The New Sudoku Players' Forum

[koushanejad74]

Benoku is a logic-based number-placement puzzle. It is distinct from but shares some properties and rules with Str8ts and Sudoku.
Rules:
• All sudoku rules apply.
• Some rows and columns are divided into two compartments by a gray cell
• Each compartment, vertically or horizontally, must contain a straight – a set of consecutive numbers, but in any order. For example: 7, 6, 4, 5 is valid, but 1, 3, 8, 7 is not.
• Unlike Str8t, every cell MUST have a number, including gray cells
• Each puzzle has a UNIQUE solution

Benoku Definition:PDF

Benoku Sample (Hard)
Hint: No guessing is required

Gray cells are shown with [ ]

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 .   .   .   |  5   .   .   |  .   .   . 
 .   .   .   |  .  [.]  .   |  .   .   . 
 .   .   .   |  3   7   .   |  .   .  [.]
----------------------------------------
[.]  .   .   |  .   .   .   |  .   .   . 
 .   .   .   | [.]  .   .   |  .   .   . 
 .   .   .   |  .   .   .   |  .  [.]  . 
----------------------------------------
 .   .   .   |  .   .   .   |  .   .   5 
 .   .   3   |  7   .   .   |  .   .   . 
 .   .  [.]  |  .   .   .   |  .   .   . 


[Leren]

Code: Select all

167594328
345218769
289376541
916452837
432187956
578963412
724831695
653729184
891645273

Solved with basics. Leren

[SpAce]

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[The full solution grid in plain text. As usual.]

Only two kinds of people do that: absolute newbies and this guy. Everyone else uses hidden blocks.

[Leren]

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  0    0    0    0    0    0    0    0    0 
  0    0    0    0   [0]   0    0    0    0 
  0    0    0    3    7    0    0    0   [0]
 [0]   0    0    0    0    0    0    0    0 
  0    0    0   [0]   0    0    0    0    0 
  0    0    7    0    0    0    0   [0]   0 
  0    0    0    0    0    0    0    0    5 
  0    0    3    7    0    0    0    0    0 
  0    0   [0]   0    0    0    4    0    0 

Leren

[SpAce]

Hi Kousha,

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 .   .   .   |  5   .   .   |  .   .   . 
 .   .   .   |  .  [.]  .   |  .   .   . 
 .   .   .   |  3   7   .   |  .   .  [.]
----------------------------------------
[.]  .   .   |  .   .   .   |  .   .   . 
 .   .   .   | [.]  .   .   |  .   .   . 
 .   .   .   |  .   .   .   |  .  [.]  . 
----------------------------------------
 .   .   .   |  .   .   .   |  .   .   5 
 .   .   3   |  7   .   .   |  .   .   . 
 .   .  [.]  |  .   .   .   |  .   .   . 


I like your presentation style. If you want to make it even prettier, you could add the outer edges as well. I would suggest something like this (Hodoku-style):

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.-------------.-------------.-------------.
|  .   .   .  |  5   .   .  |  .   .   .  |
|  .   .   .  |  .  [.]  .  |  .   .   .  |
|  .   .   .  |  3   7   .  |  .   .  [.] |
:-------------+-------------+-------------:
| [.]  .   .  |  .   .   .  |  .   .   .  |
|  .   .   .  | [.]  .   .  |  .   .   .  |
|  .   .   .  |  .   .   .  |  .  [.]  .  |
:-------------+-------------+-------------:
|  .   .   .  |  .   .   .  |  .   .   5  |
|  .   .   3  |  7   .   .  |  .   .   .  |
|  .   .  [.] |  .   .   .  |  .   .   .  |
'-------------'-------------'-------------'


In practice, I have no problem with your current style either. It's much more readable than the alternative.

[koushanejad74]

Leren,

Is there anything you cannot solve, how about this one:

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 .   6   .   |  .   .   .   |  .   .   . 
 .   .   .   |  .   3   .   |  .   .   . 
 .   .   .   |  .   .   .   | [.]  .   . 
----------------------------------------
[.]  .   .   |  .   .   2   |  .   .   . 
 .   .   .   |  .   .   .   |  .  [.]  . 
 .   8   .   |  6   .   .   |  .   .   . 
----------------------------------------
 .   .  [.]  |  .   5   .   |  .   .   . 
 .   .   .   |  .   8   .   |  .   .  [.]
 .   .   .   |  .  [.]  .   |  .   .   . 



[Leren]

I can't constructively solve either this puzzle or the last one I put up. I know the unique solutions to both because my solution counter uses guessing, but that doesn't really "count", does it.

