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Cell Forcing Chains From sudokuwiki.org, the puzzle solver's site |
These are the different types of eliminations available from Cell Forcing Chains. The diagram shows starting cells with two candidates but the same types and principles apply to cells with three (Triple) or four (Quad) candidates - just add more chains | Cell Forcing Chain Family |
In the first example, we can eliminate 3 in E2 because of the numbers left in G9, the 8 and 9 there. The small purple chain shows that with two 9s in row G and column 2 a 9 in G9 places a 9 in in E2 and therefore removes the 3. The chain is +9[G9]-9[G2]+9[E2]-3[E2]. The 8 in G9 takes a little longer but again it is just a matter of following pairs in units. We remove the 8 in A9 giving 2 which forces a 3 in A2 and that removes the 3 in E2. Since we are arguing from a single cell and the cell has two candidates, this is the simplest Forcing Chain, a Dual Cell Forcing Chain. |
Dual Cell Forcing Chain: Load Example or : From the Start |
Triple Cell Forcing Chains So if bi-value cells can work this magic, lets go up an order of complexity and test a Triple Cell Forcing Chain. This one starts on D1. The first chain, in blue, is another short one-shot. 3 removes 3 from D7. Forcing chains don't always need a nice first deduction, all three chains could be very long. The second chain +5[D1]-5[D8]+3[D8]-3[D7] is a little more involved. The 5 in D1 must remove the 3 in D7. 5 will force a 3 in D8 which does the trick. Now to force 9 in D1 to remove the 3 in D7 we have to use quite a complex chain. It reads +9[D1]-9[H1]+9[H9]-9[E9] +3{E9|D8}-3[D7]. The curly brackets is my symbol for an ALS (Almost Locked Set). Documentation forthcoming. |
Triple Cell Forcing Chain: Load Example or : From the Start |
The second triple example is nicely complex one. It's improbable that you might detect it using pen and paper, but I include it to show the reach of the Forcing Chains. We show that the content of J7 can remove the 7 in B3. The blue chain is easy. A 1 one J7 leads to a 7 in B7 and therefore no 7 in B3. Hense +1[J7]-1[B7]+7[B7]-7[B3]. The purple chain, +7[J7]-7[J1]+7[G3|H3]-7[B3] contains an interesting group, symbolised by +7[G3|H3]. All this means is that if J1 is not a 7 then G3 OR H3 must be. If one of those must be 7 we can look along their alignment - the columns, and remove 7s, eg 7 in B3. The last chain, +9[J7]-9[H7]+9[H3] -9[C3]+7{C3|A1}-7[B3] contains an ALS, which I described above. We have a {1,7,9} in two cells A1 and C3. Removing 9 from B3 leaves a locked set of 1/7 in that box. All other 1s and 7s can be removed, namely the 7 in B3. |
Complex Triple Cell Forcing Chain: Load Example or : From the Start |
In this example, the cell D3 contains {1,2,4,5}. By a miracle were any one of these the solution (and one must be) we can show that there is no 1 in F2, no really giving us a great deal of help with the rest of the puzzle. Although somewhat crowded when overlain, all four chains are relatively simple - just one ALS in the green chain. If there is a 1 in D3 it follows that +1[D3]-1[F2] If there is a 2 in D3 it follows that +2[D3]-2[D9]+2[A9]-2[A2]+1[A2]-1[F2] If there is a 4 in D3 it follows that +4[D3]-4[E1]+1[E1]-1[F2] If there is a 5 in D3 it follows that +5[D3]-5[E2]+1{E2|A2}-1[F2]. |
A Quad Cell Forcing Chain: Load Example or : From the Start |