Sources page biographical material

5.B.1. LOWERING FROM TOWER PROBLEM
The problem is for a collection of people (and objects or animals) to lower themselves from a window using a rope over a pulley, with baskets at each end. The complication is that the baskets cannot contain very different weights, i.e. there is a maximum difference in the weights, otherwise they go too fast. This is often attributed to Carroll.
Carroll-Collingwood. 1899. P. 318 (Collins: 232-233 (232 is lacking in my copy)). = Carroll-Wakeling II, prob. 4: The captive queen, pp. 8 & 65-66. 3 people of weights 195, 165, 90 and a weight of 75, with difference at most 15. He also gives a more complex form. No solutions. Although the text clearly says 165, the prevalence of the exact same problem with 165 replaced by 105 makes me wonder if this was a misprint?? Wakeling says there is no explicit evidence that Carroll invented this, and neither book assigns a date, but Carroll seems a more original source than the following and he was more active before 1890 than after.

An addition is given in both books: add three animals, weighing 60, 45, 30.

Lemon. 1890. The prisoners in the tower, no. 497, pp. 65 & 116. c= Sphinx, The escape, no. 113, pp. 19 & 100 101. Three people of weights 195, 105, 90 with a weight of 75. The difference in weights cannot be more than 15.

Hoffmann. 1893. Chap. IV, no. 28: The captives in the tower, pp. 150 & 196 = Hoffmann Hordern, p. 123. Same as Lemon.

Brandreth Puzzle Book. Brandreth's Pills (The Porous Plaster Co., NY), nd [1895]. P. 3: The captives in the tower. Same as Lemon. Identical to Hoffmann. With colour picture. No solution.

Loyd. The fire escape puzzle. Cyclopedia, 1914, pp. 71 & 348. c= MPSL2, prob. 140, pp. 98 99 & 165. = SLAHP: Saving the family, pp. 59 & 108. Simplified form of Carroll's problem. Man, wife, baby & dog, weighing a total of 390.

Williams. Home Entertainments. 1914. The escaping prisoners, pp. 126-127. Same as Lemon.

Rudin. 1936. No. 92, pp. 31-32 & 94. Same as Lemon.

Haldeman-Julius. 1937. No. 150: Fairy tale, pp. 17 & 28. Same as Lemon, except the largest weight is printed as 196, possibly an error.

Kinnaird. Op. cit. in 1 -- Loyd. 1946. Pp. 388 389 & 394. Same as Lemon.

Simon Dresner. Science World Book of Brain Teasers. Scholastic Book Services, NY, 1962. Prob. 61: Escape from the tower, pp. 29 & 99 100. Same as Lemon.

Robert Harbin [pseud. of Ned Williams]. Party Lines. Oldbourne, London, 1963. Escape, p. 29. As in Lemon.

Howard P. Dinesman. Superior Mathematical Puzzles. Allen & Unwin, London, 1968. No. 60: The tower escape, pp. 78 & 118. Same as Carroll. Answer in 15 stages. He cites Carroll, noting that Carroll did not give a solution and he asks if a shorter solution can be found.

F. Geoffrey Hartswick. In: H. O. Ripley & F. G. Hartswick; Detectograms and Other Puzzles; Scholastic Book Services, NY, 1969. No. 15: Stolen treasure puzzle, pp. 54 55 & 87. Same as Lemon.

5.B.2. CROSSING A BRIDGE WITH A TORCH
New section.

Four people have to get across a bridge which is dark and needs to be lit with the torch. The torch can serve for at most two people and the gap is too wide to throw the torch across, so the torch has to be carried back and forth. The various people are of different ages and require 5, 10, 20, 25 minutes to cross and when two cross, they have to go at the speed of the slower. But the torch (= flashlight) battery will only last an hour. Can it be done? I heard this about 1997, when it was claimed to be used by Microsoft in interviewing candidates. I never found any history of it, until I recently found a discussion on Torsten Sillke's site: Crossing the bridge in an hour (www.mathematik.uni bielefeld.de/~sillke/PUZZLES/crossing-bridge), starting in Jun 1997 and last updated in Sep 2001. This cites the 1981 source and the other references below. Denote the problem with speeds a, b, c, d and total time t by

(a, b, c, d; t), etc. t is sometimes given, sometimes not.
Saul X. Levmore & Elizabeth Early Cook. Super Strategies for Puzzles and Games. Doubleday, 1981, p. 3 -- ??NYS. (5, 10, 20, 25; 60), as in the introduction to this section..

Heinrich Hemme. Das Problem des Zwölf-Elfs. Vandenhoeck & Ruprecht, 1998. Prob. 81: Die Flucht, pp. 40 & 105-106, citing a web posting by Gunther Lientschnig on 4 Dec 1996. (2, 4, 8, 10; t).

Dick Hess. Puzzles from Around the World. Apr 1997. Prob. 107: The Bridge.

(1, 2, 5, 10; 17). Poses versions with more people: (1, 3, 4, 6, 8, 9; 31) and, with a three-person bridge, (1, 2, 6, 7, 8, 19, 10; 25).

Quantum (May/Jun 1997) 13. Brainteaser B 205: Family planning. Problem (1, 3, 8, 10; 20).

Karen Lingel. Email of 17 Sep 1997 to rec.puzzles. Careful analysis, showing that the 'trick' solution is better than the 'direct' solution if and only if a + c > 2b. [Indeed, a + c - 2b is the time saved by the 'trick' solution.] She cites (2, 3, 5, 8; 19) and (2, 2, 3, 3; 11) to Sillke and (1, 3, 6, 8, 12; 30), from an undated website. Expressing the solution for more people seems to remain an open question.

5.C. FALSE COINS WITH A BALANCE
See 5.D.3 for use of a weighing scale.

