Prob. 15: Heterogeneous squares, pp. 11 & 93. Shows there is no antimagic square of order 2 and reports that a reader found 3120 inequivalent antimagic squares of order 3. (The 8 symmetries of the squ

Prob. 15: Heterogeneous squares, pp. 11 & 93. Shows there is no antimagic square of order 2 and reports that a reader found 3120 inequivalent antimagic squares of order 3. (The 8 symmetries of the square are the equivalences.)

Prob. 16: Antimagic, pp. 12 & 94. Asserts there is no 3 x 3 consecutive antimagic square. [Translation is unclear whether he is saying none is known or none exists.] Gives examples of 4 x 4 and 5 x 5 consecutive antimagics and says there are 20 known examples for 4 x 4.

Prob. 148: Superior antimagic, pp. 81 & 184. Gives a 6 x 6 consecutive antimagic square.

M. Gardner. Puzzles from Other Worlds. Vintage (Random House), NY, 1984. Problem 8: Antimagic at the number wall, pp. 19 20, 96 97 & 142 143. Notes his examples (which are complementary) are the only rook wise connected anti magic squares of order 3 and discusses anti magic triangles.

The Notes report that one of the magic queen's tours was miscopied and that Tom Marlow has found that one of the 8 x 8 semi magic knight's tours is wrong.

C. F. de Jaenisch. Op. cit. in 5.F.1. 1862. Vol. 2, pp. 151-189. ??NYS. Semi magic knight's tour, with diagonal sums of 256 and 264. [Given in Dickins, p. 27 -- which Dickins??.]

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Magic squares in chess -- The knight's tour, pp. 101-102. Semi-magic knight's tour with diagonals of 264 & 256, taken from a BM MS: Bibl. Reg. 13, A. xviii., Plut. xx. c. -- ??NYS, but same as de Jaenisch's example. Notes that symmetric cells differ by 32. Gives a cryptic argument that this solution can be used to produce examples starting at 48 of the cells of the board.

Ball. MRE, 5th ed., 1911, p. 133 gives a magic king's tour, which is hence a magic queen's tour. The 11th ed., 1939, p. 185 refers to Ghersi, below.

Italo Ghersi. Matematica Dilettevoli e Curiosa. 2nd ed., Hoepli, 1921. Pp. 320 321, fig. 261 & 265. Magic king's tour -- different than Ball's.

H. J. R. Murray. Beverley's magic S-tour and its plan -- probs. 2106-2108. Problemist Fairy Supplement (later known as Fairy Chess Review) 2:16 (Feb 1936) 166. Discusses Beverley's 1848 semi-magic path and says the method leads to 28 solutions and many, including Beverley's example, have the property that each quarter of the board is also [semi-]magic.

H. J. R. Murray. A new magic knight's tour -- Art. 68, prob. 5226. Fairy Chess Review 5:1 (Aug 1942) 2-3. A 16 x 16 semi-magic tour. Cites Beverley and Roget. Implies that Kraitchik asserted that no such 16 x 16 tours were possible.

Joseph S. Madachy. Mathematics on Vacation. Op. cit. in 5.O, (1966), 1979. Pp. 87-89. Order 16 magic knight's tour.

Stanley Rabinowitz. A magic rook's tour. JRM 18:3 (1985 86) 203 204. Gives one. Also gives Ball's magic king's tour. Says the magic knight's tour is still unsolved.

David Marks. Knight's Tours. M500 137 (Apr 1994) 1. Brief discussion of magic knight's tours, giving a semi-magic example due to Euler? and a magic example made up of 2 tours of 32 squares due to Roget.

Kanchusen. Wakoku Chiekurabe. 1727. Pp. 5 & 18-21 show two kinds of magic circles. The first has two rings of 4 and one in the centre so that each ring adds to 22 and each diameter adds to 23. This is achieved by putting 1 in the centre and then symmetrically placing the numbers in the pairs 2-9, 3-8, 4-7, 5-6. The second example uses the same pairing principle to give three rings of six, with 1 in the centre, so each ring adds to 63 and each diameter to 64.

See Franklin, c1750, in 7.N for a magic circle.

