## solution needs a lot of AIC's and even a kraken fish

Post the puzzle or solving technique that's causing you trouble and someone will help

### solution needs a lot of AIC's and even a kraken fish help.png (15.05 KiB) Viewed 204 times
Can someone on this forum find help me to find another way of solving this sudoku which is less difficult. Perhaps I didn't saw a logical step and made it to complicated?
I can solve it, so you can ignore the basics. I am interested in different ways to solve it. Other people may have idea's which are not simular to mine.
urhegyi

Posts: 25
Joined: 13 April 2020

### Re: solution needs a lot of AIC's and even a kraken fish

Hi urhegyi,

urhegyi wrote:
help.png
Can someone on this forum find help me to find another way of solving this sudoku which is less difficult. Perhaps I didn't saw a logical step and made it to complicated?

It's not a trivial puzzle, as it probably requires several steps even from a decent player. Not very difficult ones, though. (Disclaimer: I don't have time to solve it manually, so I just used Hodoku to analyze it a bit). Hodoku claims it can solve it in a single step but it requires such a horrible net that I'm not willing to verify it (Hodoku's net solutions aren't always reliable, so I have my doubts). In any case, Hodoku's default path (with my settings) takes 12 steps (of which some are most likely unnecessary, but I haven't checked which ones):

1. 2-String Kite: -1 r4c2
2. 2-String Kite: -8 r3c2
3. AIC: -2 r1c7,r78c9 -> basics, 3 singles
4. AIC: -5 r1c2
5. AIC: -6 r3c6
6. AIC: -1 r6c7
7. AIC: -8 r6c2
8. ALS-AIC: -6 r2c6 (*) -> pointing pair
9. XY-Chain: -8 r5c6,r6c3 -> 5 singles
10. W-Wing: -6 r4c2
11. W-Wing: -6 r6c4 -> 10 singles, pointing pair
12. XY-Wing: -9 r2c2,r9c3; stte
All but one of those steps are short and basic AICs. Only step 8 is a bit complicated because it uses two relatively big ALSs. Still, it's a normal AIC too. (Btw, you have to turn on "Preferences::Steps::Allow ALS in chains" for Hodoku to find it. For some weird reason it's not on by default.)

So, no need for kraken fishes or other complicated techniques. (I haven't seen that a kraken fish is ever needed anyway, and very few human players would use them at all. Normal krakens (which Hodoku calls "Forcing Chain Verity") that use cells, rows, columns, or boxes, are usually simpler and in my experience almost always find the same eliminations.)

(*) Here's the most complicated step 8:

Code: Select all
`.-------------------.--------------------.-------------.| 1  f79     f579   |  4     59   a8  | b68   3  2  || 3  f89(6)  f89(6) |  29    1      28-6 |  7    4  5  || 2   456     4568  |  7     3568   38   | c168  9  16 |:-------------------+--------------------+-------------:| 4   356    e156   | e29   e69    e126  | d135  7  8  || 9   58      2     |  13    7      48   |  45   6  13 || 7   136     168   |  136   48     5    |  34   2  9  |:-------------------+--------------------+-------------:| 6   2       17    |  5     34     134  |  9    8  47 || 8   17      3     |  16    46     9    |  2    5  47 || 5   49      49    |  8     2      7    |  36   1  36 |'-------------------'--------------------'-------------'`

(6=8)r1c6 - r1c7 = (8-1)r3c7 = r4c7 - (1=269'5)r4c6543 - (5=798'6)b1p2356 => -6 r2c6

Or with the last node broken into two pieces:

(6=8)r1c6 - r1c7 = (8-1)r3c7 = r4c7 - (1=2695)r4c6543 - (5=79)r1c23 - (9=86)r2c23 => -6 r2c6

--

I also tried to find a shorter path with krakens (but not nets). A pretty reasonable one takes five steps with 3 simple AICs (one ALS) and 2 relatively simple krakens (cell and row):

1) ALS XY-Wing: (6=8)r1c6 - (8=4)r5c6 - (4=136)r539c9 => -6 r1c9 (-> 3 singles)
2) Kraken Cell (168)r6c3 => -4 r6c5 (-> 2 singles)

Code: Select all
`(1)r6c3 - (1=7)r7c3 - (7=4)r7c9||(6)r6c3 - r6c4 = r8c4 - (6=4)r8c5||(8)r6c3 - (8=4)r6c5=> -4 r6c5`

3) AIC: (5=8)r5c2 - (8=4)r5c6 - (4=1)r7c6 - (1=7)r7c3 - r1c3 = (7)r1c2 => -5 r1c2
4) Kraken Row (5r3) => -9 r1c2 (-> 13 singles)

Code: Select all
`(5)r3c2 - (5=8)r5c2 - (8=4)r5c6 - (4=1)r7c6 - (1=7)r7c3 - r1c3 = (7)r1c2||(5-4)r3c3 = r3c2 - (4=9)r9c2 ||(5)r3c5 - (5=9)r1c5=> -9 r1c2`

