Mathematical Recreations and Essays · W. W. Rouse Ball
In this chapter I propose to enumerate certain geometrical questions the discussion of which will not involve necessarily any considerable use of algebra or arithmetic. Unluckily no writer like Bachet has collected and classified problems of this kind, and I take the following instances from my note-books with the feeling that they represent the subject but imperfectly.
The first part of the chapter is devoted to questions which are of the nature of formal propositions: the last part contains a description of various trivial puzzles and games, which the older writers would have termed geometrical, but which the reader of to-day may omit without loss.
In accordance with the rule I laid down for myself in the preface, I exclude the detailed discussion of theorems which involve advanced mathematics. Moreover (with one possible exception) I exclude also any mention of the numerous geometrical paradoxes which depend merely on the inability of the eye to compare correctly the dimensions of figures when their relative position is changed. This apparent deception does not involve the conscious reasoning powers, but rests on the inaccurate interpretation by the mind of the sensations derived through the eyes, and I do not consider such paradoxes as coming within the domain of mathematics.
Most educated Englishmen are
acquainted with the series of logical propositions in geometry associated with the name of Euclid, but it is not known so generally that these propositions were supplemented originally by certain exercises. Of such exercises Euclid issued three series: two containing easy theorems or problems, and the third consisting of geometrical fallacies, the errors in which the student was required to find.
The collection of fallacies prepared by Euclid is lost, and tradition has not preserved any record as to the nature of the erroneous reasoning or conclusions; but, as an illustration of such questions, I append two or three demonstrations, leading to obviously impossible results, which perhaps may amuse any one to whom they are new. I leave the discovery of the errors to the ingenuity of my readers.
First Fallacy To prove that a right angle is equal to an angle which is greater than a right angle. Let ABCD be a rectangle. From A draw a line AE outside the rectangle, equal to AB or DC and making an acute angle with AB , as
indicated in the diagram. Bisect CB in H , and through H draw HO at right angles to CB. Bisect CE in K , and through K draw KO at right angles to CE. Since CB and CE are not
parallel the lines HO and KO will meet (say) at O. Join OA , OE , OC , and OD.
The triangles ODC and OAE are equal in all respects. For, since KO bisects CE and is perpendicular to it, we have OC= OE. Similarly, since HO bisects CB and DA and is perpendicular to them, we have OD = OA. Also, by construction, DC = AE. Therefore the three sides of the triangle ODC are equal respectively to the three sides of the triangle OAE. Hence, by Euc. i. 8, the triangles are equal; and therefore the angle ODC is equal to the angle OAE.
Again, since HO bisects DA and is perpendicular to it, we have the angle ODA equal to the angle OAD.
Hence the angle ADC (which is the difference of ODC and ODA ) is equal to the angle DAE (which is the difference of OAE and OAD ). But ADC is a right angle, and DAE is necessarily greater than a right angle. Thus the result is impossible.
Second Fallacy (See a note by M. Coccoz in L'Illustration, Paris, Jan. 12, 1895.) To prove that a part of a line is equal to the whole line. Let ABC be a triangle; and, to fix our ideas, let us suppose that the triangle is scalene, that the angle B is
acute, and that the angle A is greater than the angle C. From A draw AD making the angle BAD equal to the angle C , and cutting BC in D. From A draw AE perpendicular to BC.
The triangles ABC , ABD are equiangular; hence, by Euc. vi. 19,
ABC : ABD = AC^2 : AD^2.
Also the triangles ABC , ABD are of equal altitude: hence, by Euc. vi. 1, ABC : ABD = BC : BD, ; Therefore AC^2 : AD^2 = BC : BD. ; Therefore AC^2/BC = AD^2/BD. ; Hence, by Euc. ii. 13, AB^2 + BC^2 - 2BC BE/BC =AB^2 + BD^2 - 2BD BE/BD. ; Therefore AB^2/BC +BC -2BE = AB^2/BD +BD -2BE. ; Therefore AB^2/BC - BD = AB^2/BD - BC. ; Therefore AB^2 - BC BD/BC = AB^2 - BC BD/BD. ; Therefore BC = BD, a result which is impossible.
Third Fallacy To prove that every triangle is isosceles. Let ABC be any triangle. Bisect BC in D , and through D draw DO perpendicular to BC. Bisect the angle BAC by AO.
First. If DO and AO do not meet, then they are parallel. Therefore AO is at right angles to BC. Therefore AB= AC.
Second. If DO and AO meet, let them meet in O. Draw OE perpendicular to AC. Draw OF perpendicular to AB. Join OB , OC.
Let us begin by taking the case where O is inside the triangle, in which case E falls on AC and F on AB.
The triangles AOF and AOE are equal, since the side AO is common, angle OAF = angle OAE , and angle OFA= angle OEA. Hence AF=AE. Also, the triangles BOF and COE are equal. For
since OD bisects BC at right angles, we have OB=OC ; also, since the triangles AOF and AOE are equal, we have OF=OE ; lastly, the angles at F and E are right angles. Therefore, by Euc. i. 47 and i. 8, the triangles BOF and COE are equal. Hence FB=EC.
Therefore AF+FB=AE+EC , that is, AB=AC.
The same demonstration will cover the case where DO and AO meet at D , as also the case where they meet outside BC but so near it that E and F fall on AC and AB and not on AC and AB produced.
Next take the case where DO and AO meet outside the triangle, and E and F fall on AC and AB produced. Draw OE perpendicular to AC produced. Draw OF perpendicular to AB produced. Join OB , OC.
Following the same argument as before, from the equality of the triangles AOF and AOE , we obtain AF=AE ; and, from the equality of the triangles BOF and COE , we obtain FB=EC. Therefore AF-FB=AE-EC , that is, AB=AC.
Thus in all cases, whether or not DO and AO meet, and whether they meet inside or outside the triangle, we have AB = AC : and therefore every triangle is isosceles, a result which is impossible.
Fourth Fallacy I am indebted to Captain Turton for the following ingenious fallacy; it appeared for the first time in the third edition of this work.
On the hypothenuse, BC , of an isosceles right-angled
triangle, DBC , describe an equilateral triangle ABC , the vertex A being on the same side of the base as D is. On CA take a point H so that CH = CD. Bisect BD in K. Join HK and let it cut CB (produced) in L. Join DL. Bisect DL at M , and through M draw MO perpendicular to DL. Bisect HL at N , and through N draw NO perpendicular to HL. Since DL and HL intersect, therefore MO and NO will also intersect; moreover, since BDC is a right angle, MO and NO both slope away from DC and therefore they will meet on the side of DL remote from A. Join OC , OD , OH , OL.
