## SudokuP - Min Clue Project

For fans of Killer Sudoku, Samurai Sudoku and other variants

### SudokuP - Min Clue Project

I am about to embark on a rather ambitious project, namely to answer the question:
Is there a 10-clue SudokuP puzzle?

While we're at it, I hope to collect a complete set of 11-clue puzzles if possible.

I've been busy with preliminaries, assembling a toolset, scoping the problem, deciding on feasibility, etc. I'm nearly ready to begin the computational phase, so I'll describe the general approach and the initial state of play:

1. General Approach

The plan is to use the methodology employed by Gary McGuire in "There is No 16-clue Sudoku". This essentially involves testing all relevant grids, using UA set methods to determine minimum clue requirements, solving the associated Hitting Set problem to generate all possible 10-clue puzzles, and finally testing those puzzles with a solver.

All fairly straight-forward, really ...

As you know, the McGuire group solved the Sudoku problem in a mere YEAR, with a 320-node cluster, using a total of 7 million CPU-hours (give or take a few).

I have a single desktop PC with 4 hyper-threaded cores, so what makes me think I can do this? Well, mainly the reduction in complexity afforded by the SudokuP variant. We have only 200 million grids to test, and the number of 10-clue possibilities is far less than the number of 16-clue possibilities. I have run some critical scoping jobs which suggest the project is indeed feasible.

I'll get to that later.

2. Toolset

The major software tools needed are:

• a fast SudokuP solver. Brian Taylor's BB solver is not only fast, but easily modified to add the extra constraint dimension for SudokuP
• a fast UA set generator. I have combined a stand-alone UA6 function excellently coded by dobrichev, which does the basic work, and I have added my own function (less excellent, but sufficient) to generate additional UA sets, using Pittenger-style cycle generation
• a reasonably efficient Hitting Set generator to identify the 10 (or 11) clue sets that need to be tested for each grid

3. Feasibility Study

Why does this job look feasible? I'll outline my reasoning in the next post ...
Last edited by Mathimagics on Sun Feb 25, 2018 11:53 am, edited 1 time in total.

Mathimagics
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### Re: SudokuP - Min Clue Project

McGuire & friends had to deal with a staggering number of Sudoku grids. Not just the 5,472,730,538 essentially different Sudoku grids, but for each of these we have 1296 isotopes which give potentially different puzzle forms. So roughly 7 trillion grids needed to be tested by them.

For SudokuP, we have a very much smaller number of different grids to check, just 214,038,113 (see here for details). This reduces the problem by a factor of roughly 33,000.

Secondly, the cost-per-grid is notionally smaller. At the combinatorial level, we might compare the number of ways to choose 16 cells from 81 (there are 33,594,090,947,249,085) to the number of ways to choose 10 cells (1,878,392,407,320) for a rough guide, but how this translates into testing actual valid 16-clue vs valid 10-clue puzzles remains to be seen.

A complicating factor is that not all Sudoku UA sets can be used. These have to be "filtered" to determine which ones really are UA sets for the SudokuP grid in question. Each UA corresponds to a cyclic cell change (a series of cell swaps) which in most cases don't preserve the SudokuP property.

But we are clearly dealing with a much simpler problem. The reduced number of grids to be tested, all of which can be loaded into memory if needs be, allows for some simple preliminary tests to be performed.

I tested 10,000 randomly selected SudokuP grids, and looked at the minimum number of clues as determined by the associated UA set data alone. This is obtained without generating the Hitting Sets, just by looking at the UA-set clique sizes. Since there must be a clue for each UA set, then a set of N disjoint UA's (a clique) indicates (indeed, it proves) that the grid requires at least N clues for a valid puzzle.

In reality this figure is only a lower bound, and more clues are often required, but that is a more expensive test.

The results were very encouraging:

Code: Select all
`   Min Clues  UA6     UAX   ----------------------       2        1       0       3        6       0       4       40       0       5      184       1       6      403       2       7     1102       1       8     2161      30       9     3286      81      10     3016     338    ----------------------      11     1643    1036     ----------------------      12      530    2573      13       34    3191      14        3    2683      15        0    1507      16        0     524      17        0     139      18        0      31      19        0       3`

The UA6 column indicates results obtained by just using UA sets of size 4 or 6, and the UAX column gives the results on adding larger UA sets (8 to 18 typically).