Curiosities for the Ingenious selected from The most authentic Treasures of Nature, Science and Art, Biography, History, and General Literature. 2nd ed., Thomas Boys, London, 1822. Magic circle of circles, pp. 56-57 & plate IV, opp. p. 56. 12 - 75 arranged in 8 rings of 8 sectors, with another 12 in the centre, so that each ring and each radius, with the 12 in the centre, makes 360.

Rational Recreations. 1824. Exer. 4, pp. 24-25. = Curiosities for the Ingenious.

Manuel des Sorciers. 1825. Pp. 82-83. ??NX 4 x 8 semi-magic rectangle, associated. The row sums are 132 while the column sums are 66. I don't ever recall seeing a magic rectangle like this before. For an A x B rectangle, we want 1 + ... + AB  =  AB(AB+1)/2 to be divisible by both A and B which holds if and only if A  B (mod 2). Since it doesn't make sense to talk about diagonal sums, this can only give semi-magic shapes, hence they should be easier to produce. I find essentially one solution for the 2 x 4 case.

The Secret Out. 1859. The Twelve-cornered Arithmetical Star, pp. 374-375. This is the case n = 6 of the general problem of arranging 1, 2, ..., 2n around a circle so that a_i + a_i+1  = a_n+i + a_n+i+1 for each i. This immediately leads to a_i   a_n+i  =    (a_i+1   a_n+i+1) and this requires that n be odd. The given solution fails to work at several points. See Devi, 1976, and Singmaster, 1992, below.

Mittenzwey. 1880. Prob. 101, pp. 20-21 & 72; 1895?: 117, pp. 25 & 74; A

1917: 117, pp. 23 & 71. Gives the triangular form at the right and asks for B C

S = A + B + D = A + C + F = D + E + F = B + C = B + E. This easily D E F

forces D = A, C = E = 2A, B = A + F, S = 3A + F. The original pattern had A = 4, F = 5, S = 17, and asks for a solution with S = 13. The solution gives A = 2, F = 7, but there are five solutions corresponding to A = 0, 1, 2, 3, 4.

Hoffmann. 1893. Chap IV, no. 10: The 'twenty six' puzzle, pp. 146 147 & 191 = Hoffmann Hordern, pp. 118-119, with photo on p. 133. 4 x 4 square with corners deleted. Arrange 1   12 so the 4 horizontal and vertical lines of 4 give the magic sum 26, and so do the central 4 cells and hence the two sets of opposite outer cells. Gives two solutions and says there may be more. Photo shows a German version labelled with a large 26, comprising a board with square holes and 12 numbered cubes, 1880-1895.

I. G. Ouseley. Letters: Pentacle puzzle. Knowledge 19 (Mar 1896) 63 & (Apr 1896) 84. Consider a pentagram, its five points and the five interior intersections. Place the numbers 1 to 10 on these so that each line of five has the same sum and the five internal values shall add to the same sum, while the five outer values shall add to twice as much. Second letter says it seems to be unsolvable and Editorial Note points out that the sum of all the numbers is 55, which is not divisible by 3, so the problem as stated is unsolvable. [But what if we take the numbers 0 to 9 ??]

Pearson. 1907. Part I, no. 35: A magic cross, p. 35. Same pattern as Hoffmann, with numbers differently arranged. Says there are 33 combinations that add to 26.

Williams. Home Entertainments. 1914. The cross puzzle, pp. 117-118. Shape of the Ho Thu (the River Plan, see beginning of 7.N) to make have the same sum of 23 across and down. [In fact one can have magic sums of 23, 24, ..., 27.]

Wood. Oddities. 1927. Prob. 8: A magic star, pp. 9-10. Make an 8-pointed star by superimposing two concentric squares, one twisted by 45^o. There are 8 vertices of the squares and 8 points of intersection, so there are 4 points along each square edge. Arrange the numbers 1 - 16 on these points so each edge adds up to the same value. This forces the sums of the 4 vertices to be the same (which he states as a given) and the magic constant to be 34. He gives one solution and says there are 18 solutions, which I have confirmed by computer -- doing it by hand must have been tedious or I have overlooked some simplifications.

Collins. Book of Puzzles. 1927.