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 Post subject: Easter Eggs v2 Revisit
PostPosted: Tue Feb 16, 2021 6:25 am 
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Easter Eggs V2 Revisit.

After a couple of easier Revisits (I could solve them!) back to a harder looking one. Up to page 7 of archive A! This gets a score of 1.90 JSudoku uses 5 'complex intersections' and some other techniques that it doesn't use for normal Assassins so this may be over my head. Anyway, worth a try.
Code: Select, Copy & Paste into solver:
3x3::k:2304:4097:4354:4354:6916:4101:4101:2567:4360:2304:4097:4354:6916:6916:6916:4101:2567:4360:4882:4097:2068:2068:6916:3607:3607:2567:4122:4882:4882:2068:5918:2335:6176:3607:4122:4122:2852:2852:5918:5918:2335:6176:6176:2091:2091:2605:2605:3375:5918:2335:6176:2355:4404:4404:2605:5175:3375:3375:9018:2355:2355:3901:4404:1343:5175:3649:9018:9018:9018:4165:3901:1863:1343:5175:3649:3649:9018:4165:4165:3901:1863:
Solution:
+-------+-------+-------+
| 3 4 2 | 6 8 5 | 7 1 9 |
| 6 5 9 | 7 1 2 | 4 3 8 |
| 8 7 1 | 3 9 4 | 2 6 5 |
+-------+-------+-------+
| 5 6 4 | 1 3 7 | 8 9 2 |
| 9 2 8 | 5 4 6 | 3 7 1 |
| 7 1 3 | 9 2 8 | 5 4 6 |
+-------+-------+-------+
| 2 9 6 | 4 5 3 | 1 8 7 |
| 1 3 5 | 8 7 9 | 6 2 4 |
| 4 8 7 | 2 6 1 | 9 5 3 |
+-------+-------+-------+
Cheers
Ed


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PostPosted: Sat Feb 27, 2021 11:47 pm 
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Thanks Ed. That was a tough one, probably the hardest revisit which you've posted (so far); only SKX-2, which I posted as a revisit, may have been harder.

Here's how I solved it this time. As usual I haven't looked at my earlier walkthrough or how others did it, including the original 'tag'-solution.

My walkthrough for Easter Eggs v2 Revisit:
Prelims

a) R12C1 = {18/27/36/45}, no 1
b) R12C9 = {89}
c) R5C12 = {29/38/47/56}, no 1
d) R5C89 = {17/26/35}, no 4,8,9
e) R89C1 = {14/23}
f) R89C9 = {16/25/34}, no 7,8,9
g) 10(3) cage at R1C8 = {127/136/145/235}, no 8,9
h) 19(3) cage at R3C1 = {289/379/469/478/568}, no 1
i) 8(3) cage at R3C3 = {125/134}
j) 9(3) cage at R4C5 = {126/135/234}, no 7,8,9
k) 10(3) cage at R6C1 = {127/136/145/235}, no 8,9
l) 9(3) cage at R6C7 = {126/135/234}, no 7,8,9
m) 20(3) cage at R7C2 = {389/479/569/578}, no 1,2
n) 35(5) cage at R7C5 = {56789}

1a. Naked pair {89} in R12C9, locked for C9 and N3
1b. Naked quint {56789} in 35(5) cage at R7C5, locked for N8
1c. Max R9C6 = 4 -> min R89C7 = 12, no 1,2 in R89C7

2a. 45 rule on R12 3 outies R3C258 = 22 = {589/679}, 9 locked for R3
2b. 5 of {589} must be in R3C8 -> no 5 in R3C25
2c. 10(3) cage at R1C8 = {127/136/145/235}
2d. R3C8 = {567} -> no 5,6,7 in R12C8
2e. 45 rule on R89 3 outies R7C258 = 22 = {589/679}, 9 locked for R7
2f. 9 in C1 only in R45C1, locked for N4, clean-up: no 2 in R5C1

