Comments - Talk
... by: S. Lee
I love this site, but I want to point out some point.
Besides the chaining strategies, there is also a family of strategies that make use of the concept of 'wing'. Indeed, the following strategies
XY-Wing
XYZ-Wing
XYZW-Wing
Aligned Pair Exclusion (possibly partially)
Sue-De-Coq (possibly partially)
can be understand as special cases of the strategy named 'Bent-Naked subset' as is illustrated in Sudoku Assistant (http://www.stolaf.edu/people/hansonr/sudoku/explain.htm). This strategy admits two rules as follows:
Setting) Let A and B be a subset of two overlapping units (such as row-box). Let
Wa = [portion of A that does not overlap with B] = A-B
Wb = [portion of B that does not overlap with A] = B-A
U = [union of A and B] = A∪B
so that both Wa and Wb form two wings of U. Assume that U is 'bent-locked', in the sense that the number of candidates on the cells in U is equal to the number of cells in U.
Rule 1) If there is no common candidate between two wings Wa and Wb, then A behaves like a genuine naked subset with candidates of Wa in each unit, and likewise for B.
Rule 2) If there is a unique candidate k between two wings Wa and Wb, then for every cell that can see the candidate k in U, we can eliminate the candidate k from the cell.
The power of this strategy is that we do not care what happens in the intersection part A∩B, except that it affects the total number of the candidate in U = A∪B. Conceptually it involves neither the concept of bi-valued cells nor that of tri-valued cell, though they appear in many examples. Even this does not involve the notion of ALS. All the matter reduces to the problem of candidate counting in U, Wa and Wb.
Of course, the disadvantage of this strategy is that there is no guide for the choice of A and B. Thus it is of no harm to understand that both APE and Sue-De-Coq are special guidelines for choosing the pair (A, B).
In the 1st APE example, we actually have
Wa = {1,2}[F5]
Wb = {2,5,6}[E7|E8]
U = {1,2,5,6}[F5|E7|E8|F9]
Clearly U forms a bent-locked subset. Since there is a unique candidate, namely 2, in both wings Wa and Wb, we can eliminate the candidate 2 in F8 as F8 can see all the candidates 2 in the subset U by the Rule 2.
Other two examples can also be analyzed by exactly the same manner:
Extended APE example 1:
Wa = {5,6,9}[D9|F9]
Wb = {1,9}[G8]
U = {1,5,6,9}[D9|F9|G8|H9]
=> 9 in J9 is eliminated by Rule 2
Wa = {5,6,9}[D9|F9]
Wb = {3,6}[J7]
U = {3,5,6,9}[D9|F9|J7|J9]
=> 6 in H9 is eliminated by Rule 2
Extended APE example 2:
Wa = {1,5,9}[A2|A9]
Wb = {8,9}[B4]
U = {1,5,6,8,9}[A2|A9|B4|A4|A5]
=> 9 in A6 is eliminated by Rule 2
Sue-De-Coq example 1:
Wa = {2,8}[B2]
Wb = {3,5}[F3]
U = {2,3,5,8}[B2|F3|D2|E2]
=> By the Rule 1,
The triple A = [B2|F3|D2] behaves like a naked pair with candidates {2, 8} in column 2, eliminating 8 in C2, 2 in G2 and 8 in J2.
The triple B = [F3|D2|E2] behaves like a naked pair with candidates {3, 5} in box B1, eliminating 3 in E3.
Sue-De-Coq example 2:
Wa = {6,8}[E2]
Wb = {1,3,7}[D9|F9]
U = {1,3,6,8,9}[E2|D9|F9|E7|E8]
=> By the Rule 1,
The triple [E2|E7|E8] behaves like a naked pair with candidates {6,8} in the row E, eliminating 6 and 8 in the other cells in the row E.
The quadraple [D9|F9|E7|E8] behaves like a naked triple with candidates {1,3,7} in the box B3, eliminating 1, 3, 7 in the other cells in the box B3.
