The New Sudoku Players' Forum

[Leren]

koushanejad74 wrote : LO-ABC => B and F Can take 2,3

The way I see it, in the most general case, I can't see, a priori, why F can't be 1 or 4 in the LO case.

If BDEF is LO its possible range is 12345. If ABC is LO it's 123, so DE = 45, a naked pair, so F obviously can't be 4 and it can't be 1 (out of range of the naked pair), but I see these last 2 deductions as follow on moves, specific to this situation, rather than as part of the general move. Remember, at this stage we haven't considered the HI case so we don't know what DE can be in totality.

As it turns out DE are both 456 at the conclusion of the move. My solver then says that BDEF can't contain 1, since 1 of DE must be at least 5 (whether BDEF turns out to be LO or HI), and then B can't be 2 (blocked by singleton 3 in BDEF if its LO or out of range if its HI) but I see these as follow on standard Str8ts moves, rather than as part of the main move.

So I see the main move as follows :

1. Identify a compartment that is contained in a single box (it will be of size 1, 2 or 3 but at this stage I can't see 1 working but it might). We know that there will be a grey cell in its row or column.

2. Check that the compartment has no clues or solved cells ( at this stage I don't want to deal with degenerate cases - these will almost certainly be handled by standard Str8ts moves).

3. Identify a cross compartment with a common cell (its B in this case) and which also has a grey cell in its column or row and the size of both compartments is < 8 (I'll now include that in the logic as per your post - another clever observation).

4. Consider the common LO and HI cases for the compartments and identify a list of different digits that can be in the second compartment, not in the first compartment, but in the same box as the first compartment.

5. Hopefully this will lead to some eliminations in these common box cells. Any other eliminations in these compartments can hopefully be covered by other moves.

BTW I'm fine with us struggling to see our different POVs when a new move is being considered - that's how the creative process works, it's nearly always a bit messy at the start.

Leren

[Leren]

OK let's apply my proposed move to another part of the puzzle.

1. The first compartment (call it AB) is r9c12 of size 2 and is contained in Box 7. Its LO range is 12 and it must contain 12 if it's LO. its HI range is 789 and must contain 8 if it is HI.

2. It has no clues or solved cells.

3. The other compartment is r56789c1. Its size is 5 and 2 + 5 = 7 < 8, so we are OK there (love that rule - well done Koosha !). The Box 7 cells are r78c1 (call then CD). The LO range is 123456 and the HI range is 345678.

4. If AB is LO, it must contain 12 so CD are limited to 3456. If AB is HI it must contain 8, so DC are limited to 34567. So in total CD are limited to 34567. You can eliminate 128 from CD, which are, amazingly, 5 and 7 in the solution.

Works for me.

One point of interest, is that you can never get eliminations in the overlapping part of the HI / LO ranges of the second compartment.

If you read through SlowThinker's Str8ts slideshow he says a lot about split range compartments but in his examples they are never overlapping. It took me about 3 years before, one day, I realised that if the LO/HI ranges overlapped you could still treat the compartment as split range but you could only make eliminations in the non overlapping parts. I don't think that you will find that documented anywhere.

You avoid this complication in the small compartment because it is good practice to remove required digits in the large compartment from the small compartment at the start of the puzzle. So there are always missing digits in the middle digit part of the small compartment, so its LO and HI ranges are fully split and the required digits are easy to determine.

Now let's apply this move to a third part of the puzzle.

1. The small compartment is r9c12, (AB) and contained in Box 3. LO range 123, must contain 2, HI range 789 must contain 8.

2. It has no clues or solved cells.

3. The other compartment is r2c6789. Its size is 4 and 2 + 4 = 6 < 8. The Box 3 cells are r2c78 (CD). The LO range is 12345 and the HI range is 56789.

4. If AB is LO CD are limited to 1345 and if it's Hi they are limited to 5679, so in total they are limited to 1345679. So, you can eliminate 28 from CD. Again, incredibly, they are 67 in the solution !

Works for me again ! Time to start coding this up.

Leren

[Leren]

koushanejad74 wrote : LO-ABC => B and F Can take 2, HI-ABC => B and F can take 7,8

Hi kousha, I've been staring at this for a while, trying to decide if this is a general property of this technique. I think the answer, unfortunately, is no.

It's certainly true in this case, because, if you look at my post above, where I had the red digits, for both the LO and HI cases, cells DE are forced into a naked pair, from which your conclusions definitely follow, which makes this case quite effective in solving the puzzle. You can also get the extra eliminations by more standard follow-on Str8ts techniques. However I see this as a special case of a more general theorem, involving only the cells in the same Box as the crossing compartments. When you look at the other instances in the puzzle, the DE cell equivalents are not forced into a naked pair and so the eliminations are fewer and the more general technique is a less effective in demolishing the puzzle.

