## Robert's puzzles 2020-09-18

Post puzzles for others to solve here.

### Robert's puzzles 2020-09-18

Hi all,
Here is a fairly easy puzzle.
Good resolution.
Robert

....3.9.....8.1.6.34......8.....943.2.......9.192.....8......75.3.4.7.....5.6....

Puzzle: Show

Resolution: Show
After simplifying the puzzle by the basic techniques (TB), the use of two anti-tracks is enough to solve :

P'(8r56c8) : (-8r56c8) => (1r5c8 and 5r6c8)->2r3c8->2r2c5->4r2c9->4r9c8->9r8c8 => -8r8c8 => r8c7=8.

P'(6r7c2) : (-6r7c2) => 9r7c2->9r3c5->(2r2c5->4r2c9->4r9c8->9r7c8->2r8c3)->(7r2c3 and 2r1c2->8r1c3)->6r4c3 => -6r1c2, -6r45c2 => r7c2=6 stte.
Last edited by Mauriès Robert on Sat Sep 19, 2020 10:20 am, edited 6 times in total.
Mauriès Robert

Posts: 345
Joined: 07 November 2019
Location: France

### Re: Robert's puzzles 2020-09-18

Step 1. AIC (with two subchains)

Code: Select all
`.----------------------.------------------.--------------------.| 1567  c25678  c12678 | 567    3    d456 |  9     c1245   147 || 579    2579    7-2   | 8     e247   1   |  2357   6      347 || 3      4       167-2 | 5679  e279   56  |  1257  f125    8   |:----------------------+------------------+--------------------:| 567    5678    678   | 1567   178   9   |  4      3      2   || 2      5678    3     | 567    478   456 |  178   f18     9   || 4      1       9     | 2      78    3   |  5678  f58     67  |:----------------------+------------------+--------------------:| 8      69      4     | 139    19    2   |  136    7      5   || 169    3      a126   | 4      5     7   | b1268  b29-18  16  || 179    279     5     | 139    6     8   |  123   b249-1  134 |'----------------------'------------------'--------------------'a(2)r8c3 = b(2-894)b9p458 = (42)r1c832a - 4r1c6 = (42)r23c5 - (2=518)r356c8b => -2 r23c3 (a), -18 r89c8 (b)`

Step 2. Kraken Cell (259)r2c2

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`.-------------------.------------------.----------------------.| b156   2568  1268 |  567    3    456 |  9       1245   14-7 || b59   *259   7    |  8     a24   1   |  235     6      34   ||  3     4     16   | a5679  a279  56  | a125(7)  125    8    |:-------------------+------------------+----------------------:| b567   5678  68   |  1567   178  9   |  4       3      2    ||  2     5678  3    |  567    478  456 |  1-7     18     9    ||  4     1     9    |  2      78   3   |  56-7    58   bc6(7) |:-------------------+------------------+----------------------:|  8    c69    4    |  139    19   2   | c136     7      5    || b169   3    b126  |  4      5    7   |  8       29   bc16   || b179   279   5    |  139    6    8   |  123     249    134  |'-------------------'------------------'----------------------'a:(2)r2c2 - r2c5 = (29-7)r3c54 = (7)r3c7  ||b:(5)r2c2 - r12c1 = (57-1)r49c1 = r8c13 - (1=67)r86c9  ||c:(9)r2c2 - (9=6)r7c2 - r7c7 = (67)r86c9=> -7 r1c9,r56c7; stte`
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2568
Joined: 22 May 2017

### Re: Robert's puzzles 2020-09-18

Hi Space,
Interesting resolution from you in two steps, but I have difficulty with the AIC and its two sub-chains. Could you split into two AICs?
I think we can make it simpler with a resolution using only two AICs (See my resolution).
Robert
Mauriès Robert

Posts: 345
Joined: 07 November 2019
Location: France

### Re: Robert's puzzles 2020-09-18

Hi Robert,

Mauriès Robert wrote:Interesting resolution from you in two steps, but I have difficulty with the AIC and its two sub-chains. Could you split into two AICs?

