lewis carroll's puzzles (soriteses)

The boundary calculus of George Spencer-Brown is able to accomplish everything that Boolean logic can, but it is able to avoid the problems generated by occurrences of self-reference, which create paradox and, thus a termination of logical operations, in the Boolean system. In a sense, one could characterized Spencer-Brown's calculus as one "based on" self-reference in the sense that a topography of boundaries and spaces is able to withstand different descriptions based on shifting points of view without changing basic values.

The bolagram portrays the element of self-reference by "crossing itself" at the point labeled 'B'. B is given an alternate value, B-cross, and the relation between B and B-cross is called the "ij factor, symbolized by an oscillating square wave (). B/B-cross are "gratuitous" in that they seem to be "self-canceling." But, translated into riddle terminology, this means that any number of relationships can be added to the principal one (the "solution"), and in any "order," as long as the appearances of the term as an "inferior term" (as a subset of something) equals the number of its appearances as a "superior term" (the containing set).

Example:

1. Socrates is a man
2. All men are mortal
3. [therefore] Socrates is mortal.

1. Socrates (inferior term) is a subset of men (superior term)
2. men (inferior term) are a subset of mortal (superior term)
"men" cancels out because it appears as a superior and inferior term, leaving
3. Socrates (inferior term) and mortal (superior term)

Put into the form of the calculus, the conclusion can be reached in the following manner:

Because Socrates remains as an inferior term, also "mortal" as a superior term, they can combine to form the statement, "Socrates is mortal."

Here is an example showing that the "middle term" can be expanded and redistributed so that self-canceling pairings aren't immediately obvious. Lewis Carroll's famous puzzles were presented in this form (taken from Law of Form, pp. 122-23):

  1. The only animals in this house are cats;
  2. Every animal is suitable for a pet, that loves to gaze at the moon;
  3. When I detest an animal, I avoid it;
  4. No animals are carnivorous, unless they prowl at night;
  5. No cat fails to kill mice;
  6. No animals ever take to me, except what are in this house;
  7. Kangaroos are not suitable for pets;
  8. None but carnivora kill mice;
  9. I detest animals that do not take to me;
  10. Animals, that prowl at night, always love to gaze at the moon.

The solution lies in finding the two terms that only appear once (all others appear twice). One remaining term will be an inferior term, the other a superior. The answer will take the inferior term as the subset and the superior term as the containing set.

"Kangaroo" appears only once, as an inferior term. "Avoided by me" appears only once, as a superior term. The answer is "Kangaroos are avoided by me." Other terms "cancel out" because they appear once as a superior and once as an inferior term. Note that negative statements, such as "Kangaroos are not suitable for pets," turn the superior term ("suitable for pets") into an inferior term (= "not suitable for pets").

Using the following key, it's possible to see how Spencer-Brown's calculus makes quick work of the puzzle.

h house, in this
c cat
p pet, suitable for
d detested by me
a avoided by me
m moon, love to gaze at
v carnivorous
n night, prowl at
k kill mice
t take to me
r kangaroo

Translating the statements using the abbreviations, and canceling out all pairs, one finds two terms remaining.

"Kangaroos" as the inferior term (subset) and "avoided by me" as the superior term (set) yields: "All kangaroos are avoided by me," which is a statement that does not appear in the original list but is a logical conclusion based on the evidence given.

In the flip of the inside frame, any number of paired odd/even spaces can be inserted without effect as long as B is contained by two crosses. In the rules of the calculus, this permits C to be redistributed into (1) a space containing A and (2) a space containing B-cross. Like the solution to Lewis Carroll's sorites, A and B-cross work like the "left-over," although this characterization is a bit inaccurate. It is rather the spirit of "gratuitous" or "tautological" situations, scattered and combined for the sake of delay, that makes a puzzle interesting. The conclusion is not evident immediately because of the intervening and self-canceling relationships.

What is the value of the multiplication and redistribution (scattering) of self-canceling pairs? In literature and the arts in general, it is obvious that two values can be derived. First, the audience is delayed in discovering the end or conclusion. Second, the intervening process of cancellation itself provides a sense of symmetry that adds to the surprise of the conclusion. Each pair provides a symmetrical relationship that is "delayed" by providing only half of the set at any one time, or by providing an "inverted" form. The alternative would be to give away the ending prematurely, or to lack sufficient detail to make the artwork interesting. Art as a puzzle requires both that the puzzle be sufficiently complex to provide a challenge and sufficiently simple that it can be "solved" within the duration of the work. "When I detest an animal, I avoid it" is a story in miniature, yet it contains two parts that will "participate" at a higher level of the narrative and possibly figure into the conclusion, if left remaining, to combine with another "remaindered" part. The conclusion is thus implicit in the story without being literally available.

see Alexander Bogomolny's review of Lewis Carroll's soriteses