The New Sudoku Players' Forum

by **SpAce** » Wed May 22, 2019 4:07 am

Code: Select all: ..697..2.....51..6.7.23..59...427.............41.9327.....4..935...6......2......

One of the hardest puzzles I've seen in a printed publication. Took me two moves but you're free to beat that. More is acceptable too.

by **Cenoman** » Wed May 22, 2019 10:24 pm

Two steps

Code: Select all: +--------------------------+-------------------+-----------------------+ | a138 5 6 | 9 7 4 | A138 2 x18 | | 2349 239 349 | 8 5 1 | 7 B34 6 | | 148 7 48 | 2 3 6 | 148 5 9 | +--------------------------+-------------------+-----------------------+ | 3689 3689 z3589 | 4 2 7 | 13689 1368 y158 | | c23679 2369 d379-5 | C156 C18 C58 | 369 C36 4 | | b68 4 1 | 56 9 3 | 2 7 58 | +--------------------------+-------------------+-----------------------+ | b678 168 78 | 157 4 2 | 1568 9 3 | | 5 1389 34789 | 137 6 89 | 148 148 2 | | 34689 13689 2 | 135 18 589 | 14568 1468 7 | +--------------------------+-------------------+-----------------------+

1. Kraken row (8)r1c179
(8)r1c1 - (86=7)r67c1 - r5c1 = (7)r5c3
(8-3)r1c7 = r2c8 - (3168=5)r5c4568
(8-1)r1c9 = (1-5)r4c9 = (5)r4c3
=>-5r5c3; 13 placements & basics

Code: Select all: +--------------------+------------------+----------------------+ | C13 5 6 | 9 7 4 | 138 2 18 | | 2 B39 349 | 8 5 1 | 7 34 6 | | 14 7 8 | 2 3 6 | 14 5 9 | +--------------------+------------------+----------------------+ | D3-9 6 5 | 4 2 7 | 1389 138 18 | | 7 2 z39 | 15 18 58 | y369 36 4 | | 8 4 1 | 6 9 3 | 2 7 5 | +--------------------+------------------+----------------------+ | 6 18 7 | 15 4 2 | 158 9 3 | | 5 1389 349 | 7 6 89 | 148 148 2 | | a49 A189 2 | 3 18 x589 | y14568 1468 7 | +--------------------+------------------+----------------------+

2. Kraken row (9)r9c126
(9)r9c1
(9)r9c2 - (9=3)r2c2 - r1c1 = (3)r4c1
(9-5)r9c6 = (56-9)r59c7 = (9)r5c3
=>-9r4c1; ste

... but the true challenge is one step

Code: Select all: +--------------------------+-------------------+-----------------------+ | 138 5 6 | 9 7 4 | 138 2 18 | | 2349 239 349 | 8 5 1 | 7 34 6 | | 148 7 48 | 2 3 6 | 148 5 9 | +--------------------------+-------------------+-----------------------+ | 3689 3689 3589 | 4 2 7 | 13689 1368 158 | | 23679 2369 3579 | 156 18 58 | 369 36 4 | | 68 4 1 | 56 9 3 | 2 7 58 | +--------------------------+-------------------+-----------------------+ | 678 168 78 | 157 4 2 | 1568 9 3 | | 5 1389 34789 | 137 6 89 | 148 148 2 | | 34689 13689 2 | 135 18 589 | 14568 1468 7 | +--------------------------+-------------------+-----------------------+

Multi-kraken... No tags. Inferences from each kraken is colored the same, at the last line of the kraken, and at the place where it is chained.
Of course, computer generated...

