dailysudoku.com

[gsri9]

I got stuck on this one. Tried a full enumeration of all possibilities, but couldn't get past. Any hints (with a reason) would be most welcome.

214 -9- --3
--3 --4 -5-
-7- --- ---

73- 1-9 8-2
--- 2-8 ---
--2 3-7 -95

-2- 4-- -1-
-6- 9-- 5--
4-- --- 927

[geoff h]

Hi there,

Yes, personally, I found this puzzle to be VERY HARD rather than HARD.

I'd like to help, but need more details as to exactly how far you have progressed. For example, could you please supply which individual numbers you have as possibilities for each missing cell that you have not yet been able to solve. Then it will be a lot easier to direct you in the right direction.

Just a couple of quick hints that I had from memory :

1. In Box 5, I suppose you already have noticed the triplet of 4,5,6 in r4c5, r5c5 and r6c5?? Therefore, numbers 4,5,6 can NOT appear anywhere else in Column 5.

2. In Box 6, there is a tiplet of 1,4,6 in r4c8, r5c9 and r6c7. This then leads to a pair of 3,7 in r5c7 and r5c8. This of course means that 3 and 7 can NOT appear anywhere else in Row 5.

Let me know if this helps. Otherwise, please advise more detail as to how you've progressed as discussed above.

Cheers

[geoff h]

A couple more hints.

1. Consider Box 2. there is a quadruple of 5,6,7,8 in r1c4, r1c6, r2c4 and r3c4. Therefore, there can be no more 5,6,7, or 8 in Box 2.

2. Consider Box 1. Nr 5 MUST be in either r3c1 or r3c3. Therefore, there can be no more 5 s in Row 3 ( specifically r3c4 ).

3. Consider Column 7. There is a triplet of 3,6,7 in r1c7, r5c7 and r7c7. Therefore, you can exclude all other Nrs 3,6,7 from Column 7. You also now have a pair of 1,2 in Row 2 and can therefore exclude Nr1 from r2c9.

This is a very hard puzzle.

Cheers.

[Guest]

Thankyou Geoff - I too was stuck at the same stage. Your last suggestions were very obvious and should have hit between the eyes - but didn't!
After hours of agonising will have another go at it.
Thanks

[Lulu]

I thought I was getting the hang of these puzzles but I'm having trouble with this one.

I am at almost the same spot as gsri9 but can't eliminate the 8 at r1c5 to get the 9 you have.

I managed to work out the 3,7 pair in box 6 as these are the only possible places for 3 or 7 to go rather than eliminating 1,4,6.

But geoff h, how can I eliminate Nr1 from r2c9 just because I have a pair of 2,1 at r2c7 as I could have a Nr1 at r3c7 or r6c7

[geoff h]

Hi Lulu,

You can eliminate the Nr 1 from r2c9 because we are considering Row 2 at the moment. We have a pair of 1 and 7 in Row 2 ( in cells r2c5 and r2c7 ). Therefore we can eliminate all other 1s and 2s from Row 2.

I'm not sure why you are referring to r3c7 and r6c7 as they have nothing to do with eliminating the 1 from r2c9.

Also, re the Nr 9 at r1c5. You can place 9 in this cell because it is the only possible place for a 9 in Row 1 if you look carefully. There is already a 9 in column 4 ( r8c4 ), column 6 ( r4c6 ), column 7 ( r9c7 ) and column 8 ( r6c8 ). Therefore, 9 can only go in the 5th column for Row 1.

Hope this helps.

Cheers.

[Lulu]

Thanks Geoff, I've just discovered your reply after I finally managed to suss out what you meant earlier and I've completed the puzzle.
Thank you for your help

[David Bryant]

[sewandquilt]

I had figured out all the above suggestions on my own, but am still unable to proceed beyond this point:

214 -9- --3
--3 7-4 -5-
-7- --- ---

73- 1-9 8-2
--- 2-8 ---
--2 3-7 -95

-2- 4-- -1-
-6- 9-- 5--
4-- --- 927

[David Bryant]

[tavah316]

A co-worker and I have been wasting a lot of work time on this puzzle! Thanks so much for your hint about the 3s in the bottom, middle 3x3 box...that was all we needed to get us back on track! It's interesting, b/c we've eliminated numbers before by saying, this is the only row that number is in for this box so we can delete the rest from that row. But, oddly enough, we've never done the reverse (not that we can remember, anyhow!) which is what you've explained here! Thanks again for helping!!!! This was a tough "hard" puzzle!