In fact, we may extend the bent-naked subset strategy to a more general setting, though it becomes slightly harder to identify, as the naked triple is harder to apply than the naked pair.
... by: ad.joe
General structure of the APE
------------------------------------
Maybe it's not easy to put this straight, but in fact there are eliminations in 2 boxes:
g k p | _ _ _ | Y Z _
_ _ _ | _ _ _ | _ _ _
_ _ X | _ _ _ | a b c
The "wings" are X=2#, Y=2*, Z=2+
Then XYZ eliminates 2 on the aligned cells abcgkp ...
(provided that these have "not much more than #*+ on it", so you have to "cross-check")
- Doesn't the APE-structure look like an xy-wing (y-wing)? Only the function does work backwards and there is no pivot necessary!!
So you could also call it a BACKWING or a BACKSWING!
- And of course, in an X-Sudoku the 2 aligned fields need not be in the same box
(as long as one of the two is on a diagonal).
... by: ad.joe
How to tease out the APE:
-----------------------------------
First, it's not our job to solve one of the examples in another way. But it's nice that all is so understandable.
(Matt, I'd add three words: duplicate the contents of any "ONE of the" two-candidate cells they both see. )
So, as we look for the third cell of an xy-wing (having seen two "similar" ones at first), what can we do here:
Let's solve it by cross-checking: We assume that the most FREQUENT number of the bivalue cells can be eliminated in the two cells in question.
We assume this number being correct there and quickly end in error, quicker than in looking up all combinations.
(Who'd think that we would not score more than 90 percent, he/she can feel free to try with the other numbers)
... by:
I want this software. how I can have it?
Andrew Stuart writes:
... by: Bruce A Rogers
In the first example, APE strategy is not necessary. The simpler X-wing strategy, using cells C, A, and Y, will eliminate option 7 at r4c7, r5c1, and r5c2. The solution follows immediately by solving hidden singles.
... by: Phil Gooda
I actually managed to understand all this, except for one thing......why bother, there is a far simpler soultion within that box? Let me demonstrate by considering ONLY that box and naming the cells as follows:
A B C
D E F
G H I ..... which means that my C is the X mentioned, and my F is the Y mentioned.
Cells B, C and G contain 3 numbers and only 3 numbers between them. Therefore those 3 numbers cannot appear in the other cells. Which immediately gives you the paired 5/9 in F and I, which means that E has to be 4. Why bother doing all the APE stuff?
... by: Patrick Barnaby
An XY-Chain removes 7 from r8c8. A Bilocation Graph reveals a 2 at r7c2 and then you'll have a naked 7 and then a naked 8 and then quite a few hidden singles on the first and second rows. So this is easier than APE except the Bilocational graph techinque.
... by: Matt Lala
I love the site but some of the explanations need help. Or I guess I do. I think this is one of the less clear ones.
The rule you have says:
Any two cells aligned on a row or column within the same box CANNOT duplicate the contents of any two-candidate cell they both see.
If I take that literally in the first example... X in the first example can see the bivalues 25 and 37. It cannot duplicate the contents of those cells. Therefore... it can't be 2,5,3 or 7? Obviously that's not what was meant.
Or is it... if the two aligned cells see some bivalue cells, and they both mutually share a certain candidate [that's part of those bivalue cells] then that shared candidate can go? But no, that's not it either.
If either if the cells see two bivalue cells, and those two bivalue cells both share a candidate with the cell under consideration... that shared one goes?
Does one actually have to enumerate the possibilities? This seems like something that can be spotted with a glance, if only it can be made clear the exact conditions needed.
... by: Bernard Gervais
align pair type 2
I found more than one numerical solution for this example.
Best regards.
BG
... by: Bernard Gervais
Excellent site, thank you.
for example 1, I use the unicity concept which pinpoints A4 = 1.
Best regards.
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