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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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 Posted: Wed Nov 28, 2007 9:32 am Post subject: Nov 28 VH |
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Not too bad, really - a single xy-wing on 6-8-9 solved it for me.
There've been a lot of this sort of solution recently - not that I'm complaining, 'cos I can generally solve them! 
John |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Wed Nov 28, 2007 12:04 pm Post subject: |
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Lots of one step solutions in this VH.
The xy-wing mentioned by John is one of them, here is another one step,
I'm sure this is not the only one.
ER <6> in Box 7, removing <6> in R8C5.
* is the strong link on <6> in R1
| Code: |
+--------------------------+--------------------------+-------------------------+
| 2 7 *68 | 4 *36 38 | 9 1 5 |
| 89 469 3 | 5 1 29 | 24 7 468 |
| 1 5 4689 | 68 7 29 | 234 46 3468 |
+--------------------------+--------------------------+-------------------------+
| 7 49 5 | 2 49 6 | 8 3 1 |
| 6 24 1 | 38 35 358 | 7 24 9 |
| 3 8 249 | 1 49 7 | 5 246 46 |
+--------------------------+--------------------------+-------------------------+
| 5 er 1 er 69 | 367 2 345 | 34 8 347 |
| 58 26 268 | 367 35-[6] 345 | 1 9 347 |
| 4 er 3 er 7 | 9 8 1 | 6 5 2 |
+--------------------------+--------------------------+-------------------------+
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Nov 28, 2007 1:32 pm Post subject: |
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| Quote: | | ER <6> in Box 7, removing <6> in R8C5. |
That's interesting, I used the same ER in Box 7 but with the 6's in C5 to remove the <6> in R1C3 |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Wed Nov 28, 2007 2:06 pm Post subject: |
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| ...... Also there is another ER in Box 6 which using 4's in C3 removes <4> in R3C8 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Wed Nov 28, 2007 3:07 pm Post subject: |
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I found another one-step solution: a finned x-wing that removes <6> at R3C34.
Ted |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Wed Nov 28, 2007 3:25 pm Post subject: |
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| An xyz-wing with pivot at r8c3 eliminates 6 from r7c3 solving the puzzle. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Nov 28, 2007 4:43 pm Post subject: |
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| I don't understand Johan's grid with that cell apparently solved for 5 in column 1 without the eliminations therefrom. At any rate, I had a different grid and a W-Wing on 46 in boxes 1 and 3 was my one-step solution. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Wed Nov 28, 2007 5:56 pm Post subject: |
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Marty,
You're right about this one, i've made a type error R7C1 must be [59]
instead of a naked <5> in the grid, thanks for spotting this.
The grid without the type error :
| Code: |
+--------------------------+--------------------------+-------------------------+
| 2 7 *68 | 4 *36 38 | 9 1 5 |
| 89 469 3 | 5 1 29 | 24 7 468 |
| 1 5 4689 | 68 7 29 | 234 46 3468 |
+--------------------------+--------------------------+-------------------------+
| 7 49 5 | 2 49 6 | 8 3 1 |
| 6 24 1 | 38 35 358 | 7 24 9 |
| 3 8 249 | 1 49 7 | 5 246 46 |
+--------------------------+--------------------------+-------------------------+
| 59er 1 er 69 | 367 2 345 | 34 8 347 |
| 58 26 268 | 367 35-[6] 345 | 1 9 347 |
| 4 er 3 er 7 | 9 8 1 | 6 5 2 |
+--------------------------+--------------------------+-------------------------+ |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Wed Nov 28, 2007 8:13 pm Post subject: |
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| tlanglet wrote: | I found another one-step solution: a finned x-wing that removes <6> at R3C34.
