The New Sudoku Players' Forum

[urhegyi]

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87..3...9..4....2.9.....8..3.9..2.....2..69.4.8..13.6.6....1....9....4......591..

Perhaps symmetry can reduce the number of steps.

[jco]

Two steps. Nice group nodes in this puzzle!

After basics

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.-------------------------------------------------------------.
| 8     7   e156   | 2      3    *45   | 56 fa*45(1)   9      |
| 15    135  4     | 1678   9     578  | 3567  2       1357   |
| 9     2   d1356  | 167   c67  b*45(7)| 8  ba*45(17)  1357   |
|------------------+-------------------+----------------------|
| 3     6    9     | 457    47    2    | 57    8-1     18     |
| 157   15   2     | 578    78    6    | 9     3       4      |
| 4     8    57    | 9      1     3    | 257   6       257    |
|------------------+-------------------+----------------------|
| 6     345  3578  | 3478   2478  1    | 2357  9       23578  |
| 1257  9    13578 | 3678   2678  78   | 4     578     235678 |
| 27    34   378   | 34678  5     9    | 1     78      23678  |
'-------------------------------------------------------------'

1. UR(45) r13c68 using internals

(1)r13c8 == (7)r3c68-(7=6)r3c5-r3c3=(6-1)r1c3=(1) r1c8 => -1 r4c8 (& 5 placements)

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.---------------------------------------------------.
| 8    7   e156  | 2     3     45  |i56    14  9    |
|f15  f135  4    |g1678  9     578 |h3567  2   357  |
| 9    2   e1356 | 167   67    457 | 8     14  357  |
|----------------+-----------------+----------------|
| 3    6    9    | 457   47    2   |j57    8   1    |
| 157  15   2    | 578   78    6   | 9     3   4    |
| 4    8   a7-5  | 9     1     3   |j257   6  k5-7  |
|----------------+-----------------+----------------|
| 6    345 b3578 | 3478  2478  1   | 23    9   238  |
|c17   9  db1378 | 3678  2678  78  | 4     5   2368 |
| 2    34   38   | 3468  5     9   | 1     7   368  |
'---------------------------------------------------'

2. (7)r6c3=r78c3-(7=1)r8c1-r8c3=r13c3-r2c12=(1-6)r2c4=r2c7-(6=5)r1c7-r46c7=(5)r6c9
=> -5 r6c3, -7 r6c9; ste

JCO

Edit: corrected writing of eliminations for move 2 spotted by urhegyi (thanks!)

[urhegyi]

Thank you for having a look at my puzzle, but about your second step I would rather conclude that R6C3<>5 and R6C9<>7.

[jco]

Hello urhegyi,

Thank you for having a look at my puzzle, but about your second step I would rather conclude that R6C3<>5 and R6C9<>7.

Yes, sure. Thanks for mentioning. Corrected.

Regards,

[jco]

Hello,

Once (Nov 26, 2020, in: It's a Trap!) I mentioned to SpAce that solutions given explicitly
with coloring were not longer seen in the puzzle section.
This puzzle has nice group nodes available, so we have an opportunity to show just one
key elimination using Medusa coloring with 3 pairs of colors. This is for Step 2 in my previous post.
Looking at the grid, we notice the well-developed pattern for the coloured 1s (key for this elimination).
More colors could be been displayed for other digits (even for 1s) but that would not help.

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    .-----------------------------------------------------------.
    | 8      7    {:1}56  | 2     3     45  |-5+6    14  9      |
    |{.1}5 {.1}35   4     |:1+678 9     578 | 35-67  2   3{'5}7 |
    | 9      2    {:1}356 |.167   67    457 | 8      14  3{'5}7 |
    |---------------------+-----------------+-------------------|
    | 3      6     9      | 457   47    2   | {'5}7  8   1      |
    | 157    15    2      | 578   78    6   | 9      3   4      |
    | 4      8    .75     | 9     1     3   | 2{'5}7 6   2"57   |
    |---------------------+-----------------+-------------------|
    | 6      345   35{:7}8| 3478  2478  1   | 23     9   238    |
    |:1.7    9    .13{:7}8| 3678  2678  78  | 4      5   2368   |
    | 2      34    38     | 3468  5     9   | 1      7   368    |
    '-----------------------------------------------------------'