There are several related forms of this problem. Almost all of the items below deal with 12 coins with one false, either heavy or light, and its generalizations, but some other forms occur, including the following.

8 coins, 1 light: Schell, Dresner

26 coins, 1 light: Schell

8 coins, 1 light: Bath (1959)

9 coins, 1 light: Karapetoff, Meyer (1946), Meyer (1948), M. Adams, Rice

I have been sent an article by Jack Sieburg; Problem Solving by Computer Logic; Data Processing Magazine, but the date is cut off -- ??
E. D. Schell, proposer; M. Dernham, solver. Problem E651 -- Weighed and found wanting. AMM 52:1 (Jan 1945) 42 & 7 (Aug/Sep 1945) 397. 8 coins, at most one light -- determine the light one in two weighings.

Benjamin L. Schwartz. Letter: Truth about false coins. MM 51 (1978) 254. States that Schell told Michael Goldberg in 1945 that he had originated the problem.

Emil D. Schell. Letter of 17 Jul 1978 to Paul J. Campbell. Says he did NOT originate the problem, nor did he submit the version published. He first heard of it from Walter W. Jacobs about Thanksgiving 1944 in the form of finding at most one light coin among 26 good coins in three weighings. He submitted this to the AMM, with a note disclaiming originality. The AMM problem editor published the simpler version described above, under Schell's name. Schell says he has heard Eilenberg describe the puzzle as being earlier than Sep 1939. Campbell wrote Eilenberg, but had no response.

Schell's letter is making it appear that the problem derives from the use of 1, 3, 9, ... as weights. This usage leads one to discover that a light coin can be found in 3ⁿ coins using n weighings. This is the problem mentioned by Karapetoff. If there is at most one light coin, then n weighings will determine it among 3ⁿ   1 coins, which is the form described by Schell. The problem seems to have been almost immediately converted into the case with one false coin, either heavy or light.

Walter W. Jacobs. Letter of 15 Aug 1978 to Paul J. Campbell. Says he heard of the problem in 1943 (not 1944) and will try to contact the two people who might have told it to him. However, Campbell has had no further word.
V. Karapetoff. The nine coin problem and the mathematics of sorting. SM 11 (1945) 186 187. Discusses 9 coins, one light, and asks for a mathematical approach to the general problem. (?? -- Cites AMM 52, p. 314, but I cannot find anything relevant in the whole volume, except the Schell problem. Try again??)

Dwight A. Stewart, proposer; D. B. Parkinson & Lester H. Green, solvers. The counterfeit coin. In: L. A. Graham, ed.; Ingenious Mathematical Problems and Methods; Dover, 1959; pp. 37 38 & 196 198. 12 coins. First appeared in Oct 1945. Original only asks for the counterfeit, but second solver shows how to tell if it is heavy or light.

R. L. Goodstein. Note 1845: Find the penny. MG 29 (No. 287) (Dec 1945) 227 229. Non optimal solution of general problem.

Editorial Note. Note 1930: Addenda to Note 1845. Ibid. 30 (No. 291) (Oct 1946) 231. Comments on how to extend to optimal solution.

Howard D. Grossman. The twelve coin problem. SM 11:3/4 (Sep/Dec 1945) 360 361. Finds counterfeit and extends to 36 coins.

Lothrop Withington, Jr. Another solution of the 12 coin problem. Ibid., 361 362. Finds also whether heavy or light.

Donald Eves, proposer; E. D. Schell & Joseph Rosenbaum, solvers. Problem E712 -- The extended coin problem. AMM 53:3 (Mar 1946) 156 & 54:1 (Jan 1947) 46 48. 12 coins.

Jerome S. Meyer. Puzzle Paradise. Crown, NY, 1946. Prob. 132: The nine pearls, pp. 94 & 132. Nine pearls, one light, in two weighings.

N. J. Fine, proposer & solver. Problem 4203 -- The generalized coin problem. AMM 53:5 (May 1946) 278 & 54:8 (Oct 1947) 489 491. General problem.

H. D. Grossman. Generalization of the twelve coin problem. SM 12 (1946) 291 292. Discusses Goodstein's results.

F. J. Dyson. Note 1931: The Problem of the Pennies. MG 30 (No. 291) (Oct 1946) 231 234. General solution.

C. A. B. Smith. The Counterfeit Coin Problem. MG 31 (No. 293) (Feb 1947) 31 39.

C. W. Raine. Another approach to the twelve coin problem. SM 14 (1948) 66 67. 12 coins only.

K. Itkin. A generalization of the twelve coin problem. SM 14 (1948) 67 68. General solution.

Howard D. Grossman. Ternary epitaph on coin problems. SM 14 (1948) 69 71. Ternary solution of Dyson & Smith.

Jerome S. Meyer. Fun-to-do. A Book of Home Entertainment. Dutton, NY, 1948. Prob. 40: Nine pearls, pp. 41 & 188. Nine pearls, one light, in two weighings.

Blanche Descartes [pseud. of Cedric A. B. Smith]. The twelve coin problem. Eureka 13 (Oct 1950) 7 & 20. Proposal and solution in verse.

J. S. Robertson. Those twelve coins again. SM 16 (1950) 111 115. Article indicates there will be a continuation, but Schaaf I 32 doesn't cite it and I haven't found it yet.

E. V. Newberry. Note 2342: The penny problem. MG 37 (No. 320) (May 1953) 130. Says he has made a rug showing the 120 coins problems and makes comments similar to Littlewood's, below.

