# Proof of Easy Tough of February 21, 2007

The following is an illustrated proof for the Tough Sudoku of February 21, 2007. This tough puzzle is well suited for the study of the limited techniques required:

You may wish to refer to previous blog pages. Links to these pages are found to the right, under Previous Entries.

It is not the goal of this page to show every possible step.

I highly recommend that you visit the Defintions page of this blog to properly understand the following terms and symbols:

• %
• strong
• weak
• native
• house
• Unique Possibilities (UP)
• =>
Always keep in mind: weak and strong need not be mutually exlusive properties. In other words, a set that is strong may also be weak.

### Puzzle at start

• e7 = 5% column
• e3 = 2% column
• f3 = 8% row & box

### UP 25

After finding the Unique Possibilities, I prefer (and highly recommend) a search for Hidden Pairs. Finding hidden pairs is often easily accomplished before entering the possibilities. A thorough explanation of how to do this is found in previous blog pages, including the Easy Proof for the Tough? of February 13, 2007.

### Hidden Pair 34

Illustrated to the left is the justification for placing a Hidden Pair 34 at df4.

Also, one can use the Locked Candidates properties of this Hidden pair to eliminate 3 from b4 and 4 from gh4.

After limiting cells df4 to only 34, the following Unique Possibilities are available:

• d6 = 9% column & box
• f6 = 1% box
• d9 = 1% column & box

### UP 28

Here, one can continue the search for Hidden Pairs, or if you prefer, enter the possibilities and look for Naked Pairs. Either choice will lead one down similar paths.

### Hidden Pair 39

Illustrated to the left is the justification for placing Hidden Pair 39 at c15. Finding this particular Hidden Pair is not very easy. Do not be concerned, as Hidden sets are always complemented by Naked sets.

After limiting cells c15 to only 39, one has available the following cascading Unique Possibilities:

• c9 = 4% column
• h7 = 4% box & row
• g5 = 4% box, column, & row
• c6 = 8% column
• b7 = 8% column & box
• a7 = 9% box & row

### Naked Pair 12

Illustrated to the left is an alternate approach. If one does not find the Hidden Pair 39, one can find the Naked Pair 12 at c28. This justifies eliminating the circled candidates.

After making the eliminations shown above, the following cascading Unique Possibilites are available:

• c6 = 8% cell
• c9 = 4% cell
• h7 = 4% box & row
• g5 = 4% box, column, & row
• b7 = 8% column & box
• a7 = 9% box & row
As one can see, the Hidden Pair and the Naked Pair reveal exactly the same cell solutions, and furthermore eventually the exact same Possibility Matrix.

### Possibility Matrix at UP 34

At this point, the path I recommend is to look for more naked or hidden sets and to consider Locked candidates and Coloring eliminations.

Available above are:

• Naked Quad 1257 at a13,bc2 in box b2
• Equivalent to Hidden Triple 349 at b13, c1 in box b2
Please note that it is not required that all the candidates 1257 exist in those four cells in order to reveal the Quad. Instead, it is only required that those four cells be limited to no more than those four candidates. Similarly, the Hidden triple only requires that 349 be limited to no more than the three cells b13, c1. It does not matter that c1 contains no 4.

The previous blog page, Coloring: How to Find details how to search for coloring eliminations. Scrolling through the candidates, one can quickly determine:

2. Not strong enough
6. Not strong enough
7. Locked candidate at f89 forbids f1=7, then not strong enough
9. Very interesting!

### Coloring on Candidate 9

Possible 9s are highlit green. The 9s have four very strong sets:

• c1,c5
• b3,h3
• g1,h3
• g1,g4
The act of just recognizing these strong sets, (and perhaps listing them), provides all the information required (plus some) to justify the eliminations shown

Although there are many paths to take with the 9s here, the one illustrated above is:

• c5=9 => b4,h5≠9
• c5≠9 =>
• c1=9 => g1≠9 => g4=9 => b4,h5≠9
• conclude b4,h5≠9
As a Forbidding Chain, also called an Alternating Inference Chain, (AIC), the elimination can be justified as follows:
• fc on 9s: c5 == c1 -- g1 == g4 => b4,h5≠9
No matter how one views it, the strong sets represented by the *'s and black lines bridged by the weak link (red line) between c1,g1 key the eliminations.

One could call this configuration a skyscraper, although knowing that type of jargon is completely superfluous. Amongst other superfluous techniques available above are two 2 string kites. Both of these, however, only make one elimination each. Thus, the elimination shown is the more efficient one. Similarly, recongizing the hinges is of no additional value.

The reason I call these labels superfluous: Recognizing the general idea of Coloring makes knowledge of the individual, specific, and very limited sub-techniques completely unnecessary.

### Naked Pair 27

After making the eliminations of the 9s indicated previously, the possibility matrix reveals a Naked Pair.

To the left, Naked Pair 27 is circled at b46. Since this naked pair is in both column b and box b5, all the candidates circled in red are eliminated.

The puzzle now cascades with % cell type Unique Possibilities to UP 44. At this point, one need only find f1 = 6% column or f1 = 6% box. Then, the puzzle cascades again with % cell type Unique Possibilities to the end. Thus, after making the eliminations above, one would conclude: UP 81.

### Proof

1. Start at 22 filled - the given puzzle. Unique Possibilities to 25 filled. (UP 25).
2. Hidden pair 34 at df4 forbids b4=3, gh4=4, d4=679. f4=67 UP 28
3. Hidden pair 39 at c15 forbids c5=18, c1=24 UP 34
1. fc (coloring) on 9's: g4 == g1 -- c1 == c5 forbids b4, h5=9
2. Naked pair 27 at b46 forbids b128,a4=2 and forbids b1235,a45=7 UP 81
• Sets: 4(2) = 8
• Max depth 2 at each step
• Rating: 4(.03) = .12

1 Comment
 Indicate which comments you would like to be able to see GeneralJokesOtherSudoku Technique/QuestionRecipes
 The first pair I found was 48 at b7 c9. I think I solved the puzzle without the fc (unless I made a 'lucky mistake'). Maybe I'll try to replicate my solution again later. |  |

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