5) XY-Chain: (6=9)r2c2 - (9=5)r1c3 - (5=1)r4c3 - (1=6)r6c3 => -6 r23c3,r6c2; stte

--

Does that give you new ideas at all?
-SpAce-: Show
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`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"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: 2514
Joined: 22 May 2017

### Re: solution needs a lot of AIC's and even a kraken fish kraken.png (21.98 KiB) Viewed 167 times
Thank you Space for your tips. Through your explication I saw the kraken fish can be replaced through a normal kraken. What I did were your step 3 and 4 as my first 2 steps. The third was my kraken fish I replaced through this:
urhegyi

Posts: 25
Joined: 13 April 2020

### Re: solution needs a lot of AIC's and even a kraken fish

urhegyi wrote:Thank you SpAce for your tips. Through your explication I saw the kraken fish can be replaced through a normal kraken. What I did were your step 3 and 4 as my first 2 steps. The third was my kraken fish I replaced through this:

No problem. Glad that my tips helped! In fact, your new solution is better than my proposal, as it solves it in the same 5 steps with just one kraken. It can also be easily upgraded to a simple double-kraken to shave off one step. That in turn can be written easily as a nested AIC, if one wishes. Here's my updated path with that and three simple AICs of lengths [3..4] (no ALS or groups needed):

1. AIC: (2=4)r8c7 - r6c7 = (4-8)r6c5 = r3c5 - r3c7 = (8)r1c7 => -2 r1c7 (*)
2. AIC (nested): (5)r13c2 = [(8)r5c2 = r5c6 - r6c5 = (85-4)r3c53 = (4971)r3918c2 - r7c3 = (14-8)r75c6 = (8)r5c2] => -5 r5c2 (**)
3. AIC (M2-Wing): (1=6)r6c3 - r6c4 = (6-1)r8c4 = (1)r8c2 => -1 r46c2,r7c3
4. AIC (H-Wing): (5=9)r1c3 - (9=4)r9c3 - r9c2 = (4)r3c2 => -5 r3c2 (*); stte
(*) Alternate ALS moves for the first and the last step: Show
1) ALS-XZ: (6=84)r1c6 - (4=136)r539c9 => -6 r1c9
4) ALS-Z: (59)b1p35 = (635)r264c2 => -5 r3c2,r4c3; stte

(**) Step 2: uncompressed; as a double-kraken; as a matrix; as two steps: Show
1) Uncompressed:

AIC (nested): (5)r13c2 = [(8)r5c2 = r5c6 - r6c5 = (8-5)r3c5 = (5-4)r3c3 = r3c2 - (4=9)r9c2 - (9=7)r1c2 - (7=1)r8c2 - r7c3 = (1-4)r7c6 = (4-8)r5c6 = (8)r5c2] => -5 r5c2

2) As a Double-Kraken: Row (5)r3c235 & Cell (579)r1c2:

Code: Select all
`(5)r3c2||                             (5)r1c2||                             ||(5-4)r3c3 = r3c2 - (4=9)r9c2 - (9)r1c2||                             ||||                             (7)r1c2 - (7=1)r8c2 - r6c3 = (1-4)r7c6 = (4-8)r5c6 = (8)r5c2 ||(5-8)r3c5 = r6c5 - r5c6 = (8)r5c2 => -5 r5c2`

3) As a 9x9 TM (triangular matrix):

Code: Select all
` 8r5c2 8r5c6       8r6c5 8r3c5 5r3c2 ..... 5r3c5 5r3c3                   4r3c3 4r3c2                         4r9c2 9r9c2 5r1c2 ....................... 9r1c2 7r1c2                                     7r8c2 1r8c2                                           1r6c3 1r7c6       4r5c6 ................................... 4r4c7======================================================-5r5c2`

4) As two steps:

2a) AIC: (5=9)r1c5 - (9=6)r4c5 - (6=4)r8c5 - (4=7)r8c9 - r8c2 = (7)r1c2 => -5 r1c2
2b) Kraken Row (5)r3c235:

Code: Select all
`(5)r3c2||(5-4)r3c3 = r3c2 - (4=9)r9c2 - (9=7)r1c2 - (7=1)r8c2 - r7c3 = (1-4)r7c6 = (4-8)r5c6 = (8)r5c2||(5-8)r3c5 = r6c5 - r5c6 = (8)r5c2=> -5 r5c2`

(Using these two steps makes it a 5-step solution, of course.)

