Children were perhaps the first non-peasant group to be seriously studied by folklorists. In 1846, in a note in The Athenaeum, William J. Thoms proposed the "good Saxon compound, Folklore" to designate those "manners, customs, observances, ballads, proverbs, etc., of the olden time." Thoms urged the collection and study of these disparate and seemingly "trifling and insignificant" materials because of the light that they, in aggregate, might throw on the ancient past. To illustrate his point, Thoms cited a Yorkshire children's divinatory custom. Children gather around a cherry tree and sing, "Cuckoo, Cherry Tree; Come down and tell me; How many years I have to live." Each child then shakes the tree, and the number of cherries that fall are held to indicate the lifeyears remaining. Thoms points to the relation between this Yorkshire custom and the prophetic qualities of the cuckoo's song described in Jacob Grimm's Deutsche Mythologie, and to the belief that the cuckoo will not sing its prophetic song until it has eaten its fill of cherries thrice over (Thoms 1965 [1846]:4-6).
That children's custom stemmed from ancient practice became something of a rule of thumb in the nineteenth century', and folklorists set out to establish the specific relations between a host of contemporary children's expressions and the observances of the past. Certainly, there are any number of children's games and pastimes that are ancient. Thus games such as "I Spy" and "Blind Man's Buff" collected by W. W. Newell in New York at the close of the nineteenth century, are depicted in ancient paintings and are described in second century documents (Newell 1963 [1883]:160,162-63). But many of the attempts to establish connections between children's custom and ancient practice were decidedly more tenuous.
In 1888, in The Counting-Out Rhymes of Children, Henry Carrington Bolton attempted to trace the origins and demonstrate the antiquity of the custom. Bolton regarded counting-out as having two dimensions. Counting out was a form of divination by which the unknown was reckoned through casting lots. Yet counting-out also consisted of rhymes and doggerels that established the pattern of this procedure (1888:26). Bolton was determined to show connections to ancient belief for both the practice and the verse.
Citing the Bible and the Iliad, Bolton had little difficulty in demonstrating the antiquity of casting lots to identify criminals, sacrificial victims, and military champions (1888:2S34).2 The rhymes themselves Bolton believed to derive from magical incantations that accompanied the casting of lots (although he acknowledged that the rhymes were probably "of more recent date than the custom itself" [1888:35]), and Bolton cites numerous examples of incantations-many of which seem to contain nonsense words as do children's rhymes. Bolton, however, is unable to cite any ancient example of genuine counting-out, nor is he able to directly link a single magical incantation to a contemporary children's rhyme.3 Consequently, Bolton is forced to establish his linkages between children's game and ancient divinatory practice in a more roundabout fashion.4 Bolton approvingly cites T. W. Sandrey's attempt to link counting-out with the selection of sacrificial victims by ancient Britons, and his interpretation of the Cornish rhyme "Ena, mena, bora mi; Kisca, lara, mora di; Eggs, butter, cheese, bread; Stick, stock, stone dead" as a phrase of "great antiquity" which first lays a ban on a victim's chief articles of food, and foreshadows his death by beating (Bolton 1888:43). The difficulty in establishing connections between counting-out and ancient Celtic practices of human sacrifice (also argued by Charles Taylor in The Magpie: or Chatter/is of the Pica [Glasgow 1820] ) are most tellingly revealed when folklorists fell back on scenes from nineteenth-century romantic novels to illustrate their arguments. Thus Bolton cites R. D. Blackmore's Lorna Doone (1869) for a magical charm that seems reminiscent of a counting-out rhyme (Bolton 1888:50), and the Reverend Sabine Baring-Gould quotes extensively from his own novel Perpetua (1897) to illustrate how in Gaul a maiden was counted out to be sacrificed to the spirit of the spring. As Baring-Gould confidently states: "I have ventured to reproduce this, which although fiction, undoubtedly represents what actually took place" (Baring-Gould 1913:107-111). Of course, not all nineteenth-century British folklorists were enamored with the hypothesis that counting-out rhymes derived from ancient Druidic sacrificial practices. Some thought that evidence was lacking and such interpretation was hopelessly uncritical (Northall 1892:341-342).