(',"), (.,:),(-,+) are three pairs of colors with opposite parities each.
"-" denotes a color (and not that a certain digit will be eliminated) and that
"+" denotes the opposite color of "-" (and not that a certain digit will be placed).
Notice that in the above grid we have grouped nodes colored using { }.
For instance, {:7}r7c3, {:7}r8c3 means that the group (7)r78c3 has color with symbol
double dot, and no color has been given to (7)r7c4 and (7)r8c3.
Due to cell r2c4, we cannot have both colors {+,:} true, so at least one of {-,.} must be true.
In column 7, we have (-5)r1c7 and ('5)r4c7, so we cannot have both colors {-,'} true.
This implies that at least one of the colors {+,"} must be true.
So, at least one of {-,.} and at least one of {+,"} must be true.
Since {+,-} have opposite parities, we infer that at least one of {.,"} must be true.
Now, (7)r6c9 sees ("5)r6c9 and (.7)r6c3, so it must be false.
Similarly, (5)r6c3 sees ("5)r6c9 and (.7)r6c3, so it must be false.
After these eliminations, it remains only singles to the end.

Regards,
JCO

Edit: corrected grid display and text.

[denis_berthier]

.

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Resolution state after Singles:
   +----------------------+----------------------+----------------------+
   ! 8      7      156    ! 2      3      45     ! 56     145    9      !
   ! 15     135    4      ! 1678   9      578    ! 3567   2      1357   !
   ! 9      2      1356   ! 167    67     457    ! 8      1457   1357   !
   +----------------------+----------------------+----------------------+
   ! 3      6      9      ! 457    47     2      ! 57     1578   1578   !
   ! 157    15     2      ! 578    78     6      ! 9      3      4      !
   ! 4      8      57     ! 9      1      3      ! 257    6      257    !
   +----------------------+----------------------+----------------------+
   ! 6      345    3578   ! 3478   2478   1      ! 2357   9      23578  !
   ! 1257   9      13578  ! 3678   2678   78     ! 4      578    235678 !
   ! 27     34     378    ! 34678  5      9      ! 1      78     23678  !
   +----------------------+----------------------+----------------------+

1-step solution:
whip[15]: r2c1{n1 n5} - r5c1{n5 n7} - r5n1{c1 c2} - r2c2{n1 n3} - r3n3{c3 c9} - c7n3{r2 r7} - c7n2{r7 r6} - r6n7{c7 c9} - r4c7{n7 n5} - r1c7{n5 n6} - r2n6{c7 c4} - c5n6{r3 r8} - r8n2{c5 c9} - c9n5{r8 r7} - c2n5{r7 .} ==> r8c1 ≠ 1
stte

