The New Sudoku Players' Forum

by **Myth Jellies** » Sat Apr 15, 2006 7:41 pm

The Molecular Method

I tend to like solving methods which make use of the work and results from previous methods. The Molecular Method uses the work produced by the multiple colors method to feed Alternate Inference Chains. I call it the Molecular Method because it creates freeform diagrams with lines (weak inferences) and double lines (strong inferences) linking up alphanumerics, which look like depictions of large organic molecules.

You start by coloring all the candidates of your grid as completely as possible. As a first example, I am going to take Ruud's April 12 Daily Nightmare.

Code: Select all: *-----------* |..8|6..|..9| |...|.4.|...| |.9.|8.5|..1| |---+---+---| |...|..6|..2| |.39|...|16.| |8..|7..|...| |---+---+---| |7..|9.1|.3.| |...|.2.|...| |4..|..7|9..| *-----------*

Using simple coloring, multiple coloring, and either box-box coloring or finned x-wings, I got here...

Code: Select all: *--------------------------------------------------------------------* | 13 124 8 | 6 137 23 | 2457 2457 9 | | 1236 57 57 | 12 4 9 | 2368 28 368 | | 236 9 24 | 8 37 5 | 23467 247 1 | |----------------------+----------------------+----------------------| | 15 147 147 | 135 1359 6 | 3458 4589 2 | | 25 3 9 | 245 58 48 | 1 6 7 | | 8 1246 1246 | 7 1359 23 | 345 459 345 | |----------------------+----------------------+----------------------| | 7 2568 256 | 9 568 1 | 24568 3 4568 | | 9 1568 1356 | 345 2 48 | 5678 1578 568 | | 4 12568 12356 | 35 3568 7 | 9 1258 568 | *--------------------------------------------------------------------*

Now, at this point, I am expecting that some form of chain is going to be necessary to advance. But I have all of this filtered coloring work...

Code: Select all: *--------------------------* *--------------------------* | 1 1 .| . A1 .| . . .| | . 2 .| . . A2| 2 2 .| |A1 . .|a1 . .| . . .| | 2 . .|a2 . .| 2 2 .| | . . .| . . .| . . 1| | 2 . 2| . . .| 2 2 .| |--------+--------+--------| |--------+--------+--------| | 1 1 1|A1 1 .| . . .| | . . .| . . .| . . 2| | . . .| . . .| 1 . .| |a2 . .|A2 . .| . . .| | . 1 1| . 1 .| . . .| | . 2 2| . . a2| . . .| |--------+--------+--------| |--------+--------+--------| | . . .| . . 1| . . .| | . 2 2| . . .|B2 . .| | . 1 1| . . .| . b1 .| | . . .| . 2 .| . . .| | . 1 1| . . .| . B1 .| | . 2 2| . . .| . b2 .| *--------------------------* *--------------------------* *--------------------------* | 3 . .| . 3 A3| . . .| | 3 . .| . . .| 3 . B3| | 3 . .| . 3 .| 3 . .| |--------+--------+--------| | . . .| 3 3 .| 3 . .| etc. | . 3 .| . . .| . . .| | . . .| . 3 a3| 3 . b3| |--------+--------+--------| | . . .| . . .| . 3 .| | . . d3|D3 . .| . . .| | . . D3| 3 3 .| . . .| *--------------------------*

These "colors" can easily be appended after the appropriate digits in my candidate list...

Code: Select all: *----------------------------------------------------------------------* | 13 124A 8 | 6 1a37b 2A3A | 245d7 245D7 9 | | 1a236a 5a7a 5A7A | 1A2a 4 9 | 2368 28 3B68 | | 236A 9 24a | 8 37B 5 | 2346a7 247 1 | |----------------------+----------------------+------------------------| | 15b 147A 147a | 1a35 1359a 6 | 3458b 458B9A 2 | | 2a5B 3 9 | 2A4B5 58A 4b8a | 1 6 7 | | 8 1246D 1246d | 7 1359A 2a3a | 345 459a 3b4D5 | |----------------------+----------------------+------------------------| | 7 2568 256 | 9 56b8 1 | 2B4D568 3 4d568 | | 9 1568 13d56 | 3D4b5 2 4B8A | 567d8 1b57D8 568 | | 4 12568 123D56 | 35 356B8 7 | 9 1B2b58 568 | *----------------------------------------------------------------------*

Because bivalue cells are so useful in chains, you want to note weak links to other colored cells and parenthetically color the bivalue digit with its conjugate. For example the 1 in r1c1 sees the 1a in the same box. Thus we parenthetically color it 1(A). Do this for all bivalue cell candidates not already colored.

