The
general concept is to label pairs of candidate numbers which appear only twice
in a row, column, or box with letter pairs:
AB, CD, EF, etc. Similarly, if a
cell has only two candidates these candidates can be labeled with the letter
pairs. The labeling for a given letter
pair can continue as far as possible, branching out to other rows, columns, or
boxes. For a given letter pair, either
all candidates labeled with the first letter are true and all candidates
labeled with the second letter are false, or vice versa. The intertwining of these labeled candidates
will produce conclusions which otherwise are not immediately obvious.
Notation:
->
means implies
~A means A is false
~AC means that both
A and C being true is not the case
AvC
means that either A or C (maybe both) is the case
A iff C means A if and only if C
Here
are some of the simple inferences which often occur:
1)
By the way the labeling is constructed, for letter pairs we have ~AB and AvB (thus A or B is true, but not both).
2)
If two labeled candidates of the same number are in the same row, column, or
box then these two labels cannot both be true.
Similarly if two labeled candidates are in the same cell, then these two
labels cannot both be true. For example,
if we have in a given cell:
AC
567
Then
we know that ~AC.
3)
If we have ~AC and ~BE, it follows that ~CE.
Proof: C -> ~A -> B ->
~E
4)
If we have only labeled candidates of the same number in a given row, column,
or box, or only two labeled candidates in a given cell, then these labels must
be letter pairs. For example:
AC
56
If
the cell contains only these candidates, then C = B, D = A, and we can replace
all C’s and D’s with B and A respectively.
5)
If we determine that ~AC and ~BD, we can conclude that D = A, C = B, and can
replace all C’s and D’s with B and A respectively. Proof:
C -> ~A -> B. ~C -> D
-> ~B. Hence C iff B, so they are equivalent.
6)
If we determine that ~AC and ~BC, we can conclude that ~C, therefore D. So all D’s are true, and all C’s are
false. This is of course the nicest
conclusion to find.
7)
If a candidate can “see” both members of a lettered pair (i.e. is the same
number in a row, column, or box or exists in the same cell), then that
candidate can be eliminated. For
example:
A
678
B
567
Assuming
these two cells are in the same row, column, or box, we can eliminate the 6 in
the lower cell. If it were true then
neither A nor B would be true, which is impossible. This is the most common candidate elimination
which occurs with this technique.
8)
Suppose we have determined that ~AC. Therefore, BvD. So any candidate which can “see” both a B and
a D can be eliminated.
9)
If we have two labeled candidates which can “see” the members of a cell with
only two candidates, then both of these labels can’t be true. For example:
A
456
47
C
789
If
the 456 cell and the 789 cell can both see the 47 cell, then ~AC, since
otherwise it would be impossible to have an entry in the 47 cell.
The
above simple inferences are sufficient to make most deductions.
In
the following puzzle, I illustrate step by step the deductions I made from the
labeling. For each diagram, I describe
what changes I am about to make in the puzzle and why. I have made no attempt to clean up this
solution. These are the steps that I
actually took to solve the puzzle, in the order in which I found things. Thus, there will be some steps which will not
be necessary for the solution.
It
should be noted that this labeling is not necessary. Any of the conclusions I draw can be made
from appropriate inference chains or other advanced techniques. However, the
labeling makes it unnecessary to find these long chains.
