on.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">I=
s that true? Consider the definition as stated on Electowiki:<br></div><div=
dir=3D"ltr"><br></div><div dir=3D"ltr">>> LIIA requires that both of=
the following conditions always hold:<br></div><div dir=3D"ltr"><br></div>=
<div dir=3D"ltr">>> If the option that finished in last place is dele=
ted from all the<br></div><div dir=3D"ltr">>> votes, then the order o=
f finish of the remaining options must not<br></div><div dir=3D"ltr">>&g=
t; change. (The winner must not change.)<br></div><div dir=3D"ltr">>>=
If the winning option is deleted from all the votes, the order of<br></div=
><div dir=3D"ltr">>> finish of the remaining options must not change.=
(The option that<br></div><div dir=3D"ltr">>> finished in second pla=
ce must become the winner.)<br></div><div dir=3D"ltr"><br></div><div dir=3D=
"ltr">Suppose that you construct a method like the above, where the winner =
is <br></div><div dir=3D"ltr">the winner of the original method, then the s=
econd-place candidate is <br></div><div dir=3D"ltr">the winner with the ori=
ginal winner removed, etc. It is then not at all <br></div><div dir=3D"ltr"=
>clear that removing the loser (the candidate ranked last) will preserve <b=
r></div><div dir=3D"ltr">the ranking of the other candidates.<br></div><div=
dir=3D"ltr"><br></div><div dir=3D"ltr">Another reason that this seems wron=
g is, a method that satisfies <br></div><div dir=3D"ltr">majority and LIIA =
must also satisfy Condorcet, and then Smith, and then <br></div><div dir=3D=
"ltr">ISDA.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">It must sa=
tisfy the property that if A is ranked immediately ahead of B, <br></div><d=
iv dir=3D"ltr">then A beats B pairwise; suppose otherwise, then eliminate a=
ll <br></div><div dir=3D"ltr">candidates ranked below B. This shouldn't cha=
nge anything. Then <br></div><div dir=3D"ltr">eliminate all candidates rank=
ed above A. This shouldn't change anything, <br></div><div dir=3D"ltr">eith=
er. Then majority requires that A win.<br></div><div dir=3D"ltr"><br></div>=
<div dir=3D"ltr"> From this, it satisfies Condorcet because suppose A is th=
e CW but not <br></div><div dir=3D"ltr">ranked first, then the candidate ra=
nked above A must beat A pairwise, <br></div><div dir=3D"ltr">which is a co=
ntradiction.<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr">Similar r=
easoning leads to Smith (since if not Smith, then someone in <br></div><div=
dir=3D"ltr">the Smith set is ranked below someone not in it) and ISDA (bec=
ause you <br></div><div dir=3D"ltr">can eliminate everybody outside the Smi=
th set as we've established the <br></div><div dir=3D"ltr">Smith set must b=
e ranked before any non-Smith candidate).<br></div><div dir=3D"ltr"><br></d=
iv><div dir=3D"ltr">But the given construction would let you make a "LIIA" =
method that ranks <br></div><div dir=3D"ltr">any candidate first, even a Co=
ndorcet loser. Which doesn't seem right.<br></div><div dir=3D"ltr"><br></di=
v><div dir=3D"ltr">> Going off on a tangent, I've always felt that LIIA =
has somehow found its <br></div><div dir=3D"ltr">> way into the "standar=
d list" of election method criteria without any <br></div><div dir=3D"ltr">=
> proper scrutiny of its utility. It's not clear what purpose it serves.=
<br></div><div dir=3D"ltr">> It sounds good because it has "IIA" in it,=
but it doesn't really have <br></div><div dir=3D"ltr">> much, if anythi=
ng, to do with the IIA criterion. It's certainly not a <br></div><div dir=
=3D"ltr">> stepping stone towards it.<br></div><div dir=3D"ltr">> <br=
></div><div dir=3D"ltr">> I think when I mentioned this before, Kristofe=
r said that if the winner <br></div><div dir=3D"ltr">> drops out for som=
e reason, then you can just elect 2nd place as the <br></div><div dir=3D"lt=
r">> order wouldn't change if you ran the election again without the ori=
ginal <br></div><div dir=3D"ltr">> winner. But the flipside of this is t=
hat after the election, 2nd place <br></div><div dir=3D"ltr">> might be =
found to be ineligible for some reason, and there would be some <br></div><=
div dir=3D"ltr">> elections where an LIIA-failing method would save you =
from the <br></div><div dir=3D"ltr">> embarrassment of the original 3rd =
placed candidate becoming the new winner.<br></div><div dir=3D"ltr"><br></d=
iv><div dir=3D"ltr">It might also have some uses in Condorcet STV methods. =
Suppose a class <br></div><div dir=3D"ltr">of STV-like methods is construct=
ed like this:<br></div><div dir=3D"ltr"><br></div><div dir=3D"ltr"> &n=
bsp; 1. If we've filled every seat, exit.<br></div><div dir=3D"ltr">&=
nbsp; 2. If anybody has more than a Droop quota of the first pr=
eferences:<br></div><div dir=3D"ltr"> =
2.1. Elect and eliminate that candidate.<br></div><div dir=3D"ltr"> &n=
bsp; 2.2. Redistribute surpluses based on first pr=
eferences.<br></div><div dir=3D"ltr"> =
2.3. Go to 1.<br></div><div dir=3D"ltr"> 3. Otherwise:<br=
></div><div dir=3D"ltr"> 3.1. Determin=
e a winning order by some base method X<br></div><div dir=3D"ltr"> &nb=
sp; 3.2. Eliminate the loser according to X<br></d=
iv><div dir=3D"ltr"> 3.3. Go to 1.<br>=
</div><div dir=3D"ltr"><br></div><div dir=3D"ltr">There are two cases where=
small initial differences may be amplified to <br></div><div dir=3D"ltr">c=
ause widely diverging outcomes: election (where electing A instead of B <br=
></div><div dir=3D"ltr">may change who gets elected next) and elimination (=
similar to IRV's <br></div><div dir=3D"ltr">chaos). If you use a LIIA metho=
d, then the second source vanishes, <br></div><div dir=3D"ltr">because elim=
inating the loser of X doesn't change the order of victory <br></div><div d=
ir=3D"ltr">of the other candidates.<br></div><div dir=3D"ltr"><br></div><di=
v dir=3D"ltr">This might lead to a more orderly method, possibly fewer mono=
tonicity <br></div><div dir=3D"ltr">violations, etc. I don't know this for =
sure: it's just an intuitive <br></div><div dir=3D"ltr">argument.<br></div>=
<div dir=3D"ltr"><br></div><div dir=3D"ltr">-km<br></div></div>
</div>
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