The survivalist approach to counting-out assumed that it was a form of lottery; a mechanism of divination whose outcome was unknown and subject to the vagaries of chance. This perspective on counting-out is supported in more recent works on children's lore, where counting-out is regarded as "similar to roulette" which consequently explains why it is considered "a fair, impartial way of choosing" among children (Knapp and Knapp 1976:27).5 However, Kenneth Goldstein showed that counting-out was less random than previously supposed. In "Strategies of Counting-Out: An Ethnographic Folklore Field Study," Goldstein discovered that children strategically manipulate counting-out in order to exert control over outcomes. He described how children might add phrases to the counting-out rhymes if they ended on a child that the counter did not want eliminated; how a repertory of rhymes with different numbers of stresses were employed by the counter to deal with different numbers of children to be counted in order to produce favorable outcomes; or how the counter might skip himself in the count if his elimination seemed likely (1989:189192). Goldstein even reported that one player had memorized the first position to be eliminated in any group of children of ten or less. After each count, this counter would reposition himself within the remaining circle in order to eliminate whomever he wished. As Goldstein notes, "This `changing position' strategy was used by one extremely precocious nine year old boy who was considered something of a mathematical genius at school" (1989:193). A child in the play group who suspected that this "genius" was somehow manipulating the count (although he didn't know how) and voiced his concern, found himself among the first to be counted out from then on (1989:193).
Although the children who employed counting-out overwhelming agreed, on the one hand, that counting-out was "democratic" and that everyone had the "same chance" (1989:188), Goldstein documented that for some children counting-out was not entirely a matter of chance but a process that could be approached strategically with knowable, or at least partially knowable, outcomes.
In his discovery of the use of strategy in counting-out, Goldstein was unaware that there existed a long-standing concern with such strategies; a tradition dating back to the late Middle Ages. The formulation of counting-out as a mathematical "problem" seems to appear first in Europe. The problem is said to first appear in the early lOth century Codex Einsidelensis (Smith and Mikami 1914:84).6 The 12th century Jewish scholar Abraham ibn Ezra mentions the problem in his work Tachbula. Ibn Ezra's description of the problem involves being aboard ship with 15 students (S) and 15 good-for-nothings (G) during a storm. As the ship needed to reduce its load, it was agreed that every ninth person would be counted out and thrown overboard until only half of the original thirty would be left. Ibn Ezra arranged them so that every ninth person would be a good-fornothing. The arrangement was 4S, 5G" 2S, 1G, 3S, IG, IS, 2G, 2S, 3G, IS, 2G, 2S, IG (Steinschneider 1880:123-124). This formulation of the counting-out problem appears in a number of the arithmetic and algebraic treatises and problem books of the Renaissance. It appears in the manuscript of the Parisian Nicolas Chuquet which was completed in 1484.7 The problem is formulated with the same parameters as that of ibn Ezra except the voyagers are Christians and Jews. The count is also by nines and the problem is to order the thirty passengers so that the Jews are thrown overboard and only the Christians remain. The solution is also the same as that provided by ibn Ezra, although Chuquet provides a mnemonic verse by which the order of the ship's passengers might be remembered: "Populeam virgam matre regina tenebat" ("The queen held the mother's poplar rod" [?]; "The queen mother used to hold the poplar rod [?]). Each vowel is accorded a numerical value (a=l; e=2; i=3; o=4; u=5) and the thirty passengers are ordered according to the numerical value of the vowels of the verse beginning first with 4 Christians. Chuquet also suggests that the problem might be formulated with different numbers of Christians and Jews and with throwing out every lOth or 11th person, instead of the ninth person (Flegg, Hay, and Moss 1985:230).8
Variations of this problem appear in the sixteenth century in the works of mathematicians Niccolo Fontana Tartaglia (1499-1557), Gerolamo Cardano [Cardan] (1501-1576), and Petrus Ramus (1515-1572), among others (Ball 1960:220-221; Smith and Mikami 1914:84 n2). In 1612, Claude-Gaspar Bachet (1581-1638) published his first edition of Problemes plaisants et delectables. Problem XXIII is identical to that propounded by Chuquet, except that Christians and Turks are counted out rather than Christians and Jews. Bachet also offered a French mnemonic verse in place of the Latin for remembering the order of arrangement-"Mort, tu me falliras pas, En me livrant le trepas" ("Death, you will not fail to deliver my demise")-although it operated according to the same vowel cipher (Bachet 1874:119).9
