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 Post subject: Assassin 376
PostPosted: Wed May 15, 2019 8:02 am 
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Posts: 1040
Location: Sydney, Australia
Attachment:
a376.JPG
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NOTE: disjoint cage 28(5)r1c6 + r2c246+r3c3
NOTE: x-puzzle, 1-9 cannot repeat on either diagonal

Assassin 376
I found this puzzle very difficult and wasn't sure I was going to be able to solve it. Then I found one of the techniques I've recently understood so felt very excited! Very happy I stuck with it but was also very relieved to get it out. Many, many key steps. SudokuSolver gives it 1.50. JSudoku uses 6 chains.
Code:
3x3:d:k:3328:3328:3328:3585:3585:7170:6403:2308:4869:4102:4102:7170:7170:3585:7170:6403:2308:4869:4102:4359:7170:3080:3080:6403:6403:2308:4869:8212:4359:4359:3080:3080:6403:3082:3082:4363:8212:8212:3340:6669:6669:3086:3086:8463:4363:8212:3340:3340:6669:4625:4625:3086:8463:4363:8212:8978:6669:6669:4625:8463:8463:8463:2067:8212:8978:8978:6672:6672:8463:5129:5129:2067:8212:8978:8978:6672:6672:6672:5129:5129:5129:
Solution:
Code:
+-------+-------+-------+
| 5 1 7 | 8 2 9 | 3 6 4 |
| 2 6 3 | 7 4 5 | 9 1 8 |
| 8 9 4 | 3 6 1 | 5 2 7 |
+-------+-------+-------+
| 6 3 5 | 2 1 7 | 4 8 9 |
| 7 2 8 | 5 9 4 | 1 3 6 |
| 9 4 1 | 6 8 3 | 7 5 2 |
+-------+-------+-------+
| 1 5 2 | 4 7 6 | 8 9 3 |
| 4 8 9 | 1 3 2 | 6 7 5 |
| 3 7 6 | 9 5 8 | 2 4 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 376
PostPosted: Sun May 19, 2019 7:56 pm 
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Phew! I also found it tough. But on my second go through found some cool inferences I missed first time which made it easier. The attached WT describes that second go through. Indeed I don't remember much of what I did first time!

Great puzzle Ed!
Assassin 376 WT:
1. Innies c9 -> r9c9 = 1
Innies n9 -> r7c78 = {89}
-> (47) in n9 in r89c78
Also (89) in n8 in 26(5)n8
-> 7 not in 26(5)n8

2. 35(5)n7 = {56789}
-> Remaining cells n7 = {1234}

3. IOD n47 -> r3c2 = r7c3 + 7
-> [r3c2,r7c3] from [81] or [92]
-> (34) in n7 in r789c1
-> 32(7)n47 = {12349(58|67)}
-> 9 in n4 in 32(7)
Also whichever of (12) is in r7c3 is also in 32(7) in n4
Also -> r123c1 = One of the pairs (67) or (58) plus clone of r5c2

4. Outies r123 = r4c23456 = +18(5)
r3c2 from (89) -> r4c23 = +9(2) or +8(2) with no (89)
Also -> r4c456 = +9(3) or +10(3) (no 89)
-> (89) in r4 in r4c1789
-> Either 12(2)r4 = {39} and r4c19 = {78) ...
... or 12(2)r4 = {48} and r4c19 = {69}

5! Possibilities for 17(3)r3c2
Whichever of (89) is in r3c2 must go in c1 in r456c1 in n4
For [r3c2,r7c3] = [81] -> 32(7) = [{1589}{234}]
17(3)r3c2 cannot be [8{36}] since that leaves no place for 9 in r4
-> If r3c2 = 8 -> 17(3)r3c2 = [8{27}]

For [r3c2,r7c3] = [92]
-> 2 in 32(7) in n4
-> r4c23 = {17} or {35}

-> All three possibilities for 17(3)r3c2 force r4c19 = {69} and 12(2)r4 = {48}

6. -> 4 in c9 in n3
-> Since r4c19 = {69} -> 19(3)n3 = {478}
-> 1 in 9(3)n3
Also 9 in n3 in r123c7
-> r7c78 = [89]
-> 12(2)r4 = [48]
Also 1 in r56c7