J. E. Littlewood. A Mathematician's Miscellany. Methuen, London, 1953; reprinted with minor corrections, 1957 (& 1960). [All the material cited is also in the later version: Littlewood's Miscellany, ed. by B. Bollobás, CUP, 1986, but on different pages. Since the 1953 ed. is scarce, I will also cite the 1986 pages in (  ).] Pp. 9 & 135 (31 & 114). "It was said that the 'weighing pennies' problem wasted 10,000 scientist hours of war work, and that there was a proposal to drop it over Germany."

John Paul Adams. We Dare You to Solve This! Berkley Publishing, NY, nd [1957?]. [This is apparently a collection of problems used in newspapers. The copyright is given as 1955, 1956, 1957.] Prob. 18: Weighty problem, pp. 13 & 46. 9 equal diamonds but one is light, to be found in 2 weighings.

Hubert Phillips. Something to Think About. Revised ed., Max Parrish, London, 1958. Foreword, p. 6 & prob. 115: Twelve coins, pp. 81 & 127 128. Foreword says prob. 115 has been added to this edition and "was in oral circulation during the war. So far as I know, it has only appeared in print in the Law Journal, where I published both the problem and its solution." This may be an early appearance, so I should try and track this down. ??NYS

Dan Pedoe. The Gentle Art of Mathematics. (English Universities Press, 1958); Pelican (Penguin), 1963. P. 30: "We now come to a problem which is said to have been planted over here during the war by enemy agents, since Operational Research spent so many man hours on its solution."

Philip E. Bath. Fun with Figures. The Epworth Press, London, 1959. No. 7: No weights -- no guessing, pp. 8 & 40. 8 balls, including one light, to be determined in two weighings. Method actually works for  1 light.

M. R. Boothroyd & J. H. Conway. Problems drive, 1959. Eureka 22 (Oct 1959) 15-17 & 22-23. No. 9. Five boxes of sugar, but some has been taken from one box and put in another. Determine which in least number of weighings. Does by weighing each division of A, B, C, D into two pairs.

Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. The "False Coin" problem, pp. 178-182. Sketches history and solution.

Simon Dresner. Science World Book of Brain Teasers. 1962. Op. cit. in 5.B.1. Prob. 46: Dud reckoning, pp. 21 & 94. Find one light among eight in two weighings.

Philip Kaplan. More Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 55, pp. 57 & 98. Six identical appearing coins, three of which are identically heavy. In two weighings, identify two of the heavy coins.

Charlie Rice. Challenge! Hallmark Editions, Kansas City, Missouri, 1968. Prob. 7, pp. 22 & 54-55. 9 pearls, one light.

Jonathan Always. Puzzling You Again. Tandem, London, 1969. Prob. 86: Light weight contest, pp. 51 52 & 106 107. 27 weights of sizes 1, 2, ..., 27, except one is light. Find it in 3 weighings. He divides into 9 sets of three having equal weights. Using two weighings, one locates the light weight in a set of three and then weighing two of these with good weights reveals the light one. [3 weights 1, 2, 3 cannot be done in one weighing, but 9 weights 1, 2, ..., 9 can be done in two weighings.]

Robert H. Thouless. The 12 balls problem as an illustration of the application of information theory. MG 54 (No. 389) (Oct 1970) 246 249. Uses information theory to show that the solution process is essentially determined.

Ron Denyer. Letter. G&P, No. 37 (Jun 1975) 23. Asks for a mnemonic for the 12 coins puzzles. He notes that one can use three predetermined weighings and find the coin from the three answers.

Basil Mager & E. Asher. Letters: Coining a mnemonic. G&P, No. 40 (Sep 1975) 26. One mnemonic for a variable method, another for a predetermined method.

N. J. Maclean. Letter: The twelve coins. G&P, No. 45 (Feb 1976) 28-29. Exposits a ternary method for predetermined weighings for (3ⁿ-3)/2 in n weighings. Each weighing determines one ternary digit and the resulting ternary number gives both the coin and whether it is heavy or light.

Tim Sole. The Ticket to Heaven and Other Superior Puzzles. Penguin, 1988. Weighty problems -- (iii), pp. 124 & 147. Nine equal pies, except someone has removed some filling from one and inserted it in a pie, possibly the same one. Determine which, if any, are the heavy and light ones in 4 balancings.

Calvin T. Long. Magic in base 3. MG 76 (No. 477) (Nov 1992) 371-376. Good exposition of the base 3 method for 12 coins.

Ed Barbeau. After Math. Wall & Emerson, Toronto, 1995. Problems for an equal-arm balance, pp. 137-141.

1. Six balls, two of each of three colours. One of each colour is lighter than normal and all light weights are equal. Determine the light balls in three weighings.

2. Five balls, three normal, one heavy, one light, with the differences being equal, i.e. the heavy and the light weigh as much as two normals. Determine the heavy and light in three weighings.

3. Same problem with nine balls and seven normals, done in four weighings.

5.C.1 RANKING COINS WITH A BALANCE
If one weighs only one coin against another, this is the problem of sorting except that we don't actually put the objects in order. If one weighs pairs, etc., this is a more complex problem.
J. Schreier. Mathesis Polska 7 (1932) 154 160. ??NYS -- cited by Steinhaus.

Hugo Steinhaus. Mathematical Snapshots. Not in Stechert, NY, 1938, ed. OUP, NY: 1950: pp. 36 40 & 258; 1960: pp. 51 55 & 322; 1969 (1983): pp. 53 56 & 300. Shows n objects can be ranked in M(n) = 1 + kn   2^k steps where k = 1 + [log₂ n]. Gets M(5) = 8.