Code: Select all: ------------------------------------------------------------------------ 1(A)3(a) 124A 8 | 6 1a37b 2A3A | 245d7 245D7 9 1a236a 5a7a 5A7A | 1A2a 4 9 | 2368 2(A)8 3B68 236A 9 24a | 8 3(a)7B 5 | 2346a7 247 1 ------------------------+----------------------+------------------------ 1(A)5b 147A 147a | 1a35 1359a 6 | 3458b 458B9A 2 2a5B 3 9 | 2A4B5 5(b)8A 4b8a | 1 6 7 8 1246D 1246d | 7 1359A 2a3a(B)| 345 459a 3b4D5 ------------------------+----------------------+------------------------ 7 2568 256 | 9 56b8 1 | 2B4D568 3 4d568 9 1568 13d56 | 3D4b5 2 4B8A | 567d8 1b57D8 568 4 12568 123D56 | 3(d)5 356B8 7 | 9 1B2b58 568 ------------------------------------------------------------------------

Now you can start creating your molecules. You start with a strong link as your nucleus. I generally like to start with a strong color link that is combined in cells with a lot of other colored candidates. In this case I picked 1a = 1A. At this point you can start a diagram.

Code: Select all: 1a = 1A

Next I noted that r1c5 had both a 1a and a 7b. Thus I could add the following to my diagram...

Code: Select all: 7b = 7B / 1a = 1A

Hey, it is an AIC chain. Either 7B or 1A must be true. Unfortunately, no 1A cells containing a 7 see a 7B cell. Also no 7B cells contain a 1 and see a 1A cell, so no reductions. I noted that r2c1 contained both a 1a and a 6a. Add to the diagram.

Code: Select all: 6A = 6a 7b = 7B \ / 1a = 1A

Again, 1A and/or 6A must be true, but no good reductions there. In r1c1 there is a bivalue cell weakly linked with 1a.

Code: Select all: 6A = 6a 7b = 7B \ / 1a = 1A / 3(a) = 1(A)

This will connect up with the 3A = 3a conjugate colors. r1c6 is a useful cell containing a 3A = 2A inference. Add that to the diagram.

Code: Select all: 6A = 6a 7b = 7B \ / 1a = 1A / 3(a) = 1(A) / 2A = 3A

Now we have an AIC which shows that 2A and/or 1A is true. Cell r2c4 contains both 1A and 2a. We can either attempt a deduction right there, or just add it to the diagram.

Code: Select all: 6A = 6a 7b = 7B \ / 1a = 1A / \ 3(a) = 1(A) 2a = 2A / 2A = 3A

Now it is obvious. We have an AIC chain which shows 2A and/or 2A is true. Thus 2A must be true (and 2a false).

The Molecular Method is probably quite similar to Medusa and other multi-digit coloring methods. It is a slightly more mechanical way of approaching chains which uses coloring work you may have already done to keep track of those pesky bilocation conjugates that can be hard to find when you are working without computer filters to aid you.

by **Ocean** » Sat Apr 15, 2006 9:03 pm

Myth Jellies wrote:The Molecular Method

I tend to like solving methods which make use of the work and results from previous methods. The Molecular Method uses the work produced by the multiple colors method to feed Alternate Inference Chains. ...
You start by coloring all the candidates of your grid as completely as possible.

Interesting... I used an 'almost similar' approach earlier today - on this puzzle:

Code: Select all: ..9.4.7...........4..1.3..6..38.74..2.......1..84.26..9..2.6..3...........7.8.5..

(posted yesterday in the 'Help' section, thread renamed to 'Chain exercise' after solving it.)

1. After a couple of Swordfishes, and then colouring of 7s and 8s, there is a direct bivalue link (7,8) in two of the (doubly) coloured cells, which tells that one of the 7's colours is false.

2. Later (after 40 cells filled), there is again a similar techique: Colouring the 3s, then a short chain from one coloured cell to another of opposite colour, with the implication: "if the first colour is false, then also the second is false - (which is impossible) - ergo, the first colour must be true". (Not sure what to call this last chain ...).

Code: Select all: *-----------* |..9|.4.|7..| |...|...|...| |4..|1.3|..6| |---+---+---| |..3|8.7|4..| |2..|...|..1| |..8|4.2|6..| |---+---+---| |9..|2.6|..3| |...|...|...| |..7|.8.|5..| *-----------*

Code: Select all: At stage 1 (when 24 cells are filled): {13568} {12356} {9} {56} {4} {58} {7} {1235} {258} {135678} {1356} {1256} {5679} {2569} {589} {1239} {13459} {4589} {4} {578} {25} {1} {2579} {3} {289} {589} {6} {156} {1569} {3} {8} {1569} {7} {4} {259} {259} {2} {45679} {456} {3569} {3569} {59} {389} {35789} {1} {157} {159} {8} {4} {1359} {2} {6} {359} {579} {9} {1458} {145} {2} {57} {6} {18} {1478} {3} {13568} {13456} {12456} {3579} {359} {14} {129} {1469} {4789} {136} {12346} {7} {39} {8} {14} {5} {12469} {249} *-----------------------------* | . . .| . . .| . . . | |A7 . .|a7 . .| . . . | | . a7b8 .| . A7 .| . . . | |--------+---------+----------| | . . . | . . .| . . . | | . A7 . | . . .| . a7 . | |a7 . . | . . .| . . A7 | |---------+--------+----------| | . B8 . | . a7 .| . A7 . | |b8 . . |A7 . .| . . a7B8| | . . . | . . .| . . . | *-----------------------------* The 7s are coloured A and a, the 8s are coloured B and b. Only the 'colourable' subset of the digits shown. Since a7 groups with (weak inference) both b8 (in r3c3) and B8 (in r8c9), a7 is false, all a7s can be deleted, and all A7s are true.

Code: Select all: At stage 2 (when 40 cells are filled): {1356} {12356} {9} {56} {4} {58} {7} {125} {258} {7} {156} {56} {569} {2} {589} D{13} d{1345}* {458}* {4} {8} {25} {1} {7} {3} {29} {59} {6} {156} {1569} {3} {8} {19} {7} {4} {259} {259} {2} {7} {4} D{359} {6} {59} d{39} {8} {1} {15} {159} {8} {4} d{139} {2} {6} D{359} {7} {9} {4} {1} {2} {5} {6} {8} {7} {3} {8} d{356} {256} {7} D{39}* {14} {129} {1469} {49}* {36} {236} {7} d{39} {8} {14} {5} {12469} {249} Some of the 3s are coloured D/d. The chain (r8c5-r8c9-r2c9-r2c8) tells that !D=>!d, and combined with the conjugate relationship D<=>!d (from the colour chain), we have D3=true, and d3=false. Ergo r8c5=3 (as are all other cells coloured D).

O.

[Edit: Diagrams added]

by **ronk** » Sat Apr 15, 2006 9:53 pm

Myth Jellies wrote:The Molecular Method
Code: Select all
6A = 6a 7b = 7B \ / 1a = 1A / \ 3(a) = 1(A) 2a = 2A / 2A = 3A

Now it is obvious. We have an AIC chain which shows 2A and/or 2A is true. Thus 2A must be true (and 2a false).

I love the diagram. At first glance, however, the underlying method looks identical to the "vulnerable pair chains" we discussed here.

by **Myth Jellies** » Sat Apr 15, 2006 10:46 pm

I think I've added an extra wrinkle with the bivalue cells, but if you just examine the color exclusions, they certainly are identical. I actually came at it this time from the point of view of an AIC, which I just recently fleshed out. The alternating strong and weak links of an AIC seemed to mesh so naturally with the colors methods, that I was looking for an easy way to combine the two. I probably unconsciously pulled in content from several methods, including yours, in the process. Credit, where credit is due, and you certainly deserve a healthy chunk.

by **Myth Jellies** » Sun Apr 16, 2006 6:05 pm

Tested the Molecular Method on Jeff's practice puzzle for B&B plots...

Code: Select all: *-----------------------------------------------------------* | 1245 3 124 | 478 1589 4579 | 789 68 469 | | 45 7 8 | 2 59 6 | 3 1 49 | | 6 9 14 | 3478 138 47 | 78 5 2 | |-------------------+-------------------+-------------------| | 28 5 7 | 9 23 1 | 4 68 36 | | 1289 6 129 | 34 235 245 | 89 7 139 | | 149 14 3 | 6 7 8 | 5 2 19 | |-------------------+-------------------+-------------------| | 3 8 169 | 5 269 29 | 12 4 7 | | 7 2 5 | 1 4 3 | 6 9 8 | | 149 14 1469 | 78 2689 279 | 12 3 5 | *-----------------------------------------------------------*

...and after coloring and adding the bivalue implications, I ended up with this...

Code: Select all: ---------------------------------------------------------------------------- 12a45a 3 12A4 | 478 1a589 45b79 | 7a89a 6a8a 4a6A9 4a(B)5A 7 8 | 2 5a(B)9B 6 | 3 1 4A9b(Ad) 6 9 1a(B)4(A) | 3a478 1A3A8 4(d)7(ab) | 7A8(A) 5 2 -------------------------+---------------------------+---------------------- 2b(A)8a 5 7 | 9 2B3B(a) 1 | 4 6A8A 3b6a 128A9 6 12a9 | 3A(b)4d 235b 24D5B | 8a9A(d) 7 1d3B9 14b9d 1(d)4B 3 | 6 7 8 | 5 2 1D9D(aB) -------------------------+---------------------------+---------------------- 3 8 1b6b9 | 5 26B9 2(D)9 | 1B2d 4 7 7 2 5 | 1 4 3 | 6 9 8 149 1(B)4b 146B9 | 7b8b 26b8B9 27B9 | 1b2D 3 5 ----------------------------------------------------------------------------

Starting with a bivalue cell that has a lot of weak links works out really well. I started with r3c6 [4(d)=7(ab)] as my nucleus. I wound up making something like this drawing.

Code: Select all: %3a==3A==4d$ \ 3B=2B \ 2b===8a===9A | \ | 9D=1D 7a===7A===8(A) \ / 1(d)=4B %3a=3A 4(d)=7(ab) | \ / | 4b=1(B)% $4d=4D | | | | 1a(B)=4(A) *1B=1b / \ *1B=1b 4A=4a=5A

The '*', '%', and '$' in the diagram point out endpoints of AIC's which show that the 1B, 3a, and 4d colors are true. This doesn't crack the puzzle, but it gets you on your way.

[edited molecule to fix problems pointed out by ronk]

by **Ocean** » Mon Apr 17, 2006 3:54 pm

I observe that the molecule from my previous example (first step) has a different structure. It contains an illegal ring.

Code: Select all: Molecule with 'illegal ring': 7A = 7a / \ 8B = 8b

by **Myth Jellies** » Mon Apr 17, 2006 6:55 pm

Ocean wrote:I observe that the molecule from my previous example (first step) has a different structure. It contains an illegal ring.

Code: Select all
Molecule with 'illegal ring': 7A = 7a / \ 8A = 8b

Cool! Obviously the structure of an unstable substance

by **Carcul** » Tue Apr 18, 2006 9:39 am

Myth Jellies wrote:Cool! Obviously the structure of an unstable substance

That's not quite true. Many stable derivatives of methylenecyclopropene have already been synthetized, specially with electron stabilizing substituents. Cyclopropenones are also very stable compounds (due to some aromaticity), even the unsubstituted molecule (which is even more stable than cyclopropanone). However, the unsubstituted methylene-cyclopropene is probably a too reactive species to be isolated (although I am sure that it can survive at sufficiently low temperatures and be spectroscopically caractherized) and so I agree that we can call it "unstable".

Carcul

by **ronk** » Tue Apr 18, 2006 11:32 am

Myth Jellies wrote:
Code: Select all
%3a==3A==4d$ \ 3B=2B \ 2b===8a===9A | \ | 9D=1D 7a===7A===8(A) \ / 1(d)=4B %3a=3A 4(d)=7(ab) | \ / | 4b=1(B) $4d=4D | | | | 1a(B)=4(A) *1B=1b / \ *1B=1b 4A=4a=5A

[edit: changed to reflect MJ's edit of his post, but ...]

I still suggest a color only appear on the diagram twice IF it is an "endpoint" color. For this diagram, e.g., don't show 1B on the diagram but tag 1b as endpoints instead. Of course, 1b would be exclusion.

by **Myth Jellies** » Wed Apr 19, 2006 2:53 pm

Thanks ronk, I fixed the diagrams.

by **Myth Jellies** » Thu Apr 20, 2006 12:41 am

Carcul wrote:
Myth Jellies wrote:Cool! Obviously the structure of an unstable substance

That's not quite true. Many stable derivatives of methylenecyclopropene have already been synthetized, specially with electron stabilizing substituents. Cyclopropenones are also very stable compounds (due to some aromaticity), even the unsubstituted molecule (which is even more stable than cyclopropanone). However, the unsubstituted methylene-cyclopropene is probably a too reactive species to be isolated (although I am sure that it can survive at sufficiently low temperatures and be spectroscopically caractherized) and so I agree that we can call it "unstable".

Carcul

I did a little surfing after reading your post... Cyclopropenone

Code: Select all: O | | C / \ C = C

is very stable and fairly easy to create. They use it to study bonds under stress. Pretty cool...but completely unstable in a standard sudoku environment.

by **Myth Jellies** » Sat May 13, 2006 9:31 am

This recent Ruud daily nightmare...

Code: Select all: *-----------* |8..|.61|.5.| |.4.|.8.|.32| |...|7..|...| |---+---+---| |.1.|...|5..| |28.|...|.79| |..9|...|.6.| |---+---+---| |...|..7|...| |73.|.1.|.8.| |.9.|65.|..7| *-----------*

reduced to the following using basic methods

Code: Select all: *--------------------------------------------------------------------* | 8 7 3 | 2 6 1 | 9 5 4 | | 16 4 16 |#59 8 #59 | 7 3 2 | | 9 2 5 | 7 *34 *34 | 8 1 6 | |----------------------+----------------------+----------------------| | 346 1 7 | 348 9 23468 | 5 24 38 | | 2 8 46 |#1345 *34 #*3456 | 134 7 9 | | 34 5 9 | 1348 7 2348 | 1234 6 138 | |----------------------+----------------------+----------------------| | 5 6 148 | 38 2 7 | 134 9 13 | | 7 3 2 |#49 1 #49 | 6 8 5 | | 14 9 148 | 6 5 38 | 1234 24 7 | *--------------------------------------------------------------------*

A couple of uniqueness reductions are illustrated. The hashmarks denote a 59-45-49 BUG-Lite+2 pattern which eliminates the 4's from r5c46 using a type 4 reduction. The starred cells are a UR+1 which eliminates 34 from r5c6.

Code: Select all: *--------------------------------------------------------------------* | 8 7 3 | 2 6 1 | 9 5 4 | | 16 4 16 | 59 8 59 | 7 3 2 | | 9 2 5 | 7 34 34 | 8 1 6 | |----------------------+----------------------+----------------------| |*346 1 7 |*348 9 -23468 | 5 24 *38b | | 2 8 46 |#135 34 56 | 134B 7 9 | |*34 5 9 |*1348 7 -2348 | 1234B 6 *138b | |----------------------+----------------------+----------------------| |*5 6 148 |*38A 2 7 | 134 9 *13 | | 7 3 2 | 49 1 49 | 6 8 5 | | 14 9 148 | 6 5 38a | 1234A 24 7 | *--------------------------------------------------------------------*

Next is a finned (r5c4) swordfish (r467c149) on the threes. Either the fin is true or the swordfish; either way, the 3's can be removed from r46c6. Another way you can find this same deduction is to group multicolor the threes as shown above. Either 'a' or 'b' or both must be true. If a is true, then the rest of column 6 is toast. If b is true then the x-wing reductions in rows 4 & 6 would apply. Thus the threes can be removed from r46c6.

Next I took my simply multi-colored grid...

