![]() ![]() ![]() The same technique can be expanded by using information from perpendicular rows and columns. This leaves square e1 as the only possible place into which 9 can fit in. Looking at box 1 and box 3 we can see there are already 9s in row 2 and in row 3, which excludes the two bottom rows of box 2 from having 9. In our first example we will focus on box 2, which like any other box in Sudoku must contain 9. Here are some ways of using scanning techniques: 1. The scanning technique is also very useful for hard puzzles up to the point where no further progress can be made and more advanced solving techniques are required. The scanning technique is fast and usually sufficient to solve easy puzzles all the way to the end. The easiest way starting a Sudoku puzzle is to scan rows and columns within each triple-box area, eliminating numbers or squares and finding situations where only a single number can fit into a single square. The grid is also divided into nine 3x3 sub-grids named boxes which are marked box 1 through box 9. I found it by perusing other Sudoku websites.Sudoku grid consists of 81 squares divided into nine columns marked a through i, and nine rows marked 1 through 9. Between C+D+E, all possible locations of their candidates are fixed. This works because C+D+E is a locked set and D and E are mutually exclusive. The two removal rules are modified to say that the candidates from D can be removed from the cells not covered by C+D in C's row or column and the candidates from E can be removed from the cells not covered by C+E in C's box. The rule saying D and E must be a subset of C is not needed. The candidates in D and E must still be different. The second extension is that D and E can contain candidates not in C, as long as D and E each contain at least one candidate from C. If C has one or more candidates not in D and E, these candidates can be removed in the cells not covered by C+D+E in C's box and row or column. This is a possibility if C is N+3 or larger. The first extension is that C can contain candidates not in D and E. There are a couple of possible extensions if this restriction is made explicit. There is a restriction that is implied by the general terms: the size of the set of candidates in C+D+E must equal the number of cells in C+D+E. ![]() My solver uses DALS wherever an ALS can be used such in an AIC, death blossom, unique rectangle type 3 etc. There are in the same way also hidden DLS, dispersed almost locked set (DALS) etc. The eliminations follow the same logic as with an ordinary LS, resulting in the eliminations shown in the SdC example. Digits 2 and 8 are bound to column 2, digits 3 and 5 to box 4. In the first example above we have a DLS of size 4 with cells B2, D2, E2 and F3 and 4 digits 2, 3, 5 and 8. At the end all cells in the unit must be coloured. I'll demonstrate it with a colouring procedure: colour the first cell, then for each digit that is a candidate in that cell colour all other cells in the set that have that digit as candidate, repeat the procedure for all the newly coloured cells. The latter condition is a bit more complicated to explain. You have a DLS of size n when you have n cells anywhere in the grid with in total n digits, with the requirements that for each digit all candidates for that digit are bound to one unit, and that all cells are "connected". I have not seen this described anywhere else so far. In a normal LS all cell involved are bound to one unit. It is not the ordinary locked set, but what I call a "dispersed locked set". Thinking it trough I realized however that the complexity is not needed and that SdC is nothing more than a (very) special case of a simple locked set. I found it very complicated to be honest. When working on my own solver and tackling this strategy I was trying to figure out how it worked. ![]()
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