Lester R. Ford Jr. & Selmer M. Johnson. A tournament problem. AMM 66:5 (May 1959) 387 389. Note that log₂ n! = L(n) is a lower bound from information theory. Obtain a better upper bound than Steinhaus, denoted U(n), which is too complex to state here. For convenience, I give the table of these values here.
n 1 2 3 4 5 6 7 8 9 10 11 12 13

M(n) 0 1 3 5 8 11 14 17 21 25 29 33 37

U(n) 0 1 3 5 7 10 13 16 19 22 26 30 34

L(n) 0 1 3 5 7 10 13 16 19 22 26 29 33

U(n) = L(n) also holds at n = 20 and 21.
Roland Sprague. Unterhaltsame Mathematik. Op. cit. in 4.A.1. 1961. Prob. 22: Ein noch ungelöstes Problem, pp. 16 & 42 43. (= A still unsolved problem, pp. 17 & 48 49.) Sketches Steinhaus's method, then does 5 objects in 7 steps. Gives the lower bound L(n) and says the case n = 12 is still unsolved.

Kobon Fujimura, proposer; editorial comment. Another balance scale problem. RMM 10 (Aug 1962) 34 & 11 (Oct 1962) 42. Eight coins of different weights and a balance. How many weighings are needed to rank the coins? In No. 11, it says the solution will appear in No. 13, but it doesn't appear there or in the last issue, No. 14. It also doesn't appear in the proposer's Tokyo Puzzles.

Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 6: In the balance, pp. 18 & 85-86. Rank five balls in order in seven weighings.

John Cameron. Establishing a pecking order. MG 55 (No. 394) (Dec 1971) 391 395. Reduces Steinhaus's M(n) by 1 for n  5, but this is not as good as Ford & Johnson.

W. Antony Broomhead. Letter: Progress in congress? MG 56 (No. 398) (Dec 1972) 331. Comments on Cameron's article and says Cameron can be improved. States the values U(9) and U(10), but says he doesn't know how to do 9 in 19 steps. Cites Sprague for numerical values, but these don't appear in Sprague -- so Broomhead presumably computed L(9) and L(10). He gets 10 in 23 steps, which is better than Cameron.

Stanley Collings. Letter: More progress in congress. MG 57 (No. 401) (Oct 1973) 212 213. Notes the ambiguity in Broomhead's reference to Sprague. Improves Cameron by 1 (or more??) for n  10, but still not as good as Ford & Johnson.

L. J. Upton, proposer; Leroy J. Myers, solver. Problem 1138. CM 12 (1986) 79 & 13 (1987) 230 231. Rank coins weighing 1, 2, 3, 4 with a balance in four weighings.
5.D. MEASURING PROBLEMS
5.D.1. JUGS & BOTTLES
See MUS I 105-124, Tropfke 659.

NOTATION: I-(a, b, c) means we have three jugs of sizes a, b, c with a full and we want to divide a in half using b and c. We normally assume a  b  c and GCD(a, b, c) = 1. Halving a is clearly impossible if GCD(b, c) does not divide a/2 or if b+c < a/2, unless one has a further jug or one can drink some. If a  b+c  a/2 and GCD(b, c) divides a/2, then the problem is solvable.

More generally, the question is to determine what amounts can be produced, i.e. given a, b, c as above, can one measure out an amount d? We denote this by II-(a, b, c; d). Since this also produces a-d, we can assume that d  a/2. Then we must have d  b+c for a solution. When a  b+c  d, the condition GCD(b, c)  d guarantees that d can be produced. This also holds for a = b+c 1 and a = b+c 2. The simplest impossible cases are I (4, 4, 3) = II-(4, 4, 3; 2) and II (5, 5, 3; 1). Case I (a, b, c) is the same as II-(a, b, c; a/2).

If a is a large source, e.g. a stream or a big barrel, we have the problem of measuring d using b and c without any constraint on a and we denote this II-(, b, c; d). However, the solution may not use the infiniteness of the source and such a problem may be the same as II (b+c, b, c; d).

The general situation when a < b+c is more complex and really requires us to consider the most general three jug problem: III (A; a, b, c; d) means we have three jugs of sizes a, b, c, containing a total amount of liquid A (in some initial configuration) and we wish to measure out d. In our previous problems, we had A = a. Clearly we must have a+b+c  A. Again, producing d also produces A-d, so we can assume d  A/2. By considering the amounts of empty space in the containers, the problem III-(A; a, b, c; d) is isomorphic to III (a+b+c A; a, b, c; d') for several possible d'.

NOTES. I have been re-examining this problem and I am not sure if I have reached a final interpretation and formulation. Also, I have recently changed to the above notation and I may have made some errors in so doing. I have long had the problem in my list of projects for students, but no one looked at it until 1995-1996 when Nahid Erfani chose it. She has examined many cases and we have have discovered a number of properties which I do not recall seeing. E.g. in case I-(a,b,c) with a  b  c and GCD(b,c) = 1, there are two ways to obtain a/2. If we start by pouring into b, it takes b + c - 1 pourings; if we start by pouring into c, it takes b + c pourings; so it is always best to start pouring into the larger jug. A number of situations II-(a,b,c;d) are solvable for all values of d, except a/2. E.g. II (a,b,c;a/2) with b+c > a and c > a/2 is unsolvable.

From about the mid 19C, I have not recorded simple problems.
I-( 8, 5, 3): almost all the entries below

I-(10, 6, 4): Pacioli, Court

I-(10, 7, 3): Yoshida

I-(12, 7, 5): Pacioli, van Etten/Henrion, Ozanam, Bestelmeier, Jackson, Manuel des Sorciers, Boy's Own Conjuring Book

I-(12, 8, 4): Pacioli

I-(12, 8, 5): Bachet, Arago

I-(16, 9, 7): Bachet-Labosne

I-(16,11, 6): Bachet-Labosne

I-(16,12, 7): Bachet-Labosne

I-(20,13, 9): Bachet-Labosne

I-(42,27,12): Bachet-Labosne
II-(10,3,2;6) Leacock

= II(10,3,2;4)

II-(11,4,3;9): McKay

= II(11,4,3;2)

II-( ,5,3;1): Wood, Serebriakoff, Diagram Group

II-( ,5,3;4): Chuquet, Wood, Fireside Amusements,

II-( ,7,4;5): Meyer, Stein, Brandes

II-( ,8,5;11): Young World,

III-(20;19,13,7;10): Devi
General problem, usually form I, sometimes form II: Bachet Labosne, Schubert, Ahrens, Cowley, Tweedie, Grossman, Buker, Goodstein, Browne, Scott, Currie, Sawyer, Court, O'Beirne, Lawrence, McDiarmid & Alfonsin.

