The Mathematical Pastimes of Major Percy Alexander MacMahon. Part 1. Slab StackingAuthor(s): Paul GarciaSource: Mathematics in School, Vol. 34, No. 2 (Mar., 2005), pp. 23-25Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30215786 .

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The mathematical pastimes of

Major Percy Alexander MacMahon

Part 1 - slab stacking by Paul Garcia

Percy Alexander MacMahon was born in Malta in 1854, the second son of Brigadier-General P. W. MacMahon. His destiny seemed to be to follow in his father's footsteps; he entered the Royal Military Academy in Woolwich at the age of 16, and two years later was posted to India with the rank of Lieutenant. For five and a half years, his career was unremarkable, as he moved around India with various artillery units. Then fate took a hand; whilst on duty with the 1st mountain battery stationed in Kohat, on the North- West Frontier between the British Indian Empire and Afghanistan, he was taken ill.

At first, he was just moved to a little town, Muree, on the banks of the Indus, but after a few weeks it became clear that he was seriously ill. In December 1877 he was sent back to England, where he spent nearly 18 months recovering. After his convalescence, he was posted to Sheerness to help look after the stores. But this was no life for a gallant officer; unable to return to active duty in India, the thought of a lifetime counting uniform buttons was too much to bear.

So the young lieutenant took the only sensible course of action: he joined the Advanced Class in Mathematics at the Royal Military Academy, under the tutelage of Alfred George Greenhill. Greenhill (1847-1927) was second wrangler and joint winner of the Smith's prize 1868, Professor of Mathematics at the Royal Military Academy 1876-1908. He worked mainly in ballistics and elliptic functions, but also in the theory of elasticity and hydrostatics; he was LMS President 1890-1892, he won the De Morgan medal in 1902, Royal Medal in 1906 and was knighted 1908. After two years of hard work, MacMahon passed the course and was rewarded with a promotion to Captain and a job as an Instructor at the Royal Military Academy. By now it was 1882; a couple of years later, the young Captain MacMahon wrote an important paper that brought him to the forefront of Victorian mathematical endeavour. He was elected a Fellow of the Royal Society in 1890, was President of the London Mathematical Society from 1894 to 1896, and was awarded several medals and honorary degrees during his lifetime. But that's a story for another time.

What I want to tell you about in this article is MacMahon's puzzle work. With no television to distract him, he set about making his own entertainment, and later wrote a book, New Mathematical Pastimes, to help others create their own challenging puzzles. The book was first published in 1921, and reprinted in 1930, the year after his death. As part of the celebrations to mark MacMahon's 150th birthday in September 2004, the book has been reprinted a third time. In the original edition of the book, MacMahon expressed his disappointment that it had been too expensive to produce the book in colour. Things have not changed in the intervening 83 years, and a coloured print version of the book was still too expensive. But this

time we have technology that was unavailable to MacMahon; to accompany the print version, there is a CD-ROM containing a coloured version of the book, as well as some extra material. In particular, there are copies of the three patents for puzzles taken out by MacMahon. It is the first patent that I want to describe in this article.

The first puzzle MacMahon created was patented in 1892 (number 21118). It consisted of nine wooden blocks linked by flexible tapes to form a chain. The length of the links was to be such that the blocks could be formed into a stack. MacMahon claimed that this could be done in 4527 different ways - he was not quite right about this, as I shall discuss in a moment.

A design applied to the edges of the blocks would have to be reconstructed by finding the correct stacking. This patent was accepted on 8 October 1892. The diagram that went with the patent is reproduced in Figure 1. The entire patent is reproduced on the CD-ROM mentioned above.

It is not known whether the puzzle was ever manu- factured; but it is fun to make your own. I made one by cutting nine slabs of 4 cm by 1 cm wooden strip, each 8 cm long, and joining them in a chain with ribbon, as described in the patent. The ribbon is fixed to the end of each slab with a staple. The ribbons need to be long enough so that any pair of slabs could be at opposite ends of the stack. So mine are 11 cm long. I piled the slabs up in a random order and then stuck a picture on the side. By cutting the picture along the edges where the slabs meet, I separated the slabs, piled them differently and stuck a different picture on the other side. Once the slabs were separated again, I had two puzzles. You could make it more difficult by using the same picture on both sides. Photos of my version of the puzzle are shown in Figures 2-4.

This puzzle was ahead of its time, as was much of MacMahon's work. Nowadays, the puzzle is known as the Stamp Folding Puzzle. (See Eric W Weisstein. "Stamp Folding" from Mathworld - a Wolfram Web Resource. http://mathworld.wolfram.com/StampFolding.html) It was described in Martin Gardner's 1983 book, Wheels, Life and other Amusements, and he gave the earliest reference as A. Sainte-Lague's 1937 book, Avec des Nombres et des Lignes. In stamp folding, you have a strip of stamps and you have to fold it into a pile only one stamp wide. The connection with MacMahon's slab chain puzzle is clear.

How many ways can the slabs be stacked? It is clear from the original design that twisting or crossing the links is not allowed (hence making the puzzle identical to the stamp folding puzzle). This makes the puzzle much more interesting. (If you allow twisting and crossing, then all possible stacks can be made, so for 9 slabs there would be 9! = 362880 possible stacks.)

Mathematics in School, March 2005 The MA web site www.m-a.org.uk 23

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Fig. 1 The diagram from the 1892 patent

If you have one slab, then there is only one ( = 1!) stack. Two slabs give you two (=2!) stacks, and three slabs will make six (=3!) stacks. You can easily check this by drawing. Here is a three slab chain.