That's not to say that a constructive solution is not available, but I may need to put more Sudoku style moves into my Benuko solver.

Also, with the first puzzle in this thread, which for me solved easily, if you tell me the point at which you required guessing, I might be able to help you out .

Leren

[koushanejad74]

I can't constructively solve either this puzzle or the last one I put up. I know the unique solutions to both because my solution counter uses guessing, but that doesn't really "count", does it.

That's not to say that a constructive solution is not available, but I may need to put more Sudoku style moves into my Benuko solver.

Also, with the first puzzle in this thread, which for me solved easily, if you tell me the point at which you required guessing, I might be able to help you out .

Leren

Leren,

As I mentioned, the first puzzle I put in this thread can be solved with no guessing

[koushanejad74]

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  0    0    0    0    0    0    0    0    0 
  0    0    0    0   [0]   0    0    0    0 
  0    0    0    3    7    0    0    0   [0]
 [0]   0    0    0    0    0    0    0    0 
  0    0    0   [0]   0    0    0    0    0 
  0    0    7    0    0    0    0   [0]   0 
  0    0    0    0    0    0    0    0    5 
  0    0    3    7    0    0    0    0    0 
  0    0   [0]   0    0    0    4    0    0 

Leren

I don't quite get it, isn't this the same as the one I posted, the only difference I see is that the number 5 in the first row is gone, could you please clarify?

[Leren]

Actually there are three differences. After the puzzle you posted at the top of the thread solved easily, I looked around for a variation that might be more challenging. I removed the clue 5 from r1c4 and added two clues 7 in r6c3 and 4 in r9c7, so the number of clues is one more, and the solution is different and unique, but I couldn't solve it without guessing. BTW it's quite common for Str8ts type puzzles to be extremely difficult to solve, some require what's called "heavy chaining" and even "testing" (that's a polite way of saying guessing, but they sometimes do that in Str8ts).

Leren

[Leren]

I'll explain, in part, the start of the solution of your puzzle, which will show you why it solved so easily. (I won't go through the whole thing, Str8ts solutions can be very tedious.) To avoid cluttering the page I'll put the details in a Show/Hide box.

Hidden Text: Show

The key to the start was in Column 4. You can instantly see that r1234c4 must be 2345, because of the 3 and 5 clues, which leaves a naked pair 24 in r24c4. The other thing about Column 4 is that it contains twin 4 cell compartments, with the black cell being in Row 5. What is well known about this situation is that, irrespective of any other clues you may have, one of these compartments must have a LO range 1-5 and the other a HI range 5-9. So we can instantly see that r6789 must have the HI range, and it can't contain 5, so it must be 6-9. That leaves only one place for the 1 to go - in the black cell r5c4 ! So you can remove a whole bunch of 1's from Row 5 and Box 5.

The next well known thing about twin 4 cell compartments is that the LO one must contain 234 and the Hi one 678. So if one cell in them is reduced to either of these ranges then you can decide the LO/HI for that situation.

So in Row 2, we know that r2c4 contains just 24, so r2c1234 is LO (remove 6789) and r2c6789 is HI (remove 1234). There are a few intervening moves but the next thing of importance is that we know that r2c1 is 12345, but r56789c1 is a 5 cell compartment and must contain 5, so r2c1 is reduced to 1234 and there is no 5 in the compartment r123c1, which is reduced to range 1234, so you can remove 6789 from r13c1.

Now comes our first Sudoku type move. You now have a hidden quad 6789 in Box 1 r13c23, so you can remove 1245 from these cells. That leaves a pointing pair of 5's in Box 1 r2c23.

The 5's remove 5 from r2c6789, which becomes a naked quad 6789, and you instantly see that black cell r2c5 can't be 9, so it's 1, so exit some 1's from Row2, Column 5 and Box 2, which then leaves a pointing pair of 1's in r13c1.

The pointing pair of 1's removes a whole lot of 1's from the rest of Column 1, including in the black cell r5c1, which is then 9. Exit a whole bunch of 9's from Row 5, Column 1 and Box 4.

So you can see that the first 3 cells that you solve are all black cells !

This might look a bit daunting when you first see it, but trust me, any competent and experienced Str8ts/ Benuko player will have this all sorted in a few seconds, sometimes without even writing down any candidates (they say that sort of thing on the Str8ts forum, and I'm not going to dispute their claims, although it's beyond me).

Removing the clue 5 in my variant puzzle removed this puzzle crib and made my variant a lot harder, at least for me.

