Four Uniqueness Techniques Examples in one Puzzle

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 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.

• Locked Candidates
• Pairs Triples Quads
• Forbidding Chains 101 The Theory
• Forbidding Chains 102 The Practice
• Advanced Forbidding Chains, Forbidding chains are now referred to as Alternating Inference Chains (AIC)
• Definitions

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

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!