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Don't see it as a technique. It's just an observation, which completely depends on the cage totals. I mean it only happens if the person who made the puzzle designs it that way on purpose. This also means you have to have a certain looking solution. That is also why udosuk has a puzzle with the same solution and a different cage pattern instead of a different solution in the same cage pattern. (well this is my assumption of course).

Okay, here is a brief graphical explanation:

Firstly, in the solution grid, label each cell value A11 to A99 as the following:



Now, subtract each of these values from 10, and rotate the whole grid 180 degrees:



We know the cage structure in the original grid also applies in this new grid, because if two cells belong in the same cage in the original grid, they must also belong to the same cage in the new grid.

Now what do you notice about the cage sums in the new grid?

Yes, we can work them all out algebraically. For example, the cage sum at r1c456 in the new grid = (10-A96)+(10-A95)+(10-A94) = 30-(A94+A95+A96) = 30-24 = 6.

(We know A94+A95+A96 = 24 from cage at r9c456 in the original grid.)

So, working all the new cage sums out, we obtain the following grid:



Just take a look at all the cage sums, they are the exact copy of the original puzzle!

Because the original puzzle is a valid puzzle, we can immediately determine all the 81 cell values of this new grid. For example:

10-A99 = A11
10-A98 = A12
...

In particular:
10-A55 = A55
=> 2 A55 = 10
=> A55 = 5

Also this is an established technique in Vanilla Sudoku. Check out this thread.

_________________
ADYFNC HJPLI BVSM GgK Oa m

OK, hmm. Thanks for all the work to make up the pics udosuk. I don't really understand it (or the link) yet... from what I can work out, as Para says and udosuk has implied, more like an interesting feature rather than a technique as such.

Don't know if this is related or not, but "45" rule on r1 + r9 + c1 + c9 = r2c2 + r2c8 + r8c2 + r8c8 = 180. Combined with the 45(9) cage that runs along the diagonals, does this imply that 1-9 cannot repeat on the diagonals? Does this make it an implied Diagonals puzzle? I can't see that it does. Don't see how this would help get r5c5 placed for example. (Note: it is "diagonal" since V2X has the same solution).

My main reason for this post is to follow up to my earlier rating of Afmob's step 2 in his Assassin 123 Walkthrough. By PM, Afmob asked, quite rightly, what I found difficult in his step 2. In actual fact, it's not as difficult as I first thought. So, I'll withdraw that rating. Though steps 2 e & j still seem above the average Weekly Assassin and I never found them. Which is also why I called my solution (not udosuk's trick above!) an Easy 1.50. My key move is more like a one-off for this puzzle and could easily be missed. Of course, if most people found it, then it's average!! And my rating's wrong. But, it is finding this sort of trick that keeps me comin' back for more (though won't be trying A123 V2: lots of T&E according to the SS log). Bring it on Frank... A nice Canadian Thanksgiving puzzle perhaps...?

OT: Talking about Canada, there was a bit of news coverage here recently of the now Canadian PM (Steven Harper?) plagiarizing our then PM's pre-Iraq war speech from 2 days earlier.

Cheers
Ed

Hopefully others can see it better. I'm not sure if I can explain it more clearly.


With my symmetry property, r28c28 always sums to 20. And no it doesn't (directly) imply the diagonals must have no repeat. You only know r19c19 & r28c28 both consist of two pairs from {19|28|37|46|55}. Initially you can't even eliminate the possibilities of say, \19=\28. I think you're on the totally wrong track if you want to "see" the fact r5c5=5 from the X property.


I already mentioned, all the moves in step 2 can be understood (fairly) easily, but it's a matter of whether you actually want to apply them. I think (at least for me) elegancy is an important factor. For example, the relationship 7 = R1469C3 - R5C2 (and the subsequent crunching deductions) feels not as elegant as your establishment one of r37c7 = r4c5 (I dubbed this move "cell cloning" although you've outdone me in using a "one of two cells cloning" ). Of course, it's all a matter of personal taste. I always think cracking moves involving fewer cells are more elegant than ones involving more cells. In this case the "cells involved" ratio is 5:3. However, if for some the process of sitting in front of the computer for a few minutes to crunch on a 5-cell group gives them the hots, I have nothing against that.


