Wrong, you overcounted.

送交者: james_hussein_bond 于 2008-04-30, 10:41:50:

回答: 昨天那个3个BOOLEANS组合的问题我用集合论解决了由 Enlighten 于 2008-04-30, 10:30:51:

I'm not sure what the rules are in your original question. But if according to your earlier post, two BOO's give 4 combinations (A, B, A&B, A|B), then three cannot give 128.

所有跟贴:

2 booleans should be 8 combinations if NOT is allowed (无内容) - Enlighten (0 bytes) 2008-04-30, 10:44:47 (222498)
- Change the question to fit the answer? :-) (无内容) - james_hussein_bond (0 bytes) 2008-04-30, 10:46:43 (222500)
  - the question wasnt well defined. now it is. big deal. (无内容) - Enlighten (0 bytes) 2008-04-30, 10:49:34 (222502)
    - You're still wrong. if NOT is allowed, there are at least 12 for 2 - james_hussein_bond (161 bytes) 2008-04-30, 10:54:38 (222507)
      - forgot one constraint: A+B+C=full set (无内容) - Enlighten (0 bytes) 2008-04-30, 10:57:07 (222510)
        
        What is a "full set"? Can you give a definition? (无内容) - james_hussein_bond (0 bytes) 2008-04-30, 11:04:32 (222515)
- might need to minus 1 from the solution, nonetheless - Enlighten (16 bytes) 2008-04-30, 10:46:16 (222499)

加跟贴

everything must be in A or B or C. (无内容) - Enlighten (0 bytes) 2008-04-30, 11:08:03 (222516)

What does that mean? What is "everything"? (无内容) - james_hussein_bond (0 bytes) 2008-04-30, 11:11:50 (222519)
- !A&!B&!C=null,empty, zip, zero, nada... (无内容) - Enlighten (0 bytes) 2008-04-30, 11:15:26 (222522)
  - Do you mean to remove !A, !B, !(A|B)? But what about !(A&B)? - james_hussein_bond (55 bytes) 2008-04-30, 11:29:20 (222529)