dailysudoku.com

[alanr555]

This is a place holder to establish the topic and to avoid the
sort of duplication that occurred on 11th when the same puzzle
had two thread due to contemporaneous composing of posts.

[alanr555]

This is an example of a puzzle where Mandatory Pairs can solve a LOT
of the preliminary work.

Before deriving any pencil mark profiles and using purely the M/Pairs
techniques the number of unresolved cells has been reduced to 28
by solving 27 in addition to the 26 original values set. That reduces
the pencil marks by around 50%

Of the 28 unresolved cells ten are in mutual reception pairs (ie two
cells within the same region which can be occupied only by one of
two identified digits). Of the 18 remaining six are in what one might
term "Mandatory Triples" (the first use of this term!) in that they are
in a set of three cells within a region with only three possible values
between them and each digit constrained to just two of the three cells.
This leaves only twelve cells which at this stage are not in a strong
relationship with another cell.

The next stage is to derive the profiles and then to discern the
patterns which will enable the full solution.

As it is now the early hours of the morning in UK, I may need to sleep
on this one before posting further. The protocol on this site is one
benefit of being on the American continent!

[alanr555]

There is A resolution using implication chains. The challenge is then
to refine these into a more elegant solution.

Cell r4c9 can take values 1 and 5.
Positing value 1 for this cell leads to the following

a) r6c9 must be 6 and so r6c6 must be 2.
b) r4c1 must be 5 and so either r4c4 or r4c5 must be 2
This would result in two cells in region five having value 2.

Using this "reductio ad absurdem" approach means that the
value in r4c9 must be 5. Given this the rest of the puzzle
resolves easily.


+++

This puzzle appears to have only ONE "crunch point" and so
the VH grading must rely upon the quality of the techniques
required rather than the old method of crunch-point quantity.

The technique of reductio ad absurdem (or contradictory
implications) was argued by David Bryant (who seems not
to have posted since October 2006) as very much NOT being
"trial and error".

This leaves two possibilities.

a) Either the solver uses this method and ipso facto demonstrates
that the method is not trial and error
or
b) There is a more elegant technique included in the program of
the solver and I have not found it!

Other contributors to this forum may be able to advise whether
Reductio ad absurdem is a valid method within the protocols
of this site.

Now, having solved this one I can retire to bed!

+++
Postscript:

It is pertinent to note that a reductio ad absurdem will almost
certainly involve an odd number of "in-region" relationships
as the rectilinear relationships (using just rows and columns)
will generally be congruent in binary link terms. This is more
a comment on solution methods than this puzzle.

Here having the double cell r4c4/r4c5 adds complexity to the
implication chain but assuming just one of these cells (and
they are functionally equivalent in this case) would lead to
a closed chain from r4c9 back to itself using links as
HHVHV where HV are horizontal/vertical and D is diagonal.
If this is a rule (and I have not the skills to prove it) then it may
prove an aid to "spotting" possible implication chains. For this
puzzle, the efficacy of the M/Pairs preliminary work resulted in
a lot of two-element cell profiles and so it was relatively easy to
spot the potential for an implication chain.

[storm_norm]

[storm_norm]

put me down for a xy-wing, {2,4,6}

[Marty R.]

[Marty R.]

[nataraj]

Came to that same 2,4,6 and it solved the puzzle.

____

quo usque tandem abutere, Catalana, patientia nostra?

[andras]

I found the 146, but my wife found a simple x-wing on 1, and that broke the puzzle as well. No doubt other solutions are also possible.

John

[alanr555]

[alanr555]

[nataraj]

[Johan]

The [246] xy-wing did it for me, but looking for some other steps to solve the grid I came across this one. There

is a potential [46] DP* in R59C45, only three ways for avoiding that DP :

1. R9C4=7
2. R5C5=3
3. R9C4=7 and R5C5=3

Suppose R9C4=7 => R5C4=6 => R5C9=5 => R5C1=3 => R6C1=2 => R6C6=6 => contradiction in Box 5 (R5C4 and R6C6 are both <6>)

This means that R5C5 must be <3>, solving the puzzle.

[cgordon]

[alanr555]

[alanr555]

[Asellus]

Alan,

DP is Deadly Pattern, the general term for any arrangement of digit placements that result in a non-unique solution. A UR is just the simplest form of DP.

Elsewhere (I don't currently have the link) there is a description of 6 Types of UR solutions. Usually, one of these established Type patterns is used when exploiting a UR. But, not always. Sometimes it is possible to note a particular pattern in a UR instance that allows it to be exploited. That was the case here. By noting that the <7> placement results in a contradiction it could be discarded, leaving a Type 1 UR: three matching bivalue corners where the UR digits can be removed from the 4th corner. Here, that left only the <3>, which was determined.

[tjones99999]

Can someone post which cells are involved in this xy wing?
Thanks

[cgordon]

[Johan]