I decided I'm not commenting on any personal rating anymore myself, but I think you need to make up your mind here. Which do you want to choose as your measurement, easiness/difficulty of the spotting of cracking moves, or simplicity/elegancy of the logic involved?

_________________
ADYFNC HJPLI BVSM GgK Oa m

Thanks udosuk for a very clear explanation about how Rotational Symmetry works for this puzzle. It clearly only works for the very few cases where the cage pattern is symmetrical about the centre cell and all the cage totals are complementary except, of course, the cage containing the centre cell. I take this to mean that the totals of a complementary pair of cages, for example 17(3) cage at R2C3 and 13(3) cage at R7C6, must total 10 times the number of cells in each of those cages. This brief description appears to be a way that it can be recognised by a software solver, if the programmer considers it worth including for the very occasional cases where it appears; however it could well be a situation that will appear more in the future. Edit. However after looking at Gurth's Symmetrical Placement on the linked thread, maybe it's not as simple as I thought above, unless there is something specific for Killers that makes what I've said above correct. udosuk, please tell me whether I was right or wrong. In any case I'll leave my above statement as it is, even if it is wrong.

Unlike Ed, I've got no problem in accepting it as a technique. It's clearly logical. udosuk has told us that it's an established technique for Vanilla Sudoku. My first reaction to that was that it is an advanced technique but then I realised that most/all advanced techniques for Vanilla Sudoku start with a "what if" assumption whereas Rotational Symmetry is completely logical.

To put this in context, from what I've read after this story blew up during the Canadian election campaign, the plagiarising was actually done by one of his speech writers who was being lazy. Maybe it was also lazy of Stephen Harper, then leader of the opposition, not to check what had been said by John Howard but I doubt that most politicians check everything that has been written by their speech writers; after all they are paying those guys to do a job.

Wow! V2 was really strange since like Messy One 6 it opened up itself without much trouble and all of a sudden I couldn't progress without using forcing chains.

It's also interesting that all of my chains didn't use the Killer property so they could have been applied to an X Sudoku with the same candidates left. The most difficult moves were step 10i and its symmetrical twin step 10j which were forcing chains that used forcing chains.
JSudoku used a different move to get the same result:
I could have used the symmetry technique here but since it requires uniqueness I refused to do so.

One thing about the rating, despite my rating of 2.0 it felt a lot easier than A74 Brickwall. If I had used Gurth's symmetrical present then I would have rated it my wt around 1.75.

A123 V2 Walkthrough:

1. N2468
a) Innies N2 = 27(4) <> 1,2
b) Outies N2 = 6(2) = {15/24}
c) Outies N4 = 12(2) = [75/84/93]
d) Outies N6 = 8(2) = [53/62/71]
e) Innies N8 = 13(4) <> 8,9
f) Outies N8 = 14(2) = {59/68}

2. R456
a) Innies R1234 = 8(3) = 1{25/34} -> 1 locked for R4
b) Outies R6789 = 22(3) = 9{58/67} -> 9 locked for R6
c) 17(5) must have 1 -> 1 locked for N1
d) 33(5) must have 9 -> 9 locked for N9

3. N28
a) Outies N2 = 6(2): R2C7 <> 5
b) 16(3): R23C6 <> 4 since R2C7 = (124)
c) Outies N8 = 14(2): R8C3 <> 5
d) 14(3): R78C4 <> 6 since R8C3 = (689)

4. N1379
a) Innies+Outies N7: -6 = R6C4 - (R7C2+R8C3): R6C4 <> 1,2 since R7C2+R8C3 >= 9
b) Innies+Outies N9: -2 = R6C6 - (R7C8+R8C7): R6C6 <> 3 since R7C8+R8C7 >= 6
c) Innies+Outies N1: -8 = R4C4 - (R2C3+R3C2): R4C4 <> 7 since R2C3+R3C2 <= 14
d) Innies+Outies N3: -4 = R4C6 - (R2C7+R3C8): R4C6 <> 8,9 since R2C7+R3C8 <= 11

5. R456
a) 1,2 in R6 locked in 9(3) and 8(3) and none of them can have two of (12) @ R6
-> Both cages must have one of (12) @ R6
b) 8(3): R6C78 <> 3 because 4 only possible @ R6
c) 8,9 in R4 locked in 22(3) + 21(3) and none of them can have two of (89) @ R4
-> Both cages must have one of (89) @ R4
d) 22(3): R4C23 <> 7 because 6 only possible there
e) Outies N89 = 20(3+1): R6C6 <> 5 since R6C78 @ 8(3) cannot be {24}
f) Outies N12 = 20(3+1): R4C4 <> 5 since R4C23 @ 22(3) cannot be {68}