Sure. Here you are:

SpAce wrote:a(2)r8c3 = b(2-894)b9p458 = (42)r1c832a - 4r1c6 = (42)r23c5 - (2=518)r356c8b => -2 r23c3 (a), -18 r89c8 (b)

As two AICs:
Code: Select all
`a: (2)r8c3 = (289-4)b9p458 = (4-2)r1c8 = (2)r1c23 => -2 r23c3b: (94)r89c8 = 4r1c8 - r1c6 = (4-2)r2c5 = r3c5 - (2=518)r356c8 => -18 r89c8`

Subchain b (and the original) could also be written with a different last node to make it more symmetrical:

Code: Select all
`b: (94)r89c8 = 4r1c8 - r1c6 = (4-2)r2c5 = r3c5 - r3c8 = (249)r198c8 => -18 r89c8`

Note that the original b is practically equivalent to your first anti-track reversed:

Robert wrote:P'(8r56c8) : (-8r56c8) => (1r5c8 and 5r6c8)->2r3c8->2r2c5->4r2c9->4r9c8->9r8c8 => -8r8c8 => r8c7=8

As an AIC:
Code: Select all
`(8=15)r56c8 - (1|5=2)r3c8 - r3c5 = (2-4)r2c5 = r2c9 - r9c9 = (4-9)r9c8 = (9)r8c8 => -8 r8c8`

The difference is that the way I wrote it allowed to eliminate also 1r89c8, though that's not really necessary.

I think we can make it simpler with a resolution using only two AICs (See my resolution).

Yours is an interesting solution, but I can't agree that it's simpler in any way. For one, I can't write your second step as a reasonable AIC at all. It's clearly a net, and if properly analyzed, it's actually way more complex than my second step (which I can easily write as an AIC, but I think it's simpler and clearer as a kraken). The least horrible chain representation I could come up with for yours is a memory chain (which is not a valid AIC):

Robert wrote:P'(6r7c2) : (-6r7c2) => 9r7c2->9r3c5->(2r2c5->4r2c9->4r9c8->9r7c8->2r8c3)->(7r2c3 and 2r1c2->8r1c3)->6r4c3 => -6r1c2, -6r45c2 => r7c2=6 stte.

As a memory chain:
Code: Select all
`(6=9)r7c2 - r7c5 = (9-2)r3c5 = (2*-4)r2c5 = r2c9 - r9c9 = (4-9)r9c8 = (9-2)r8c8 = r8c3 - b1p369*5 = (2^87)b1p236 - (8|7=6)r4c3 - r45^1c2 = (6)r7c2 => +6 r7c2`

The complexity is also easily seen in the matrix form:

12x12 TM: Show
Code: Select all
` 6r7c2 9r7c2       9r7c5 9r3c5             2r3c5 2r2c5                   4r2c5 4r2c9                         4r9c9 4r9c8                               9r9c8 9r8c8                                     2r8c8 2r8c3                   2r2c2 ................. 2r123c3 2r1c2                                                   2r2c3 7r2c3                                                    8r1c2 ..... 8r1c3                                                         7r4c3 8r4c3 6r4c3 6r7c2 ........................................... 6r1c2 ........... 6r45c2===========================================================================+6r7c2`

It has three complex strong inference sets (SIS): 2b1, r4c3, 6c2, and two memories: 2r2c5, 2r1c2. My kraken only has one 3-SIS: r2c2, and one memory: 6r6c9:

10x10 TM: Show
Code: Select all
` 7r6c9 6r6c9        6r8c9 1r8c9             1r8c13 1r9c1                    7r9c1 7r4c1                          5r4c1 5r12c1       6r8c9 ......................... 6r7c7                                       6r7c2 9r7c2                                5r2c2  ..... 9r2c2 2r2c2                                                   2r2c5 2r3c5 7r3c7 ................................................. 79r3c45================================================================-7r1c9-7r56c7`

