When there are 2 or more predictive cells in a column that becomes active are both chosen as winner cells?

You definitely want to activate all of the predictive cells, because the prediction is ambiguous, so either of the two contexts might be accurate. Usually a couple time steps after this happens, the ambiguity will be resolved, and you’ll be confidently on one of the two potential branches.

One special case is repeating inputs. When you first implement the TM algorithm, this usually causes some head scratching at first. The vanilla TM algorithm will reach the end of the repeating sequence, burst, add one more element to the end, and then cycle back though it, reach the end, burst, and add one more, etc. Indefinitely. Each cycle through will add additional cells per minicolumn, until all the cells have been used. I wrote some analysis about this behavior on another thread here and some thoughts on how it might be addressed if it is a concern for you.

–EDIT-- Sorry, I misread the title, you were talking about which should be considered “winner”. I think you still have one “winner” from the perspective of the next input (i.e. if next input bursts, you would only form new connections with one of them). I haven’t looked at this in a while though, so let me look into it a bit and update you

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