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CORUJA
Joined: 16 Jun 2007
Posts: 15
Location: BRUMADINHO - MG; BRAZIL
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 Posted: Mon Jun 18, 2007 7:24 pm Post subject: 29 May, VH |
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Earl,
I'm stuck exactly at the point yo were on May 30: can't get rid of the 2 in R1.
Please help me! Thanks. |
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dulaby
Joined: 02 May 2007
Posts: 13
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| Posted: Tue Jun 19, 2007 7:47 pm Post subject: |
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Coruja,tudo bem? .
Repare que existe uma xy-wing em r1c3(7 ,r4c3(37) e r4c5(3 e o <8> sai de r1c5 ,portanto r1c4 tem que ser <2>.Isso resolve a sua duvida? |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Wed Jun 20, 2007 11:12 am Post subject: |
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You do need those sunglasses in sunny Brazil !
Dulaby, the 8 followed by a ) produces that smiley. You're better off using square [] brackets. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Wed Jun 20, 2007 5:31 pm Post subject: May 29 |
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Coruga,
If you show me the grid, I might be able to help.
Without the particular situation, my memory is untrustworthy.
Earl |
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CORUJA
Joined: 16 Jun 2007
Posts: 15
Location: BRUMADINHO - MG; BRAZIL
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| Posted: Wed Jun 20, 2007 6:37 pm Post subject: MAY 29 VH |
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Earl,
I'm still at this point, and can't see how to get rid of the 2 in R1C3 , in R1C5 and in R1C8.
Thanks again.
| Code: |
+------------+--------------+------------+
| 4 6 278 | 257 258 3 | 9 125 15 |
| 3 9 278 | 1 258 78 | 678 256 4 |
| 5 278 1 | 6 4 9 | 378 23 37 |
+------------+--------------+------------+
| 9 4 37 | 357 358 678 | 2 136 16 |
| 27 1 6 | 2379 239 4 | 5 8 37 |
| 8 237 5 | 237 1 67 | 367 4 9 |
+------------+--------------+------------+
| 6 38 389 | 4 7 5 | 1 39 2 |
| 1 5 39 | 39 6 2 | 4 7 8 |
| 27 27 4 | 8 39 1 | 36 356 56 |
+------------+--------------+------------+
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Jun 20, 2007 9:31 pm Post subject: Re: MAY 29 VH |
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| CORUJA wrote: | Earl,
I'm still at this point, and can't see how to get rid of the 2 in R1C3 , in R1C5 and in R1C8.
Thanks again.
| Code: |
+------------+--------------+------------+
| 4 6 278 | 257 258 3 | 9 125 15 |
| 3 9 278 | 1 258 78 | 678 256 4 |
| 5 278 1 | 6 4 9 | 378 23 37 |
+------------+--------------+------------+
| 9 4 37 | 357 358 678 | 2 136 16 |
| 27 1 6 | 2379 239 4 | 5 8 37 |
| 8 237 5 | 237 1 67 | 367 4 9 |
+------------+--------------+------------+
| 6 38 389 | 4 7 5 | 1 39 2 |
| 1 5 39 | 39 6 2 | 4 7 8 |
| 27 27 4 | 8 39 1 | 36 356 56 |
+------------+--------------+------------+
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I don't know if this specifically answers the question, but there is a basic move that needs to be made. In column 3 the only place for a 2 is in box 1. Thus the 2 must be removed from r3c2. This leaves a 378 triple (or a sole 2) in row 3 which sets off a chain of additional moves. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Wed Jun 20, 2007 9:36 pm Post subject: May 29 |
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Coruja,
Now I recall.
First the 9 in r9c5 is unique which leads to a solution of all 9's.
Then the 2's in r1c3, r2c3 are unique to c3, eliminating the 2 from r3c2.
That makes the 2 in r3c8 unique and opens up the puzzle.
You can eliminate the 3 in r4c5 by the xy wing r4c3, r5c1 r5c5.
Rest is straight forward.
Earl |
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CORUJA
Joined: 16 Jun 2007
Posts: 15
Location: BRUMADINHO - MG; BRAZIL
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| Posted: Thu Jun 21, 2007 3:44 pm Post subject: 29 May, VH |
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Earl and Marty,
thank you sooooo much!! I really overlooked the basic move that eliminates the 2 from R3C2.
Coruja.
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CORUJA
Joined: 16 Jun 2007
Posts: 15
Location: BRUMADINHO - MG; BRAZIL
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| Posted: Thu Jun 21, 2007 3:48 pm Post subject: 29 MAY, VH |
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Olá dulaby,
muito obrigada pela sua sugestão, que utilizei após eliminar o 2 da R3C2 com a ajuda do Earl e do Marty.
Coruja.
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