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«Contents Ü Foreword Elwyn Berlekamp and Tom Rodgers ½ I Personal Magic ¿ Martin Gardner: A “Documentary” Dana Richards ½¿ Ambrose, Gardner, ...»

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Silhouettes are often used in, e.g., mechanical engineering (working drawings) and robotics (pattern recognition). One particular mathematical problem reads: “Which solid object has a circular, a triangular, and a square silhouette?” The solution is the well-known Wedge of Wallis (Figure 1).

Figure 1. Wedge of Wallis.

A silhouette defines a cylindrical region in space, perpendicular to the silhouette surface. With this meaning the word no longer refers to an object, but only to a region of a flat surface (Figure 2).

Two or more silhouettes (not necessarily perpendicular to each other) define unambiguously a region or object in space that is the cross-section of two or more cylinders.

Dewdney introduced the term projective cast for an object defined by a number of silhouettes. Not every object can be defined as the projective 214 M. O. VAN DEVENTER Figure 2. (a) Negative of a silhouette; (b) The cylindrical region in space defined by it.

cast of a (limited) number of silhouettes: One cannot cast a hollow cube or a solid sphere.

There are two interesting topological properties that apply to silhouettes, to projective casts, and to mazes and graphs in general. First, a graph may have either cycles or no cycles. Second, a graph may consist of either one part or multiple parts (Figure 3).

Figure 3. Topological properties of mazes and graphs: (a) multiple parts with cycles; (b) one part with cycles; (c) multiple parts, no cycles; (d) one part, no cycles.

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Figure 4. Three viable silhouettes yielding a projective cast with one cycle.

Figure 5. Three viable silhouettes yielding a projective cast with two parts.

and synthesis of silhouettes and projective casts: How can the properties of a projective cast be found without constructing it from its silhouettes, and how can a projective cast be given certain properties by its silhouettes? I do not know whether there is a general answer.

Hollow Mazes

A hollow maze is a rectangular box with six sides (Figure 6). Each side is a two-dimensional control maze: a surface into which slots have been cut.

A cursor consisting of three mutually perpendicular spars registers one’s position in the hollow maze. Each spar passes from one side of the box to the other, sliding along the slots of the control maze on each side. The two control mazes on opposite sides of the box are identical. In this way, a single three-dimensional maze is produced from three pairs of two-dimensional mazes. The resemblance between multiple silhouettes and hollow mazes is evident.

Control mazes are always viable because of their construction. There can be no cycles since the center of a cycle would fall out. Control mazes cannot consist of multiple parts, since a spar cannot jump from one part to another.

216 M. O. VAN DEVENTER

Figure 6. A simple hollow maze.

I have developed several restrictions for the investigation of hollow mazes:

¯ There are three (pairs of) perpendicular control mazes;

¯ the control mazes are viable;

¯ the control mazes are line mazes (i.e., their paths have zero width);

¯ a square or cubic grid is used for control mazes and their projective casts.

Now we can take another look at the topological properties of the projective cast. It seems that the projective cast cannot have cycles anymore, because the control mazes are line mazes. (I’m sure that there must be a simple proof of this, but I lack the mathematical tools.) It can easily be tested whether or not a projective cast is viable, assuming that it cannot have cycles: We count the number of grid points of the projective cast. For example, a ¿ ¢¿¢¿ projective cast has 27 grid points. Then there must be exactly 26 “links” between these points for a viable cast. The number of links can be counted from the control mazes without constructing the projective cast.

Count the links in the hollow mazes of Figure 7. The dotted cross-section of Figure 7(a) has six links through it; the total number of links is found after taking all cross-sections.

–  –  –

student, showed me that not all viable hollow mazes can be constructed by this method. Figure 9 is a counterexample.

Figure 8. The cut and connect method for constructing hollow mazes.

–  –  –

Figure 9. A viable hollow maze that cannot be constructed by the “cut and connect” method.

¯ Recursion. Replace each point by a hollow maze (see Figure 11).

