Tuesday 12-Apr-2011
... by: rocannon
At your first 'Triple Cell Forcing Chains' example,9's chain is not necessary. If D1<>9 than {D1, D8}=3/5.
Thanks your site! (from Korea)
 |
| Print Version |
| Solvers |
| Puzzles |
| Basic Strategies |
| Tough Strategies |
| Diabolical Strategies |
| Extreme Strategies |
| Depreciated Strategies |
| Str8ts |
| Other |
|
 Order Now! |
 Order Now! |
| Forcing Chains are a powerful extension of chain strategies. Given that a cell or unit contains certain numbers, and that these are the only alternatives, we can show that certain candidates must be eliminated. Specifically we show that *all* of the alternatives must eliminate the target. The chains force the removal as no alternative exists. Note: to see this strategy in the solver in most cases requires Digit Forcing Chains to be turned off Dual Cell Forcing Chains |
In the first example, we can eliminate 3 in E2 because of the numbers left in A2, the 2 and 3 there. It is uncontroversial that a 3 in A2 removes the 3 in E2. Hence the short chain +3[A2]-3[E2]. The 2 in A2 must now be considered. If A2 really is 2 then A9 will be 8, which forces G8 to be 9. This removes 9 from G2 and the only remaining 9 in column 2 is in E2. Therefore 2 removes 3 from 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 |
| In red, the chain starts with a 9 in D1 (+9). Removing the 9 in H1 means the 9 must be in H9 as its the last in the row. If that is the case we must remove the 9 from E9. E9 is interesting as it forms part of an Almost Locked Set with D8. An ALS is a set of N cells with N+1 candidates. In this case {3,5,9}. But by removing the 9 from E9 we have reduced and Almost Locked Set to a Locked Set. A 2-cell locked set is a Naked Pair. 3 and 5 must exist in D8 and E9 - we just don't know which way round yet. If this is the case then 3 can't exist in D8. Viola, we have forced 3 from D8 using the contents of D1. |
| 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 |
| Quad Cell Forcing Chains Logic relentlessly obliges us to consider the Quad formation, although I have not and will not look for higher order forcing chains, although they are possible. Quads are pretty rare. I found 6 in a random stock of forty thousand puzzles. |
| In this example, the cell B9 contains {2,5,6,7}. By a miracle were any one of these the solution (and one must be) we can show that there is no 5 in G2, giving us the solution of 4 in that cell. Although somewhat crowded when overlain, all four chains are simple - no ALS or grouped components. If there is a 2 in B9 it follows that +2[B9]-2[G9]+5[G9]-5[G2] If there is a 5 in B9 it follows that +5[B9]-5[C7]+5[C2]-5[G2] If there is a 6 in B9 it follows that +6[B9]-6[H9]+6[H1]-8[H1]+8[H2] -1[H2]+1[G2]-5[G2] If there is a 7 in B9 it follows that +7[B9]-7[F9]+2[F9]-2[G9]+5[G9] -5[G2]. |
 : Load Example or : From the Start |
| Back to Digit Forcing Chains Continue to Unit Forcing Chains.... |
Comments
9's chain is not necessary. If D1<>9 than {D1, D8}=3/5. Thanks your site! (from Korea) Saturday 16-Oct-2010... by: Bob RutanTRIPLE FORCING CHAINS-4TH PARAGRAPHNote your typo in second sentence. Believe you meant H9 as cell last in the row. You are terrific. Loved your book. Bob "In red, the chain starts with a 9 in D1 (+9). Removing the 9 in H1 means the 9 must be in H1 as its the last in the row. If that is the case we must remove the 9 from E9. E9 is interesting as it forms part of an Almost Locked Set with D8. An ALS is a set of N cells with N+1 candidates. In this case {3,5,9}. But by removing the 9 from E9 we have reduced and Almost Locked Set to a Locked Set. A 2-cell locked set is a Naked Pair. 3 and 5 must exist in D8 and E9 - we just don't know which way round yet. If this is the case then 3 can't exist in D8. " Andrew Stuart writes: Fixed thx! |
Post a Comment using Facebook... |