PureBasic Forums - English

Trond wrote:

Logic proceeds this way, if a room's sign contains fluctuating truthness (i patented that term ) then the room must be empty. This applies to the group of rooms {2, 5, 6}, {1, 4}, and {3, 8}.

You can logically prove that there must be a tiger in room 3 and 8. Thus your assumption that these rooms are empty can not be right.

• 8. This room contains a tiger and sign 3 is false. == If (#8 contains tiger) and (#3 is false) then #8 is true, else #8 is false.
  3. Room 2 and 7 are not empty. == If (#2 <> empty) and (#7 <> empty) then #3 is true, else #3 is false.

• If #8 is true the room contains a tiger -> if #8 contains a tiger it has to be False -> if #8 is False it cannot contain a tiger or sign #3 is True ( via !(A and B) = (!A or !B) ).
  
  If #3 is True then room #2 is empty, but we know that room #2's sign is both True and False ( and thus contains neither the princess (True) nor the tiger (False) ) -> if #3 is False then either #2 is empty or #7 is empty ( via !(C and D) = (!C or !D) ) -> Room #3 sign is False because #7 contains the princess AND #2 contains nothing.
  
  Since #3 is False, none of the conditions on the sign for #8 are True and #8 is thus completely False -> because #8 is completely False it must contain either a tiger or sign #3 is True -> #8 is empty because it cannot contain a tiger and also because it cannot be True and False, regarding sign #3, at the same time (thus no princess and no tiger).

s3 = sign #3, s8 = sign #8, c2 = contents of #2, c7 = contents of #7,  c8 = contents of #8

sign 3:        
                A                 B
         If (c2 = #Empty) And (c7 = #Empty)
           s3 = #True
         Else
           s3 = #False
         Endif
        
        
      Condition B is True because the princess is in room #7.
      Condition A fluctuates between True ane False.
     
                    s3
         A      B   result     
         --     --  ------
      1. T  And T   TRUE 
      3. F  And T   FALSE
      
      s3 fluctuates between True and False which means that room #3 contains nothing.
      

sign 8:       
                C                 D
         If (c8 = #Tiger) And (s3 = #False)
           s8 = #True
         Else
           s8 = #False
         Endif      

      If Condition C = #True then s8 = #False and either C or D must be False.
      C cannot be False and True at the same time.
      D is a fluctuating condition and cannot be False and True all the time.       
                    s8
         C      D   result     
         --     --  ------
      1. F  And T   FALSE
      2. F  And F   FALSE
      3. T  And T   TRUE    
      4. T  And F   TRUE
         
      
      s8 fluctuates between True and False which means that room #8 contains nothing and results in condition C always being False.

blueznl wrote:You're right, my coding for rule 8 is wrong. I will change it and try again. It should be something like:

Code: Select all

if room(8) <> #tiger or room(3) = #tiger
  found = #false
endif

... but I will check later.

if room(8) <> #tiger or room(3) = #tiger
  found = #false
endif

Given: If a room contains a tiger, it's sign must always be true; if a room contains the princess it's sign must always be false.

Trond wrote:

Arctic Fox wrote:

Trond wrote:H. If the truth value of sign 2 is true, the truth value of sign 6 must be true, but sign 6 can only be true if sign 2 is false. It thus leads that 2, 5 and 6 cannot be true, they must be false.

If sign 6 is false, and it says that sign 2 is false (which is a false statement), then sign 2 must be true? And then sign 6 must be true, too - I am confused

Yes, an unresolvable conflict occurs if sign 2, 5 or 6 is true. So we know all of them must be false. Because we can't have any unresolvable conflicts in our logic.

RASHAD wrote: So Sign 2 is False and Sign 7 is True

Thorium wrote:

Trond wrote:

Arctic Fox wrote:

Trond wrote:H. If the truth value of sign 2 is true, the truth value of sign 6 must be true, but sign 6 can only be true if sign 2 is false. It thus leads that 2, 5 and 6 cannot be true, they must be false.

If sign 6 is false, and it says that sign 2 is false (which is a false statement), then sign 2 must be true? And then sign 6 must be true, too - I am confused

Yes, an unresolvable conflict occurs if sign 2, 5 or 6 is true. So we know all of them must be false. Because we can't have any unresolvable conflicts in our logic.

That cant be. If sign 2 is false the princess is in room 2, because the sign sais so. But that cant be because the sign of the princess have to be true.

RASHAD wrote:The Sign of an empty Room can be True or False but still empty
Sign 7 is True The Princess must be in an odd No. Room (Room 3) but Room 7 still empty

Trond wrote: If sign 5 is false and the princess is not in room 2, sign 2 will be false, but the princess won't be in there either.

Thorium wrote:

Trond wrote: If sign 5 is false and the princess is not in room 2, sign 2 will be false, but the princess won't be in there either.

Thats unlogical. So the first part of the sign is false and the second is true? That makes no sense.

Little John wrote:It looks to me that here is some disagreement concerning the truth table of a logical conjunction.