3a. Hidden killer pair 8,9 in 15(3) cage at R7C8 and 16(3) cage at R8C7 for N9, neither can hold both of 8,9 -> they must each contain one of 8,9
3b. 15(3) cage = {168/249/258/348} (cannot be {159} which clashes with 10(3) cage at R1C8), no 7
3c. 10(3) cage at R1C8 = {127/136/235} (cannot be {145} which clashes with 15(3) cage), no 4
3d. 45 rule on C8 3 innies R456C8 = 20 = {479/569/578} (cannot be {389} which clashes with 15(3) cage), no 1,2,3, clean-up: no 5,6,7 in R5C9
3e. 45 rule on C2 3 innies R456C2 = 9 = {126/135/234}, no 7,8,9, clean-up: no 3,4 in R5C1
3f. 1 of {126/135} must be in R6C2 -> no 5,6 in R6C2
3g. 19(3) cage at R3C1 = {289/379/469/478/568}
3h. 2,3 of {289/379} must be in R4C2 -> no 2,3 in R34C1
3i. 10(3) cage at R6C1 = {127/136/145/235}
3j. R6C2 = {1234}, 10(3) cage contains two of 1,2,3,4 -> R67C1 must contain one of 1,2,3,4
3k. Killer quad 1,2,3,4 in R12C1, R67C1 and R89C1, locked for C1

4a. 45 rule on C1234 2 innies R28C4 = 15 = {69/78}
4b. 45 rule on C6789 2 innies R28C6 = 11 = [29/38/47/56/65]

5a. 45 rule on R1234 3 innies R4C456 = 11 = {128/137/146/236/245}, no 9
5b. 45 rule on R6789 3 innies R6C456 = 19 = {289/379/469/478/568}, no 1
5c. 2,3 of {289/379} must be in R6C5 -> no 2,3 in R6C46
5d. R4C456 = 11, R6C456 = 19 -> R5C456 = 15
5e. 45 rule on N5 2 outies R5C37 = 11 = [29]/{38/47/56}, no 1, no 2 in R5C7
5f. 9(3) cage at R4C5 = {126/135/234}
5g. 1 in R5 only in R5C456 = 15 = {159/168} or R5C89 = [71] -> R5C456 = 15 = {159/168/249/258/348/456} (cannot be {267/357}, blocking cages), no 7
5h. 1 in N5 only in R4C456 or in R5C456 = 15 = {159/168}
5i. R4C456 = 11 = {128/137/146/245} (cannot be {236} which clashes with 9(3) cage = {126/234} CCC (combination crossover clash) while after R4C456 = {236}, R5C456 = {159} clashes with 9(3) cage = {135}, CCC)
5j. R5C456 = {168/249/258/456} (cannot be {159/348} which clash with R4C456), no 3
5k. R6C456 = 19 = {289/379/469/568} (cannot be {478} which clashes with R4C456)
5l. 3 in R5 only in R5C12 = {38} or R5C37 = {38} or R5C89 = [53] -> R5C456 = {168/249/456} (cannot be {258} blocking cages)
5m. R4C456 = {128/137/245} (cannot be {146} which clashes with R5C456), no 6
5n. R6C456 = {289/379/568} (cannot be {469} which clashes with R5C456), no 4
5o. 1 in R5 only in R5C456 = 15 = {168} or R5C89 = [71] -> R5C89 = [53/71] (cannot be [62], blocking cages), no 2,6
5p. Consider combinations for 9(3) cage = {126/135/234}
9(3) cage = {126/135}, 1 locked for N5
or 9(3) cage = {234} => R5C456 = {456} (cannot be {249} which clashes with {234}, CCC), 5 locked for R5 => R5C89 = [71]
-> no 1 in R5C46
5q. Consider combinations for R4C456 = {128/137/245}
R4C456 = {128/137}
or R4C456 = {245} => R5C456 = {168} => R5C5 = 1 => 9(3) cage = [513] (cannot be [216] which clashes with R5C456, CCC)
-> no 5 in R4C46, no 4 in R4C5
[Taking this a bit further …]
5r. R4C456 = {128/137}, 1 locked for N5 => R5C456 = {249/456}
or R4C456 = {245} => 9(3) cage = [513] => R5C46 = {68}, R6C46 = {79}, R5C89 = [53], R45C6 both even, R6C6 odd => R5C7 odd = {79}, R5C7 + R6C6 = {79} = 16 => R45C6 = 8 = [26]
-> no 4 in R4C6, no 8 in R5C6

