推理遊戲9 - 硬幣, Weighing again

Similar to the one posted before (http://hksan.net/forum/index.php?showtopic=15677)

有27枚硬幣, 26枚的重量是10g, 1枚的重量不同 (可以是9g 或 11g). 

用平衡秤, 最少要量多少次才能找出不同重量的一枚?  (要找出不同的一枚是較輕或較重)

26次

4次?

4次?

Just back home and here is my approach: (in white)

Similar to the one posted before:

1. split them into three equal piles (9-9-9: named as pile A,B,C; 1~9,10~18,19~27) and weigh two piles (let's say piles A,B) first. [first weighing]

2. if equal, then examine the remaining pile in the similar approach (so hereafter I will focus on most of the unbalanced scenarios, the case of balanced weight could be left as exercise for interested readers)

3. so assume left is lighter than right, then put 7~9,16~18 aside; weigh 1~3+10~12 on the left and 4~6+13~15 on the right [second weighing]

4. if left is still lighter, the odd is in 1~3,13~15; otherwise it is in 4~6,10~12;

5. then we have only 6 suspects here and two weighings left. 

6. assume the left is still lighter, so the odd is in 1~3,13~15; put 3,15 aside; weigh 1,14 on the left and 2,13 on the right [third weighing]

7. if it is balanced, then the odd one is either 3 or 15. 

if the left is still lighter, then the odd one is either 1 or 13. (either 1 lighter or 13 heavier)

if the balance shifts, then the odd one is either 2 or 14. (either 2 lighter or 14 heavier)

8. So, we have two coins left and one weighing left. Just weigh any one of them VS a normal one and see if it is heavier or lighter. [final weighing]

Any comments or corrections are most welcome!

26次

好吧，只要不是27次還算可以 

4次?

Just back home and here is my approach: (in white)

I had the same approach! (Even grouping them into A/B/C and numbering them from 1-27!)
One of the solutions was the easiest and most efficient but literally drove me nuts! 
Hint: Weigh twice to find out whether the odd one is lighter or heavier.

也許是54次吧，逐枚秤，而且要覆檢。

那要用27P2 =702才夠準確呀 

本篇文章已被 Pearltea 於 Jan 29 2016, 00:24 編輯過

4次?

Just back home and here is my approach: (in white)

I had the same approach! (Even grouping them into A/B/C and numbering them from 1-27!)
One of the solutions was the easiest and most efficient but literally drove me nuts! 
Hint: Weigh twice to find out whether the odd one is lighter or heavier.

OMG... is it weighing pile A VS pile B and then weighing pile A vs pile C...

Yes   (Smiley smile is invincible but not invisible)