天平平分真假币--This one is good[:-M][:-M] - 珍珠湾ART

有32枚外观完全一样的金币，其中2枚是假币，另外30枚是真币。两者的重量都不知，只知真币重量都相同，假币重量也相同，真币与假币的重量则不同，也不知真币较重还是假币较重。现有一架无砝码的天平，它只能测量置于左右两托盘中的物品孰重孰轻或者两者相等。限称四次，请问如何才能把这32枚金币分成重量相等的两堆。

将金币分成4组, 每组8枚: A, B, C, Dwww.ddhw.com

称A, B
如 A = B
将C, D 分成4组, 每组4枚: C1, C2, D1, D2
(问题是：如果A≠B怎么办，如果只有一枚假币在A或B中另外一枚假币在C或D中，那么如何可以称四次就可以把32枚硬币分出两份重量相等的两堆呢？)www.ddhw.com

称C1, D1
如 C1 = D1
将C2, D2 分成4组, 每组2枚: C21, C22, D21, D22www.ddhw.com

称C21, D21
如 C21 = D21
将C22, D22 分成4组, 每组1枚: C221, C222, D221, D222www.ddhw.com

称C221, D221
如 C221 = D221
将A,C放在一堆, B,D放在一堆www.ddhw.com

如 C221<>D221
那么一定C222<>D222
将C221,D221放在一堆, C222,D222放在一堆, 再用其他补足

Label each coin using binary digits, from 00000 to 11111. Put all the coins with first digit = 0 on one side of the scale, and remaining coins on the other, if get equal weight, done, otherwise, repeat the process with the second digit...

If all the first 4 trials fail, then we know that the two fake coins differ in the last digit, and the 5th division according to the last digit must be successful, and we don't have to weigh them.

将32等分为4组，A1~8, B1~8, C1~8, D1~8www.ddhw.com

www.ddhw.com

1、称A组与B组，若相等www.ddhw.com

2、称A+C与B+D，若相等，问题已解；若不相等，则假币必然同在C组或D组当中www.ddhw.com

3、称（C1234+D1234）与（C5678+D5678），若相等，问题已解；若不相等，(注：C1234代表C1,C2,C3,C4四枚币）www.ddhw.com

4、称（C1256+D1256）与（C3478+D3478），若相等，问题已解；若不相等，则题解为：（C1357+D1357+A组）与（C2468+D2468+B组）www.ddhw.com

(问题是：假设C1与D1就是假币的话，那么到最后你的答案是（C1357+D1357+A组）与（C2468+D2468+B组）那么你还是把C1与D1两个假币放在同一组，两个假币放在同一组的话两堆金币的重量肯定不相等)www.ddhw.com

www.ddhw.com

1、称A组与B组，若不相等www.ddhw.com

2、称A+B与C+D，若相等，问题已解；若不相等，则假币必然同在A组或B组当中www.ddhw.com

3、称（A1234+B1234）与（A5678+B5678），若相等，问题已解；若不相等，www.ddhw.com

4、称（A1256+B1256）与（A3478+B3478），若相等，问题已解；若不相等，则题解为：（A1357+B1357+C组）与（A24688+B2468+D组）

该问题的关键是要把两枚假币分到两堆金币中，一堆一个。明白这一点，问题就简单了。

第一次分，每边16个，若相等，你很lucky，假币每边一个，不用再量了。若不等，两个假币必在同一盘中，但是不知是轻还是重。

任意取轻的或者重的一边，先假定重的一边进行第二次分，分成8个，若相等，再量轻的一边（第三次称量），若也相等，你很幸运，虽不知假币是轻是重，中要把第二、三两次的左边放作一堆，右边作放一堆，也成了。

若第二次称量，左右不等，则假币重于真币。取重的一侧量第三次每边4个，若相等，则假币必定左右每边一个，只要把与其他的等分与左右两边就行。第三次若不等，取重侧分每边两个量第四次，若相等，则假币必定左右每边一个。若不等，两假币必在重盘中。

同样也适于假币情于真币的情况。

不知说清楚了没有，呵呵！

First of all, divide all these 32 coins into four groups, say A1, A2,
A3, A4, each group has 8 coins. Bottom line is, the two bad coins can
either be in two separate groups, or in a same group. And we will find
it out.

