[EM] is the Raynaud method Smith-efficient?

[Chris Benham]

James,
I probably have a less "academic bent"  than you; but to me it is just 
*obvious* that as we only ever eliminate
pairwise losers it is impossible for all the members of the Smith set to 
be eliminated.
Therefore I'm happy to believe Rob Le Grand and Blake Cretney when they 
tell us on their web pages that
Raynaud meets Smith.

I gave some thought to Raynaud a few months ago:

"It seems to me that  three versions of  Raynaud are possible ( one that
meets Symmetric Completion and two that don't).
The obvious one that does is  Raynaud (Margins).  The other two  could
be called  Raynaud (Pairwise Opposition),
which eliminates the candidate which loses the pairwise comparison in
which the winner  has the highest gross score
(explicit winning votes);  and  Raynaud (Gross Loser), which eliminates
the candidate with the lowest gross score in any
pairwise comparison.

In this example of  Kevin Venzke's:

49: A
24: B
27: C>B

A>C  49-27 (m 22)
C>B  27-24  (m 3)
B>A  51-49  (m 2)

The three versions each give a different winner.  PO eliminates A and
elects C, failing Plurality.
GL eliminates  B and elects A, failing  Minimal Defense.  Margins
eliminates C  and elects B."

At the time, I preferred  Raynaud (Gross Loser).

Chris Benham