In the famous
"The Odd Coin" problem you are given twelve coins, exactly one of which is lighter or heavier than the other coins. You are to determine the counterfeit coin, and whether it is lighter or heavier than the other coins, in just three weighings with a balance.
Can you solve this problem with the additional restriction that you must decide what coins go on each pan for all three weighings before any weighing takes place?
(In reply to
Another way to answer it by Jennifer)
You're not assured that the odd one is lighter. The odd one might be heavier.
Here's a table of the results of the weighings Ken Haley gives in his solution:
1 light rlr
2 light r==
3 light rrl
4 light rr=
5 light lrr
6 light lr=
7 light l=r
8 light l=l
9 light =lr
10 light =l=
11 light =ll
12 light ==l
1 heavy lrl
2 heavy l==
3 heavy llr
4 heavy ll=
5 heavy rll
6 heavy rl=
7 heavy r=l
8 heavy r=r
9 heavy =rl
10 heavy =r=
11 heavy =rr
12 heavy ==r
where l means the left pan goes down; r, the right; and = both pans balance.
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Posted by Charlie
on 2005-06-05 17:43:52 |
re: Another way to answer it
