![]() ![]() But since I did it on paper, the "colors" remained on the sheet. (but if you load this in the solver, you'll have to go through some pointing pairs and an X-wing to eliminate some candidates, which were already eliminated in the original process, to get to the point I'm talking about)Īt this point (actually right before this point but the order doesn't matter) I tried to use simple colouring on 2s - unsuccessfully, nothing was eliminated. The interesting part comes if you run it through the solver to the point where it uses simple colouring on 9s. I found a sudoku, which can demonstrate what I find an interesting twist to the colouring strategies: ![]() I really like your solvers and they are very helpful in learning the various strategies. I just want to know if I should be using singles chaining for this type of Sudoku puzzle and if your solvers can be used to solve them. ![]() Maybe I can use the killer solver, but I think it works differently. I'm thinking that either I am missing something, but cannot use your solver to find what I've missed. Hyper-Sudoku has a set of four shaded square areas overlaying the board (B2-D4, B6-D8, F2-H4, F6-H8) and they must be also filled in accordance with the normal rules.Īnyway, I find that sometimes singles chaining will indicate a number can be eliminated from a cell and the program indicates that it cannot. I can check my board and the app will tell where i have made mistakes and let me back up, so I know when I have made a mistake. I find that I can use singles chains when I have exhausted easier strategies. I recently found they are also called Windoku. I am trying to get through a series on fiendish Hyper-Sudoku puzzles (from an Android app - Super Sudoku). by: Can I use singles chains in Windoku puzzles? This would reflect the full meaning of Rule 2 and not just half of it as now. This needs to be enhanced with "and all blue coloured candidates become solutions". The fundamental problem is in the thinking - the message in the results window says "all yellow coloured candidates can be removed". its membership of the chain is forgotten and its fate is decided by later scanning. But the 5 in E9 becomes a Hidden Single and doesn't get promoted to a solution until two steps later, i.e. Six of them become Naked Singles and become solutions in the next step because they are singles. In the example for Rule 2 there are seven ON candidates. The beauty of Simple Colouring Rule 2 is that we know immediately the fate of every candidate in the chain. I'm sure that you know what you mean and not what you say in your reply to my comment of 2 March - you're not going to "remove all candidates on the ON cells" are you? What I would like to see is all ON candidates become solutions either in the same or the next step of the solver. Thanks for the challenging and interesting site. In which case the strategy of 'Simple Colouring/Singles Chains' becomes entirely redundant.įrom the humanities side, I would point out that the apostrophe in "Single's" isn't needed, you're trying to mark plurality not possession. Having noted this, even greater simplification of strategies can be achieved by noting that all instances of Rule 4 Singles Chains are in fact examples of X-cycles with an odd number of nodes - the candidate that sees both ends of a chain that begins and ends on strong links will be the odd one out, linked by a W-W connection. 'off chain' distinction, which is perhaps more confusing than enlightening? If this is the case, then Rule 2 is redundant, and there would be no need for the 'off-chain' vs. (But I can't provide a mathetmatical proof). I'm pretty sure that any eliminations found by applying Rule 2 can also be found by applying Rule 4, simply by starting at different points and following different paths through the network of connections. ![]()
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