Help with my first very hard please.

Advanced methods and approaches for solving Sudoku puzzles

Postby Animator » Fri Jun 03, 2005 2:50 pm

Actually Andrew, it was your reply that confused me...

First of all, there are two distinct groups: 4, 6, 7 and 2, 3, 8.

The group 4, 5, 6, 7 is incorrect.

Why? there are three cells that share the values 4, 6 and 7 (r1c3, r8c3, r9c3). This allows you to remove 4, 6 and 7 from r7c3 leaving only the number 5 as a candidate.

To Goldie, you should be looking for two things, just as you should for finding pairs...

First is a group of cells, which share the same candidates and where there are exactly the same number as cells as the number of candidates. In this case that would be the 4, 6, 7 group. In these cells you may not have any other candidates.

The other group of cells you should look for, is a group of cells which are the only possiblities for certain numbers. Again, the number of cells has to match the number of candidates. In this puzzle this would be the 2, 3, 8 group. These cells can have more candidates, but those are irrelevant, since these cells need to have the numbers 2, 3 8 (in some order).


Here is another example with only two numbers:
{1}, {2,3,4}, {3,4}, {3,4}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}

In this example there are cleary two cells that can only have the numbers 3 and 4. Therefor you eliminate them from the other cells, right? (or haven't you applied that technique yet?) (this would be the same as the first group I explained)

Another example:
{1}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7,8,9}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}, {2,3,4,5,6,7}

In this second example there are only two cells that can have the numbers 8 and 9, but they have other candidates too... but because you know that these are the only two valid cells you know you can remove the other candidates from the cell. (this is the same as the second group)

Does this make sense to you?
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Postby abailes » Fri Jun 03, 2005 2:57 pm

Absolutely on the right lines, however, there are two levels to it.

The first level, as you noted (and as applies to the 2, 3, 8 group) is to eliminate other candidates alongside the pattern candidates within the same cells.

The second, as applies to the 4,5,6,7 group is to note that none of those numbers can appear in any other cell in that particular row/column/box. This is following exactly the same logic as the above - as you state 'the number of candidates equals the number of different cells', so the candidate numbers are unable to appear anywhere else.

One last point:
You could use this process identify a single cell that can only hold a single candidate as you note. However, as in the problem above, sometimes you are using this process so that a particular candidate only has one option left in a row/column/box. Furthermore, other (harder) problems eliminate possibilities, so that you can then eliminate further possibilities in an intersecting second row/column/box (e.g. eliminating some numbers from one cell allows you to eliminate numbers from another cell before you actually get to put in another 'hard' number).

Nobody ever said Sudoku was easy!
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Postby abailes » Fri Jun 03, 2005 3:05 pm

Animator,

Sorry I confused you, there are indeed two ways of looking at the same problem. I guess it is one of these things that gets fixed in your head once you have seen it one way.

If you have a group of cells for column three that is:

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

Then you can either see the 4,6,7 group of three (cells one, eight and nine) or the 4,5,6,7 group of four (cells one, seven, eight and nine). The same rules apply to either option.

The first allows you to eliminate the 467 from cell seven giving five as the only option for that cell (amongst other eliminations).

The second allows you to eliminate 456 from cell four (also amongst other eliminations), leaving the five in cell seven as the only option left in the column.

Cheers


Andrew
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Postby goldie5218 » Fri Jun 03, 2005 4:43 pm

oh sorry guys, and further to my previous post, once i have finished the easy part of identifying all the candidates, and have inserted all the big numbers identified during this phase - i then start the real tricky bit of this seeking of patterns ??? and i assume this seeking extends to columns/rows/blocks--- must it be tackled in a fixed order? ie columns/rows/blocks or do i proceed as to wherever i find a pattern(s)?
and the most important parts of my query are --- do i start with the highest/largest pattern first, irrespective of where i find it ? ---and do i revise/amend the entire puzzle as soon as i am able to insert a big number as a result of my reducing of the candidates to allow this? AND this real mckoy is how do i find the darn patterns in the first place???
seriously though i get the theory now (i think) and now need to try the practice! again i have a theory it may be a 'boys thing' that you guys can just see these patterns and we have to try and use intuition? cheers
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Postby paulf2127 » Fri Jun 03, 2005 5:59 pm

Goldie,

Definitely not a boys think although I suspect most would argue that the logic of the ladies is a little different to ours!

My best trick is to look at the rows and columns of a 3 by 3 box. If there are say for example 7's only in the middle row, then you can eliminate all the sevens elsewhere in the larger row.

That's not very well explained.

if for example I had the following "box 1";

752
138
xxx

say the candidates for the bottom row are [4, 6], [4,9], [6,9]

Thus the 9 can only go in box 1, row 3. This means I can eliminate any 9 candidates from the rest of row 3.

