The following is page three of my campaign versus the Easter Monster.
If this is your first visit to this blog, WELCOME!!
Previous blog pages may be helpful. Links to these pages are found to the right,
under Sudoku Techniques. Specifically, it may be helpful to have visited
the following four pages first:
The illustrations of steps shown in this proof will share this key:
- black line = strong inference performed upon a set (strong link)
- red line = weak inference performed upon a set (weak link)
- black containers define a partioning of a strong set(s)
- candidates crossed out in red = candidates proven false
- Orange labels mark derived inferences
Please be aware that, for me, strong and weak need not be mutually exclusive properties
Previously, the following derived strong and weak inference groups were proven.
- Each set in Group Z below contains exactly one truth:
- (27)r2c13, (27)r2c56, (16)r2c56,(16)r2c79
- Group X is a strong inference set(sis): (2)[r8c5,r5c2,r8c7]
Before making the next elimination, I will add two more such groups.
Group U - using an Almost Almost Two String Kite
The following sets are derived sis U
By now, hopefully the idea is clear. Write the conclusion and the appendages that could prevent
the conclusion as a strong inference set. The proof could look like:
- (1): [r2c5=r6c2]=[r8c5=r4c5-r6c6=r6c3]-r8c3=r7c2 => [(1)r2c5,r67c2] is a sis.
- Using Z:(1)r8c3=(6)r8c1 => [(1)r2c5,(1)r6c2,(6)r8c1] is a sis.
Group T - using an AAA Y Wing Style.
The following set is derived sis T
This one seems quite a stretch - 3 almostings. The previous deduction U
made looking for this one seem logical at the time.
The AIC justification reads:
- sis[(4):r3c2,r8c17]=[(4)r8c3=(4)r8c9-(4)r3c9=(4-5)r3c1=(5)r1c3]-(5)r8c3=(14)r8c3 =>
13x13 TM using derived and native strong inference sets
The graphic above does not dispay all the information used. I graphed some of the main ideas.
Again, a Triangular Matrix below will serve as both the proof and illustration of the elimination.
Long, messy, and a tad convoluted. I suspect 'tis the nature of this beast. r2c5≠2 lets
us make some easy eliminations.
Steps 2d, 2e.
Above we have two steps illustrated:
- 2d: Using Z, (2=7)r2c56, note r2c5 now is void of  => r2c6≠16
- 2e: (2):r4c5=r8c5-r8c7=r7c8 => r4c8<>2
Step 2f. Single candidate fish.
This elimination is symmetric to V done on candidate 1. The appendage has been eliminated,
so the derivation makes the elimination. As AIC:
- (6): [r2c9=r2c5-[r6c1,r4c5]=[r4c89=r4c1-r56c2=r9c2-r9c6=r6c6] =>
- sis(6):[r2c9,r4c89,r6c6] => r6c9≠6
At this point, the Almost Almost Naked Pair 39 at r6c79 caught my attention. Because of
the super cell-like relationship (27)at r45c8, the potential for an odd Sue de Coq using
super cells reared its head. If I could have found such an animal, it would have been very cool.
Unfortunately, I was unequal to the task. Nevertheless, looking in that direction lead me directly
to a fairly simple elimination (for this puzzle!).
Step 2g - a Three way pincer with candidate 2.
Above, I have graphed all the strong inference sets considered to make this elimination. After
finding this elimination, I found numerous ways to reach the same elimination. Some are a bit
quicker, but none as cool - imho. Placing all the weak links into the diagram proved difficult for me.
The logic, as AIC, looks like this:
- [2:r4c5=r8c5-r8c7=r7c6] =
- = Z(7)r4c8=(pair39)r6c79 &
- = (1)r7c2-(1=6)r2c5-(6)r1c6=[(6):r6c6=r9c6-r9c2=r1c8]]
- reducing nicely to sis2g 2:[r4c5,r7c6,r6c1] => r6c6≠2
This is a tad messy to follow as AIC. It should be crystal clear as a Mixed Block Matrix.
The matrix below is called Mixed Block only becuase of the use of a contained Pigeonhole
matrix - pair 39. To shorten the matrix, I treat this as one Boolean. This
reduces the Matrix to a simple Triangular Matrix.
As noted previously, this derives sis 2g 2[r4c5,r6c7,r4c1] => r6c6≠2.
This sis may be useful later.
One could justify the same elimination, considering the same strong inference sets, but thinking
roughly in terms of
- [naked triple 239]r6c179=
- [(2=7)r56c8 - (7)r6c9=(6)r6c1-(6's) ... =(1)r4c7]
- -(2)r4c7 = [skyscraper with 2's]
- => sis2g => r6c6≠2
There are, as I noted, some other paths that consider only a few of the same strong inference sets.
Those are also fairly interesting, but I lack the time to expose every possible elimination path.
This concludes this page. There remains many difficult eliminations, unless I find
a simpler path before I publish. Hope springs eternal!.
If you have read this far, thanks for your patience!