I have received several letters on the subjects of Knots II. and VI., which lead me to think some further explanation desirable.
In Knot II., I had intended the numbering of the houses to begin at one corner of the Square, and this was assumed by most, if not all, of the competitors. Trojanus however says "assuming, in default of any information, that the street enters the square in the middle of each side, it may be supposed that the numbering begins at a street." But surely the other is the more natural assumption?
In Knot VI., the first Problem was of course a mere jeu de mots, whose presence I thought excusable in a series of Problems whose aim is to entertain rather than to instruct: but it has not escaped the contemptuous criticisms of two of my correspondents, who seem to think that Apollo is in duty bound to keep his bow always on the stretch. Neither of them has guessed it: and this is true human nature. Only the other day—the 31st of September, to be quite exact—I met my old friend Brown, and gave him a riddle I had just heard. With one great effort of his colossal mind, Brown guessed it. "Right!" said I. "Ah," said he, "it's very neat—very neat. And it isn't an answer that would occur to everybody. Very neat indeed." A few yards further on, I fell in with Smith and to him I propounded the same riddle. He frowned over it for a minute, and then gave it up. Meekly I faltered out the answer. "A poor thing, sir!" Smith growled, as he turned away. "A very poor thing! I wonder you care to repeat such rubbish!" Yet Smith's mind is, if possible, even more colossal than Brown's.
The second Problem of Knot VI. is an example in ordinary Double Rule of Three, whose essential feature is that the result depends on the variation of several elements, which are so related to it that, if all but one be constant, it varies as that one: hence, if none be constant, it varies as their product. Thus, for example, the cubical contents of a rectangular tank vary as its length, if breadth and depth be constant, and so on; hence, if none be constant, it varies as the product of the length, breadth, and depth.
When the result is not thus connected with the varying elements, the Problem ceases to be Double Rule of Three and often becomes one of great complexity.
To illustrate this, let us take two candidates for a prize, A and B, who are to compete in French, German, and Italian:
(a) Let it be laid down that the result is to depend on their relative knowledge of each subject, so that, whether their marks, for French, be "1, 2" or "100, 200," the result will be the same: and let it also be laid down that, if they get equal marks on 2 papers, the final marks are to have the same ratio as those of the 3rd paper. This is a case of ordinary Double Rule of Three. We multiply A's 3 marks together, and do the same for B. Note that, if A gets a single "0," his final mark is "0," even if he gets full marks for 2 papers while B gets only one mark for each paper. This of course would be very unfair on A, though a correct solution under the given conditions.
(b) The result is to depend, as before, on relative knowledge; but French is to have twice as much weight as German or Italian. This is an unusual form of question. I should be inclined to say "the resulting ratio is to be nearer to the French ratio than if we multiplied as in (a), and so much nearer that it would be necessary to use the other multipliers twice to produce the same result as in (a):" e.g. if the French Ratio were 2⁄10, and the others 2⁄9, 1⁄9 so that the ultimate ratio, by method (a), would be 2⁄45, I should multiply instead by 2⁄3, 1⁄3, giving the result, 1⁄3 which is nearer to 2⁄10 than if he had used method (a).
(c) The result is to depend on actual amount of knowledge of the 3 subjects collectively. Here we have to ask two questions. (1) What is to be the "unit" (i.e. "standard to measure by") in each subject? (2) Are these units to be of equal, or unequal value? The usual "unit" is the knowledge shown by answering the whole paper correctly; calling this "100," all lower amounts are represented by numbers between "0" and "100." Then, if these units are to be of equal value, we simply add A's 3 marks together, and do the same for B.
(d) The conditions are the same as (c), but French is to have double weight. Here we simply double the French marks, and add as before.