Before trying to understand headless, franken and so on I thought it might help me to define a finned fish. This is done in two stages and, for brevity, based on columns:
An m-fin base for a candidate, X, is a set of m (> 0) columns (the base columns), a non-empty set of boxes (the target boxes) and a non- empty set of rows (the target rows) subject to the following constraints:
(a) X remains to be placed in each base column. The cells in the base columns which may admit X are referred to as the base cells.
(b) Each of the target rows contains a base cell which is not in a target box.
(c) Each of the target boxes contains a base cell which is not in a target row.
Condition (c) ensures that the boxes contain at least one fin.
A finned m-fish is an m-fin base with tb target boxes and tr target rows such that tb + tr = m or tb + tr = m + 1.
(1) If F is a finned m-fish for which tb + tr = m, X may be eliminated from its target boxes and rows except for the base cells.
The base column must hold m Xs, each in a base cell. There are precisely m targets so each receives one X, again in a base cell. No other cell in a target may contain X.
(2) If F is a finned m-fish for which tb + tr = m + 1, X may be eliminated from the intersection of any target box and any target row except for the base cells. (It is not asserted that a non-empty intersection is bound to be present.)
Suppose, if possible, that X does occupy such a cell. m Xs remain to be placed because the new insertion is not in a base cell. However, the number of boxes available is reduced from tb to tb 1 and the number of rows from tr to tr 1. The m Xs have m 1 recipients and the contradiction establishes (2).
It is noteworthy that the propositions represent nothing more than counting in finite subsets of the grid. There is no requirement that the boxes are of a particular shape or size. As a matter of fact there is no requirement for columns and rows to run in straight lines.
Duality is much easier to understand in square n x n grids with square boxes, however. Suppose p Xs have been placed when a finned m-fish, F, is found. Then
- the n p tr rows which are void of X and are not target rows of F
- the tb boxes of F and
- the n p m columns which are void of X and are not base columns of F
form a finned (n p tr)-fish, F*, based on rows. This is the fish dual to F.
From the construction:
(3) F and F* make precisely the same eliminations.
(4) The dual of F* is F.
It is interesting to inspect Havards splendid discovery against this background:
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Cannibalistic Franken-Whale in columns: 3 4 5 6 7 9
19 159 37 | 1368X 1578X 4 | 2 356789 1358X
8 125 37 | 1236X 1257X 9 | 1346X 3567 1345X
129 6 4 | 1238X 1578X 178X | 1389X 35789 1358X
---------------------+----------------------+---------------------
3 4 128X | 7 1268X 5 | 168X 268 9
5 7 128X | 9 4 16X- | 1368X 2368 138X
129 289 6 | 128X 128X 3 | 5 4 7
---------------------+----------------------+---------------------
4 18 9 | 5 368 168X | 7 38 2
6 3 128X | 18X 1789X 278 | 489 589 458
7 28 5 | 4 39 28 | 39 1 6
I am not convinced it is a finned 6-fish because it seems to me as that the elimination follows if column 6 is omitted from the recipe.
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| . . . | X X . | . . X |
| . . . | X X . | X . X |
| . . . | X X . | X . X |
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| . . X | . X . | X . . |
| . . X | . . - | X . X |
| . . . | X X . | . . . |
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| . . . | . . . | . . . |
| . . X | X X . | . . . |
| . . . | . . . | . . . |
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This has the look of a finned 5-fish
- base columns 3, 4, 5, 7 and 9 (m = 5)
- target boxes 2, 3, and 5 (tb = 3)
- target rows 4, 5 and 8 (tr = 3)
According to the theory above, its dual should also be a finned 5-fish because p = 1 and m p tr = 9 1 3 = 5.
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| X X . | X X . | . . X |
| . X . | X X . | X . X |
| X . . | X X X | X . X |
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| . . . | . . . | . . . |
| . . . | . . - | . . . |
| X . . | X X . | . . . |
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| . X . | . . X | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
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Finned 5-fish
- base rows 1, 2, 3, 6 and 7
- target boxes 2, 3, and 5
- target columns 1, 2 and 6
If there is a smaller fish there I cannot find it.
Of course there will be other ways of solving the puzzle, perhaps easier ways. At the same time, unless the eliminations are false, I find it difficult to deny Havard his catch.
Steve