The triangles OMD and OML are equal, hence OD = OL. Similarly the triangles ONL and ONH are equal, hence OL = OH. Therefore OD = OH. Now in the triangles OCD and OCH , we have OD= OH , CD = CH (by construction), and OC common, hence (by Euc. i. 8) the angle OCD is equal to the angle OCH , which is absurd.
Fifth Fallacy (Mathesis, October, 1893, series 2, vol. iii, p. 224.) To prove that, if two opposite sides of a quadrilateral are equal, the other two sides must be parallel. Let ABCD be a quadrilateral such that AB is equal to DC. Bisect AD in M , and through M draw MO at right angles to AD. Bisect BC in N , and draw NO at right angles to BC.
If MO and NO are parallel, then AD and BC (which are at right angles to them) are also parallel.
If MO and NO are not parallel, let them meet in O ; then O must be either inside the quadrilateral as in the left-hand
diagram or outside the quadrilateral as in the right-hand diagram. Join OA , OB , OC , OD.
Since OM bisects AD and is perpendicular to it, we have OA = OD , and the angle OAM equal to the angle ODM. Similarly OB = OC , and the angle OBN equal to the angle OCN. Also by hypothesis AB = DC , hence, by Euc. i. 8, the triangles OAB and ODC are equal in all respects, and therefore the angle AOB is equal to the angle DOC.
Hence in the left-hand diagram the sum of the angles AOM , AOB is equal to the sum of the angles DOM , DOC ; and in the right-hand diagram the difference of the angles AOM , AOB is equal to the difference of the angles DOM , DOC ; and therefore in both cases the angle MOB is equal to the angle MOC , OM (or OM produced) bisects the angle BOC. But the angle NOB is equal to the angle NOC , ON bisects the angle BOC ; hence OM and ON coincide in direction. Therefore AD and BC , which are perpendicular to this direction, must be parallel. This result is not universally true, and the above demonstration contains a flaw.
Sixth Fallacy The following argument is taken from a text-book on electricity, published in 1889 by two distinguished mathematicians, in which it was presented as valid. A given vector OP of length l can be resolved in an infinite number of ways into two vectors OM , MP , of lengths l' , l" , and we can make l'/l" have any value we please from nothing to infinity. Suppose that the system is referred to rectangular axes Ox , Oy ; and that OP , OM , MP make respectively angles θ , θ' , θ" with Ox. Hence, by projection on Oy and on Ox , we have lθ = l'θ' + l"θ",; lθ = l'θ' + l"θ".; Therefore θ = nθ' + θ"/nθ' + θ",; where n = l'/l". This result is true whatever be the value of n. But n may have any value ( n = ∞ , or n = 0 ), hence θ = θ' = θ" , which obviously is impossible.
Seventh Fallacy Here is a fallacious investigation, to
which Mr Chartres first called my attention, of the value of π : it is founded on well-known quadratures. The area of the semi-ellipse bounded by the minor axis is (in the usual notation) equal to 1/2π ab. If the centre is moved off to an indefinitely great distance along the major axis, the ellipse degenerates into a parabola, and therefore in this particular limiting position the area is equal to two-thirds of the circumscribing rectangle. But the first result is true whatever be the dimensions of the curve. Therefore 1/2π ab = 2/3a × 2b,; Therefore π = 8/3, a result which is obviously untrue
.
To the above examples I may add the following questions, which, though not exactly fallacious, lead to results which at a hasty glance appear impossible.
First Paradox The first is a problem, sent to me by Mr Renton, to rotate a plane lamina (say, for instance, a sheet of paper) through four right angles so that the effect is equivalent to turning it through only one right angle.
If it is desired that the effect shall be equivalent to turning it through a right angle about a point O , the solution is as follows. Describe on the lamina a square OABC. Rotate the lamina successively through two right angles about the diagonal OB as axis and through two right angles about the side OA as axis, and the required result will be attained.
Second Paradox As in arithmetic, so in geometry, the theory of probability lends itself to numerous paradoxes. Here is a very simple illustration. A stick is broken at random into three pieces. It is possible to put them together into the shape of a triangle provided the length of the longest piece is less than the sum of the other two pieces (cf. Euc. i. 20), that is, provided the length of the longest piece is less than half the length of the stick. But the probability that a fragment of a stick shall be half the
original length of the stick is 1/2. Hence the probability that a triangle can be constructed out of the three pieces into which the stick is broken would appear to be 1/2. This is not true, for actually the probability is 1/4.
Third Paradox The following example illustrates how easily the eye may be deceived in demonstrations obtained by actually dissecting the figures and re-arranging the parts. In fact proofs by superposition should be regarded with considerable distrust unless they are supplemented by mathematical reasoning. The well-known proofs of the propositions Euclid i. 32 and Euclid i. 47 can be so supplemented and are valid. On the other hand, as an illustration of how deceptive a non-mathematical proof may be, I here mention the familiar paradox that a square of paper, subdivided like a chessboard into 64 small squares, can be cut into four pieces
which being put together form a figure containing 65 such small squares (I do not know who discovered this paradox. It is given in various modern books, but I cannot find an earlier reference to it than one by Prof. G.H. Darwin, Messenger of Mathematics, 1877, vol. vi, p. 87.). This is effected by cutting the original square into four pieces in the manner indicated by the thick lines in the illus066first figure. If these four pieces are put together in the shape of a rectangle in the way shown in the illus066second figure it will appear as if this rectangle contains 65 of the small squares.
This phenomenon, which in my experience non-mathematicians find perplexing, is due to the fact that the edges of the four pieces of paper, which in the second figure lie along the diagonal AB , do not coincide exactly in direction. In reality they include a small lozenge or diamond-shaped figure, whose area is equal to that of one of the 64 small squares in the original square, but whose length AB is much greater than its breadth. The diagrams show that the angle between the two sides of this lozenge which meet at A is ^-12/5 - ^-13/8 , that is, is ^-11/46 , which is less than 11/4^circ. To enable the eye to distinguish so small an angle as this the dividing lines in the first figure would have to be cut with extreme accuracy and the pieces placed together with great care.
The paradox depends upon the relation 5 × 13 - 8^2 = 1. Similar results can be obtained from the formulae 13 × 34 - 21^2 = 1 , 34 × 89 - 55^2 = 1 ,; or from the formulae 5^2 - 3 × 8 = 1 , 13^2 - 8 × 21 = 1 , 34^2 - 21 × 55 = 1. These numbers are obtained by finding convergents to the continued fraction 1 + 1/1 0pt+ 1/1 0pt+ 1/1 0pt+....