Normal, much easier solution: only very short chains
hidden-pairs-in-a-row: r4{n1 n8}{c8 c9} ==> r4c9 ≠ 7, r4c9 ≠ 5, r4c8 ≠ 7, r4c8 ≠ 5
finned-x-wing-in-columns: n3{c7 c2}{r2 r7} ==> r7c3 ≠ 3
biv-chain[3]: r1c7{n5 n6} - b1n6{r1c3 r3c3} - r3n3{c3 c9} ==> r3c9 ≠ 5
biv-chain[3]: r1n1{c8 c3} - b1n6{r1c3 r3c3} - r3n3{c3 c9} ==> r3c9 ≠ 1
biv-chain[3]: r3c9{n3 n7} - r3c5{n7 n6} - r2n6{c4 c7} ==> r2c7 ≠ 3
singles ==> r7c7 = 3, r6c7 = 2
finned-x-wing-in-rows: n5{r6 r7}{c9 c3} ==> r8c3 ≠ 5
whip[2]: r6n5{c9 c3} - b1n5{r1c3 .} ==> r2c9 ≠ 5
biv-chain[3]: c7n7{r4 r2} - r2n6{c7 c4} - r3c5{n6 n7} ==> r4c5 ≠ 7
naked-single ==> r4c5 = 4
whip[3]: r8c6{n7 n8} - c5n8{r8 r5} - b5n7{r5c5 .} ==> r7c4 ≠ 7
whip[3]: r8c6{n7 n8} - c5n8{r8 r5} - b5n7{r5c5 .} ==> r8c4 ≠ 7
whip[3]: r8c6{n7 n8} - c5n8{r8 r5} - b5n7{r5c5 .} ==> r9c4 ≠ 7
biv-chain[4]: r3c5{n7 n6} - b1n6{r3c3 r1c3} - r1n1{c3 c8} - b3n4{r1c8 r3c8} ==> r3c8 ≠ 7
whip[1]: c8n7{r9 .} ==> r7c9 ≠ 7, r8c9 ≠ 7, r9c9 ≠ 7
biv-chain[2]: r7n7{c5 c3} - b4n7{r6c3 r5c1} ==> r5c5 ≠ 7
naked-single ==> r5c5 = 8
whip[1]: b5n7{r5c4 .} ==> r2c4 ≠ 7, r3c4 ≠ 7
biv-chain[3]: r9c1{n2 n7} - r7n7{c3 c5} - r7n2{c5 c9} ==> r9c9 ≠ 2
hidden-single-in-a-row ==> r9c1 = 2
biv-chain[3]: r8n2{c9 c5} - r7c5{n2 n7} - r8c6{n7 n8} ==> r8c9 ≠ 8
biv-chain[3]: r3c4{n1 n6} - c3n6{r3 r1} - r1n1{c3 c8} ==> r3c8 ≠ 1
biv-chain[4]: c3n6{r1 r3} - r3c5{n6 n7} - r7n7{c5 c3} - r6c3{n7 n5} ==> r1c3 ≠ 5
biv-chain[3]: r2c1{n5 n1} - r1c3{n1 n6} - b3n6{r1c7 r2c7} ==> r2c7 ≠ 5
biv-chain[4]: r7n7{c3 c5} - r8c6{n7 n8} - r7c4{n8 n4} - r7c2{n4 n5} ==> r7c3 ≠ 5
finned-x-wing-in-rows: n5{r7 r6}{c9 c2} ==> r5c2 ≠ 5
naked-single ==> r5c2 = 1
biv-chain[3]: r3n3{c9 c3} - c3n5{r3 r6} - r6n7{c3 c9} ==> r3c9 ≠ 7
singles ==> r3c9 = 3, r2c2 = 3, r9c2 = 4, r7c2 = 5, r7c4 = 4
whip[1]: r3n7{c6 .} ==> r2c6 ≠ 7
biv-chain[3]: r7n8{c9 c3} - r7n7{c3 c5} - r8c6{n7 n8} ==> r8c8 ≠ 8
biv-chain[3]: r3c8{n4 n5} - r8c8{n5 n7} - c6n7{r8 r3} ==> r3c6 ≠ 4
singles ==> r1c6 = 4, r3c8 = 4
biv-chain[3]: r6c3{n7 n5} - r3n5{c3 c6} - c6n7{r3 r8} ==> r8c3 ≠ 7
biv-chain[4]: r2c1{n1 n5} - b4n5{r5c1 r6c3} - r6c9{n5 n7} - r2c9{n7 n1} ==> r2c4 ≠ 1
hidden-single-in-a-block ==> r3c4 = 1
biv-chain[4]: r8c1{n7 n1} - r2c1{n1 n5} - r3c3{n5 n6} - c5n6{r3 r8} ==> r8c5 ≠ 7
biv-chain[4]: b8n7{r7c5 r8c6} - r8c8{n7 n5} - c9n5{r8 r6} - r6n7{c9 c3} ==> r7c3 ≠ 7
stte