Sudoku strategies explained with graphics

Strategies


Overview Singles Naked Pairs Naked Triples Hidden Pairs Hidden Triples Naked Quads Hidden Quads Pointing Pair Pointing Triple Box Reduction X-Wing Finned X-Wing Sashimi Finned X-Wing Franken X-Wing Finned Mutant X-Wing Skyscraper Chute Remote Pairs Simple Coloring Y-Wing W-Wing Swordfish Finned Swordfish Sashimi Finned Swordfish Franken Swordfish Mutant Swordfish Finned Mutant Swordfish Sashimi Finned Mutant Swordfish Sue De Coq XYZ-Wing X-Cycle Bi-Value Universal Grave XY-Chain 3D Medusa Jellyfish Jellyfish Jellyfish Avoidable Rectangle Unique Rectangle Hidden Unique Rectangle WXYZ-Wing Firework Subset Exclusion Empty Rectangle Sue De Coq Extended SK Loop Exocet Almost Locked Sets Alternating Inference Chain Digit Forcing Chains Nishio Forcing Chains Cell Forcing Chains Unit Forcing Chains Almost Locked Sets Forcing Chain Death Blossom Pattern Overlay Bowman Bingo



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Almost Locked Set Forcing Chain


If two ALS have one Candidate in common and if every occurrence of that Candidate in the first ALS can see all occurrences of that Candidate in the second ALS, then it can be a solution in only one of these two ALS. Such a Candidate is called a Restricted Common Candidate (rcc in short).

Let us consider a Chain of ALS where each ALS is connected to the previous ALS via an rcc and to the next ALS via a different rcc, and where the first ALS and the last ALS in the chain share a common Candidate. If that common Candidate is not the rcc connecting the first ALS to the second ALS and is also not the rcc connecting the last but one ALS to the last ALS, then this Candidate cannot be the solution in any Cell outside of these ALS that see all occurrences of that Candidate in the first ALS and in the last ALS of the chain.

Indeed, if it were the solution in one of these Cells outside of the ALS, then it would be eliminated in the first and in the last ALS of the chain. This implies also that all the other Candidates in the first ALS must be the solution in this ALS. In particular, the rcc connecting the first ALS to the second ALS would be a solution in the first ALS, hence eliminating it in the second ALS, which in turn implies that the rcc connecting the second ALS to the third ALS would be a solution in the second ALS, and so on until the last ALS in the chain. As the rcc connecting the last but one ALS to the last ALS would be eliminated in the last ALS, all the other Candidates in the last ALS would have to be the solution in the last ALS, in particular the one that is common with the first ALS, which is not possible.


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In this example, if Candidate 6 were the solution in C9, then Candidate 6 would be eliminated in ALS {A8} where Candidate 4 would have to be the solution. This would eliminate Candidate 4 in ALS {A2,A3,B1,C1} and Candidate 3 would have to be a solution in that ALS. This would imply that Candidate 3 would be eliminated in ALS {C2,C4,C6}. But Candidate 6 in C9 has already eliminated Candidate 6 in that ALS, so that this ALS would have 2 Candidates left (1 and 4) in 3 Cells. Which is not possible. Hence Candidate 6 can not be the solution in C9.



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