6. N1 + R456 !
a) ! 20(3) = 9{38/47/56} because {578} blocked by Killer pair (57) of 17(5)
-> 9 locked for N1
b) Outies N4 = 12(2) <> 3
c) 22(3) = 9{58/67} -> 9 locked for R4+N4
d) 9(3) = 3{15/24} -> 3 locked for R6+N4; R6C23 <> 4,5
e) 26(5) = 478{16/25} -> 8 locked for N4
f) 21(3) = {678} -> 8 locked for N6
g) Outies N6 = 8(2) <> 3
h) 8(3) = {125} -> 5 locked for R6+N6
i) Outies R6789 = 22(3) = {679} locked for R6
j) Naked pair (48) locked in R6C46 for N5

7. N9 + R456 !
a) 33(5) must have 7 -> 7 locked for N9
b) ! 10(3) = 1{36/45} because {35} is a Killer pair of 33(5) -> 1 locked for N9
c) R7C8 = 2
d) Outie N6 = R3C8 = 6
e) 8(3) = {125} -> 1 locked for R6+N6
f) 9(3) = {234} -> R7C2 = 4; 2 locked for N4
g) Outie N4 = R3C2 = 8
h) 22(3) = {589} -> 5 locked for R4+N4
i) Outies R1234 = 8(3) = {134} locked for R4

8. R789
a) Killer pair (56) locked in 33(5) + 10(3) for N9
b) R8C7 = 8, R4C7 = 7, R4C8 = 8
c) Outie N8 = R8C3 = 6
d) Outie N7 = R6C4 = 4
e) R6C6 = 8
f) 13(3) = [14/32]8
g) 14(3) = 6{17/35}
h) Killer pair (13) locked in 14(3) + R7C6 for N8

9. R123
a) Innies+Outies N3: 2 = R4C6 - R2C7 -> R4C6 = 6, R2C7 = 4
b) Outie N2 = R2C3 = 2
c) 17(3) = 2[69/87]
d) R4C4 = 2, R6C3 = 3, R6C2 = 2
e) 17(5) must have 3 -> 3 locked for N1
f) Killer pair (79) locked in R3C4 + 16(3) for N2

10. D\/ !
a) 1 in R2 locked in R2C258 -> CPE: R5C5 <> 1
b) 9 in R8 locked in R8C258 -> CPE: R5C5 <> 9
c) 17(5): R2C2+R3C1 <> 5 since (36) only possible there
d) 33(5): R7C9+R8C8 <> 5 since (47) only possible there
e) 20(3) @ N1: R1C1 <> 7 because 4 only possible there
f) 10(3) @ N9: R9C9 <> 3 because 6 only possible there
g) Consider placement of 7 in D\ -> R2C8 <> 7
- i) 7 in R2C2, R5C5 or R8C8 -> CPE: R2C8 <> 7
- ii) 7 in R3C3 -> R2C6 = 7 (HS @ N2)
h) Consider placement of 3 in D\ -> R8C2 <> 3
- i) 3 in R2C2, R5C5 or R8C8 -> CPE: R8C2 <> 3
- ii) 3 in R7C7 -> R8C4 = 3 (HS @ N8)

i) ! Consider placement of 3 in C2 -> R5C5 <> 3
- i) 3 in R2C2
- ii) 3 in R9C2 -> 3 in N9:
- ii) a) Either 3 in R7C7+R8C8 locked for D\ or
- ii) b) 3 in R78C9 locked for C9 -> 3 locked R5C78 @ N6 for R5
j) ! Consider placement of 7 in C8 -> R5C5 <> 7
- i) 7 in R8C8
- ii) 7 in R1C8 -> 7 in N1:
- ii) a) Either 7 in R2C2+R3C3 locked for D\ or
- ii) b) 7 in R23C1 locked for C1 -> 7 locked in R5C23 @ N4 for R5