Ted |
I like it. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Wed Nov 28, 2007 10:24 pm Post subject: |
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| Code: | +--------------------------+--------------------------+-------------------------+
| 2 7 *68 | 4 *36 38 | 9 1 5 |
| 89 469 3 | 5 1 29 | 24 7 468 |
| 1 5 4689 | 68 7 29 | 234 46 3468 |
+--------------------------+--------------------------+-------------------------+
| 7 49 5 | 2 49 6 | 8 3 1 |
| 6 24 1 | 38 35 358 | 7 24 9 |
| 3 8 249 | 1 49 7 | 5 246 46 |
+--------------------------+--------------------------+-------------------------+
| 59er 1 er 69 | 367 2 345 | 34 8 347 |
| 58 26 268 | 367 35-[6] 345 | 1 9 347 |
| 4 er 3 er 7 | 9 8 1 | 6 5 2 |
+--------------------------+--------------------------+-------------------------+ |
Just for more fun, the W-Wing in boxes 1 and 2 removes the 8 from r1c6 which I believe solves the puzzle. |
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mckeann
Joined: 29 Nov 2007
Posts: 4
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| Posted: Thu Nov 29, 2007 3:11 am Post subject: |
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Hi folks, first time poster so be gentle.
I have solved upto the grid posted by Johan. I have also read all the techniques forum in a bid to try and understand it but i cant.
Could someone please explain to an absolute numpty (please dont use technical terms, my head is buzzing  ) how to get the next number.
thanks
Neil |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 29, 2007 4:59 am Post subject: |
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| mckeann wrote: | Hi folks, first time poster so be gentle.
I have solved upto the grid posted by Johan. I have also read all the techniques forum in a bid to try and understand it but i cant.
Could someone please explain to an absolute numpty (please dont use technical terms, my head is buzzing ) how to get the next number.
thanks
Neil |
Hi Neil,
Welcome to the forum. Of the techniques discussed in this thread, probably the most well known and commonly used is the XY-Wing. In this case, 68-69-89, pivoted in r1c3. If you're not familiar with the technique, just google it (I'm pressed for time at the moment) and you'll find plenty to read. In the meantime, don't hesitate to ask all the questions you want, there are lots of helpful people here.
Good luck. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Nov 29, 2007 5:37 am Post subject: |
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Neil,
this is probably the perfect puzzle to ask for help.
all I want to do is coach your eyes on finding a very nice xy-wing example in this puzzle.
if you are looking at johan's example, concentrate on row 7. there are only two places in that row that a 9 can be placed. this is crucial because those two cells happen to have two candidates. cells with two candidates are commonly called bi-value cells. this is a good starting place.
r7c1={5,9} r7c3={6,9}
this is key: there really isn't any trick. keep in mind the three candidates you have in these cells: 5, 6, and 9.
with this in mind. let your eyes scan the same two columns 1 and 3 for cells containing these same candidates and, very crucial, are also bi-value.
you will find: r2c1 {8,9} and r1c3 {6,8}
now you have four cells.
this is where the beauty of sudoku comes in. you already know that row 7 can only get one 9. so lets plug 9 into r7c1 and lets see what happens:
r7c1 is a 9
that means r2c1 will be an 8
and that makes r1c3 a 6
and that in turn makes r7c3 a 9 .... oh !!! what's this, a contradiction.
now we know that r7c1 can't be a 9. but there is more. now that we know that r7c3 is a nine,
this eliminates the 9 in r3c3.
in these very hards. its not so much finding the next number, but more about eliminating that one last candidate to make it fall apart. |
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andras
Joined: 31 Oct 2007
Posts: 56
Location: Mid Wales
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| Posted: Thu Nov 29, 2007 4:42 pm Post subject: |
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That's a nice explanation, Norm.
Now, if some kind soul here can explain ERs in a way I can understand I'll feel that my ignorance is a little less exposed. I sort of get the idea, but I'm not at all sure that I really grasp the principle behind them, and it's quite beyond me to spot them.
John |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 29, 2007 5:08 pm Post subject: |
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| andras wrote: | That's a nice explanation, Norm.
Now, if some kind soul here can explain ERs in a way I can understand I'll feel that my ignorance is a little less exposed. I sort of get the idea, but I'm not at all sure that I really grasp the principle behind them, and it's quite beyond me to spot them.
