The following is an illustrated proof for the
Tough Sudoku of 01/24/07.
Since this proof uses forbidding chains, you may need to refer to previous blog pages to
understand this proof. Links to these pages, including *Forbidding Chains 101, The
Theory*, are found to the right, under **Previous Entries**.

### Puzzle at start

A few Unique Possibilities:

- i4 = 3% box & column
- b5 = 3% row
- a9 = 3% row, box & column
- i8 = 5% column

Now at UP 26, (or at the start),
note also Hidden pair 17 at e2f3. Thus e2f3 are limited to only 17.

### Possibility Matrix at 26 filled

In column i, both 1s and 4s are locked in box h2. Forbids:

3 strong sets with the 7's:

- a67 = 7
- h72 = 7
- gf3 = 7

Beg this forbidding chain on 7's:

- a6 == a7 -- h7 == h2 -- g3 == f3 forbids f6=7

Yielding this new possibility matrix:

### Possibility Matrix again at 26 filled

Most *normal* techniques fail to advance this puzzle further. In order to properly
analyze this puzzle at this point, I print it out and mark it up. Often times, the act of
marking the puzzle makes a few forbidding chains obvious.

### Puzzle Mark-up at 26 cells solved

Refer to the page,
Forbidding Chains 102 The Practice for an in depth explanation of
my puzzle markings at this stage. I have added the black V at g8 because of the 26 at a8 and
69 at b9.

Of special interest are the endpoints of strong links in the following cells: a6, c7, c9.
Two of these three beg the following:

### Depth 4 Forbidding Chain

Key:

- Black circles = endpoints of strong links
- Black lines = strong links
- Red lines = weak links
- Green circles = elimination targets

This forbidding chain is an excellent example of mixing the use of strength in location and
strength in cells. It can be written as follows:

- c9=8 == c7=8 -- c7=1 == f7=1 -- f3=1 == f3=7 -- g3=7 == g9=7
- forbids c9=7 and
- forbids g9=8

One can now solve the following cells:

- g9 = 7% row
- h2 = 7% box & column
- f3 = 7% box & row
- e2 = 1% cell
- i3 = 1% box, row, & column
- i2 = 4% cell

Yielding Unique Possibilities to 32 cells filled (UP 32).

###
New Possibility Matrix at UP 32

Here we have the following Locked candidate eliminations:

- Locked 4's at ab3
- Locked 6's at ac2

There are a few forbidding chains on 8's possible. One that gets all the eliminations in one
step is commonly called a

*finned X wing*. Note these strong sets:

- d3, g3 = 8
- g8, ef8 = 8

Thus the following forbidding chain on 8's:

- d3 == g3 -- g8 == ef8 forbids d79=8

Once again, most *normal* techniques fail to advance this puzzle further. One could
do another puzzle mark-up, but not that many changes have occurred. I often just use my old
mark-up at this point. Happily, the possibly pivotal 6 at g8 yields some eliminations.

###
Y wing style elimination #1

Key as as before.

The following strong sets are considered:

- g8, ab8 = 6
- b9 = 6, 9
- b3, g3 = 9

The strong link:

is a common trick.

b9 = 69 is the vertex of the Y.

Forbidding chain representation of this step:

- g8=6 == ab8=6 -- b9=6 == b9=9 -- b3=9 == g3=9 forbids g8=9

After this step, we have the additional eliminations:

- Locked 9's at hi7 forbids cd7=9

### Y wing style eliminations #2

This elimination type I find to be *very common*. The 6's at g8h7 form the vertex of
this Y. Here is this step as a forbidding chain:

- a8=2 == a8=6 -- g8=6 == h7=6 -- d7=6 == d6=2

### Y wing style elimination #3

Here, f8=18 is the vertex. Notice again that a candidate is grouped within a strong set.
In this case, the 8's in box h8 are grouped according to rows. Here is this step as a
forbidding chain:

- c7=1 == f7=1 -- f8=1 == f8=8 -- g8=8 == hi7=8

Now we can solve some more cells again:

- c9 = 8% box & column
- c1 = 9% column
- i1 = 8% cell
- i7 = 9% cell
- d3 = 8% row & box
- g3 = 9% row & box

### Possibility Matrix at UP 38

From this point, the puzzle solves quickly. One way to finish:

Then:

- d2 = 2% row
- d7 = 6% cell
- % cells to the end

### Solved Sudoku

### Proof

Here is the proof, pared down to required steps and presented in my usual style:

- Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Hidden pair 17 at e2f3 forbids e2=245, f3=248
- c9=8 == c7=8 -- c7=1 == f7=1 -- f3=1 == f3=7 -- g3=7 == g9=7
forbids g9=8,c9=7 UP 32

- fc on 8's:d3 == g3 -- g8 == ef8 forbids d79=8
- g8=6 == ab8=6 -- b9=6 == b9=9 -- b3=9 == g3=9 forbids g8=9
- Locked 9's at hi7 forbids cd7=9
- a8=2 == a8=6 -- g8=6 == i7=6 -- d7=6 == d7=2 forbids ac7,ef8=2
- c7=1 == f7=1 -- f8=1 == f8=8 -- g8=8 == hi7=8 forbids c7=8 UP 38

- Locked 2's at ab3 forbids ac2=2 UP 81

- Sets: 2 + 4 + 2 + 3 + 1 + 3 + 3 + 1 = 19
- Max depth 4 at step 2.2
- Rating: 2(.01) + 2(.03) + 3(.07) + .15 = .37

Like most puzzles, there are many ways to tackle this one. Below find two alternate approaches.

### Alternate Proof 1

Step 3.2 of this proof uses a forbidding chain element explained in
Advanced Forbidding Chains.

- Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Hidden pair 17 at e2f3 forbids e2=245, f3=248
- c9=8 == c7=8 -- c7=1 == f7=1 -- f3=1 == f3=7 -- g3=7 == g9=7
forbids g9=8,c9=7 UP 32

- Locked 6's at ac2 forbids c1=6
- b9=9 == b9=6 -- a78=6 == a2=6 -- c2=6 =={pair 25 at c25} -- c1=25 == c1=9
forbids b3,c79=9 UP37

- Locked 2's at ab3 forbids ac2=2 UP 81

- Sets: 2 + 4 + 1 + 5 + 1 = 13
- Max depth 5 at step 3.2
- Rating: 2(.01) + .03 + .15 + .31 = .51

### Alternate Proof 2

Step 3.2 of this proof also uses a forbidding chain as a Boolean variable

- Start at 22 filled - the given puzzle. Unique Possibilities to 26 filled. (UP 26).
- Hidden pair 17 at e2f3 forbids e2=245, f3=248
- c9=8 == c7=8 -- c7=1 == f7=1 -- f3=1 == f3=7 -- g3=7 == g9=7
forbids g9=8,c9=7 UP 32

- c5=2 == c5=5 -- a6=5 == a2=5 -- d2=5 == d2=2 forbids c2=2
- Very common Y wing style above

- {pair 24 at ab3} == b3=9 -- b9=9 == b9=6 -- a8=6 == a8=2 forbids a2=2 UP 33

- Triple 689 at dhi7 (dh7=689,i7=89) forbids ac7=6, af7=8, c7=9 UP 81

- Sets: 2 + 4 + 3 + 4 + 3 = 16
- Max depth 4 at steps 2.2 and 3.2
- Rating: .03 + 2(.07) + 2(.15) = .47

### Proof choices

For every tough sudoku puzzle, there are many manners to come to the solution, and therefor
many possible proofs. The choice as to which proof is *better* is a completely
**arbitrary** one. I prefer proofs of less maximum depth, but as is illustrated
above, adding a little bit of depth may reduce both the total strong sets considered
and the total steps used in a proof. The ratings that I assign are my clumsy attempt at comparing
proof complexity.

The choice of rating system is also completely arbitrary. For this reason, I
like to provide the entire vector: Total sets, Maximum depth, rating. Of course, more factors
could be considered. I am always hoping for someone to suggest a better, yet not overly
complicated, way to evaluate puzzle proofs.

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