11. D\/ !
a) R5C5 = 5
b) 21(5): R3C9 <> 5 since 7 only possible there
c) 29(5): R7C1 <> 5 since 3 only possible there
d) Hidden Single: R3C6 = 5 @ R3 -> R2C6 = 7
e) R3C4 = 9 -> R2C4 = 6
f) Hidden Single: R7C4 = 5 @ R7 -> R8C4 = 3
g) R7C6 = 1 -> R8C4 = 4
h) 33(5) = {36789} -> 3,6 locked for N9
i) 17(5) = {12347} -> 4,7 locked for N1

12. R456 + D\/
a) 24(5): R5C79 <> 3 since R5C789 = {239} blocked by R5C6 = (39)
b) ! Consider placement of 6 in R6 -> R5C6 <> 6
- i) R6C1 = 6 -> R7C7 = 6 (HS @ D\)
- ii) R6C9 = 6
c) ! Consider placement of 2 in D/ -> R5C6 <> 9
- i) R3C7 = 2 -> R5C7 = 9
- ii) R9C1 = 2 -> R9C6 = 9
d) R5C6 = 3, R4C5 = 1, R5C4 = 7, R6C5 = 9
e) R4C1 = 4, R4C9 = 3, R6C9 = 6, R6C1 = 7
f) R1C6 = 2, R9C6 = 9, R9C4 = 8, R1C4 = 1
g) 33(5) = {36789} -> 3 locked for C7

13. R123
a) Naked triple (569) locked in R1C127 for R1
b) 20(3) @ N3 = 8{39/57} -> R2C9 = (59)
c) 7 locked in R13C3 for C3
d) Hidden Single: R1C9 = 8 @ N3

14. Rest is singles without considering diagonals.

Rating: 2.0. I used lots of forcing chains and two were chains in chains.

Thanks for confirming that I've explained it at least clear enough for some. You're right that this technique only works for the few puzzles that display symmetry. However if the property is there it doesn't hurt to use it. For example, for a puzzle that isn't specified as an X puzzle, if you can somehow deduce that one or both of the diagonals must have no repeat, I don't see any problem to apply diagonal techniques (e.g. CPE) on your solving path. However, most randomly generated puzzles don't have non-repeating diagonals, just as most randomly generated puzzles don't have symmetry. If one decides to only use techniques that can apply to *all* puzzles, that's his/her personal choice but I'd think such a mindset is too narrow-sighted.


Spot on!


Yes, of course it's not difficult at all to be implemented into a program. It's up to the programmers (e.g. Richard, JC) if they want to implement it as it is rarely occured in randomly generated puzzles. As a matter of fact there are 5 main categories of symmetry (aka automorphism) in Sudoku puzzles (rotation is just one of them) and they each have quite different characteristics. There is a lot of info available in the sudoku.com forums but unless someone shows more interest I'll spare you all the (quite intimidating) details.


What you said above is perfectly correct. However, in vanilla puzzles you can freely permute the 9 digits so the "opposite cells must sum to 10" constraint is no longer necessary. For example, you can have {13|25|49|68|77} as the opposite pairs. In this case r5c5 must be the self-mapping digit, i.e. 7. But in killer puzzles I don't see any obvious pairing other than {19|28|37|46|55} (I call this the "orthodox" pairing) to easily transfer the symmetry to the cage sums.


Thanks for accepting the technique! Everyone has his/her own taste and limitation of what techniques/methods he/she would choose to use. Some people simply refuse to use any techniques related to the uniqueness/validity of the puzzle and consider it a disgrace to use the extra help from this information. Some people don't use elimination solving and only solve the puzzle from a completely blank grid. Some people even refuse to use pencilmarks and only do it using pen and paper. But there are too many puzzles out there to suit everyone's own scope of acceptable puzzles so if it happens to be out of your scope just feel free to ignore it. Life is too short to waste time on every puzzle out there!

_________________
ADYFNC HJPLI BVSM GgK Oa m

Having got stuck on A123 I went back to it about 3-4 weeks later and managed to solve it then. However I've only just gone through Afmob's and Ed's posted walkthroughs.

My solving path was much more like Afmob's than Ed's but I took much longer to find the breakthrough moves so there are some heavier moves before each of them.

Ed's breakthrough step 7 was neat! I'd quickly spotted that one of R37C7 must be the same as R4C5 but hadn't made the mental leap that R2C7 and R4C5 must therefore be different. I felt that Ed's walkthrough was easier because of that step although I've no idea what rating ought to be given to step 7. On a human level it's an easy step but only if one spots it.

