dailysudoku.com

[techschool]

I successfully made the first 17 moves, but fail to see the logic in the 18th move; it looks like a guess to me (which the authors say should not happen). I do not believe they are mistaken, but that I can not see it.

[someone_somewhere]

Hi,

I suppose you mean this puzzle:

[techschool]

I must be dumb, for I do not see the elimination here: "- 8 it is in r7c3 or r8c3, so you can eliminate it from r7c1, r7c2 and r9c2" (after 17 moves I have one 8 located in r1c9 and therefore can only eliminate 8s from r1 or c9.

[David Bryant]

I worked through this puzzle without making any extraneous pencil marks, and got hung up at exactly the point our guest is talking about.

[techschool]

I follow "At this point I should have been able to spot the second piece of the {7, 9} pair in row 7, but for some reason I didn't see it. It's obvious enough now, because of the {7, 9} in the bottom left 3x3 box, the {7, 9} in column 4, and the {7, 9} in column 8. But I did find another way to proceed beyond this point.

What struck me was the fact that this puzzle could only have a _unique_ solution if the "7" in column 5 lies in the bottom center 3x3 box. If there were a "7" at r1c5 then there would be a {7, 9} pair in r7c6 & r8c6 -- this would match up with the {7, 9} pair in r7c9 & r8c9, making at least two solutions possible.

Armed with confidence in Samgj's devotion to unique solutions, I placed a "7" at r8c5 and went merrily on my way. dcb" but disagree with the logic because r1c5; r1c6; and r3c6 could also contain a 7 and I see a {7, 9} at r7c6; r7c9; r8c6; and r8c9. I think the problem lies in the fact that there is one 8 located in a corner box which leaves too many variables to solve the puzzle without guessing.

[David Bryant]

OK, I guess this hasn't been explained clearly enough yet. I suppose that the position you arrived at is the one illustrated in my previous post.

[David Bryant]

[techschool]

Thank you David, I see it now! Knew there was something I was missing. My patience was in short supply yesterday.

[Guest]

im stuck at the same place david bryant is at with his first post. i did not read any further because i want to solve this without any help or guessing.

[Booya]

WOW, I was stuck at the EXACT same point too...

I too missed the hidden pair 7,9 That led to the unique 2's. Isn't it funny how logic works?
________
weed vaporizer

[alanr555]

> There must be an "8" in column 3.
> It must lie in r7c3 or r8c3, as asserted.

This also is the conclusion of Mandatory Pairs.

However, what broke the back of this one for me was
not the 79 in row 7 but (after setting the Candidate Profiles!)
the 1458 quadruple in row 7. I supposed that it being "naked"
made the difference!

Having spotted the quadruple, it was the subsequent eliminations
which exposed the 79 pair.

+++

I treat each row or column as an exercise in congruence. This
requires that the row be determinable as a collection of subsets
of which any with just one member will be the resolved cells and
can safely be ignored.

In this case we started with

(3)(6) (1245789)
The 3,6 are the resolved cells and so can be ignored.
Spotting the quadruple transforms the subsets to
(3)(6) (1458) (279)
but then the eliminations expose the 2 in r7c2
and the subsets become
(2)(3)(6) (1458) (79)

The congruence rule requires that any subset must contain
exactly as many digits as the number of cells that it covers.

This is a valuable check on eliminations. Here, in what was
initially the (279) subset the cell profiles were

12458
5789
79

The latter 79 had been reduced previously because of the
mutual reception between r7c9 and r8c9. Mutual Reception
"ties together" the two cells, thereby excluding interlopers.

Clearly the above profiles contain more than the three
distinct digits implied by the length of the subset (279)
and so any digits that occur in the other subset must be
eliminated - meaning removal of 1458 wherever they
occur. This left 2, 79, 79 - which IS a congruent set!
However it is trivial to spot the further division into
(2) (79) whilst maintaining congruence.

In practice, I ignore the "single-member" subsets when
testing for congruence but in theory they should be
included so that the "line" profile always contains exactly
nine members.

The congruence principle applies equally to rows, columns
and regions. Indeed, one technique is inspection of the
derived profiles to ensure that congruence is maintained
after resolving one or more cells. If INcongruence is found
then further eliminations are required!

Alan Rayner BS23 2QT