Kraken column (4)r238c3
(4)r2c3 - (4=3)r2c8 - (3=6)r5c8 - (6)r5c4
(4-8)r3c3 = r13c1 - (86=7)r67c1 - (7=2396)r5c1278 - (6)r5c4
(4)r8c3
=> (6)r5c4 => (4)r8c3

Kraken cell (239)r2c2
(2)r2c2 - r5c2 = (2-7)r5c1 = r5c3 - (7=8)r7c3 - (8=4)r3c3 - (4)r3c7
(3)r2c2 - (3=4)r2c8 - (4)r3c7
(9)r2c2
=> (4)r3c7 => (9)r2c2

Kraken cell (13689)r9c2
(1)r9c2 - (18=9)b8p68 - (9)r9c6
(3)r9c2 - r9c4 = (3-7)r8c4 = r7c4 - (7=8)r7c3 - r3c3 = r13c1 - (8=6)r6c1 - r6c4 = (6-1)r5c4 = r5c5 - (18=9)b8p68 - (9)r9c6
(6)r9c2 - (67=8)r7c13 - r789c2 = r4c2 - (8=6)r6c1 - r6c4 = (6-1)r5c4 = r5c5 - (18=9)b8p68 - (9)r9c6
(9)r9c2 - (9)r9c6
(8)r9c2
=> (9)r9c6 => (8)r9c2

Kraken column (4)r238c3
(4)r2c3 - (4=3)r2c8 - (314=8)r138c7 - (8)r7c7
(4-8)r3c3 = r13c1 - r6c1 = r6c9 - r4c8 = (8)r89c8 - (8)r7c7
(4)r8c3
=> (8)r7c7 => (4)r8c3

Kraken row (9)r9c126
(9)r9c1 - (4)r9c1
(9)r9c6 => (8)r9c2 - r7c123 = r7c7 => (4)r8c3 - (4)r9c1
(9)r9c2
=> - (9)r9c2 => - (4)r9c1

Kraken row (4)r3c137
(4)r3c1 - (4)r9c1
(4-8)r3c3 = r13c1 - (8=6)r6c1 - r6c4 = r5c4 => (4)r8c3 - (4)r9c1
(4)r3c7 => (9)r2c2 - r9c2 => - (4)r9c1
=> -4 r9c1; ste

This could be represented in a BTM (Block Triangular Matrix) 35x35 (count no guaranteed !)

by **SpAce** » Thu May 23, 2019 2:07 am

That's awesome, Cenoman!

Cenoman wrote:Two steps

Even that beats me, because your solution made me notice that I'd screwed up with mine!! It seems that I never had a two-step solution, or a solution at all actually!

1. Kraken row (8)r1c179 =>-5r5c3; 13 placements & basics

Yep, that works nicely. I had a different first step with fewer placements, but for some unknown reason (a misclick probably) I'd ended up with the same resulting grid as yours when I started working on the second step. So, there's a clear gap between my original first and second steps, and I need to change one or the other to get to two working steps (or put something in between them for a three-stepper, which I did for now). An embarrassing but interesting mistake! Glad I at least caught it before someone else.

Here's my original first step anyway:

Code: Select all: .-------------------------.----------------.------------------. | 13-8 5 6 | 9 7 4 | 138 2 18 | | 2349 239 349 | 8 5 1 | 7 34 6 | | 14-8 7 48 | 2 3 6 | 148 5 9 | :-------------------------+----------------+------------------: | 369-8 3689 3589 | 4 2 7 | 13689 1368 158 | | c23679 2369 c3579 | b156 18 b58 | 369 36 4 | | ad(6[8]) 4 1 | a56 9 3 | 2 7 58 | :-------------------------+----------------+------------------: | d(68)7 168 78 | 157 4 2 | 1568 9 3 | | 5 1389 34789 | 137 6 89 | 148 148 2 | | 3469-8 13689 2 | 135 18 589 | 14568 1468 7 | '-------------------------'----------------'------------------'

Step 1: (8=65)r6c14 - r5c46 = (57)r5c31 - (7=68) => -8 r1349c1 (=> 7 placements)

So... after that I'd mistakenly ended up with your grid state and thought I had a two-step solution. Here's a quick fix with an added middle step (until I figure out something else):

Code: Select all: .------------------.---------------.---------------------. | 13 5 6 | 9 7 4 | 138 2 e1(8) | | 2 39 349 | 8 5 1 | 7 e34 6 | | 14 7 8 | 2 3 6 | e14 5 9 | :------------------+---------------+---------------------: | 39 68 359 | 4 2 7 | 13689 1368 158 | | 7 2 359 | c156 18 58 | 369 d36 4 | | a6[8] 4 1 | b56 9 3 | 2 7 5-8 | :------------------+---------------+---------------------: | 68 168 7 | 15 4 2 | 1568 9 3 | | 5 1389 349 | 7 6 89 | 148 148 2 | | 49 1689 2 | 3 18 589 | 14568 1468 7 | '------------------'---------------'---------------------'