I decided to post my walkthrough because I think some of the steps are interesting, for example I got more out of the IOUs in steps 6 to 9, but I'm not going to give a rating for it since it's clearly harder than either of the earlier ones.

Here is my walkthrough for A123. Steps 15 to 18 would probably have been unnecessary if I'd spotted the common cells (steps 19 and 20) after step 14. Even after that I got into some harder work until step 29 when I eventually revisited step 8, this time writing out the combinations and spotting that they provided a clash that gave the key breakthrough.

Prelims

a) R1C456 = {123}, locked for R1 and N2
b) 22(3) cage at R3C2 = {589/679}, CPE no 9 in R56C2
c) 21(3) cage at R3C8 = {489/579/678}, no 1,2,3
d) R5C234 = {489/579/678}, no 1,2,3
e) R5C678 = {126/135/234}, no 7,8,9
f) 9(3) cage at R6C2 = {126/135/234}, no 7,8,9
g) 8(3) cage at R6C7 = {125/134}, CPE no 1 in R5C8
h) R9C456 = {789}, locked for R9 and N8
i) 34(6) cage in N3 must contain 9, locked for N3
j) 26(6) cage in N7 must contain 1, locked for N7

1. Min R23C5 = {45} = 9 -> max R4C5 = 3 because R234 cannot be {45}4
1a. Max R23C5 = 12, no 9
1b. 45 rule on N2 3(2+1) outies R2C37 + R4C5 = 7, R2C37 = {12345}

2. Max R78C5 = {56} = 11 -> min R6C5 = 7 because R678C5 cannot be 6{56}
2a. Min R78C5 = 8, no 1
2b. 45 rule on N8 3(2+1) outies R6C5 + R8C37 = 23 -> min R8C37 = 14, no 1,2,3,4

3. 45 rule on C5 3 innies R159C5 = 15
3a. All combinations must contain at least one of 4,5,6 -> R5C5 = {456}
3b. Killer triple 4,5,6 in R5C234, R5C5 and R5C678, locked for R5

4. Hidden killer triple 7,8,9 in R23C5, R6C5 and R9C5 for C5 -> R23C5 must contain one of 7,8
4a. R234C5 cannot contain 6 and one of 7,8 -> no 6 in R23C5
4b. Min R23C5 containing one of 7,8 = 11 -> max R4C5 = 2

5. Hidden killer triple 1,2,3 in R1C5, R4C5 and R78C5 for C5 -> R78C5 must contain one of 2,3
5a. R678C5 cannot contain 4 and one of 2,3 -> no 4 in R78C5
5b. Max R78C5 containing one of 2,3 = 9 -> min R6C5 = 8

6. 45 rule on C89 4 outies R1469C7 = 1 innie R5C8 + 23
6a. Min R5C8 = 2 -> min R1469C7 = 25
6b. Max R14C7 = 17 -> min R69C7 = 8, no 1 in R6C7, no 1,2 in R9C7
6c. All combinations for R1469C7 totalling at least 25 need at least two of 7,8,9 -> R14C7 = {789}
6d. IOU R149C7 cannot total 23 and no 7 in R9C7 -> max R149C7 = 22 -> R6C7 must be greater than R5C8 -> max R5C8 = 4, min R6C7 = 3

[I only spotted steps 7 and 9, which are logical extensions of steps 6d and 8e, after step 14. They have been moved here for clarity.]

7. R5C8 less than R6C7 (step 6d) -> R567C8 = 6,7 = {123/124}, no 5, 1,2 locked for C8

8. 45 rule on C12 4 outies R1469C3 = 1 innie R5C2 + 7
8a. IOU R169C3 cannot total 7 and no 3 in R1C3 -> min R169C3 = 8
8b. Min R1469C3 = 13 -> min R5C2 = 6
8c. Max R5C2 = 8 -> max R1469C3 = 15, min R69C3 = 3 -> max R14C3 = 12, no 8,9 in R1C3, no 9 in R4C3
8d. Max R1469C3 = 15, all combinations for R1469C3 totalling 15 or less need at least two of 1,2,3 -> R69C3 = {123}
8e. Min R169C3 = 8 -> R5C2 must be greater than R4C3 -> no 8 in R4C3

