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Familiar Letters about the Art of Reasoning

15 May 1890

Houghton Library

Stagira, May 15, 1890.

My dear Barbara:

The University of Cracow once conferred upon a very good fellow a degree for having taught the philosophical faculty to play cards. I cannot tell you in what year this happened,—perhaps it was 1499. The graduate was Thomas Murner, of whose writings Lessing said that they illustrated all the qualities of the German language; and so they do if those qualities are energy, rudeness, indecency, and a wealth of words suited to unbridled satire and unmannered invective. The diploma of the university is given in his book called Chartiludium, one of the numerous illustrations to which is copied to form the title page of the second book of a renowned encyclopaedia, the Margarita Philosophica.¹ Murner’s pack contained 51 cards. There were seven unequal suits; 3 hearts, 4 clubs (or acorns), 8 diamonds (or bells), 8 crowns, 7 scorpions, 8 fish, 6 crabs. The remaining seven cards were jokers, or unattached to suits; for such cards formed a feature of all old packs. The object of Murner’s cards was to teach the art of reasoning, and a very successful pedagogical instrument they no doubt proved.

If you will provide yourself, my dear Barbara, with a complete pack of cards with a joker, 53 in all, I will make a little lesson in mathematics go down like castor-oil in milk. Take, if you will be so kind, the 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 of spades, and arrange these ten cards in their proper order. I mean by this that the ace, or 1, is to be at the back of the pack, the 2 next, and so on, the 10 alone showing its face. I call this the “proper order,” because I propose always to begin the count of cards in a pack at the back, so that, in the pack of ten cards you have just been so obliging as to arrange, every card is in its proper place, that is the number it bears on its face is equal to the number of its place from the back of the pack. The face-value of the 2^nd card is 2, that of the 3^rd card, 3, and so on.

Now let us add 3 to the face-value of each card in the pack. How shall we do that without a printing-press? Why, by simply taking 3 cards from the back of the pack of ten and carrying them to the face. The face-value of card number 1 is now 3 + 1, or 4; that of card 2 is 5, and so on up to card 7 which is 10. Card 8 is 1; but 1 and 11 are the same for us. Since we have only ten cards to distinguish, ten different numbers are enough. We, therefore, treat 1, 11, 21, 31, as equal, because we count round and round the 10, thus:

We say 13 and 23 are equal, meaning their remainders after division by 10 are equal. This sort of equality of remainders after division is called congruence by mathematicians and they write it with three lines, thus

13 ≡ 23 (mod. 10).

The number 10 is said to be the modulus, that is, the divisor, or the smallest number congruent to zero, or the number of numbers in the cycle.

Instead of ten cards you may take the whole suit of thirteen, and then, imagining a system of numeration in which the base is thirteen and in which we count

1 2 3 4 5 6 7 8 9 10 Jack Queen King

we have a similar result. Fourteen, or king-ace, is congruent with 1; fifteen, or king-two, with 2, etc.

It makes no difference how many cards there are in a pack. To cut it, when arranged in its proper order, and transpose the two parts, is to add a constant amount to the face-value of every card. So much for addition.

Now how shall we multiply? Suppose we have the pack of ten in its proper order, and wish to multiply the face-value of the cards by 3. We deal out the cards one by one from first to last, into 3 piles, laying them face up upon the table. We first take up the pile the 10, or 0, falls upon, then the next pile, last the third. Putting each pile after the first at the back of that last taken.

We now find in place 1 card 3, or 3 times 1;

in place 2 card 6, or 3 times 2;

in place 3 card 9, or 3 times 3;

in place 4 card 2, congruent to 3 times 4;

in place 5 card 5, congruent to 3 times 5;

in place 6 card 8, congruent to 3 times 6;

in place 7 card 1, congruent to 3 times 7;

in place 8 card 4, congruent to 3 times 8;

in place 9 card 7, congruent to 3 times 9;

in place 10 card 0, congruent to 3 times 10.

Take this pack and multiply again by 3. Multiplying by 3 twice is multiplying by 9. But 9 ≡ − 1.

Accordingly we shall now find

in place 1 card − 1 or 9,

in place 2 card − 2 or 8,

in place 3 card − 3 or 7,

etc.

Multiply again by 3, and since 3 × 9 ≡ 7, we shall find

in place 1 card 7 × 1 ≡ 7,

in place 2 card 7 × 2 ≡ 4,

in place 3 card 7 × 3 ≡ 1,

in place 4 card 7 × 4 ≡ 8,

in place 5 card 7 × 5 ≡ 5,

in place 6 card 7 × 6 ≡ 2,

etc.

