De Robert Mauriès
(Publié le 08/02/2021)
Pas de commentaire pour l'instant.
De François C
(Publié le 10/02/2021)
Bonjour,
Grille de niveau TDP <= 4
Voici une résolution avec 4 pistes invalides de 10 candidats maximum :
Candidats uniques: 4L1C7, 4L2C1, 1L5C5, 4L9C6, 4L7C9, 2L8C5, 6L8C3, 6L7C6, 6L3C2, 6L1C5, 2L3C3, 2L4C7, 2L5C2, 2L7C8, 7L9C3, 1L9C7, 1L3C9, 1L7C3
Alignement: 3C3B1 => -3L3C1
P(3L2C5) = 3L2C5,9L2C9,3L6C7,3L8C9,9L8C1,3L9C4,9L5C8,9L9C5,
=> 9B5 vide => -3L2C5
P(8L2C5) = 8L2C5,5L4C5,5L5C1,8L3C7,9L3C1,7L8C7,9L8C9,9L2C7,3L6C7,3L9C5,
=> 3B9 vide => -8L2C5
Paire nue: 59C5L24 => -9L6C5 -9L9C5
Alignement: 9B8C4 => -9L1C4 -9L3C4 -9L5C4
P(8L5C1) = 8L5C1,9L6C3,3L6C7,3L9C5,3L7C1,5L3C1,5L5C4,5L1C8,7L1C4,3L3C4,
=> 3C8 vide => -8L5C1
Candidats uniques: 8L6C3, 3L6C5, 9L6C7, 8L9C5, 8L2C7
Alignement: 3C6B2 => -3L1C4 -3L3C4
P(9L2C5) = 9L2C5,3L2C9,5L2C3,3L1C3,8L1C6,9L1C2,9L3C8,9L9C4,3L9C8,
=> 3C7 vide => -9L2C5
Candidats uniques jusqu’à la solution.
De François C
(Publié le 10/02/2021)
Autre résolution : avec 10 pistes invalides de 4 candidats maximum :
Candidats uniques: 4L1C7, 4L2C1, 1L5C5, 4L9C6, 4L7C9, 2L8C5, 6L8C3, 6L7C6, 6L3C2, 6L1C5, 2L3C3, 2L4C7, 2L5C2, 2L7C8, 7L9C3, 1L9C7, 1L3C9, 1L7C3
Alignement: 3C3B1 => -3L3C1
1) P(9L9C5) = 9L9C5,9L4C3,9L6C7,9L2C9, => 9C8 vide => -9L9C5
Alignement: 9B8C4 => -9L1C4 -9L3C4 -9L5C4
2) P(9L5C9) = 9L5C9,3L2C9,3L6C7,3L9C5, => 3C8 vide => -9L5C9
3) P(9L8C7) = 9L8C7,3L6C7,7L5C9, => 7L8 vide => -9L8C7
4) P(9L9C8) = 9L9C8,9L8C1,9L5C6,9L3C7, => 9C9 vide => -9L9C8
5) P(3L8C7) = 3L8C7,9L6C7,5L7C7, => L9C8 vide => -3L8C7
Candidats uniques: 7L8C7, 7L5C9
6) P(3L9C8) = 3L9C8,3L2C9,3L6C5, => 3L5 vide => -3L9C8
Candidat unique: 5L9C8
Alignement: 3L9B8 => -3L7C4
Paire nue: 39C7L67 => -3L2C7 -9L2C7 -3L3C7 -9L3C7
7) P(5L3C1) = 5L3C1,5L1C4, => 5L5 vide => -5L3C1
8) P(8L6C5) = 8L6C5,8L5C1,3L6C7,3L7C1, => 5C1 vide => -8L6C5
Candidat unique: 8L6C3
9) P(8L9C2) = 8L9C2,8L2C5, => 8L1 vide => -8L9C2
Candidats uniques: 9L9C2, 3L8C1, 9L8C9, 3L2C9, 3L7C7, 9L6C7, 3L5C8, 3L6C5, 8L9C5, 9L7C4, 3L9C4, 8L2C7, 5L3C7, 3L3C6, 3L1C3
10) P(5L7C2) = 5L7C2,8L1C2,9L1C6,5L2C5, => 5L1 vide => -5L7C2
Candidats uniques jusqu’à la solution.