The following is an illustrated proof for the most diabolical
puzzle that I have ever tackled. It has been named the Easter Monster. Its moniker is
well deserved.
If this is your first visit to this blog, WELCOME!!
It has been quite some time since I have written in this blog. I suppose I have had more
important things to do! Nevertheless, hopefully some of you find this new series of pages interesting.
Previous blog pages may be helpful. Links to these pages are found to the right,
under Sudoku Techniques.
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
Please be aware that, for me,
strong and weak need not be mutually exclusive properties.
Easter Monster
One may wish to look at this puzzle before placing the possibilities. The first step
that I shall illustrate is available now, without looking at the possibility matrix. However, it
is instructive to look at the possibility matrix. It tells us quickly that some outside
of the box thinking maybe required.
The Possibility Matrix
Above, note that not a single bivalue cell exists. In fact, there are but few bilocation
sets. I count 15: 1 each with candidates 1,2,6. 2 each with candidates 4,5,9. 6 with candidate 7.
This is not promising, and makes the normal puzzle mark-up that I employ almost useless. When I first
looked at this puzzle, I quickly shelved it. There was nothing that I could find to attack.
After some time, I decided to look again. This time, some symmetries jumped right out at me.
Here is what I found:
Hidden Pair Loop - or - Hidden Quad Intersections
The Inspiration
For clarity (I hope!!), I have shaded above the unfilled cells in columns 2,8 and rows 2,8. I have
divided them into four groups, column 2, column 8, row 2, row 8. The logic that follows relies
only upon the ability to count:
- Consider only candidates 1267 in each group above. There are thus four true locations
for candidates 1267 in each of the four groups. This is a total of 16 truths.
- Note that in each corner box, only two of 1267 can exist. Note further the weak
links shown above. Thus, in each corner box, there can be no more than 2 truths 1267 in
the highlit cells. This is then: 8 truths maximum in the corner boxes.
- Note that in each midde box (except the center one), there are only 2 locations in
each group. Thus, each of boxes 2,4,6,8 can contain no more than 2 truths 1267 in the highlit cells.
This is then:8 truths maximum in the remaining boxes being considered.
- So, we have exactly 16 truths, and we know the distribution cannot be less than 2 in
any box, as it cannot be more than 2 in any box. In other words, we need exactly 2 of
1267 to be true in each box amongst the highlit cells.
This leads immediately to 13 eliminations.
Eliminations made by counting accurately to 16
Above, I have illustrated precisely which eliminations are justified. This leads us to a pair of easy steps!
Naked triple 126 in row 5
The naked triple that is revealed in row 5 produces the indicated eliminations. Note we
get to solve a cell! We also get one more easy step!
Naked triple 126 in column 5
The naked triple 126 in column 5 above justifies the indicated eliminations. Now, however, the
puzzle is still quite difficult. In order to continue, I thought it best to investigate a bit
more into the group of 16 strong inferences considered in the first step. Although I cannot
eliminate any pencil marks, I certainly can restrict the puzzle solution group. In other words,
I can prove some relationships which hopefully will make the puzzle easier to handle.
Prelude
Note that because the center cell, r5c5 is given 7, there is a bit of asymmetry in this almost
too symmetrical puzzle. This makes the weak relationships graphed above very significant, as they
serve to substantially restrict the possible puzzle solutions. Consider for a moment the original
4 mega sets of candidates 1267 in rows 2&8, columns 2&8. We can say a bit more about them!
Here is the logic for one little piece of the puzzle:
- note: (2)r5c2-(2)r5c8 =>(16)r5c2=(16)r5c8
- (16)r5c2: (1)r8c3=(1)r7c2-(1=6)r5c2-(6)r9c2=(6)r8c1 => (1=6)r8B7 and (1-6)c2B7
- (16)r5c8:
- [(7)r8c4=(7)r8c9-(7)r9c8=(7-2)r4c8=(2)r7c8-(2)r8c7=(2)r8c45]
- -(16)@r8c4ORc5=[(1)r8c3=(1-6)r8(c5ORc4)=(6)r8c1]
=> (1=6)r8B7 and (1-6)c2B7
- Clearly, we have proven (1=6)r8B7 and (1-6)c2B7
- Note (2)r2c5-(2)r8c5=>(16)r5c2=(16)r5c8 (note extreme puzzle symmetry)
- We can continue with symmetric conclusions.
- Specifically, the following is easily proven:
- All the sets in this list contain exacly one truth:
- (27)r2c13, (27)r2c56, (16)r2c56,(16)r2c79
- (16)c8r13,(16)c8r45,(27)c8r45,(27)c8r79
- (27)r8c79,(27)r8c45,(16)r8c45,(16)r8c13
- (16)c2r79,(16)c2r56,(27)c2r56,(27)c2r13
We can now use these relationships prn.
Also, we can expect that there will occasionally be symmetric eliminations. This will
be illustrated on some following pages.
This puzzle is still a bear. However, there exists multiple logical attacks. Over the
next few days, I hope to illustrate one such path to tame this beast.