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It's a good idea, but as I said, freak's brute forced solutions are wrong.blueznl wrote:Gotta' love brute force
Yes, i misread the sign 8 when i did my code. If i correct it however, i get no solutions at all. Here is why:Trond wrote:Actually, I think the reason my answers are different from yours, is that some of yours are wrong. It's impossible to do without a tiger.freak wrote:Note that the whole thing works without any tigers at all. Good for the king in case there is a tiger shortage in the country
See:
0. Assume we know the princess is in room 7
A. Sign 2 is true if sign 5 is true
B. Sign 2 is false if sign 5 is false
C. The truth value of sign 2 and 5 are equal.
D. Sign 5 is true if sign 6 is true.
E. Sign 5 is false if sign 6 is false.
F. The truth value of sign 5 and 6 are equal.
G. The truth value of sign 2, 5 and 6 are all equal.
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.
I. For sign 6 to be false when sign 2 is false, the statement "room 8 is empty" must be false.
J. If the statement "room 8 is empty" is false, something must be in room 8.
K. If the princess is in room 7, she cannot be in room 8.
L. There must be a tiger in room 8.
brute force solution before mine, but here's mine, and it returns room 7.Code: Select all
Enumeration
#princess
#tiger
#empty
EndEnumeration
;
Dim rule(8)
Dim room(8)
;
For princess = 1 To 8
;
; fill up the rooms
;
For tiger = 0 To 255
For n = 1 To 8
If princess = n
room(n) = #princess
rule(n) = #True
ElseIf tiger & (1 << (n-1) )
room(n) = #tiger
rule(n) = #False
Else
room(n) = #empty
rule(n) = #True
EndIf
Next n
;
; now check if the rules are valid
;
found = #True
;
; 1. Room 4 is Not empty.
;
If room(1) = #tiger
If room(4) <> #empty
found = #False
EndIf
Else
If room(4) = #empty
found = #False
EndIf
EndIf
;
; 2. Sign 5 is true And the princess is Not in this room
;
If room(2) = #princess
found = #False
EndIf
If room(2) = #tiger
If room(5) <> #tiger And room(2) <> #princess
found = #False
EndIf
EndIf
;
; 3. Room 2 And 7 are Not empty.
;
If room(3) = #tiger
If room(2) <> #empty And room(7) <> #empty
found = #False
EndIf
Else
If room(2) = #empty Or room(7) = #empty
found = #False
EndIf
EndIf
;
; 4. Room 1 is Not empty.
;
If room(4) = #tiger
If room(1) <> #empty
found = #False
EndIf
Else
If room(1) = #empty
found = #False
EndIf
EndIf
;
; 6. Sign 2 is false And room 8 is empty.
;
If room(6) = #tiger
If room(2) <> #True And room(8) = #empty
found = #False
EndIf
Else
If room(2) <> #tiger
found = #False
EndIf
If room(8) <> #empty
found = #False
EndIf
EndIf
;
; 5. Sign 6 is true.
;
If room(5) = #tiger And room(6) = #tiger
ElseIf room(5) <> #tiger And room(6) <> #tiger
Else
found = #False
EndIf
;
; 7. The princess is in a room With an odd number.
;
If room(7) = #tiger
If Mod(princess,2) = 1
found = #False
EndIf
Else
If Mod(princess,2) = 1
Else
found = #False
EndIf
EndIf
;
; 8. This room contains a tiger And sign 3 is false.
;
If room(8) <> #tiger
found = #False
EndIf
;
; report result
;
; s.s = ""
; For n = 1 To 8
; If room(n) = #tiger
; s = "1"+s
; ElseIf room(n) = #princess
; s = "x"+s
; Else
; s = "0"+s
; EndIf
; Next n
; Debug s+" "+Str(tiger)
;
If found
Debug "princess in room "+Str(princess)
For n = 1 To 8
If room(n) = #tiger
Debug " tiger in room "+Str(n)
EndIf
Next n
EndIf
;
Next tiger
;
Next princess
;
) then the room must be empty. This applies to the group of rooms {2, 5, 6}, {1, 4}, and {3, 8}. Note if a sign mentions a conjunctive condition ('and') all conditions must be true or the sign is false. This leaves only room seven for the princess and its sign says the princess is in an odd numbered room, and thus is true.
)
Code: Select all
; ... more wrong code ...blueznl wrote:Well, who says there's a difference between the princess and the tiger?
No, think again.Arctic Fox wrote:@Trond
The sign of room 8 says that this room contains a tiger. But if it contains a tiger the sign must be false. And thus it cannot contain a tiger which the sign claims, right?
Nope, the signs on empty rooms can be either true or false.A. If the princess is in room 7, then all other signs must be false.
Yup!Damn, i see my error: I overread the part where empty rooms mean either true or false. Well...
Your code prints several solution, of which many are wrong, so there must be a flaw in your logic.Frankly, I tried my solution and it seems to qualify, could anyone find a flaw in the logic?
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.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}.
Code: Select all
if room(8) <> #tiger or room(3) = #tiger
found = #false
endif