9. R5C2 greater than R4C3 (step 8e) -> R345C2 = 23,24 = {689/789}, no 5, 8,9 locked for C2

10. 45 rule on N1 3 innies R2C3 + R3C23 = 17
10a. Max R2C3 + R3C2 = 14 -> min R3C3 = 3

11. 45 rule on N3 3 innies R2C7 + R3C78 = 11
11a. Min R2C7 + R3C8 = 5 -> max R3C7 = 6

12. 45 rule on N7 3 innies R7C23 + R8C3 = 19
12a. Max R7C2 + R8C3 = 15 -> min R7C3 = 4

13. 45 rule on N9 3 innies R7C78 + R8C7 = 13
13a. Min R7C8 + R8C7 = 6 -> max R7C7 = 7
13b. Hidden killer triple 7,8,9 in R1C7, R4C7 and R78C7 for C7 -> R78C7 must contain one of 7,8,9
13c. R7C78 + R8C7 cannot contain 6 and one of 7,8,9 -> no 6 in R78C7

14. Hidden killer triple 1,2,3 in R23C3, R6C3 and R9C3 for C3 -> R23C3 must contain one of 1,2,3
14a. R2C3 + R3C23 cannot contain 4 and one of 1,2,3 -> no 4 in R23C3

15. 6,9 in N2 locked in R23C46
15a. 45 rule on N2 4 innies R23C46 = 1 outie R4C5 + 26
15b. R4C5 = {12} -> R23C46 = 27,28 = {4689/5679/5689}
15c. R23C4 cannot be {79} (because 16(3) cage at R2C6 cannot be {556}) -> no 1 in R2C3
15d. Min R2C3 + R4C5 = 3 -> no 5 in R2C7 (step 1b)
15e. R23C4 and R23C6 cannot be {78} (which clashes with R23C5) or {79} -> R23C4 and R23C6 must each contain one of 7,8,9

16. R23C46 = {4689/5679/5689} (step 15b)
16a. R23C4 cannot total 1 more than R23C6 because that would make R2C3 equal R2C7
16b. R23C46 = {4689} can only be R23C4 = {48} and R23C6 = {69} (R23C4 cannot be {49} because 17(3) cage at R2C3 cannot be {449}, R23C6 cannot be {48} because 16(3) cage at R2C6 cannot be {448}, R23C6 cannot be {49} because R23C4 = {68} would be 1 more than R23C6) -> no 4 in R23C6
16c. R23C46 = {5679} can only be R23C4 = {69} and R23C6 = {57} (R23C4 cannot be {57} because 17(3) cage at R2C3 cannot be {557}, R23C4 cannot be {59} because it would be 1 more than R23C6 = {67}, R23C4 cannot be {67} because R2C3 doesn’t contain 4) -> no 7 in R23C4
16d. R23C46 = {5689} can be R23C4 = {59/68/69} (R23C4 cannot be {58} because no 4 in R2C3), R23C6 = {58/59/68}

17. 1,4 in N8 locked in R78C46
17a. 45 rule on N8 4 innies R78C46 = 1 outie R6C5 + 4
17b. R6C5 = {89} -> R78C46 = 12,13 = {1245/1246/1345}
17c. R78C6 cannot be {13} (because 14(3) cage at R7C4 cannot be {455}) -> no 9 in R8C7
17d. Max R6C5 + R8C7 = 17 -> min R8C3 = 6 (step 2b)
17e. R78C4 and R78C6 cannot be {12}, {13} or {23} which clashes with R78C5 -> R78C4 and R78C6 must each contain one of 1,2,3

18. R78C46 = {1245/1246/1345} (step 17b)
18a. R78C4 cannot total 1 more than R78C6 because that would make R8C3 equal R8C7
18b. R78C46 = {1246} can only be R78C4 = {14} and R78C6 = {26} (R78C4 cannot be {16} because it would be 1 more than R78C6 = {24}, R78C4 cannot be {24} because 13(3) cage at R7C6 cannot be {166} even if there was still a 6 in R8C7, R78C4 cannot be {26} because 14(3) cage at R78C4 cannot be {266}) -> no 6 in R78C4
18c. R78C46 = {1345} can only be R78C4 = {35} and R78C6 = {14} (R78C4 cannot be {14} because 13(3) cage at R7C6 cannot be {355}, R78C4 cannot be {34} which would be 1 more than R78C6 = {15}, R78C6 cannot be {34} because no 6 in R8C7 -> R78C4 cannot be {15}) -> no 3 in R78C6
18d. R78C46 = {1245} can be R78C6 = {14/15/24} (cannot be {25} because no 6 in R8C7), R78C4 = {15/24/25}

[I’ve just spotted that the total of the outer cages is 180 which means that R19C19 = R28C28 but it’s no help at this stage and might only help in the final mop-up, if at all.]