6a. 45 rule on N1 2 innies R3C13 = 1 outie R1C4 + 3
6b. Min R3C13 = 6 -> min R1C4 = 3
6c. 10(3) cage at R1C8 (step 3c) = {127/136/235}
6d. 45 rule on N3 2 innies R3C79 = 1 outie R1C6 + 2
6e. Min R3C79 = 6 = {24} (cannot be {12/13/15/23} which clash with 10(3) cage, cannot be {14} which combined with 10(3) cage = {23}5 clash with 8(3) cage at R3C3) -> min R1C6 = 4
6f. 45 rule on N9 2 innies R7C79 = 1 outie R9C6 + 7
6g. Min R7C79 = 8 -> no 1 in R7C9

7a. 15(3) cage at R7C8 (step 3b) = {168/249/258/348}, 10(3) cage at R1C8 (step 3c) = {127/136/235}, R5C37 = 11 (step 5e)
7b. Consider combinations for R456C8 (step 3d) = {479/569/578}
R456C8 = {479/578} contains 7
or R456C8 = {569), locked for C8 => R5C89 = [53], no 8 in R5C7, 15(3) cage = {348}, locked for N9, 9 in N9 only in 16(3) cage at R8C7 = {169/259} (cannot be {349} because 3,4 only in R9C6) => R7C9 = 7 (hidden cage in N9) => R45C7 = [87] (hidden pair in N6)
-> 7 in N6 only in R456C8 + R5C7, locked for N6
[Just spotted.]
7b. 4 in N5 only in R4C4 + R5C456, CPE no 4 in R5C3, clean-up: no 7 in R5C7
7c. 7 in N6 only in R456C8 = {479/578}, no 6, 7 locked for C8
7d. 10(3) cage at R1C8 = {136/235}, 3 locked for C8 and N3
7e. R3C258 (step 2a) = {589/679}
7f. R3C8 = {56} -> no 6 in R3C25

[Now unexpected progress, thanks to the hard work in step 5.]
8a. R4C456 (step 5m) = {128/137/245}, R5C456 (step 5l) = {168/249/456}, R28C6 = 11 (step 4b)
8b. Consider placement for 5 in 35(5) cage at R7C5
5 in R789C5, locked for C5 => R4C456 = {128/137}
or R8C6 = 5 => R2C6 = 6 => R5C456 = {249/456} (cannot be {168} = [816]), 4 locked for N5
-> R4C456 = {128/137}, no 4,5, 1 locked for R4 and N5
[Cracked. The rest is fairly straightforward.]
8c. R5C456 = {249/456}, no 8, 4 locked for R5
8d. R5C9 = 1 (hidden single in R5) -> R5C8 = 7, clean-up: no 6 in R89C9
8e. 8(3) cage at R3C3 = {125/134}, 1 locked for R3
8f. 15(3) cage at R7C8 (step 3b) = {168/258} (cannot be {249} which clashes with R89C9), no 4,9, 8 locked for C8 and N9
8g. R46C8 = {49} (hidden pair in C8), locked for N6, clean-up: no 2 in R5C3 (step 5e)
8h. 9 in N9 only in 16(3) cage at R8C7 = {169/259/349}, no 7
8i. R7C9 = 7 (hidden single in N9) -> R6C89 = 10 = [46], R4C8 = 9, clean-up: no 5 in R5C3 (step 5e)
8j. R7C258 (step 2e) = {589}, no 6, 5,8 locked for R7
8k. 45 rule on N9 1 remaining innie R7C7 = 1 outie R9C6 = {1234}
8l. 6 in R7 only in R7C13, locked for N7
8m. 20(3) cage at R7C2 = {389/479/578}
8n. Killer triple 7,8,9 in R3C2 and 20(3) cage, locked for C2
8o. 9 in R6 only in R6C456, locked for N5
8p. R5C456 = {456}, 5,6 locked for R5, 5 locked for N5
8q. R5C12 = [92] (hidden pair in R5)
8r. Max R7C34 = 10 -> min R6C3 = 3
8s. 1 in R6 only in R6C12, locked for 10(3) cage at R6C1
8t. 10(3) cage = {127/136/145}
8u. 6 of {136} only in R7C1 -> no 3 in R7C1
8v. R456C2 (step 3e) = 9, R5C2 = 2 -> R46C2 = 7 = [43/61]
8w. 19(3) cage at R3C1 = {478/568}
8x. 6 of {568} must be in R4C2 -> no 6 in R34C1
8y. R4C2 = 6 (hidden single in N4) -> R6C2 = 1, R34C1 = {58}, locked for C1, clean-up: no 1,4 in R12C1
8z. 1 in C1 only in R89C1 = {14}, locked for N7