I managed to code this up today, and the three instances of the pattern that I saw were the only three seen by the silicon monster, so you would have to say that it's easy to spot for a manual solver.

I also put the other unsolved puzzle in, and the computer found three instances of the pattern in that puzzle, but the resulting eliminations were not enough to solve that one.

I think that's a good thing. It means that it still may be possible to have tough Benuko puzzles, even with this new move. I can't say for sure because I haven't transported most of the Sudoku techniques into my Benuko solver, so one of these may work on the one remaining unsolvable (without guessing) puzzle.

Leren

[koushanejad74]

Leren,

You asked: "The way I see it, in the most general case, I can't see, a priori, why F can't be 1 or 4 in the LO case."

Just try it out and you will see why,

Compartment BDEF must contain 2 and 3; If F is neither 2 nor 3, then you must place 2 and 3 in BDE, then A cannot be 2,3, and C cannot be 2,3, which makes the compartment ABC unsolvable;
Note that ABC MUST contain 2 and 3;



Are we on the same page?

[koushanejad74]

Leren,

Are you a software programmer? if yes, maybe we can collaborate on coding stuff.

[Leren]

Hi Kousha, I agree with your conclusions with the original case in cells r1c123 (ABC) and r2c234 (DEF) but my reasoning was more general.

For the LO case I showed that DE had to 45 and so BF must be 23, as you said, and in the HI case, DE had to 45, so BF had to be 78. Since there is no 78 in F, that alone would rule out the HI case.

But what I have done was focus on DE alone, to see what eliminations can be made there. I concluded that DE had to be 456, considering both the LO and HI cases.

So, I was left with B = 123678, D = 456, E = 456 and F = 1245. That rules out the HI case straight away, because you have only one 7 and one 8 and they are in the same cell (B).

So, in one more move you can reduce BDEF to 123, 45, 45 and 1245, with only digits 5 - 9 in the right hand 4 cell compartment, which must be HI.

3 in cell B and 2 in cell F are a singles in Row 2, so you can place them there. So you see that you get to the same conclusions very quickly without making any direct eliminations in F.

On the right hand side there is a similar pattern, but the vertical compartment has only 2 cells, so, if you like, there is no cell C, Only AB and BDEF. In this pattern, for the LO case, the only thing you can say is that AB must contain 2.

You could have A = 2, B = 1 or 3 and B = 2, A = 1 or 3, so the inferences for this pattern are not so strong. Nevertheless, You can still make some eliminations in D and E.

So, my approach is to make only eliminations in cells D and E, and if there are stronger inferences (as in your original pattern), they will be quickly picked up by other standard St8ts moves, (or Sudoku moves that I haven't included in my Benuko solver yet)

I'd recommend that your look at the similar pattern on the right and side of the puzzle, and see what you can make of it. If you can convince me that there are direct eliminations in the cells of the pattern other than in the D & E equivalents, I'll be impressed.

Leren

[koushanejad74]

Leren,

I can't impressed you as you covered all the bases beautifully

[Leren]

How about we call this construction a Kousha, as it was your idea . Leren

[Leren]

Hi Kousha, I've got further with that last unsolved puzzle - almost but not quite solved.

Code: Select all

*--------------------------------------------------------------------------------------------------*
| 134789    6         1345789    | 145789    2         145789     | 13789     134579    1345789    |
| 124789    124579    1245789    | 145789    3         145789     | 2789      145679    12456789   |
| 1234789   1234579   12345789   | 145789    6         145789     | 19        2789      13789      |
|--------------------------------+--------------------------------+--------------------------------|
| 19        134579    1345679    | 13589     1479      2          | 13456789  13456789  13456789   |
| 234579    1234579   12345679   | 13589     1479      13589      | 123456789 19        1289       |
| 234579    8         1234579    | 6         1479      1359       | 1234579   1234579   1234579    |
|--------------------------------+--------------------------------+--------------------------------|
| 238       1379      19         | 123479    5         134679     | 12346789  12346789  12346789   |
| 3456-79   1345-279  12379      | 123479    8         134679     | 12345679  12345679  19         |
| 3456      1345-79   1238       | 123479    19        134679     | 123456789 123456789 1289       |
*--------------------------------------------------------------------------------------------------*

Look at Box 7. There is an interaction between the 2 cell compartments r7c12 and r89c3. One must be LO (123) and must contain 2 and the other must be HI and contain 8 and 7 or 9

That means that 2 and 79 are contained to those 2 compartments and you can make eliminations of 279 as shown.