Diagram 1
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9 |
8 |
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7 |
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6 |
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8 |
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9 |
2 |
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1 |
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7 |
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5 |
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4 |
3 |
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6 |
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4 |
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5 |
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7 |
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5 |
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4 |
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1 |
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9 |
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8 |
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3 |
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2 |
6 |
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9 |
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2 |
3 |
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8 |
4 |
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1 |
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1 |
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5 |
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8 |
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A |
B |
C |
D |
E |
F |
G |
H |
I |
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This
is problem 1426 from Paul’s
pages. Toughness rating: 1846
Notation: Columns are labeled a to
i. Rows are
labeled 1 to 9. Cells are a letter and a
number, e.g. c8 stands for the cell in column c and row 8. Boxes are denoted TMB (top, middle bottom)
and LCR (left, center, right) as below:
TL |
TC |
TR |
ML |
MC |
MR |
BL |
BC |
BR |
C8(3) means the candidate 3 in c8
c8
= 3 means that 3 is in cell c8
c8
<> 3 means that c8 does not contain the number 3
Diagram 2
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|
9 |
8 |
2 |
34 |
1 9 139 |
1 13 |
7 |
5 59 |
6 |
5 459 |
|
8 |
467 |
367 |
9 |
2 |
6 36 |
5 |
8 |
47 |
1 |
|
7 |
1 |
5 |
67 |
8 9 89 |
8 6 68 |
4 |
3 |
2 27 |
2 279 |
|
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
|
5 |
29 25679 |
367 |
8 35678 |
4 13478 |
1378 |
26 1236 |
12569 |
12345 |
234569 |
|
4 |
2 2567 |
367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
|
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
|
2 |
3 |
9 |
567 |
17 |
2 127 |
8 |
4 |
1257 |
2567 |
|
1 |
467 |
1 |
467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
|
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Here
is the board after the usual preliminary work.
I list the candidates at the bottom of the cell. In addition, if a particular candidate can
appear in a cell only in one row or one column, I designate that by putting the
candidate in the upper left corner of the possible cells if they are in one row, or in the upper right corner of the possible cells if
they are in one column.
For
example, look at the candidate 1 in the TC box.
1 can appear only in d9 or e9. So
in these two cells, I put a 1 in the upper left corner.
Similarly,
in the same
I
find this helps me see patterns later on.
Diagram 3
|
9 |
|
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
34 |
1 9 139 |
1 13 |
7 |
5 59 |
6 |
5 459 |
145 |
8 |
467 |
367 |
9 |
2 |
6 36 |
5 |
8 |
47 |
1 |
34
|
7 |
1 |
5 |
67 |
8 9 89 |
8 6 68 |
4 |
3 |
2 27 |
2 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 25679 |
367 |
8 35678 |
4 13478 |
1378 |
26 1236 |
12569 |
12345 |
234569 |
|
4 |
2 2567 |
367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
4 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
4 |
2 |
3 |
9 |
567 |
17 |
2 127 |
8 |
4 |
1257 |
2567 |
6 |
1 |
467 |
1 |
467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
In addition, I list on top of each column every candidate which
appears only twice in that column, and on the right of each row every candidate
which appears only twice in that row. I do the same for each box as illustrated
below. The candidates which
are in boldface are those which are paired only in that row, column, or
box. These turn out to be more valuable,
since they let me jump around more.
Doing
this helps in several ways.
1) It may pick up any errors which I have
made.
2) I may find last minute naked
singles, pairs, or triplets which I had overlooked.
3) It helps me judge which candidates
to label.
4) It cuts down on labeling mistakes.
34 |
1689 |
245 |
89 |
4 |
|
5 |
24 |
136 |
|
Diagram 4
|
9 |
|
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
34 |
1 9 139 |
1 13 |
7 |
5 59 |
6 |
5 459 |
145 |
8 |
467 |
367 |
9 |
2 |
6 AB 36 |
5 |
8 |
47 |
1 |
34
|
7 |
1 |
5 |
67 |
8 9 89 |
8 6 68 |
4 |
3 |
2 27 |
2 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 25679 |
367 |
8 35678 |
4 13478 |
1378 |
26 1236 |
12569 |
12345 |
234569 |
|
4 |
2 2567 |
367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
4 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
4 |
2 |
3 |
9 |
567 |
17 |
2 127 |
8 |
4 |
1257 |
2567 |
6 |
1 |
467 |
1 |
467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Now I start my labeling. The idea is to label candidates which either
are all true or all false with the same letter, say A, and to label
corresponding candidates B which will be true iff A
is false. I use pairs of letter to
denote this – AB, CD, EF and so on.