John |
John, this is the explanation that allowed me to start using them. Click on "Solving", then just scroll down a little and click on "Empty Rectangles."
http://www.intosudoku.com/ |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Nov 29, 2007 6:14 pm Post subject: |
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My personal experience is that they're only confusing and slow to absorb if you have a confusing explanation. I didn't use an ER for a year and one-half because the explanation I had said it started with the empty rectangle within a box and that was difficult for me to spot and grasp.
Then I stumbled across the explanation that said it started in a box where a number was present in only two lines, and based on that explanation, ERs became very simple and mechanical. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Nov 29, 2007 8:59 pm Post subject: |
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| Quote: | | ERs are, admittedly, a bit confusing when first encountered. It takes a while to absorb the concept. But, once you get your head around them, you'll never be the same |
I love this quote.
ER's are probably the ultimate in candidate searches, much like x-wings or swordfish.
talk about never being the same !! did everyone else's head explode when you reasoned out a skyscraper?? just the name "skyscraper" is enough to make you look for more and more. I agree with marty and Asellus that ER's aren't easy, but all the same ER's are definitely the next big step in solving.
my question for nataraj and Asellus is when I start a puzzle which I know has more than lets say 3 or 4 xy-wings and an x-wing and the UR thrown in, is it better to start looking for an ER toward the beginning of the solving process or after I got some candidates out of the way?? because Asellus has pointed me towards ER's early in some grids and in the middle in others.
Last edited by storm_norm on Thu Nov 29, 2007 9:03 pm; edited 1 time in total |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Nov 29, 2007 9:55 pm Post subject: |
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| storm_norm wrote: |
my question for nataraj and Asellus is when I start a puzzle [...] is it better to start looking for an ER toward the beginning of the solving process or after I got some candidates out of the way?? |
Aside from the more or less obvious answer that I start the puzzle by squeezing (look for naked or hidden singles), naked pairs, box/line, the usual stuff, ... 
... as a general rule, the fewer candidates left the better. That said, those methods that work on single numbers tend to be more applicable earlier in the process because they remove candidates even from rather crowded (in terms of PMs) cells. Therefore ...
... for me the timing is quite simple. When looking for patterns in single numbers ("1"s, "2"s etc.), i.e. looking for xwings, skyscrapers, kites, fish (you name it) I make drawings of all the positions of let's say number "2" on the grid, plus lines between two candidates when there is a strong link. That is when I spot ERs (or the occasional finned creature - or nothing :boo:)
URs, xy-wings, x-wings, w-wings, xy-chains all reside in a different part of the brain (they all have to do with bi-value cells) - I look for them at a different time, sometimes before (because I hope for a "quick win"), sometimes after the coloring / fishy stuff |
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mckeann
Joined: 29 Nov 2007
Posts: 4
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| Posted: Fri Nov 30, 2007 1:09 am Post subject: |
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| storm_norm wrote: | Neil,
this is probably the perfect puzzle to ask for help.
all I want to do is coach your eyes on finding a very nice xy-wing example in this puzzle.
if you are looking at johan's example, concentrate on row 7. there are only two places in that row that a 9 can be placed. this is crucial because those two cells happen to have two candidates. cells with two candidates are commonly called bi-value cells. this is a good starting place.
r7c1={5,9} r7c3={6,9}
this is key: there really isn't any trick. keep in mind the three candidates you have in these cells: 5, 6, and 9.
with this in mind. let your eyes scan the same two columns 1 and 3 for cells containing these same candidates and, very crucial, are also bi-value.
you will find: r2c1 {8,9} and r1c3 {6,8}
now you have four cells.
this is where the beauty of sudoku comes in. you already know that row 7 can only get one 9. so lets plug 9 into r7c1 and lets see what happens:
r7c1 is a 9
that means r2c1 will be an 8
and that makes r1c3 a 6
and that in turn makes r7c3 a 9 .... oh !!! what's this, a contradiction.
now we know that r7c1 can't be a 9. but there is more. now that we know that r7c3 is a nine,
this eliminates the 9 in r3c3.
in these very hards. its not so much finding the next number, but more about eliminating that one last candidate to make it fall apart. |
THANK YOU. thats absolutely brilliant, i think i finally get it. I'm going to go back through the archives and try a few to get my head round it.
Thanks again |
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