## Takabisha

Post puzzles for others to solve here.

### Takabisha

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`+-------+-------+-------+| . . . | 6 . 1 | . . . || . . 2 | . 3 . | 7 . . || . 8 . | . . . | . 4 . |+-------+-------+-------+| 1 . . | . . 4 | . . 5 || . 7 . | 1 . . | . 9 . || 3 . . | 9 6 . | . . 1 |+-------+-------+-------+| . 6 . | . . . | . 8 . || . . 9 | . 5 . | 2 . . || . . . | 7 . 6 | . . . |+-------+-------+-------+...6.1.....2.3.7...8.....4.1....4..5.7.1...9.3..96...1.6.....8...9.5.2.....7.6...`

estimated rating: 8.3
been a while since i posted! busy lately, but it should be calming down now (っˆwˆς)

shye

Posts: 158
Joined: 12 June 2021

### Re: Takabisha

Code: Select all
`   +---------------------+---------------------+---------------------+   | 4579   345    3457  | 6     *24789  1     |*3589  *235   *28–39 |   | 4569   145    2     | 458    3      89    | 7      156   *689   |   | 5679   8      13567 | 25     79–2   279   | 13569  4     *2369  |   +---------------------+---------------------+---------------------+   | 1      9      68    | 238    7–28   4     | 368    2367   5     |   |*24568  7      456–8 | 1     *28     35–28 | 346–8  9     *23468 |   | 3      245    458   | 9      6      2578  | 48     27     1     |   +---------------------+---------------------+---------------------+   |*2457   6      13457 | 234    149–2  239   | 13459  8      3479  |   |*478    134    9     | 348    5      38    | 2      136    3467  |   |*28–45 *12345 *13458 | 7     *12489  6     | 13459  135    349   |   +---------------------+---------------------+---------------------+`
9 Truths = {28R19 28C19 5N5}
10 Links = {28r5 28c5 28b37 1n9 9n1}
5N5 is covered twice and can therefore be double-counted, which proves the pattern rank0. stte

Marek
marek stefanik

Posts: 250
Joined: 05 May 2021

### Re: Takabisha

.
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`Resolution state after Singles and whips[1]:   +-------------------+-------------------+-------------------+    ! 4579  345   3457  ! 6     24789 1     ! 3589  235   2389  !    ! 4569  145   2     ! 458   3     89    ! 7     156   689   !    ! 5679  8     13567 ! 25    279   279   ! 13569 4     2369  !    +-------------------+-------------------+-------------------+    ! 1     9     68    ! 238   278   4     ! 368   2367  5     !    ! 24568 7     4568  ! 1     28    2358  ! 3468  9     23468 !    ! 3     245   458   ! 9     6     2578  ! 48    27    1     !    +-------------------+-------------------+-------------------+    ! 2457  6     13457 ! 234   1249  239   ! 13459 8     3479  !    ! 478   134   9     ! 348   5     38    ! 2     136   3467  !    ! 2458  12345 13458 ! 7     12489 6     ! 13459 135   349   !    +-------------------+-------------------+-------------------+ 200 candidates.`

