Тёмный

The New York Times hard sudoku from May 25, 2024. 

NMalteC
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You can try this puzzle here:
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13 окт 2024

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Комментарии : 7   
@DavidShatz-hw6jl
@DavidShatz-hw6jl 4 месяца назад
Box 5 was analyzed cleverly, creating several blocks! Thence, C7 digits were unearthed and the solve went smoothly! Nicely done and enjoy the weekend!
@pvuor
@pvuor 4 месяца назад
This took some work. Clearing out block 5 helped, but still, one had to push this around to open it.
@AnonimityAssured
@AnonimityAssured 4 месяца назад
When I saw only 21 givens, I was expecting something good, and I wasn't disappointed. The hidden singles and elusive 5's made this fun to solve (76 = row 7 column 6): 76, 86, 36, 63, 44, 95, 85, 99, 94, 75, 24, 28, 97, 79, 87, 58, 37, 47, 67, 54, 64, 55, 45, 39, 29, 31, 93, 41, 32, 61, 83, 98, 89, 48, 38, 19, 68, 49, 59, 42, 73, 72, 81, 23, 35, 52, 51, 92, 91, 12, 53, 11, 21, 15, 16, 13, 26, 25, 66, 65.
@grahamfisher307
@grahamfisher307 4 месяца назад
Nice one - there were a lot of pairs to store in the brain box that had to wait for resolution, so I wrote them in! Did I hear NYT occasionally repeat a puzzle? Perhaps you may find on a previous video you had a slight different approach the second time. Anyway, your swim wouldn't have been delayed much by this one.
@NMalteC
@NMalteC 4 месяца назад
Could be
@vidyanandsinha8879
@vidyanandsinha8879 4 месяца назад
Very hard
@luishectormejiaflores
@luishectormejiaflores 4 месяца назад
(7,6)=2 (2,8)=9 (6,3)=8; par (2,9) en 5a col dentro del Sector 5 ent (4,4)=8 (8,6)=9 (9,7)=9 (3,6)=8; tercia (1,3,6) en 6a col; pares (1,3) y (4,6) en el Sector 5 ent (9,5)=6 (3,5)=8 (9,4)=5 (7,5)=7 (2,4)=7; tercia (3,4,9) en 7a fila; (9,9)=8 único lugar (8,7)=4 (7,9)=3; tercia (1,5,7) en 8a fila; cuat(1,2,3,7) en 9a fila; cuat(4,5,6,7) en 6a fila ent (6,7)=6 (6,4)=4 (5,4)=6 (4,1)=6 (3,7)=3 (4,7)=2 (4,5)=9 (5,5)=2; cuat(2,4,5,6) en el Sector 3, (5,8)=3 tercia (1,4,7) en 4a fila; cuat(1,4,5,9) en 5a fila; como el #6 quedó alineado en 1a y 2a fila en los Sectores 1 y 2 ent (3,9)=6 (2,9)=2 (3,1)=2 (9,3)=2; Rectángulo Especial UR(1,3) en 2 y 5 Sector; (3,2)=9 (6,1)=7 (8,3)=7...
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