denis_berthier wrote:Your resolution path is a sequence of eliminations of candidates C1, C2, ... (forgetting the Singles).
Each elimination Ci is done by a call to a T&E(X3,6) procedure based on Ci, i.e. in which Ci is assumed to be true - where X3,6 is some set of patterns restricted in some way by two parameters 3 and 6 (how the restriction is done is not the topic of this question).
Do you agree with this? Possible answers: True/False
If False, forget the sequel, I'm completely lost and I give up.
Truedenis_berthier wrote:My next question bears on ONE of these eliminations: Ci, i fixed.
This Ci is itself a sequence of eliminations Ci,j (again forgetting the Singles) [of course, these eliminations don't apply to the initial puzzle; they are local to Ci].
When you say that patterns used for the elimination of Ci (i.e. for each Ci,j step) are restricted to 3 "planes", do you mean that the 3 "planes" are the same for all the Ci,j (i fixed) or that they can be different for different Ci,j (i.e. different j's, i fixed).
Possible answers: same/different.
sameBut now I am confused because the alternative answer "different" makes no sense to me. Moreover you said
denis_berthier wrote:This Ci is itself a sequence of eliminations Ci,j
Although this is not wrong, it does not conform to the leading idea. In the in initial post I gave arguments that all (constraint driven) eliminations can be expressed by a set of baselines. So each step Ci should be specified by some base lines. T&E is one method among others to find the baselines of an elimination. Also the listing shows no T&E(2), its all T&E(1) including all failed sequences. So I feel there is still something odd.
You can apply eliminations derived from planes or plane sets at the global level or on T&E(1) level. Its always the same procedure.
I give an example at global level to simplify the context. The solver uses only planes and plane pairs.
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000400100080020090007005006400001700090000030002000008000090020005300000010007004
APPLY ROOT: S2_R4C49_N29R4 E={544,644,844,549}
APPLY ROOT: PLANE_N2 B3=[2r1|2r4|2r9] T={231} D={}
+------------------------+------------------------+------------------------+
|23569 2356 369 |4 3678 3689 |1 578 2357 |
|1356 8 1346 |167 2 36 |345 9 357 |
|139 234 7 |189 138 5 |2348 48 6 |
+------------------------+------------------------+------------------------+
|4 356 368 |29 3568 1 |7 56 29 |
|15678 9 168 |25678 45678 2468 |2456 3 125 |
|13567 3567 2 |5679 34567 3469 |4569 1456 8 |
+------------------------+------------------------+------------------------+
|3678 3467 3468 |1568 9 468 |3568 2 1357 |
|26789 2467 5 |3 1468 2468 |689 1678 179 |
|23689 1 3689 |2568 568 7 |35689 568 4 |
+------------------------+------------------------+------------------------+
+------------------------+------------------------+------------------------+
| |4 |1 |
|1 14(*) |1 |4 |
|1 4 |1 1 |(4) 4 |
+------------------------+------------------------+------------------------+
|4 | 1 | |
|(1) 1 | 4 4 |4 1 |
|1 | 4 4 |4 14(*) |
+------------------------+------------------------+------------------------+
| 4 4 |1 4 | 1 |
| 4 | 14* 4 | 1 1 |
| 1 | | 4 |
+------------------------+------------------------+------------------------+
APPLY ROOT: PLANE_N1+N4 B4=[1r6|4r2|1c3|4b6] T={151} D={}
APPLY ROOT: PLANE_N1+N4 B4=[1r6|4r2|1c3|4b6] T={437} D={}
APPLY ROOT: CONN_N1+N4 B4=[1r6|4r2|1c3|4b6] T={323,623} D={}
APPLY ROOT: CONN_N1+N4 B4=[1r6|4r2|1c3|4b6] T={568,668} D={}
Skipping the first two eliminations, a pair and a 3-queue, the first diagram shows the situation afterwards. The second shows the plane set N1+N4 with only numbers 1,4 and *. The star symbol indicates that the cell contains more candidates, otherwise the contraint in the cells containing both 1 and 4 would be wrong. The second diagram constitutes a sudoku-like puzzle where the usual methods can be applied to find eliminations.
From the initialisation phase (skipping how this is done) I know that the N1+N4 puzzle has 16 "multiple solutions" from the start. The two eliminations until now did not change this. To avoid confusion with the global solution they are called
valid permutations. One of the 16 valid permutations is matching the unique final solution of the whole sudoku. The candidates 1r5c1 and 4r3c7 are not part of
any of the 16 valid permutations and therefore not part of the final solution. This check is done by simple comparison (without any T&E) and very cheap even if the number of permutations is larger. The candidates marked by stars in the cells r2c3 and r6c8 are also elimination candidates, because another permutation check verifies that either 1 or 4 is true for all valid permutations. So the N1+N4 planes generates 6 eliminations (put in parenthesis). Theres is
no elimination for all other candididates inside N1+N4.
Up to now nothing is known about the
size of these eliminations. As all eliminations are determined by a set of baselines the size is determined after finding the base set. This can be accomplished by T&E after setting the eliminmation candidate true or like in this case by mapping on a traditional pattern. All six eliminations are bound in a
nice loop of size 4.
There are various possible base sets: B4=[1r6|4r2|1c3|4b6] or [1r6|4r2|1c3|4c8] or [1r6|4r2|1b1|4c8] or [1r6|4r2|1b1|4b6] ...
Next planes:
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APPLY ROOT: PLANE_N2+N9 B3=[r4c4|2r9|9r3] T={991} D={544,644,844}
APPLY ROOT: PLANE_C7+C8 B7=[r3c8|r4c8|r9c8|2c7|4c7|6c7|8c7] T={688} D={437}
The plane pair N2+N9 has only 9 valid permutations and produces 1 elimination of size 3. The plain pair C3+C8 has 73 valid permutations and and elimination of size 7. In this case the baselines are determined by a T&E sequence.
After this point none of the 36 single plains and none of the 126 plaine pair can justify any more eliminations. One can now switch to plane triples or to T&E(1) using the same permutation tables. All valid permutations are generated only once and reused with all sequences. The number of valid permutations of plane sets can only shrink with candidates eliminated. In this example the total number of valid permutations of all plane sets is 251355.