Versions with 4 or more jugs: Tartaglia, Anon: Problems drive (1958), Anon (1961), O'Beirne.

Impossible versions: Pacioli, Bachet, Anon: Problems drive (1958).
Abbot Albert. c1240. Prob. 4, p. 333. I-(8,5,3) -- one solution.

Columbia Algorism. c1350. Chap. 123: I-(8,5,3). Cowley 402 403 & plate opposite 403. The plate shows the text and three jars. I have a colour slide of the three jars from the MS.

Munich 14684. 14C. Prob. XVIII & XXIX, pp. 80 & 83. I-(8,5,3).

Folkerts. Aufgabensammlungen. 13-15C. 16 sources with I-(8,5,3).

Pseudo-dell'Abbaco. c1440. Prob. 66, p.62. I-(8,5,3) -- one solution. "This problem is of little utility ...." I have a colour slide of this.

Chuquet. 1484. Prob. 165. Measure 4 from a cask using 5 and 3. You can pour back into the cask, i.e. this is II-(,5,3;4). FHM 233 calls this the tavern-keeper's problem.

HB.XI.22. 1488. P. 55 (= Rath 248). Same as Abbot Albert.

Pacioli. De Viribus. c1500.

Ff. 97r - 97v. LIII. C(apitolo). apartire una botte de vino fra doi (To divide a bottle of wine between two). = Peirani 137-138. I-(8,5,3). One solution.

Ff. 97v - 98v. LIIII. C(apitolo). a partire unaltra botte fra doi (to divide another bottle between two). = Peirani 138-139. I-(12,7,5). Dario Uri points out that the solution is confused and he repeats himself so it takes him 18 pourings instead of the usual 11. He then says one can divide 18 among three brothers who have containers of sizes 5, 6, 7, which he does by filling the 6 and then the problem is reduced to the previous problem. [He could do it rather more easily by pouring the 6 into the 7 and then refilling the 6!]

Ff. 98v - 99r. LV. (Capitolo) de doi altri sotili divisioni. de botti co'me se dira (Of two other subtle divisions of bottles as described). = Peirani 139-140. I (10,6,4) and I-(12,8,4). Pacioli suggests giving these to idiots.

Ghaligai. Practica D'Arithmetica. 1521. Prob. 20, ff. 64v-65r. I (8,5,3). One solution.

Cardan. Practica Arithmetice. 1539. Chap. 66, section 33, f. DD.iiii.v (p. 145). I-(8,5,3). Gives one solution and says one can go the other way.

H&S 51 says I-(8,5,3) case is also in Trenchant (1566). ??NYS

Tartaglia. General Trattato, 1556, art. 132 & 133, p. 255v 256r.

Art. 132: I-(8,5,3).

Art. 133: divide 24 in thirds, using 5, 11, 13.

Buteo. Logistica. 1559. Prob. 73, pp. 282-283. I-(8,5,3).

Gori. Libro di arimetricha. 1571. Ff. 71r 71v (p. 76). I-(8,5,3).

Bachet. Problemes. 1612. Addl. prob. III: Deux bons compagnons ont 8 pintes de vin à partager entre eux également, ..., 1612: 134-139; 1624: 206-211; 1884: 138 147. I (8,5,3)  -- both solutions; I-(12,8,5) (omitted by Labosne). Labosne adds I (16,9,7); I (16,11,6); I (42,27,12); I-(20,13,9); I-(16,12,7) (an impossible case!) and discusses general case. (This seems to be the first discussion of the general case.)

van Etten. 1624. Prob. 9 (9), pp. 11 & fig. opp. p. 1 (pp. 22 23). I (8,5,3) -- one solution. Henrion's Nottes, 1630, pp. 11 13, gives the second solution and poses and solves I (12,7,5).

Hunt. 1631 (1651). P. 270 (262). I-(8,5,3). One solution.

Yoshida (Shichibei) Kōyū (= Mitsuyoshi Yoshida) (1598-1672). Jinkō ki. 2nd ed., 1634 or 1641??. ??NYS The recreational problems are discussed in Kazuo Shimodaira; Recreative Problems on "Jingōki", a 15 pp booklet sent by Shigeo Takagi. [This has no details, but Takagi says it is a paper that Shimodaira read at the 15th International Conference for the History of Science, Edinburgh, Aug 1977 and that it appeared in Japanese Studies in the History of Science 16 (1977) 95-103. I suspect this is a copy of a preprint.] This gives both Jingōki and Jinkōki as English versions of the title and says the recreational problems did not appear in the first edition, 4 vols., 1627, but did appear in the second edition of 5 vols. (which may be the first use of coloured wood cuts in Japan), with the recreational problems occurring in vol. 5. He doesn't give a date, but Mikami, p. 179, indicates that it is 1634, with further editions in 1641, 1675, though an earlier work by Mikami (1910) says 2nd ed. is 1641. Yoshida (or Suminokura) is the family name. Shimodaira refers to the current year as the 350th anniversary of the edition and says copies of it were published then. I have a recent transcription of some of Yoshida into modern Japanese and a more recent translation into English, ??NYR, but I don't know if it is the work mentioned by Shimodaira.