9a. R4C3 = 4 (hidden single in N4) -> R3C34 = 4 = {13}, 3 locked for R3
9b. 20(3) cage at R7C2 (step 8m) = {389/578}, 8 locked for C2 and N7
9c. R3C258 (step 2a) = {679} (cannot be {589} which clashes with R3C1) -> R3C8 = 6, R3C25 = {79}, 7 locked for R3
9d. R3C8 = 6 -> R12C8 = 4 = {13}, 1 locked for C8 and N3
9e. Naked triple {258} in 15(3) cage at R7C8, 2,5 locked for N9
9f. Naked pair {34} in R89C9, locked for C9 and N9
9g. R89C7 = {69} -> R9C6 = 1 (cage sum), R89C1 = [14], R89C9 = [43]
9h. R9C4 = 2 -> R89C3 = 12 = [39/57/75], no 9 in R8C3
9i. 7 in R4 only in R4C456 (step 8b) = {137}, 3,7 locked for N5, 3 locked for R5
9j. R7C7 = 1 -> R6C7 + R7C6 = 8 = [53], R34C9 = [52], R34C1 = [85], R4C67 = [78], clean-up: no 4 in R2C6, no 8 in R8C6 (both step 4b)
9k. R6C5 = 2 -> R45C5 = 7 = [16/34]
9l. R4C6 = 7, R5C7 = 3 -> R56C6 = 14 = [59/68]
9m. R28C6 (step 4b) = [29] (cannot be {56} which clashes with R56C6) -> R56C6 = [68], R3C67 = [42], R1C6 = 5, R45C5 = [34]
9n. R6C4 = 9, clean-up: no 6 in R28C4 (step 4a)
9o. R1C4 = 6 (hidden single in C4) -> R12C3 = 11 = [29]

and the rest is naked singles.


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PostPosted: Sun Feb 28, 2021 6:19 pm 
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Somewhat similar conclusions in my steps as for Andrew's, but individual ways to get those conclusions quite different as usual.
Here's how I did it.
Easter Eggs v2 Revisit WT:
1. Innies/outies
Outies r12 = r3c258 = =22(3) = {589} or {679}
Innies r1234 = r4c456 = +11(3)
Innies r6789 = r6c456 = +20(3)
-> r5c456 = +15(3)
Outies n5 = r5c37 = +11(2)
Outies r89 = r6c258 = +22(3) = {589} or {679}
35(5)n8 = {56789} -> Innies n8 = r79c46 = +10(4) = {1234}
Innies c1234 = r28c4 = +15(2) -> 5 not in r8c4
Innies c6789 = r28c6 = +11(2)
Innies c2 = r456c2 = +9(3)
Innies c8 = r456c8 = +20(3)

2. 17(2)n3 = {89}
Of (89) in n9 - at most one in c7 and at most one in c8
-> One each of (89) in c7 and c8 in n9
-> One each of (89) in c7 and c8 in n6

3! Max r56c5 = +8(2) -> Min r56c46 = +26(4) -> 1 not in r56c46
-> 1 in r5 only in r5c5 or r5c9

Trying r5c5 = 1
(a) r5c456 = <519> puts r4c456 = {236} which leaves no solution for 9(3)n5
(b) r5c456 = <618> puts 8(2)n6 = {35} and r46c5 = {35} which puts r28c6 = [65]
which leaves no solution for 24(4)r4c6

-> 1 in r5 only in r5c9

4. -> 8(2)n6 = [71]
-> 7(2)n9 from {25} or {34}
Also r46c8 from {49} or {58}
But the latter case leaves no place for 9 in c8
-> r46c8 = {49}