7! 9(3)n3 = {126} or {135}
-> At least one of (26) or (35) from the 9(3) must be in 17(3)n6
-> Both of (26) or (35) from the 9(3) in 17(3)n6
-> Both of (26) or (35) from the 9(3) in r89c7
-> r89c8 = {47}
-> r56c7 = {17}
-> r5c6 = 4

8! Outies n3 -> r34c6 = +8(2)
Those cannot be the same as the (26) or (35) in r123c7
Remaining outies n69 = r78c6 also = +8(2)
Whichever of (26) or (35) is in r123c7 is in n6 in r56c8 (and in n9 in r78c9)
-> r78c6 also cannot be the same as the (26) or (35) in r123c7
-> Neither r34c6 or r78c6 can be the same as the (26) or (35) in r123c7
-> One of r34c6 and r78c6 = {17}

9! Given:
r7c2 from (567)
r7c3 from (12)
IOD n5689 -> r4c456 = r7c3 + 8
r78c6 = +8(2) -> r7c45 = +11(2) = {47} or {56}
One of r34c6 and r78c6 = {17}

a) r34c6 = [71] -> r7c3 = 2 (on D/)
b) r34c6 = [17] -> Min r456c4 = +10 -> Min r7c3 = 2
c) r78c6 = [71] -> r7c45 = {56} contradicts r7c2 from (567)
d) r78c6 = [17] -> r7c3 = 2

-> In all cases r7c3 = 2

10. -> r3c2 = 9
Also 2 in 32(7) in n4
Remaining innies n1 = r23c3 = +7(2) = {16} or {34}
If the former -> 6 in c1 in n4 -> 32(7) = [{2679}{134}]
If the latter -> r46c2 = [34] -> r456c3 = [5{18}]
and again -> 32(7) = [{2679}{134}]
-> 17(3)r3c2 = [9{35}]
-> 13(3)n4 = {148} with 4 in r6c23
-> (58) in c1 in r123c1

11. -> r4c456 can only be [217]
-> r3c6 = 1
Also -> r7c45 = {47} (Since 7 not in 26(5)n8)
Also HS 1 in D/ -> r2c8 = 1
-> 1 in n1 in r1c23
-> r23c3 = {34}
-> r4c23 = [35]
Also -> r6c2 = 4 and r56c3 = {18}
-> 13(3)n1 = [517] (Since 5 must be in c1 in n1)
-> r89c3 = {69}
Also HS 7 in D\ -> r89c8 = [74]
-> r789c2 = [587]
Also -> r789c1 = [143]

12. Remaining cells in disjoint 28(5) in 2 = +21(3) without a 4
-> Must be {579} or {678} with 7 in r2c4
-> r7c45 = [47]

13. 9 in D/ in n5 and 9 in D\ in n5 -> r5c5 = 9
HS 4 in D/ -> 19(3)n3 = [487]
-> 16(3)n1 = [268] (2 already on D\)
NP on D/ r3c7,r6c4 = {56}
-> r56c4 = {56}
-> r6c56 = [83] (8 already on D\)
-> r56c3 = [81]
Also r23c3 = [34]
-> r3c45 = [36]
-> r3c7,r6c4 = [56]
etc. Phew!


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 Post subject: Re: Assassin 376
PostPosted: Sat May 25, 2019 6:50 pm 
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Joined: Wed Apr 16, 2008 1:16 am
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Great solution wellbeback with some really fun inferences. I had to work harder and mostly in different places in the puzzle. Thanks again Andrew for going through my WT and for the corrections.

A376 WT:
Preliminaries courtesy of SudokuSolver
Cage 8(2) n9 - cells do not use 489
Cage 12(2) n6 - cells do not use 126
Cage 9(3) n3 - cells do not use 789
Cage 19(3) n3 - cells do not use 1
Cage 12(4) n25 - cells do not use 789
Cage 35(5) n7 - cells ={56789}

1. "45" on n9: 2 innies r7c78 = 17 = {89} only: both locked for r7, n9 and for 33(6) cage

2. hidden pair 8,9 in c8 -> r4c8 = (89)
2a. r4c7 = (34)

3. "45" on r56789: 2 outies r4c19 = 15 = {69/78}(no 1,2,3,4,5)