A similar paradox for a square of 17 cells, by which it was shown that 289 was equal to 288 , was alluded to by Ozanam (Ozanam, 1803 edition, vol. i, p. 299.) who gave also the diagram for dividing a rectangle of 11 by 3 into two rectangles whose dimensions appear to be 5 by 4 and 7 by 2.
Turton's Seventy-Seven Puzzle A far better dissection puzzle was invented by Captain Turton. In this a piece of cardboard, 11 inches by 7 inches, subdivided into 77 small equal squares, each 1 inch by 1 inch, can be cut up and re-arranged so as to give 78 such equal squares, each 1 inch by 1 inch, of which 77 are arranged in a rectangle of the same dimensions as the original rectangle from one side of which projects a small additional square. The construction is ingenious,
but cannot be described without the use of a model. The trick consists in utilizing the fact that cardboard has a sensible thickness. Hence the edges of the cuts can be bevelled, but in the model the bevelling is so slight as to be imperceptible save on a very close scrutiny. The play thus given in fitting the pieces together permits the apparent production of an additional square.
[Colouring Maps][The Four-Colour Theorem.]Colouring Maps I proceed next to mention the
geometrical proposition that not more than four colours are necessary in order to colour a map of a country (divided into districts) in such a way that no two contiguous districts shall be of the same colour. By contiguous districts are meant districts having a common line as part of their boundaries: districts which touch only at points are not contiguous in this sense.
The problem was mentioned by A.F. Möbius (Leipzig Transactions (Math.-phys. Classe), 1885, vol. xxxvii, pp. 1--6.) in his Lectures in 1840, but it was not until Francis Guthrie (Proceedings of the Royal Society of Edinburgh, July 19, 1880, vol. x, p. 728.) communicated it to De Morgan about 1850 that attention was generally called to it: it is said that the fact had been familiar to practical map-makers for a long time previously. Through De Morgan the proposition became generally known; and in 1878 Cayley (Proceedings of the London Mathematical Society, 1878, vol. ix, p. 148, and Proceedings of the Royal Geographical Society, 1879, N.S., vol. i, p. 259.) recalled attention to it by stating that he did not know of any rigorous proof of it.
Probably the following argument, though not a formal demonstration, will satisfy the reader that the result is true.
Let A , B , C be three contiguous districts, and let X be any other district contiguous with all of them. Then X must
lie either wholly outside the external boundary of the area ABC or wholly inside the internal boundary, that is, it must occupy a position either like X or like X'. In either case every remaining occupied area in the figure is enclosed by the boundaries of not more than three districts: hence there is no possible way of drawing another area Y which shall be contiguous with A , B , C , and X. In other words, it is possible to draw on a plane four areas which are contiguous, but it is not possible to draw five such areas.
If A , B , C are not contiguous, each with the other, or if X is not contiguous with A , B , and C , it is not necessary to colour them all differently, and thus the most unfavourable case is that already treated. Moreover any of the above areas may diminish to a point and finally disappear without affecting the argument.
That we may require at least four colours is obvious from the diagram illus:069, since in that case the areas A , B , C , and X would have to be coloured differently.
A proof of the proposition involves difficulties of a high order, which as yet have baffled all attempts to surmount them.
The argument by which the truth of the proposition was formerly supposed to be demonstrated was given by A.B. Kempe (He sent his first demonstration across the Atlantic to the American Journal of Mathematics, 1879, vol. ii, pp. 193--200; but subsequently he communicated it in simplified forms to the London Mathematical Society, Transactions, 1879, vol. x, pp. 229--231, and to Nature, Feb. 26, 1880, vol. xxi, pp. 399--400.) in 1879, but there is a flaw (See articles by P.J. Heawood in the Quarterly Journal of Mathematics, London, 1890, vol. xxiv, pp. 332--338; and 1897, vol. xxxi, pp. 270--285.) in it.
In 1880, Tait published a solution (Proceedings of the Royal Society of Edinburgh, July 19, 1880, vol. x, p. 729; and Philosophical Magazine, January, 1884, series 5, vol. xvii, p. 41.) depending on the theorem that if a closed network of lines joining an even number of points is such that three and only three lines meet at each point then three colours are sufficient to colour the lines in such a way that no two lines meeting at a point are of the same colour; a closed network being supposed to exclude the case where the lines can be divided into two groups between which there is but one connecting line. His deduction therefrom that four colours will suffice for a map was given in the last edition of this work. The demonstration appeared so straightforward that at first it was generally accepted, but it would seem that it too involves a fallacy (See J. Peterson of Copenhagen, L'Intermédiaire des mathématiciens, vol. v, 1898, pp. 225--227; and vol. vi, 1899, pp. 36--38.). The proof however leads to the interesting corollary that four colours may not suffice for a map drawn on a multiply-connected surface such as an anchor ring.
Although a proof of the theorem is still wantingAppel and Haken's controversial computer-assisted proof appeared in 1977: Illinois Journal of Mathematics, vol. xxi, pp. 429--490 and 491--567., no one has succeeded in constructing a plane map which requires more than four tints to colour it, and there is no reason to doubt the correctness of the statement that it is not necessary to have more than four colours for any plane map. The number of ways which such a map can be coloured with four
tints has been also considered (See A.C. Dixon, Messenger of Mathematics, Cambridge, 1902--3, vol. xxxii, pp. 81--83.), but the results are not sufficiently interesting to require mention here
.
[Physical Geography][Hills and Dales.] Physical Configuration of a Country As I have been alluding to maps, I may here mention that the theory of
the representation of the physical configuration of a country by means of lines drawn on a map was discussed, by Cayley and Clerk Maxwell (Cayley on 'Contour and Slope Lines,' Philosophical Magazine, London, October, 1859, series 4, vol. xviii, pp. 264--268; Collected Works, vol. iv, pp. 108--111. J. Clerk Maxwell on 'Hills and Dales,' Philosophical Magazine, December, 1870, series 4, vol. xl, pp. 421--427; Collected Works, vol. ii, pp. 233--240.). They showed that a certain relation exists between the number of hills, dales, passes, &c. which can co-exist on the earth or on an island. I proceed to give a summary of their nomenclature and conclusions.
All places whose heights above the mean sea level are equal are on the same level. The locus of such points on a map is indicated by a contour-line. Roughly speaking, an island is bounded by a contour-line. It is usual to draw the successive contour-lines on a map so that the difference between the heights of any two successive lines is the same, and thus the closer the contour-lines the steeper is the slope, but the heights are measured dynamically by the amount of work to be done to go from one level to the other and not by linear distances.
