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| 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;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
Versions with 4 or more jugs: Tartaglia, Anon: Problems drive (1958), Anon (1961), O'Beirne. Impossible versions: Pacioli, Bachet, Anon: Problems drive (1958).
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.
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