Now suppose in the LO case either r9c2 is 1 or r78c3 is 1. Then r7c3 is 9. Now suppose either r9c2 is 1 or r78c3 is not 1. Then we must have r7c23 = 23 or r78c23 = 23. Then you have a naked quad 1456 in r89c12.

Either way. r7c3 is 9. We've solved that grey cell ! Marking that off and with some simple follow on moves the PM reduces to

Code: Select all

*--------------------------------------------------------------------------------------------------*
| 134789    6         134578     | 145789    2         145789     | 13789     134579    1345789    |
| 124789    124579    124578     | 145789    3         145789     | 2789      145679    12456789   |
| 1234789   1234579   1234578    | 145789    6         145789     | 19        2789      13789      |
|--------------------------------+--------------------------------+--------------------------------|
| 19        134579    134567     | 13589     1479      2          | 13456789  13456789  13456789   |
| 34579     1234579   1234567    | 13589     1479      13589      | 123456789 19        1289       |
| 34579     8         123457     | 6         1479      1359       | 1234579   1234579   1234579    |
|--------------------------------+--------------------------------+--------------------------------|
| 28        17        9          | 12347     5         13467      | 1234678   1234678   1234678    |
| 3456      345       127        | 123479    8         134679     | 12345679  12345679  19         |
| 3456      345       128        | 12347     19        134679     | 123456789 123456789 1289       |
*--------------------------------------------------------------------------------------------------*

Leren

[koushanejad74]

Hi Leren,

Spent several hours, didn't get further than you did, will keep trying

[Leren]

Haven't solved that puzzle yet but there is a similar construction in Box 3 involving the digits 13 & 2 and grey cell r1c7, can't be 9. Also, its easy to show that r8c1 <> 5 and r8c2 <> 4. The the elimination PM for all this is

Code: Select all

*--------------------------------------------------------------------------------------------------*
| 134789    6         1345789    | 145789    2         145789     | 13789     4579-13   45789-13   |
| 124789    124579    1245789    | 145789    3         145789     | 2789      45679-1   456789-12  |
| 1234789   1234579   12345789   | 145789    6         145789     | 1-9       2789      13789      |
|--------------------------------+--------------------------------+--------------------------------|
| 19        134579    1345679    | 13589     1479      2          | 13456789  13456789  13456789   |
| 234579    1234579   12345679   | 13589     1479      13589      | 123456789 19        1289       |
| 234579    8         1234579    | 6         1479      1359       | 1234579   1234579   1234579    |
|--------------------------------+--------------------------------+--------------------------------|
| 238       12379     9-1        | 123479    5         134679     | 12346789  12346789  12346789   |
| 346-579   135-2479  12379      | 123479    8         134679     | 12345679  12345679  19         |
| 3456      1345-79   1238       | 123479    19        134679     | 123456789 123456789 1289       |
*--------------------------------------------------------------------------------------------------*

and with follow on eliminations this reduces a bit to

Code: Select all

*--------------------------------------------------------------------------------------------------*
| 134789    6         134578     | 145789    2         145789     | 389       457       457        |
| 124789    124579    124578     | 145789    3         145789     | 289       4567      4567       |
| 234789    234579    234578     | 45789     6         45789      | 1         289       389        |
|--------------------------------+--------------------------------+--------------------------------|
| 19        134579    134567     | 13589     1479      2          | 3456789   356789    13456789   |
| 34579     1234579   1234567    | 13589     1479      13589      | 23456789  19        1289       |
| 34579     8         123457     | 6         1479      1359       | 234579    1234579   1234579    |
|--------------------------------+--------------------------------+--------------------------------|
| 28        17        9          | 12347     5         13467      | 234678    1234678   1234678    |
| 46        35        127        | 123479    8         134679     | 2345679   12345679  19         |
| 3456      345       128        | 12347     19        134679     | 23456789  123456789 1289       |
*--------------------------------------------------------------------------------------------------*

Must be close. eg if r7c1 <> 8, r8c1 <> 6 or r8c2<> 5 the puzzle will solve easily.

Leren

[Leren]

I've tried putting the wrong value into a cell, and see what my solver does before a contradiction is reached. Unfortunately, I've tried a couple of different positions and in each case, many cells are solved until an empty cell is reached.