Adding UAX sets is critical here, since the result is that, in all probability, we can eliminate up to 95% of the grids by this method alone. This calculation took only a few minutes, and can possibly be improved, so with a bit of luck we can do this "cherry picking" elimination process for the entire grid set in less than 14 days, which gives me ample time to work on the Hitting Set generator.

It also does tend to suggest that a 10-clue SudokuP is, in fact, extremely unlikely. But we are seeking to prove whether or not that is the case.

Mathimagics
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### Re: SudokuP - Min Clue Project

Hi, Mathimagics!
Interesting! I'll be waiting for the results.
Could you post some simple UA sets for SudokuP?

Serg
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### Re: SudokuP - Min Clue Project

Hi Serg,

Yes, I will provide some examples of UA's for SudokuP.

By the way, how is that cross-checking exercise coming along? (Serg was going to try and verify the number of SF-different SudokuP grids that I obtained (30,603,892) in the SudokuP Analysis thread.)

Mathimagics
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### Re: SudokuP - Min Clue Project

Hi, Mathimagics!
Mathimagics wrote:By the way, how is that cross-checking exercise coming along? (Serg was going to try and verify the number of SF-different SudokuP grids that I obtained (30,603,892) in the SudokuP Analysis thread.)

At the moment I am trying to count number of "SudokuP grids using the restricted P-equivalence, ie 214,038,113" SudokuP grids. My method is dobrichev's "first approach".
dobrichev wrote:Let talk about the first approach - by generating sudokuP from scratch using templates (or whatever you named them).
You should operate only with the p-templates (some counted them earlier in this forum, they are far less than 46K).
You may choose a suitable canonical form. Say, if you order all possible templates, then the grid representation could be a ordered set of the 9 template indexes. The canonical form may be chosen to have minimal such set.
You should write the respective sudokuP-to-sudokuP canonicalization procedure, which can terminate once it founds that the given (sub)grid has ANY better (lesser, closer to the canonical) P-morph.
Then you can generate all sudokuP grids in this way, by joining each p-compatible template (respectively p-subgrid) with those that are later in the chosen order.
Choosing a good templates order and canonical form (not necessarily the one above, it is just an example that first came in my mind) could benefit from the order of joining the templates and respectively minimize the canonicalization work.
You may think for applying the canonicalization before all 9 p-templates are joined, even pre-build a table with all p-template morphs, and apply logic for non-canonical morphs elimination as earlier as possible in the generation process.

I had good starting position, because I had my own SudokuP grids generator developed for counting all-different SudokuP solution grids. I added to that generator restricted p-equivalence checker (at the output), but that checker was too "heavy" - it slowed down grid's generator by 100 times! So, I am forced to use tabulating functions (arrays), which were described by Mladen.

The work is not too fast, but it goes on though.

Serg
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### Re: SudokuP - Min Clue Project

Hi Serg,

That sounds like fun!

Perhaps you could move that post to the other thread? Meanwhile I have a set of UA examples for SudokuP which I will proceed to give below.

Cheers!
Last edited by Mathimagics on Sun Feb 25, 2018 8:37 pm, edited 1 time in total.

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### SudokuP - UA Primer

A quick primer on UA theory (prompted by a reader who suggested that I try and make my posts more intelligible to a wider audience, so hats off to David Bird).

The term UA refers to an UnAvoidable set. That is, a set of grid positions in a Sudoku grid whose values can be exchanged in such a way that the resulting grid remains valid. Thus any puzzle for that grid must include at least one clue from each such set, otherwise the puzzle will have more than one solution.

UA theory is fundamentally linked with cycle theory. The idea of a cycle is quite simple. Choose any pair of grid cells with different values, call these values A and B. So we have {ra,ca} = A and {rb,cb} = B. Now let ck be the column in row a that has value B. So {ra,ck} = B, now find B in column ck, and repeat this process until we arrive back at our original position. We have traced the unique cycle that joins {ra,ca} with {rb,cb}.

If ra = rb, then we simply start with finding B in column ca and trace the cycle that way.