. .2.. Now list all the possible permutations of the numbers 1, 2, 3:

1 1 2 2 3 3 2 3 1 3 1 2 3 2 3 1 2 1

Now join the numbers in order:

|3a2 M3

2 3 3

Notice that if you reduce the slabs in the diagrams to points, you get only two essentially different diagrams, called 'spirals' by Sainte-Lague, and known as 'meanders'. If

Fig. 2 My nine slab chain unfolded

Fig. 3 My puzzle partly done

the slabs were not numbered or identified any way, then these are the only two stackings you could get.

Sainte-Lague tried to use these spirals to create a formula for calculating Nx,

as I shall describe later.

When we get to four slabs, it gets more interesting. There are 4! = 24 permutations of the slabs, but only sixteen are possible stacks. For example, with the slabs in an unfolded chain numbered 1, 2, 3, 4 from left to right, then the stack (from the top down) 2431 is possible, but 1423 is not.

If you draw out all the possibilities, as described above for three slabs, you can check this result. You will also find that there are only five different spirals.

Sainte-Lague introduced some notation to help with the enumeration. For x slabs, the number of possible stackings

24 Mathematics in School, March 2005 The MA web site www.m-a.org.uk

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Fig. 4 The completed puzzle

is written as Nx. So the results mentioned above are N1

= 1, N2 - 2, N3 = 6, and N4 = 16. The notation Nx(p) means the number of possible stackings starting with p, Nx(p, q) the number of stackings beginning with pq. For example, below are listed all 16 stackings of a 4 slab chain. So you can see that N4(1) = 4 and N4(1, 2) = 2.

1234, 1243, 1342, 1432 4123, 3124, 2134, 2143 3412, 4312, 4213, 3214 2341, 2431, 3421, 4321

These can be grouped in two different ways. The first is a grouping into cyclic permutations. For example, starting with 1234, you get 4123, 3412 and 2341 by moving the last digit to the front successively. All four permutations are possible stackings. If a stacking is not possible (say, 1324) then none of its cyclic permutations are possible. Once you have found all the stackings beginning with a 1, then all the other possible stackings are generated by the cyclic permutations. The table above is arranged with the cycles in columns.

The second grouping is into complementary pairs of stackings. For example, if we take 3412, and subtract each digit from 5, we get 2143, which is also a possible stacking.

Sainte-Lague states some 'self-evident' formulas:

Nx = Nx(1)

+ Nx(2)

+ Nx(3) + ...+ Nx(X) N (1) =

Nx(1, 2) + Nx(1, 3) + Nx(1, 4) + ...+ N(1, x) N = x

Nx(1).

That the first two are true can be seen with just a little thought. The third, whilst it can be seen to be true for x = 1, 2, 3 and 4, is not obviously always true. The justification relies on the spirals mentioned above. If, for example withx = 4, you take the four stackings beginning with a 1 at the top, then each will correspond to a different spiral, and each will generate four other stackings by the cyclic method.

The spirals also provide a recursive formula, Nx(1, 2) N _(1). Start with the spirals representing the stackings for x-1 slabs starting 1 at the top, add an extra point and loop, label the new point 1 and relabel all the remaining points by adding 1. The result is all the stackings for x slabs starting 15 25...

The effect of all this is that we can enumerate possible stackings by considering only the spirals that start with a 1, and the order in which points on the spiral are joined.

So for x = 4, we consider only these four spirals:

~ I~ ~ir

4- q Lt- Considering now the order in which the points are joined

gives 1234, 1243, 1423 and 1432. These are not stacking orders (recall 1423 is impossible), but all the results above still apply.

In addition, there are now some symmetry formulas:

Nx(1, 2) = Nx(1, x) x xc

Nx(1, 3) Nx(1, x-1) x x

Nx(1, 4) Nx(1, x-2) Nx(1,p) N(1, x+2-p)

N (1.5p) -N (1, x+ 2-p)

Finally, using the revised system, note that if x is even, spirals starting 1, 3,... 1, 5,... 1, 7,... etc. do not exist. That

is, N2q(l, 3) = N2q (1, 5) = N2q (1, 7) = ...= 0.

Now we can calculate Ns.

Ns = 5.N,(1), Ns(1) = Ns(1, 2) + N,(1, 3) + Ns(1, 4) + Ns(1, 5)

We know N,(1, 2) = N4(1) = 4, N,(1, 2) = N,(1, 5) and N5(1, 3) = Ns(1, 4).

So we only have to find N,(1, 3). To do this we have to find all the spirals on five points that start at 1. There are ten of these, but only one starts 1, 3,...

Thus Ns(1) = 4 +1 + 1 + 4 = 10, and N, = 5 x 10 = 50.

Using this method, it is not too difficult to calculate the results for 6, 7, 8 and 9 slabs (144, 462, 1392 and 4536, respectively). So MacMahon was pretty close with his figure of 4527.

A Challenge Problem

The existence of the cyclic stackings suggests an interesting design problem. Is there an image that could go on the side of the slabs that would make four different pictures when the slabs are cycled through the stacking? R

References

Gardner, M. 1983 Wheels, Life and other Amusements, Freeman. MacMahon, P. A. 2004 New Mathematical Pastimes CD-ROM, Tarquin. Sainte-Lague, A. 1937 Avec des Nombres et des Lignes.

Keywords: History; Puzzles; Combinatorics.

Author Paul Garcia, 6 Westgate Terrace, Westgate Street, Long Melford, Suffolk CO10 9DW. e-mail: paul@marybj.cix.co.uk

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