A contour-line in general will be a closed curve. This curve may enclose a region of elevation: if two such regions meet at a point, that point will be a crunode ( a real double point) on the contour-line through it, and such a point is called a pass. The contour-line may enclose a region of depression: if two such regions meet at a point, that point will be a crunode on the contour-line through it, and such a point is called a fork or bar. As the heights of the corresponding level surfaces become greater, the areas of the regions
of elevation become smaller, and at last become reduced to points: these points are the summits of the corresponding mountains. Similarly as the level surface sinks the regions of depression contract, and at last are reduced to points: these points are the bottoms (or immits) of the corresponding valleys.
Lines drawn so as to be everywhere at right angles to the contour-lines are called lines of slope. If we go up a line of slope generally we shall reach a summit, and if we go down such a line generally we shall reach a bottom: we may come however in particular cases either to a pass or to a fork. Districts whose lines of slope run to the same summit are hills. Those whose lines of slope run to the same bottom are dales. A watershed is the line of slope from a summit to a pass or a fork, and it separates two dales. A watercourse is the line of slope from a pass or a fork to a bottom, and it separates two hills.
If n + 1 regions of elevation or of depression meet at a point, the point is a multiple point on the contour-line drawn through it; such a point is called a pass or a fork of the n th order, and must be counted as n separate passes (or forks). If one region of depression meets another in several places at once, one of these must be taken as a fork and the rest as passes.
Having now a definite geographical terminology we can apply geometrical propositions to the subject. Let h be the number of hills on the earth (or an island), then there will be also h summits; let d be the number of dales, then there will be also d bottoms; let p be the whole number of passes, p_1 that of single passes, p_2 of double passes, and so on; let f be the whole number of forks, f_1 that of single forks, f_2 of double forks, and so on; let w be the number of watercourses, then there will be also w watersheds. Hence, by the theorems of Cauchy and Euler, h =1 + p_1 + 2p_2 +..., ; d =1 + f_1 + 2f_2 +..., ; and w =2(p_1 + f_1) + 3(p_2 + f_2)) +....;
The above results can be extended to the case of a multiply-connected closed surface
.
Leaving now the question of formal geometrical propositions, I proceed to enumerate a few games or puzzles which depend mainly on the relative position of things, but I postpone to chapter iv the discussion of such amusements of this kind as necessitate any considerable use of arithmetic or algebra. Some writers regard draughts, solitaire, chess and such like games as subjects for geometrical treatment in the same way as they treat dominoes, backgammon, and games with dice in connection with arithmetic: but these discussions require too many artificial assumptions to correspond with the games as actually played or to be interesting.
The amusements to which I refer are of a more trivial description, and it is possible that a mathematician may like to omit the remainder of the chapter. In some cases it is difficult to say whether they should be classified as mainly arithmetical or geometrical, but the point is of no importance.
Statical Games of Position Of the innumerable statical
games involving geometry of position I shall mention only three or four.
[Three-in-a-row. Extension to p-in-a-row] [Three-in-a-row.]Three-in-a-row First, I may mention the game of three-in-a-row , of which noughts and crosses, one form of merrilees, and go-bang are well-known examples. These games are played on a board--generally in the form of a square containing n^2 small squares or cells. The common practice is for one player to place a white counter or piece or to make a cross on each small square or cell which he occupies: his opponent similarly uses black counters or pieces or makes a nought on each square which he occupies. Whoever first gets three (or any other assigned number) of his pieces in three adjacent cells and in a straight line wins. The mathematical theory for a board of 9 cells has been worked out completely, and there is no difficulty in extending it to one of 16 cells: but the analysis is lengthy and not particularly interesting. Most of these
games were known to the ancients (Becq de Fouquières , Les jeux des anciens, second edition, Paris, 1873, chap. xviii.), and it is for that reason I mention them here.
Three-in-a-row. Extension I may, however, add an elegant but difficult extension which has not previously found its way, so far as I am aware, into any book of mathematical recreations. The problem is to place n counters on a plane so as to form as many rows as possible, each of which shall contain three and only three counters (Educational Times Reprints, 1868, vol. viii, p. 106; 1886, vol. xlv, pp. 127--128.).
It is easy to arrange the counters in a number of rows equal to the integral part of 1/8(n-1)^2. This can be effected by the following construction. Let P be any point on a cubic. Let the tangent at P cut the curve again in Q. Let the tangent at Q cut the curve in A. Let PA cut the curve in B , QB cut it in C , PC cut it in D , QD cut it in E , and so on. Then the counters must be placed at the points P, Q, A, B.... Thus 9 counters can be placed in 8 such rows; 10 counters in 10 rows; 15 counters in 24 rows; 81 counters in 800 rows; and so on.
If however the point P is a pluperfect point of the n th order on the cubic, then Sylvester proved that the above construction gives a number of rows equal to the integral part of 1/6(n-1)(n-2). Thus 9 counters can be arranged in 9 rows, which is a well-known and easy puzzle; 10 counters in 12 rows; 15 counters in 30 rows; and so on.
Even this however is an inferior limit and may be exceeded--for instance, Sylvester stated that 9 counters can be placed in 10 rows, each containing three counters; I do not know how he placed them, but one way of so arranging them is by putting them at points whose coordinates are (2, 0) , (2, 2) , (2, 4) , (4, 0) , (4, 2) , (4, 4) , (0, 0) , (3, 2) , (6, 4) ; another way is by putting them at the points (0, 0) , (0, 2) , (0, 4) , (2, 1) , (2, 2) , (2, 3) , (4, 0) , (4, 2) , (4, 4) ; more generally, the angular points of a regular hexagon and the three points of intersection
of opposite sides form such a group, and therefore any projection of that figure will give a solution.
Thus at present it is not possible to say what is the maximum number of rows of three which can be formed from n counters placed on a plane.
p -in-a-row Extension to p -in-a-row The problem mentioned above at once suggests the extension of placing n counters so as to form as many rows as possible, each of which shall contain p and only p counters. Such problems can be often solved immediately by placing at infinity the points of intersection of some of the lines, and (if it is so desired) subsequently projecting the diagram thus formed so as to bring these points to a finite distance. One instance of such a solution is given above.
As easy examples I may give the arrangement of 16 counters in 15 rows'13' corrected to '15' as per errata sheet, each containing 4 counters; and the arrangement of 19 counters in 10 rows, each containing 5 counters. A solution of the second of these problems can be obtained by placing counters at the 19 points of intersection of the 10 lines x=± a , x=± b , y=± a , y=± b , y=± x : of these points two are at infinity. The first problem I leave to the ingenuity of my readers
.
