A New Logical Machine
一台新的逻辑机
艾伦·马昆德 Allan Marquand（1853–1924）
1885 年 11 月 11 日宣读

冷冻图书馆 · 开架书库 · S-007215 · https://coldlibrary.com/s/007215
权利：1885 年宣读，1886 年刊印；马昆德卒于 1924 年。已按澳大利亚、欧盟、英国、中国大陆和美国的规则核对（截至 2026 年 10 月 11 日）。仅供参考，不是法律意见。
来源：Internet Archive (JSTOR Early Journal Content), A New Logical Machine，https://archive.org/details/jstor-25129824（取于 2026-10-11）
中文译文：Claude（AI），2026 · CC0 1.0
生成于 2026-10-14，来自 https://coldlibrary.com/open-stacks/marquand-a-new-logical-machine

----

During the year 1881 I constructed a logical machine somewhat similar to the well-known machine of Prof. Jevons, and printed logical diagrams for problems involving as many as ten terms. This earlier instrument and the logical diagrams formed the basis of the machine illustrated on the accompanying plate. The new machine was constructed in Princeton during the winter of 1881–82, by my friend Prof. C. G. Rockwood, Jr., whose mechanical skill and untiring patience gave me invaluable assistance. The machine was made from the wood of a red-cedar post, which once formed part of the enclosure of Princeton's oldest homestead. It measures 32 cm. high by 21 cm. wide and 15 cm. deep. Like the instrument of Prof. Jevons, and that of Prof. Venn, it is constructed for problems involving only four terms, but more readily than either of those instruments admits of being extended for problems involving a larger number of terms.

The face of the machine (Fig. I.) presents to view sixteen small pointers representing the sixteen logical combinations of the symbols A, B, C, and D, with their negatives a, b, c, d. These combinations are so arranged that all the A combinations are found in the two vertical columns to our left, the a combinations in the two vertical columns to the right. The A combinations are subdivided vertically into the B and b combinations, and the a series in like manner. Both are also subdivided horizontally into the C and c combinations, and each of these again into the D and d combinations. Thus the uppermost pointer to the left represents the combination A B C D, the horizontally adjoining one A b C D, the next a B C D, and so on until we reach the lowermost to the right, which is a b c d. Below the pointers may be seen the two operation keys marked 1 and 0, and the four positive and four negative letter keys under their respective symbols.

The 1 key may be called the *restoration key*. It prepares the machine for a new problem, by raising all the pointers to the horizontal position. The 0 key may be called the *destruction key*, as when pressed down it allows the pointers to fall. The function of a *letter key* is to sustain in the horizontal position the pointers representing the corresponding negative combinations. Thus, if the A key is pressed, the a pointers are sustained; if the b key is pressed, the B pointers are sustained; and similarly for all the remaining letter keys.

Turning now to Fig. II., the mechanism by which this is effected will be easily seen. On the inner face of the machine, and corresponding to the pointers, are sixteen drops, some of which are pictured in the horizontal position, and some as fallen. The two operation keys move rectangular frameworks, each consisting of four vertical brass rods fastened together, and carrying pins which reach all of the sixteen drops. The framework of the 0 key moves on the inner, that of the 1 key on the outer side of the drops, both horizontally. If we wish to bring all the pointers to the horizontal position and hold them there, both operation keys are pressed down. The 1 key raises all the drops. Releasing first the 0 key, its framework is drawn back by a fixed spring, and by means of its pins holds the drops in position; on releasing the 1 key, its framework is drawn back so as not to interfere with subsequent operations. Each letter key operates two vertical or two horizontal rods, free to revolve on their axes. In each rod are set four pins, conveniently bent, so that when the rods are partially revolved, by pressure of the letter keys, the pins are made to sustain the drops in the horizontal position without raising those which have fallen. The rods return to their original position, by means of small spiral springs, as soon as the pressure of the letter keys is released. The desired motion is communicated to the rods by means of cat-gut strings, there being two such strings from each letter key. Thus the A key operates the two vertical rods which are to hold the a drops; the b key the rods for holding the B drops; and the other letter keys in like manner. By this device it will be seen that, if we should press the A key (this holds the a combinations), and then the 0 key, only the A pointers will fall; or, if both the A and B keys are depressed (this holds the a and b combinations), and then the 0 key, only the A B pointers will fall; and similarly for the other combinations.

To utilize the instrument for the solution of logical problems, we first raise all the pointers to the horizontal position. This will indicate the state of a logical universe of four terms before the introduction of premises. Now, since the establishment of any combination means the negation of some other, we may express our premises in negative form. Thus, in general, A −< B (every A is B) may be expressed A b −< 0 (A's which are b do not exist). This we express upon our machine by pressing down the letter keys A and b, and then the destruction key. The falling of the A b pointers indicates exactly the change effected in the logical universe by the introduction of the premise A −< B. We may then continue to impress as many premises as we please, until all the pointers have fallen. The following formulæ will suffice to illustrate the manner in which premises may be reduced from the positive to the negative form.

