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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Thu May 04, 2017 10:35 pm Post subject: May 5 VH |
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Code: |
+-------------+---------------+------------+
| 9 4 2 | 3 8 7 | 15 6 15 |
| 13 7 136 | 15 9 5-6 | 4 8 2 |
| 5 8 16 | 12 26 4 | 9 7 3 |
+-------------+---------------+------------+
| 128 29 189 | 4 123-6 1236 | 7 5 1689 |
| 4 3 1789 | 79 5 16 | 18 2 1689 |
| 127 6 5 | 79 12* 8 | 3 4 19 |
+-------------+---------------+------------+
| 238 5 4 | 6 123 123 | 28 9 7 |
| 6 29 389 | 258 7 235 | 258 1 4 |
| 278 1 78 | 258 4 9 | 6 3 58 |
+-------------+---------------+------------+
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XY-Wing 12-6 with pivot in r6c5; r2c6, r4c5 <> 6; stte |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Fri May 05, 2017 12:06 am Post subject: |
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Hi everyone
If R3C5=1-->R5C6=Ø-->R3C5<1>R5C6=1 and R6C5=1 impossible-->R3C5<>2 solution basic strategies
Ciao Paolo
Last edited by Ajò Dimonios on Fri May 05, 2017 12:39 am; edited 1 time in total |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Fri May 05, 2017 12:32 am Post subject: |
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Hi everyone
Sorry I correct my previous post
If R3C5=1 --> R5C6=Ø-->R3C5≠1
If R3C5=2--> R5C6=1 and R6C5=1 impossible-->R3C5≠ 2 solution with basic strategies.
Ciao Paolo |
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ZeroAssoluto
Joined: 05 Feb 2017 Posts: 942 Location: Rimini, Italy
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Posted: Fri May 05, 2017 7:50 am Post subject: |
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Hi everyone,
Another possible solution
Quote: | XY-Chain r2c4 1,5 - r2c6 5,6 - r5c6 6,1 - r6c5 1,2 - r3c5 2,6 - r3c3 6,1 and -1 in r2c13 and r3c4. |
Ciao Gianni |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri May 05, 2017 5:31 pm Post subject: |
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Ajò Dimonios wrote: | Hi everyone
Sorry I correct my previous post
If R3C5=1 --> R5C6=Ø-->R3C5≠1
If R3C5=2--> R5C6=1 and R6C5=1 impossible-->R3C5≠ 2 solution with basic strategies.
Ciao Paolo |
I could use an explanation. Agree, r3c5=2=>r5c6=1; but I don't understand the next term, and r6c5=1 |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Fri May 05, 2017 9:47 pm Post subject: |
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Dear Marty R.
Sorry for my English. Try to explain.
When in a box there are two certain digits in two cells, the sudoku is invalid. In this situation, if you make sure that the number in cell R3C6 is 2 it is obtained as an immediate consequence that the cell r5C6 and cell R6c5 are at the same time the number 1. This makes the sudoku invalid because these cells are in box 5. For this Why the initial hypothesis is false and the alternative hypothesis that number 2 should not belong to cell R3C5 is valid. Then it is legitimate to delete the number 2 in that cell. The same reasoning can be applied for value 1 in R3C5. In this case, the error that is created is due to the fact that the cell R5C6 becomes empty (no number can be entered).
Ciao Paolo |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat May 06, 2017 1:54 am Post subject: |
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Quote: | same reasoning can be applied for value 1 in R3C5. |
This is what has been confusing me because there is no 1 in r3c5. And I am just unable to see how two 1s got into box 5. It may be my problem in not understanding. How did you arrive at to 1s in box 5? |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Sat May 06, 2017 7:00 am Post subject: |
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At the beginning, before Clement's situation, just after the basic technique the puzzle is in this situation. There is 1 in R3C5.
Code: |
+-------------+---------------+------------+
| 9 4 2 | 3 8 7 | 15 6 15 |
| 13 7 136 | 15 9 15-6 | 4 8 2 |
| 5 8 16 | 12 126 4 | 9 7 3 |
+-------------+---------------+------------+
| 128 29 189 | 4 123-6 1236 | 7 5 1689 |
| 4 3 1789 | 179 5 16 | 18 2 1689 |
| 127 6 5 | 179 12* 8 | 3 4 19 |
+-------------+---------------+------------+
| 238 5 4 | 6 123 123 | 28 9 7 |
| 6 29 389 | 258 7 235 | 258 1 4 |
| 278 1 78 | 258 4 9 | 6 3 58 |
+-------------+---------------+------------+ |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Sat May 06, 2017 7:19 am Post subject: |
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R3C5 =2 deletes 2 in R6C5 and 6 in R5C6 for locked candidates (there is only one 6 in box 2 in R2C6). For this reason there are two 1 in box 5 |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat May 06, 2017 7:19 pm Post subject: |
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Ajò Dimonios wrote: | At the beginning, before Clement's situation, just after the basic technique the puzzle is in this situation. There is 1 in R3C5.
