10 candidate tie example

[DEMOREP1 at aol.com]

Bruce Anderson wrote:
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As an aside, I recommend against using the term "circular tie" without
definition.  To see why, consider the following example that Mike and I
worked out.  Suppose that there are 10 candidates, A through J.  Suppose that
A beats B and C, and ties everyone else.  Suppose that B beats C who beats D
who beats B, and they tie everyone else (except for B and C losing to A).
 Suppose that E beats F who beats G who beats E and that they tie everyone
else, except that G beats H, and that H ties everyone else.  Finally suppose
that I beats J and they tie everyone else.  Is there a circular tie here and,
if so, who is in it?
(All 10 candidates are Smith winners here.)
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The math of the above would be nice to see. 

If an X-Y array is produced (winner at top, loser at left) and adding W, L or
T for win, loss or tie, then one sees A-D, E-H and I-J  groupings of wins and
losses along the array diagonal.

An *ordinary* circular tie exists if X>Y>Z>X (B>C>D>B and E>F>G>E in the
above)

Since each group is independent the phrase *circular tie* might not be
*technically* correct when applied to the entire example. 
Since no candidate loses to each other candidate they all tie. (Candidates H
and J fail to beat anyone.)

Thus, perhaps the phrase should be just *tie* (i.e. Condorcet tie).