This is really simplistic and I apologise if you already know this. sometimes though even with a completely empty box, you may only have say, 7's in the 1 st column.

This means you can remove the 7's from the rest of column 1.

Any help?

PaulF
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Postby abailes » Sat Jun 04, 2005 6:58 am

My advice on how to sort your groups would be to see the effects of any and all you can find. As soon as you find one put the effects of it into the puzzle.

If you cannot find any then I would start looking logically, each row, column and block in turn. The beauty of Sudoku puzzles is that none of row/column/block is more important than another. It is just as likely that you will find a group in a block as a row as a column.

Scan first and then look in a logical manner if (when) that doesn't work.
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Postby goldie5218 » Sat Jun 04, 2005 8:55 am

ok guys, first let me put on record my very very very heartfelt thanks to all who offered help and particularly examples to try and help me see the wood from the trees so to speak - am really appreciative of your time and trouble! secondly, i can see that i need to back off from the really hard puzzles until ive had some time to study the given examples and see if i can recognise the "pattern groups" on my own. LOL! im still convinced its a 'boys thing' to enable you guys to spot these darn things but i am gonna give it a go again! many thanks and cheers goldie
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Postby Roz » Sat Jun 04, 2005 2:13 pm

goldie5218 wrote:LOL! im still convinced its a 'boys thing' to enable you guys to spot these darn things but i am gonna give it a go again! many thanks and cheers goldie


It isn't just a boys thing at all Goldie, don't bring our gender into disrepute lol:( .
You do need to work through the levels though, before trying a very hard. I've only recently completed my first at that level ..........feeling a tad smug at the moment ;-) ...........

Onward and upward's the way to go Goldie. You'll do it!:D
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And this is where I'm having a problem as well

Postby Guest » Sun Jun 05, 2005 5:40 am

I'm having the same problem as Goldie, but Animator's example picks up what I think is my 'mental block'.

abailes wrote:If you have a group of cells for column three that is:

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

Then you can either see the 4,6,7 group of three (cells one, eight and nine) or the 4,5,6,7 group of four (cells one, seven, eight and nine). The same rules apply to either option.


Lets take the 4,6,7 group in cells 1, 8 and 9. Given that I'm looking for (say) 3 values and therefore looking for 3 cells, cells 8 and 9 look good, but what is the reasoning for me to then add in cell 1? Doesn't it need to have the same 3 values?

Alternativly, why should I not pick cell 3 on the basis that it has the 4 and the 6, just like cell 7?

Cheers,

Don
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Postby abailes » Sun Jun 05, 2005 7:09 am

Don,

I should probably confess - I have a Maths degree from Cambridge, so I pretty much see these sort of things straight away. It is interesting to me that my previous posts are not sufficiently clear, there must be something wrong with my explanations. It would be nice to be able to write a definitive post explaining groups.

I think that part of the problem may be that Animator and I have been talking about two different logical concepts as if they are one. This only really dawned on me reading your post (this is why I like going through things as thoroughly as possible). Let me explain given the example here:

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

Logical way One. One 'way' of finding a group, using the 4,6,7 group in cells one, eight and nine is to find (say) three cells with only three possible values. It does not matter that the first of these three cells only has two of those possible three options. The fact remains that the only possible options for cells one, eight and nine is some combination of 4, 6 and 7 - excluding those combinations where 6 is in cell one (in cell order, the only possible combinations here are 467, 476, 746 and 764). No other number can ever be placed in those three cells because all others have already been eliminated. Furthermore (and this is the key bit) the numbers 4, 6 and 7 cannot go anywhere else in that column either - if they did then there would not be 'enough' of them to fill up the three cells in question. We can therefore eliminate them from the possibilities in the other cells in the column (this is as discussed previously and has the immediate effect of being able to put a 5 in cell seven). [Extra note - In this option, it does not matter that 4,6 and 7 can be seen in other cells, we know they must go in cells one, eight and nine in order that we can eventually fill one, eight and nine. In fact, if they were not present in any other cells, there would be no other options to eliminate, so finding the group would not actually be helpful.]

Logical way Two. The second 'way' of finding groups is to find (say) three numbers that are only present in three cells. This is subtely different from before. Before we were looking for three cells that only contained (some combination of) three numbers. Now we are looking for three numbers that - in some combination - only appear in three cells. This happens with the numbers 2,3 and 8 in cells three, four and five. Examination of the list of groups shows you that none of the numbers 2, 3 or 8 appears anywhere else. Given that we have to fit each of the nine numbers in these nine cells somehow, we can therefore logically state that the cells three, four and five can only contain some combination of 2, 3, or 8. We have already found that 2, 3 and 8 do not appear anywhere else in the column and if any one of those three cells contained a different number, there would not be enough 'places' to put 2, 3, and 8 in this column. It does not matter that cell three does not contain an 8 - again this merely restricts our combinations further. The combinations are (in cell order) 238, 283, 328 and 382. We can therefore eliminate any other possibilities from these three cells. Cell three becomes (23), cell four (238) and cell five (238). Again this immediately gives us a 5 in cell seven as this is then the only place left in the column that a 5 can go. [Extra note. As a corollary to my extra note in the previous option, in this option it does not matter that cells three four and five contain additional options. The fact remains that 2, 3 and 8 must go into those three cells in some combination. If there were no other options in these three cells, there would be no other numbers to eliminate and this group would not help us]

