I was no quite happy with the path given so far.
Here my best path for this puzzle rated 11.5 by sudoku explainer
- Code: Select all
9 8 13 |7 1245 12345 |6 245 125
5 17 4 |6 8 12 |3 279 1279
1367 2 136 |134 9 1345 |157 4578 1578
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2346 5 2369 |2349 2467 8 |279 1 23679
12368 1369 7 |5 126 1239 |4 289 23689
123468 13469 12369 |12349 12467 12349 |2579 25789 2356789
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1247 1479 8 |1249 1245 6 |12579 3 12579
1237 1379 12359 |8 125 1259 |12579 6 4
1246 1469 12569 |1249 3 7 |8 259 1259
We have a partial DJE for the digits 259,
R8c56;r7c7,r9c3
R9c89;r7c4,r9c3
As usual in a DJE, each base sees the targets of the other base.
We cannot have more than once the 3 digits '259' in the 2 bases, so we have once or twice the digit ‘1’
Having twice the digit ‘1’ leads to a contradiction
- Code: Select all
9 8 13 |7 1245 12345 |6 245 125
5 17 4 |6 8 12 |3 279 1279
1367 2 136 |134 9 1345 |157 4578 1578
----------------------------------------------------------
2346 5 2369 |2349 2467 8 |279 1 23679
12368 1369 7 |5 126 1239 |4 289 23689
123468 13469 12369 |12349 12467 12349 |2579 25789 2356789
----------------------------------------------------------
1247 1479 8 |249 245 6 |2579 3 2579
1237 1379 2359 |8 125 1259 |2579 6 4
1246 1469 2569 |249 3 7 |8 259 1259
I use here the single chain of Blue
column single 1r3c7
column single 1r6c4
column single 1r1c3
row single 3r1c6
cell single 4r3c4
cell single 5r3c6
box single 4r7c5
column single 5r8c5
--> no candidate for 1 in c5
Having only once the digit ‘1’, we have a pure DJE in bases
R7c7=7; r8c4=4
And a new PM
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9 8 13 |7 1245 12345 |6 245 125
5 17 4 |6 8 12 |3 279 1279
1367 2 136 |13 9 1345 |15 4578 1578
--------------------------------------------------------
2346 5 2369 |239 2467 8 |29 1 23679
12368 1369 7 |5 126 1239 |4 289 23689
123468 13469 12369 |1239 12467 12349 |259 25789 2356789
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47 47 8 |129 125 6 |1259 3 1259
123 139 1259 |8 125 1259 |7 6 4
126 169 1259 |4 3 7 |8 259 1259
Here the best seems to consider the impossible pattern threat 29r3c347
with 3r1c3<->3r3c4
any candidate forcing 36r13c3 can be cleared
-1 r2c2,r3c1,r3c6,r3c7
r2c2=7; r3c7=5=> r7c7=1After basic plus DJE
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9 8 13 |7 125 1235 |6 4 12
5 7 4 |6 8 12 |3 29 129
36 2 136 |13 9 4 |5 7 8
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4 5 239 |239 7 8 |29 1 6
12 19 7 |5 6 129 |4 8 3
8 1369 12369 |1239 4 1239 |29 5 7
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7 4 8 |29 25 6 |1 3 259
23 39 259 |8 125 1259 |7 6 4
126 169 2569 |4 3 7 |8 29 259
The 29 threat in r4 gives now -3 r1c3,r3c4
r1c3=1 stte