Shimodaira discusses a jug problem on p. 14: I-(10,7,3) -- solution in 10 moves. Shimodaira thinks Yoshida heard about such puzzles from European contacts, but without numerical values, then made up the numbers. I certainly can see no other example of these numbers. The recent transcription includes this material as prob. 7 on pp. 69-70.

Wingate/Kersey. 1678?. Prob. 7, pp. 543-544. I-(8,5,3). Says there is a second way to do it.

Witgeest. Het Natuurlyk Tover-Boek. 1686. Prob. 38, p. 308. I-(8,5,3).

Ozanam. 1694.

Prob. 36, 1696: 91-92; 1708: 82 83. Prob. 42, 1725: 238 240. Prob. 21, 1778: 175 177; 1803: 174-176; 1814: 153-154. Prob. 20, 1840: 79. I-(8,5,3) -- both solutions.

Prob. 43, 1725: 240 241. Prob. 22, 1778: 177-178; 1803: 176-177; 1814: 154-155. Prob. 21, 1840: 79 80. I-(12,7,5) -- one solution.

Dilworth. Schoolmaster's Assistant. 1743. Part IV: Questions: A short Collection of pleasant and diverting Questions, p. 168. Problem 8. I-(8,5,3). (Dilworth cites Wingate for this -- cf in 5.B.) = D. Adams; Scholar's Arithmetic; 1801, p. 200, no. 10.

Les Amusemens. 1749. Prob. 17, p. 139: Partages égaux avec des Vases inégaux. I-(8,5,3) -- both solutions.

Bestelmeier. 1801. Item 416: Die 3 Maas Gefäss. I-(12,7,5).

Badcock. Philosophical Recreations, or, Winter Amusements. [1820]. Pp. 48-49, no. 75: How to part an eight gallon bottle of wine, equally between two persons, using only two other bottles, one of five gallons, and the other of three. Gives both solutions.

Jackson. Rational Amusement. 1821. Arithmetical Puzzles.

No. 14, pp. 4 & 54. I-( 8,5,3). One solution.

No. 52, pp. 12 & 67. I-(12,7,5). One solution.

Rational Recreations. 1824. Exer. 10, p. 55. I-(8,5,3) one way.

Manuel des Sorciers. 1825. ??NX

Pp. 55-56, art. 27-28. I-(8,5,3) two ways.

P. 56, art. 29. I-(12,7,5).

Endless Amusement II. 1826? Prob. 7, pp. 193-194. I-(8,5,3). One solution. = New Sphinx, c1840, p. 133.

Nuts to Crack III (1834), no. 212. I-(8,5,3). 8 gallons of spirits.

Young Man's Book. 1839. Pp. 43-44. I-(8,5,3). Identical to Wingate/Kersey.

The New Sphinx. c1840. P. 133. I-(8,5,3). One solution.

Boy's Own Book. 1843 (Paris): 436 & 441, no. 7. The can of ale: 1855: 395; 1868: 432. I (8,5,3). One solution. The 1843 (Paris) reads as though the owners of the 3 and 5 kegs both want to get 4, which would be a problem for the owner of the 3. = Boy's Treasury, 1844, pp. 425 & 429.

Fireside Amusements. 1850. Prob. 9, pp. 132 & 184. II-(,5,3;4). One solution.

Arago. [Biographie de] Poisson (16 Dec 1850). Oeuvres, Gide & Baudry, Paris, vol. 2, 1854, pp. 593 ??? P. 596 gives the story of Poisson's being fascinated by the problem I (12,8,5). "Poisson résolut à l'instant cette question et d'autres dont on lui donna l'énoncé. Il venait de trouver sa véritable vocation." No solution given by Arago.

Parlour Pastime, 1857. = Indoor & Outdoor, c1859, Part 1. = Parlour Pastimes, 1868. Arithmetical puzzles, no. 8, pp. 174-175 (1868: 185-186). I-(8,5,3). Milkmaid with eight quarts of milk.

Magician's Own Book. 1857.

P. 223-224: Dividing the beer: I-(8,5,3).

P. 224: The difficult case of wine: I-(12,7,5).

Pp. 235-236: The two travellers: I-(8,5,3) posed in verse.

Each problem gives just one solution.

Boy's Own Conjuring Book. 1860.

P. 193: Dividing the beer: I-(8,5,3).

P. 194: The difficult case of wine: I-(12,7,5).

Pp. 202 203: The two travellers: I-(8,5,3) posed in verse.

Each problem gives just one solution.

Illustrated Boy's Own Treasury. 1860. Prob. 21, pp. 428-429 & 433. I (8,5,3). "A man coming from the Lochrin distillery with an 8-pint jar full of spirits, ...."

Vinot. 1860. Art. XXXVIII: Les cadeaux difficiles, pp. 57-58. I-(8,5,3). Two solutions.

The Secret Out (UK). c1860. To divide equally eight pints of wine ..., pp. 12-13.

Bachet-Labosne. 1874. For details, see Bachet, 1612. Labosne adds a consideration of the general case which seems to be the first such.

Kamp. Op. cit. in 5.B. 1877. No. 17, p. 326: I-(8,5,3).

Mittenzwey. 1880. Prob. 106, pp. 22 & 73-74; 1895?: 123, pp. 26 & 75-76; 1917: 123, pp. 24 & 73-74. I-(8,5,3). One solution.

Don Lemon. Everybody's Pocket Cyclopedia. Revised 8th ed., 1890. Op. cit. in 5.A. P. 135, no. 1. I-(8,5,3). No solution.

