The New Sudoku Players' Forum

[ArkieTech]

I started here

300080004090007010000400200080705400200090006005302080001006800040100060500070003

and wound up here

Code: Select all

 *--------------------------------------------------------------------*
 | 3      1257   27     | 2569   8      19     | 5679   579    4      |
 | 48     9      248    | 256    235    7      | 356    1      58     |
 | 1678   567    678    | 4      135    39     | 2      3579   5789   |
 |----------------------+----------------------+----------------------|
 | 169    8      369    | 7      16     5      | 4      239    129    |
 | 2      137    347    | 8      9      14     | 357    357    6      |
 | 14679  67     5      | 3      146    2      | 79     8      179    |
 |----------------------+----------------------+----------------------|
 | 79     237    1      | 259    2345   6      | 8      24579  2579   |
 | 789    4      23789  | 1      235    389    | 579    6      2579   |
 | 5      26     2689   | 29     7      489    | 1      249    3      |
 *--------------------------------------------------------------------*


I see a couple of ALS's that go nowhere. Can anyone help?

dan

[eleven]

r2c1=4 -> r5c3=4 -> r6c5=4 -> r9c6=4 -> r9c3=8
So r23c3 and r8c1 cannot be 8.

Sorry, there were 2 typos, corrected now.

[ArkieTech]

eleven said

r1c2=4 -> r4c3=4 -> r6c5=4 -> r9c6=4 -> r9c3=8
So r23c3 and r8c1 cannot be 8.

4 is not a candidate in r1c2. What am I missing?
dan

[Jean-Christophe]

How about this:
(8=4)r2c1-(4)r6c1=(4)r6c5-(4)r5c6=(4)r9c6-(8)r9c6=(8)r9c3 -> r23c3, r8c1 <> 8
r78c1 = naked pair {79}
r4c3 = 9
...

[ArkieTech]

Jean_Christope said

(8=4)r2c1-(4)r6c1=(4)r6c5-(4)r5c6=(4)r9c6-(8)r9c6=(8)r9c3 -> r23c3, r8c1 <> 8
r78c1 = naked pair {79}
r4c3 = 9

Excellent! Something for me to study today. Thanks

dan

[eleven]

Sorry for the typo, should say r2c1=4 instead of r1c2=4.

It is the same AIC like the one by Jean_Christope, he just took a shortcut (r1c2=4 -> r6c5=4 instead of r1c2=4 -> r5c3=4 -> r6c5=4). The way i found it was to look, where the 4's can be, and there are only 2 possibilities in 5 boxes.
Since there is the bivalue 48. i also looked for strong links for 8. There are only 2 (one in col 6, which also eliminates 8 in r8c1) and the one in row 9.

[ArkieTech]

eleven said

r2c1=4 instead of r1c2=4

Thanks eleven. Some day I will learn how to "see" these things. Getting there sure is fun.

dan

[StrmCkr]

DEATH BLOSSOM: ([ALS:r2c1|r2c9] and [ALS:r1c2|r1c3|r1c6|r1c8]) + Stem:r2c3 means 5 can be removed from r1c7

DEATH BLOSSOM: ([ALS:r7c1|r8c1] and [ALS:r1c3|r2c1|r2c3|r3c3]) + Stem:r9c3 means 7 can be removed from r3c1

[ArkieTech]

strmckr said

DEATH BLOSSOM: ([ALS:r2c1|r2c9] and [ALS:r1c2|r1c3|r1c6|r1c8]) + Stem:r2c3 means 5 can be removed from r1c7

DEATH BLOSSOM: ([ALS:r7c1|r8c1] and [ALS:r1c3|r2c1|r2c3|r3c3]) + Stem:r9c3 means 7 can be removed from r3c1

That gets you

Code: Select all

 *--------------------------------------------------------------------*
 | 3      1257   27     | 2569   8      19     | 679    579    4      |
 | 48     9      248    | 256    235    7      | 356    1      58     |
 | 168    567    678    | 4      135    39     | 2      3579   5789   |
 |----------------------+----------------------+----------------------|
 | 169    8      369    | 7      16     5      | 4      239    129    |
 | 2      137    347    | 8      9      14     | 357    357    6      |
 | 14679  67     5      | 3      146    2      | 79     8      179    |
 |----------------------+----------------------+----------------------|
 | 79     237    1      | 259    2345   6      | 8      24579  2579   |
 | 789    4      23789  | 1      235    389    | 579    6      2579   |
 | 5      26     2689   | 29     7      489    | 1      249    3      |
 *--------------------------------------------------------------------*

What is next?

dan

[StrmCkr]

heres everything i can spot. after the two death blossoms.
(death blossom is an extension of als combined)

ALS/ALS: [r1c6|r3c5|r3c6] and [r2c1|r2c3|r2c5|r2c9], 3 is restricted common, other common candidate 5 can be removed from r2c4

AIC Rule 2, on 8 (Alternating Inference Chain, length 6):
4[r2c3]=4[r5c3]-4[r5c6]=4[r9c6]-8[r9c6]=8[r9c3]-
- Chain ends r2c3 cannot be 8 and r9c3 cannot be 4

ALS/ALS: [r1c3|r2c3] and [r2c1|r7c1|r8c1], 4 is restricted common, other common candidate 7 can be removed from r8c3

ALS/ALS: [r1c6|r3c5|r3c6] and [r2c1|r2c3|r2c5|r2c9], 3 is restricted common, other common candidate 5 can be removed from r2c4

AIC Rule 1, 8 taken off r3c3 - chain ends: r2c1 r9c3
AIC Rule 1, 8 taken off r8c1 - chain ends: r2c1 r9c3
AIC on 8 (Alternating Inference Chain, length 8):
8[r2c1]=4[r2c1]-4[r2c3]=4[r5c3]-4[r5c6]=4[r9c6]-8[r9c6]=8[r9c3]-

puzzle cracked to singles at this point.