1) Resolution path in W4:
Code: Select all
`x-wing-in-rows: n1{r2 r8}{c2 c8} ==> r9c8≠1, r9c2≠1z-chain[3]: r4n7{c5 c8} - r4n2{c8 c4} - r5c5{n2 .} ==> r4c5≠8t-whip[4]: r5c5{n8 n2} - r4c5{n2 n7} - r3c5{n7 n9} - r2c6{n9 .} ==> r1c5≠8, r5c6≠8, r6c6≠8whip[1]: r1n8{c9 .} ==> r2c9≠8finned-x-wing-in-columns: n8{c5 c1}{r5 r9} ==> r9c3≠8whip[1]: c3n8{r6 .} ==> r5c1≠8biv-chain[3]: r2n8{c4 c6} - r8c6{n8 n3} - b5n3{r5c6 r4c4} ==> r4c4≠8singles ==> r5c5=8, r1c9=8, r9c1=8biv-chain[3]: r9n2{c5 c2} - b4n2{r6c2 r5c1} - c9n2{r5 r3} ==> r3c5≠2biv-chain[3]: c9n2{r3 r5} - r6c8{n2 n7} - c6n7{r6 r3} ==> r3c6≠2naked-pairs-in-a-block: b2{r3c5 r3c6}{n7 n9} ==> r2c6≠9, r1c5≠9, r1c5≠7singles ==> r2c6=8, r8c6=3, r4c4=3, r8c4=8whip[1]: b2n7{r3c6 .} ==> r3c1≠7, r3c3≠7whip[1]: b2n9{r3c6 .} ==> r3c1≠9, r3c7≠9, r3c9≠9naked-pairs-in-a-row: r4{c3 c7}{n6 n8} ==> r4c8≠6naked-pairs-in-a-block: b6{r4c8 r6c8}{n2 n7} ==> r5c9≠2stte`

2) There is no 1-step solution in W8 (which is already allowing ridiculously hard patterns for such an easy puzzle)

3) There are several 2-step solutions in W6 (which is also allowing too hard patterns)
They all have the same first step, a whip[6]:
Code: Select all
`whip[6]: r4n2{c5 c8} - r6c8{n2 n7} - c6n7{r6 r3} - r3c5{n7 n9} - c6n9{r3 r7} - c6n2{r7 .} ==> r5c5≠2naked-single ==> r5c5=8whip[1]: r1n8{c9 .} ==> r2c9≠8hidden-single-in-a-column ==> r1c9=8whip[1]: c1n8{r9 .} ==> r9c3≠8hidden-single-in-a-row ==> r9c1=8`

There are several possibilities for the second step. Here are the simplest two:
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`z-chain[5]: r1n2{c5 c8} - r4n2{c8 c4} - b2n2{r3c4 r3c6} - c6n7{r3 r6} - r4c5{n7 .} ==> r9c5≠2stte`

Code: Select all
`whip[5]: b4n2{r5c1 r6c2} - r6c8{n2 n7} - c6n7{r6 r3} - c6n2{r3 r7} - r9n2{c5 .} ==> r5c9≠2stte`
denis_berthier
2010 Supporter

Posts: 3176
Joined: 19 June 2007
Location: Paris

### Re: Takabisha

marek stefanik wrote:9 Truths = {28R19 28C19 5N5}
10 Links = {28r5 28c5 28b37 1n9 9n1}
5N5 is covered twice and can therefore be double-counted, which proves the pattern rank0. stte

exactly my solution, nice!
another way to view it, whatever value is placed in r5c5 places that candidate in the perimeter rows and columns, then a CNL on the other digit appears. r5 and c5 have common elims for 2s and 8s therefore (-28r347c5 and -28r5c367), which is enough for singles

shye

Posts: 158
Joined: 12 June 2021

### Re: Takabisha

This is a self-convincing exercise. I'm sure to understand any logic that I can put into a matrix.
Marek's logic can be put into the following symmetric PM 10x10 (with r5c5 truth repeated, to get the row -column balance)

Code: Select all
`       2c5     2b3   r1c9     8c5    8b3    r9c1   2b7    8b7    2r5    8r5----------------------------------------------------------------------------- 2r1| 2r1c5   2r1c8  2r1c9 8r1|                8r1c9   8r1c5  8r1c7 2r9| 2r9c5                                2r9c1  2r9c2  8r9|                        8r9c5         8r9c1         8r9c3 2c1|                                             2r79c1        2r5c1   8c1|                                                    8r89c1        8r5c1 2c9|         2r13c9                                            2r5c9 8c9|                               8r12c9                             8r5c9r5c5| 2r5c5                  8r5c5                          r5c5|                                                           2r5c5  8r5c5------------------------------------------------------------------------------=>   -2r347c5       -39r1c9 -8r4c5        -45r9c1              -2r5c6 -8r5c367; ste`