Loyd. Problem 11: "Two thieves of Damascus". Tit Bits 31 (19 Dec 1896 & 16 Jan 1897) 211 & 287. Thieves found with 2 & 2 quarts in pails of size 3 & 5. They claim the merchant measured the amounts out from a fresh hogshead. Solution is that this could only be done if the merchant drained the hogshead, which is unreasonable!

Loyd. Problem 13: The Oriental problem. Tit Bits 31 (19 Jan, 30 Jan & 6 Feb 1897) 269, 325 & 343. = Cyclopedia, 1914, pp. 188 & 364: The merchant of Bagdad. Complex problem with hogshead of water, barrel of honey, three 10 gallon jugs to be filled with 3 gallons of water, of honey and of half and half honey & water. There are a 2 and a 4 gallon measure and also 13 camels to receive 3 gallons of water each. Solution takes 521 steps. 6 Feb reports solutions in 516 and 513 steps. Cyclopedia gives solution in 506 steps.

Dudeney. The host's puzzle. London Magazine 8 (No. 46) (May 1902) 370 & 8 (No. 47) (Jun 1902) 481 482 (= CP, prob. 6, pp. 28 29 & 166 167). Use 5 and 3 to obtain 1 and 1 from a cask. One must drink some!

H. Schubert. Mathematische Mussestunden, 3rd ed., Göschen, Leipzig, 1907. Vol. 1, chap. 6, Umfüllungs Aufgaben, pp. 48 56. Studies general case and obtains some results. (The material appeared earlier in Zwölf Geduldspiele, 1895, op. cit. in 5.A, Chap. IX, pp. 110-119. The 13th ed. (De Gruyter, Berlin, 1967), Chap. 9, pp. 62 70, seems to be a bit more general (??re-read).)

Ahrens. MUS I, 1910, chap. 4, Umfüllungsaufgaben, pp. 105 124. Pp. 106 107 is Arago's story of Poisson and this problem. He also extends and corrects Schubert's work.

Dudeney. Perplexities: No. 141: New measuring puzzle. Strand Magazine 45 (Jun 1913) 710 & 46 (Jul 1913) 110. (= AM, prob. 365, pp. 110 & 235.) Two 10 quart vessels of wine with 5 and 4 quart measures. He wants 3 quarts in each measure. (Dudeney gives numerous other versions in AM.)

Loyd. Cyclopedia. 1914. Milkman's puzzle, pp. 52 & 345. (= MPSL2, prob. 23, pp. 17 & 127 128 = SLAHP: Honest John, the milkman, pp. 21 & 90.) Milkman has two full 40 quart containers and two customers with 5 and 4 quart pails, but both want 2 quarts. (Loyd Jr. says "I first published [this] in 1900...")

Williams. Home Entertainments. 1914. The measures puzzle, p. 125. I-(8,5,3).

Hummerston. Fun, Mirth & Mystery. 1924. A shortage of milk, Puzzle no. 75, pp. 164 & 183. I-(8,5,3), one solution.

Elizabeth B. Cowley. Note on a linear diophantine equation. AMM 33 (1926) 379 381. Presents a technique for resolving I-(a,b,c), which gives the result when a = b+c. If a < b+c, she only seems to determine whether the method gets to a point with A empty and neither B nor C full and it is not clear to me that this implies impossibility. She mentions a graphical method of Laisant (Assoc. Franç. Avance. Sci, 1887, pp. 218-235) ??NYS.

Wood. Oddities. 1927.

Prob. 15: A problem in pints, pp. 16-17. Small cask and measures of size 5 and 3, measure out 1 in each measure. Starts by filling the 5 and the 3 and then emptying the cask, so this becomes a variant of II-(,5,3;1).

Prob. 26: The water-boy's problem, pp. 28-29. II-(;,5,3;4).

Ernest K. Chapin. Scientific Problems and Puzzles. In: S. Loyd Jr.; Tricks and Puzzles, Vol. 1 (only volume to appear); Experimenter Publishing Co., NY, nd [1927] and Answers to Sam Loyd's Tricks and Puzzles, nd [1927]. [This book is a selection of pages from the Cyclopedia, supplemented with about 20 pages by Chapin and some other material.] P. 89 & Answers p. 8. You have a tablet that has to be dissolved in 7½ quarts of water, though you only need 5 quarts of the resulting mixture. You have 3 and 5 quart measures and a tap.

Stephen Leacock. Model Memoirs and Other Sketches from Simple to Serious. John Lane, The Bodley Head, 1939, p. 298. "He's trying to think how a farmer with a ten-gallon can and a three-gallon can and a two-gallon can, manages to measure out six gallons of milk." II-(10,3,2;6) = II-(10,3,2;4).

M. C. K. Tweedie. A graphical method of solving Tartaglian measuring puzzles. MG 23 (1939) 278 282. The elegant solution method using triangular coordinates.

H. D. Grossman. A generalization of the water fetching puzzle. AMM 47 (1940) 374 375. Shows II-(,b,c;d) with GCD(b,c) = 1 is solvable.

McKay. Party Night. 1940.

No. 18, p. 179. II-(11,4,3;9).

No. 19, pp. 179-180. I-(8,5,3).

Meyer. Big Fun Book. 1940. No. 10, pp. 165 & 753. II-(,7,4,5).

W. E. Buker, proposer. Problem E451. AMM 48 (1941) 65. ??NX. General problem of what amounts are obtainable using three jugs, one full to start with, i.e. I-(a,b,c). See Browne, Scott, Currie below.

Eric Goodstein. Note 153: The measuring problem. MG 25 (No. 263) (Feb 1941) 49 51. Shows II-(,b,c;d) with GCD(b,c) = 1 is solvable.