I try to choose my labeling to branch
out to the most candidates possible.
For this puzzle, I choose to start
with:
A: e8(3)
B: e8(6)
I can then continue labeling A’s and
B’s as long as I can find pairs. For
example, since there are only two 6’s in the e-column, I can label e7(6) = A. And then
since there are only two candidates in e7, I can label e7(8)
= B.
34 |
1689 |
245 |
89 |
4 |
|
5 |
24 |
136 |
Diagram 5
|
9 |
|
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
467 |
BA 367 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
34
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 25679 |
367 |
8 35678 |
4 13478 |
A 1378 |
26 1236 |
12569 |
A 12345 |
234569 |
|
4 |
2 2567 |
367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
4 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
4 |
2 |
3 |
9 |
567 |
17 |
2 127 |
8 |
4 |
1257 |
2567 |
6 |
1 |
467 |
1 |
467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
I have completed the AB labeling as far
as I can go. Now look at b8. b8(3) = B and b8(6)
= A. Since A or B is true, it is clear
that I can eliminate b8(7).
The logic chain would be as
follows: b8 = 7 -> e8 = 3 -> e7 =
6 -> c7 = 7 -> b8 <> 7
By similar reasoning, since b8(6) = A and e8(6) = B, a8(6) can be eliminated.
The beauty of this approach is that it
isn’t necessary to recognize these patterns or to follow the inference
chains. Simply seeing the AB in b8
immediately tells us that b8(7) can be eliminated.
In the next diagram I eliminate b8(7) and a8(6), which allows me to do more labeling. E.g. a8(7) = B,
since there are now only two 7’s in the 8-row and h8(7) = A.
34 |
1689 |
245 |
89 |
4 |
|
5 |
24 |
136 |
Diagram 6
|
9 |
7 |
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 2569 |
7 367 |
8 3568 |
4 13478 |
A 1378 |
26 1236 |
12569 |
A 12345 |
234569 |
|
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
4 |
2 |
3 |
9 |
567 |
17 |
2 127 |
8 |
4 |
1257 |
2567 |
6 |
1 |
467 |
1 |
A 467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
In addition to the elimination and further
labeling, I try to keep my bookkeeping of pairings up-to-date. For example, now there are 7’s only in b4 or
b5 on the b-column. Hence I put 7’s in
the upper right corner of these cells, and can thus eliminate any other 7’s in
the ML box.
At this point I happened to notice that
the 5’s in c2, c5, h2, and h5 form an x-wing, so I can eliminate all other 5’s
in the 2-row and 5-row. But if I hadn’t noticed this, or even if I had never
heard of an x-wing, my approach would have caught this as soon as I labeled
these. Let’s say I label c5(5) = C and c2(5) = D (which I will, in fact). C -> h5 <> 5 and D -> h2 <>
5 -> h5 = 5. Thus h5(5)
= D and h2(5) = C. For example i5 <> 5 from c5(5)
= C and h5(5) = D.
3467 |
1689 |
245 |
789 |
4 |
|
5 |
24 |
136 |
Diagram 7
|
9 |
7 |
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 C 3568 |
4 13478 |
A 1378 |
26 1236 |
1269 |
AD 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
15 |
9 |
357 |
4 |
2 |
3 |
9 |
D 567 |
17 |
2 127 |
8 |
4 |
C 1257 |
267 |
56 |
1 |
467 |
1 |
A 467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
From h5 I have the important piece of
information that ~AD. Therefore, we must
have BvC.
Consequently, any candidate which can see both a B and a C can be
eliminated. In this case I can eliminate
c5(6) since it can see c5(5) = C and c7(6) = B.