[alanr555]

> we've eliminated numbers before by saying, this is the only row that
> number is in for this box so we can delete the rest from that row. But,
> oddly enough, we've never done the reverse.

The post on sudokusolver refers to six solution methods used by the
programmers. Whilst people do not think like computers (non-nerds
anyway!) consideration of those methods can aid understanding.

I started off applying computer methods manually and it was taking me
too long to solve things. Subsequently I have adopted a methodology
using 'human' inspection first and then resorting to "solution grid" ways
only when the inspection seems to be exhausted.

Today I was able to solve the Daily Mail puzzle whilst taking tea in the
supermarket and to do it without any "pencil marks". For me that was
a mark of progress - to stop thinking in computer grind terms and just
to rely on logic to narrow down possibilities rather than the mechanical
task of spotting patterns in a solution grid.

Alan Rayner BS23 2QT

[Guest]

That's very encouraging from us Non-Nerd people who enjoy solving Sudoku by simple logic!!!!

[Guest]

[gsri9]

Thank you all very much. It clarifies matters so much. Sorry, I couldn't post a thanks note earlier as my internet connection was a bit unstable.

[Guest]

Yet another thought on this one. In box 8 (middle bottom) you must have 6 & 3 in the bottom row. (They can't go in columns 2 or 3.) That gives 8 numbers 'impinging' on r7c6: it must be 5, and the rest's easy.

[alanr555]

I agree that this one was a REAL challenge. However, it did not contain
unduly complex logic - just the need to see patterns at a much earlier
stage in the solution than usual.

I got stuck on a solution grid looking like

-/-/- 568/-/56 67/678/-
689/89/- -/12/- 12/-/689
5689/-/5689 68/123/123 124/468/14689

-/-/56 -/456/- -/46/-
1569/459/1569 -/456/- 37/37/146
168/48/- -/46/- 14/-/-

3589/-/5789 -/378/356 36/-/68
138/-/178 -/12378/123 -/348/48
-/58/158 568/138/1356 -/-/-

where the hyphens represent resolved cells.

I find the concept of "congruence" useful when checking a solution grid.

Essentially this requires that every row/box/column has as many cells
unresolved as there are unallocated digits within it.

For example row 1 above has -/-/- 568/-/56 67/678/-
which has FOUR cells unresolved and four unique digits (5,6,7,8).

If a solution grid contains any row, box or column that is IN-congruent
there is clearly scope for further work to make it congruent. The problem
starts when (like the grid above) there is congruence in all rows, boxes
and columns. Then there is a need for a more sophisticated endeavour.

In this case the key has already been identified by others as being in
box 8. Some values may be eliminated outside the confines of simple
congruence.

Two useful rules apply:

A) If the only possible locations for a particular digit within a box lie in
a single row or column then that digit CANNOT be possible in any
cell of that row/column outside the box.

Example: In Box 9 the digit '6' must be in row 7.
This removes '6' from r7c6.

B) If the only possible locations for a particular digit within a row or a
column lie in a single box that digit CANNOT be possible in any other
row/column INside the box.

Example: In column 5 the digit '8' must be in box 8.
( -/12/123 456/456/46 378/12378/138)
This removes '8' from r9c4.

Example: In row 9 the digit '3' must be in box 8.
(-/58/158 568/138/1356 -/-/-)
This removes '3' from r7c5, r7c6, r8c5, r8c6.

Applying these two rules leaves r7c6 as '5' and immediately afterwards
reduces r9c4 to '6'. It is a useful croos check then to confirm that the
grid is still congruent. It will not be unless one removes the consequences
of resolving the two cells so box 8 has FIVE unresolved cells (containing
1,2,3,7,8) instead of the seven previously.

If one does not reduce r1c6 from 56 to 6 (ie resolve it!) then column 6
will have FIVE unique digits (1,2,3,5,6) but only FOUR cells. This is, of
course, the clarion call for work to remove the incongruity!!

Having resolved r7c6 and r9c4 the remainder of the puzzle resolved by
means of congruity checks (although simple logical consequences were
used in practice rather than tortuous congruity checks).

I have added this note as it highlights the use of congruity checking.
However, one always hopes that such checks will not be necessary
as they can be laborious - especially in cases such as this one where
they are needed at such an early stage.

Alan Rayner BS23 2QT