In answer to your question from your post, there is no 'group' that is helpful by looking at the numbers 4, 6 and 7 and choosing cells eight, nine and three together. That selection then becomes a strange mixture of the two 'ways' I have described above. It does not fit into either Logical way One (which is looking for three cells that only contain three numbers - our selection of cells eight, nine and three contains 2, 3, and 5 as well as 4, 6, 7), nor does it fit into Logical way Two (which is looking for three numbers that only appear in three cells - the numbers 4, 6 and 7 also appear in cells one, four, five and seven in this column).

Please ignore this last paragraph if it confuses you, it is not important, but just a matter of interest:
I'm thinking that in any combination of cells that contains a group of logical way One, there must also be another combination of cells that contains logical way Two and visa versa. If I had written the explanation for logical way One above based on the four-number group of 4,5,6,7 (cells one, seven, eight and nine as I had stuck in my head before Animator pointed out the simpler 4,6,7 group) then the result would have been exactly the same as the 2,3,8 group. The result of both groups would be to eliminate the options 4,5,6 from cell three, 7 from cell four and 6,7 from cell five. Furthermore, by combining the numbers of the two groups {4,5,6,7 and 2,3,8} you are left with all the possibilities for numbers in this column that are not fixed (i.e. everything except 1 and 9 which are already fixed in cells two and six). What this implies is that you do not need to use both logical way One and logical way Two. If you get one stuck in your head and become a master at spotting it in the puzzles, you do not need to worry about the other one at all. I would be interested in thoughts on this. I'm not sure of its applicability to columns with three groups in, for example.

Do let me know if there is more I can say...

Cheers,


Andrew
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Re: And this is where I'm having a problem as well

Postby Roz » Sun Jun 05, 2005 6:26 pm

Gennadog wrote:I'm having the same problem as Goldie, but Animator's example picks up what I think is my 'mental block'.

abailes wrote:If you have a group of cells for column three that is:

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

Then you can either see the 4,6,7 group of three (cells one, eight and nine) or the 4,5,6,7 group of four (cells one, seven, eight and nine). The same rules apply to either option.


Lets take the 4,6,7 group in cells 1, 8 and 9. Given that I'm looking for (say) 3 values and therefore looking for 3 cells, cells 8 and 9 look good, but what is the reasoning for me to then add in cell 1?

Don



I don't have a maths degree, and am a visual learner, so for others who might be the same this might help..................

(4,7) (1) (23456) (2378) (23678) (9) (4567) (467) (467)

The cells coloured blue cannot have any other number in except one of the three in question, whereas cell seven (red), could possibly be hold the number five.
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Postby goldie5218 » Mon Jun 06, 2005 10:13 am

hi guys, many thanks to abailes for the clear explanation re the two 'types' of logics used to explain the spotting of the groups - it makes sense to me at last - also thanks to roz for the clever use of colour in his explanation - us learners are lucky to have such help - thanks a ton
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Postby Roz » Mon Jun 06, 2005 3:04 pm

goldie5218 wrote:- also thanks to roz for the clever use of colour in his explanation - us learners are lucky to have such help - thanks a ton


Her, not his Goldie please lol ;-):) ......................Wish we had a wink emoticon on here:( .
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Postby clara5218 » Mon Jun 06, 2005 5:39 pm

hi forum people,am new to sudoku but am learning fast. v easy ok easy almost ok but harder puzzles leave me with a bunch of candidates in most cells and there i sit! obviously there must be a way forward but the logic i dont get! herewith three rows from a recent puzzle

13569 - 4 - 1679 - 5678 - 578 -567 - 2 - 36789 - 13789 (row one)

1569 - 156789 - 2 - 378 - 789 - 179 - 4 - 3678 - 1378 (row five)

23469 - 23689 - 5 - 467 - 479 - 4679 - 2378 - 1 - 234789 (row seven)

balance of rows somewhat similar but my plea is how do i make some sense out of all these candidates? i know i have to eliminate candidates to end up with a single candidate to a particular cell and then use that 'hard ' number to eliminate further candidates in the block/row/column but its how to do this first step that has me stymied - any kind person willing to help a raw rookie plse ?
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Postby paulf2127 » Mon Jun 06, 2005 5:42 pm

I think you'd need to cross match with the columns. Post your grid as you have it. don't worry about the pencil marks.
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