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March 16 very hard

 
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LT



Joined: 22 Mar 2006
Posts: 4
PostPosted: Wed Mar 22, 2006 5:25 am    Post subject: March 16 very hard Reply with quote
(467) (9) ( 46) | (5) (478) (278) | (28) (3) (1)
(37 ) (2) (8) | (9) (137) (17) | (4) (6) (5)
(345) (35) (1) | (6) (34) (28) | (289) (7) (29)
- - - - - - - - - - - - - - - - - - - - - - - - - - -
(1) (356) (2356)| (348) (9) (568) | (7) (24) (2346)
(8) (4) ( 236) | (137) (1367)(167) | (1269) (5) (2369)
(356 ) (7) (9) | (134) (2) (156) | (16) (14) (8)
- - - - - - - - - - - - - - - - - - - - - - - - - - -
(256) (1) (256)| (278) (5678) (4) | (3) (9) (267)
(2346) (36) (7) | (12) (16) (9) | (5) (8) (246)
(9) (8) (2456)| (127) (1567) (3) | (126)(124) (2467)

I'm stuck. Any hints?
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Wed Mar 22, 2006 11:48 am    Post subject: Intersections Reply with quote
Take a look at the row / block / column intersections.

For example, the <1> in B9 is in R9, so R9C4 and R9C5 cannot be <1>.

The <3> in C4 is in B5, so R5C5 cannot be <3>.

There are more: Look at <6> in C6, <2> in C1, <3> in C3, then you should be on your way with pinned and forced moves.

Alternatively, there is an X-wing on <3> in C3 and C9. So, R4C2, R4C4, R5C4 an R5C5 cannot be <3>. Then, R6C4 is pinned to be <3>, etc.

Keith
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
PostPosted: Wed Mar 22, 2006 4:30 pm    Post subject: Here's another explanation Reply with quote
Hi, LT! Welcome to the forum.

Keith's explanation is accurate. But since you're a newcomer, you may benefit from another explanation that uses different words. Here it is. (Oh -- I'm not going to get into the "X-Wing" -- that's probably a level of complexity you don't need yet.)

Start by looking at the possible locations for a "2" in the first column. There can't be a "2" in the first three rows because of the "2" in the top left 3x3 box. And there can't be a "2" in the sixth row, because of the "2" in the middle center 3x3 box. So the "2" in column 1 must appear in the bottom left 3x3 box.

But this also means that the "2" in the bottom left 3x3 box must appear in column 1. So the lists of possibilities you gave for r7c3 & r9c3 (256 and 2456, respectively) are incorrect. The only values that can appear at r7c3 are {5, 6}, and the only values that can appear at r9c3 are {4, 5, 6}. Does that make sense?

Now the matrix looks like this.
Code:
 467    9    46     5    478   278   28     3     1
 37     2     8     9    137   17     4     6     5
 345   35     1     6    348   28    289    7    29
  1    356  2356   348    9    568    7    24   2346
  8     4    236   137  1367   167  1269    5   2369
 356    7     9    134    2    156   16    14     8
 256    1    56    278  5678    4     3     9    267
2346   36     7    12    16     9     5     8    246
  9     8    456   127  1567    3    126   124  2467

Now we can see a "triplet" in column 3 -- the values {4, 5, 6} must occupy the cells r1c3, r7c3, and r9c3, in some order. So we can identify a "hidden pair" {2, 3} in the two remaining unresolved cells in column 3, or r4c3 & r5c3. And since the values {2, 3} must occupy r4c3 & r5c3, there cannot be a "3" at r4c2 or at r6c1. So let's update the matrix with that information.
Code:
 467    9    46     5    478   278   28     3     1
 37     2     8     9    137   17     4     6     5
 345   35     1     6    348   28    289    7    29
  1    56    23    348    9    568    7    24   2346
  8     4    23    137  1367   167  1269    5   2369
 56     7     9    134    2    156   16    14     8
 256    1    56    278  5678    4     3     9    267
2346   36     7    12    16     9     5     8    246
  9     8    456   127  1567    3    126   124  2467

There are more eliminations like these that can be made to clean up the grid. For instance, the "1" in the bottom right 3x3 box must lie in row 9, which means there cannot be a "1" at either r9c4 or r9c5. But you can place a couple of values in the puzzle now without doing all that cleanup. Take a look at row 6.

Clearly the only place to put a "3" in row 6 is at r6c4. And then the only place left for a "4" in that row is at r6c8, leaving an unresolved "triplet" {1, 5, 6} at r6c1, r6c6, & r6c7.

I hope that's enough help to get you rolling again. Have fun! dcb
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LT



Joined: 22 Mar 2006
Posts: 4
PostPosted: Wed Mar 22, 2006 10:27 pm    Post subject: Re: March 16 very hard Reply with quote
Got it! Thank you both. I don't know why I couldn't see those things on my own ...

-- LT
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