(-2r3c5, -28r4c5 are the effective eliminations for ste finish)
Last edited by Cenoman on Mon Aug 30, 2021 10:50 pm, edited 1 time in total.
Cenoman
Cenoman

Posts: 1873
Joined: 21 November 2016
Location: Paris, France

### Re: Takabisha

shye wrote:
marek stefanik wrote:9 Truths = {28R19 28C19 5N5}
10 Links = {28r5 28c5 28b37 1n9 9n1}
5N5 is covered twice and can therefore be double-counted, which proves the pattern rank0. stte

exactly my solution, nice!
another way to view it, whatever value is placed in r5c5 places that candidate in the perimeter rows and columns, then a CNL on the other digit appears. r5 and c5 have common elims for 2s and 8s therefore (-28r347c5 and -28r5c367), which is enough for singles

For those, who - like me - neither understand the XSudo notation nor would search for that without the program, shye's explanation gives the hint to understand it.
This is a long version.
Code: Select all
` *----------------------------------------------------------------------------* |  4579    345     3457    |  6    a24789   1      | b3589   b235   *2389    | |  4569    145     2       |  458   3       89     |  7       156   a689     | |  5679    8       13567   |  25    279     279    |  13569   4     a2369    | |--------------------------+-----------------------+-------------------------| |  1       9       68      |  238   278     4      |  368     2367   5       | | a24568   7       4568    |  1    *28      2358   |  3468    9     b23468   | |  3       245     458     |  9     6       2578   |  48      27     1       | |--------------------------+-----------------------+-------------------------| | b2457    6       13457   |  234   1249    239    |  13459   8      3479    | | b478     134     9       |  348   5       38     |  2       136    3467    | | *2458   a12345  a13458   |  7    b12489   6      |  13459   135    349     | *----------------------------------------------------------------------------*`

The digit x in r5c5 cannot be in r5c1/r9c5 or r1c5/r5c9.
Since both 28 are not elsewhere, it then must be in r789c1 and r9c123, i.e. in r9c1, and in r1c789 and r123c9, i.e. in r1c9. => -39r1c9,-45r9c1.
For the other digit y (one of 28) then you have
r1c5 = r1c78 - r23c9 = r5c9 - r5c1 = r78c1 - r9c23 = r9c5, loop
So one of 28 must be in r5c5 (and r1c9 and r9c1), and the other in one of r19c5 and one of r5c19.
=> -28r347c5,r5c367
eleven

Posts: 2788
Joined: 10 February 2008

### Re: Takabisha

Seeing that many people struggle to get their heads around the pattern, I'll explain the way I found it (it's a bit different from shye's intended way).

The first thing I noticed when looking at the puzzle (after filling in 9r4c2) was the pattern of givens in r19c19b1379.
If no digit were given both in the lines and the boxes, we would get something very similar to an SK-loop (only the givens that usually are in the corners would now be in the minilines in b2468).
Since many digits do appear in both sets, I decided to at least check out 2 and 8 – they appear only in one set, twice each and they're given in the same boxes.

In r1 and c9, one of them can take the intersection, the other must appear at least once in r1c5, r5c9. By symmetry, there is a 2 or 8 in r5c1, r9c5.
Now I looked for ways to limit the number of 28s in these cells and it turned out that r5c5 did exactly that. There cannot be more than two of them, one in r5, one in c5.
This is enough to prove the pattern rank0, so if you just remember the links and carry out the eliminations, you're fine. Worst-case scenario, you forget to do the eliminations in r1c9 and r9c1.

I hope someone will find this explanation helpful and maybe spot the pattern next time if they ever encounter it in another puzzle (which maybe could be called Shellraiser ).

Marek
marek stefanik

Posts: 250
Joined: 05 May 2021