Example of a cycle with A=9 and B=5 :
Code: Select all
` 4  7  5 | 1  8  9 | 3  2  6 8  9a 6 | 2  7  3 | 1  5b 4 1  3  2 | 4  5  6 | 8  7  9 --------------------------- 2  5b 4 | 6  3  8 | 9a 1  7 6  1  7 | 9  2  5 | 4  3  8  9  8  3 | 7  1  4 | 2  6  5 --------------------------- 7  6  1 | 8  4  2 | 5b 9a 3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6  4  1`

Each cycle has the property that swapping the values A and B in each row or col of the cycle preserves the Latin Square property of the grid. For Latin Square puzzles it follows that every such cycle forms an unavoidable set, and since the cycle has 6 cells we would call that a UA6 for an LS puzzle onn this grid.

However that particular cycle includes only one cell each from box 4 and box 6. So the result of exchanging A and B along this cycle is NOT be a valid Sudoku Square. Let's pick another pair:
Code: Select all
` 4  7  5a| 1  8  9 | 3  2  6a 8  9  6b| 2  7  3 | 1  5b 4 1  3  2 | 4  5  6 | 8  7  9 --------------------------- 2  5  4 | 6  3  8 | 9  1  7 6  1  7 | 9  2  5 | 4  3  8  9  8  3 | 7  1  4 | 2  6b 5a --------------------------- 7  6  1 | 8  4  2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6  4  1`

This cycle hits 2 cells in boxes 1,3 and 6, the swapping of A and B in those positions is A valid Sudoku, and so this cycle is a genuine UA6 for Sudoku.

But not for SudokuP! For a UA to apply to SudokuP, assuming of course that our grid is SudokuP to begin with (which it is), we need additionally that the pair swaps in each box have pairwise positional correspondence. For each A in the cycle, there must be a B in the cycle at the same box position.

This one does the trick, and constitutes a UA10 for this grid as a SudokuP:
Code: Select all
` 4a 7  5 | 1  8b 9 | 3  2  6 8b 9  6 | 2  7  3 | 1  5  4a 1  3  2 | 4a 5  6 | 8b 7  9 --------------------------- 2  5  4 | 6  3  8 | 9  1  7 6  1  7 | 9  2  5 | 4a 3  8b 9  8  3 | 7  1  4 | 2  6  5 --------------------------- 7  6  1 | 8b 4a 2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6  4  1`

Finally, I should point out that all of the cycles/UA's described above involve just 2 grid values, A and B (they are 2-symbol cycles). There are more exotic cycles involving 3 or more symbols, and thus a greater variety of UA's than I have described here. Nevertheless, the underlying principle is still that any cyclic permutation of cells that preserves the grid properties (Latin Square, Sudoku Square, SudokuP) is a UA for that grid.
Last edited by Mathimagics on Sun Feb 25, 2018 8:44 pm, edited 2 times in total.

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### SudokuP - UA4 Examples

The simplest of all UA's is the case where {ra, ca} = {rb,cb} and {ra, cb} = {rb, ca}. This is the shortest possible cycle, it has 4 cells, which are the corners of a rectangle in the grid.

For Sudoku, since the 4 cells obviously can't all be in the same box, we can say that a 4-cycle is a UA4 whenever it spans exactly 2 boxes.

For SudokuP, it follows that the cell pairs in each of the two boxes must be in the same box positions.

Using the same grid as in the examples above, I have labelled 3 different 4-cycles with labels (a,b) (e,f) and (p,q):
Code: Select all
` 4a 7  5 | 1  8  9b| 3  2  6 8  9  6 | 2  7p 3 | 1  5q 4 1  3  2 | 4  5q 6 | 8  7p 9 --------------------------- 2e 5f 4 | 6  3  8 | 9  1  7 6  1  7 | 9  2  5 | 4  3  8  9b 8  3 | 7  1  4a| 2  6  5 --------------------------- 7  6  1 | 8  4  2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5f 2e 8 | 3  9  7 | 6  4  1`

The pair (a,b) are in 4 different boxes so can't be UA4's for either Sudoku or SudokuP.

The pair (e,f) are in 2 boxes, so are a UA4 for Sudoku, but not for SudokuP. The box positions of the 4 cells don't match.
'
The pair (p, q) is just right, so is a UA4 for SudokuP on this grid.

