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Sep 20 Sudoku

 
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gsri9
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PostPosted: Wed Sep 21, 2005 3:13 am    Post subject: Sep 20 Sudoku Reply with quote

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
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geoff h



Joined: 07 Aug 2005
Posts: 58
Location: Sydney

PostPosted: Wed Sep 21, 2005 6:45 am    Post subject: Reply with quote

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
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geoff h



Joined: 07 Aug 2005
Posts: 58
Location: Sydney

PostPosted: Wed Sep 21, 2005 9:13 am    Post subject: Reply with quote

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.
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Guest
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PostPosted: Wed Sep 21, 2005 9:30 am    Post subject: Reply with quote

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 Very Happy
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Lulu



Joined: 21 Sep 2005
Posts: 11
Location: Manchester, England

PostPosted: Wed Sep 21, 2005 12:10 pm    Post subject: Reply with quote

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 Confused
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geoff h



Joined: 07 Aug 2005
Posts: 58
Location: Sydney

PostPosted: Wed Sep 21, 2005 12:45 pm    Post subject: Reply with quote

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.
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Lulu



Joined: 21 Sep 2005
Posts: 11
Location: Manchester, England

PostPosted: Wed Sep 21, 2005 1:29 pm    Post subject: Reply with quote

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 Very Happy
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Wed Sep 21, 2005 3:12 pm    Post subject: Difficulty Reply with quote

geoff h wrote:
... snip ...

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

... snip ...


I have to agree with Geoff. I messed this puzzle up twice before I finally got it done. Several of the patterns Samgj built in were not easy to discern. Great puzzle, Sam! dcb Smile
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sewandquilt
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PostPosted: Wed Sep 21, 2005 4:26 pm    Post subject: Sept 20th Reply with quote

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
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Wed Sep 21, 2005 6:03 pm    Post subject: Re: Sept 20th Reply with quote

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

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


Hi, SewandQuilt!

It would have been helpful if you had explained a bit more about the possibiliities you've already ruled out. I'll do my best to explain this clearly by proceeding from the position shown. Your (not so obvious) next move is to place a "6" in row 9, column 4 and a "5" in row 7, column 6. Let me explain why.

First, look at r1c6. I thnk it's obvious that this cell must contain either a "5" or a "6".

Now look at the ninth row. Since there's already a "3" in column 2, in column 3, and in column 4, it's clear that the "3" in row 9 must appear in either r9c5 or in r9c6 -- all the other cells in row 9 are blocked. This means that we can eliminate the possibility of a "3" appearing in any of the remaining cells in the bottom middle 3x3 box.

Now look at r7c6. By direct elimination it's clear that this cell must contain a "5" or a "6" -- {1, 2, 4} appear in row 7 already, {7, 8, 9} appear in column 6, and we can rule out "3" by the analysis presented above.

But this simplifies the contents of column 6 -- we know that the pair {5, 6} must occupy the two cells r1c6 and r7c6, leaving the triplet [1, 2, 3] to occupy r3c6, r8c6, and r9c6.

But we also know that the triplet {4, 5, 6} must appear in the middle 3x3 box -- specifically, in r4c5, r5c5, and r6c5. So the cells r7c5, r8c5, and r9c5 cannot contain either a "5" or a "6". This means that the pair [5, 6] MUST occupy the cells r9c4 and r7c6 -- there's nowhere else they can fit in the bottom middle 3x3 box.

Now look at the bottom right 3x3 box. Clearly the value "6" must appear in the seventh row in this box -- the ninth row is filled already, and the eighth row is blocked by the "6" in r8c2. But this means the pair {5, 6} at r7c6 is resolved to the value "5", and thus a "6" must appear at r9c4.

Is that helpful? I think the rest of the puzzle is pretty easily solved once you've placed these two values. dcb Smile

PS This was the "AHA!" moment in this puzzle for me. Even though I spent a lot of time on it, and had worked it out correctly, it took quite a bit of thought to get back to the position you illustrated. As I've observed before, this is a great puzzle.
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tavah316
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PostPosted: Wed Sep 21, 2005 7:03 pm    Post subject: David, you are a genius!! Reply with quote

A co-worker and I have been wasting a lot of work time on this puzzle! Wink 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!
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alanr555



Joined: 01 Aug 2005
Posts: 194
Location: Bideford Devon EX39

PostPosted: Thu Sep 22, 2005 1:18 am    Post subject: Re: David, you are a genius!! Reply with quote

> 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
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Guest
Guest





PostPosted: Thu Sep 22, 2005 6:31 am    Post subject: Reply with quote

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





PostPosted: Thu Sep 22, 2005 6:34 am    Post subject: Re: Sep 20 Sudoku Reply with quote

gsri9 wrote:
I got stuck on this one.


gsri9 started all this off tto the benefit of some of us - has he too found the answers helpful???
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gsri9
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PostPosted: Fri Sep 23, 2005 2:42 am    Post subject: Thank you very much Reply with quote

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.
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PostPosted: Sat Sep 24, 2005 8:26 pm    Post subject: Reply with quote

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.
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alanr555



Joined: 01 Aug 2005
Posts: 194
Location: Bideford Devon EX39

PostPosted: Sun Sep 25, 2005 10:24 pm    Post subject: Reply with quote

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
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