Writings of Charles S. Peirce: A Chronological Edition, Volume 8. Charles S. Peirce

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      This restores the original order because 12 × 31 = 372, and 53 into 372 goes 7 times and 1 over; so that

      12 × 31 ≡ 1 (mod 53);

      that is, the two dealings are equivalent to multiplying by 1; that is, they leave the cards in their original order.

      You, Barbara, come from an ancient and a proud family. Conscious of being raised above the necessity of using ideas, you scorn them in your own exalted circle, while excusing them in common heads. Your cousins Baroco and Bocardo were always looked upon askance in the family, because they were suspected of harboring ideas,—a quite baseless suspicion, I am sure. But do you know that the unremitting study of years has tempted me to favor a belief subversive of your kindred’s supremacy, and of those principles of logic that are accepted upon all hands, I mean a belief that one secret of the art of reasoning is to think? In this matter of card-multiplication, instead of conceiving the dealing out into piles as one operation and the gathering in as another, I would prefer a general formula which shall describe both processes as one. At the outset, the cards being in no matter what order, we may conceive them as spread out into a row of 53 piles of 1 card each. If the cards are in their proper order, the last card is the Joker. In any case, you will permit me to call any pile that it may head the Ultima. The dealing out of the cards may be conceived to begin by our taking piles (single cards, at first) from the beginning of the row and putting them down in successive places following the ultima, until we reach the pile which we propose to make the final one, and which is destined to receive all the cards. When in this proceeding, we have reached the final pile, let us say that we have completed the first “round.” Thereupon we go back to the pile after the ultima as the next one upon which we will deposit a pile. We may complete a number of rounds each ending with placing a pile (a single card) on the final pile. We make as many as the number of cards in the pack will permit, and we will call these the rounds of the “first set.” It will be found useful, by the way, to note their number. Having completed them, we go on just as if we were beginning another; but when we have moved the ultima, let us say that we have completed the first round of the second set. Every round of the first set ends by placing a pile on the final pile. Let us call such a round “a round of the odd kind.” Every round of the second set ends by moving the ultima. Let us call such a round a round of the even kind. We make as many rounds of this kind as the whole number of places after the ultima enables us to complete. We call these the rounds of the second set. We then return to making rounds of the odd kind and make as many as the number of piles before the ultima enables us to make. So we go alternating sets of rounds of the odd and the even kind, until finally the ultima is placed upon the final pile; and then the multiplication process is finished.

      I will now explain to you the object of counting the rounds. But first let me remark that the last round, which consists in placing the ultima upon the final pile, should always be considered as a round of the odd kind. When you dealt into 12 piles and gathered them up, with the first 48 cards you performed 4 rounds of 12 cards each, and had 5 cards left over. These five you dealt out, making the first round of the second set; and then you transferred these five piles over to the tops of the second five, making another round of the second set. Then from these five piles you dealt to the other two piles twice, making two rounds of the third set. Next the ultima was placed upon the next pile, making a round of the fourth set. Finally the ultima was placed on the last pile which, being a round of an odd set, belonged to the fifth set. So the numbers of rounds were

      4, 2, 2, 1, 1.

      From this row of numbers, which we will call the M’s, we make a second row, which we will call the N’s. The first two N’s are 0, 1, the rest are formed by multiplying the last by the first M not already used and adding to the product the last N but one. Thus the N’s are

      0, 1, 4, 9, 22, 31, 53.

      The last N is 53. It will always be the number of cards in the pack. Reversing the order of the M’s

      1, 1, 2, 2, 4

      will make no difference in the last N. Thus, the N’s will be

      0, 1, 1, 2, 5, 12, 53.

      Leave off the first M, and the last N will be the number of piles. Thus from

      2, 2, 1, 1

      we get

      0, 1, 2, 5, 7, 12.

      Leaving off the last, will give the number of piles into which you must deal to restore the order. Thus from

      4, 2, 2, 1

      we get

      0, 1, 4, 9, 22, 31.

      If you deal 53 cards into 37 piles, the numbers of rounds will be

      1, 2, 3, 4, 1.

      If you deal into 34 piles the numbers will be

      1, 1, 1, 3, 1, 2, 1.

      If you deal into 33 piles, the numbers will be

      1, 1, 1, 1, 1, 5, 1.

      If you deal into 32 piles, the numbers will be

      1, 1, 1, 1, 10, 1.

      If you deal into 30 piles, the numbers will be

      1, 1, 3, 3, 2.

      You perceive that the object of counting the rounds is to find out how many piles you must deal into to restore the proper order, and consequently by multiplication how many piles you must deal into to make any given card the first.

      Going back to 10 cards, if we were to deal them into 5 piles or 2 piles, the piles could not be taken up so as to conform to the rule. The reason is that 5 and 2 exactly divide 10; so that the last card falls on the last pile, and there is no pile to the right of the last card upon which to pile the others. To avoid that inconvenience, we had best deal only with packs having a prime number of cards, or one less than a prime number; for, in the last case, we can imagine an additional last card which remains in the zero place, as long as there is only multiplication, no addition; that is, as long as the pack is not cut.

      If we deal a pack of 10 cards into 3 piles twice or into 7 piles twice, we multiply by − 1; for 3 × 3 = 9 and 7 × 7 = 9, and 9 is one less than 0 or 10. Suppose, then, starting with 10 cards in their proper order we deal them into 3 piles (or 7 piles) and, taking them up according to the rule, next lay them down backs up in a circle, thus:—