Step 2 (added): (8=6)r6c1 - r6c4 = r5c4 - (6=3)r5c8 - (3=418)b3p573 => -8 r6c9 (=> 6 placements)

2. Kraken row (9)r9c126 =>-9r4c1; ste

That also works nicely. My (now third) step for the same grid state (which I hadn't originally earned):

Code: Select all: .--------------------.---------------.-------------------. | 13 5 6 | 9 7 4 | 138 2 18 | | 2 39 349 | 8 5 1 | 7 34 6 | | 14 7 8 | 2 3 6 | 14 5 9 | :--------------------+---------------+-------------------: | b39 6 5 | 4 2 7 | c1389 138 18 | | 7 2 ef(3)9 | 15 18 58 | d369 36 4 | | 8 4 1 | 6 9 3 | 2 7 5 | :--------------------+---------------+-------------------: | 6 18 7 | 15 4 2 | 158 9 3 | | 5 1389 a[4]-39 | 7 6 f8(9) | 148 148 2 | | b49 189 2 | 3 18 f589 | de14568 1468 7 | '--------------------'---------------'-------------------'

Step 3: (4)r8c3 = (49)r94c1 - r4c7 = (96)r59c7 - (9|5)r5c3,r9c7 = (359)r5c3,r98c6 => -39 r8c3; stte

Let's see if I'll bother looking for a real two-step solution after this educating failure

(This wouldn't have happened if I'd kept solving it on paper as planned. I switched to Hodoku once I noticed that the puzzle was much harder than expected, because coloring is so much easier with it.)

[Added. Here's a two-stepper using my original first step and a new second step:

Step 1 (original): (8=65)r6c14 - r5c46 = (57)r5c31 - (7=68) => -8 r1349c1 (=> 7 placements)

Code: Select all: .----------------------.----------------.-------------------. | 1-3 5 6 | 9 7 4 | 138 2 18 | | 2 k(3)9 k349 | 8 5 1 | 7 34 6 | | 14 7 8 | 2 3 6 | 14 5 9 | :----------------------+----------------+-------------------: | al(3)9 d68 l359 | 4 2 7 | c13689 1368 158 | | 7 2 al359 | 156 18 f58 | b369 36 4 | | e68 4 1 | e56 9 3 | 2 7 58 | :----------------------+----------------+-------------------: | d68 168 7 | 15 4 2 | c568 9 3 | | 5 1389 j349 | 7 6 fI89 | 148 148 2 | | 49 1689 2 | 3 18 H589 | G14568 1468 7 | '----------------------'----------------'-------------------'

Step 2 (new): Double Kraken 6c7 & 9c3

Code: Select all: (6)r47c7 - r4c2&r7c1 = (65)r6c14 - (5=89)r58c6 - (93)r2c32 || \ || (3,9)b4p16 = (9-6)r5c7 - (9)r8c3 || / || (6-5)r9c7 = (59)r98c6 -------------------------- (9,3)b4p361 => -3 r1c1; stte

]
[Added 2. Here's another two-stepper with a new first step and my original last step:

Code: Select all: .-----------------------.-----------------.---------------------. | f138 5 6 | 9 7 4 | 138 2 h1(8) | | 2349 239 349 | 8 5 1 | 7 fg34 6 | | f148 7 g48 | 2 3 6 | g148 5 9 | :-----------------------+-----------------+---------------------: | 3689 3689 3589 | 4 2 7 | 13689 1368 158 | | c23679 2369 c3579 | c156 c18 c58 | 369 de36 4 | | ae6[8] 4 1 | b56 9 3 | 2 7 5-8 | :-----------------------+-----------------+---------------------: | de678 168 78 | 157 4 2 | 1568 9 3 | | 5 1389 34789 | 137 6 89 | 148 148 2 | | 34689 13689 2 | 135 18 589 | 14568 1468 7 | '-----------------------'-----------------'---------------------'