19. R2C3 + R3C23 = 17 (step 10) and 17(3) cage at R2C3 share a common cell at R2C3
19a. R23C4 can only be {48/59/68/69} (steps 16b to 16d)
19b. R2C3 + R3C23 = {278/359/368} (cannot be {269} which clashes with 17(3) cage at R2C3 = {269})
19c. 9 of {359} must be in R3C2 -> no 9 in R3C3
19d. R3C23 = {68/78}/[93/95]

20. R7C78 + R8C7 = 13 (step 13) and 13(3) cage at R7C6 share a common cell at R8C7
20a. R78C6 can only be {14/15/24/26} (steps 18b to 18d)
20b. R7C78 + R8C7 = {157/238/247} (cannot be {148} which clashes with 13(3) cage at R7C6 = {148})
20c. 1 of {157} must be in R7C8 -> no 1 in R7C7
20d. R7C78 = {23/24}/[51/71]

21. R2C3 + R3C23 = 17 (step 10), R2C7 + R3C78 = 11 (step 11) -> R3C23 cannot be 6 more than R3C78 because that would make R2C3 equal to R2C7, also R3C78 cannot be greater than 10
21a. Hidden killer triple 1,2,3 in R3C1, R3C37 and R3C9 for R3 -> R3C37 must contain at least one of 1,2,3
21b. 45 rule on R123 3 innies R3C2378 = 1 outie R4C5 + 21, R4C5 = {12} -> R3C2378 = 22,23
21c. R3C2378 = 22 = {1678/2389/2578/3469} (cannot be {1489/2479/4567} which clash with R3C23, cannot be {1579/2569/3568} because R3C23 would be 6 more than R3C78, cannot be {3478} because R2C7 + R3C78 cannot be [434])
21d. R3C2378 = 23 = {1589/2678/3569} (cannot be {1679/2489} which clash with R3C23, cannot be {2579} because R2C7 + R3C78 cannot be [227], cannot be {3479} because R3C78 would be greater than 10, cannot be {3578} because R2C7 + R3C78 cannot be [335], cannot be {4568} which doesn’t contain any of 1,2,3)
21e. R3C23 (step 19d) = {78}/[93/95] ({1678} cannot be {68}[17] because R23C23 would be 6 more than R3C78, {2678} cannot be {68}[27] because R2C7 + R3C78 cannot be [227]), no 6
21f. R3C78 = [16/18/25/26/28/36/46/64] ({1678} cannot be {68}[17] because R23C23 would be 6 more than R3C78, {2678} cannot be {68}[27] because R2C7 + R3C78 cannot be [227], {3569} cannot be [93]{56} because R3C78 would be greater than 10), no 5 in R3C7, no 7 in R3C8

22. 21(3) cage at R3C8 = {489/579/678}
22a. 5 of {579} must be in R3C8 -> no 5 in R4C8

23. R7C23 + R8C3 = 19 (step 12), R7C78 + R8C7 = 13 -> R7C23 cannot be 6 more than R7C78 because that would make R8C3 equal to R8C7, also R7C23 must be at least 10
23a. Hidden killer triple 7,8,9 in R7C1, R7C37 and R7C9 for R7 -> R7C37 must contain at least one of 7,8,9
23b. 45 rule on R789 4 innies R7C2378 = 1 outie R6C5 + 9, R6C5 = {89} -> R7C2378 = 17,18
23c. R7C2378 = 17 = {1259/1457/2348} (cannot be {1268/1349} which clash with R7C78, cannot be {1358} because R7C23 + R8C3 cannot be [388], cannot be {1367} because R7C23 would be less than 10, cannot be {2357} because R7C23 + R8C3 cannot be [577], cannot be {2456} which doesn’t contain any of 7,8,9)
23d. R7C2378 = 18 = {1278/1467/2349/2358} (cannot be {1269/1368} which clash with R7C78, cannot be {1359/1458/2457} because R7C23 would be 6 more than R7C78, cannot be {2367} because R7C23 + R8C3 cannot be [676], cannot be {3456} which doesn’t contain any of 7,8,9)
23e. R7C23 = [28/29/46/47/48/49/58/64] ({1457} cannot be {45}[71] because R7C23 would be less than 10, {2348} cannot be [38]{24} because R7C23 + R8C3 cannot be [388] , {2349} cannot be [39]{24} because R7C23 would be 6 more than R7C78), no 3 in R7C2, no 5 in R7C3
23f. R7C78 (step 20d) = {23}/[51/71] ({2348} cannot be [38]{24} because R7C23 + R8C3 cannot be [388], {2349} cannot be [39]{24} because R7C23 would be 6 more than R7C78), no 4

