Ken's POTW 050121

Fibonacci Dials

Can you create a set of dials that meet the following requirements?

There are six dials, each with four values. Each value is a unique positive integer (for a total of 24 unique values.)
The dials are arranged left to right. The first two dials spin freely. The other dials may only turn to a new value if the two previous dials sum to that value.
There are no unused values. There exists some way for each dial to turn to all four of its values, and the values on the first two dials can all be used in a sum. (The first dial's values can all be used with values from the second dial to obtain a sum on the third dial. The values on the second dial can be used to obtain a sum on either the third or fourth dial. For example, the first three dials could be: {1,2,3,4}; {5,6,7,8}; {9,10,11,12}. The {1,2,3,4} can be summed with {8,7,6,5} to obtain "9" on the third dial, and the {10,11,12} can all be obtained through other sums.)
The largest value (of all 24) is as small as possible.

Source: Original.

Solutions were received from Alan O'Donnell, Kirk Bresniker, Joseph DeVincentis, Dan Chirica. A valid, perfect solution, using the numbers 1-24 on the dials was found by Joseph DeVincentis. His analysis follows. He proposes that, theoretically, there should be a perfect solution for 7 dials, but not for more. Can someone find a perfect solution for 7 dials? Note: "Perfect" means that every number can be used in a sum, either as a summand, or as the result.

For 6 dials:
{2,5,7,12}, {1,9,10,11}, {3,6,8,13}, {4,14,15,16}, {17,18,19,20}, {21,22,23,24}

It is impossible to make a perfect solution for 8 or more dials.

The smallest number you can put on each dial (besides the first) is given by the following sequence:

2 1 3 4 7 11 18 29 47

Thus, the 9th dial must have numbers larger than 36.
The 8th dial can only form a perfect 8-dial solution if it contains 29, 30, 31, and 32, AND all the other numbers from the sequence are forced.

Now, note that 1 through 4 are already placed and no pair of adjacent dials adds to 5. So 5 must go on the 1st or 2nd dial. And 6 is also not formed by any pair, unless 5 goes on the 1st dial and 6 on the 3rd. The smallest remaining numbers are 8 and 9, and while it is possible to get either of these on the 4th dial by placing 5 or 6 on the 2nd, there is no way to get either of these on the 5th dial.

For the given forced values, the 2nd-smallest possible number on the 4th dial is 8 and the 2nd-smallest possible number on the 5th dial is 10. Therefore, the 2nd-smallest possible number on the 6th dial is 14 and the 2nd-smallest possible number on the 7th dial is 21. This means there is no way to make the sums 30 and 31 on the 8th dial.

So if a 6-dial solution is easy and an 8 is impossible, a perfect 7-dial solution should be just right! I have not found one yet, though having coming just one number short four times without much difficulty suggests that it should be possible.

Joe

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