It can also be written as a fairly simple and valid AIC, either with a nested chain (a) or with split-nodes (b):

Code: Select all
`a: (7)r3c7 = (79-2)r3c78 = r2c5 - 2r2c2 = [(76=1)r68c9 - r8c13 = (17-5)r94c1 = r12c1 - (5=96)r27c2 - r7c7 = (67)r86c9] => -7 r1c1,r56c7b: (7=6)r6c9 - r8c9 = (1,6)b9p61 - 1r8c13|6r7c2 = (17,5)r9421,(9)r7c2 - (5|9=2)r2c2 - r2c5 = (29-7)r3c45 = (7)r3c7 => -7 r1c1,r56c7`

...but I chose to present it as a kraken because it's much easier to read that way. Yours is difficult to write even as a multi-kraken:

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`           (8)r4c3 - r1c3 = (8)r1c2 -.           ||                         )           (7)r4c3 - (7=2)r2c3 -.    /           ||                    )  /(6)r45c2 - (6)r4c3     .--------'- '||                    ((6)r1c2 -------------- (2)r1c2||                     ||||                     (2)r123c3 - r8c3 = (2-9)r8c8 = (9-4)r9c8 = r9c9 - r2c9 = (4-2)r2c5 = (2-9)r3c5 = r7c5 - (9=6)r7c2 *||                     ||                                                         /||                     (2)r2c2 --------------------------------------------------'||(6)r7c2 *=> +6 r7c2`

Would you still say that's a simpler approach?

--
Added. FWIW, I think both second steps (yours and mine) would translate to braids instead of whips in Denis' system, so no difference there.

SpAce

Posts: 2568
Joined: 22 May 2017

### Re: Robert's puzzles 2020-09-18

One step:
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` +-------------------------+---------------------+----------------------+ |  1567   25678   12678   |  567    3     456   |  9      1245   147   |  |  579    2579    27      |  8      247   1     |  2357   6      347   |  |  3      4       1267    |  5679   279   56    |  1257   125    8     |  +-------------------------+---------------------+----------------------+ |  567    5678    678     |  1567   178   9     |  4      3      2     |  |  2      5678    3       |  567    478   456   |  178    18     9     |  |  4      1       9       |  2      78    3     |  5678   58     67    |  +-------------------------+---------------------+----------------------+ |  8      69      4       |  139    19    2     |  136    7      5     |  |  169    3       126     |  4      5     7     |  1268   1289   16    |  |  179    279     5       |  139    6     8     |  123    1249   134   |  +-------------------------+---------------------+----------------------+`

Multi-krakens: cell (136)r7c7, column (8)r568c8, cell (169)r8c1, cell (1567)r1c1
Code: Select all
`(1)r7c7 - [(1)r5c7 = (1-8)r5c8 = r6c8 - (8=7)r6c5 - (7=61)r68c9] = (8)r8c8 - (8=1263)b9p1467 ||(3)r7c7 ||(6)r7c7 - [(6=9)r7c2 - (9=1)r8c1 - (1=6)r8c9] = (6)r8c1 - (6)r1c1                                                           ||                                                          (1)r1c1 - r13c3 = (1-2)r8c3 = r9c2 - (2=163)b9p167                                                           ||                                                          (5)r1c1-r2c12=(5)r2c7                                                           ||                                                          (7)r1c1-(7=2)r2c3-r8c3=r9c2-(2=163)b9p167---------------- => -3 r2c7; ste`

Added: matrices of the net above.