The logic chain is easy to follow with
the labeling. c5 = 6 -> c5 <> 5
-> h5 = 5 -> h5 <> 4 -> h8 = 4 -> a8 = 7 -> c7 = 6 ->
c5 <> 6.
This chain wouldn’t be immediately
apparent, at least not to me. But I
don’t need to spot the chain. Simply
noticing that c5(6) can see both B and C is all I
need.
I now choose to label
g3(1) = E
g3(5) = F
Next diagram
3467 |
1689 |
245 |
789 |
4 |
|
5 |
24 |
136 |
~AD
Diagram 8
|
9 |
7 |
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 C 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AD 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
D 567 |
17 |
2 127 |
8 |
4 |
F C 1257 |
267 |
56 |
1 |
467 |
1 |
A 467 |
5 |
24 2347 |
9 |
26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
From h2 I immediately have ~CF. And since ~AD, it follows that ~AF. The reasoning is as follows:
A
-> ~D -> C -> ~F therefore ~AF.
This logical inference comes up all the
time with this approach.
I now choose to label
g1(2) = G
g1(6) = H
Next diagram
3467 |
1689 |
245 |
789 |
4 |
|
5 |
24 |
136 |
~AD
~AF
~CF
Diagram 9
|
9 |
7 |
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 C 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AD 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
DH 567 |
17 |
2 127 |
8 |
4 |
F C 1257 |
G 267 |
56 |
1 |
467 |
1 |
A 467 |
5 |
24 2347 |
9 |
GH 26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
~DH from c2. Since ~CF, therefore ~FH.
~BH from c7(6)
and c2(6).
A -> c7 = 7 -> c2 <> 7
A -> c2 <> 5 since ~AD
Therefore, A -> c2 = 6, so ~AG
Since ~AG and ~BH, it follows that G =
B and H = A. (next diagram)
I didn’t really have to label GH to get
to this conclusion. The key is that
~AD. So B -> c2 <> 6 and A
-> c2 = 6, which means I can label c2(6) = A and
the rest of the labeling would follow.
3467 |
1689 |
245 |
789 |
4 |
|
5 |
24 |
136 |
~AD
~AF
~CF
~DH
~FH
~BH
Diagram 10
|
9 |
7 |
458 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 368 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 C 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AD 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
DA 567 |
17 |
2 127 |
8 |
4 |
F C 1257 |
B 267 |
56 |
1 |
467 |
1 |
A 467 |
5 |
24 2347 |
9 |
BA 26 |
237 |
8 |
3 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
c1 <> 6 from c2(6)
and c7(6).
Thus, I can label c1(7)
= B.
Also, c6 <> 6 by same reasoning
(next diagram)
3467 |
1689 |
245 |
789 |
4 |
|
5 |
24 |
136 |
~AD
~AF
~CF
Diagram 11
|
9 |
7 |
4568 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 C 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AD 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
DA 567 |
17 |
2 127 |
8 |
4 |
F C 1257 |
B 267 |
56 |
1 |
B 467 |
1 |
AB 47 |
5 |
24 2347 |
9 |
BA 26 |
237 |
8 |
36 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Now I have that c2 <> 7 from c1(7) and c7(7). That
leaves only 2 entries in c2. Since these
are labeled D and A, it follows that D = B and therefore C = A. I can thus replace all C’s and D’s with A and
B respectively (next diagram).
3467 |
1689 |
245 |
789 |
4 |
|
56 |
24 |
136 |
~AD
~AF
~CF
Diagram 12
|
9 |
7 |
45678 |
49 |
2468 |
|
19 |
45 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AB 12345 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
BA 56 |
17 |
2 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
B 467 |
1 |
AB 47 |
5 |
24 2347 |
9 |
BA 26 |
237 |
8 |
36 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
From h5(4) and
h5(5), I can eliminate h5(1), h5(2), and h5(3).
That lets me label h6(1) = E (next diagram).