Step 1 (new): (8=6)r6c1 - r6c4 = (18657)r5c65431 - (6)r5c8|(7)r7c1 = (3)r5c8,(68)r76c1 - (3)r2c8|(8)r13c1 = (481)r2c8,r3c37 - (1=8)r1c9
=> -8 r6c9 (=> 13 placements)

Step 2 (original): (4)r8c3 = (49)r94c1 - r4c7 = (96)r59c7 - (9|5)r5c3,r9c7 = (359)r5c3,r98c6 => -39 r8c3; stte
]

... but the true challenge is one step

Wow... that's something else! I didn't really believe anyone would come up with a one-stepper, but I'm glad that my lack of faith was misplaced

I can't quite fathom how that was found even with computer help. That's a very interesting and impressive piece of logic! It'll take some time to study it.

by **Leren** » Thu May 23, 2019 3:40 am

Not a solution but a couple of observations.

1. stte opportunities exist at 1 r3c1 & 4 r9c1. I suspect that proving either of these would be problematic (for me).

2. There is a relatively straightforward move that clears the decks quit a bit.

Code: Select all: *------------------------------------------------------* |b138 5 6 | 9 7 4 | 138 2 18 | |b2349 b239 b349 | 8 5 1 | 7 34 6 | |b148 7 c48 | 2 3 6 | 148 5 9 | |-----------------------+-------------+----------------| | 39-68 3689 359-8 | 4 2 7 | 13689 1368 158 | | 2379-6 2369 3579 | 156 18 58 | 369 36 4 | |a68 4 1 | 56 9 3 | 2 7 58 | |-----------------------+-------------+----------------| |a678 168 c78 | 15-7 4 2 | 1568 9 3 | | 5 1389 349-78 | 137 6 89 | 148 148 2 | | 349-68 13689 2 | 135 18 589 | 14568 1468 7 | *------------------------------------------------------*

ALS XY Wing Loop: (7=8) r67c1 - (8=4) r1c1, r2c123, r3c1 - (4=7) r37c3 => - 68 r4c1, - 8 r4c3, - 6 r5c1, - 7 r7c4, - 78 r8c3, - 68 r9c1

And the second step is : haven't found it yet.

Leren

by **SpAce** » Thu May 23, 2019 1:52 pm

Leren wrote:2. There is a relatively straightforward move that clears the decks quit a bit.

ALS XY Wing Loop: (7=8) r67c1 - (8=4) r1c1, r2c123, r3c1 - (4=7) r37c3 => - 68 r4c1, - 8 r4c3, - 6 r5c1, - 7 r7c4, - 78 r8c3, - 68 r9c1

Yes, it's a nice move! Too bad I didn't see it. However, despite the additional immediate eliminations, the end result (after basics) is the same as that of my first step (7 placements). Cenoman's first kraken is still the king with its 13 placements.

A couple of stylistic notes. When writing loops with ALS nodes it's essential to include the bystander digits. In normal chains it's technically correct to skip them (although I don't like that then either), but with loops it's a really bad style. As written, your loop doesn't justify the eliminations resulting from the locked bystanders because they're not listed. I would write it:

(7=6'8)r76c1 - (8=1239'4)b1p17456 - (4=8'7)r37c3 - loop => -6 r459c1, -8 r1349c1,r48c3, -7 r7c4,r8c3

That makes it easy to see which digits are used for the weak links and which ones are the locked bystanders, both kinds having their own eliminations.

I would also simplify it a bit, because the large ALS node in box 1 is an unnecessary complication:

(7=6'8)r76c1 - r13c1 = (8'7)r37c3 - loop => -6 r459c1, -8 r1349c1,r48c3, -7 r7c4,r8c3

Lastly:

(8=4) r1c1, r2c123, r3c1

I'd prefer the bXpY notation because it's both shorter and clearer. The comma notation implies true split-nodes in different houses, but here everything is in the same box. If you insist on doing that, why not simply: "r13c1,r2c123" or "r123c1,r2c23"?

And the second step is : haven't found it yet.

