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I read about the following puzzle thirty-five years ago or so, and I still do not know the answer.

One gives an integer $n\ge1$ and asks to place the integers $1,2,\ldots,N=\frac{n(n+1)}{2}$ in a triangle according to the following rules. Each integer is used exactly once. There are $n$ integers on the first row, $n-1$ on the second one, ... and finally one in the $n$th row (the last one). The integers of the $j$th row are placed below the middle of intervals of the $(j-1)$th row. Finally, when $a$ and $b$ are neighbours in the $(j-1)$th row, and $c$ lies in $j$-th row, below the middle of $(a,b)$ (I say that $a$ and $b$ dominate $c$), then $c=|b-a|$.

Here is an example, with $n=4$. $$\begin{matrix} 6 & & 10 & & 1 & & 8 \\\\ & 4 & & 9 & & 7 \\\\ & & 5 & & 2 & & \\\\ & & & 3 & & & \end{matrix}$$

Does every know about this ? Is it related to something classical in mathematics ? Maybe eigenvalues of Hermitian matrices and their principal submatrices.

If I remember well, the author claimed that there are solutions for $n=1,2,3,4,5$, but not for $6$, and the existence was an open question when $n\ge7$. Can anyone confirm this ?

Trying to solve this problem, I soon was able to prove the following.

If a solution exists, then among the numbers $1,\ldots,n$, exactly one lies on each line, which is obviously the smallest in the line. In addition, the smallest of a line is a neighbour of the highest, and they dominate the highest of the next line.

The article perhaps appeared in the Revue du Palais de la Découverte.

Edit. Thanks to G. Myerson's answer, we know that these objects are called Exact difference triangles in the literature.

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This reminds me of graceful labeling of a graph: en.wikipedia.org/wiki/Graceful_labeling . It seems your problem could be rephrased as gracefully labeling a particular graph. Not clear this would yield any insights, though... –  Joseph O'Rourke Dec 1 '10 at 14:19
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The problem reminds me of Ducci sequences : see en.wikipedia.org/wiki/Ducci_sequence –  François Brunault Dec 1 '10 at 18:47
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in all examples in every row the values never grow/fall twice. Can one proof this or is it just wrong ? –  HenrikRüping Dec 2 '10 at 0:15
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@HenrikRüping: the n=5 example with first row [6,14,15,3,13] shows this doesn't hold always. –  ndkrempel Dec 2 '10 at 1:08
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4 Answers

up vote 17 down vote accepted

This is the first problem in Chapter 9 of Martin Gardner, Penrose Tiles to Trapdoor Ciphers. In the addendum to the chapter, he writes that Herbert Taylor has proved it can't be done for $n\gt5$. Unfortunately, he gives no reference.

There may be something about the problem in Solomon W Golomb and Herbert Taylor, Cyclic projective planes, perfect circular rulers, and good spanning rulers, in Sequences and their applications (Bergen, 2001), 166–181, Discrete Math. Theor. Comput. Sci. (Lond.), Springer, London, 2002, MR1916130 (2003f:51016).

See also http://www.research.ibm.com/people/s/shearer/dts.html and the literature on difference matrices and difference triangles.

EDIT. Reading a little farther into the Gardner essay, I see he writes,

The only published proof known to me that the conjecture is true is given by G. J. Chang, M. C. Hu, K. W. Lih and T. C. Shieh in "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Taipei, Taiwan (vol. 5, June 1977, pages 191- 197).

This paper can be found at http://www.math.sinica.edu.tw/bulletin/bulletin_old/51/5120.pdf and the review is MR0491218 (58 #10483).

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For small values of $n$, there is a relatively small state space to search.

In the most naive way possible, I found the following (showing the top row of triangle only):

1: 1 way: [1]
2: 4 ways: [1,3], [2,3], [3,1], [3,2]
3: 8 ways: [1,6,4], [2,6,5], [4,1,6], [4,6,1], [5,2,6], [5,6,2], [6,1,4], [6,2,5]
4: 8 ways: [6,1,10,8], [6,10,1,8], [8,1,10,6], [8,3,10,9], [8,10,1,6], [8,10,3,9] [9,3,10,8], [9,10,3,8]
5: 2 ways: [6,14,15,3,13], [13,3,15,14,6]
6: no ways

In particular, it is possible for $n = 5$, but not possible for $n = 6$.

The computation for $n = 7$ seems entirely feasible, and I'm happy to carry it out.

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Ok, using a slightly less naive search, I've shown there are none for n = 7 either. –  ndkrempel Dec 1 '10 at 22:08
    
Update: none for n = 8. That took about half an hour (using Ruby 1.9), so a more clever approach or faster language/computer may be required soon. –  ndkrempel Dec 1 '10 at 22:48
    
Thanks. I mixed up 5 and 6. I make a correction. –  Denis Serre Dec 2 '10 at 0:42
    
Since I set it running before more definitive answers came in, I may as well report there are none for n = 9 either, although that took about 6 hours (without improving the method). –  ndkrempel Dec 2 '10 at 5:56
    
Noting, that the biggest number must always be placed in the first row, the second biggest number in the first row or below a bigger number etc. should speed up the program a lot (if u didn't use that already). –  HenrikRüping Dec 2 '10 at 10:14
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Not an answer, but a question about constraints mod 2. There you are just taking differences, and the number 0s and 1s must be the same for n that is 0 or 3 mod 4 (one different in the other two cases). On the face of it the left edge of the triangle could be any binary sequence. Does this give anything?

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It does indeed. For instance, for $n=5$, I got only two distinct (up to rotation) $(0,1)$ patterns. –  Denis Serre Dec 2 '10 at 0:44
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From what I remember there is now an actual proof by 6 authors (one is Andy Liu); I think the paper was presented at g4g9. The paper might have not been published yet.

It is my understanding that Herbert Taylor was true but way to complicated, I think that Liu & et used his basic idea but simplified a lot the proof, which is now at an elementary level.

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For the non-cognoscenti, g4g9 is the 9th Gathering for Gardner, held earlier this year. I couldn't find anything about this paper on their webpage, nor on Andy Liu's. I'm not suggesting Nick S is wrong, just reporting on the difficulty of finding anything. –  Gerry Myerson Dec 2 '10 at 22:50
    
Yea I couldn't find anything on Andy's page either and the g4g9 page doesn't contain any info on the speakers (they have the list from last year). But Andy gave a talk on this during the October 21st Celebration of Mind (see the Edmonton event). –  Nick S Dec 3 '10 at 6:14
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