It was Bachet who drew attention to the relation between this countingout problem and a situation described in The Jewish War by Flavius Josephus. Josephus (Joseph ben Matthias) was a Jewish commander who was assigned the defense of the Galilee during the Jewish Revolt against Rome. He garrisoned Galilean towns, and in 67 A.D. he directed the defense of the town of Jotapata against the Romans under the command of Vespesian. When these Roman forces breached the walls of the town, and began a massacre of the inhabitants and defenders, Josephus sought refuge in a cavern under the city where he encountered forty other defenders also in hiding. These defenders had decided to take their own lives, and after Josephus failed to dissuade them, he suggested:
Since we are resolved to die, come let us leave the lot to decide the order in which we are to kill ourselves; let him who draws the first lot fall by the hand who comes next; fortune will thus take her course through the whole number and we shall be spared from taking our lives with our own hands. For it would be unjust that, when the rest were gone, any should repent and escape. This proposal inspired confidence; his advice was taken, and he drew lots with the rest. Each man thus selected presented his throat to his neighbor, in the assurance that his general was forthwith to share his fate.... He, however (should one say by fortune or by the providence of God?) was left alone with one other; and anxious neither to be condemned by the lot nor, should he be left to the last, to stain his hand with the blood of a fellow-countryman, he persuaded this man also, under a pledge, to remain alive (Josephus 1927:685-687).
Josephus makes no mention of the type of lot employed, and there is no indication that counting-out was used. Nor is there any suggestion that Josephus manipulated the outcome in order to remain among the survivors. In fact, Josephus attributes his survival only to "fortune" or the "providence of God" Josephus 1927:687).
Josephus's Greek text of The Jewish War was not the original draft of the work. There was an earlier Aramaic edition now lost Josephus 1927: ix) . There are scholars who have held, however, that the Slavonic version of The Jewish War may derive from the lost Aramaic draft Josephus 1927: xi). It seems that the Slavonic version originated in Lithuania somewhere around 1250 A.D. and was later copied in a number of manuscripts in the fifteenth century. There are some who believe that the Slavonic version was translated from a lost Greek text which may have been based on the lost Aramaic version josephus 1927: xi; Josephus 1927: xi;Josephus 1970:470).
The Slavonic version of The Jewish War abbreviates a number of events in the standard Greek text, but it also includes a number of passages not found there. One of these passages pertains to the selection of the victims in the cave at Jotapata:
And he, commending his salvation to God the Protector, said, "Since it is well pleasing to God that we should die, let us be killed in turn. Let him whose turn comes last be killed by the second." And when he had thus spoken, he counted the numbers with cunning, and thereby misled them all. And they were all killed, one by another, except one; and, anxious not to stain his right hand with the blood of a fellow-countryman, he besought this one, and they both went out alive (Josephus 1928:654).
This passage suggests that the method employed to establish the order of victims was a form of counting-out, and that Josephus somehow could foresee and manipulate the order of selection so that he would be the last to remain. But it is not clear how well-known the Slavonic version was in Western Europe and whether it had any influence on European mathematical tradition.
It would seem that Bachet independently attributed Josephus's salvation to careful reckoning, rather than providence or fate (Bachet 1879:9). Bachet also went on to formulate the story in Josephus in mathematical terms. Bachet speculated that with a total of 41 men in the cave, and if they had counted out by threes, Josephus would have had to place himself in the 31st position to be the last remaining. And if he had wanted to save two companions, they would have to be in the 16th and 35th positions, as these would be the second- and third-to-last respectively (Bachet 1879:120-21). Thus it would seem that Bachet is responsible for conceptualizing what has come to be known in contemporary mathematics as the "Josephus Problem,"11 the Slavonic version of The Jewish War notwithstanding.
