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I'm following an equation from a published paper in order to calculate probabilities using a Markov Chain. The equation says:

Construct a set $U$ that consists of all items that appear in the top-$k$ in at least one list.

For each pair of items $i$ and $j$ in $U$, let the preference for $i$ over $j$, $m_i{_j}$, equal $1$ if the majority of the lists ($>=50$%) that rank both $i$ and $j$ rank $j$ above $i$ and $0$ otherwise. Let $m_i{_j} = m_j{_i} = 0.5$ if items $i$ and $j$ are never directly compared in any list.

My problem with the above is: $U$ is composed of the top-k items in at least one list. Wouldn't this mean that $m_i{_j} = 0.5$ would never be possible because if $U$ is composed of items that must all appear in one list, each item is directly compared in at least one list.

I just want someone to confirm my reading of this.

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    @did after reading your point and going over my data again, with a pen and paper for more than an hour!, I can see that you are correct and that there are in fact a few instances of this state. Thanks for that!2012-07-13

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If there are at least two lists, the convention is necessary if item $i$ appears in some lists, item $j$ appears in some other lists, and no list contains both items $i$ and $j$.

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    Do you think they care? Anyway, this is a mistake to erase your comment where you gave the reference of the paper, and you should put this information back, in a new comment or, better still, in the post.2012-07-13