# Almost Finned SwordFish example

The Tough Sudoku of February 26, 2011 is another potentially instructive Sudoku puzzle. In this particular puzzle, grouping together a single candidate technique as an argument in an AIC can be a nifty Sudoku technique, trick or tip.

If this is your first visit to this blog, welcome! Unfortunately, if you are a first time visitor, this page may seem like it is written in a different language. Well, it is!! Previous blog pages may be helpful. Links to these pages are found to the right, under Sudoku Techniques. The earliest posts are at the bottom, and if you have never perused the intricacies of our special coded language here, you may wish to start close to beginning. The list is rather large, so below find a list of links that may be pertinent to this particular puzzle.

This blog page will spend but scant time with the easier sudoku techniques. I will not explain most of the steps which are part of the Simple Sudoku Technique Set - SSTS.

### The Puzzle

Some easy moves are available here:

• (1)g1 % Row,Box - (Hidden Single (1) at g1)
• HP(69)df8 => df8 ≠ 1357 - (Hidden Pair)
• (7) h8 % Row , Thus UP 25 - Unique Possibilities to 25 filled cells)
• HP(25)d6f4 => d6f4 ≠ 1679
• LC(1)df5 => ahi5 ≠ 1 - (Locked Candidate)
• LC(3)df1 => df2 ≠ 3
• LC(7)c12 => b12 ≠ 7
• SKYscraper(8)g39e91 => f3 ≠ 8 not required

Simple Sudoku can find no further deductions to make at this point. The puzzle with pencil marks (also called possibilities) is shown below.

### Puzzle at UP 25 and Basic SSTS

Above, the first thing that struck me was the following continuous loop:

• (9=7)e4 - (7=8)e9 - (8)g9 = (8-9)g3 = (9)g4 LOOP
Although this loop certainly produces some valid eliminations, the step I illustrate below solves one of the SIS' (Strong Inference Sets) used in the loop. When this occurs, the loop becomes superfluous. If it is a bivalue/bilocal loop, each SIS in the loop solves when any SIS in the loop solves.

### An Almost Finned SwordFish with candidate 9

Below, the cells marked in Blue would be a SwordFish. The green cell, when added to the group of Blue cells, represent a potential Finned SwordFish with candidate (9). It matters not that (9) cannot exist in some of the cells - in this case g58. b5 is the only cell such that (9) is a possibility outside of the pattern. Thus, it is valid to say:

• FSF(9)dfg358,i5 = (9)b5
because clearly, at least one of FSF(9)dfg358,i5 OR (9)b5 must be TRUE.

The Finned SwordFish has (9)g4 as a potential target. The black lines represent endpoints of Strong Inference Sets. The red lines are weak links between the SIS. The step can be written as the following AIC:

• FSF(9)dfg358,i5 = (9-7)b5 = (7)b4 - (7=9)e4 => g4 ≠ 9
Since eliminating (9) from g4 solves (9) in column g, and since this SIS, (9)column g, is part of the loop noted previously, a cascade of at least some singles must follow. In this case, the following easy steps are newly available after the elimination of (9)g4:
• Singles to UP 33
• Hidden Triple(689)i256 => i256 ≠ 135
• (3)h2 % Row - thus UP 34
• LC(3)ab(6) => a4 ≠ 3
• Two String Kite(5) f9=f4 - d6=h6 => h9 ≠ 5
• LC(5) => i4 ≠ 5 not required

### Two String Kite on candidate (5)

Below the two string kite is illustrated (before the Locked Candidate(3) elimination.)

After the LC(3) elimination, the puzzle can solve with just one more advanced step

### A Locked Set Partitioning used in a chain

The elimination below can be argued without using Locked Sets or Almost Locked Sets, but the manner below minimizes the use of the "=" symbol, which seems to be currently preferred on the tough pages here.

Illustrated above, one can see that by partitioning the cell c2 into (24) = (79), one is also creating a SIS involving two Locked Sets: (124)c279 = (579)c2bc1. This partitioning works well in an AIC:

• (241)c279 = (579)c2bc1 - (5)a1 = (5)a8 => a8 ≠ 1
Note that within the potential Naked Triplet (124)c279, candidate (1) is locked into c79 in box b8. Thus,
regardless of whether c2 contains (24) => (1) locked at c79 => a8 ≠ 1
or c2 contains (79) => a1 ≠5 => (5)a8 => a8 ≠ 1

The puzzle is reduced to a cascade of singles until solved.

### The Solution and solution path

Just for Kicks, I use a Finned Jellyfish rather than the Finned Swordfish in the path below

Solution Path

1. Start 23, SSTS to UP 25, SSTS
2. FJF(9)bcdi1246,i5 = (9-7)b5 = NP(79)be4 =>
• g4≠9
• UP 33, SSTS
• UP 34, SSTS
3. (241)c279 = (579)c2bc1 - (5)a1 = (5)a8 =>
• a8 ≠ 1
• UP 81
• AIC count: 2
• Max AIC Search Depth 2, I believe?