Counting-out was also of concern to Japanese mathematicians of the seventeenth century. The classical formulation of the question in Japan was the Mameko-date or "step-children problem." It appears in the Jinko-ki of Yoshida (Yoshida Shichibei Koyu, or Mitsuyoshi) in 1627, and in the Mantoku Jinko-ri of Muramatsu Kudayu Mosei in 1665. Seki Shinsuke Kowa, the successor to Yoshida, included it in his Sanda.tsu Kempu where he characterizes it as an "old tradition" (Smith and Mikami 1914:80-84). The problem is more elaborate than its formulations in the West. The problem, following the version in Seki, concerns a wealthy farmer who had thirty sons, fifteen by his first wife, and fifteen by his second. The second wife wanted one of her sons to inherit all his father's wealth so she proposed that all the sons be arranged in a circle and be counted out by tens until only one remained. That one would be the father's sole heir. The husband agreed, and the second wife arranged the sons in such a way that fourteen of fifteen of her step-children were immediately counted out Feeling confident of her success, she suggested that the direction of the count be reversed beginning with a child of her choosing. This was also agreed to, but all her fifteen children were thus counted out leaving the sole remaining son of the first wife to inherit his father's fortune (Smith and Mikami 1914:82-83).12
It is not clear whether the problem originated in the East and made its way West, or whether it originated in the West, and made its way East, but as the problem in both the East and the West describes two competing groups of fifteen members each, one of which is to be eliminated, it suggests that some relationship exists between the traditions. Curiously, the problem does not seem to appear in the Chinese mathematical literature (Needham with Ling 1979:61-62).
For centuries, counting-out problems continued to appear in the mathematical treatises and puzzle books of both East and West, yet no advancement took place in the solution of the problem. The problems, it would seem, were solved by brute force physically drawing a diagram and counting-out the requisite number until a result was obtained-and then, as in the case in the West, the solution was recalled through some mnemonic device. It was not until the very end of the nineteenth century that a generalization of the problem was proposed by mathematician Peter Guthrie Tait. Tait focused on the problem created by Bachet around the predicament of Josephus in the cave at Jotapata (Tait 1898-1900, 2:432435) . Bachet had shown that with 41 men, and counting by three, the last remaining would stand in the 31st position. The question is, what would be the last position in a group of any size counting by any number. While Tait did not provide a formula by which this last position could be computed, he outlined a method that could serve to compute such a position for a variety of cases, even if the number of persons to be counted were very large.
Tait began with a simple observation.'3 If there is a group of seven people and we count by three (that is, let n equal the number of people in the group, and let m equal the number by which people are counted out; so that n = 7 and m = 3), the order of elimination is indicated in the first circle below. (Inside numbers are the order of people in the circle. Outside numbers represent the order in which they are eliminated. The arrow points to the last person eliminated.) A group of eight people (n = 8 and m = 3) is represented by the second circle, and a group of nine by the third.
Thus when n = 7, the position of the last person remaining when all the others have been counted out (a position we will call p) is the fourth. When n=8 then p=7.When n=9,p=1.
It can be seen each time we add one more person to the group, the position of the last person will increase by 3 (since we count by threes; if we counted by some other m, p would increase by whatever that m is). So if p = 4 when n = 7, p = (4 + 3) = 7 when the group equals n + 1 or 8. P should increase to 10 when n = 9, but since there are only nine people in the group, there is no person number 10. The tenth person in a circle of nine is in actuality number 1. (This we call cycling, and it will be addressed below.)
It is easy to show that if p is the last position in a circle of any group of n people, that the last position in a group of n + 1 people is p + 3. (It is not just true for the examples of n = 7, 8, or 9 in the diagrams.) To see this, we must work backwards. Start with a circle of n + 1 people (as numbered inside the circle). As soon as we start to count out by 3, the third person is eliminated. If we stop counting at this point, we now have a circle of only n people. Let us renumber the remaining circle of n people so that the person who was 4 is numbered 1', the person who was 5 is numbered 2', and so on until all the members of this new circle are renumbered. Each position in the circle of n members is 3 less than in the circle of n + 1 members. Reversing this relation, if we know p for group n, we know that for group n + 1 the last position will be p + 3. The exceptions will be when p = (n -1)' and p = n'. For in a group of size n, if p = (n -1), then in a group of (n + 1) members p will be 1. And if p = n in a group size n, then in group size (n + 1) p will be 2. (Cycling again!)
