# Page 1 of Diabolic Proof for 03/11/07

The following is an illustrated proof for the Tough Sudoku of March 11, 2007. This is a very difficult puzzle to solve, unless either one resorts to guessing, or chooses to make Uniqueness arguments.

This is page one of a multi-page proof for this puzzle. A link to the next page is provided at the bottom of this page.

There are many ways to tackle this one. The primary focus of this proof is to illustrate using Forbidding Chains, also called Alternating Inference Chains or AIC. Some of the Chains that are presented are complex forbidding chains - or advanced forbidding chains. Such chains merely use arguments that are Boolean variables (something that is either TRUE, or False). Some of these Booleans represent native puzzle conditions that are neither bivalue in a cell, nor candidates that are bilocation.

You may need to refer to previous blog pages to understand this proof. Links to these pages are found to the right, under Sudoku Techniques.

At many times during this illustration, there are other steps available. It is not the goal of this page to show every possible step, but rather to illustrate steps that, taken together, unlock this puzzle. Some steps that are not required will be illustrated. The purpose of illustrating such steps is primarily to clue the potential solver on how to decide upon a plan of attack for truly diabolical puzzles.

The information on the following blog pages is required to understand this page:

The illustrations of forbidding chains used in this proof will share the same key:

• black lines = strong links
• red lines = weak links
• black circles connected with black lines = multi-part strong sets
• candidates crossed out in red = candidates proven false

### Puzzle at start A few Unique Possibilities are available here.

• d5 = 1% column
• d4 = 7% column & box
• d8 = 5% column & box
• i8 = 4% box
• c6 = 8% column & box
The puzzle, now advanced to UP 27, starts without any real indication of its relative difficulty.

### Puzzle at UP 27 Completely illustrating each of the relatively easy steps will consume much space. Therefor, easy steps will be referenced, without complete illustrations.

To advance this puzzle a very small bit further, here is what I found:

1. Locked 2s at b46 forbids b789=2
2. Locked 3s at a45 forbids a8=3
3. Pair 69 at ab8 forbids ab9,b7,fg8=6 and forbids b7,g9=9
• => g8 = 8% cell, thus UP 28

### Puzzle at UP 28 At this point, there are many easy techniques available. Unfortunately, none of them advance the puzzle significantly. I found the following:

1. Locked 8's at d13 forbids e13=8
2. Locked 2's at d23 forbids d3f2=2
3. Hidden pair 28 at dh3 forbids d3=6, h3=4569
4. Locked 1's at c12 forbids c79=1
5. Locked 2's at hi7 forbids cef7=2
6. Locked 7's at g79 forbids h7=7
7. Locked 6's at d12 forbids e13f2=6
8. Hidden pair 69 at b38 forbids b3=45
• AUR 69 at ab38 forbids a3=69
I have chosen to not use the AUR step in this proof. If one chooses to use that step, the puzzle is much easier to solve. It is, however, then impossible to prove a unique solution.

If any of the nomenclature above is unfamiliar, this blog has previously detailed the manner used to describe each step listed.

### Y wing style Illustrated above is a relatively easy Y wing style elimination. Standard puzzle marks, discussed in quite some detail on previous blog pages, will indicate to look for this pattern:

• i2=3 == i5=3 -- i5=8 == i1=8 -- h3=8 == h3=2 => i2≠2
• i7 = 2% column

### The first of many depth 4 forbidding chains Illustrated above is a standard forbidding chain using 4 native strong sets. Again, standard puzzle marks will indicate to search in this area. The puzzle advances as follows:

• e3=3 == e7=3 -- e7=1 == b7=1 -- b7=5 == c7=5 -- c3=5 == c3=4 => e3≠4
• e1 = 4% box & column
• h2 = 4% box & column
• h3 = 2% box & column
• % cells: d3 = 8, d1 = 6, d2 = 2
• Thus, UP 35
• Locked 9's at ghi1 forbids gi2,g3=9

### Another forbidding chain using 4 native strong sets Illustrated above is another chain beating up on column i:

• i5=3 == i2=3 -- i2=6 == i6=6 -- gh4=6 == e4=6 -- e4=8 == h4=8 => i5≠8
• i1 = 8% column
• Thus UP 36
The chain above is not as strongly indicated by the standard puzzle marks. The more advanced marks will underline and V the 6 at e4, making this chain a bit easier to uncover.

The proof continues on the next page. 