3467 |
1689 |
245 |
789 |
4 |
|
56 |
24 |
136 |
~AF
Diagram 13
|
9 |
7 |
45678 |
49 |
2468 |
|
19 |
1345 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
E 123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AB 45 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
357 |
4 |
2 |
3 |
9 |
BA 56 |
17 |
2 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
B 467 |
1 |
AB 47 |
5 |
24 2347 |
9 |
BA 26 |
237 |
8 |
36 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
I now choose to label
h6(3) = I
h1(3) = J
Next diagram
3467 |
1689 |
245 |
789 |
4 |
1 |
56 |
24 |
136 |
~AF
Diagram 14
|
9 |
7 |
45678 |
49 |
2468 |
|
19 |
1345 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
E I 123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AB 45 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
I 357 |
4 |
2 |
3 |
9 |
BA 56 |
17 |
2 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
B 467 |
1 |
AB 47 |
5 |
24 I 2347 |
9 |
BA 26 |
J 237 |
8 |
36 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
~EI from h6.
Since ~AF, it follows that ~AI.
Since
~AI, BvJ.
Therefore h1 <> 2, since it can see both B and J in g1(2) and h1(3) respectively.
This sort of relationship is very easy
to spot. As soon as ~AI is determined, a
quick check of the B’s and J’s will find any candidate which can see both
letters and can thus be eliminated. The
chain which proves this is probably pretty long, but I don’t have to worry
about it.
This elimination of h1(2)
and appropriate labeling is reflected in the next diagram
3467 |
1689 |
245 |
789 |
4 |
1 |
56 |
24 |
136 |
~AF
~EI
~AI
Diagram 15
|
9 |
7 |
45678 |
49 |
2468 |
|
19 |
1345 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 59 |
6 |
5 A 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
E I 123 |
2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F 1269 |
AB 45 |
23469 |
5 |
4 |
2 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
23456 |
47 |
3 |
457 |
8 |
2 |
6 |
4 1347 |
13 |
EF 15 |
9 |
I 357 |
4 |
2 |
3 |
9 |
BA 56 |
17 |
2 B 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
B 467 |
1 |
AB 47 |
5 |
24 AI 2347 |
9 |
BA 26 |
JI 37 |
8 |
236 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
a1 <> 4 from a1(6)
and c1(4). Obviously I could have done
this a while ago had I seen it. This
will label a1(7) = A, since only two candidates remain
in a1. That eliminates e1(7) and h1(7) due to a1(7) = A and c1(7) = B. Also, a3(7) is eliminated
due to a1(7) = A and a8(7) = B Since h1
<> 7, I is false and J is true.
All this new information is reflected in the next diagram.
3467 |
1689 |
245 |
789 |
4 |
1 |
56 |
24 |
136 |
~AF
~EI
~AI
Diagram 16
|
4579 |
7 |
45678 |
49 |
2468 |
|
19 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 KL 59 |
6 |
5 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F K 1269 |
AB 45 |
3 23469 |
5 |
4 |
2 B 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
256 |
8 |
3 23456 |
47 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
17 |
2 B 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
I choose to label
g9(5) = K
g9(9) = L
~
~FK from g5
Since ~AL, BvK,
therefore g5 <> 2 from g5(9) and g1(2). This allows me to label g4(2)
= A (next diagram)
3467 |
1689 |
245 |
5789 |
4 |
1 |
4567 |
234 |
16 |
~AF
~
~FK
Diagram 17
|
4579 |
7 |
45678 |
49 |
2468 |
|
129 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 KL 59 |
6 |
5 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 269 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
89 |
5 |
29 269 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F K 169 |
AB 45 |
3 23469 |
5 |
4 |
2 B 256 |
7 367 |
1 |
4 347 |
9 |
26 236 |
A 256 |
8 |
3 23456 |
47 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
17 |
2 B 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Now a4 <> 2 from a4(5) and g4(2). That
means that (a5a6) = (29), a naked pair, so 6 can be eliminated from a5 and a6
(next diagram).