The current record holders for number of UA4's that I have found in a Sudoku is 36, and for SudokuP we have this example with 30 UA4's:

Code: Select all
` 1 2 3 | 4 8 5 | 7 9 6 4 5 6 | 1 7 9 | 2 8 3 7 8 9 | 3 2 6 | 1 5 4 --------------------- 2 1 7 | 6 5 4 | 8 3 9 6 3 4 | 8 9 7 | 5 2 1 5 9 8 | 2 3 1 | 6 4 7 --------------------- 3 4 1 | 5 6 2 | 9 7 8 9 6 5 | 7 4 8 | 3 1 2 8 7 2 | 9 1 3 | 4 6 5`

Complete UA4 list for this grid (32 Sudoku UA4's, of which 30 are SudokuP UA4's):
Hidden Text: Show
Code: Select all
` UA4   1: P {11,14,21,24} UA4   2: P {11,12,41,42} UA4   3: P {13,19,23,29} UA4   4:   {17,19,67,69} UA4   5: P {23,26,33,36} UA4   6: P {21,23,51,53} UA4   7: P {25,26,55,56} UA4   8: P {22,28,32,38} UA4   9: P {12,15,32,35} UA4  10: P {32,33,62,63} UA4  11: P {34,35,64,65} UA4  12: P {11,17,31,37} UA4  13: P {43,46,53,56} UA4  14: P {44,45,74,75} UA4  15: P {47,49,77,79} UA4  16: P {52,55,62,65} UA4  17: P {56,59,66,69} UA4  18: P {54,56,84,86} UA4  19: P {51,57,61,67} UA4  20: P {58,59,88,89} UA4  21: P {65,66,95,96} UA4  22: P {67,68,97,98} UA4  23: P {72,75,82,85} UA4  24: P {11,13,71,73} UA4  25:   {35,36,75,76} UA4  26: P {76,79,86,89} UA4  27: P {71,77,81,87} UA4  28: P {17,18,77,78} UA4  29: P {22,23,82,83} UA4  30: P {83,89,93,99} UA4  31: P {27,29,87,89} UA4  32: P {31,32,91,92}`

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### SudokuP - UA Primer II

All of the UA example described above involve just 2 symbols (A and B), which when swapped in rows/cols along the same cycle give a valid Latin Square, which might be a valid Sudoku (given the right box-hitting conditions), and maybe a SudokuP (if we begin with a SudokuP, and the box-hitting conditions are met, and the box/position hitting conditions are met).

It is also possible to have UA's with more than 2 symbols, and I will now describe the most important case, using 3 symbols. I will call the 3rd symbol X.

The simplest cases of 3-symbol UA's are the UA6's (ie 6-cell UA's) which occur when 3 values (A, B, X) occur in the same 3 columns of 2 different rows, or the same 3 rows of 2 columns. But these are just the smallest members of a much larger set of potential UA's of varying size from 6 to 18.

Here is the same grid as in the examples above, in which I have marked an (A, B) cycle with A = 4, B = 6, and have also marked two particular cells with matching values, X = 9:
Code: Select all
` 4a 7  5 | 1  8  9 | 3  2  6b 8  9  6b| 2  7  3 | 1  5  4a 1  3  2 | 4a 5  6b| 8  7  9 --------------------------- 2  5  4b| 6b 3  8 | 9x 1  7 6b 1  7 | 9x 2  5 | 4a 3  8  9  8  3 | 7  1  4a| 2  6b 5 --------------------------- 7  6  1 | 8  4  2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6b 4a 1`

Notice that the X cells are the corners of a rectangle with an (A, B) pair. This property allows us to produce two different Latin Squares as follows:

• Beginning with corner cell A, and the B cell in the same column, follow the (A,B) cycle until we arrive at the opposite column, ie the one containing X and B
• Beginning with corner cell A, and the B cell in the same row, follow the (A,B) cycle until we arrive at the opposite corner row, ie the row containing X and B

The first (A, B, X) cycle thus produced is:
Code: Select all
` 4  7  5 | 1  8  9 | 3  2  6    8  9  6 | 2  7  3 | 1  5  4    1  3  2 | 4a 5  6b| 8  7  9  --------------------------- 2  5  4 | 6b 3  8 | 9x 1  7  6  1  7 | 9x 2  5 | 4a 3  8  9  8  3 | 7  1  4a| 2  6b 5  --------------------------- 7  6  1 | 8  4  2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6b 4a 1 `