4. 35(5)n7 = {56789}: all locked for n7

5. "45" on n47: 1 outie r3c2 - 7 = 1 innie r7c3 = [81/92]
5a. 17(3)r3c2 can't have both 8 & 9 -> no 8,9 in r4c23

6. 3 & 4 in n7 only in c1: both locked for c1 and locked for 32(7)
6a. 32(7) = {12349}{58/67}: note, must have one of {58/67} missing in n4

Key steps. Not sure I would have found 7a. (or 9) before Andrew's WT for A366
7. 32(7) must have 2 -> if [92] in r3c2 + r7c3 -> 2 in 32(7) in n4 -> [9]{26} blocked from 17(3)r3c2
7a. 8 in r3c2 -> 8 in c1 in 32(7) = {58} in n4 (step 6a) -> the missing {67} from the 32(7) must be in both the 13(3) and r4c23 for n4 since neither spot can have both 6 & 7 (cage sums) -> 8 in r3c2 must have 6 or 7 in r4c23 -> [8]{45} blocked
7b. 17(3)r3c2 = [9]{17/35}/[8]{27/36}(no 4): note = 3 or 7 in r4c23

8. h15(2)r4c19 + 12(2)r4c7 cannot be {78}[39] since r4c23 = 3 or 7
8a. -> h15(2) = {69} only, both locked for r4
8b. r4c78 = [48], r7c78 = [89] (8 placed for d\)

9. if 8 in r3c2 -> 8 in c1 in 32(7) = {58} in n4 -> the missing {67} from the 32(7) must be in both the 13(3) and r4c23 for n4
9a. -> generalised x-wing 6 & 7 with 35(5)r7c2 -> no other 6 or 7 in c23 (no eliminations yet)
9b. "45" on n1: 3 innies r23c3 + r3c2 = 16
9c. but 8 in r3c2 -> {17/26} blocked from r23c3 (step 9a)
9d. if 9 in r3c2 -> 2 in r7c3 (IODn47=-7) -> {25}[9] blocked from r23c3
9e. -> h16(3)n1 = {16/34}[9]/{35}[8](no 2,7; no 8,9 in r23c3)
9f. max. r23c3 = 8 (hcage sum) -> min. r12c6 + r2c4 = 20 (no 1,2)

10. "45" on c9: 1 innie r9c9 = 1 (placed for D\)

This step took a long, long time to find
11. 28(5)r1c6 = must have [16]/{34}/{35} for r23c3 (step 9e)
11a. -> {13789} blocked
11b. = {14689/15679/34579/34678}
11c. 12(4)r3c4 = {1236/1245} = 4 or 6 in r3c45
11d. -> {14689} as [16]{489} only, blocked by r3c45
11e. 28(5) = {15679/34579/34678}
11f. must have 7: 7 locked for n2

Loved finding this one. Always prefer x-puzzles to be a required part of my solution rather than just for uniqueness.
12. "45" on n47: 3 innies r4c23 + r7c3 = 10.
12a. 32(7)r4c1 = {12349}{58/67}: must have 1 & 2
12b. -> can't have repeats of 1 or 2 in 3 innies n47
12c. 3 innies n47 = 10 = {17/35}[2]/{27}[1]
12d. must have 2 -> no 2 in r4c6 (through D/)
12e. and two locked for n47 with 32(8) (generalized x-wing) -> no 2 in 13(3)n4

Very slow to see this "45".
13. "45" on n689: 1 outie r5c6 + 7 = 2 innies r7c45
13a. max. r7c45 = 13 -> max. r5c6 = 6

14. "45" on n3: 2 outies r34c6 = 8 = [17]/{35}(no 2,4,6,8,9;in r4c6 no 1 ) = 1 or 5, 3 or 7

15. "45" on n69: 3 outies r578c6 = 12
15a. but {156/237} blocked by r34c6
15b. = {147/246/345}: note if it has 1, must also have 7
15c. must have 4: 4 locked for c6

16. 1 in c6 in h8(2) = [17] or h12(3) = {147} -> 7 locked for c6

17. r2c4 = 7 (hsingle n2)