[Tesselation. Cross-Fours] [Tesselation. Cross-Fours.]Tesselation Another of these statical recreations is known as tesselation and consists in the formation of geometrical designs or mosaics by means of tesselated tiles.
To those who have never looked into the matter it may be surprising that patterns formed by the use of square tiles (of which one-half bounded by a diagonal is white and the other half black) should be subject to mathematical analysis. In view of the discussion of this subject by Montucla (See Ozanam, 1803 edition, vol. i, p. 100; 1840 edition, p. 46.), Lucas (Lucas, Récréations Mathématiques, Paris, 1882--3, vol. ii, part 4: hereafter I shall refer to this work by the name of the author.), and other writers it would be hard to refuse to call the formation of such patterns a mathematical amusement, but the treatment is (perhaps necessarily) somewhat empirical,
and though there are some interesting puzzles of this kind, I do not propose to describe them here.
Sylvester (see the Educational Times Reprints, London, 1868, vol. x, pp. 74--76, 112: see also vol. xlv, p. 127; vol. lvi, pp. 97--99.) proposed a modified tesselation problem which consists in forming anallagmatic squares, that is, squares such that in every row and every column the number of changes of colour or the number of permanences is constant, the tiles used being square white tiles and square black tiles.
If more than two colours are used, the problems become increasingly difficult. As a simple instance of this class of problems I may mention one, sent to me by a correspondent who termed it Cross-Fours, wherein sixteen square counters are used, the upper half of each being yellow, red, pink, or blue, and the lower half being gold, green, black, or white, no two counters being coloured alike. Such counters can be arranged in the form of a square so that in each vertical, horizontal, and diagonal line there shall be 8 colours and no more: they can be also arranged so that in each of these ten lines there shall be 6 colours and no more, or 5 colours and no more, or 4 colours and no more. Puzzles of this kind are but little known; they are however not uninstructive.
Colour-Cube Problem As an example of a recreation
analogous to tesselation I will mention the colour-cube problem; I select this partly because it is one of the most difficult of such puzzles, but chiefly because it has been subjected (By Major MacMahon; an abstract of his paper, read before the London Mathematical Society on Feb. 9, 1893, was given in Nature, Feb. 23, 1893, vol. xlvii, p. 406.) to mathematical analysis.
Stripped of mathematical technicalities the problem may be enunciated as follows. A cube has six faces, and if six colours are chosen we can paint each face with a different colour. By permuting the order of the colours we can obtain thirty such cubes, no two of which are coloured alike. Take any one of these cubes, K , then it is desired to select eight
out of the remaining twenty-nine cubes, such that they can be arranged in the form of a cube (whose linear dimensions are double those of any of the separate cubes) coloured like the cube K , and placed so that where any two cubes touch each other the faces in contact are coloured alike.
Only one collection of eight cubes can be found to satisfy these conditions. To pick out these eight cubes empirically would be out of the question, but the mathematical analysis enables us to select them by the following rule. Take any face of the cube K : it has four angles, and at each angle three colours meet. By permuting the colours cyclically we can obtain from each angle two other cubes, and the eight cubes so obtained are those required.
For instance suppose that the six colours are indicated by the letters a , b , c , d , e , f. Let the cube K be put on a table, and to fix our ideas suppose that the face coloured f is at the bottom, the face coloured a is at the top, and the faces coloured b , c , d , and e front respectively the east, north, west, and south points of the compass. I may denote such an arrangement by (f; a; b, c, d, e). One cyclical permutation of the colours which meet at the north-east corner of the top face gives the cube (f; c; a, b, d, e), and a second cyclical permutation gives the cube (f; b; c, a, d, e). Similarly cyclical permutations of the colours which meet at the north-west corner of the top face of K give the cubes (f; d; b, a, c, e) and (f; c; b, d, a, e). Similarly from the top south-west corner of K we get the cubes (f; e; b, c, a, d) and (f; d; b, c, e, a): and from the top south-east corner we get the cubes (f; e; a, c, d, b) and (f; b; e, c, d, a).
The eight cubes being thus determined it is not difficult to arrange them in the form of a cube coloured similarly to K , and subject to the condition that faces in contact are coloured alike; in fact they can be arranged in two ways to satisfy these conditions. One such way, taking the cubes in the numerical order given above, is to put the cubes 3 , 6 , 8 , and 2 at the SE, NE, NW, and SW corners of the bottom face; of
course each placed with the colour f at the bottom, while 3 and 6 have the colour b to the east, and 2 and 8 have the colour d to the west: the cubes 7 , 1 , 4 , and 5 will then form the SE, NE, NW, and SW corners of the top face; of course each placed with the colour a at the top, while 7 and 1 have the colour b to the east, and 5 and 4 have the colour d to the west. If however K is not given, then, without the aid of mathematical analysis, it is a difficult puzzle to arrange the eight cubes in the form of a cube coloured similarly to one of the other twenty-two cubes and subject to the condition that faces in contact are coloured alike
.
It is easy to make similar puzzles in two dimensions which are fairly difficult; it is somewhat surprising that none are to be bought, but I have never seen any except those that I have made myself.
[Dynamical Games of Position][Dynamical Geometrical Games.] Dynamical Games of Position Games which are played by moving pieces on boards of various shapes
--such as merrilees, fox and geese, solitaire, backgammon, draughts, and chess--present more interest. In general, however, they permit of so many movements of the pieces that any mathematical analysis of them becomes too intricate to follow out completely. Probably this is obvious, but it may emphasize the impossibility of discussing such games effectively if I add that it has been shown that in a game of chess there may be as many as 197299 ways of playing the first four moves, and nearly 72000 different positions at the end of the first four moves (two on each side), of which 16556 arise when the players move pawns only (L'Intermédiaire des mathématiciens, Paris, December, 1903, vol. x, pp. 305--308: also Royal Engineers Journal, London, August--November, 1889; or British Association Transactions, 1890, p. 745.).
Games in which the possible movements are very limited may be susceptible of mathematical treatment. One or two of these are given in the next chapter: here I shall confine myself mainly to puzzles and simple amusements.
Shunting Problems The first I will mention is a little
puzzle which I bought some years ago and which was described as the "Great Northern puzzle
." It is typical of a good many problems connected with the shunting of trains, and though it rests on a most improbable hypothesis, I give it as a specimen of its kind.
The puzzle shows a railway, DEF , with two sidings, DBA and FCA , connected at A. The portion of the rails at A which is common to the two sidings is long enough to permit of a single wagon, like P or Q , running in or out of it; but is too short to contain the whole of an engine, like R. Hence, if an engine runs up one siding, such as DBA , it must come back the same way.
