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Re: A Weighty ball problem?
Now I've had time to take lunch, and write it up here's a solution
I think It's more or less what Danielf had, but with more detail.
STEP 1
Weigh 2 v 2
Possible results
A) pairs are unequal (heavier or lighter not known)
B) pairs are equal
Thus we have 4 balls known to be "good" and 4 balls still "suspect".
STEP 2)
Weigh 2 "suspect" balls v 2 "known good" balls
Possible results
C) pairs are unequal (heavier or lighter is now known)
D) pairs are equal (discard these)
STEP 3C
Weigh one of the "suspect" balls tested in STEP 2 against a "known good" ball
Possible Results
E) Balance equal (the other suspect is guilty)
F) Balance unequal (the chosen suspect is guilty)
STEP 3D
Weigh one of the "suspect" balls not tested in STEP 2 against a "known good" ball
G) Balance equal (the other suspect is guilty)
H) Balance unequal (the chosen suspect is guilty)
NOTE
In G) it is impossible to tell if the guilty ball is heaver or lighter without a further weighing
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