First matrix, 12x12 TM with two condensed nodes (ALS b9p1476,Y-Wing r7c2,r8c19)
Hidden Text: Show
Code: Select all
` 3r79c7 1628b9p1476          2r9c2     2r8c3                    2r2c3 7r2c3                    1r8c3       1r13c35r2c7                                  5r2c12                                            7r1c1 1r1c1  5r1c1  6r1c1                                              6r8c1 Y-Wing(r7c2,r8c19)3r7c7                                                     6r7c7         1r7c7                                                                        1r5c7  1r5c8           8r8c8                                                               8r5c8 8r6c8                                                                                     8r6c5 7r6c5                                                                       16r68c9             7r6c9---------------- => -3 r2c7; ste`

Matrix 2, almost everything unfolded, notably the two embedded AIC's, 17x17 BTM
Hidden Text: Show
Code: Select all
`3r7c7 16b9p163r9c7  1r9c7  2r9c7      1|6r8c7 2r8c7 8r8c7              2r9c2       2r8c3                          2r2c3 7r2c3                          1r8c3       1r13c35r2c7                                        5r2c12                                                  7r1c1 1r1c1  5r1c1  6r1c13r7c7                                                     6r7c7              1r7c7                                                          6r8c9 1r8c9                                                          6r7c2       9r7c2                                                     6r8c1       1r8c1 9r8c1                                                                             1r5c7 1r5c8                    8r8c8                                                          8r5c8 8r6c8                                                                                         8r6c5 7r6c5                                                                                               7r6c9 6r6c9                                                                             1r8c9                   6r8c9---------------- => -3 r2c7; ste`

Matrix 3, everything unfolded, 18x18 (BTM)
Hidden Text: Show
Note the repetion of truth (136)r7c7 in rows 1 & 10, to be considered for the BTM property proof.
Code: Select all
`                            3r7c7 1r7c7 6r7c7      1r8c9 6r8c93r9c7 1r9c7       2r9c7      1r8c7 6r8c7 2r8c7 8r8c7                  2r9c2       2r8c3                              2r2c3 7r2c3                              1r8c3       1r13c35r2c7                                            5r2c12                                    7r1c1  1r1c1  5r1c1 6r1c13r7c7                                                         6r7c7              1r7c7                                                              6r8c9 1r8c9                                                              6r7c2       9r7c2                                                         6r8c1       1r8c1 9r8c1                                                                                 1r5c7 1r5c8                        8r8c8                                                          8r5c8 8r6c8                                                                                             8r6c5 7r6c5                                                                                                   7r6c9 6r6c9                                                                                 1r8c9                   6r8c9---------------- => -3 r2c7; ste`

For this matrix, hereafter triangular sub-matrices TMa and TMb for top entries 1r7c7 and 6r7c7 resp, proving matrix 3 to be a valid BTM.
Code: Select all
`TMa 10x103r7c7 1r7c7       1r8c9 6r8c93r9c7 1r9c7       2r9c7      1r8c7 6r8c7 2r8c7 8r8c73r7c7                         1r7c7                              1r5c7 1r5c8                        8r8c8       8r5c8 8r6c8                                          8r6c5 7r6c5                                                7r6c9 6r6c9                              1r8c9                   6r8c9---------------- => -3 r2c7TMb 12x123r7c7 6r7c7      6r8c9 1r8c93r9c7       1r9c7 2r9c7                  2r9c2 2r8c3                        2r2c3 7r2c3                        1r8c3       1r13c35r2c7                                      5r2c12                              7r1c1  1r1c1  5r1c1 6r1c13r7c7                                                   6r7c7                                                                      6r8c9 1r8c9                                                        6r7c2       9r7c2                                                   6r8c1       1r8c1 9r8c1---------------- => -3 r2c7`

Added: sub matrices TMa and TMb checking BTM validity of matrix 3.
Last edited by Cenoman on Sun Sep 20, 2020 10:19 pm, edited 2 times in total.
Cenoman
Cenoman

Posts: 1517
Joined: 21 November 2016
Location: Paris, France

### Re: Robert's puzzles 2020-09-18

Hi Cenoman,

Cenoman wrote:Multi-krakens: cell (136)r7c7, column (8)r568c8, cell (169)r8c1, cell (1567)r1c1