D. H. Browne & Editors. Partial solution of Problem E451. AMM 49 (1942) 125 127.

W. Scott. Partial solution of E451 -- The generalized water fetching puzzle. AMM 51 (1944) 592. Counterexample to conjecture in previous entry.

J. C. Currie. Partial solution of Problem E451. AMM 53 (1946) 36 40. Technical and not complete.

W. W. Sawyer. On a well known puzzle. SM 16 (1950) 107 110. Shows that I-(b+c,b,c) is solvable if b & c are relatively prime.

David Stein. Party and Indoor Games. Op. cit. in 5.B. c1950. Prob. 13, pp. 79 80. Obtain 5 from a spring using measures 7 and 4, i.e. II-(,7,4,5).

Anonymous. Problems drive, 1958. Eureka 21 (Oct 1958) 14-16 & 30. No. 8. Given an infinite source, use: 6, 10, 15 to obtain 1, 6, 7 simultaneously; 4, 6, 9, 12 to obtain 1, 2, 3, 4 simultaneously; 6, 9, 12, 15, 21 to obtain 1, 3, 6, 8, 9 simultaneously. Answer simply says the first two are possible (the second being easy) and the third is impossible.

Young World. c1960. P. 58: The 11 pint problem. II-(,8,5;11). This is the same as II (13,8,5;11) or II-(13,8,5,2).

Anonymous. Moonshine sharing. RMM 2 (Apr 1961) 31 & 3 (Jun 1961) 46. Divide 24 in thirds using cylindrical containers holding 10, 11, 13. Solution in No. 3 uses the cylindricity of a container to get it half full.

Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. "Pouring" problems -- The "robot" method. General description of the problem. Attributes Tweedie's triangular 'bouncing ball' method to Perelman, with no reference. Does I (8,5,3) two ways, also I-(12,7,5) and I-(16,9,7), then considers type II questions. Considers the problem with II-(10,6,4;d) and extends to II-(a,6,4;d) for a > 10, leaving it to the reader to "try to formulate some rule about the results." He then considers II (7,6,4;d), noting that the parallelogram has a corner trimmed off. Then considers II-(12,9,7;d) and II-(9,6,3;d).

Lloyd Jim Steiger. Letter. RMM 4 (Aug 1961) 62. Solves the RMM 2 problem by putting the 10 inside the 13 to measure 3.

Irving & Peggy Adler. The Adler Book of Puzzles and Riddles. Or Sam Loyd Up-To-Date. John Day, NY, 1962. Pp. 32 & 46. Farmer has two full 10-gallon cans. Girls come with 5-quart and 4-quart cans and each wants 2 quarts.

Philip Kaplan. More Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 80, pp. 81 & 109. Tavern has a barrel with 15 pints of beer. Two customers, with 3 pint and 5 pint jugs appear and ask for 1 pint in each jug. Bartender finds it necessary to drink the other 13 pints!

T. H. O'Beirne. Puzzles and Paradoxes. OUP, 1965. Chap. 4: Jug and bottle department, pp. 49 75. This gives an extensive discussion of Tweedie's method and various extensions to four containers, a barrel of unknown size, etc.

P. M. Lawrence. An algebraic approach to some pouring problems. MG 56 (No. 395) (Feb 1972) 13 14. Shows II-(,b,c,d) with d  b+c and GCD(b,c) = 1 is possible and extends to more jugs.

Louis Grant Brandes. The Math. Wizard. revised ed., J. Weston Walch, Portland, Maine, 1975. Prob. 5: Getting five gallons of water: II (,7,4,5).

Shakuntala Devi. Puzzles to Puzzle You. Orient Paperbacks (Vision Press), Delhi, 1976.

Prob. 53: The three containers, pp. 57 & 110. III-(20;19,13,7;10). Solution in 15 steps. Looking at the triangular coordinates diagram of this, one sees that it is actually isomorphic to II-(19,13,7;10) and this can be seen by considering the amounts of empty space in the containers.

Prob. 132: Mr. Portchester's problem, pp. 82 & 132. Same as Dudeney (1913).

Victor Serebriakoff. A Mensa Puzzle Book. Muller, London, 1982. (Later combined with A Second Mensa Puzzle Book, 1985, Muller, London, as: The Mensa Puzzle Book, Treasure Press, London, 1991.) Problem T.16: Pouring puttonos, part b, pp. 19-20 (1991: 37-38) & Answer 19, pp. 102-103 (1991: 118-119). II-( ,5,3;1).

The Diagram Group. The Family Book of Puzzles. The Leisure Circle Ltd., Wembley, Middlesex, 1984. Problem 161, with Solution at the back of the book. II-(,5,3;1), which can be done as II-(8,5,3;1).

D. St. P. Barnard. 50 Daily Telegraph Brain Twisters. 1985. Op. cit. in 4.A.4. Prob. 4: Measure for measure, pp. 15, 79 80, 103. Given 10 pints of milk, an 8 pint bowl, a jug and a flask. He describes how he divides the milk in halves and you must deduce the size of the jug and the flask.

Colin J. H. McDiarmid & Jorge Ramirez Alfonsin. Sharing jugs of wine. Discrete Mathematics 125 (1994) 279-287. Solves I-(b+c,b,c) and discusses the problem of getting from one state of the problem to another in a given number of steps, showing that GCD(b,c) = 1 guarantees the graph is connected. indeed essentially cyclic. Considers GCD(b,c)  1. Notes that the work done easily extends to a > b + c. Says the second author's PhD at Oxford, 1993, deals with more cases.

John P. Ashley. Arithmetickle. Arithmetic Curiosities, Challenges, Games and Groaners for all Ages. Keystone Agencies, Radnor, Ohio, 1997. P. 11: The spoon and the bottle. Given a 160 ml bottle and a 30 ml spoon, measure 230 ml into a bucket.