Puzzle 5 (from Swordfish compendium)

Animator

I've got a problem solving it...

Where I got:

rubylips · Posted: Wed Apr 27, 2005 11:19 pm Post subject:

Thanks for the posting.

The first wing occurs in Column 1, where the two candidates for the value 3 are the cells r1c1 and r8c1.

The second (3-leg) wing occurs across Box 7 (or [3,2]), Column 5 and Box 2 (or [1,2]). There are two candidates for the value 3 in each sector, which means that exactly one of the cells r8c4 and r1c6 must contain a 3.

When we put these two facts together, we see that 3s must appear in the cells r1c1 and r8c4 or the cells r1c6 and r8c1, which enables us to remove 3 as a candidate for the cells r1c3, r1c7 and r8c3. In particular, now that r1c7 is no longer a candidate, the cell r6c7 is left as the sole candidate for the value 3 in Column 7.

I'll think about the swordfish in the 5s ...

Animator · Posted: Thu Apr 28, 2005 7:38 am Post subject:

Hmm, I find this to be a different swordfish then the one I've seen (although I can't say I've seen that many)

The 5-swordfish for example allows you to write down (as in only 6 cells):

Tempbow · Posted: Sat Apr 30, 2005 1:39 am Post subject:

IJ · Posted: Sat Apr 30, 2005 1:43 pm Post subject:

Personally, I wouldn't call this Swordfish, as it is another order of complexity, and therefore should perhaps have another name? Up to Rubylips, of course.

However, I would describe this example by saying:

3 units have two candidate cells (in this case Col 1, Boxes 2 and 7)

3 other units (in this case Rows 1 & 9, and col 5) also have 2 of the 6 candidate cells.

The second set of units can be cleared, except for the intersections with the first 3 (i.e. the first 6 defining cells)

This definition also works for the standard Swordfish model, where all the first units are rows (or columns) and the second set are columns (or rows).

You can "N it up" to:

Look for N units that have only 2 candidate cells for a given digit, and a second descreet set of N intersecting units that also contain 2 of the candidate cells. The second set of N units can be cleared of the digit in question.

Rubylips - do you have a better, or more succinct description?

rubylips · Posted: Tue May 03, 2005 6:34 pm Post subject:

Here's an attempt at a more precise definition of a Swordfish:

A sector is the generic term for a row, column or box.

A unit-length string comprises two cells within a single sector that are the only possible candidate positions for some given value within that sector.

Two strings are connected if they share a common-cell and have been defined using the same value. Of course, connected strings will have to occur across different sector types - e.g., one string in a column could connect to another in a box.

An n-leg string comprises n connected unit-length stings.

Some logical observations:

Provided that a string has an odd number of legs, exactly one of its end cells must contain the given value.

When we find two strings, each with an odd number of legs, such that their end points lie on the same row (or column), we know that exactly one of the two end-points on each row (or column) must contain the given value. This allows us to eliminate the given value as a candidate for the remaining cells in each row (or column).

More definitions:

When the two strings are of unit length, the pattern is called an X-Wings.

When one or more of the strings has length greater than one, the pattern is known as a Swordfish.

Tempbow · Posted: Tue May 03, 2005 10:05 pm Post subject:

Tempbow · Posted: Tue May 03, 2005 11:27 pm Post subject:

Thanks, that is a good definition!

AMcK · Posted: Fri May 20, 2005 9:58 am Post subject: Colouring solution

Starting from

* 4 * | 6 * * | * * 9
8 * * | 7 * 4 | 6 5 *
6 * 1 | 9 * 8 | 2 4 *
---------------------
1 * 6 | * 9 * | 4 * 8
2 * * | * 4 * | 5 * 6
5 * 4 | * 6 * | * * *
---------------------
4 * 8 | 2 * 9 | 7 6 *
* 6 * | * * 1 | * * 4
9 * * | 4 * 6 | * 3 *

The 3-rules take us to

Code:

37 4 2357 6 12 35 138 178 9
8 239 239 7 12 4 6 5 13
6 357 1 9 35 8 2 4 37
1 37 6 35 9 2357 4 27 8
2 3789 379 18 4 37 5 179 6
5 789 4 18 6 27 139 1279 1237
4 135 8 2 35 9 7 6 15
37 6 2357 35 78 1 89 289 4
9 1257 257 4 78 6 18 3 125

Colouring digit 3 then yields:

Conjugate row 6: digit 3 cells {6,7} {6,9}
Colour: {6,7} (a) {6,9} (b)
Conjugate row 7: digit 3 cells {7,2} {7,5}
Colour: {7,2} (c) {7,5} (d)
Conjugate column 1: digit 3 cells {1,1} {8,1}
Colour: {1,1} (e) {8,1} (f)
Conjugate column 4: digit 3 cells {4,4} {8,4}
Colour: {4,4} (g) {8,4} (h)
Conjugate column 5: digit 3 cells {3,5} {7,5}
Colour: {3,5} (c) {7,5} (d)
Conjugate column 7: digit 3 cells {1,7} {6,7}
Colour: {1,7} (b) {6,7} (a)
Conjugate box 2: digit 3 cells {1,6} {3,5}
Colour: {1,6} (d) {3,5} (c)
Conjugate box 6: digit 3 cells {6,7} {6,9}
Colour: {6,7} (a) {6,9} (b)
Conjugate box 8: digit 3 cells {7,5} {8,4}
Recolour: {4,4} (g->d)
Recolour: {8,4} (h->c)
Colour: {7,5} (d) {8,4} (c)
Exclusion in row 1: colours e d
Exclusion in row 1: colours e b
Exclusion in row 1: colours d b
Exclusion in row 8: colours f c
Exclusion in box 7: colours c f
Colour exclusion: c/d excludes f/e
Recolour: {1,1} (e->c)
Recolour: {8,1} (f->d)

With the colour map