The New Sudoku Players' Forum

[jovi_al01]

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54.....69
7..1....5
....3....
.2...4...
..6...4..
...9...2.
....5....
8....1..6
47.....38

54.....697..1....5....3.....2...4.....6...4.....9...2.....5....8....1..647.....38

this puzzle probably will not be interesting to most of you, but i'd like to test the waters here on this forum a little bit.
for my fellow manual solvers, i hope you find this to be an enjoyable little treat!

[Cenoman]

Don't know if this is your expectation, I found it funny:

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 +----------------------+-------------------------+----------------------+
 |  5     4      13*    |  278     278    278     |  13*    6      9     |
 |  7     369    2389   |  1       469    69      |  238    48     5     |
 |  26*   169    89-12  |  456     3      569     |  78-12  1478   24*   |
 +----------------------+-------------------------+----------------------+
 |  139   2      7      |  3568    168    4       |  5689   589    13    |
 |  139   58     6      |  23578   1278   23578   |  4      589    137   |
 |  13    58     4      |  9       1678   35678   |  568    2      137   |
 +----------------------+-------------------------+----------------------+
 |  26*   1369   3-129  |  3478    5      378     |  79-12  1479   24*   |
 |  8     39     2359   |  347     47     1       |  2579   4579   6     |
 |  4     7      15*    |  26      269    269     |  15*    3      8     |
 +----------------------+-------------------------+----------------------+

Cross X-Wings (1)r19\c37, (2)c19\r37 => -12 r37c37, enabling Y-Wing (9=8)r3c3 - (8=7)r3c7 - (7=9)r7c7 => -9r7c3; ste

[jovi_al01]

Don't know if this is your expectation, I found it funny:

looks exactly right to me!

[denis_berthier]

.

Code: Select all

Resolution state after Singles and whips[1]:
   +----------------------+----------------------+----------------------+
   ! 5      4      1238   ! 278    278    278    ! 12378  6      9      !
   ! 7      369    2389   ! 1      24689  2689   ! 238    48     5      !
   ! 1269   169    1289   ! 245678 3      256789 ! 1278   1478   1247   !
   +----------------------+----------------------+----------------------+
   ! 139    2      7      ! 3568   168    4      ! 15689  1589   13     !
   ! 139    13589  6      ! 23578  1278   23578  ! 4      15789  137    !
   ! 13     1358   4      ! 9      1678   35678  ! 15678  2      137    !
   +----------------------+----------------------+----------------------+
   ! 12369  1369   1239   ! 23478  5      23789  ! 1279   1479   1247   !
   ! 8      39     2359   ! 2347   2479   1      ! 2579   4579   6      !
   ! 4      7      1259   ! 26     269    269    ! 1259   3      8      !
   +----------------------+----------------------+----------------------+
221 candidates.

Notice that this is a lot of candidates, so that we might expect the puzzle to be hard. But we know that the increasing relation between SER and the number of candidates is only statistical and it has a very large variance.
Indeed, SER = 4.2 and the puzzle can be solved in BC3, using the simplest first strategy:

Code: Select all

hidden-pairs-in-a-row: r9{n1 n5}{c3 c7} ==> r9c7≠9, r9c7≠2, r9c3≠9, r9c3≠2
whip[1]: r9n2{c6 .} ==> r7c4≠2, r7c6≠2, r8c4≠2, r8c5≠2
whip[1]: r9n9{c6 .} ==> r7c6≠9, r8c5≠9
hidden-pairs-in-a-column: c2{n5 n8}{r5 r6} ==> r6c2≠3, r6c2≠1, r5c2≠9, r5c2≠3, r5c2≠1
whip[1]: b4n1{r6c1 .} ==> r3c1≠1, r7c1≠1
whip[1]: b4n3{r6c1 .} ==> r7c1≠3
whip[1]: b4n9{r5c1 .} ==> r3c1≠9, r7c1≠9
hidden-pairs-in-a-column: c9{n2 n4}{r3 r7} ==> r7c9≠7, r7c9≠1, r3c9≠7, r3c9≠1
whip[1]: c9n1{r6 .} ==> r4c7≠1, r4c8≠1, r5c8≠1, r6c7≠1
whip[1]: c9n7{r6 .} ==> r5c8≠7, r6c7≠7
hidden-pairs-in-a-row: r1{n1 n3}{c3 c7} ==> r1c7≠8, r1c7≠7, r1c7≠2, r1c3≠8, r1c3≠2
whip[1]: r1n2{c6 .} ==> r2c5≠2, r2c6≠2, r3c4≠2, r3c6≠2
whip[1]: r1n8{c6 .} ==> r2c5≠8, r2c6≠8, r3c4≠8, r3c6≠8
whip[1]: b3n7{r3c8 .} ==> r3c4≠7, r3c6≠7
x-wing-in-columns: n1{c2 c8}{r3 r7} ==> r7c7≠1, r7c3≠1, r3c7≠1, r3c3≠1
x-wing-in-columns: n2{c1 c9}{r3 r7} ==> r7c7≠2, r7c3≠2, r3c7≠2, r3c3≠2
naked-pairs-in-a-block: b7{r7c3 r8c2}{n3 n9} ==> r8c3≠9, r8c3≠3, r7c2≠9, r7c2≠3
biv-chain[3]: r2c8{n8 n4} - r3c9{n4 n2} - b1n2{r3c1 r2c3} ==> r2c3≠8
stte

which is 1-step for people who consider Subsets[2] as "basics".