Initially a small block of wood, P , coloured to represent a wagon, is placed at B ; a similar block, Q , is placed at C ; and a longer block of wood, R , representing an engine, is placed at E. The problem is to use the engine R to interchange the wagons P and Q , without allowing any flying shunts.
This is effected thus. (i) R pushes P into A. (ii) R returns, pushes Q up to P in A , couples Q to P , draws them both out to F , and then pushes them to E. (iii) P is now uncoupled, R takes Q back to A , and leaves it there. (iv) R returns to P , pulls P back to C , and leaves it there. (v) R running successively through F , D , B comes to A , draws Q out, and leaves it at B.
A somewhat similar puzzle, on sale in the streets in 1905, is made as follows. A loop-line BGE connects two points B and E on a railway track AF , which is supposed blocked at both ends, as shown in the diagram. In the model, the track
AF is 9 inches long, AB = EF = 15/6 inches, and AH = FK = BC = DE = 1/4 inch. On the track and loop are eight wagons,
numbered successively 1 to 8 , each one inch long and one-quarter of an inch broad, and an engine of the same dimensions. Originally the wagons are on the track from A to F and in the order 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , and the engine is on the loop. The construction and the initial arrangement ensure that at any one time there cannot be more than eight vehicles on the track. Also if eight vehicles are on it only the penultimate vehicle at either end can be moved on to the loop, but if less than eight are on the track then the last two vehicles at either end can be moved on to the loop. If the points at each end of the loop-line are clear, it will hold four, but not more than four, vehicles. The object is to reverse the order of the wagons on the track, so that from A to F they will be numbered successively 8 to 1 ; and to do this by means which will involve as few transferences of the engine or a wagon to or from the loop as is possible.
Other shunting problems are not uncommon, but these two examples will suffice
.
Ferry-Boat Problems Everybody is familiar with the story of the showman who was travelling with a wolf, a goat, and a basket of cabbages; and for obvious reasons was unable to leave the wolf alone with the goat, or the goat alone with the cabbages. The only means of transporting them across a river was a boat so small that he could take in it only one of them at a time. The problem is to show how the passage could be effected (Ozanam, 1803 edition, vol. i, p. 171; 1840 edition, p. 77.).
A similar problem, given by Alcuin, Tartaglia, and others, is as follows (Bachet, Appendix, problem iv, p. 212.). Three beautiful ladies have for husbands three men, who are as jealous as they are young, handsome, and
gallant. The party are travelling, and find on the bank of a river, over which they have to pass, a small boat which can hold no more than two persons. How can they pass, it being agreed that, in order to avoid scandal, no woman shall be left in the society of a man unless her husband is present?
The method of transportation to be used in the above cases is obvious, and can be illustrated practically by using six court cards out of a pack. Another problem similar to the one last mentioned is the case of n married couples who have to cross a river by means of a boat which can be rowed by one person and will carry n - 1 people, but not more, with the condition that no woman is to be in the society of a man unless her husband is present. Alcuin's problem is the case of n = 3. Let y denote the number of passages from one bank to the other which will be necessary. Then it has been shown that if n = 3 , y = 11 ; if n = 4 , y = 9 ; and if n > 4 , y = 7 ; the demonstration presents no difficulty.
The following analogous problem is due to the late Prof. Lucas (Lucas, vol. i, pp. 15--18, 237--238.). To find the smallest number x of persons that a boat must be able to carry in order that n married couples may by its aid cross a river in such a manner that no woman shall remain in the company of any man unless her husband is present; it being assumed that the boat can be rowed by one person only. Also to find the least number of passages, say y , from one bank to the other which will be required. M. Delannoy has shown that if n = 2 , then x = 2 , and y = 5. If n = 3 , then x = 2 , and y = 11. If n = 4 , then x = 3 , and y = 9. If n = 5 , then x = 3 , and y = 11. And finally if n > 5 , then x = 4 , and y = 2n - 1.
M. De Fonteney has remarked that, if there was an island in the middle of the river, the passage might be always effected by the aid of a boat which could carry only two persons. If there are only two or only three couples the island is unnecessary, and the case is covered by the preceding method. If n > 3 then the least number of passages from land to land which will be required is 8 (n - 1).
His solution is as follows. The first nine passages will be the same, no matter how many couples there may be: the result is to transfer one couple to the island and one couple to the second bank. The result of the next eight passages is to transfer one couple from the first bank to the second bank: this series of eight operations must be repeated as often as necessary until there is left only one couple on the first bank, only one couple on the island, and all the rest on the second bank. The result of the last seven passages is to transfer all the couples to the second bank.
The solution for the case when there are four couples may be represented as follows. Let A and a , B and b , C and c , D and d , be the four couples. The letters in the successive lines indicate the positions of the men and their respective wives after different passages of the boat.
Prof. G. Tarry has suggested an extension of the problem, which still further complicates its solution. He supposes that each husband travels with a harem of m wives or concubines; moreover, as Mohammedan women are brought up in seclusion, it is reasonable to suppose that they would be unable to row a boat by themselves without the aid of a man. But perhaps the difficulties attendant on the travels of one wife may be deemed sufficient for Christians, and I content myself with merely mentioning the increased anxieties experienced by Mohammedans in similar circumstances.
[Geodesic Problems][Geodesics.]Geodesics Geometrical problems connected with finding the shortest routes from one point to another on a curved surface are often difficult, but geodesics on a flat surface or flat surfaces are in general readily determinable.
I append an instance (I heard a similar question propounded at Cambridge in 1903, but the only place where I have seen it in print is the Daily Mail, London, February 1, 1905.), but I should have hesitated to do so had not experience shown that some readers do not readily see the solution. It is as follows: A room is 30 feet long, 12 feet wide, and 12 feet high. On the middle line of one of the smaller side walls and one foot from the ceiling is a wasp. On the middle line of the opposite wall and 11 feet from the ceiling is a fly. The wasp catches the fly by crawling all the way to it: the fly, paralysed by fear, remaining still. The problem is to find the shortest route that the wasp can follow.
To obtain a solution we observe that we can cut a sheet of paper so that, when folded properly, it will make a model to
scale of the room. This can be done in several ways. If, when the paper is again spread out flat, we can join the points representing the wasp and the fly by a straight line lying wholly on the paper we shall obtain a geodesic route between them. Thus the problem is reduced to finding the way of cutting out the paper which gives the shortest route of the kind.
illus084Here is the diagram corresponding to a solution of the above question, where A represents the floor, B and D the longer side-walls, C the ceiling, and W and F the positions on the two smaller side-walls occupied initially by the wasp and fly. In the diagram the square of the distance between W and F is (32)^2 + (24)^2 ; hence the distance is 40 feet.