Note that this deduction could also
have been made from a4(6) = A after a4(2) is
eliminated, since a1(6) = B.
3467 |
1689 |
245 |
5789 |
4 |
1 |
4567 |
234 |
16 |
~AF
~
~FK
Diagram 18
|
24579 |
7 |
45678 |
49 |
2468 |
|
129 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 KL 59 |
6 |
5 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 29 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
89 |
5 |
29 29 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F K 169 |
AB 45 |
3 23469 |
5 |
4 |
BA 56 |
7 367 |
1 |
4 347 |
9 |
26 236 |
A 256 |
8 |
3 23456 |
47 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
17 |
2 B 127 |
8 |
4 |
F A 1257 |
B 267 |
56 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Look at the e-column.
A -> e5 <> 7 from e5(8)
A -> e3 <> 7 from e3(4)
Therefore, A -> e2 = 7 since that is
the only 7 remaining on the e-column.
Since B-> e2 <> 7 from e2(2), I can
label e2(7) = A, thus eliminating e2(1).
The results of this elimination are reflected in the next diagram
3467 |
1689 |
245 |
25789 |
4 |
1 |
4567 |
234 |
16 |
~AF
~
~FK
Diagram 19
|
24579 |
7 |
45678 |
49 |
2468 |
|
129 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 KL 59 |
6 |
5 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 29 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
89 |
5 |
29 29 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F K 169 |
AB 45 |
3 23469 |
5 |
4 |
BA 56 |
7 367 |
1 |
4 347 |
9 |
26 236 |
A 256 |
8 |
3 23456 |
47 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
F A 1257 |
B 267 |
156 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
i4 <> 6 from a4(6)
and i2(6).
g4 <> 5 from a4(5)
and g4(2).
Next diagram
3467 |
1689 |
245 |
25789 |
4 |
1 |
4567 |
234 |
16 |
~AF
~
~FK
Diagram 20
|
24579 |
7 |
45678 |
49 |
2468 |
|
1259 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 KL 59 |
6 |
5 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 29 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
89 |
5 |
29 29 |
7 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
F K 169 |
AB 45 |
3 23469 |
5 |
4 |
BA 56 |
7 367 |
1 |
4 347 |
9 |
26 236 |
AB 26 |
8 |
3 A 2345 |
457 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
F A 1257 |
B 267 |
156 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
Since g3(5)
and g9(5) are the only 5’s in the g-column, it must be that K = E and L = F.
f4 <> 6 and b4 <> 6 from a4(6) and g4(6)
g5 <> 6 from g1(6)
and g4(6)
Next diagram reflects this new
information
3467 |
1689 |
245 |
25789 |
4 |
15 |
4567 |
234 |
16 |
~AF
~
~FK
Diagram 21
|
24579 |
67 |
45678 |
49 |
2468 |
6 |
12569 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 EF 59 |
6 |
5 AF 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
29 29 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 2369 |
689 |
5 |
29 29 |
7 B 367 |
8 A 358 |
4 13478 |
A 1378 |
26 1236 |
FE 19 |
AB 45 |
3 23469 |
5 |
4 |
BA 56 |
7 37 |
1 |
4 347 |
9 |
2 23 |
AB 26 |
8 |
3 A 2345 |
4657 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
F A 1257 |
B 267 |
156 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
F -> h6 = 2 -> a6 = 9 -> i6
<> 9
E -> g5 = 9 -> i6 <> 9
Therefore,
i6 <> 9.
Astute readers will note that this is a
classic xy-wing. But I never had to
recognize it as such.