        Positive Form.                  Negative Form.
(1.)  A −< B . . . . . . . . . .       A b −< 0.
(2.)  A + B −< C . . . . . . . .     { A c −< 0.
                                     { B c −< 0.
(3.)  A −< B + C . . . . . . . .       A b c −< 0.
(4.)  A −< b + C D . . . . . . .     { A B c −< 0.
                                     { A B d −< 0.
(5.)  A B −< C . . . . . . . . .       A B c −< 0.
(6.)  A −< B C . . . . . . . . .     { A b −< 0.
                                     { A c −< 0.
(7.)  A (B + C) −< D . . . . . .     { A B d −< 0.
                                     { A C d −< 0.
(8.)  A −< B (C + D) . . . . . .     { A b −< 0.
                                     { A c d −< 0.

Having expressed our premises upon the machine, and their effect being recorded by the pointers, it only remains for us to read off the conclusion. The entire conclusion is represented by the fallen pointers, and might be expressed as their joint sum; or it may be viewed as the logical sum of the combinations represented by the horizontal pointers. Thus, the premises of Barbara, A −< B and B −< C, give as the entire conclusion read negatively,

A b C }
A b c }
A B c }  −< 0.
a B c }

Read positively, the conclusion is,

A B C + a B C + a b C + a b c = 1;

or, more briefly,

B C + a b = 1.

Ordinarily, the conclusion called for is part only of the total conclusion. Thus, syllogism with the above premises asks for a conclusion involving only A and C. An inspection of the dial-plate will show us the conclusion A −< C, and also other conclusions involving relations between other terms than A and C; thus,

c (A + B) −< 0;   b −< a + c, etc.

Nor is it necessary that our conclusions should be limited to relations between terms given in the premises, as may be seen in the solution of the following problems.


Problem I.

Let us suppose that there are four girls at school, Anna, Bertha, Cora, and Dora, and that some one had observed that

(1.) Whenever either Anna or Bertha (or both) remained at home, Cora was at home; and

(2.) When Bertha was out, Anna was out; and

(3.) Whenever Cora was at home, Anna was at home.

What information is here conveyed concerning Dora?

Indicating by the capital letters the fact of *remaining at home*, and by the small letters that of *going out*, our premises are

A + B −< C  =  A c }  −< 0
               B c }
b −< a      =  b A −< 0
C −< A      =  C a −< 0

and, impressing them upon the machine, there will result the state of things indicated by Fig. 1. From this we may read off the conclusion,

D −< A B C + a b c.
d −< A B C + a b c.

Or, if Dora remain at home, her three sisters will be all at home or all out; and the same will be true if Dora goes out.


Problem II.

If A = B and B = C, what may be said of D?

Impressing upon the machine our premises,

(A = B)  =  A −< B }  =  A b −< 0
            B −< A }     B a −< 0
(B = C)  =  B −< C }  =  B c −< 0
            C −< B }     C b −< 0

the same state of the logical universe is produced as by the premises of the preceding problem. Hence,

D −< A B C + a b c.

----

中文译文 · Claude（AI）译

1881 年，我制作了一台逻辑机，与杰文斯教授那台著名的机器有些相似，还印制了可处理多达十个项的问题的逻辑图。这台早先的仪器和这些逻辑图，构成了附图所示这台机器的基础。新机器是 1881 至 1882 年冬天在普林斯顿由我的朋友小 C. G. 罗克伍德教授制作的，他的机械技巧和不倦的耐心给了我极其宝贵的帮助。机器用一根红雪松木桩的木料做成，这根木桩曾是普林斯顿最古老的宅院围栏的一部分。机器高 32 厘米，宽 21 厘米，深 15 厘米。和杰文斯教授的仪器以及维恩教授的仪器一样，它是为只涉及四个项的问题而造的，但比那两台仪器更容易扩展，用于涉及更多项的问题。

机器的正面（图一）显示出十六根小指针，代表符号 A、B、C、D 及其否定 a、b、c、d 的十六种逻辑组合。这些组合的排列方式是：所有 A 组合都在我们左边的两竖列里，a 组合在右边的两竖列里。A 组合在竖直方向上又分为 B 组合和 b 组合，a 系列也照此划分。两者在水平方向上又都分为 C 组合和 c 组合，每一种再分为 D 组合和 d 组合。因此，左边最上面的指针代表组合 A B C D，横向相邻的一根代表 A b C D，再下一根代表 a B C D，依此类推，直到右边最下面的一根，即 a b c d。指针下面可以看到两个标着 1 和 0 的操作键，以及在各自符号下的四个肯定字母键和四个否定字母键。

1 键可以称为*复原键*。它把所有指针抬到水平位置，使机器准备好解一个新问题。0 键可以称为*消去键*，因为按下它时，指针会落下。*字母键*的功能，是把代表相应否定组合的指针撑在水平位置。因此，如果按下 A 键，a 指针就被撑住；如果按下 b 键，B 指针就被撑住；其余字母键也是如此。