–  –  –

There is still much research to be done on multiple silhouettes and hollow mazes. I keep receiving letters from readers of Scientific American, and perhaps some of the problems mentioned above will be solved in the future.

Some Diophantine Recreations David Singmaster There are a number of recreational problems that lead to equations to be solved.

Sometimes the difficulty is in setting up the equations, with the solution then being easy. Other times the equations are straightforward, but the solution requires some ingenuity, such as exploiting the symmetry of the problem. In a third class of problems, the equations and their solution are relatively straightforward, but there is an additional diophantine requirement that the data and/or the answers should be integral. In this case, it is not always straightforward to determine when the problem has solutions or to find them all or to determine the number of solutions. I discuss three examples of such diophantine recreations. The first is a simple age problem, done to illustrate the ideas. The second is the Ass and Mule Problem, for which I have just found a reasonably simple condition for integer data to produce an integer solution. The third is the problem of Selling Different Amounts at the Same Price Yielding the Same, for which I give an algorithm for finding all solutions and a new simple formula forthe number of solutions.





–  –  –

Between these are a number of types of problem. The type that I want to examine is illustrated by the following, which is the earliest I have found, reportedly from The American Tutor’s Assistant, 1791.½

–  –  –

A more typical, more comprehensible, albeit less poetic, version, but with the same numbers, appears at about the same time in Bonnycastle, 1824 [5].

A person, at the time he was married, was 3 times as old as his wife; but after they had lived together 15 years, he was only twice as old; what were their ages on their wedding day?

We can state the general form of this problem as follows.

Problem ´ µ; is now times as old as ; after years, is times as old as.

Thus the problem in The American Tutor’s Assistant and Bonnycastle is Problem (3, 15, 2). It is easily seen that the problem leads to the equations

–  –  –

My daughter Jessica is 16 and very conscious of her age.Our neighbour Helen is just 8, and I teased Jessica by saying “Seven years ago, you were 9 times as old as Helen; six years ago, you were 5 times her age; four years ago, you were 3 times her age; and now you are only twice her age. If you are not careful, soon you’ll be the same age!” Jessica seemed a bit worried, and went off muttering. I saw her doing a lot of scribbling.

The next day, she said to me, “Dad, that’s just the limit! By the way, did you ever consider when I would be half as old as Helen?” Now it was my turn to be worried, and I began muttering — “That can’t be, you’re always older than Helen.” “Don’t be so positive,” said Jessica, as she stomped off to school.

Can you help me out?

*** It is also possible to find a situation in which one person’s age is an integral multiple of another’s during six consecutive years.

Ass and Mule Problems

–  –  –

History of the Ass and Mule Problem. The earliest known version of this problem has an ass and a mule with parameters (1, 2; 1, 1) and is attributed to Euclid, from about 300 B.C. Heiberg’s edition [10] gives the problem in Greek and Latin verses.

Diophantus studied the problem in general. He proposes “to find two numbers such that each after receiving from the other may bear to the remainder a given ratio.” He gives (30, 2; 50, 3) as an example. He also covers similar problems with three and four persons.

Two versions, (10, 3; 10, 5) and (2, 2; 2, 4), appear in The Greek Anthology [16]. Alcuin gives an unusual variant of the problem in that he assumes the second person starts from the situation after the transfer mentioned by the first person takes place. That is, the second equation of (1) would be replaced by Ý  ·  µ ´Ü ·   µ.

This is our problem ´ By about the ninth century, the problem was a standard, not only in Europe but also in India and the Arab world, and it has remained a standard problem ever since.

The problem is readily generalised to more than two participants, but then there are two forms, depending on whether ‘you’ refers to all the othµ for ers or just the next in cyclic sequence. I denote these as I ´ µ for the case the case where ‘you’ means all the others and II ´ where ‘you’ means just the next person in a cyclic sequence. The earliest three-person version that I know of appears in India, about A.D. 850, in the work of Mahavira [15]; the problem is of type I. The earliest four-person version I have seen is Arabic, in al-Karkhi (aka al-Karaghi) from about 1010 [13], and the problem is of type II.

