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