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Example 2 (simple)
9
2
3
7
2
8
7
9
6
8 5
9
2
0 0 0
3 1
H
0 0
4
7
1
6 2
3
Example 3 (advanced) after two presses on the HINT button.
1
0
6
2
7
3
6
2
5
7
4
6
2
8
1
4
6
0
6
5
8
7
3
8
2
5
6
4
9
2
6
A
B
C
D
E
F
G
The next two displays follow on from this after another press on the HINT
button (the machine shows them alternating with each other). Apart from the 2,
the other three missing numbers in the E-column are: 1, 3 and 9.
1
2
3
6
7
3
6
2
5
7
4
6
2
8
1
4
6
3
6
5
8
7
3
8
0
5
6
4
9
0
6
A B C D E F G H I
So there must be a 3 in one of these cells. The next press on HINT produces
the following display:
1
6
2
7
3
6
2
4
6
2
5
8
7
1
4
6
6
5
8
7
3
8
H
0
5
6
0
4
0
9
0
6
A
B
C
D
E
F
G
or Fb, as it would conflict with the 3 at Hb. Therefore the H cell is the only
possible cell in this block for a 3.
The cells indicated by 0s cannot contain the "flashing"
4
5
number (in this case 4). Therefore the H cell is the only
place in this row where a 4 can be inserted.
6
1
4
0
1
4
There must be a 2 at Ea or Ec, so there cannot be a 2
7
i
5
h
further up the same column (at Ef or Ei).
9
g
3
1
f
9
e
6
d
6
c
3
2
b
8
a
H
I
i
7
1
h
5
7
3
g
9
3
4
6
2
f
1
4
e
9
6
d
6
8
c
6
8
b
3
2
6
a
8
A B C D E F G H I
7
i
5
h
The H flashing with the 0s shows you that one
9
3
g
particular number has been eliminated as a
1
f
9
"candidate" for all the 0 cells and must therefore be in
e
6
d
the H cell. Which number do you think this is?
6
c
3
2
b
8
a
We know that there must be a 3 at Ef or Ei, so there
H
I
cannot be a 3 at Ea or Ec. Nor can there be a 3 at Db
i
2
9
6
7
h
6
2
5
g
5
7
9
8
1
3
f
6
9
1
e
5
9
d
7
3
6
c
0
5
6
b
4
3
2
a
9
0
6
8
- 7 -
These alternating displays
show that the only possible
numbers for Ef and Ei are 3
and 9. (We know that there
cannot be a 2 in either of
these cells, and there is
already a 1 in row i and
another 1 in row f.)
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