And the second case is this:
Code: Select all
` 4a 7  5 | 1  8  9 | 3  2  6b 8  9  6b| 2  7  3 | 1  5  4a 1  3  2 | 4  5  6 | 8  7  9 --------------------------- 2  5  4a| 6b 3  8 | 9x 1  7 6b 1  7 | 9x 2  5 | 4a 3  8 9  8  3 | 7  1  4 | 2  6  5 --------------------------- 7  6  1 | 8  4  2 | 5  9  3 3  4  9 | 5  6  1 | 7  8  2 5  2  8 | 3  9  7 | 6  4  1`

The 2 Latin Squares are obtained by swapping each marked pair in the 5 rows of the first case, and the 5 columns of the second case. Notice also that in both cases, the cycle trace hits 2 values in 5 different boxes, so the result will also be a valid Sudoku Square, and thus we have 2 UA's of size 10 for Sudoku.

Using our box-position checking method, we can see that case 1 is a UA10 for SudokuP, as the cells correspond to 5 different positions, each position hit twice.

Case 2 does not produce a SudokuP, as it only hits 4 different box-positions (2 are hit 3 times, and 2 are hit twice).

Any pair (A, B) on the same 2-symbol cycle and on different rows/columns will produce 2 different Latin Squares for every case where the corresponding corner pairs have the same value X.

This is in fact the basis of the famous Pittenger Random Latin Square Generator. Pittenger proved that the entire space of Latin Squares is connected by using combinations of 2S-cycle swaps and 3S-cycle swaps.

So we can form even more exotic UA's (of 4 or more symbols) by using the same method, ie by combining multiple 2S and 3S cycle swap operations, so long as the result of each is a valid Sudoku/SudokuP.

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### SudokuP - Cycle Validity Checking Simplified

For any potential UA we have a cycle, that is, a list of cells. The cycle list will always contain an even number of cells, 2N, and these cells will form N pairs, with each pair being in the same row, OR (in the 3-symbol cases) the same column.

• we always get a valid Latin Square by applying the cycle, ie swapping the cells pairwise

• Given we have a Sudoku grid to begin with, let NB(b) be the number of cells in the cycle that are in box b. We will get another valid Sudoku grid if and only if there is no box with NB(b) = 1

• Given we have a SudokuP grid to begin with, let NP(p) be the number of cells in the cycle that are in box-position p. We will get another valid SudokuP grid if and only if there is no box with NB(b) = 1, AND no box-position with NP(p) = 1

The Sudoku test follows from the observation that, if any box has only one representative in the cycle, then it must necessarily be changed to an existing value.

The SudokuP test follows by applying the same logic to box-position-sets.

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### SudokuP - A pathological case

.
The actual process of finding 10 clue SudokuP puzzles is deceptively simple.

First we need to select the grids need testing. We need only test those SudokuP grids which are "different", ie not equivalent. Here "equivalence" means P-equivalence, where two grids are equivalent if one can be obtained from the other by any of the 2592 permutations that preserve the SudokuP property (see the "SudokuP Analysis" thread for for more details).

I suggest here (without proof!) that P-equivalent SudokuP grids are also "puzzle-equivalent" - ie, they have exactly the same UA sets structure. Thus we only need to test each of the 214,038,113 P-different grids.

For each grid we need to:

• obtain a lower bound N for the minimum number of SudokuP clues (MPC) from the UA list for this grid

• if N <= 10, then we need to generate all possible 10-clue puzzles for this grid, and test each puzzle for uniqueness of solution

The latter step is a "Hitting Set Problem" - we need to find all combinations of 10 clues that "hit" each and every known UA for the grid in question.

A problem which appears to be unique to SudokuP is that we sometimes find that N is very small. In fact it sometimes completely vanishes!

Here is a case in point. This grid is one of the 73 SudokuP isotopes of the {MC} grid:

Code: Select all
` 8 9 7 | 1 2 3 | 6 4 5 2 3 1 | 4 5 6 | 9 7 8 5 6 4 | 7 8 9 | 3 1 2 --------------------- 3 1 2 | 5 6 4 | 7 8 9 6 4 5 | 8 9 7 | 1 2 3 9 7 8 | 2 3 1 | 4 5 6 --------------------- 4 5 6 | 9 7 8 | 2 3 1 7 8 9 | 3 1 2 | 5 6 4 1 2 3 | 6 4 5 | 8 9 7`

As a Sudoku grid, this has 378 UA's (using the 2S and 3S method), however not a single one is valid for SudokuP (independent verification of this would be useful here, should anyone care to do so. The list is given below).