I just did (it's now in the edited version of my previous post). It obviously works for your first step too, assuming I avoided mistakes this time. Please report if you find something else!

by **Wecoc** » Fri May 24, 2019 12:45 am

Code: Select all: +-------------------------------------------------+ | 138 5 6 | 9 7 4 | 138 2 18 | | 2349 239 349 | 8 5 1 | 7 34 6 | | 148 7 48 | 2 3 6 | 148 5 9 | +-------------------------------------------------+ | 3689 3689 3589 | 4 2 7 | 13689 1368 158 | | 23679 2369 3579 | 156 18 58 | 369 36 4 | | 68 4 1 | 56 9 3 | 2 7 58 | +-------------------------------------------------+ | 678 168 78 | 157 4 2 | 1568 9 3 | | 5 1389 34789 | 137 6 89 | 148 148 2 | | 34689 13689 2 | 135 18 589 | 14568 1468 7 | +-------------------------------------------------+

r8c6 (89) can't be 8 because it forces a 3 in r2c8 and a 6 in r9c8, therefore r5c8 would be empty
r1c9 (18) eliminates candidate 8 both in r3c1 and r8c3
r9c6 (58) can only be 5, otherwise results in a 4 both in r3c7 and r8c7
r5c4 (56) can't be 6 because it forces a 1 in r3c7 and an 8 in r6c9, therefore r1c9 would be empty
r1c1 (13) can't be 3 because it forces a 3 in r8c2 and a 4 in r9c1, therefore r8c3 would be empty
From there only singles till the end

by **SpAce** » Fri May 24, 2019 1:04 am

Cenoman wrote:... but the true challenge is one step
Multi-kraken... No tags. Inferences from each kraken is colored the same, at the last line of the kraken, and at the place where it is chained.

Hi Cenoman! Does this network diagram depict your logic accurately?

Code: Select all: @(4)r3c1 @(4)r8c3 || || || (6=34)r52c8 ------- (4)r2c3 || / || (4-8)r3c3 =*= (6)r5c4 || || \ || || (6=3927)r5c8721 -*- (4)r3c3 || || || @(9)r9c1 @(4)r8c3 || || || || (4=87)r37c3 -*- (2)r2c2 || ------ (1)r9c2 (8=413)r138c7 --*-- (4)r2c3 || / || || / || / || (4)r3c7 (9)r2c2 - (9)r9c2 (9=81)b8p68 (8)r9c2 - r7c123 = (8)r7c7 || \ || || / \ || \ || (4=3)r2c8 ----- (3)r2c2 (9)r9c6 - \ --*-- (3)r9c2 (8)r89c8 = r4c8 -*- (4)r3c3 \ \ || \ -*-- (6)r9c2 \ || ---------- (9)r9c2 <=> (4r3c1 | 4r8c3 | 9r9c1) => -4 r9c1; stte ('*' implies an omitted single-threaded chain fragment)

This could be represented in a BTM (Block Triangular Matrix) 35x35 (count no guaranteed !)

I haven't tried to write a matrix, but it's easy to see that it must be (at least) a BTM because there's no binary starting point (all end-points lie in 3-SISs). Perhaps this could be a good exercise for that?

Edit: fixed notational mistakes (two non-alternating links) thanks to Cenoman.

by **SpAce** » Fri May 24, 2019 2:22 am

Wecoc wrote:r8c6 (89) can't be 8 because it forces a 3 in r2c8 and a 6 in r9c8, therefore r5c8 would be empty
r1c9 (18) eliminates candidate 8 both in r3c1 and r8c3
r9c6 (58) can only be 5, otherwise results in a 4 both in r3c7 and r8c7
r5c4 (56) can't be 6 because it forces a 1 in r3c7 and an 8 in r6c9, therefore r1c9 would be empty
r1c1 (13) can't be 3 because it forces a 3 in r8c2 and a 4 in r9c1, therefore r8c3 would be empty
From there only singles till the end

Hi Wecoc! Thanks for participating. Sure, that would work (I assume, I haven't checked except the first step). However, randomly applied and most likely computer-executed T&E is not exactly the kind of solution we're looking for

Could you represent any of those eliminations as a detailed step? I bet not. Your first elimination is not only extremely tedious (and low strategic value) as a T&E step, but practically impossible to prove more elegantly (except by Cenoman). Here's how Hodoku would do it:

Hodoku net image: Show

Hodoku output puke: Show

I haven't tried to check if that even makes sense, but I bet it doesn't as is often the case with complex Hodoku nets (its chains sometimes make weird jumps for no apparent reason). The "UNKNOWN ALS" (I haven't seen that before!) is also a hint that its logic is probably messed up.

