The
Tough Sudoku of February 24, 2011 is a potentially instructive
Sudoku puzzle. In this particular puzzle, using the fact that the puzzle has exactly one solution can serve to
shorten the solution path. Uniqueness is a slightly controversial, but certainly powerful 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 **Previous Entries**. 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 presumes that the reader is familiar with most * easy* sudoku techniques. I will not list any of the steps
which are part of the Simple Sudoku Technique Set - SSTS.

### The Puzzle

Some easy moves are available here:

- Two singles
- Some SSTS
- => UP 24 (Unique Possibilities to 24 filled and given cells)

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 24 and initial SSTS - a simple Unique Rectangle avoidance deduction.

Note that most of the things I show later are available already below.

Above, the puzzle has NP(78) e46 (naked pair (78) at e46).
Candidate (8) is locked at c46. If candidate (7) were also locked at c46, then
we would also have HP(78)c46 (hidden pair (78) at c46). The
following would thus all simultaneously exist:

- Pair(78)c46 and box b5
- Pair(78)e46 and box e5
- Pair(78)ce6
- Pair(78)ce4
- XWing(7)ce46
- XWing(8)ce46

These conditions conspire to form an impenetrable loop. The rest of
the puzzle cannot effect cells ce46. Those cells cannot uniquely resolve
themselves without an arbitrary assignment. Thus, if this condition were to occur,
the puzzle could not be resolved uniquely. (In this case, there would be at least
two solutions to the puzzle). Hence, if the puzzle has exactly one solution, we
can safely remove candidate(7) from c46. We could write:

- AUR(78)ce46 => c46 ≠7
- UP 27
- Some more SSTS

### Puzzle at UP 27 plus SSTS

Note that the list of steps I find here is not exhaustive.

I find it helpful to brainstorm a bit with a puzzle. I simply look for a collection
of steps or partial steps that are available. This often reveals a way to significantly advance
the puzzle.

Above, I can quickly see a few easy snippets that involve uniqueness

- AUR(26)bh45 => (4)bh4 = (9)b5.
- AUR(26)bh45 => (4)bh4 = (6)i5 => (4)bh4 = (264)hi5h4 => cg4 ≠4
- AUR(68)ai13 => (8)d1 = (6)g13
- MUG(246)bh456 => (9)b56 = (3)h6

There are a few more, but their SIS's

(Strong Inference Sets) are
not as easily managed.

The MUG(246),

Multivalue Unique Grave, - can be seen as at least one of
several possible BUG lites. That is not my preferred manner of proof, but it does
suffice.

Also available above is an easy ALS xz, rc(3) using cells ce3, g13. That particular
step does not seem very useful, but it does help one speculate about possible
continuous loops in that area. I see three ways to present essentially the same loop:

- Note (24)c3,(24)e1. But for (24) i2, one would have a WWing Ring
- But, (24)i2 - (3)i2 = (3)d2 - (3=24)e3
- Since (24)e3 conflicts with the Almost WRing, we have a continuous network
- => a3,abd1 ≠24; i2 ≠8

- Note ALS(2468)egi1
- Note (6)a3 = (6)gi3 and (8)a3 = (8)gi3
- This forms a typical HSR with
- HUB a3
- Spokes (68)agi3
- Rim ALS (2468)egi1

- One can write this as an AIC:
- (6)a3 = (6)gi3 - NT(246=248)egi1 - (8)i3 = (8)a3 LOOP
- => the same eliminations as the continous loop using (234) above

- Similarily, we have this:
- NQ(2468)egi1,(8)d1 = (8)df2 - (8)a2 = HP(68)a13 LOOP
- => the same eliminations again

Although these loops do not really knock out the puzzle, they imply that
solving any candidate in the bottom three boxes will significantly advance the puzzle.

The next loop I see ends up being more useful in terms of letting me know where
to look than it is in making any elinations:

- Remember the MUG SIS: (9)b56 = (3)h6
- One can find the following easy continuous loop:
- (9)b56 = MUG(246)bh456 = (3)h6 - (3)h89 = HP(37-79)gi8 = (9)a8 - (9)a1 = (9)b1 Loop
- => a79 ≠9; gi(8) ≠4
- Exactly one triplet(246)b456,h456 is true.

This loop opens my eyes to this potential step:

- Remember the AUR SIS: (6)g13 = (8)d1
- (6)g13 = HP(81-19)bd1 = (9)b56 - NT(246)b456 = see above ... NT(246)h456
- =>g4<>6

Since we had a target at g4 with the (26) UR, the next snippet is interesting:

- (9)b56 = (9-8)c6 = (8)c4 - (8=7)e4 - (7=5 almost)g4 - (5)f4 = (5)f2 - (5=1)d3 - (1)d1 = (1)b1
- It should be obvious that using g4, one can elminate(9)b1.
- Also, that doing so will advance the puzzle helpfully

### Puzzle again at UP 27 plus SSTS

Note that the list of steps I find here is not exhaustive.

Below, I will borrow the presentation style of the great sudoku solver, ttt.
He popularized the Kraken Diagrams. This one below now writes itself:

(4)g4 - (4)bh4 = AUR(26)bh45 = (9)b5

||

(6)g4 - (6)g13 = AUR(68)ai13 = HP(81)db1

||

(5)g4 - (5)f4 = (5)f2 - (5=1)d3 - (1)d1 = (1)b1

||

(7)g4 - (7=8)e4 - (8)c4 = (8-9)c6 = (9)b56

=> b1 ≠9 => UP 35

### Puzzle at UP 35 before any further SSTS

The step below is, imo, easier than the YWing. It should be fairly automatic for
good solvers to find the WWing below.

Above, (24)b2, (24)e1, (2)ce3 => WWing => f2 ≠4

One might write it as an AIC:

- (4=2)b2 - (2)c3 = (2)e3 - (2=4)e1 => f2 ≠4

SSTS finishes the puzzle from there.

### Solution

I look forward to reading some of the creative solutions that the clever solvers
on the tough page will put forth for this puzzle!

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