Take a pack of 11 cards. We shall now have,

11 ≡ 0

12 ≡ 1

23 ≡ 1

and, in short, to find what any card will be, having performed the necessary arithmetical operation, we subtract the number in the tens place from the number in the units place, and repay anything we borrow in the addition. Thus, suppose we deal into 5 piles, and take up the piles from left to right putting each one at the back of the pile that was at the left of it. We shall now have

in place 1, since 5 × 1 = 5, card 5;

in place 2, since 5 × 2 = 10, card 10;

in place 3, since 5 × 3 = 15 and 1 from 5 leaves 4, card 4;

in place 4, since 5 × 4 = 20 and 2 from 10 leaves 8, and repaying 1 borrowed we have 9, card 9;

in place 5, since 5 × 5 = 25 and 2 from 5 leaves 3, card 3;

in place 6, since 5 × 6 = 30 and 3 from 10 leaves 7, and repaying 1 we get 8, card 8;

in place 7, since 5 × 7 = 35 and 3 from 5 leaves 2, card 2;

in place 8, since 5 × 8 = 40 and 4 from 10 leaves 6, and repaying 1 we get 7, card 7;

in place 9, since 5 × 9 = 45 and 4 from 5 leaves 1, card 1;

in place 10, since 5 × 10 = 50 and 5 from 10 leaves 5, and repaying 1 we get 6, card 6;

in place 11, since 5 × 11 = 55 and 5 from 15 leaves 10, and repaying 1 we get 11, card 11 (≡ 0).

Suppose we now deal again into 9 piles. Now, the last card falls on the second pile. How are we to take up the piles? Answer: After the cards are exhausted, go on dealing in rotation upon the piles to the right of the last single card dealt no longer single cards but whole piles, always taking the extreme left hand one. Thus, in the present case, after the piles are all dealt out, put the left hand pile upon the pile to the right of the Jack, the last single card dealt; that is, put the pile headed by the 6 on that headed by the 4. Then, on the pile one further to the right, that headed by the 9, put the extreme left one headed by the Jack. Next, on the one headed by the 3 put the one headed by the 6, and so on until the piles are reduced to one. You will then find the proper order restored. Why? Because you have multiplied by 5 and by 9, that is, by 45, and 4 from 5 leaves 1, so that you have multiplied the cards in their proper order by 1, which leaves them in their proper order.

I now beg you, my dear Barbara, to take the full pack of 53 cards, and arrange them in their proper order, first the spades, second the diamonds, third the clubs, and fourth the hearts, each suit in its proper order,

1 2 3 4 5 6 7 8 9 X J Q K

with the Joker at the face. Deal them out into 12 piles and take up the piles according to the rule. Namely, denoting the Joker by O,

Next deal the cards out again into 31 piles, and take up the piles according to the rule. Namely,

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:—

Then, my dear Barbara, you can say to your little friend Celarent, who is so fond of denying everything, “Celarent, what number do you want to find?” Suppose she says 6. Then, you count 6 places from the 0, say in the right-handed direction. You turn up the 6^th card, which is the 8; and you say: “If the 8 is in the 6^th place clockwise, then the 6 is in the 8^th place counter-clockwise.” Thereupon, you count 8 places from the 0 to the left and turn up the 8^th card, and lo, it is the 6. Or you might have counted, at first, 6 places to the left and turning up the 6^th card, have found the 2. Then you would say “If the 2 is in the 6^th place counterclockwise, then the 6 is in the 2^nd place clockwise.” And counting 2 places from the 0 to the right, you would again find the 6. The same would hold good if Celarent were to call for any other number.

If you want to do this little trick with 13 cards, you must deal them into 5 or 8 piles. You might begin by asking Celarent how many piles she would like the cards dealt into. If she says 2 (or 11), deal them as she commands, and having done so, ask her whether she would now like them dealt into 4 or 9 piles. If she makes you first deal them into 3 or 10 piles, give her her choice afterward between 6 and 7 piles. If she makes you first deal into 4 or 9 piles, give her then a choice between 2 and 11. If she makes you first deal into 6 or 7 piles, give her her choice afterwards between 3 and 10 piles. If she makes you deal them into 5 or 8 piles, lay them down in a circle at once. In doing so let all be face down except the king, which you place face up. The order will be

You ask: “What spade would you like to find?” If she says “The knave,” reply “Then we count to the knave place.” You count and turn up the 3. Then you say “If the 3 is in the place of the Jack counting clockwise, then the Jack is in the place of the 3 counting counter-clockwise.” You can then count round to this place clockwise, and find it is the 10^th. So you continue: “And if the Jack is in the place of the 10 counting clockwise, then the 10 is in the place of the Jack counting counter-clockwise.” You count, turn up and find it so. Then you count up to this card clockwise, and go on “And if the 10 is in the place of the 2 counting clockwise, then the 2 is in the place of the 10, counting counter-clockwise.”