Interestingly, each and every in the list is a UA6, and only 2 of the 73 isotopes in question have this problem, in fact all of the other 71 have ample quantities of valid SudokuP UA's.

How are we to deal with this grid? Either we have to find more exotic UA's, or we need to test every one of the 1,878,392,407,320 ways to choose 10 cells (not to mention the various clue value combinations!).

We can presumably reduce this number by establishing minimum values for the number of rows/columns/boxes that need to be covered for a valid Sudoku puzzle (so if anybody can point me to any known results for normal Sudoku that would help greatly!). Then we can hopefully use that knowledge to reduce further for SudokuP cases.

===========================================================
The UA list for the grid above:
Hidden Text: Show
Code: Select all
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Mathimagics
2017 Supporter

Posts: 1926
Joined: 27 May 2015
Location: Canberra

### Re: SudokuP - Min Clue Project

Indeed a challenge - and yet again the MC grid throws up curve balls.

I cant find the post where a 11-clue sudoku-DG puzzle was published ....

And I'm just wondering what the clues per band are ....
.... and the way blue/champagne postulated an easy way to prove the no-16 problem might be employed here

ie
If there is a 10-clue - then it means that somewhere in the puzzle there has to be at least 4 clues in a band.[eg 433 or 532] [? or 622]
Which means you could show that there is no non-minimal 11 - which would have to have a clue-in-band distribution of 5xx [ 533 or 542 or ?551] or 6xx [632 or ?641] or 7xx [ 722]

bands-and-low-clue-puzzles-t30218.html
scan-solution-grids-for-17-clues-as-of-blue-t34012.html
Last edited by coloin on Tue Feb 27, 2018 11:35 am, edited 1 time in total.
coloin

Posts: 2328
Joined: 05 May 2005
Location: Devon

### SudokuP - UA's in the MC grid

All 73 SudokuP isotopes of the {MC} grid have 378 UA6's. Typically the number of these which are valid SudokuP UA's is around 200. But there are two exceptional cases where none of them are valid (one is shown above).

And there is one case in which all 378 are valid:

Code: Select all
` 1 2 3 | 4 5 6 | 7 8 9 4 5 6 | 7 8 9 | 1 2 3 7 8 9 | 1 2 3 | 4 5 6 --------------------- 2 3 1 | 5 6 4 | 8 9 7 5 6 4 | 8 9 7 | 2 3 1 8 9 7 | 2 3 1 | 5 6 4 --------------------- 3 1 2 | 6 4 5 | 9 7 8 6 4 5 | 9 7 8 | 3 1 2 9 7 8 | 3 1 2 | 6 4 5`

The UA list:
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Mathimagics
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Location: Canberra

### Re: SudokuP - Min Clue Project

Hi coloin,

I think it was Gordon Royle who posted the 11-clue examples. I can no longer find that post but here they are:

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`1.2........3..............4.4..5.....6..............1..7...........8..........7..1.2........3..............4.4..5.....6..7...........2..8......................8..1.2........3..4...........5.5..6.....7..............1..8......................8..1.2........3..4...........5.5..6.....7..............2..8......................8..1.2......3..4...........5...6..7.....7..............2..8......................8..1.2..3....................4.4..5.....6..............1..7...........8..........7..1.2..3....................4.4..5.....6..7...........2..8......................8..1.2..3........4...........5.5..6.....7..............1..8......................8..1.2..3........4...........5.5..6.....7..............2..8......................8..1.23........4...........5...6..7.....7..............2..8......................8..`

Mathimagics
2017 Supporter

Posts: 1926
Joined: 27 May 2015
Location: Canberra

### Re: SudokuP - Min Clue Project

and a further thought....

presumably the clue frequency count in a 11-clue would have to be
{411111110} or {321111110} or {222111110}
all the above look like {222111110}
a 10 clue would be particularly sparse
{311111110} or {221111110}
coloin

Posts: 2328
Joined: 05 May 2005
Location: Devon

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