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 The 5 jugs (Posted on 2005-08-18)
You have five different jugs, without any marks at all, with capacities of 3, 4, 5, 6 and 7 litres. Initially, the 3, 5 and 7-jugs are completely filled with liquid (so we have 15 litres of the liquid; in the diagram below, an "x" stands for 1 litre) and the other two are empty. The jugs are arranged in the circular manner shown below, which can't be changed:
```                           | |
+-+ +-+
| o o |
| o o |
+-----+
| |                        | |
+-+ +-+                  +---+ +---+
|  x  |                  | x x x x |
| x x |                  |  x x x  |
+-----+                  +---------+

| |           | |
+--+ +--+     +--+ +--+
| x x x |     | o o o |
|  x x  |     | o o o |
+-------+     +-------+```
The goal is to achieve 3 litres in each of the 5 jugs, only by pouring liquid from one jug into an adjacent one. (That is, at any time, you can pour liquid from the 3-jug only into the 4-jug or to the 5-jug; or from the 7-jug only into the 4-jug or to the 6-jug, etc...)

 See The Solution Submitted by pcbouhid Rating: 2.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: symetrical bottles | Comment 16 of 36 |
(In reply to symetrical bottles by Sir Percivale)

The hypothesis also assumes that the capacity includes the use of the neck of the bottle as part of the storage, so the capacity is reached only when filled to the actual rim.  Also, since pourings take place, one must replace (or initially place) that foil top when needed.  The caveat also is that the jug must be full at the given stage when this is used.

But to answer your question, only 5 steps are needed if it's possible to pour half the contents:

3    0    7    0    5
0    3    7    0    5
3    3    7    0    2
3    3    7    2    0
3    3    3    6    0
3    3    3    3    3

The half rule was applied only in the last step.

The program, with the half rule feature in bold:

DECLARE SUB try (lvl!)
DIM SHARED amt(3 TO 7)
DIM SHARED hist(15, 3 TO 7), ansCt, best

OPEN "fivejug2.txt" FOR OUTPUT AS #2

amt(3) = 3
amt(5) = 5
amt(7) = 7
best = 999

try 1

CLOSE

END

SUB try (lvl)
good = 1
FOR jug = 3 TO 7
hist(lvl, jug) = amt(jug)
IF amt(jug) <> 3 THEN good = 0
NEXT
IF good THEN
ansCt = ansCt + 1
IF lvl < best THEN best = lvl
FOR i = 1 TO lvl
FOR j = 3 TO 7
j2 = j
IF j = 5 THEN j2 = 7
IF j = 7 THEN j2 = 5
PRINT TAB((j - 2) * 5); hist(i, j2);
PRINT #2, TAB((j - 2) * 5); hist(i, j2);
NEXT
IF i = 1 THEN
PRINT ansCt, lvl, best
PRINT #2, ansCt, lvl
ELSE
PRINT
PRINT #2,
END IF
NEXT
PRINT
PRINT #2,
ELSE
IF lvl < 6 THEN
FOR j1 = 3 TO 7
IF j1 < 7 THEN j2 = j1 + 1:  ELSE j2 = 4
IF j1 = 4 THEN j2 = 7
IF amt(j1) > 0 AND amt(j2) < j2 THEN
IF amt(j1) > j2 - amt(j2) THEN
trfr = j2 - amt(j2)
ELSE
trfr = amt(j1)
END IF
amt(j2) = amt(j2) + trfr
amt(j1) = amt(j1) - trfr
try lvl + 1
amt(j2) = amt(j2) - trfr
amt(j1) = amt(j1) + trfr
IF amt(j1) = j1 AND j1 / 2 < j2 - amt(j2) THEN
amt(j2) = amt(j2) + j1 / 2
amt(j1) = amt(j1) - j1 / 2
try lvl + 1
amt(j2) = amt(j2) - j1 / 2
amt(j1) = amt(j1) + j1 / 2
END IF
END IF
IF j1 > 3 THEN j2 = j1 - 1:  ELSE j2 = 5
IF j1 = 5 THEN j2 = 3
IF amt(j1) > 0 AND amt(j2) < j2 THEN
IF amt(j1) > j2 - amt(j2) THEN
trfr = j2 - amt(j2)
ELSE
trfr = amt(j1)
END IF
amt(j2) = amt(j2) + trfr
amt(j1) = amt(j1) - trfr
try lvl + 1
amt(j2) = amt(j2) - trfr
amt(j1) = amt(j1) + trfr
IF amt(j1) = j1 AND j1 / 2 < j2 - amt(j2) THEN
amt(j2) = amt(j2) + j1 / 2
amt(j1) = amt(j1) - j1 / 2
try lvl + 1
amt(j2) = amt(j2) - j1 / 2
amt(j1) = amt(j1) + j1 / 2
END IF
END IF
NEXT
END IF
END IF
END SUB

 Posted by Charlie on 2005-08-19 14:59:59

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