Condorcet Basics

[DEMOREP1 at aol.com]

DEMOREP1 wrote:
Of some interest in one tiebreaker is the number of votes that each choice
gets at the N/2 (N even) or (N+1)/2 (N odd) horizontal choices level (both
left to right and right to left).

Donald wrote-
I am interested in this tiebreaker, but it is not clear how it works.
Could you give us a simple example?
----
D-adding some clarifying words--
Of some interest in one tiebreaker is the number of votes that each choice
gets at first (or last) through the N/2 (N even) or (N+1)/2 (N odd) horizontal
choices level (both left to right and right to left).
     1 2 3 4 5 
22 A B C D E
21 B C D E A
20 C D E A B
19 D E A B C
18 E A B C D
100

Votes in 1-3 places 
A 59
B 61
C 63 max
D 60
E 57
 300

Votes in 5- 3 places
A 60
B 57
C 59
D 61
E 63 max
 300

The theory- If there is no Condorcet winner, then the compromise winner should
get the highest majority in the first N/2 places.

If there is no Condorcet winner and if elimination of candidates is to be
done, then the highest majority in the last N/2 places should lose.

Such a tiebreaker assumes that each vote is an Approval type YES vote.  I note
again that many choices will not have a YES majority in a specific YES/NO
vote.  

I would prefer that 
A. both a YES/NO and a number vote be done, 
B. do the Condorcet math for the majority YES choices, 
C. if there is no Condorcet winner, then look at all the ballots and drop the
last numbered YES vote on each ballot.

In the above Example (assuming that each choice gets a YES majority)--
Each YES vote for number 5 choices (if any) would be dropped on all ballots. 
If a choice no longer had a YES majority, then the Condorcet math would be
rechecked for the remaining choices.
If necessary, then repeat for the number 4 YES votes, the number 3 YES votes,
etc. on all ballots.

How many times will there be 3 or more candidates in a public election each
having YES majorities in a circular tie ?