Code: Select all: ---------------------------------------------------------------------------------- 8 7 3 | 2 6 1 | 9 5 4 1a6a 4 1A6A | 5a9a 8 5A9A | 7 3 2 9 2 5 | 7 3a4E 3A4e(F) | 8 1 6 ------------------------+----------------------------+---------------------------- 3b4(Bd)6A 1 7 |-348 9 -2a46a8 | 5 %2A4D(b) 3(B)8b 2 8 4b(AE)6a |*1b35A %3A4e(B) *5a6A |-1B(d)34 7 9 3B4(B) 5 9 | 1B(d)348 7 2A48 | 12a34 6 1D(b)38B ------------------------+----------------------------+---------------------------- 5 6 14a(B)8a | 3A8A 2 7 | 134A(D) 9 1d3(a) 7 3 2 | 4F9A 1 4f(E)9a | 6 8 5 1A4b(AD) 9 148A | 6 5 3a8a | 12A3A4 2a4d(aB) 7 ----------------------------------------------------------------------------------

...and starting from the 2a=4d(AB) nucleus in r9c8, I generate the following enzyme via the molecular method...

Code: Select all: 4b(AD)=1A / / *6A==6a---2a==4d(AB) / \ / 4A(D)=4a(B) %4D(b)=2A | | 8A==8a / %4e(FB)==3A==3a / \ *5A=9A / 4b==4B \ / 9a==4f==4F

%: 4e and/or 4D must be true, thus any cells which see both (such as r4c46 & r5c7) cannot be a 4.
*: 6A and/or 5A must be true, so r5c6 cannot be a 5.

...and the puzzle solved simply from there.

by **Carcul** » Sat May 13, 2006 10:18 am

What an unnecessarily long solution Myth Jellies.

Sometimes, we have just to look more carefully to a grid in order to find a more straightforward (and still simple) solution. Consider the grid again:

Code: Select all: *-----------------------------------------------* | 8 7 3 | 2 6 1 | 9 5 4 | | 16 4 16 | 59 8 59 | 7 3 2 | | 9 2 5 | 7 34 34 | 8 1 6 | |-------------+-----------------+---------------| | 346 1 7 | 348 9 23468 | 5 24 38 | | 2 8 46 | 1345 34 3456 | 134 7 9 | | 34 5 9 | 1348 7 2348 | 1234 6 138 | |-------------+-----------------+---------------| | 5 6 148 | 38 2 7 | 134 9 13 | | 7 3 2 | 49 1 49 | 6 8 5 | | 14 9 148 | 6 5 38 | 1234 24 7 | *-----------------------------------------------*

Consider the following ALSs A = r7c479 and B = r456c4/r5c5, and the UR in cells r35c56. Now we can write

[r9c8]-4-[r7c7]=4|8=[r7c4]-8-[r46c4]=8|5=[r5c4]-5-[r5c6]-6-[r5c3]-4-[r79c3]=4=[r9c1]-4-[r9c8],

which implies r9c8<>4 and the puzzle is solved.

Carcul

by **ravel** » Sat May 13, 2006 4:33 pm

See here for some other solutions for the puzzle.

by **Myth Jellies** » Sat May 13, 2006 10:28 pm

Nice solutions, all. I tend to use uniqueness and single-digit methods first, followed by chains and then ALS (unless the ALS is pretty obvious). I guess I didn't need the finned swordfish or the uniqueness reductions. The molecular method enzyme chain works for the r5c6 <> 5 reduction, which is all you need to crack the puzzle. Funny, but you don't realize how long the chain is until you show it on the grid

Code: Select all: *--------------------------------------------------------------------* | 8 7 3 | 2 6 1 | 9 5 4 | | 16 4 16 | 59 8 K59 | 7 3 2 | | 9 2 5 | 7 34 I34 | 8 1 6 | |----------------------+----------------------+----------------------| | 346 1 7 | 348 9 B23468 | 5 C24 38 | | 2 8 46 | 1345 34 A3456 | 134 7 9 | | 34 5 9 | 1348 7 2348 | 1234 6 138 | |----------------------+----------------------+----------------------| | 5 6 F148 |G38 2 7 |E134 9 13 | | 7 3 2 | 49 1 J49 | 6 8 5 | | 14 9 148 | 6 5 H38 | 1234 D24 7 | *--------------------------------------------------------------------*

A6=B6-B2=C2-D2=D4-E4=F4-F8=G8-G3=H3-I3=I4-J4=J9-K9=K5 :
Chain implies A=6 and/or K=5, therefore A <> 5.

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