PS. Next time it would be nice if you posted the puzzle string like this, so it doesn't have to be manually entered into a solver:

Code: Select all
`1..4.....3...1.7.5.......9.4.......89.2.7..6......5.296......8.8.3..9.......27...`

(You can get it from Hodoku::Edit::Copy Givens.) SpAce

Posts: 2514
Joined: 22 May 2017

### Re: solution needs a lot of AIC's and even a kraken fish

urhegyi wrote:
help.png
Can someone on this forum find help me to find another way of solving this sudoku which is less difficult. Perhaps I didn't saw a logical step and made it to complicated?
I can solve it, so you can ignore the basics. I am interested in different ways to solve it. Other people may have idea's which are not simular to mine.

A solution with no chain with length longer than 5:

Hidden Text: Show
singles
whip: r9n6{c9 .} ==> r8c9 ≠ 6, r8c7 ≠ 6
whip: r9n3{c9 .} ==> r7c9 ≠ 3, r7c7 ≠ 3
whip: c4n3{r6 .} ==> r6c5 ≠ 3, r4c5 ≠ 3, r4c6 ≠ 3, r5c6 ≠ 3
whip: r2n2{c6 .} ==> r1c6 ≠ 2
naked-pairs-in-a-row: r9{c2 c3}{n4 n9} ==> r9c9 ≠ 4, r9c7 ≠ 9, r9c7 ≠ 4
hidden-single-in-a-block ==> r7c7 = 9
whip: r9n4{c3 .} ==> r7c2 ≠ 4, r7c3 ≠ 4, r8c2 ≠ 4
hidden-pairs-in-a-block: b5{r5c6 r6c5}{n4 n8} ==> r6c5 ≠ 6, r5c6 ≠ 1
hidden-pairs-in-a-column: c4{n2 n9}{r2 r4} ==> r4c4 ≠ 6, r4c4 ≠ 3, r4c4 ≠ 1, r2c4 ≠ 6
finned-x-wing-in-columns: n1{c6 c3}{r7 r4} ==> r4c2 ≠ 1
finned-x-wing-in-rows: n8{r5 r2}{c6 c2} ==> r3c2 ≠ 8, r1c2 ≠ 8
naked-triplets-in-a-row: r1{c6 c7 c9}{n6 n8 n2} ==> r1c5 ≠ 8, r1c5 ≠ 6, r1c3 ≠ 8, r1c3 ≠ 6, r1c2 ≠ 6
biv-chain: r5c4{n3 n1} - b8n1{r8c4 r7c6} - c6n4{r7 r5} - r5n8{c6 c2} ==> r5c2 ≠ 3
biv-chain: r8c7{n2 n4} - r6n4{c7 c5} - c5n8{r6 r3} - b3n8{r3c7 r1c7} ==> r1c7 ≠ 2
singles ==> r1c9 = 2, r8c7 = 2, r7c2 = 2, r5c9 ≠ 4
naked-pairs-in-a-row: r5{c4 c9}{n1 n3} ==> r5c7 ≠ 3, r5c7 ≠ 1, r5c2 ≠ 1
biv-chain: r5n8{c2 c6} - c6n4{r5 r7} - r7n1{c6 c3} - c2n1{r8 r6} ==> r6c2 ≠ 8
biv-chain: r6n4{c7 c5} - c6n4{r5 r7} - r7n1{c6 c3} - c2n1{r8 r6} ==> r6c7 ≠ 1
biv-chain: c6n3{r3 r7} - c6n1{r7 r4} - c7n1{r4 r3} - r3c9{n1 n6} ==> r3c6 ≠ 6
z-chain: c2n3{r4 r6} - c2n1{r6 r8} - r8c4{n1 n6} - r6n6{c4 .} ==> r4c2 ≠ 6
biv-chain: c2n7{r1 r8} - r8c9{n7 n4} - r8c5{n4 n6} - r4c5{n6 n9} - r1c5{n9 n5} ==> r1c2 ≠ 5
z-chain: r3n4{c3 c2} - b1n5{r3c2 r1c3} - r4c3{n5 n1} - b6n1{r4c7 r5c9} - r3c9{n1 .} ==> r3c3 ≠ 6
t-whip: r3c9{n6 n1} - c7n1{r3 r4} - r4n3{c7 c2} - r4n5{c2 c3} - c2n5{r5 .} ==> r3c2 ≠ 6
whip: b1n6{r2c3 .} ==> r2c6 ≠ 6
biv-chain: b2n9{r2c4 r1c5} - r1n5{c5 c3} - r3c2{n5 n4} - r9c2{n4 n9} ==> r2c2 ≠ 9
biv-chain: c6n4{r5 r7} - b8n1{r7c6 r8c4} - c2n1{r8 r6} - c2n6{r6 r2} - c2n8{r2 r5} ==> r5c6 ≠ 8
singles ==> r5c6 = 4, r5c7 = 5, r5c2 = 8, r2c2 = 6, r6c5 = 8, r6c7 = 4
biv-chain: c2n1{r8 r6} - r6n3{c2 c4} - c4n6{r6 r8} ==> r8c4 ≠ 1
stte
denis_berthier
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