Lot of combinations in this step so difficult.
18. "45" on n12: 4 innies r3c2456 = 19 and must have 4 or 6 for r3c45 (step 11c) and 8 or 9 for r3c2
18a. = {1369/1459/1468/2368/2458} = 5 or 6
18b. 28(5)r1c6 must have [16]/{34}/{35} for r23c3 (step 9e) = {15679/34579/34678}
18b. but {15679} must have [16] in r23c3 and {59} in n2, but this clashes with 5 or 6 in h19(4)r3
18c. 28(5)r1c6 = {34579/34678}(no 1)
18d. must have 4 -> r23c3 = {34} only (step 18b): both locked for c3, n1 and 28(5) cage
18e. r23c3 = 7 -> r3c2 = 9 (h16(3))
18f. -> r7c3 = 2 (1 innie n47), 2 placed for D/

Finally cracked it!
19. 16(3)n1: {178} blocked by no 1,7,8 in r2c2
19a. = {268} only: all locked for n1, 8 locked for c1

20. 13(3)n1 = {157}: all locked for r1

21. 1 must be in r12(4)r3c4 only in r3c34 or r4c5 -> no 1 in r2c5 (Common Peer Elimination CPE)

22. 28(5)r1c6: r23c3 + r2c4 = 14 -> r12c6 = 14 = {68}/[95] = 5 or 6

23. 14(3)n2: {356} blocked by r12c6
23a. = {239/248}(no 5,6)
23b. must have 2: locked for n2

24. 12(4)r3c4 must have 2 which is only in r4: locked for r4 and n5

25. 8(2)n9 = [62]{35}(no 7) = 5 or 6

26. 17(3)n6 = {269/359}(no 7)
26a. must have 9: locked for c9 and n6

27. 19(3)n3: must have 4 & 7 for c9 = {478}: all locked for n3 (alternatively, Andrew noticed it is hidden triple 4,7,8 for c9 = {48}[7]. So gets rid of step 28)

28. r3c9 = 7 (hsingle r3)

29. r3c1 = 8 (hsingle r3)
29a. r2c12 = {26}: both locked for r2

30. 9(3)n2 = {126/135}
30a. must have 1: locked for c8 and n3

31. r2c8 = 1 (hsingle r2); 1 placed for D/

32. r3c8 = 2 (hsingle r3), r1c8 = 6 (cage sum)

33. naked triple {359} in r123c7: locked for c7 and 25(5) cage
33a. r34c6 = [17], 7 placed for D/

34. r4c45 = [21] (hsingles in 12(4)), 2 placed for D\
34a. r2c2 = 6, placed for D\, r2c1 = 2, r4c23 = [35]

35. hidden single 5 in c1 -> r1c1 = 5: placed for D\

36. naked triple {578} in r789c2: 7 & 8 both locked for c2 and n7

37. r8c8 = 7 (hsingle D\)

38. r12c6 = 14 (cage sum) = [95] only
etc
Cheers
Ed


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 Post subject: Re: Assassin 376
PostPosted: Sun Jun 02, 2019 8:49 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
I was very busy in the last week of May, then came back to this puzzle and thought that I'd solved it but found, on checking my WT, that I'd made a careless mistake so was unable to post this before A377 appeared. Fortunately I then found step 15 fairly quickly; apart from that I've used most of my original step with only minor changes to them.

Thanks Ed for pointing out that my step 9c was flawed. I've now reworked it by moving one of my later steps forward, so essentially the same solving path with slight simplification.
Here is my walkthrough for Assassin 376:
Prelims

a) R4C78 = {39/48/57}, no 1,2,6
b) R78C9 = {17/26/35}, no 4,8,9
c) 9(3) cage at R1C8 = {126/135/234}, no 7,8,9
d) 19(3) cage at R1C9 = {289/379/469/478/568}, no 1
e) 12(4) cage at R3C4 = {1236/1245}, no 7,8,9
f) 35(5) cage at R7C2 = {56789}

1a. 35(5) cage at R7C2 = {56789}, locked for N7
1b. 45 rule on C9 1 innie R9C9 = 1, placed for D\, clean-up: no 7 in R78C9
1c. 45 rule on N9 2 innies R7C78 = 17 = {89}, locked for R7, N9 and 33(6) cage at R5C8
1d. 45 rule on N3 2 outies R34C6 = 8 = {17/26/35}, no 4,8,9
1e. R47C8 = {89} (hidden pair in C8) -> R4C7 = {34}
1f. 19(3) cage at R1C9 = {289/379/469/478} (cannot be {568} which clashes with R78C9), no 5
1g. 17(3) cage at R4C9 = {269/278/359/458/467} (cannot be {368} which clashes with R78C9)

2a. 45 rule on N47 1 outie R3C2 = 1 innie R7C3 +7 -> R3C2 = {89}, R7C3 = {12}
2b. 17(3) cage at R3C2 cannot contain both of 8,9, R3C2 = {89} -> no 8,9 in R4C23
2c. 3,4 in N7 only in R789C1, locked for C1 and 32(7) cage at R4C1
2d. 32(7) cage contains 3,4 = {1234589/1234679}, 9 locked for N4
2e. 45 rule on C123 3 innies R237C3 = 9 = {126/135/234}, no 7,8,9

3. 45 rule on R1234 2 innies R4C19 = 15 = {69/78}

4a. 45 rule on C123 3 outies R1C4 + R2C46 = 1 innie R7C3 + 19
4b. R7C3 = {12} -> R1C4 + R2C46 = 20,21, no 1,2

5. 8,9 in N8 only in 26(5) cage at R8C4 = {12689/13589/23489}, no 7

6. 45 rule on N689 2 innies R7C45 = 1 outie R5C6 + 7
6a. Max R7C45 = 13 -> max R5C6 = 6

7. 45 rule on N6 4 innies R56C78 = 16 must contain 1 for N6 = {1258/1267/1357/1456} (cannot be {1249/1348} which clash with R4C78), no 9

8. Hidden killer pair 3,4 in R4C23 and 13(3) cage at R5C3 for N4, R4C23 cannot contain both of 3,4 which would clash with R4C78 (alternatively 17(3) cage at R3C2 cannot contain both of 3,4) -> 13(3) cage must contain at least one of 3,4 -> 13(3) cage = {148/238/247/346}, no 5

9. 45 rule on N3 3 innies R123C7 = 17 = {269/278/359/458/467} (cannot be {179/368} which clash with 19(3) cage at R1C9), no 1
9a. 1 in N3 only in 9(3) cage at R1C8, locked for C8
9b. 9(3) cage = {126/135}, no 4
Rest of original step 9 deleted.

10. 1 in N6 only in R56C7, locked for 12(3) cage at R5C6, no 1 in R5C6
10a. 12(3) cage = {138/147/156}, no 2
10b. 3,4 of {138/147} must be in R5C6 -> no 3,4 in R56C7
10c. R7C45 = R5C6 + 7 (step 6)
10d. Min R5C6 = 3 -> min R7C45 = 10, no 1,2 in R7C45
10e. 18(3) cage at R6C5 = {279/369/378/459/468/567} (cannot be {189} because 1,8,9 only in R6C56), no 1

11. 45 rule on N69 3 outies R578C6 = 12 = {147/237/246/345} (cannot be {156} which clashes with R34C6)
11a. 1 in C6 only in R34C6 = {17} or R578C6 = {147}, 7 locked for C6 (locking chain)
11b. 1 in C6 only in R34C6 = {17} or R578C6 = {147} -> R578C6 = {147/246/345} (cannot be {237}, locking-out chain), 4 locked for C6

[Moving up my original step 15 and slightly modifying it.]
12. Consider placements for {17} in C6 (step 11)
R34C6 = {17}, locked for 25(5) cage at R1C7 => 7 in N3 only in 19(3) cage at R1C9
or R78C6 = {17} => R5C6 = 4 (hidden single in C6) => R56C7 = 8 contains 1 = {17}, locked for C7 => 7 in N3 only in 19(3) cage at R1C9
-> 19(3) cage at R1C9 (step 1f) = {379/478}, no 2,6, 7 locked for C9 and N3, clean-up: no 8 in R4C1 (step 3)
12a. 17(3) cage at R4C9 (step 1g) = {269/359/458}

