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## This case will only happend if subset[j] has already contains partAs[i]
## if subset[j] does not have partBs[i] , add partBs[i] to subset[j]
## FIXME what if subset[j] already contains partBs[i]
if (subset[j][indexB] != partBs[i]):
subset[j][indexB] = partBs[i]
subset[j][-1] += 1
subset[j][-2] += candidate[partBs[i].astype(int), 2] + connection_all[k][i][2]
after i check the mapIdx, there will be a problem ,as joint 6 was put into the subset by [2,6] in the 2nd iteration, [16,18] will put 18 in the subset in the 17th iteration, however, [6,18] may put another 18 joint into subset at the 19th iteration. From my understanding, [6,18] and [16,18] connections are independent of each other, then, one subset may have two different joint 18.
i think it will be great if we can make sure each subset will not have more than 1 each joint for some application.
to do that, we can compare two different joint 18's score to decide which one to put into the subset.
how do you think about that ? and anyway, it is a great work, i have read the whole implementation, it is great. thank you.
The text was updated successfully, but these errors were encountered:
after i check the mapIdx, there will be a problem ,as joint 6 was put into the subset by [2,6] in the 2nd iteration, [16,18] will put 18 in the subset in the 17th iteration, however, [6,18] may put another 18 joint into subset at the 19th iteration. From my understanding, [6,18] and [16,18] connections are independent of each other, then, one subset may have two different joint 18.
i think it will be great if we can make sure each subset will not have more than 1 each joint for some application.
to do that, we can compare two different joint 18's score to decide which one to put into the subset.
how do you think about that ? and anyway, it is a great work, i have read the whole implementation, it is great. thank you.
The text was updated successfully, but these errors were encountered: