Ah, thanks. In that case… wouldn’t the meet of A and B often end up being the entire space?
For that matter, why this coarsening operation rather than the set of all the possible pairwise intersections between members of I and members of J?
ie, why coarsening instead if “fineing” (what’s the appropriate word there anyways?)
When two rationalists exchange information, shouldn’t their conclusions then sometimes be finer rather than coarser since they have, well, each gained information they didn’t have previously?
When two rationalists exchange all information, their new partition is the ‘join’ of the two old partitions, where the join is the “coarsest common fining”. If you plot omega as the rectangle with corners at (-1,-1) and (1,1) and the initial partitions are the x axis for agent A and the Y axis for agent B, then they share information and ‘join’ and then their common partition separates all 4 quadrants.
“common knowledge” is the set of questions that they can both answer before sharing information. This is the ‘meet’ which is the coarsest common fining. In the previous example, there is no information that they both share, so the meet becomes the whole quadrant.
If you extend omega down to y = −2 and modify the original partitions to both fence off this new piece on its own, then the join would be the original four squares plus this lower rectangle, while the meet would be the square from (-1,1) to (1,1) plus this lower rectangle (since they now have this as common knowledge).
wait, what? is it coarsest common fining or finest common coarsening that we’re interested in here?
And isn’t common knowledge the set of questions that not only they can both answer, but that they both know that both can answer, and both know that both know, etc etc etc?
Actually, maybe I need to reread this a bit more, but now am more confused.
Actually, on rereading, I think I’m starting to get the idea about meet and common knowledge (given that before exchanging info, they do know each other’s partitioning, but not which particular partition the other has observed to be the current one).
Ah, thanks. In that case… wouldn’t the meet of A and B often end up being the entire space?
For that matter, why this coarsening operation rather than the set of all the possible pairwise intersections between members of I and members of J?
ie, why coarsening instead if “fineing” (what’s the appropriate word there anyways?)
When two rationalists exchange information, shouldn’t their conclusions then sometimes be finer rather than coarser since they have, well, each gained information they didn’t have previously?
If I’ve got this right...
When two rationalists exchange all information, their new partition is the ‘join’ of the two old partitions, where the join is the “coarsest common fining”. If you plot omega as the rectangle with corners at (-1,-1) and (1,1) and the initial partitions are the x axis for agent A and the Y axis for agent B, then they share information and ‘join’ and then their common partition separates all 4 quadrants.
“common knowledge” is the set of questions that they can both answer before sharing information. This is the ‘meet’ which is the coarsest common fining. In the previous example, there is no information that they both share, so the meet becomes the whole quadrant.
If you extend omega down to y = −2 and modify the original partitions to both fence off this new piece on its own, then the join would be the original four squares plus this lower rectangle, while the meet would be the square from (-1,1) to (1,1) plus this lower rectangle (since they now have this as common knowledge).
Does this help?
wait, what? is it coarsest common fining or finest common coarsening that we’re interested in here?
And isn’t common knowledge the set of questions that not only they can both answer, but that they both know that both can answer, and both know that both know, etc etc etc?
Actually, maybe I need to reread this a bit more, but now am more confused.
Actually, on rereading, I think I’m starting to get the idea about meet and common knowledge (given that before exchanging info, they do know each other’s partitioning, but not which particular partition the other has observed to be the current one).
Thanks!