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I am able to pick out finned swordfishes but am unable to tell exactly which number is to be eliminated. Sometimes I can pick out two sets of finned swordfishes on one board inclusive of the same number. Don't know when this happens which one is the one to follow for the right solution. So is there a simple answer as to which number is to be eliminated?

Mowpar, welcome to the New Sudoku Player's Forum!

For Finned Swordfish you eliminate the subset of the regular Swordfish eliminations that share a block with the fin(s).

If there are two Finned Swordfish, you can make the eliminations for both of them, even if making the eliminations for one of them destroys the other. That is assuming you have identified the Swordfish and their eliminations correctly of course.

Code: Select all

*------------------------------------------------------------------------*
|x12678  x12678  267     | 5      129    79      | 4      1267   3       |
|*1237   *1237   9       |*237    6      4       | 5      8      17      |
| 5       4      2367    |x237    12     8       | 9      1267   167     |
|------------------------+-----------------------+-----------------------|
| 368     5      36      | 9      7      2       | 1368   136    4       |
| 9       78     1       | 6      4      3       | 78     5      2       |
|*2367   *2367   4       | 8      5      1       | 367    367    9       |
|------------------------+-----------------------+-----------------------|
| 67      9      5       | 1      3      67      | 2      4      8       |
| 4      *12367 f2367    |*27     8      679     | 1367   13679  5       |
|x1367-2 x1367-2 8       | 4      29     5       | 1367   13679  167     |
*------------------------------------------------------------------------*

Here is a typical finned Swordfish move : Finned Swordfish in 2's r268 c124 with a fin Cell r8c3 => - 2 r9c12.

Jason was quite correct, but here is one explanation of why. Suppose the 2 in the fin cell was False. Then there would be an ordinary (un-finned) Swordfish r268 c124 and you could eliminate 2's in c124 except for those in r268.

This would result in 5 eliminations - 2 r1c12, r3c4 and r9c12. Now suppose 2 in the fin cell was True.

Then you could still eliminate the 2's in r9c12 (since they can see the fin cell) but you couldn't say anything about the 2's in r1c12 or r3c4 (since they can't see the fin cell).

Since the 2 in the fin cell can only be True or False, then - 2 r9c12 is a common outcome and can be made, but there is no common outcome for the 2's in r1c12 or r3c4, so they have to stay.

I've marked all the 5 Swordfish Potential Eliminations (PE's) with x in their cells, but only those in r9c12 are marked with -2 to indicate that they have to go.

So, in general, you can eliminate Swordfish PE's in any cell that can see the fin. This usually means that they share a box (block) with the fin.

If the Swordfish has 2 fins, a PE must see both of the fins to be eliminated.

Leren

I can't put the explanation, with the puzzle given, guess I am thick!

All right, let's take it one step at a time.

Do you agree, that if there was no 2 in the fin cell (r8c3) there would be a Swordfish in 2's in Rows 2, 6 and 8 Columns 1, 2 and 4 in the eight cells I've marked with *'s and you could eliminate the 2's in the five cells I've marked with x's ?

Leren