Strategies for Number Puzzles of all kinds
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Sunday 26-Mar-2017

... by: ray

1. Use only basic strategies and the forcing bi-location cell set method.Start
with E7[467} - bi-location and tri-location cell for 7. Progressively form the
2-element bi-location set E7B2,forcing only the 1-cell to 2-cells run out values until E7B2 = (4,6) which gives the solution. Other bi-location cell
solutions found were: E7H1J3 = (4,6,1), C8H1J3 = (4,6,1),B2E8J3 = (6,6,1).
2. Using this solution,with basic strategies,the following non bi-location cell
set solutions were found: B2C8 = (6,4), H14J3 = (6,7,1), D5E56 = (8,71),
A56C6 = (5,92). N.B; using basic to extreme strategies(no Exocet) E6 = 1
a 1-cell solution.

Saturday 25-Mar-2017

... by: jkl

"Tarek golden nugget" to # 243 (1 to 9 becomes):
Rot.90°clockwise 0
Renumb: 8 7 9 2 4 6 1 3 5
Columns: 8 9 7 2 3 1 4 5 6
& Rows: 8 9 7 6 4 5 2 3 1

Saturday 25-Mar-2017

... by: JC Van Hay

Analysis of the puzzle from the Exocet (1278){r89c4, R267} -> -{9r2c6}

NP(12)r89c4
|
V
-(12)r3c5, HP(12)r23c6, L2-Wing{2R2, 1R27}; 10 singles
HP(12)r89c3 [!?]; 0 solution

NP(17)r89c4
|
V
-(17)r5c6, HP(17)r56c5, L2-Wing{7R6, 1R67}; 12 singles
HP(17)r89c7 [!?]; 0 solution

NP(18)r89c4
|
V
1r2c6 -> 0 solution; 10 singles
2r2c4 -> 0 solution; 4 singles
LC{5B27}; L3C1=5
The 2 solutions of R3 -> 0 solution

NP(27)r89c4 + 2r2c1
|
V
9 singles
1r7c6 -> 0 solution
1r5c6 -> 1 solution


NP(27)r89c4 + 2r2c6
|
V
9 singles
8r2c4 -> 0 solution
8r2c9 -> 0 solution via LC{6B6}, 3 singles, NP(89)r4c36, XWing{8C24}

NP(28)r89c4 -> 0 solution

NP(78)r89c4 + 7r2c2
|
V
9 singles
AIC{1R6, 1C3, 2C3, 2R2} -> r2c8=1, r6c9=2
HP(12)r8c17; 0 solution

NP(78)r89c4 + 7r2c6
|
V
9 singles
AIC{1R2, 1C7, 2C7, 2R6} -> r2c1=2, r6c1=1
0 solution

Saturday 25-Mar-2017

... by: Paolo Calaresu

I use single and Nishio with only basic strategies.
IF E7=4--> A9 ≠ (6,8,9); B1 ≠ 9; B2≠(8,9);
B4 ≠ 9; B8 ≠ (1,6); A1 ≠ (6,9);
A3 ≠(6,8,9); A4 ≠(5,9); A5 ≠ 3;
B6 ≠ 1-->solution with only basic strategies.

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