[Problems with Counters placed in a row][Problems with Counters.] Problems with Counters placed in a row Numerous dynamical
problems and puzzles may be illustrated with a box of counters, especially if there are counters of two colours. Of course coins or pawns or cards will serve equally well. I proceed to enumerate a few of these played with counters placed in a row.
First Problem with Counters The following problem must be familiar to many of my readers. Ten counters (or coins) are placed in a row. Any counter may be moved over two of
those adjacent to it on the counter next beyond them. It is required to move the counters according to the above rule so that they shall be arranged in five equidistant couples.
If we denote the counters in their initial positions by the numbers 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , we proceed as follows. Put 7 on 10 , then 5 on 2 , then 3 on 8 , then 1 on 4 , and lastly 9 on 6. Thus they are arranged in pairs on the places originally occupied by the counters 2 , 4 , 6 , 8 , 10.
Similarly by putting 4 on 1 , then 6 on 9 , then 8 on 3 , then 10 on 7 , and lastly 2 on 5 , they are arranged in pairs on the places originally occupied by the counters 1 , 3 , 5 , 7 , 9.
If two superposed counters are reckoned as only one, solutions analogous to those given above will be obtained by putting 7 on 10 , then 5 on 2 , then 3 on 8 , then 1 on 6 , and lastly 9 on 4 ; or by putting 4 on 1 , then 6 on 9 , then 8 on 3 , then 10 on 5 , and lastly 2 on 7 (Note by J. Fitzpatrick to a French translation of the third edition of this work, Paris, 1898.).
There is a somewhat similar game played with eight counters, but in this case the four couples finally formed are not equidistant. Here the transformation will be effected if we move 5 on 2 , then 3 on 7 , then 4 on 1 , and lastly 6 on 8. This form of the game is applicable equally to (8 + 2n) counters, for if we move 4 on 1 we have left on one side of this couple a row of (8 + 2n - 2) counters. This again can be reduced to one of (8 + 2n - 4) counters, and in this way finally we have left 8 counters which can be moved in the way explained above.
A more complete generalization would be the case of n counters, where each counter might be moved over the m counters adjacent to it on to the one beyond them.
Second Problem with Counters Another problem of a somewhat similar kind is due to Tait (Philosophical Magazine, London, January, 1884, series 5, vol. xvii, p. 39.). Place four florins (or white counters) and four halfpence (or black counters) alternately in a line in contact with one another. It is required
in four moves, each of a pair of two contiguous pieces, without altering the relative position of the pair, to form a continuous line of four halfpence followed by four florins.
His solution is as follows. Let a florin be denoted by a and a halfpenny by b , and let × × denote two contiguous blank spaces. Then the successive positions of the pieces may be represented thus: 4pt 0
The operation is conducted according to the following rule. Suppose the pieces to be arranged originally in circular order, with two contiguous blank spaces, then we always move to the blank space for the time being that pair of coins which occupies the places next but one and next but two to the blank space on one assigned side of it.
A similar problem with 2n counters-- n of them being white and n black--will at once suggest itself, and, if n is greater than 4 , it can be solved in n moves. I have however failed to find a simple rule which covers all cases alike, but solutions, due to M. Delannoy, have been given (La Nature, June, 1887, p. 10.) for the four cases where n is of the form 4m , 4m + 2 , 4m + 1 , or 4m + 3 ; in the first two cases the first 1/2n moves are of pairs of dissimilar counters and the last 1/2n moves are of pairs of similar counters; in the last two cases, the first move is similar to that given
above, namely, of the penultimate and antepenultimate counters to the beginning of the row, the next 1/2 (n - 1) moves are of pairs of dissimilar counters, and the final 1/2 (n - 1) moves are of similar counters.
The problem is also capable of solution if we substitute the restriction that at each move the pair of counters taken up
must be moved to one of the two ends of the row instead of the condition that the final arrangement is to be continuous.
Tait suggested a variation of the problem by making it a condition that the two coins to be moved shall also be made to interchange places; in this form it would seem that 5 moves are required; or, in the general case, n + 1 moves are required.
[Problems on a Chess-board with Counters or Pawns] [Problems with Counters or Pawns] Problems on a Chess-board with Counters or Pawns The following three problems require the use of a chess-board as well
as of counters or pieces of two colours. It is more convenient to move a pawn than a counter, and if therefore I describe them as played with pawns it is only as a matter of convenience and not that they have any connection with chess. The first is characterized by the fact that in every position not more than two moves are possible; in the second and third problems not more than four moves are possible in any position. With these limitations, analysis is possible. I shall not discuss the similar problems in which more moves are possible.
First Problem with Pawns (Lucas, vol. ii, part 5, pp. 141-143.) On a row of seven squares on a chess-board 3 white pawns (or counters), denoted in the diagram by " a "s, are placed on the 3 squares at one end, and 3 black pawns (or counters), denoted by " b "s, are placed on the 3 squares at the other end--the middle square being left vacant. Each piece can move only in one direction; the " a " pieces can move from left to right, and the " b " pieces from right to left. If the square next to a piece is unoccupied, it
can move on to that; or if the square next to it is occupied by a piece of the opposite colour and the square beyond that is unoccupied, then it can, like a queen in draughts, leap over that piece on to the unoccupied square beyond it. The object is to get all the white pawns in the places occupied initially by the black pawns and vice versa.
The solution requires 15 moves. It may be effected by moving first a white pawn, then successively two black pawns
then three white pawns, then three black pawns, then three white pawns, then two black pawns, and then one white pawn. We can express this solution by saying that if we number the cells (a term used to describe each of the small squares on a chess-board) consecutively, then initially the vacant space occupies the cell 4 and in the successive moves it will occupy the cells 3 , 5 , 6 , 4 , 2 , 1 , 3 , 5 , 7 , 6 , 4 , 2 , 3 , 5 , 4. Of these moves, six are simple and nine are leaps.
Similarly if we have n white pawns at one end of a row of 2n + 1 cells, and n black pawns at the other end, they can be interchanged in n (n + 2) moves, by moving in succession 1 pawn, 2 pawns, 3 pawns, ..., n - 1 pawns, n pawns, n pawns, n pawns, n - 1 pawns, ..., 2 pawns, and 1 pawn--all the pawns in each group being of the same colour and different from that of the pawns in the group preceding it. Of these moves 2n are simple and n^2 are leaps.