To summarize, if n is the number of people in a circle, and if p is the last person counted out in that group, and if m is the number we count by; then the last counted out in a circle of size n + 1 is p + m, or in the Josephus problem, p + 3. If we use subscripts to simplify our notation (such that P(n) refers to the p in a group of n persons, and P(n+l) is p in a group of n + 1 persons) then P(n+l) = P(n) + 3.
If we increase group n by some number other than 1, x let us say, then P(n+x) = P(n) + mx or P(n) + 3x. Thus we can add 3x to any P(n) to compute the new P(n+x), but this is only true so long as cycling does not take place. For example, when n = 9, p = 1. If we increase n by 4 [let x = 4], then p(9+4) = p(9) + 3(4) = 1 + 12 = 13. And this is in fact correct. The last person remaining when n = 13 is the 13th. But if we try to increase n = 9 by x = 5 (so n = 14 altogether), the same computation does not work: P(14) X P(9) + 3(5). That is because 1 + 15 = 16, and the new p cannot be 16 in a group with only 14 members. Cycling has occurred.
Thus 3x can be added to pP(nP to compute Pn+x) only so long as cycling does not occur between n and n + x. If cycling does occur between n and n + x, then one needs to know the n at which cycling has last occurred and compute Pn+x) from that point. Since we know that cycling has occurred when n = 14, and that P(14) cannot equal 16, we can easily compute what p should be for n = 14. All we need do is subtract 14 [i.e. n + x] from 16 [p(9) + 3(5)] and we get P(14) = 2 [since 16 -14 = 2]. Thus when n = 14, then P(14) = 2. Note that when cycling occurs, we need to subtract (n + x) from (p + 3x) to compute P(n+x).
Since we now know what p is when n is 14, we can compute the next time that cycling takes place. We know that when cycling occurs, the new p must either equal 1 or 2. To compute the next cycling, simply solve the following equation for x. (p + 3x) - (n + x) = 1 or 2
We know that p = 2 when n = 14, so the computation is simple. (2 + 3x) - 14-x= 1 or 2 2x- 12 = 1 or 2
Since the answer must be a whole number, the expression cannot equal 1. So we set it equal to 2: 2x -12 = 2; then 2x = 14; and x = 7. Then (n + x) = 21.
What this shows is that starting with n = 14 and p = 2, the next time that cycling occurs is when n = 21 and p = 2.
If we wish to solve the Josephus problem as posed by Bachet, where there are 41 men, we must compute pP(41) or who p is when n = 41. We must first determine whether cycling occurs again between n = 21 and n = 41. We repeat the computation: P(21) + 3x - (n + x) = 1 or 2. Substituting and simplifying we find: 2 + 3x - 21 - x 1 or 2. 2x = 20 and x = 10. So the next time cycling takes place is when n = 31 (i.e., 21 + 10) and p = 1. If we compute the next time cycling takes place we find it is at an n greater than the 41 men in our problem. So knowing that when n = 31, p = 1, and knowing that if we increase the size of group n = 31 by 10 to get n = 41, all we have to do is solve the following equation P(41) = P(31) + 3x, where x = 10 and pP(31) = 1. The solution p(41) = 1 + 3(10) = 31. The last person remaining when counting-out by threes in a group of 41 people is the 31st. This is exactly the solution that Bachet arrived at by brute force.
Someone might suppose that it would be easier to do the actual counting-out than engage in the computation, but this would be true only when n was very small. When n = 2,000,000, then the last remaining person is the 685,043rd. If one were able to count out on average one person per second (and assuming no errors), it would take over twentythree days and nights of continuous counting to arrive at the same conclusion. (It would actually take much longer since one would first have to write out two million numbers in order to count them.) Tait's method has value.
Where does this all lead? What is its significance? This inquiry was first conceived after I asked several mathematicians whether it would be possible to write a formula that would determine the last counted out for any group n counting by any m. No theoretical concern prompted my question; only idle curiosity. When one mathematician recognized my question as a variant of the `Josephus problem," the literature-including Tait's generalization-became accessible.14
