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.

Some defintions required to understand this page

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

Three cascading Unique Possibilities:

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:

Dead ended

Not strong enough

Dead ended

Dead ended

Dead ended

Not strong enough

Locked candidate at f89 forbids f1=7, then not strong enough

Dead ended

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

Start at 22 filled - the given puzzle. Unique Possibilities to 25 filled. (UP 25).

Hidden pair 34 at df4 forbids b4=3, gh4=4, d4=679. f4=67 UP 28

Hidden pair 39 at c15 forbids c5=18, c1=24 UP 34

fc (coloring) on 9's: g4 == g1 -- c1 == c5 forbids b4, h5=9

Naked pair 27 at b46 forbids b128,a4=2 and forbids b1235,a45=7 UP 81

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.