I believe the key is that you have to remember which side is heavier
across different weighs.

Weigh A2 and A3:
If A2 == A3 then: (1st time)

It is possible that one of each bad coins is in one of these two
groups, or none of the bad coins is in any of these two groups.

Then weigh A1 and A4: (2nd time)

if A1 == A4 :
perfect, things are figured out. A1 + A2 == A3 + A4 (FIGURED OUT
AT 2ND WEIGHS)www.ddhw.com

MAIN PROCEDURE:
if A1 > A4:
     then, both bad coins are in either A1 or A4:
     Now split A1 and A4 into four sub-gorups (A11, A12, A41 and A42),
     each group has 4 coins. Now weigh A42 + A12 against A41 + A11 (3rd time),

if (A42 + A12 == A41 + A11):
(perfect, FIGURED OUT AT 2ND WEIGHS)www.ddhw.com

          if (A42 + A12 > A41 + A11):
             then the two coins must be in either A12 or A41, because
             by switching the place of A11 and A42 didn't change the
             direction of the inequality, therefore neither A11 nor A42 has
             any bad coins.www.ddhw.com

             Now split A12 and A41 into four smaller groups (A121,
             A122, A411, A412), each has two coins. Now weigh A412 +
             A122 against A411 + A121:(4th time)
                     if (A412 + A122 > A411 + A121):
                        then the bad coins are both in either A122 or
                        A411 (the direction of inequality doesn't
                        change). Because now A122 and A411 only have
                        two coins each, simply split these two groups
                        in half and add the filtered out groups into
                        each half, you will get the result. (FIGURED
                        OUT AT 4th WEIGHS)www.ddhw.com

                     if (A412 + A122 == A411 + A121):
                        then problem solved (FIGURED OUT AT 4th
                        WEIGHS)
                     if (A412 + A122 < A411 and A121):
                        same as the first case, just now either the
                        A412 or A121 has both bad coins, now split
                        these two groups and add the filtered out
                        groups into each half. (FIGURED OUT AT 4th
                        WEIGHS)www.ddhw.com

           if (A42 + A12 < A41 + A11):
             then the two coins must be in either A42 or A11, follow
             the above procedures.www.ddhw.com

If A2 > A3 then:

   At this point, there are two possibilities:

   1. both bad coins can be in either A2 or A3
   2. one in A2 or A3 and one in A1 or A4

Weigh A1 and A4:www.ddhw.com

         if (A1 > A4):
            Then you know the following possibilities:
            1. both A1 and A2 have one bad coin;
            2. or both A3 and A4 have one bad coin;
            Anyways, A1 + A3 == A2 + A4, problem solved (FIGURED OUT
            AT 2nd WEIGHS)www.ddhw.com

         if (A1 < A4):
            Then you know the following possibilities:
            1. both A1 and A3 have one bad coin;
            2. or both A2 and A4 have one bad coin;
            Anyways, A1 + A2 == A3 + A4, problem solved (2FIGURED OUT
            AT 2nd WEIGHS)

         if (A1 == A4):
            Then you know for sure both bad coins will be either A2 or
            A3. Go to the text labled "MAIN PROCEDURE" and apply the
            same process on A2 and A3,

每次分为4组，每组相同数目，任取其中两组，称之：

如果等重，则将另两组合在一起，再分成4组。。。

如果不等，就将这两组喝在一起，再分成4组。。。

如此4次，即得。其中原由，各位可自己考证

上面的答案都不正确，我觉得这个题没有答案。