So a6 = 9 since only 9 left in 6-row,
and a5 = 2. Results in next diagram
3467 |
1689 |
245 |
256789 |
46 |
15 |
4567 |
234 |
16 |
~AF
Diagram 22
|
457 |
67 |
45678 |
49 |
2468 |
26 |
12569 |
145 |
|
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 EF 59 |
6 |
5 AF 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
9 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 236 |
68 |
5 |
2 |
7 B 367 |
8 A 358 |
4 13478 |
A 1378 |
6 136 |
9 FE 19 |
AB 45 |
9 3 3469 |
59 |
4 |
BA 56 |
7 37 |
1 |
4 347 |
9 |
2 23 |
AB 26 |
8 |
3 A 2345 |
4657 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
F A 1257 |
B 267 |
156 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
i2 <> 7 from e2(7)
and i2(6). Results of this in next
diagram
3467 |
1689 |
245 |
5678 |
246 |
159 |
4567 |
234 |
16 |
~AF
Diagram 23
|
457 |
67 |
45678 |
49 |
2468 |
26 |
12569 |
145 |
7 |
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 EF 59 |
6 |
5 AF 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 27 |
2 A 279 |
2689 |
6 |
9 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 236 |
68 |
5 |
2 |
7 B 367 |
8 A 358 |
4 13478 |
A 1378 |
6 136 |
9 FE 19 |
AB 45 |
9 3 3469 |
59 |
4 |
BA 56 |
7 37 |
1 |
4 347 |
9 |
2 23 |
AB 26 |
8 |
3 A 2345 |
4657 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
F A 1257 |
AB 26 |
156 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
h2 <> 2 from e2(2)
and i2(2). So h7(2) = E, since only two 2’s left in
h-column and h7(7) = F. Results next
diagram
3467 |
1689 |
245 |
5678 |
246 |
159 |
4567 |
234 |
167 |
~AF
Diagram 24
|
457 |
67 |
45678 |
49 |
2468 |
26 |
12569 |
1245 |
7 |
|
9 |
8 |
2 |
AB 34 |
1 9 A 139 |
1 13 |
7 |
5 EF 59 |
6 |
5 AF 459 |
145 |
8 |
AB 47 |
BA 36 |
9 |
2 |
6 AB 36 |
5 |
8 |
BA 47 |
1 |
3467
|
7 |
1 |
5 |
BA 67 |
8 9 AB 89 |
8 6 AB 68 |
4 |
3 |
2 EF 27 |
2 F A 279 |
2689 |
6 |
9 |
4 |
8 38 |
138 |
5 |
26 1236 |
7 |
EF 12 |
3 236 |
68 |
5 |
2 |
7 B 367 |
8 A 358 |
4 13478 |
A 1378 |
6 136 |
9 FE 19 |
AB 45 |
9 3 3469 |
59 |
4 |
BA 56 |
7 37 |
1 |
4 347 |
9 |
2 23 |
AB 26 |
8 |
3 A 2345 |
4657 |
3 |
BA 45 |
8 |
2 |
6 |
3 4 A 1347 |
3 13 |
EF 15 |
9 |
57 |
347 |
2 |
3 |
9 |
BA 56 |
EF 17 |
2 BA 27 |
8 |
4 |
FA 157 |
AB 26 |
1256 |
1 |
BA 67 |
1 |
AB 47 |
5 |
24 AB 24 |
9 |
BA 26 |
3 |
8 |
2467 |
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
A -> e3 = 4 -> e3 <> 3
-> f3 = 3 -> f4 = 2 -> g4 <> 2 –> ~A, which is a
contradiction. Hence, A must be false,
and B true. This lets me fill in a lot
of cells. Results next
diagram.
