Quantum Example in Tough Sudoku for 09/02/2009

Wed, 02 Sep 2009 00:00:00 GMT

The Tough Sudoku of September 2, 2009 contains yet another excellent example of the power of using Quantums. Unlike the previous example, this one does not utilyze uniqueness. However, it was during a search for uniqueness quantums that I found the step. Below, I have only sparsely illustrated the proof for this puzzle. The main intent of this post is to further investigate the sudoku tip, trick or technique of using quantum sets.

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.

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)
Tough Puzzle solution for August 7, 2009, a less complex introduction to Quantums
Quantum Examples in August 8, 2009 Tough
Quantum Examples in August 10, 2009 Tough
Definitions: This page has been recently updated, finally

The Puzzle

Above, find the puzzle with the possibilities, or pencil marks, already filled in. One way to proceed from here:

1) Start 22 UP 23: (6)c1 % column & box (hidden single).
2)(NakedTriple258)abc4 => a6,bc5,efi4≠258
- A cascade of Unique Possibilities (UPs) until 35 cells are uniquely identified - UP 35.

Some more easy steps

Some more easy steps are available here. The following are sufficient for this solution:

3a) (LockedCandidate8)h78 => g7≠8
3b) (NT134)gi7,h9 => g9,h78,i89≠134
3c&d)(LC14)e789 => e1234≠14
3e&f) (TwoStringKites1&3):if4,d51 => i1≠13
3g) (LC7)b12 => b8≠7

Although there remain some more relatively easy steps, the following quantum step is available, and it serves, imo, to significantly uncomplicate the puzzle.

Almost Quantum Naked Triple 147

Below, find a precursor to the actual step.

Above, three apparently disparite items can be considered as one possibility. HP is shorthand for Hidden Pair:

(HP17)b12 = (1)b3
- Meaning at least one of (1)b3, (HP17)b12 are true
(HP14)f23 = (1)f4
(QuantumNakedTriple147)bfg2.b1.f3 = [(1)b3 = (1)f4]

In short, because of the almost Hidden pairs at b12, f23 and the contents of cell g2, at least one of the following three items must be true:[(QNT147)bfg2, (1)b3, (1)f4].

This, by itself, does not eliminate anything. However, just a small amount of additional information is required.

Almost Two String Kite with candidate (1)

Above, but for the possibility of (1) at f3, we would have the following Two String Kite with candidate 1:

h5 = d5 - f4 = f2

Since this Almost Kite and the Almost Quantum Naked Triple share a target, we can now easily write:

3h) (QNT147)bfg2.b1.f3 = (1)b3f4 - (1)f3 = (Kite1)hd5,f42
=> h2 ≠1

Although this elimination seems rather tame, imo, it does unlock this puzzle significantly.

Below, find a path that finishes this puzzle:

3i) (NP45)h26 => h9≠4
3j) (LC4)gi7 => ace7≠4
3k) (4)h2 = (4)h6 - (4=6)g6 - (6=7)g9 - (7)g12 = (7)i1 => i1≠4
3l) (7)c8 = (7)i8 - (7=5)i1 - (5=3)e1 - (3=1)d1 - (1)d5 = (1)h5 - (1)h9 = (1-4)e9 = (4)e8 => c8≠4 , UP 56
4) (NP57)bi1 => aeg1≠57, UP 81

Solution

The following spectrum of difficulty applies to the path published above: