Ex 117 starts with G7=2, E7=6, E6=8, E9=7, F6=9, C1=3, C2=2, C6=5, C9=4, B6=6 and J1=1. Now will follow Single D4=9s, Pair D1=G1=56, F1=4p, Triple B9+D9+G9=123, F9=6t and Single G3=9s. After clearing of col 5 you get Pair G1=G5=56 and G2=8. Pair B5=B7=78 leads to another Pair B4=B8=45.

I’s possible, that there are hidden some X-Wings and Fishes, but before searching it you should know, which digits are necessary, possible or impossible. Therefore it could be usefull to take a look at black cells, making a Black Cell Analysis BCA.
Picture "Str8ts Ex117 #24" shows the known impossible digits in black colour in the near of right and bottom border. Possible digits are in red colour at left and top border.

Comparing red digits shows no 4 and 5 on top and no 7 und 9 at left border. This shows that no other 4, 5 und 7 could be hidden in black cells. Therefore red 4, 5 and 7 can be removed. Now candidates 4, 5 and 7 are necessary in white cells. After this is done, left red digits 16 near row A and left red digits 123 near row D will stay allone. Because candidate 4 becomes cecessary in row D, D8 must be 4!

Now candidate 5 is necessary in row E too and for candidate 1 there is no longer free any white cell in row E1-4=1-5. Digit 1 becomes black and changes sites. E1-4=2-5. The same procedure as with red digit 1 is done with red digit 9 on top of col 7. BC7=78.

As the matter of facts two black 9 stay at right and lower border. Only one lonely red 9 is standing over col 5. It can be removed similar to red 7 and candidate 9 becomes necessary in col 5! BCD5=789, FGHJ5=2-6.
With picture "Str8ts Ex117 #25" starts the end of solving, because Ex127 solves quite easy now. D9=3, G9=1, B9=2, B8=5, A8=6, G8=3, B4=4, B3=3, A3=2, A4=3, A2=4, E4=5, E3=4, E2=3, J4=2, J6=3, J8=7, C8=8, H6=2, H8=1, F8=2, H4=8, F4=1, F5=5, G5=6, F2=7, F3=8, G1=5, J5=4, D1=6, H5=3, J7=5, H7=4, J3=6, D2=5, H2=6, H3=5, D3=7, D5=8, B5=7, C5=9, B7=8 und C7=7.

## Solving Ex 117 with Black Cell Analysis BCA

### Solving Ex 117 with Black Cell Analysis BCA

Gruß von Jens

### Add: Solving Ex 117 with Black Cell Analysis BCA

Please eliminate the previous text after Picture Str8ts Ex117 #24 and use the following changed text. The green marked text is new.

Picture "Str8ts Ex117 #24" shows the known impossible digits in black colour in the near of right and bottom border. Possible digits are added in red colour at left and top border.

Comparing red digits shows no 4 and 5 on top and no 7 und 9 at left border. This shows that no other 4, 5 und 7 could be hidden in black cells. Therefore red 4, 5 and 7 can be removed. Now candidates 4, 5 and 7 are necessary in white cells. After this is done, left red digits 16 near row A and left red digits 123 near row D will stay allone. Because candidate 4 becomes cecessary in row D, D8 must be 4!

Now candidate 5 is necessary in row E too and for candidate 1 there is no longer free any white cell in row E1-4=1-5. Digit 1 becomes black and changes sites. E1-4=2-5. A similar procedure as with red digits 15 is done with red digits 79 on top of col 7. Red digit 9 over column 7 changes colour and sites. BC7=78.

As the matter of facts two black 9 stay at right and two black 9 stay at lower border. Only one lonely red 9 is standing over col 5. It can be removed similar to red 7 and candidate 9 becomes necessary in col 5! BCD5=789, FGHJ5=2-6. With picture "Str8ts Ex117 #25" starts the end of solving, because Ex117 solves quite easy now. D9=3, G9=1, B9=2, B8=5, A8=6, G8=3, B4=4, B3=3, A3=2, A4=3, A2=4, E4=5, E3=4, E2=3, J4=2, J6=3, J8=7, C8=8, H6=2, H8=1, F8=2, H4=8, F4=1, F5=5, G5=6, F2=7, F3=8, G1=5, J5=4, D1=6, H5=3, J7=5, H7=4, J3=6, D2=5, H2=6, H3=5, D3=7, D5=8, B5=7, C5=9, B7=8 und C7=7.

Picture "Str8ts Ex117 #24" shows the known impossible digits in black colour in the near of right and bottom border. Possible digits are added in red colour at left and top border.

Comparing red digits shows no 4 and 5 on top and no 7 und 9 at left border. This shows that no other 4, 5 und 7 could be hidden in black cells. Therefore red 4, 5 and 7 can be removed. Now candidates 4, 5 and 7 are necessary in white cells. After this is done, left red digits 16 near row A and left red digits 123 near row D will stay allone. Because candidate 4 becomes cecessary in row D, D8 must be 4!

Now candidate 5 is necessary in row E too and for candidate 1 there is no longer free any white cell in row E1-4=1-5. Digit 1 becomes black and changes sites. E1-4=2-5. A similar procedure as with red digits 15 is done with red digits 79 on top of col 7. Red digit 9 over column 7 changes colour and sites. BC7=78.

As the matter of facts two black 9 stay at right and two black 9 stay at lower border. Only one lonely red 9 is standing over col 5. It can be removed similar to red 7 and candidate 9 becomes necessary in col 5! BCD5=789, FGHJ5=2-6. With picture "Str8ts Ex117 #25" starts the end of solving, because Ex117 solves quite easy now. D9=3, G9=1, B9=2, B8=5, A8=6, G8=3, B4=4, B3=3, A3=2, A4=3, A2=4, E4=5, E3=4, E2=3, J4=2, J6=3, J8=7, C8=8, H6=2, H8=1, F8=2, H4=8, F4=1, F5=5, G5=6, F2=7, F3=8, G1=5, J5=4, D1=6, H5=3, J7=5, H7=4, J3=6, D2=5, H2=6, H3=5, D3=7, D5=8, B5=7, C5=9, B7=8 und C7=7.

Gruß von Jens