Code: |
+-------------+---------------+------------+
| 9 4 2 | 3 8 7 | 15 6 15 |
| 13 7 136 | 15 9 15-6 | 4 8 2 |
| 5 8 16 | 12 126 4 | 9 7 3 |
+-------------+---------------+------------+
| 128 29 189 | 4 123-6 1236 | 7 5 1689 |
| 4 3 1789 | 179 5 16 | 18 2 1689 |
| 127 6 5 | 179 12* 8 | 3 4 19 |
+-------------+---------------+------------+
| 238 5 4 | 6 123 123 | 28 9 7 |
| 6 29 389 | 258 7 235 | 258 1 4 |
| 278 1 78 | 258 4 9 | 6 3 58 |
+-------------+---------------+------------+ |
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Hello, I'm not sure of your name; is it Ajo or Paolo?
Quote: | just after the basic technique the puzzle is in this situation. There is 1 in R3C5. |
I'm sorry, but that is not correct. This is the grid after basics, same as what Clement shows. Unfortunately, I can't see what the specific problem is with your grid, but I hope someone sharper than me can point it out.
Code: |
+-------------+---------------+------------+
| 9 4 2 | 3 8 7 | 15 6 15 |
| 13 7 136 | 15 9 56 | 4 8 2 |
| 5 8 16 | 12 26 4 | 9 7 3 |
+-------------+---------------+------------+
| 128 29 189 | 4 1236 1236 | 7 5 1689 |
| 4 3 1789 | 79 5 16 | 18 2 1689 |
| 127 6 5 | 79 12 8 | 3 4 19 |
+-------------+---------------+------------+
| 238 5 4 | 6 123 123 | 28 9 7 |
| 6 29 389 | 258 7 235 | 258 1 4 |
| 278 1 78 | 258 4 9 | 6 3 58 |
+-------------+---------------+------------+
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Play this puzzle online at the Daily Sudoku site |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Sat May 06, 2017 9:00 pm Post subject: |
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I'm sorry if I insist but the problem is just a different definition of basic strategies that takes nothing away from the resolution of the problem. I, as I have already written in a previous post, define basic strategies only nacked single, hidden single and locked candidates.For delete 1 in R3C5 you have to use Hidden pair (7,9) in R5C4 and R6C6 which first removes 1 in R5C4 and 1 and 2 in R6C6. Only at this point can you delete 1 in R3C5. (Locked candidates). However you see the problem I can delete the 1 in R3C5 or as I just described or with the Nisho I described in the first post. The important thing is to describe a method of solution. For the rest In Italy we say that it is a question of goat wool or the sex of angels (lana caprina o sesso degli angeli).
My name is Paolo |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat May 06, 2017 11:46 pm Post subject: |
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Paolo,
I think we ought to discontinue this conversation since it seems to be going nowhere. If you defined your basics in your first post, that's too much for this old man to remember. On both Sudoku forums, it is understood that basics consist of subsets (pairs, triples, quads) and locked candidates. Everything else is considered advanced. If we have different grids, both claiming to be after basics, things are going to be confusing, as they are here. Maybe this is a case of sex of angels (I didn't know that angels had sex).
Don't get me wrong, I'm delighted to have you on the forum. But if we talk with two voices, expect more confusion ahead.
Cheers,
Marty |
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rmireland
Joined: 21 Sep 2013 Posts: 33 Location: New Orleans
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Posted: Sun May 07, 2017 1:50 am Post subject: |
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Hi Marty and all,
I think all Paolo is saying, regardless of whether his basics are complete, is that a 2 in r3c5 leads to a contradiction, in that r6c5 (in box 5) must be a 1, while at the same time the 2 in r3c5, via the required 6 in r2c6, makes r5c6 (also in box 5) also a 1. These two ones in box 5 can't be, so r3c5 cannot be 2.
This contradiction exists regardless of whether there is a 1 candidate in r3c5, so whether or not the basics are complete is irrelevant.
I don't know if this is helpful, so if it's not, please disregard.
Rick
PS I have nothing to contribute about goat sex or angel wool. |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Sun May 07, 2017 5:42 am Post subject: |
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Hi everyone
I agree with Rick. I do not think a logical discourse creates confusion. On the contrary, in the presence of logical reasoning increases clarity.
Ciao Paolo |
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Ajò Dimonios
Joined: 01 May 2017 Posts: 339 Location: Sassari Italy
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Posted: Sun May 07, 2017 2:32 pm Post subject: |
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A discussion on the sex of angels “discussione sul sesso degli angeli “ is the metaphor that perhaps in English is described as “how many angels can dance on the head of a pin”.
A question of goat wool “una questione di lana caprina” is a Latin-speaking way that was used by the poet Orazio in the "Epistles" (I, XVIII, 15) on arguments not very important similar to the value in that historical period of goat's wool that had not no commercial value. Maybe in English you can translate “To split hairs”. |
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