Second Problem with Pawns (Lucas, vol. ii, part 5, p. 144.) A similar game may be played on a rectangular or square board. The case of a square board containing 49 cells, or small squares, will illustrate this sufficiently: in this case the initial position is shown in the annexed diagram where the " a "s denote the pawns or pieces
of one colour, and the " b "s those of the other colour. The " a " pieces can move horizontally from left to right or vertically down, and the " b " pieces can move horizontally from right to left or vertically up, according to the same rules as before.
The solution reduces to the preceding case. The pieces in the middle column can be interchanged in 15 moves. In the course of these moves every one of the seven cells in that column is at some time or other vacant, and whenever that is the case the pieces in the row containing the vacant cell can be interchanged. To interchange the pieces in each of the seven rows will require 15 moves. Hence to interchange all the pieces will require 15 + (7 × 15) moves, that is, 120 moves.
If we place 2n(n + 1) white pawns and 2n(n + 1) black pawns in a similar way on a square board of (2n + 1)^2 cells, we can transpose them in 2n (n + 1)(n + 2) moves: of these 4n(n + 1) are simple and 2n^2 (n + 1) are leaps.
Third Problem with Pawns The following analogous, though somewhat more complicated, game was I believe originally published in the first edition of this work: but I find
that it has been since widely distributed in connexion with an advertisement and probably now is well-known. On a square board of 25 cells, place eight white pawns or counters on the cells denoted by small letters in the annexed diagram, and eight black pawns or counters on the cells denoted by capital letters: the cell marked with an asterisk ( * ) being left blank. Each pawn can move according to the laws already explained--the white pawns being able to move only horizontally from left to right or vertically downwards, and the black pawns being able to move only horizontally from right to left or vertically upwards. The object is to get all the white pawns in the places initially occupied by the black pawns and vice versa. No moves outside the dark line are permitted.
Since there is only one cell on the board which is unoccupied, and since no diagonal moves and no backward moves are permitted, it follows that at each move not more than two pieces of either colour are capable of moving. There are however a very large number of solutions. The following empirical solution in forty-eight moves is one way of effecting the transfer--the letters indicating the cells from which the pieces are successively moved: 2pt 1 It will be noticed that the first twenty-four moves lead to a symmetrical position, and that the next twenty-three moves can be at once obtained by writing the first twenty-three moves in reverse order and interchanging small and capital letters
.
Probably, were it worth the trouble, the mathematical theory of games such as that just described might be worked out by the use of Vandermonde's notation, described later in chapter vi, or by the analogous method employed in the theory of the game of solitaire (On the theory of the solitaire, see Reiss, 'Beiträge zur Theorie des Solitär-Spiels,' Crelle's Journal, Berlin, 1858, vol. liv, pp. 344--379; and Lucas, vol. i, part v, pp. 89--141.). I believe that this has not been done, and I do not think it would repay the labour involved.
Problems with Chess-pieces Problems on a Chess-board with Chess-pieces There are several mathematical recreations with chess-pieces, other than pawns, somewhat similar to those given above. One of these, on the determination of the ways in which eight queens can be placed on a board so that no queen can take any other, is given later in chapter iv. Another, on the path to be followed by a knight which is moved on a chess-board so that it shall occupy every cell once and only once, is given in chapter vi. Here I will mention one of the simplest of such problems, which is interesting from the fact that it is given in Guarini's manuscript
written in 1512; it was quoted by Lucas, but so far as I know has not been otherwise published.
Guarini's Problem tocsubsectionGuarini's Problem On a board of nine cells, such as that drawn below, the two white knights are placed on the two top
corner cells ( a , d ), and the two black knights on the two bottom corner cells ( b , c ): the other cells are left vacant. It is required to move the knights so that the white knights shall occupy the cells b and c , while the black shall occupy the cells a and d.
The solution is tolerably obvious. First, move the pieces from a to A , from b to B , from c to C , and from d to D. Next, move the pieces from A to d , from B to a , from C to b , and from D to c. The effect of these eight moves is the same as if the original square had been rotated through one right angle. Repeat the above process, that is, move the pieces successively from a to A , from b to B , from c to C , from d to D ; from A to d , from B to a , from C to b , and from D to c. The required result is then attained
.
[Geometrical Puzzles (rods, strings, &c.)] [Geometrical Puzzles]Geometrical Puzzles with Rods, etc Another species of geometrical puzzles, to which here I will do no more than allude, are made of steel rods, or of wire, or of wire and string. Numbers of these are often sold in the streets of London for a penny each, and some of them afford ingenious problems in the geometry of position. Most of them could hardly be discussed without the aid of diagrams, but they are inexpensive to construct, and in fact innumerable puzzles on geometry of position can be made with a couple of stout sticks and a ball of string, or even with only a box of matches: several examples are given in the appendix to the fourth volume of the 1723 edition of Ozanam's work.
I will mention, as an easy example, analogous to one group of the string puzzles, that any one can take off his waistcoat (which may be unbuttoned) without taking off his coat, and without pulling the waistcoat over the head like a jersey.
This last feat may serve to show the difficulty of mentally realizing the effect of geometrical alterations in a figure unless they are of the simplest character.
The fact just stated is illustrated by the familiar experiment of making paradromic rings by cutting a paper ring prepared in the following manner.
Take a strip of paper or piece of tape, say, for convenience, an inch or two wide and at least nine or ten inches long, rule a line in the middle down the length AB of the strip, gum one end over the other end B , and we get a ring like a section of a cylinder. If this ring is cut by a pair of scissors along the ruled line we obtain two rings exactly like the first, except that they are only half the width. Next suppose that the end A is twisted through two right angles before it is gummed to B (the result of which is that the back of the strip at A is gummed over the front of the strip at B ), then a cut along the line will produce only one ring. Next suppose that the end A is twisted once completely round ( through four right angles) before it is gummed to B , then a similar cut produces two interlaced rings. If any of my readers think that these results could be predicted off-hand, it may be interesting to them to see if they can predict correctly the effect of again cutting the rings formed in the second and third experiments down their middle lines in a manner similar to that above described.
The theory is due to J.B. Listing (Vorstudien zur Topologie, Die Studien, Göttingen, 1847, part x.) who discussed the case when the end A receives m half-twists, that is, is twisted through mπ , before it is gummed to B.
If m is even we obtain a surface which has two sides and
two edges, which are termed paradromic. If the ring is cut along a line midway between the edges, we obtain two rings, each of which has m half-twists, and which are linked together 1/2m times.
If m is odd we obtain a surface having only one side and one edge. If this ring is cut along its mid-line, we obtain only one ring, but it has 2m half-twists, and if m is greater than unity it is knotted.
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End of part 4 of 16
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