When I went to the race track in Racing Town, a town made up only of Knights which always tell the truth, Knaves which tell truths and lies in an alternating pattern, and Liars which always lie, a race between 6 citizens of that town had just finished.
I went to the 6 citizens and asked each of them the order that all 6 finished. They all gave me different responses, each thinking themselves as winning, displayed here left to right as first to last.
A: A C D E B F
B: B D F E C A
C: C D E F A B
D: D E F B A C
E: E B A D F C
F: F C B A E D
From what they said, I was able to figure out what the correct order was. What is it?
The answer is also provided by running:
DECLARE FUNCTION eval$ (s1$, s2$)
DATA ACDEBF
DATA BDFECA
DATA CDEFAB
DATA DEFBAC
DATA EBADFC
DATA FCBAED
CLS
FOR i = 1 TO 6
READ order$(i)
NEXT
FOR winner = 1 TO 6
good = 1
FOR speaker = 1 TO 6 ' evaluate if winner is knight
id$ = eval$(order$(speaker), order$(winner))
IF id$ = "-" THEN good = 0: EXIT FOR
NEXT speaker
IF good THEN PRINT order$(winner)
' evaluate if winner is knave
var$ = MID$(order$(winner), 2, 1) + MID$(order$(winner), 4, 1) + MID$(order$(winner), 6, 1)
winsays$ = order$(winner)
FOR perm = 1 TO 2
var$ = MID$(var$, 2) + LEFT$(var$, 1)
FOR psn = 1 TO 3
MID$(winsays$, 2 * psn, 1) = MID$(var$, psn, 1)
NEXT psn
good = 1
FOR speaker = 1 TO 6
id$ = eval$(order$(speaker), winsays$)
IF id$ = "-" THEN good = 0: EXIT FOR
NEXT speaker
IF good THEN PRINT winsays$
NEXT perm
NEXT winner
FUNCTION eval$ (s1$, s2$)
e$ = "-": evenCt = 0: oddCt = 0
FOR i = 1 TO 3
IF MID$(s1$, 2 * i - 1, 1) = MID$(s2$, 2 * i - 1, 1) THEN oddCt = oddCt + 1
IF MID$(s1$, 2 * i, 1) = MID$(s2$, 2 * i, 1) THEN evenCt = evenCt + 1
NEXT i
IF evenCt = 3 THEN
IF oddCt = 3 THEN
e$ = "knight"
ELSEIF oddCt = 0 THEN
e$ = "knave"
END IF
ELSEIF evenCt = 0 THEN
IF oddCt = 3 THEN
e$ = "knave"
ELSEIF oddCt = 0 THEN
e$ = "liar"
END IF
END IF
eval$ = e$
END FUNCTION
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Posted by Charlie
on 2003-12-15 09:28:25 |
... and the computer way.