I suggest you learn the ways of the Force so you don't have to use brute force

It's much more fun.

by **Cenoman** » Fri May 24, 2019 4:11 pm

SpAce wrote:Hi Cenoman! Does this network diagram depict your logic accurately?

Hidden Text: Show

Hi SpAce, first thank you for your deep reading of my cumbersome net !

I chose to use colors to display the links between the six krakens. But consequently, I could'nt use the html tags "code", "/code".
Here is another presentation with kraken indents, showing the hierarchy.

Hidden Text: Show

And here is the net I had drawn on my paper.

Code: Select all: (8)r9c2 - (8)r7c123 = (8)r7c7 -*- (4)r2c3 || \ || || ---*--- (4)r3c3 || || (1)r9c2 ----- @(4)r8c3 || \ (3)r9c2 --*-- (18=9)b8p68 || / \ (6)r9c2 --*-- - (9)r9c6 || / || (9)r9c2----------- @(9)r9c1 || (9)r9c2 - (9)r2c2 @(4)r3c1 || || (2)r2c2 --*-- (4)r3c7 ---*--- (4)r3c3 || / || / || (3)r2c2 -*- (4-8)r3c3 =*=(6)r5c4 -*- (4)r2c3 || @(4)r8c3

Very similar to yours ! I have adopted your conventions: '*' for omitted single threaded chain fragments, '@' for end-points
Only a slight difference. I find your node '(8)r9c2 =*= (8)r7c7' ambiguous. (8)r9c2 should be weakly linked on its right side and (8)r7c7 should be strongly linked on its left side. '=*=' suggests a derived strong link (the kind I usually write '==') Indeed, (8)r9c2 is strongly linked as an element of the r9c2-SIS and (8)r7c7 is weakly linked twice on its right side. In my own net, I have simply used the full chain (8)r9c2 - (8)r7c123 = (8)r7c7.
You could use the '=>' symbol, correct in this context, but I would not like (8)r9c2 -*= (8)r7c7 !

SpAce wrote:I haven't tried to write a matrix, but it's easy to see that it must be (at least) a BTM because there's no binary starting point (all end-points lie in 3-SISs). Perhaps this could be a good exercise for that?

I agree that it must be a BTM because all end-points lie in 3-SIS. I think it is a good exercise !
When posting my solution I had tried to write a matrix, but I missed time to check if it was actually a BTM. I failed to meet Steve K's last but one requirement:

If there are two top entries in a single row, then they each form a block. Each of these blocks must form triangular matrices.

I have tried again, and unless I made a mistake somewhere, I guess this matrix is a BTM:

Hidden Text: Show

Original matrix
Some lines encompass non native strong links, derived from single threaded chains. These strong links are tagged '=='
Should they be developed, they would add as many rows as columns, containing each two linked elements (strongly or weakly resp.) without changing the rationale below.
The matrix is thus reduced to size 18x18.

Code: Select all: 4r8c3 4r3c3 4r2c3 8r3c3 == 6r5c4 6r5c8 == 4r2c8 6r5c1278 == 8r3c3 4r3c1 4r3c3 4r3c7 4r2c8 3r2c8 4r3c3 == 2r5c2 3r2c2 2r2c2 9r2c2 9r9c1 9r9c2 9r9c6 9b8p68 1b8p68 1r5c5 == 8r6c1 8r13c1 == 3r9c4 8r4c2 == 6r7c13 9r9c2 1r9c2 3r9c2 6r9c2 8r9c2 8r7c123 8r7c7 8r138c7 == 4r2c8 8r89c8 == 8r3c3 4r8c3 4r2c3 4r3c3

Note that matrix first row and last row are the same SIS (4)c3. Deleting last row and placing 4r2c8 and 8r3c3 weak links in columns 4 and 2 resp. would bring the matrix to size 17x16.

