Condorect sub-cycle rule

David Marsay djmarsay at
Fri Sep 26 04:18:45 PDT 1997

Dear All, particularly Markus and Mike

The Arrow and Gibbard-Satterthwaite thread contains, from Markus 
Schulze (18 Sep 97):
>Smith//Condorcet[EM] with the subcycle rule fails
>to meet Pareto.

Mike Ossipoff also wrote (21 Sep 97):
>Certainly the
>subcycle rule improves Smith//Condorcet, especially when a provision
>is added so that it will be used for a subcycle only when nothing in
>that subcycle would win otherwise (the wording probably needs

Does anyone have an authoritative English reference for the sub-cycle 

My version would be
1) consider least strong majorities first
2) for all links of the strength being considered, discount all 
those that contribute to any cycle.
This gives a unique answer, independent of the the order in which 
cycles are considered. (To implement, for a given strength, mark 
links for breaking first, then break all links when all links have 
been considered).

The result has no cycles. One never needs to break a consensus. 
(Since they can't form cycles). Hence this rule is Pareto.

Have I really invented a new rule?

An alternative implementation is to accept the highest majorities 
first, and carry on until it would make a cycle. Which is quickest 
depends on how complicated the example is.

Markus gives the following:

>25 voters vote BCDFEA.
>24 voters vote CDFEAB.
>20 voters vote ABFECD.
>15 voters vote EABCDF.
>8 voters vote EBCADF.
>4 voters vote ECADBF.
>4 voters vote ECABDF.

(To the following table, I add the majorities on the right)

>A:B=67:33	34
>A:C=35:65	-30
>A:D=51:49	2
>A:E=20:80	-60
>A:F=51:49	2
>B:C=68:32	36
>B:D=72:28	44
>B:E=45:55	-10
>B:F=76:24	52
>C:D=100:0	100
>C:E=49:51	-2
>C:F=80:20	60
>D:E=49:51	-2
>D:F=80:20	60
>E:F=31:69	-38

He breaks the ABC cycle first, which leads to problems. Note that the 
method if not 'fair', in the sense that it depends on the order in 
which the options are listed.

For 'my' method:
I note that cycles exists, and that the least preference is 2.
AD, AF, CE and DE all have strength +/-2 and make cycles. Discount 
them. I still have at least one cycle (EBCFE).
The lowest remaining majority is now BE, 10. Since this contributes 
to the above cycle, I discount it.
Markus's ABC cycle remains, so I carry on.
The least remaining preference is now AC, -30.
Since this breaks the above cycle, I discount it.
I still have ABFEA.
The least remaining preference is now AB, 34.
Since this breaks the above cycle, I discount it.
I now have BCDFEA.

Alternatively, taking the strongest preference I have CD, 100.
The next strongest are CF, DF and EA (60),  so I have CDF+EA. (could 
be ECADF).
The next strongest is BF, 52. Hence BF+CDF+EA.
The next is FE, 38. Hence CDFEA+BF.
The next is BC, 36. Hence BCDFEA. Done!

This example is a typical. It can happen that a given strength of 
majority is redundant. It can also happen that some links of a given 
strength make a cycle, others do not. It is important that if two or 
links make a cycle, they are both counted.

Comments, anyone?

(See also Condorcet//FPP thread).

Sorry folks, but apparently I have to do this. :-(
The views expressed above are entirely those of the writer
and do not represent the views, policy or understanding of
any other person or official body.

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