3467 |
1689 |
245 |
5678 |
246 |
159 |
4567 |
234 |
1267 |
~AF
Diagram 25
|
|
|
38 |
478 |
7 |
13 |
159 |
127 |
24579 |
|
9 |
8 |
2 |
4 |
13 13 |
13 13 |
7 |
59 EF 59 |
6 |
59 FE 59 |
1359 |
8 |
7 |
3 |
9 |
2 |
6 |
5 |
8 |
4 |
1 |
|
7 |
1 |
5 |
6 |
9 |
8 |
4 |
3 |
27 EF 27 |
27 FE 27 |
27 |
6 |
9 |
4 |
38 38 |
8 F 138 |
5 |
6 |
7 |
2 EF 12 |
2 3 EF 23 |
128 |
5 |
2 |
6 |
38 38 |
7 48 E 13478 |
7 F 137 |
13 |
9 FE 19 |
5 |
9 34 F 349 |
4789 |
4 |
5 |
7 |
1 |
4 34 |
9 |
2 |
6 |
8 |
34 34 |
34 |
3 |
4 |
8 |
2 |
6 |
3 E 137 |
3 13 |
5 EF 15 |
9 |
5 EF 57 |
357 |
2 |
3 |
9 |
5 |
EF 17 |
2 |
8 |
4 |
FE 17 |
6 |
17 |
1 |
6 |
1 |
7 |
5 |
4 |
9 |
2 |
3 |
8 |
|
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
d5 <> 1 from d2(1)
and d6(1)
Also d9 <> 1 for same reason, so
d9 = 3 and e9 = 1
Next diagram
|
13 |
2579 |
38 |
478 |
1249 |
|
37 |
157 |
Diagram 26
|
|
|
38 |
1478 |
17 |
13 |
159 |
127 |
24579 |
|
9 |
8 |
2 |
4 |
3 |
1 |
7 |
59 EF 59 |
6 |
59 FE 59 |
59 |
8 |
7 |
3 |
9 |
2 |
6 |
5 |
8 |
4 |
1 |
|
7 |
1 |
5 |
6 |
9 |
8 |
4 |
3 |
27 EF 27 |
27 FE 27 |
27 |
6 |
9 |
4 |
38 38 |
8 F 138 |
5 |
6 |
7 |
2 EF 12 |
2 3 EF 23 |
128 |
5 |
2 |
6 |
38 38 |
7 48 E 3478 |
7 F 137 |
13 |
9 FE 19 |
5 |
9 34 F 349 |
4789 |
4 |
5 |
7 |
1 |
4 34 |
9 |
2 |
6 |
8 |
34 34 |
34 |
3 |
4 |
8 |
2 |
6 |
3 E 137 |
3 13 |
5 EF 15 |
9 |
5 EF 57 |
357 |
2 |
3 |
9 |
5 |
EF 17 |
2 |
8 |
4 |
FE 17 |
6 |
17 |
1 |
6 |
1 |
7 |
5 |
4 |
9 |
2 |
3 |
8 |
|
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|
From d9 I have d4 = 4 -> i4 = 3
-> i6 <> 3. So F is false, E is
true, and from this the puzzle can be solved by straightforward means (next
diagram)
|
|
2579 |
38 |
478 |
1249 |
|
37 |
157 |
Diagram 27
|
|
|
|
|
|
|
|
|
|
|
9 |
8 |
2 |
4 |
3 |
1 |
7 |
5 |
6 |
9 |
|
8 |
7 |
3 |
9 |
2 |
6 |
5 |
8 |
4 |
1 |
|
7 |
1 |
5 |
6 |
9 |
8 |
4 |
3 |
2 |
7 |
|
6 |
9 |
4 |
3 |
8 |
5 |
6 |
7 |
1 |
2 |
|
5 |
2 |
6 |
8 |
7 |
3 |
1 |
9 |
5 |
4 |
|
4 |
5 |
7 |
1 |
4 |
9 |
2 |
6 |
8 |
3 |
|
3 |
4 |
8 |
2 |
6 |
7 |
3 |
1 |
9 |
5 |
|
2 |
3 |
9 |
5 |
1 |
2 |
8 |
4 |
7 |
6 |
|
1 |
6 |
1 |
7 |
5 |
4 |
9 |
2 |
3 |
8 |
|
|
A |
B |
C |
D |
E |
F |
G |
H |
I |
|