Unfortunately (or perhaps fortunately) this means that a long chain may be necessary to determine the solution without guessing. This is not uncommon in Str8ts, and Benuko may also have similarly difficult puzzles, although rarer than in Str8ts.

The real trick, will be to come up with puzzles that are not too easy and not too difficult, and it's a tricky exercise to find these.

Leren

[koushanejad74]

Wasn't easy, but I believe I got it, starting from where you left off:

General Rule (R4):
if two compartments of size 4 ,which have the same range ,intersect then puzzle will be UNSOLVABLE.

For this specific puzzle:

chart.png (12.33 KiB) Viewed 873 times

(1) Red compartments and green compartments must be of the same type
(2) The yellow compartment and the green compartment in box 1 cannot be both LO
(3) If the blue compartment is HI, then cell r9c9 cannot be 8

Now the game begins:

(4) If green compartment in box 1 is LO then:

(5) based on (2): The Yellow compartment is HI
(6) based on (5): The black compartment is LO
(7) based on (6): The blue compartment is HI with range 6-9
(8) based on (4): The red compartment in box 8 and box 9 is HI
(9) based on (8),(6) and (3): cell r9c9 is 9
(10) based on (9) and (8): The red compartment in box 8 and box 9 has range 6-9
(11) based on (7) and (10): Apply rule R4 => puzzle is unsolvable
Conclusion: Assumption (4) is invalid, therefor Green compartments are HI

The rest is easy,

[Leren]

Hi Kousha, I'd also suspected that for a Benuko Puzzle with lots of grey cells, you could devise a HI/LO map according to a set of rules and check to see if one assumption doesn't work, so the other must be the solution.

So, all that is required is to carry out the initial grey cell row/column interactions, apply your map conclusions, and set all the two compartment row/columns to the map deduced HI/LO values, if possible.

This, by itself, will virtually solve the puzzle, which may make many Benuko puzzles somewhat trivial if you know all the HI/LO rules.

I think what you should have also said is that the red and green compartments must be of the opposite range ("range", meaning HI/LO, may be a better word than "type").

In Box 3, the two 2 cell compartments are of the opposite colour. According to your "Rule of 8" it seems to me that r3c89 is black, and r12c7 is blue. I think that would also make the yellow compartment blue.

Also I don't follow (2) and (3). Perhaps you could explain these a bit more.

In any event congratulations on having the perseverance to carry out a HI/LO map colouring analysis.

The only negative comment I could make is that your deduction (4) to (11) is a form of complex chaining, involving compartment ranges as nodes, rather than candidates, and is ultimately a form of trial and error.

I could have done a similar thing with individual candidates and solved the puzzle in a similar fashion.

In Str8ts they call this sort of thing "testing" and it's regarded as a perfectly acceptable form of solving. In the world of Sudoku I think it would be not unreasonable to say that there is some controversy about this.

It would be more in keeping with constructive solving if you could start off by assuming that a compartment is both HI and LO and conclude that one or more other compartments must have the same range (HI or LO).

Leren

PS - Just noticed a linguistic quirk in your Rule R4 statement. What you mean is that if two crossing 4 cell compartments have exactly the same range, the puzzle is unsolvable.

eg The crossing 4 cell compartments in r9c6789 and r6789c8 are both LO but their ranges are not exactly the same.

PPS. It Looks like a corollary of your Rule R4 is that if two 4 cell compartments cross, the common cell can't be 1 or 9, otherwise they would be forced into exactly the same range. So in this puzzle r9c8 can't be 1 or 9.

PPPS. A second corollary is that (unless the common cell is 5) their ranges can only be 1234/2345 or 5678/6789. if the common cell is 5 they can only be 2345/5678. Can't see an immediate application in this puzzle - yet !

PPPPS. I think there may be a third corollary. For crossing 4 cell compartments ..... One of the grey cells must be 1 and the other must be 9 !!!!! Every case I've tried so far seems to support this. If this turns out to be generally true, then the intersection cell of the grey cells, which is always r5c5, can't be 1 or 9.

[koushanejad74]

explaining (2)

(2) The yellow compartment and the green compartment in box 1 cannot be both LO

If the Green compartment in box 1 is LO, then it has to take 2 and 3, on the other hand, the part of the yellow compartment (assuming it's LO) which is is box two cannot take either 2 or 3, so the part of the yellow compartment which is in box one must take 2 and 3; this simply not possible

explaining (3)
(3) If the blue compartment is HI, then cell r9c9 cannot be 8
Blue compartment must take 8, it cannot be in r6c8, so if r9c9 is 8, then all the possibilities of placing 8 in the blue compartment will be eliminated