Puzzle 6:
With singles and box-line for 6 in c4 you get here:
- Code: Select all
. 5 3 | - . 9 | 7 8 #
7 1 9 | 3 . . | . 6 5
x x x | - 7 5 | 3 9 1
------+-------+------
1 7 5 | # 9 3 | 6 # 8
. 9 . | 5 6 7 | 1 3 #
3 . . | . . . | 9 5 7
------+-------+------
9 . 7 | . 5 . | . . 3
5 2 1 | 9 3 4 | 8 7 6
. . 4 | 7 . . | 5 . 9
The 4 # marked cells only can be 24, thus there is a remote pair in r4c4 and r1c9, eliminating 24 in r1c4 (leaving 16).
But there is more: 2/4 in r1c9 also implies 2/4 in r3c123. So you can eliminate 24 also in r3c4, leaving 6 alone, because the 8 also is locked in r3c123.
Singles after that move.
Puzzle 7 was a long one for me.
After singles:
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9 6 2|8 5 3|1 7 4
5 8 .|. . 4|3 2 6
# 3 4|6 2 @|p p 8
-----+-----+-----
. . .|7 8 5|6 . .
. . .|. 6 .|. . .
. . 6|3 4 2|° p .
-----+-----+-----
4 . .|5 1 6|. 8 .
- 1 5|2 @ 8|4 6 .
6 2 8|4 3 #|. 1 5
In row 3 17 is left in the marked cells. With the strong link for 7 in box 8 we have a kite eliminating 7 in r8c1, leaving 3.
Also note the UR 59 (p) in r36c78 leaving 78 in r6c7.
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9 6 2|8 5 3|1 7 4
5 8 @|. # 4|3 2 6
# 3 4|6 2 @|. . 8
-----+-----+-----
12 . .|7 8 5|6 . .
78 . .|. 6 .|. . 27
. . 6|3 4 2|78. .
-----+-----+-----
4 . .|5 1 6|2 8 3
3 1 5|2 @ 8|4 6 #
6 2 8|4 3 #|@ 1 5
Now we can find more with those 7's. With the strong links either all # cells must be 7 or all @ cells.
So one of r1c3 and r9c7 must be 7, implying r5c1=8 and r6c7=8 resp. Since one leads to the other, both must be true.
Since r3c1=17, either r3c1=1 implying r4c1=2 or (r3c1 and) r8c9=7 implying r5c9=2. Again one 2 leads to the other.
This gives
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9 6 2|8 5 3|1 7 4
5 8 @|. # 4|3 2 6
# 3 4|6 2 @|. . 8
-----+-----+-----
2 . .|7 8 5|6 . .
8 . .|. 6 .|# . 2
@ . 6|3 4 2|8 . @
-----+-----+-----
4 . .|5 1 6|2 8 3
3 1 5|2 @ 8|4 6 #
6 2 8|4 3 #|@ 1 5
and new strong links for 7 in columns 1, 7 and 9. Now the @-marked cells cannot be 7, because there are two in row 6.