5. -> 8 in n9 in 15(3)n9 and 9 in n9 in r89c7
-> HS 7 in n9 -> r7c9 = 7
-> r6c89 = [46] and r4c8 = 9
Also Outies r89 = r7c258 = {589} with 9 in r7c25

6. Since 7 already on r5 -> 4 not in 11(2)n4 or outies n5 = r5c37 = +11(2)
-> 4 in r5 in r5c456
Also 1 in n5 in r4c456
-> Innies r1234 = r4c456 = +11(3) = {128} or {137}
-> 6 in n5 in r5c456
-> r5c456 = {456}
-> 6 in n4 in r4c12 and 4 in n4 in r4c123

7! Outies r12 = r3c258 = +22(3) contains a 9
-> 9 not in 19(3)r3c1
-> 19(3)r3c1 = {568} with r3c1 = 5 or 8
-> r4c3 = 4
Also -> Outies r12 = r3c258 = +22(3) = [{79}6]
-> 10(3)n3 = [{13}6]
-> 15(3)n9 = {258}
-> (Since r9c6 is max 4) -> r89c7 = {69} and r7c7 = 1
-> r9c7 = 1

8. r456c2 = +9(3) (No 789)
-> 8 not in 11(2)n4
-> 11(2)n4 = [92]
-> r456c2 = [621]
-> r34c1 = {58}
Also 9(2)n1 = {27} or {36}
-> 5(2)n7 = {14}

9. Since r7c6 is max 4 -> 9(3)r6c7 = [531]
-> r34c9 = [52]
-> r4c456 = +11(3) = {137}
-> r4c7 = 8
-> r5c37 = [83]
-> r6c456 = [928]
-> r45c4 = [15], r45c5 = [34], r45c6 = [76]
etc.


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PostPosted: Wed Mar 03, 2021 5:40 am 
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Very tough. Had to use the n5 black-hole which I was desperately trying to avoid since I make mistakes very easily in that situation. And did. Hence, this WT took a long time.
EEv2 start:
Preliminaries
Cage 17(2) n3 - cells ={89}
Cage 5(2) n7 - cells only uses 1234
Cage 7(2) n9 - cells do not use 789
Cage 8(2) n6 - cells do not use 489
Cage 9(2) n1 - cells do not use 9
Cage 11(2) n4 - cells do not use 1
Cage 8(3) n124 - cells do not use 6789
Cage 9(3) n5 - cells do not use 789
Cage 9(3) n689 - cells do not use 789
Cage 10(3) n47 - cells do not use 89
Cage 10(3) n3 - cells do not use 89
Cage 20(3) n7 - cells do not use 12
Cage 19(3) n14 - cells do not use 1
Cage 35(5) n8 - cells ={56789}

note: no clean-up done unless stated
1. 17(2)n3 = {89}: both locked for c9 and n3

2. "45" on r12: 3 outies r3c258 = 22 = {589/679}(no 1,2,3,4)
2a. must have 9: locked for r3

3. 9 in c1 only in r45c1: locked for n4

4. "45" on c34: 2 innies r28c4 = 15 = {69/78}
4a. -> r5c3 + r456c4 = [2]{489/579/678/} all blocked
4b. -> no 2 in r5c3

5. "45" on n5: 2 outies r5c37 = 11 = {38/47/56}(no 1,2,9)

6. "45" on r6789: 3 innies r6c456 = 19 (no 1)
6a. "45" on r5: 3 remaining innies r5c456 = 15
6b. only combos with 1 are {159/168} (no eliminations yet)

first key step. Very tough
7. 9(3)n5 = {126/135/234}
7a. and "45" on r1234: 3 innies r4c456 = 11
7b. but {236} can only have {135} in 9(3) since {126/234} have two common digits with {236} (Combo crossover clash, CCC)
7c. -> {236} + 9(3) = {26}[3]{26} + [315]
7d. but then h15(3)r5c456 (step 6b) can't make {159/168}
7e. -> {236} blocked
7f. -> r4c456 = {128/137/146/245} = 1 or 5,
7g. -> {159} blocked from h15(3)r5c456