Tropfke [20] discusses the problem and cites some Chinese, Indian, Arabic, Byzantine, and Western sources. The examples cited in the Chiu Chang Suan Shu [7] state, e.g., “If I had half of what you have, I’d have 50.” This is a different type of problem and belongs in Tropfke’s previous section.

The other Chinese reference is a work of about A.D. 485 that has only been translated into Russian and may well be similar to the earlier Chinese example, so that, surprisingly, the Ass and Mule problem may not be in any Chinese source.

Below I give a summary of versions that I have noted up through Fibonacci, who typically gives many versions, including some with five persons, an inconsistent example, and extended versions where a person says, e.g., “If you give me 7, I’d have 5 times you plus 1 more.” This includes all the relevant sources cited by Tropfke except Abu al-Wafa (only available in Arabic) and “Abraham,” who is elsewhere listed as probably being

SOME DIOPHANTINE RECREATIONS 225

from the early fourteenth century and hence after Fibonacci, although the material is believed to be based on older Arabic sources.

–  –  –

Selling Different Amounts at the Same Price Yielding the Same Although this problem is quite old, there seems to be no common name for it, hence the above title, which is rather a mouthful. The problem is treated in one form by Mahavira [15], Sridhara [18], and Bhaskara [4], but this form is less clearly expressed and has infinitely many solutions, so I will first describe the later form which is first found in Fibonacci [11]. I quote a version that appears as No. 50 in the first printed English riddle collection “The Demaundes Joyous,” printed by Wynken de Worde in 1511 [9].

–  –  –

‘In like many’ means ‘the same number’ and ‘in like much money’ means ‘the same sum of money.’ Anyone meeting this problem for the first time soon realises there must be some trick to it. Fibonacci has no truck with trick questions and gives the necessary information — the sellers sell part of their stock at one price and then the remainders at another price. This may happen by the sellers going to different markets, or simply changing prices after lunch, or lowering prices late in the day in an attempt to sell their remainders, or raising them late in the day when few items are left and new buyers arrive. Although the objects are usually eggs, or sometimes fruit, which normally are all of the same value, Tartaglia [19] justifies the different prices by having large and small pearls. Even with this trick exposed, it can take a fair amount of time to work out a solution, and it is not immediately obvious how to find or count all the solutions.

by ´ ½ ¾ µ, so that Let us denote the problem with numbers ½ ¾ the above is (10, 30, 50). This is historically by far the most common version.

Fibonacci only gives two-person versions: (10, 30); (12, 32);

(12, 33). Fibonacci is one of the few to give more than a single solution. He gives a fairly general rule for generating solutions [12], which would produce the positive solutions with the smaller price equal to 1, and he gives five solutions for his first version — but there are 55, of which 36 are positive. Fibonacci goes on to consider solutions with certain properties — e.g., if the prices are also given, is there a solution; find all solutions with one price fixed; find solutions where the amounts received are in some ratio, such as the first receives twice the second.

The question of finding all solutions for a three-person problem seems to have first been tackled in the 1600s. According to Glaisher [12], the 1612 and 1624 editions of Bachet [3] ¾ give a fairly general rule for this case and apply it to (20, 30, 40). In 1874, Labosne revised the material, dropping Bachet’s rule and replacing it with some rather vague algebra; he added several more versions as well, for which he often gives fractional answers, distinctly against the spirit of the problem. The 1708 English edition of Ozanam [17], which is essentially the same as the first edition of 1694, considers (10, 25, 30) and gives two solutions. The extensively revised and enlarged edition of 1725 (or 1723) outlines a fairly general method and correctly says there are ten solutions, contrary to De Lagny’s statement that there are six solutions. Neither Glaisher nor I have seen the relevant work of De Lagny, but we both conjecture that he was counting the positive solutions, of which there are indeed six.



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