To check the BTM requirement If there are two top entries in a single row, then they each form a block. Each of these blocks must form triangular matrices.

let's assume 4r2c3 = True (in first row AND in last row)
The resulting matrix (TM a) is as follows (emptied rows and columns kept for a better comparison to the matrix above):

Code: Select all: 4r8c3 4r3c3 8r3c3 == 6r5c4 == 6r5c1278 == 8r3c3 4r3c1 4r3c3 4r3c7 4r2c8 3r2c8 4r3c3 == 2r5c2 3r2c2 2r2c2 9r2c2 9r9c1 9r9c2 9r9c6 9b8p68 1b8p68 1r5c5 == 8r6c1 8r13c1 == 3r9c4 8r4c2 == 6r7c13 9r9c2 1r9c2 3r9c2 6r9c2 8r9c2 8r7c123 8r7c7 == 8r89c8 == 8r3c3 4r8c3 4r3c3

TM a is a valid triangular matrix 16x16, having 4r9c1 as an elimination.

Now, let's assume 4r3c3 = True (in first row AND in last row)
The resulting matrix (TM b) is as follows (emptied rows and columns kept for a better comparison to the matrix above):

Code: Select all: 4r8c3 4r2c3 == == == 4r2c8 3r2c8 4r2c8 == 2r5c2 3r2c2 2r2c2 9r2c2 9r9c1 9r9c2 9r9c6 9b8p68 1b8p68 1r5c5 == 8r6c1 8r13c1 == 3r9c4 8r4c2 == 6r7c13 9r9c2 1r9c2 3r9c2 6r9c2 8r9c2 8r7c123 8r7c7 8r138c7 == 4r2c8 == 4r8c3 4r2c3

TM b is a valid triangular matrix 13x13, having 4r9c1 as an elimination.

Overall conclusion: the matrix above is a valid BTM demonstrating the elimination of 4r9c1

by **SpAce** » Fri May 24, 2019 5:47 pm

Cenoman wrote:Hi SpAce, first thank you for your deep reading of my cumbersome net !

It was fun! Thanks for providing the source material for it!

I chose to use colors to display the links between the six krakens. But consequently, I could'nt use the html tags "code", "/code".

Yeah, it's too bad that colors can't be used within code blocks. I guess it would require some escape mechanism which would complicate things.

Here is another presentation with kraken indents, showing the hierarchy.

Thanks!

And here is the net I had drawn on my paper.
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Code: Select all
(8)r9c2 - (8)r7c123 = (8)r7c7 -*- (4)r2c3 || \ || || ---*--- (4)r3c3 || || (1)r9c2 ----- @(4)r8c3 || \ (3)r9c2 --*-- (18=9)b8p68 || / \ (6)r9c2 --*-- - (9)r9c6 || / || (9)r9c2----------- @(9)r9c1 || (9)r9c2 - (9)r2c2 @(4)r3c1 || || (2)r2c2 --*-- (4)r3c7 ---*--- (4)r3c3 || / || / || (3)r2c2 -*- (4-8)r3c3 =*=(6)r5c4 -*- (4)r2c3 || @(4)r8c3

Very nice! I agree that the largest SIS is the better starting point. I thought about that afterwards but didn't want to start over.

Only a slight difference. I find your node '(8)r9c2 =*= (8)r7c7' ambiguous.

Yes, of course! Thanks for pointing out that newbie mistake (one way to see it is as a discontinuity in link alternation)!

I don't know how it slipped through. There was a similar one with the double weak link around (8)r89c8. They both should now be corrected (one way -- yours would have been shorter, but I added the missing strongly-linked node because there was one on the other side already).

'=*=' suggests a derived strong link (the kind I usually write '==')

Exactly, and '-*-' is similarly a derived weak link (normally written as '--'). The normal marks would obviously be ambiguous in a net diagram, so I added the '*' for the derived links.

You could use the '=>' symbol, correct in this context, but I would not like (8)r9c2 -*= (8)r7c7 !

I agree on both accounts. However, using the '=>' within Eureka-like chains bothers my sense of aesthetics somehow, so I rather stick to the link markers (standard or derived). I agree that it's logically fully correct, of course.

I agree that it must be a BTM because all end-points lie in 3-SIS. I think it is a good exercise !

Great! I'll try my hand at that later. I don't know what will come of it, but many thanks for providing a reference if (and when) I get lost!

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