Very nice! I love the way you used those almost-AICs. It gave me an idea to write Robert's net as a relatively clean AIC:

(6)r7c2 = [(6)r1c2 = r45c2 - (6=8)r4c3 - r1c3 = (8)r1c2]|(72)r42c3 - 2r1c2 = [(2)r2c2 = r123c3 - r8c3 = (2-9)r8c8 = (9-4)r9c8 = r9c9 - r2c9 = (4)r2c5] - 2r2c5 = (2-9)r3c5 = (9)r7c5 => -9 r7c2

Obviously it's horribly long and takes some deciphering, but at least it's a valid AIC. This would be a slightly shorter compromise with a kludge node:

(9)r7c5 = (9-2)r3c5 = 2r2c5 - [(4)r2c5 = r2c9 - r9c9 = (4-9)r9c8 = (9-2)r8c8 = r8c3 - r123c3 = (2)r2c2] = (2,87,6)b1p2,r124c3 - 6r145c2 = (6)r7c2 => -9 r7c2

Btw, how would (or probably did) you write yours a matrix? I'm still incapable of writing valid BTMs (and a bit unwilling too), despite your best effort to help me (which I greatly appreciate), so this is what I came up with:

15x15 ?M: Show
Code: Select all
` 3r79c7 8216b9p4716        8r8c8 ..... 8r5c8 8r6c8                          8r6c5 7r6c5                                7r6c9 6r6c9                                      6r8c9 |1r8c9                    1r5c8 ................. |1r5c7 3r7c7 ....................................  1r7c7  6r7c7                                                   |6r8c9 1r8c9                                                   |6r7c2 ..... 9r7c2                                                          1r8c1 9r8c1 6r8c1        2r9c2 ............................................................. 2r8c3                                                                            1r8c3 1r13c3                                                                            2r2c3 ...... 7r2c3                                                                      6r1c1 ..... 1r1c1  7r1c1 5r1c1 5r2c7 ....................................................................................... 5r2c12 =====================================================================================================-3r2c7`

It has the same top entry problem as before (marked with the first set of '|'; the other is just for symmetry), but personally I don't mind. The way I see it is that it's readable both ways using the same exact rules. That way it actually makes both nested AICs quite visible.

--
Btw, if one wants to go overboard with nested AICs, I guess it could combine two branches in the 4-kraken:

Code: Select all
`- (6)r1c1  ||  (5)r1c1 - r2c12 = (5)r2c7  ||  [(1)r8c3 = r13c3 - (1=7)r1c1 - (7=2)r2c3] - 2r8c3 = r9c2 - (2=163)b9p167`

(Of course I'm not saying it should.)

SpAce

Posts: 2568
Joined: 22 May 2017

### Re: Robert's puzzles 2020-09-18

Hi Space,
SpAce wrote:Yours is an interesting solution, but I can't agree that it's simpler in any way. For one, I can't write your second step as a reasonable AIC at all. It's clearly a net, and if properly analyzed, it's actually way more complex than my second step (which I can easily write as an AIC, but I think it's simpler and clearer as a kraken). The least horrible chain representation I could come up with for yours is a memory chain (which is not a valid AIC):

The notion of simplicity is relative to the experience one has of the techniques one applies. For me, that of AICs is not familiar so I thought that my second step could be written as an AIC as easily as the track P(9r7c2) quickly leads to a sequence of 2 quite visible as soon as you have placed the 8r8c7, hence the importance of having previously studied the 8s.
As for multi-kraken, they are for me bifurcations (or) therefore more complex than direct sequences (and).
Thank you for all your explanations.
Cordially
Robert
Mauriès Robert

Posts: 345
Joined: 07 November 2019
Location: France

### Re: Robert's puzzles 2020-09-18

Mauriès Robert wrote:The notion of simplicity is relative to the experience one has of the techniques one applies.