The same thing can be done with a full pack of 52 or 53 cards.

We have thus far considered addition and multiplication separately. Now let us study them combined. Take a pack of 11 cards in their proper order. Cut it so as to carry 3 cards from back to face of the pack. That adds 3 to the face-value of the card in any given place.

Now deal them into 5 piles and gather up the piles according to rule. This by itself would multiply the face-value of the card in any given place by 5. But acting after the other operation, if x be the place and y the face-value (or original place) we have

y = 5x + 3.

On the other hand, starting again with the cards in their proper order, if we first deal into 5 piles and then carry 3 cards from back to face, we have

y = 5(x + 3).

In short, the order in which the operations are to be taken in the calculation of the face-values is just the reverse of that of their actual performance. The reason is too obvious to require explanation.

It is easy to see that before dealing the cards out in the little trick I proposed your showing Celarent you can perfectly well allow her to cut the pack first, provided that after the dealing, or at any time, you recut so as to bring the zero card to the face of the pack. This will annul the effect of the cutting.

I want to call your attention, Barbara, to the fact that there is another way of effecting multiplication besides dealing out into piles and gathering in. Suppose for instance you hold in your hand the first 11 spades in their proper order while the first 11 diamonds in their proper order are lying in a pack face down upon the table. We will now effect upon the spades the operation

5x + 3

and simultaneously upon the diamonds the inverse operation

For this purpose begin by bringing 3 spades from the back to the face of the pack. Then bring 5 spades from back to face, lay the face card down on the table face up and in its place put the top diamond. Bring 5 more spades from back of pack to its face, lay the face card down face up upon the other card lying on the table face up, and replace it by the top card in the pile of diamonds. Repeat this process until it can be repeated no more owing to the exhaustion of the pile of diamonds. You will now hold all the diamonds in your hand. Carry 3 cards from the face of the pile to the back, and the whole double operation will be complete.

You can now say “If the 7 of diamonds is the 5^th card in the pack of diamonds, then the 5 of spades is the 7^th card in the pack of spades,” and, in short, each pack serves as an index to the other.

From the point of view of this proceeding, multiplication appears as a continually repeated addition. Now let us ask what will result from continually repeating multiplication. As before lay the 11 diamonds in their proper order face down on the table, and take the 11 spades in their proper order in your hand. Deal the spades into 2 piles and gather them up. Put the back card, the 2, down on the table and replace it by the top diamond. Again deal the cards in your hand into 2 piles and gather them up, and put the back card (the 4) upon the one lying face up, and replace it by the top diamond. Proceed in this way until you have laid down all your spades except the knave which you never can get rid of in this way. You will now find that the spades run in geometrical progression, each the double of the preceding

2 4 8 5(≡ 16) 10 9(≡ 20) 7(≡ 18) 3(≡ 14) 6 1(≡ 12).

In fact, if, as before, x be the place, y the face-value,

y = 2^x.

Then, in the other pack we ought to have

x = log y/log 2.

In fact, for these the face-value is increased by 1 when the place is doubled. For the order is

10 1 8 2 4 9 7 3 6 5.

You may now do this surprising trick. Ask Celarent to cut the pack of diamonds (with the knave of spades but without the knave of diamonds). Then, ask how many piles she would like to have the diamonds dealt into. Suppose, to fix our ideas, she says 5. You obey and gather up the cards according to rule. You then cut so as to bring the knave of spades to the face; and in doing so you notice the face-value of the card carried to the back. In the case supposed it will be 4. Then carry as many cards (i.e. in this case, 4) from the face to the back of the pile of spades. Then ask Celarent what diamond she would like to find. Suppose she says the 3. Count to the 3^rd card in the pack of spades. It will be the 6. Then say, “If the 6 of spades is the 3^rd card, then the 3 of diamonds is the 6^th card,” and so it will be found to be.

I have not given any reason for anything, my Barbara, in this letter. In your family you are very high in reasons and in principles. But if you think I have said anything not true, it will be a nice exercise in the art of reasoning to make sure whether it is true or not.

1. Published at Heidelberg in 1496, at Freiberg in 1503, in Strassburg by Grüninger in 1504, in Strassburg by Schott in 1504, in Basle in 1508; etc.