现在看图二，就很容易看出实现这一点的机构。在机器的内壁上，与指针相对应，有十六个落片，图中有些处于水平位置，有些已经落下。两个操作键各推动一个长方形框架，每个框架由四根固定在一起的竖直铜杆组成，上面装着能够到全部十六个落片的销钉。0 键的框架在落片的内侧移动，1 键的框架在外侧移动，两者都是水平移动。如果想把所有指针抬到水平位置并保持住，就同时按下两个操作键。1 键把所有落片抬起。先松开 0 键，它的框架被一根固定的弹簧拉回，并用它的销钉把落片撑在原位；再松开 1 键，它的框架被拉回，以免妨碍后续的操作。每个字母键操纵两根竖直的或两根水平的杆，杆可以绕自身的轴转动。每根杆上装有四根弯成合适形状的销钉，当字母键按下、杆转过一定角度时，这些销钉就把落片撑在水平位置，而不会把已经落下的抬起来。字母键一松开，杆就靠小螺旋弹簧回到原来的位置。所需的运动通过羊肠线传给这些杆，每个字母键有两根这样的线。因此，A 键操纵用来撑住 a 落片的两根竖杆；b 键操纵撑住 B 落片的杆；其余字母键也是如此。由这个装置可以看出，如果我们先按 A 键（它撑住 a 组合），再按 0 键，就只有 A 指针会落下；或者，如果同时按下 A 键和 B 键（它们撑住 a 组合和 b 组合），再按 0 键，就只有 A B 指针会落下；其他组合也是如此。

用这台仪器解逻辑问题时，我们先把所有指针抬到水平位置。这表示引入前提之前一个四项的逻辑宇宙的状态。由于确立任何一种组合都意味着否定另外某种组合，我们可以把前提表示成否定形式。例如，一般地，A −< B（每个 A 都是 B）可以表示为 A b −< 0（是 b 的 A 不存在）。在机器上，我们按下字母键 A 和 b，再按消去键，就表示了这一点。A b 指针的落下，恰好表示引入前提 A −< B 给逻辑宇宙带来的变化。然后我们可以继续输入任意多个前提，直到所有指针都落下。下面的公式足以说明如何把前提从肯定形式化为否定形式。

        肯定式                          否定式
(1.)  A −< B . . . . . . . . . .       A b −< 0.
(2.)  A + B −< C . . . . . . . .     { A c −< 0.
                                     { B c −< 0.
(3.)  A −< B + C . . . . . . . .       A b c −< 0.
(4.)  A −< b + C D . . . . . . .     { A B c −< 0.
                                     { A B d −< 0.
(5.)  A B −< C . . . . . . . . .       A B c −< 0.
(6.)  A −< B C . . . . . . . . .     { A b −< 0.
                                     { A c −< 0.
(7.)  A (B + C) −< D . . . . . .     { A B d −< 0.
                                     { A C d −< 0.
(8.)  A −< B (C + D) . . . . . .     { A b −< 0.
                                     { A c d −< 0.

前提在机器上表示出来、它们的作用由指针记录下来之后，剩下的只是读出结论。全部结论由落下的指针表示，可以表示为它们的联合之和；也可以看作由水平指针所代表的那些组合的逻辑和。例如，Barbara 式的两个前提 A −< B 和 B −< C，按否定形式读出的全部结论是：

A b C }
A b c }
A B c }  −< 0.
a B c }

按肯定形式读出，结论是：

A B C + a B C + a b C + a b c = 1;

或者更简短地写成：

B C + a b = 1.

通常所要求的结论只是全部结论的一部分。例如，以上述前提作三段论，所求的是只涉及 A 和 C 的结论。查看表盘，就能看出结论 A −< C，以及涉及 A、C 以外其他项之间关系的其他结论，例如：

c (A + B) −< 0;   b −< a + c, etc.

我们的结论也不必局限于前提中给出的项之间的关系，这从下面两个问题的解法中可以看出。


问题一

假设学校里有四个女孩：安娜、贝尔塔、科拉和多拉；有人观察到：

（1）每当安娜或贝尔塔（或两人都）留在家里时，科拉都在家；并且

（2）贝尔塔出门时，安娜也出门；并且

（3）每当科拉在家时，安娜都在家。

这里传达了关于多拉的什么信息？

用大写字母表示*留在家里*，用小写字母表示*出门*，我们的前提是：

A + B −< C  =  A c }  −< 0
               B c }
b −< a      =  b A −< 0
C −< A      =  C a −< 0

把它们输入机器，就会得到图一所示的状态。从中我们可以读出结论：

D −< A B C + a b c.
d −< A B C + a b c.

也就是说，如果多拉留在家里，她的三个姐妹要么都在家，要么都出门；如果多拉出门，情况也是一样。


问题二

如果 A = B，并且 B = C，关于 D 能说些什么？

把我们的前提输入机器：

(A = B)  =  A −< B }  =  A b −< 0
            B −< A }     B a −< 0
(B = C)  =  B −< C }  =  B c −< 0
            C −< B }     C b −< 0

所产生的逻辑宇宙的状态与上一个问题的前提所产生的相同。因此：

D −< A B C + a b c.