True, but it doesn't mean there aren't objective measures of complexity as well. Those can only be seen and analyzed with a proper notation. Both your and Denis' notations lack the details necessary for such analysis, which makes moves expressed with them appear simpler than they are. Matrices and AICs/Krakens/Nets expressed with Eureka don't lack those details, so they represent the reality much more accurately.

I thought that my second step could be written as an AIC as easily

Like I just said -- when details are omitted, many things can appear simpler than they are.

As for multi-kraken, they are for me bifurcations (or) therefore more complex than direct sequences (and).

It's just a different point of view of the same exact logic, so it's not inherently any more or less complex. If you read a kraken or a multi-kraken backwards you have "AND-logic". Just try it with the one I wrote for your logic -- you'll see that it has your logic precisely when read from right to left. (If seen as pure Boolean logic there's no directionality at all, because ANDs, ORs, and NANDs are all commutative operations.)

That's the beauty of bidirectional (or non-directional) notations -- they're always reversible in that sense. (Not by Denis' definition of reversibility, but I don't use that.) The only time when you don't get to choose between AND/OR points of view is when both are needed going either way, but those are examples of more complex logic. (Ironically, Denis' definition counts such mixed cases as reversible, but not the simpler ones.)

Anything that can be written as a whip or a braid (AND-logic) is a kraken or a multi-kraken (OR-logic) going backwards, and vice versa. Denis' notation just doesn't allow seeing that duality (TDP even less) because it lacks the details necessary for reading them backwards (and forwards, as far as I'm concerned). He does acknowledge the duality itself, though -- just not that they're equivalent points of view.

Why do we normally prefer to write our net moves as krakens/multi-krakens instead of the admittedly more sequential AND-logic? One advantage is that it allows writing and following each branch linearly, though possibly introducing repetition if written that way. That's often easier to read than complex nets requiring join points.

Another benefit is that it produces natural verities. When more than two end points (z-candidates in Denis' parlance) are needed, the only way to write pure AND-logic is to start with the elimination and end up with a contradiction (as whips and braids do). As you know, we don't like contradictions, so we rather flip it around and use OR-logic to produce a verity with multiple end points. The end result is the same.

SpAce

Posts: 2568
Joined: 22 May 2017

### Re: Robert's puzzles 2020-09-18

SpAce wrote:Btw, how would (or probably did) you write yours a matrix?

Hi SpAce,
I missed time to check the validity of my matrices. Now completed. I have added three of them (from compacted to unfolded) in my initial post.
Still short of time, I will add later the sub-matrices proving matrix 3 being actually a BTM (Edit: carried out)

PS my preference is matrix 2.
Cenoman
Cenoman

Posts: 1517
Joined: 21 November 2016
Location: Paris, France

### Re: Robert's puzzles 2020-09-18

Hi Cenoman,

Thank you very much for the matrix demonstration, once again! I agree that 2 and 3 are valid BTMs (and the first one a valid TM, of course). You proved that it can be done here without any special tricks. Nice job!

Personally I'm still inclined to break the rules in this case, because for me it produces a much more natural logic flow. I think there's room for both styles, as long as they have different names to avoid confusion.

Since matrices are more like personal analysis tools than actual presentations, I think it's best to use whatever style works for oneself -- as long as one doesn't claim it to be something that it isn't (like in my case a valid BTM). I can still consider it a fun challenge to try to write valid BTMs, but I don't think I'm going to obsess about it the same way I obsess about valid AICs

PS. One suggestion. I don't know about you and others, but I find such large matrices really difficult to read without any line helpers (like the dots I use). The SISs get spread out so widely that it's almost impossible to see which nodes are in the same rows. I don't know if the dots are the best solution, but I think something is needed.

SpAce

Posts: 2568
Joined: 22 May 2017

### Re: Robert's puzzles 2020-09-18

SpAce wrote:Thank you very much for the matrix demonstration, once again!