13. 9(3) cage (step 9b) = {126/135}, 19(3) cage at R1C9 (step 12) = {289/379/478}, 17(3) cage at R4C9 (step 12a) = {269/359/458}
13a. Consider placements for 2 in N6
2 in R56C8, locked for C8 => 9(3) cage = {135} => 19(3) cage = {478}
or 2 in 17(3) cage = {269}, locked for C9 => 19(3) cage = {478}
-> 19(3) cage = {478}, 4,8 locked for C9 and N3, clean-up: no 7 in R4C1 (step 3)
13b. Naked pair {69} in R4C19, locked for R4, R4C8 = 8 -> R4C7 = 4, R7C78 = [89], 8 placed for D\, clean-up: no 2 in R3C6 (step 1d)
13c. R56C78 (step 7) = {1267/1357}
13d. Killer pair 2,3 in 9(3) cage at R1C8 and R56C8, locked for C8
13e. 12(3) cage at R5C6 (step 10a) = {147/156}, no 3
13f. R7C45 = R5C6 + 7 (step 6)
13g. Min R5C6 = 4 -> min R7C45 = 11, no 3

14. R1C4 + R2C46 = R7C3 + 19 (step 4a), R7C3 = {12} -> R1C4 + R2C46 = 20, 21 = {389/569/578}{489/579/678} (cannot be {479} because 4,7 only in R2C4)
14a. R237C3 (step 2e) = {126/234} (cannot be {135} = {35}1 which clashes with R1C4 + R2C46 = 20 = {389/569/578}), no 5, 2 locked for C3
14b. R1C4 + R2C46 = {389/578}{489/579/678} (cannot be {569} which clashes with R237C3 = {26}1)
14c. 7 of {678} must be in R2C4 -> no 6 in R2C4

15. Moved to become the start of new step 12.

16. 4 in N4 only in 13(3) cage at R5C3 = {148/247/346}
16a. Consider combinations for R237C3 (step 14a) = {126/234}
R237C3 = {126}, 6 locked for N1 => 6 in C1 only in R456C1, locked for N4 => 13(3) cage = {148/247}
or R237C3 = {234}, locked for C3 => 13(3) cage = {148/247} (cannot be {346} because 3,4 only then in R6C2)
-> 13(3) cage = {148/247}, no 3,6
16b. 6,9 in N4 only in 32(7) cage at R4C1 = {1234679}, no 5,8, 7 locked for N4
16c. 13(3) cage = {148} (only remaining combination), 1 locked for N4
16d. R4C23 = {35} (hidden pair in N4), locked for R4, R3C2 = 9 (cage sum) -> R7C3 = 2 (step 2a), placed for D/, clean-up: no 6 in R8C9
[Cracked at last, the rest is fairly straightforward.]
16e. R4C45 = [21], 2 placed for D\, R4C6 = 7, placed for D/ -> R3C6 = 1 (step 1d)
16f. R2C8 = 1 (hidden single on D/)
16g. R237C3 = {234} (only remaining combination) -> R23C3 = {34}, locked for C3, N1 and 28(5) disjoint cage at R1C6 -> R4C23 = [35]
16h. Naked pair {18} in R56C3, locked for C3 and N4 -> R6C2 = 4

17a. R1C2 = 1 (hidden single in N1) -> R1C13 = 12 = [57], 5 placed for D\
17b. R2C2 = 6, placed for D\, R5C2 = 2 (hidden single in C2)
17c. R3C9 = 7 (hidden single in R3)
17d. R3C1 = 8 (hidden single in R3) -> R2C1 = 2
17e. R3C8 = 2 (hidden single in R3) -> R1C8 = 6 (cage sum)
17f. 6 in R3 only in R3C45 -> 12(4) cage at R3C4 = {1236}, 3 locked for R3 and N2
17g. R3C3 = 4, placed for D\, R2C3 = 3
17h. R123C7 = [395], 5 placed for D/
17i. R8C2 = 8, placed for D/ -> R1C9 = 4, placed for D/ -> R9C1 = 3, placed for D/

18a. R5C5 = 9, R6C4 = 6, R6C6 = 3 -> R67C5 = 15 = [87], R7C2 = 5, R7C46 = [46], R7C9 = 3 -> R8C9 = 5
18b. R23C3 = {34} = 7 -> R1C6 + R2C46 = 21 = {579} (only remaining combination) -> R1C6 = 9, R2C46 = [75]

and the rest is naked singles, without using the diagonals.

Rating Comment:
I think I'll rate my walkthrough for A376 at 1.5, rather than Easy 1.5. I used locking and locking-out chains, then three forcing chains. Step 15 wasn't too easy to find.


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