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Message Forums : News : Crossword metapuzzle
6 total messages. For real this time.

Associated URL: /news/847.html

Hydrogen
06/01/2010 @ 22:45:22 EDT

Definitions
Let W represent the set that contains the words to use. Assuming W is unordered, Wi will represent any given word in W, and it is understood that Wi is a distinct word from Wj.

Let the function L(x) represent the length of the word x, where x ∈ W.

Let Ci,j be the set of common letters between i and j, where i, j ∈ W.

Let AR represent the area of the minimal containing rectangle for the puzzle.

Conjecture 1
If no words are given, then AR is undefined.

Reasoning:
If no words are given, then no grid can be defined that contains the words. If there is no grid then there is no rectangle that can contain it.

Lemma 1
The trivial case of |W| = 1 has AR = L(W1)

Proof:
After W1 is placed on the grid, there are no more words to place. No block squares were previously placed. The word must read either horizontally or vertically, not both, so the squares in which the letters of the word were placed define the resulting grid, and the containing rectangle has a length = L(W1) and a height of 1, since all letters must go along the same axis as the others in the word. Thus, AR = 1 * L(W1).

Lemma 2
If |W| = 2 and |C1,2| = 0, AR = L(W1) + L(W2) + 1

Proof:
Since neither of the two words has a letter in common, they cannot intersect. Also, since consecutive letters must form a word in W, the two words cannot be directly adjacent or they will form words which are not in W, as all words in W will have already been on the grid.

Since lemma 1 proves that AR for W with a single word = L(W1), then AR for W where |W| = 2 must be at least L(W1) + L(W2). Also, since the words cannot be adjacent, there must be at least one blocked grid square separating them. Placing the words on the same line of squares, "side-by-side", would require only a single blocked grid square to separate them. Since the words cannot be adjacent, any other arrangement of the words would render the same area or greater. Thus, AR = L(W1) + L(W2) + 1 for W where |W| = 2 and |C1,2| = 0.

Hafnium
06/03/2010 @ 11:39:14 EDT

Um.. I stopped remembering how stuff like that worked (I think you're using absolute values?) around the time I got my degree.

But I wanted to point out that in "real" crossword puzzles the placement of the blocks are symmetrical.

Hydrogen
06/09/2010 @ 08:29:49 EDT

me_tew and I were talking about this the other day. He noted that crossword puzzles are typically symmetrical and also square. Should we focus on this case or focus on the general case?

Also, in set theory, given a set S, |S| represents the cardinality of a set, or number of elements in the set. So if S = {A, B, C}, |S| = 3.

Hafnium
08/29/2010 @ 18:01:31 EDT

Um... so how is this crossword going? Did you have an actual list of words? I'm pretty sure most crossword puzzles are already as condensed as they can be, giving the typical requirements of being square and symmetrical.

Hydrogen
08/30/2010 @ 21:27:12 EDT

Meh. Too many variables to think about. Thought it would be an interesting mental exercise, but I think it's just turned out to be a bad koan.

Hassium
09/01/2010 @ 09:26:26 EDT

The actual solution to the metapuzzle is left as an exercise for the reader.

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