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Ghost Post
Icarian Member

 Posted: Fri Mar 16, 2001 6:29 am    Post subject: 1 Take 1 bar from Room 1, 2 from Room 2, 3 from Room 3, 5 from Room 4, 8 from Room 5, 13 from Room 6 and 21 from Room 7 (53 bars total). NB: This is the Fibonacci sequence. Let W = total weight of the bars, and Wm = 477 (the minimum weight 53 * 9) W - Wm = Wx, or the Excess Weight. Solve for Wx such that Wx = the sum of 4 of the terms 1,2,3,5,8,13,21. There is only one answer. E.g. if Wx = 35, we determine by iterative search that 1,5,8 & 21 are the four terms. You can generate a number in "Fibonarry" - like binary, but with each digit representing a term (R -> L) in the Fibonacci sequence instead of the function x = 2^n e.g. as follows: (21) (13) ( 8) ( 5) ( 3) ( 2) ( 1) 1011001 This number shows us that in this case Rooms 1, 4, 5 & 7 contain gold bars, whereas Rooms 2, 3 & 6 contain fake bars. With one weighing of 53 bars we have reached the most efficient solution. The reason why this is the most economical answer is because the Fibonacci sequence is the lowest exponential sequence composed of whole numbers. An exponential sequence is what we need to use here to avoid the addition problem inherent in arithmetic sequences. Because F(x) = F(x-1) + F(x-2), you don't have the problem as in the arithmetic sequence (1,2,3,4,5,6,7) where for instance the 4th & 5th terms are equal to the 3rd & 6th terms. As others have suggested, the binary sequence would work just fine but it is not quite as efficient. The Fibonacci sequence has the special property of being both an arithmetic sequence and an exponential sequence (x to the power of the Golden Ratio, or approx. x^1.61, plus an error term). Lucky for us it kicks out whole numbers! Don't know if I explained that well, but there it is! Regards, Joe P.S. Would you kindly forward my gold bar reward by air mail - I could use it!
mwf
Daedalian Member

 Posted: Fri Mar 16, 2001 2:55 pm    Post subject: 2 Sorry, we could not ship you your bar. After we took the corect number of bars from each of the vault and weighed them. We came out 24K short. Now the mathmaticans ran into a roadblock. Do they use 21+2+1 or 13+8+3 to get 24? This should have been posted under the thread "Solid gold - not adequate for real life situations"
Ghost Post
Icarian Member

 Posted: Sat Mar 17, 2001 12:08 am    Post subject: 3 The solution is to weigh together the bars in the following quantities: Vault Bars 1 0 2 1 3 2 4 4 5 8 6 16 7 32 For a total of 63 bars. This combination with give a unique weight value, where the weight difference can be used to determine which vaults have fake bars and which have real ones. It also minimizes the number of bars being weighed.
QuikSand
Daedalian Member

Posted: Mon Mar 19, 2001 10:32 pm    Post subject: 4

 Quote: This combination with give a unique weight value, where the weight difference can be used to determine which vaults have fake bars and which have real ones. It also minimizes the number of bars being weighed.

Right on the first count, wrong on the second. A quick look through the long threads on this topic will reveal a solution with fewer bricks.

Using the powers of two is the lowest number of bricks that will yield singular answers with any number of vaults full of fakes. However, since we know that there are exactly three vauilts with fakes-- we can use that information to further contract the number of bricks used.

The total number needed is a fair bit below 63...
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