8. 1 in r5 is only in h15(3)r5c456 = {168} or in 8(2)n6
8a. -> {26} blocked from 8(2)n6 (Locking-out cages)
8b. -> 8(2)n6 = {17/35}

9. if 1 in r5 in 8(2)n6 = {17} -> 4 blocked from the two 11(2) in r5 -> 4 in r5c456: locked for n5
9a. or 1 in r5 in h15(3) = {168} only -> h11(3)r4c456 = {245} only
9b. -> 4 locked for n5 in r45c456
9c. and {168} in h15(3) -> 9(3)r4c5 = {135} = [513] (can't be {126} because of CCC)
9d. -> no 4 in r4c5
9e. and no 1 in r5c46

10. h19(3)r6c456 = {289/379/568}
10a. 2 and 3 must be in r6c5 -> no 2,3 in r6c46
10b. h15(3)r5c456: {267/348} blocked by h19(3)
10c. {357} blocked by 8(2)n6
10d. {258} blocked by two 11(2) in r5
10e. = {168/249/456}(no 3,7)

11. h11(3)r4c456 = {128/137/146/245}
11a. but {14}[6]{14} blocked by 9(3)r4c5 = [612] (CCC)
11b. -> no 6 in r4c5
11c. {46}[1]{46} blocked by 9(3) can only be [153] but 5 in r5c5 can only have {456} in r5c456 (step 10d)
11d. -> h11(3)r4c456 = {128/137/245}

Another tough one to see
12. 4 in n5 only in r4c46 + r5c456
12a. -> 4 in r5c7 sees all 4 in n5 apart from r4c4 so must repeat there
12b. -> 23(4)r4c4 must have {47} = 11
12c. but then r56c4 = 12: blocked since no 3 in r56c4
12d. -> no 4 in r5c7

and the other side
12. 4 in r5c3 sees all 4 in n5 apart from r4c6 so must repeat there
12a. -> 24(4)r4c6 must have {47} = 11
12b. -> r56c6 = 13 = {58} only
12c. but 4 in r4c6 has {25} in r4c45 (step 11d)
12d. -> no 4 in r5c7

13. h11(2)r5c37 = {38/56}(no 7) = 3 or 5

14. 8(2)n6: {35} blocked by r5c37
14a. = {17} only: both locked for r5 and n6

15. "45" on c9: 3 outies r456c8 = 20 (no 1,2)
15a. -> r5c89 = [71]
15b. -> r46c8 = 13 = {49/58}(no 3,6)

16. 7(2)n9 = {25/34}(no 6)

17. 17(3)r6c8: {359} as [9]{35} only, blocked by 7(2)
17a. {458} as [8]{45} only, blocked by 7(2)
17b. = {269/278/368/467}(no 5)
17c. no 8 in r4c8

18. 16(3)r3c9: must have 4,5,9 for r4c8
18a. = {259/349/457}(no 6)

19. h15(3)r5c456 = {249/456}(no 8)
19a. must have 4, locked for r5 and n5

20. 1 in n5 only in h11(3) = {128/137}(no 5,6,9)

21. 8(3)r3c3 must have 1 which is only in r3: locked for r3

22. "45" on c1, 3 outies r456c2 = 9 = {126/135/234}(no 7,8
22a. -> r6c2 = (1,2,3,4)

23. 19(3)r3c1 = {289/379/469/478/568}
23a. note: if it has 5, must also have 8
23b. -> h9(3)r456c2: {135} as [531] blocked since it will have two 8s in c1
23c. -> no 5 in r4c2
23d. -> no 6 in r4c1 in 19(3) (no valid permutations)

24. killer quad 1,2,3,4 in 9(2)n1, 10(3)r6c1 and 5(2)n7: 2,3,4 locked for c1

25. 6 in r4 only in r4c2 or r4c7
25a. if in r4c2 -> r34c1 = {58} only (step 23)
25b. if in r4c7 -> r3c67 = 8 = {35}
25c. -> r3c167 = 5 or 8

26. h22(3)r3c25: {589} blocked by r3c167
26a. -> = {679} only: all locked for r3
26a. -> r3c8 = 6
26b. -> r128 = 4 = {13}: both locked for n3 and c8

Much easier now
Cheers
Ed


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