`--order-from-ccs` orders on number solved, disregarding tiebreaker number submitted

[vmcj]

We got the slides as: A: tried(20), solved(8) B: tried(10), solved(8)

Where I would say A should have been regarded harder and the order should have been B,A instead of A,B.

[mzuenni]

Should we only consider num_judged or also num_pending?

[mpsijm]

We had a similar issue in Delft, where problems C and K both had one solve before the freeze (scoreboard). Problem K had many tries from other teams, problem C had no tries from other teams. I would like to argue that problem K should be considered to be easier than C, because it looked like it was the easier of the two from the perspective of the participants (hence the many tries) :stuck_out_tongue:

Maybe this is a personal preference thing, that can be resolved manually using the --order flag if needed? I'm fine with the existing ordering, i.e. using alphabetical order as tiebreaker when the number of solves is equal.

[mzuenni]

I would say considering this as a tie breaker is a good idea/better than just alphabetically (and its easy to add). IMO less WA should be considered easier?

[vmcj]

I would put the one with the most attempts (while easy) more to the middle to keep the attention available. Something which is solved often but not tried often (so almost ratio of 1) is interesting to tell in the beginning but people are waiting for the one they tried.

[mzuenni]

hmm I am a bit sceptical about that ^^'