dailysudoku.com

[alanr555]

Preliminary work with Mandatory Pairs resolves fourteen cells
and identifies two pairs in mutual reception - plus a remote pair
in the remaining two cells of a column.

Having done that work, the grade is still Very Hard. When preparing
commentaries on the application of the Mandatory Pairs approach
to solving I use the draw/play facility on this site to confirm how
far I have got by "fixing" the cells resolved so far. This also enables
a check on the profiles by using the "sweep" facility - although I will
usually have shorter strings in a few cells as a result of using the
M/Pair detail that the sweep cannot recognise.

Today the puzzle reduces to

6-r/-58/--4
--9/34-/675
-45/--6/8-3
--7/1--/-46
-9d/d--/-5-
-2d/d-5/3--
1-r/569/48-
968/--4/53-
--4/8--/-69

where d represents downward mutual reception
a is across mutual reception and r is remote grouping

Seeing no obvious patterns in the profiles, the next step was to
consider cells with only two elements in the profile and which
look to have some connections with two element cells on BOTH
the axes through the cell.

A candidate is r7c2 holding profile 37.
This is part of a triple in row seven and offers some possibility
of gaining information from column 2.

I do not like "trial and error" but sometimes it seems necessary
to posit each of two possible values for a cell and look to see if
this results in either a paradox or a unique value for another cell.

Here positing 7 in r7c2 eliminates both 3 and 7 from the region
and resolves all cells within it - including r9c2.
Positing 3 in r2c7 eliminates 2 and 3 from the region and sets up
a mutual reception of 5 and 7 in row nine. This does NOT lead to
a resolution and so I was inclined to look elsewhere.

However the hint brought me back to r9c2 - suggesting that there
must be a clue in there somewhere to the resolution of the pair
in r9c1 and r9c2 determining a unique value for r9c2.

My aim is for simplicity and I find the advanced techniques somewhat
mind boggling (I am psychologically challenged by a sleep disorder
and mental effort puts me to sleep!). Thus I have reached the point
of getting stuck here. Of course I could try placing values in ALL the
cells until a paradox arises but this is not in the spirit of this site and
I would expect that a specific technique exists which will lead to the
solution on this point - indeed the computer solver was satisfied!

This is NOT any shadow on the Mandatory Pairs method as M/Pairs
would never claim to be able to solve Hard or Very Hard puzzles
(but ought to be able to solve Easy or Medium).

I look forward to another participant in this forum explaining the
route to the full solution.

[alanr555]

Following on from the previous posting, it appears that I
did not pursue the implications far enough of the mutual
reception 57 in r9c1,r9c2.

This removes the other sevens from row nine and fixes
a 7 in r5c6 which leaves a 12 pair in column 7 and then
a paradox of 19 and 29 in the same column as the 12 pair.

This disproves the 57 M/R in row 9 and therefore sets the
alternative scenario which has 5 in r9c2.

However, I am not happy about having such an elaborate
implication chain. This very much has the feel of trial and
error and I would not have found this approach had it not
been for the hint or a clue from another forum member.

Column seven has a quartet.
Removing a value from the four possibilities would leave
only three values to fill four spaces and thus disprove
any hypothesis on which it was based.

The difficulty here is that the removal of all possibility of
a particular digit in the column arises only in ONE option
for the "starting" cell (r7c3). That is why I do not like
this solution - but perhaps my dislike is just to the length
of the implication chain!

Clearly the computer solver contains some code to identify
this situation. How complex is it? Is it amenable for the
human brain to follow a parallel course?

[andras]

It's certainly a more challenging one than some recent VH puzzles.

My own route involved an x-wing on 3, an xyz on 123, and a final x-wing on 7, after which the puzzle broke nicely. But there must be an easier way!

John

[alanr555]

Thinking about it in recent years, I find that I made it FAR
too complicated in my earlier posts.

The mandatory Pairs work was fine.
The next step should have been recognition of a rectangle
in r5c6, r5c7, r9c6,r9c7
which contains the ONLY occurrences of a particular digit
in columns six and seven.

This eliminates all other occurrences of that digit in rows
five and nine - because the two placements of the digit
in the rectangle MUST occupy both rows and columns.

This results in the elimination of the relevant digit from r9c2
which is then reduced to a single candidate.

After this the puzzle resolves simply.

+++

The danger as ever is to home in on a "wrong" aspect of
a puzzle - building complexity upon complexity when there
is a simpler way forward. This is akin to the way that some
scientists have posited up to eleven dimensions in order
to make the equations of their string theory consistent.
However if they had but considered the possibility of
"negative mass" the observable three dimensions would
suffice and the whole theoretical string theory edifice then
would become unnecessary. The added advantage of this
negative mass is that the resultant negative gravity could
explain the expansion of the universe!

As with the universe, so with Sudoku. It is just a matter of
finding the right clue - and revealing the implicit simplicity.

+++

I apologise for falling into the trap this time.

[Marty R.]

[Clement]

I agree with Marty R. The X-Wing r5c6, r5c7, r9c6, r9c7 solves the puzzle.

Clement.

[alanr555]

[Marty R.]