SudokuSolver Forum

Richard
Correction
I hadn't considered that the longest rod might not have the most shafts. So starting with 56 for the longest of 6 rods can't be conjectured either. Even worse whereas my 2 solutions had total lengths 25 yours have (2x25,26,2x27) a pleasing symmetry so perhaps that is all there is.

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I have 81 brain cells left, I think.

Not being a mathematician (I was a Physics student) I don't know whether there are any special properties of Tetrahedral number that can be employed here.

[Edit]Glyn,

There might be something in the symmetry, given that these sequences (4,10,20,35,56 for the required measurement and 2, 6, 10, 15, 21 the number of shafts) both can be found in Pascal's triangle . . .

You guys are brilliant! Keep pressing on!

Glyn, the longest rod is definitely 56 units, because you have to use the longest rod as a whole to measure something (after all you only have 56 ways to use the rods), and the longest distance to measure is 56 units. However you're right that the longest rod does not necessary have 6 shafts. It can theoretically even be the single-shaft-rod.

Just some more insights: of the 21 shafts, two of them must be 1 & 2 respectively. And one or both of {3,4} must also be among the shafts. So as one or more from {5,6,7,8}. (What happens if it goes 1,2,3,4,9,...?) I wonder if it helps a lot, or even if one knows the exact lengths of the 21 shafts would it necessarily become much easier...

I think the solution is unique but am not 100% sure... Googling surely doesn't help much here...

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From Richards work I would guess there are multiple solutions. The optimal in total length is probably biased towards a highly composite longest rod. In the 3 rod problem the total length was 25 with a 3-sectioned rod of length 10, a shaft of length 10 or 9 led to the greatest length solutions.

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I have 81 brain cells left, I think.

Interesting suggestions.

My initial thought was that the longest rod would indeed be 56 long I also thought that this would be split with a 28 rod and then composites to make up the other 28 - with the 28 long rod being embedded somewhere in the middle.

From what I've seen of the solutions for simpler case, this does not necessarily seem to mean this will be the longest rod.

Given that one of the 10/6 solutions was a rod of length 10, this does lend some sympathy to a single length of 56 being one of the rods, altought that might not be a unique solution.

Furrther, I think Glyn's estimate of the number of combinations is optmistic. As well as choosing the combination of differetn shaft lengths, there is the consideraion of the permutation of such lengths to give various combined lengths. I think the actual brute force calcutions will be signficantly larrger

I'm keen to seee if there is some emerging pattern from the cases of 1,2,3 rods but nothing is currently leaping out at me. I've examined the Pascal triangle to see if there is an obvious correlation to any if the precendent digits, again without any success.

I'll keep plugging away at pattern identification from what we know - currently trying out various options for the 4 rod scenario to se if it helps - but don't hold out a lot hope that there will be anything obvious there.

Yes the number of permutations is definitely higher my figure only deals with subproblem of the longest being made of the maximum number of shafts. In that case it is the starting point for the decision as to the lengths of the next rod. In short there are 781,920 permutations of 6 different integers such that their sum is 56.

The figures for different compositions of the longest rod are:- 113,880 for 5 shafts, 11,040 for 4 shafts, 702 for 3 shafts, 27 for 2 shafts and last and in this case least 1 for 1 shaft.

Each of these is a starting point for the decision as to the length of the next rod. Of course reducing the shafts in the longest rod makes that one easier, but increases the combinations/permutations in the shorter rods.

If we restricted the 6 rods in descending order of number of shafts to descending lengths 56 thru 51, a worst case for the distribution of rod length, the number of permutations to examine is 12519946706042880000 (I think). Then there is another huge factor to account for the choices of other rod lengths with lower numbers of permutations..

The wizard is truly amazing.

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I have 81 brain cells left, I think.

OK - picking up the cryptic hint, I'd like to pursue a view of some pattern emerging.

For the 2 rod problem we have {1 3}{2}
For the 3 rod problem we have (one solution) that is {1 3 6}{2 5}{8}

So the pattern I see emerging from this is to have the base rods on a sequence of
{1 3 6 ? ? ? }
{2 5 ? ? ?}
{8 ? ? ? }
{? ? ?}
{? ?}
{?}

Is there a sequence emerging 1, 2, 8, . . . for the start of each rod? Is there a sequence emerging 1, 3, 6, 12? for the 6-shafted rod. More investigation required . . .

Edit: Maybe not. Can't find any way to get the 4 rod problem to work out with {1 3 6 x} as a starting point

Well - a little more progress. I've worked out a couple of possibilities for the 4 rod problem - I wanted to go with the idea of the single shaft rod measuring the maximum length.

{1 4 11 2}
{7 3 9}
{6 8}
{20}

{1 5 7 3}
{2 9 8}
{4 14}
{20}

This throws my sequencing ideas out of the window . . . fixing now on having a solution with a single rod of the maximum length shaft

2 rods
{1 2} {4}

3 rods
{4 2 3} {7 1} {10}

4 rods
{1 4 11 2} {7 3 9} {6 8} {20}
{1 5 7 3} {2 9 8} {4 14} {20}
...

5 rods
{?????} {????} {???} {??} {35}
...

6 rods
{??????} {?????} {????} {???} {??} {56}
...

I hope I didn't give you the mis-impression that the longest rod is necessarily a single-shaft one. But please go ahead because I definitely remember the solution for this is unique (despite multiple solutions in the 3-rod & 4-rod cases) and would like to see you guys prove or disprove it.

Still searching for a reliable source on web. This must be a good programming assignment in the university...

_________________
ADYFNC HJPLI BVSM GgK Oa m

Not really.

For instance, the 4 rod example has solutions such as

{1, 2, 10, 5}
{4, 7, 9}
{6, 8}
{19}

{1, 2, 11, 6}
{4, 5, 7}
{8, 10}
{15}

I was just trying to narrow down the parameters for the big one . . .