Leren

[koushanejad74]

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  0    0    0    0    0    0    0    0    0 
  0    0    0    0   [0]   0    0    0    0 
  0    0    0    3    7    0    0    0   [0]
 [0]   0    0    0    0    0    0    0    0 
  0    0    0   [0]   0    0    0    0    0 
  0    0    7    0    0    0    0   [0]   0 
  0    0    0    0    0    0    0    0    5 
  0    0    3    7    0    0    0    0    0 
  0    0   [0]   0    0    0    4    0    0 

Leren

Leren, this can be solved without guessing
Terminology:
Low-Compartment: compartment which takes the lower numbers
High-Compartment: compartment which takes the higher numbers
Every row or column which has a gray cell not in the first or last position, divides it to a low-compartment, a high-compartment, and a gray cell.
My argument for a case like the one shown is that the only possibilities for cell F is 2,3,7,8, and here’s why:
Assumption: (ABC) is low
(BDEF) must be low as well,
(ABC) must contain numbers 2 and 3
(BDEF) must contain 2 and 3 as well
Cells D and E cannot take either 2 or 3
Cells B and F must take 2 and 3
Using the same logic when ABC is high => F can take either 7 or 8

Conclusion: Cell F can contain 2,3,7,or 8 no matter what
In the example given, we know cell F cannot take 3,7,8 => it has to take 2,
The rest would be a piece of cake for you.

LerenPuzzle.png (8.87 KiB) Viewed 1270 times

[Leren]

Your idea is brilliant, but I don't think you have the details quite right but nevertheless, with my suggested amendments, it still solves the puzzle.

This move I would call Dual Split Compartment Range /Box Interaction, a true Benuko only move (only the second one discovered, the first being the limitations of black cell candidates). It's a combination of St8ts ideas and Sudoku ideas.

OK, let's see if I've got this right. In the low range, Cells BDEF can all be 12345 and in the HI range 56789 (in the most general case).

Now, turning to ABC. Using some standard Str8ts moves, in its LO range it can only hold 123 (note the removal of 4 by a separate standard move) and in its HI range 6789. So, in the LO range it must hold 123 and in the HI range it must hold 78.

Let's see what this does to cells DE. Your brilliant observation was that Cells DE are in the same box as ABC, so they are affected by ABC's LO and HI ranges.

So, in the LO range we have ABC = 123 and D = 12345, E = 12345.

The red colour indicates the digits that can be removed from cells DE.

Now the HI range. ABC must contain 78, so D = 5678, E = 5678.

The nett result of all this is that D and E can be reduced to at most all of the different black digits ie 456. So, the result of the move is to eliminate 12378 from D & E.

Luckily, I have a feature in my solver, that caters for testing new moves. I can hand-dress eliminations to see what the effect of some new move might be. If it works, I can then set about coding it into my solver.

The result ? The puzzle solves easily, with just basics apart from this move.

I think what you may have missed in your post was that the LO range (in the most general case) for BDEF includes 1 and 5.

Does this look right to you ?

Finally, if you want to learn more about Str8ts solving techniques, you can look here and here.

You can ignore anything with the label Setti. Whilst hugely important in Str8ts, you won't need to use it at all in Benoku.

Leren

PS. I think I can see another similar move on the other side.

Consider r12c9. If it's LO it must contain 2 and r2c6789 is LO, so r2c78 can be 1345. If it's HI it contains 8 and r2c6789 is HI, so r2c78 can be 5679, so together r2c78 can be 1345679 ie you can remove 28 from r2c78.

Leren

[koushanejad74]

Hi Leren,

Sorry if I didn't explain it clearly, so I try again:

Like Sudoku, there are techniques you can develop to solve Benoku, so first I developed a technique and then I applied it.

General rule:
for A LO-compartment of size N, where N > 1 => must includes numbers from 2 to N; must contain either 1 or N+1
for A HI-compartment of size N, where N > 1 => must includes numbers from (9-N+1) to 8; must contain either (9-N) or 9

Then I applied this rule to the given situation
A . .
B D E F
C . .
[.]

LO-ABC => B and F Can take 2,3
HI-ABC => B and F can take 7,8

We can call this, as you said, "Dual Split Compartment" technique;

Then if "Dual Split Compartment" technique is applied to the given puzzle, Cell F will be 2

[koushanejad74]

Another rule which could be helpful:

Crossing Small Compartments:
If two compartments share a cell and the total size is less than 8, then the two compartments must be of the same type (LO or HIGH)