PS. One suggestion. I don't know about you and others, but I find such large matrices really difficult to read without any line helpers (like the dots I use). The SISs get spread out so widely that it's almost impossible to see which nodes are in the same rows. I don't know if the dots are the best solution, but I think something is needed.

Something like that ?
Hidden Text: Show
Code: Select all
`3r7c7 1r7c7 6r7c7 .  . 1r8c9 6r8c93r9c7 1r9c7  .  . 2r9c7 .  . 1r8c7 6r8c7 2r8c7 8r8c7 .  .  .  .  .  . 2r9c2  .  . 2r8c3 .  .  .  .  .  .  .  .  .  . 2r2c3 7r2c3 .  .  .  .  .  .  .  .  .  . 1r8c3  .  . 1r13c35r2c7  .  .  .  .  .  .  .  .  .  .  .  .  .  .  5r2c12 .  .  .  .  .  .  .  .  .  .  .  . 7r1c1  1r1c1  5r1c1 6r1c13r7c7  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .   6r7c7   .  .  .  . 1r7c7 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .   6r8c9 1r8c9 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .   6r7c2   .   9r7c2  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .   6r8c1   .   1r8c1 9r8c1 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 1r5c7 1r5c8 .  .  .  .  .  .  .  . 8r8c8  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 8r5c8 8r6c8 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 8r6c5 7r6c5 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 7r6c9 6r6c9 .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  . 1r8c9  .  .  .  .  .  . 6r8c9`

I have kept a density of one dot out of three blanks (easy to do with "replace all") and made some manual re-alignments (maybe dispensable)
Cenoman
Cenoman

Posts: 1517
Joined: 21 November 2016
Location: Paris, France

### Re: Robert's puzzles 2020-09-18

Always!

Something like that ?
...
I have kept a density of one dot out of three blanks (easy to do with "replace all") and made some manual re-alignments (maybe dispensable)

That's a great idea! Works like magic and looks good too. Thanks for the tip! It actually provides column helpers as well, though they're not so crucial.
--

Added. If I flip the upper part of my matrix upside down, wouldn't it result in a valid BTM?

15x15 BTM: Show
Code: Select all
` 3r79c7 8216b9p4716 3r7c7  ........... 1r7c7 ....................... 6r7c7                    1r5c7 1r5c8                     .                    1r8c9 ..... 6r8c9               .                                6r6c9 7r6c9         .                                      7r6c5 8r6c5   .        8r8c8 ........... 8r5c8 ........... 8r6c8   .                                                  6r8c9 1r8c9                                                  6r7c2 ..... 9r7c2                                                        1r8c1 9r8c1 6r8c1        2r9c2 ........................................................... 2r8c3                                                                          1r8c3 1r13c3                                                                          2r2c3 ...... 7r2c3                                                                    6r1c1 ..... 1r1c1  7r1c1 5r1c1 5r2c7 ..................................................................................... 5r2c12 ===================================================================================================-3r2c7`

7x7 TM:

Code: Select all
` 3r79c7 8216b9p4716 3r7c7  ........... 1r7c7                    1r5c7 1r5c8                    1r8c9 ..... 6r8c9                                6r6c9 7r6c9                                      7r6c5 8r6c5        8r8c8 ........... 8r5c8 ........... 8r6c8=================================================-3r2c7 `

10x10 TM

Code: Select all
` 3r79c7 8216b9p4716 3r7c7  ........... 6r7c7                    6r8c9 1r8c9                    6r7c2 ..... 9r7c2                          1r8c1 9r8c1 6r8c1        2r9c2 ............................. 2r8c3                                            1r8c3 1r13c3                                            2r2c3 ...... 7r2c3                                      6r1c1 ..... 1r1c1  7r1c1 5r1c1 5r2c7 ....................................................... 5r2c12 =====================================================================-3r2c7`

SpAce

Posts: 2568
Joined: 22 May 2017