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6.AO. CONFIGURATION PROBLEMS
NOTATION: (a, b, c) denotes the configuration of a points in b rows of c each. The index below covers articles other than the surveys of Burr et al. and Gardner.
( 5, 2, 3): Sylvester

( 6, 3, 3): Mittenzwey

( 7, 6, 3): Criton

( 9, 8, 3): Sylvester; Carroll; Criton

( 9, 9, 3): Carroll; Bridges; Criton

( 9, 10, 3): Jackson; Family Friend; Parlour Pastime; Magician's Own Book; The Sociable; Book of 500 Puzzles; Charades etc.; Boy's Own Conjuring Book; Hanky Panky; Carroll; Crompton; Berkeley & Rowland; Hoffmann; Dudeney (1908); Wehman; Williams; Loyd Jr; Blyth; Rudin; Young World; Brooke; Putnam; Criton

(10, 5, 4): The Sociable; Book of 500 Puzzles; Carroll; Hoffmann; Dudeney (1908); Wehman; Williams; Dudeney (1917); Blyth; King; Rudin; Young World; Hutchings  & Blake; Putnam

(10, 10, 3): Sylvester

(11, 11, 3): The Sociable; Book of 500 Puzzles; Wehman

(11, 12, 3): Hoffmann; Williams; Young World

(11, 13, 3): Prout

(11, 16, 3): Wilkinson -- in Dudeney (1908 & 1917); Macmillan

(12, 4, 5) -- Trick version of a hollow 3 x 3 square with doubled corners, as in 7.Q: Family Friend (1858); Secret Out; Illustrated Boy's Own Treasury;

(12, 6, 4): Endless Amusement II; The Sociable; Book of 500 Puzzles; Boy's Own Book; Cassell's; Hoffmann; Wehman; Rudin; Criton

(12, 7, 4) -- Trick version of a 3 x 3 square with doubled diagonal: Secret Out; Hoffmann (1876); Mittenzwey; Hoffmann (1893), no. 8

(12, 7, 4): Dudeney (1917); Putnam

(12, 19, 3): Macmillan

(13, 9, 4): Criton

(13, 12, 3): Criton

(13, 18, 3): Sylvester

(13, 22, 3): Criton

(15, 15, 3): Jackson

(15, 16, 3): The Sociable; Book of 500 Puzzles; H. D. Northrop; Wehman

(15, 23, 3): Jackson

(15, 26, 3): Woolhouse

(16, 10, 4): The Sociable; Book of 500 Puzzles; Hoffmann; Wehman

(16, 12, 4): Criton

(16, 15, 4): Dudeney (1899, 1902, 1908); Brooke; Putnam; Criton

(17, 24, 3): Jackson

(17, 28, 3): Endless Amusement II; Pearson

(17, 32, 3): Sylvester

(17, 7, 5): Ripley's

(18, 18, 4): Macmillan

(19, 19, 4): Criton

(19, 9, 5): Endless Amusement II; The Sociable; Book of 500 Puzzles; Proctor; Hoffmann; Clark; Wehman; Ripley; Rudin; Putnam; Criton

(19, 10, 5): Proctor

(20, 12, 5): trick method: Doubleday - 3

(20, 18, 4): Loyd Jr

(20, 21, 4): Criton

(21, 9, 5): Magician's Own Book; Book of 500 Puzzles; Boy's Own Conjuring Book; Blyth; Depew

(21, 10, 5): Mittenzwey

(21, 11, 5): Putnam

(21, 12, 5): Dudeney (1917); Criton

(21, 30, 3): Secret Out; Hoffmann

(21, 50, 3): Sylvester

(22, 15, 5): Macmillan

(22, 20, 4): Dudeney (1899)

(22, 21, 4): Dudeney (1917); Putnam

(24, 28, 3): Jackson; Parlour Pastime

(24, 28, 4): Jackson; Héraud; Benson; Macmillan

(24, 28, 5): Jackson

(25, 12, 5): Endless Amusement II; Young Man's Book; Proctor; Criton

(25, 18, 5): Bridges

(25, 30, 4): Macmillan

(25, 72, 3): Sylvester

(26, 21, 5): Macmillan

(27, 9, 6): The Sociable; Book of 500 Puzzles; Hoffmann; Wehman

(27, 10, 6): The Sociable; Book of 500 Puzzles; Wehman

(27, 15, 5): Jackson

(29, 98, 3): Sylvester

(30, 12, 7): Criton

(30, 22, 5): Criton

(30, 26, 5): Macmillan

(31, 6, 6) -- with 7 circles of 6: The Sociable; Book of 500 Puzzles; Magician's Own Book (UK version); Wehman

(31, 15, 5): Proctor

(36, 55, 4): Macmillan

(37, 18, 5): Proctor

(37, 20, 5): The Sociable; Book of 500 Puzzles; Illustrated Boy's Own Treasury; Hanky Panky; Wehman

(49, 16, 7): Criton
Trick versions -- with doubled counters: Family Friend (1858); Secret Out; Illustrated Boy's Own Treasury; Hoffmann (1876); Mittenzwey; Hoffmann (1893), nos. 8 & 9; Pearson; Home Book ....; Doubleday - 3. These could also be considered as in 7.Q.2 or 7.Q.

A different type of configuration problem is considered by Shepherd, 1947.

Jackson. Rational Amusement. 1821. Trees Planted in Rows, nos. 1-10, pp. 33-34 & 99-100 and plate IV, figs. 1-9. [Brooke and others say this is the earliest statement of such problems.]

1. (9, 10, 3). Quoted in Burr, below.

"Your aid I want, nine trees to plant

In rows just half a score;

And let there be in each row three.

Solve this: I ask no more."

2. (n, n, 3), He does the case n = 15.

3. (15, 23, 3).

4. (17, 24, 3).

5. (24, 24, 3) with a pond in the middle.

6. (24, 28, 4).

7. (27, 15, 5)

8. (25, 28, c) with c = 3, 4, 5.

9. (90, 10, 10) with equal spacing -- decagon with 10 trees on each side.

10. Leads to drawing square lattice in perspective with two vanishing points, so the diagonals of the resulting parallelograms are perpendicular.

Endless Amusement II. 1826?

Prob. 13, p. 197. (19, 9, 5). = New Sphinx, c1840, p. 135.

Prob. 14, p. 197. (12, 6, 4). = New Sphinx, c1840, p. 135.

Prob. 26, p. 202. (25, 12, 5). Answer is a 5 x 5 square array.

Ingenious artists, how may I dispose

Of five-and-twenty trees, in just twelve rows;

That every row five lofty trees may grace,

Explain the scheme -- the trees completely place.

Prob. 35, p. 212. (17, 28, 3). [This is the problem that is replaced in the 1837 ed.]

Young Man's Book. 1839. P. 239. Identical to Endless Amusement II.

Crambrook. 1843. P. 5, no. 15: The Puzzle of the Steward and his Trees. This may be a configuration problem -- ??

Boy's Own Book. 1843 (Paris): 438 & 442, no. 15: "Is it possible to place twelve pieces of money in six rows, so as to have four in each row?" I. e. (12, 6, 5). = Boy's Treasury, 1844, pp. 426 & 429, no. 13. = de Savigny, 1846, pp. 355 & 358, no. 11.

Family Friend 1 (1849) 148 & 177. Family Pastime -- Practical Puzzles -- 1. The puzzle of the stars. (9, 10, 3).

Friends of the Family Friend, pray show

How you nine stars would so bestow