## The Molecular Method

Advanced methods and approaches for solving Sudoku puzzles

### The Molecular Method

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.
Myth Jellies

Posts: 593
Joined: 19 September 2005

### Re: The Molecular Method

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.

Last edited by Ocean on Sun Apr 16, 2006 12:05 pm, edited 1 time in total.
Ocean

Posts: 442
Joined: 29 August 2005

### Re: The Molecular Method

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.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

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.
Myth Jellies

Posts: 593
Joined: 19 September 2005

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]
Last edited by Myth Jellies on Wed Apr 19, 2006 10:51 am, edited 1 time in total.
Myth Jellies

Posts: 593
Joined: 19 September 2005

### Ring Molecules

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
Last edited by Ocean on Mon Apr 17, 2006 4:20 pm, edited 1 time in total.
Ocean

Posts: 442
Joined: 29 August 2005

### Re: Ring Molecules

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
Myth Jellies

Posts: 593
Joined: 19 September 2005

### Organic Chemistry Elucidation

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
Carcul

Posts: 724
Joined: 04 November 2005

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.
Last edited by ronk on Wed Apr 19, 2006 2:09 pm, edited 1 time in total.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Thanks ronk, I fixed the diagrams.
Myth Jellies

Posts: 593
Joined: 19 September 2005

### Re: Organic Chemistry Elucidation

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

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.
Myth Jellies

Posts: 593
Joined: 19 September 2005

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
/
/
*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.
Myth Jellies

Posts: 593
Joined: 19 September 2005

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
Carcul

Posts: 724
Joined: 04 November 2005

See here for some other solutions for the puzzle.
ravel

Posts: 998
Joined: 21 February 2006

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.
Myth Jellies

Posts: 593
Joined: 19 September 2005

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