Calcudoku puzzle forumhttps://www.calcudoku.org/forum/ Step by step solution of a 6x6 "difficult"https://www.calcudoku.org/forum/viewtopic.php?f=3&t=58 Page 1 of 1

 Author: clm  [ Thu Sep 22, 2011 12:12 am ] Post subject: Re: Step by step solution of a 6x6 "difficult" Step by step solution of a 6x6 “difficult”. (This is a retransmision of the topic with larger graphics and enhanced resolution). This topic is mainly oriented to “help” beginners and some “medium” players to finish puzzles like this one that in fact only requires simple analysis (however only about the 30% of the active puzzlers got the solution). For this example I will use the saturday Aug 06, 2011, 6x6 difficult puzzle (Puzzle id: 333856): I will use pairs of numbers (in small size in the lower part of the cell) when those are the final candidates for those cells, in black and big size (in the center of the cell) for the final solution for that cell, and I will use colors in the different steps (and when in the left side of the cages to show the different options discussed). The small numbered ellipses refer to the steps along the procedure. Different methods may be followed, but perhaps a good strategy is determining the “big” products and the “small” additive cages. Step 1. The cage “96x” in B4-B5-C5, since 96 = 3 x 2 x 2 x 2 x 2 x 2, has the only possible combination 6-4-4 so we fill the cage with those numbers. The cage “7+” in E4-D5-E5 could have these four combinations: 1-1-5, 2-2-3, 3-3-1 (with a duplicated number) and 1-2-4 (all different). The 1-1-5 (numbers in blue) is not a valid option because no number (from 1 to 6) could be located in A5. The 2-2-3 (numbers in red) is not a valid option because no number (from 1 to 6) could be located in F5. Obviously the 1-2-4 is not valid due to 4’s in cage “96x” so we write 3-3-1 as the solution. Step 2. We can place now a 2 in F5 and a 5 in A5 (in green) considering A1 and F1. Let’s now consider the cage “0-“ in D2-E2-E3. This is an “even” cage with a sum that runs from 4 to 12 and a “higher” number of 2, 3, 4, 5 or 6. But 1-1-2 is not possible (1 in E5), 1-2-3 not possible (3’s in E4 and D5); 4-2-2 or 4-1-3 are not valid since no 4 is possible inside the cage (observe that 4’s must be either in D1 or E1 and D6 or E6 so simultaneously “erasing” columns D and E, or alternatively, either in A2 or A3 and F2 or F3 so simultaneously “erasing” rows 2 and 3); with a 5 as the higher number (total of 10 inside the cage), the pairs 2-3 (3’s in E4 and D5) or 1-4 (just discussed) are not posible. Then the higher number will be a 6, but 3-3 and 2-4 are not possible inside so the other two numbers are 1 and 5. Then we conclude that a 1 is in D2 and the pair 5-6 in E2-E3. Considering A1 we place a 2 in E6 and a 4 in E1. And now the 4 of row 6 goes to D6 (and since a 6 cann’t go inside the cage “7+” in D3-D4 due to the 1 in D2) we write a 6 in D1. Step 3 (numbers in violet color). The 6 of row 6 must go to C6 and the 5 to B6. Now the 5 required in the cage “30x” in B2-C2-B3 (30 = 2 x 3 x 5) goes to C2 and the pair B2-B3 is 2-3 since 1-6 is not possible (6 in B5). Then B1 =1 and C1 = 3. Also E3 = 5 and E2 = 6. The final steps 4 and 5 complete the puzzle: Step 4 (numbers in orange color). “7+” in D3-D4 is 2-5 so D4 = 5, D3 = 2; “1-“ in C3-C4 is 1-2 so C4 = 2 and C3 = 1. Also B3 = 3 and B2 = 2. Considering A6, F2 = 3 and A2 = 4. Step 5 (numbers in green color). Considering the 4 in B4, F3 = 4 and A3 must be 6. And finally A4 = 1 (due to F6) and F4 = 6 (also in opposition to A3 = 6). Conclusions: We knew that the total addition value of the inner part was 53 (see my topic on this subject, “Why in a 4x4 numbers in corners are those in inner area (1)”), the addition of the four corners plus 2 lines in this puzzle, that is, 2 + 5 + 3 + 1 + 2 x 21 = 11 + 42, or simply in this case the total puzzle minus the border, 6 x 21 - 2 - 14 - 5 - 15 - 1 - 17 - 3 - 16 = 126 - 73 = 53. In these conditions, once the cages “96x”, “7+” and “0-“ have been defined in the inner area, the sum of cages “30x” and “1-“ equals 13 (53 - 14 - 7 - 12 - 7) so (see my topic “The use of maximums and minimums in the sum of cages”) since the minimum value of “1-“ is 3 (1-2) the maximum value of “30x” would be 10 and the combination 5-1-6 impossible. The conclusion that the cage “30x” is 5-2-3 could have been raised a little befote, however it has not been necessary to take this into consideration (or to break into prime factors the “144x” that is directly given, etc.), or going to any complex calculation for this puzzle. It is quite like solving a “sudoku”. And here is the official solution:

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