Affirmative, sir. Promote me up a rank if mission successful.

I offer this new and improved solution, I hope it’s also a final solution, if so, I’m indebted to all your encouragement.

Weighting 1:

Split into 3 equal groups and weight any 2 groups.

If the result is balance, then the odd ball must be in the unweighted group. Pick 2 balls out of that group and weight them.

If the result is balance, then the odd ball must be among the other 2 balls. Weight those 2 balls and we’re done.

However, if the result is inbalance, then the odd ball must be among the 2 balls. Weight any of these 2 balls against any of the other 2 balls and we’re done.

Weighting 2:

Proceed into this stage if the result of the very first weighting is inbalance.

The table below describes the current nasty situation.

Heavygroup…Lightgroup…Auxiliarygroup

H1…L1…A1

H2…L2…A2

H3…L3…A3

H4…L4…A4

Swope H3 and H4 with A3 and A4, thus we formed Newauxiliarygroup containing A1, A2, H3 and H4. Weight Lightgroup against Newauxilarygroup.

a. The result of this weighting can be balanced. If so, it implies that H3 and H4 are innocent balls. It follows that the odd ball must be among H1 and H2, also, it must be heavy, otherwise Heavygroup wouldn’t be heavier than Lightgroup in the first place. So weight H1 and H2, vola.

b. The result of this weighting can also be Lightgroup < Newauxiliarygroup.

c. The result of this weighting can’t possibly be Lightgroup > Newauxiliarygroup.

So follow situation b, which would imply that the odd ball is among H3 and H4, also, it is heavier. Weight H3 and H4 and fill in arendt’s blank.

Now tell us, Ed3, shall we screw the cork out?

One last thing, a thing which could potentially send Congressional Medal Of Honor to my mother. If you choose to form your Newauxiliarygroup by selecting any 2 balls from Lightgroup, then repeat the steps with the same logics mutatis mutandis.