24. 9(3) cage at R6C2 = {126/135/234}
24a. 5 of {135} must be in R7C2 -> no 5 in R6C2

25. R2C3 + R3C23 = 17 and 17(3) cage at R2C3 share a common cell at R2C3 (step 19)
25a. R3C23 (step 21e) = {78}/[93/95]
25b. R2C3 + R3C23 (step 19b) = {278/359}
25c. R23C4 (step 19a) = {48/68/69} (cannot be {59} which clashes with R3C23 = [95]), no 5

26. R7C78 + R8C7 = 13 and 13(3) cage at R7C6 share a common cell at R8C7 (step 20a)
26a. R7C78 (step 23f) = {23}/[51/71]
26b. R7C78 + R8C7 (step 20b) = {157/238}
26c. R78C6 (step 20a) = {14/24/26} (cannot be {15} which clashes with R7C78 = [51]), no 5

27. R456C1 = {129/147/237/246/345} (cannot be {138} because R6C23 cannot be [42/62] since no 1,3 in R7C2, cannot be {156} because R4C23 + R5C2 cannot be [97]8 since no 6 in R3C2), no 8
27a. Cannot be {237} => R6C23 = [61] (R6C23 cannot be [41] because 8(3) cage cannot be [414]) => R45C3 = [54] => R45C2 = [98], R4C23 = [59] => R3C2 = 8 clashes with R5C2
27b. -> R456C1 = {129/147/246/345}

28. R456C9 = {189/369/378/468/567} (cannot be {279} because R4C78 cannot be {48/68} since no 7,9 in R3C8, cannot be {459} because R6C78 cannot be [31/32] since no 4 in R7C8 and 8(3) cage cannot be [323]), no 2
28a. Cannot be {378} => R4C78 = [94] (R4C78 cannot be [96] because 21(3) cage at R3C8 cannot be [696]) => R56C7 = [65] => R5C8 = 2 clashes with R67C8 = [12]
28b. -> R456C9 = {189/369/468/567}

29. R1469C3 = R5C2 + 7 (step 8), min R169C3 = 8 (step 8a)
29a. R5C2 = {678} -> R1469C3 = 13,14,15 = {1345/1256/1346/2345/1257/1347/1356/2346} (cannot be {1237/1246/1247} because min R169C3 = 8)
29b. R2C3 + R3C23 (step 25b) = {278} (cannot be {359} which clashes with R1469C3) -> R2C3 = 2, R3C23 = {78}, locked for R3 and N1
29c. R2C3 = 2 -> R23C4 = 15 = {69}, locked for C4 and N2 -> R3C6 = 5, R3C5 = 4, R3C8 = 6, R23C4 = [69]
29d. Max R2C3 + R4C5 = 4 -> min R2C7 = 3 (step 1b)

30. R3C8 = 6 -> R4C78 = 15 = {78}, locked for R4 and N6
30a. R4C2 = 9 (hidden single in C2, and in 22(3) cage at R3C2), clean-up: no 4,5 in R5C34 (prelim d)
30b. Naked triple {678} in R5C234, locked for R5, 6 locked in R5C23 for N4 -> R5C5 = 5, clean-up: no 1 in R5C67 (prelim e)
30c. Naked triple {234} in R5C678, locked for R5 -> R5C19 = [19], R3C1 = 3, R69C3 = [31]

31. R5C1 = 1 -> R46C1 = [47] (step 27)
31a. Naked pair {68} in R5C23, locked for R5 and N4 -> R4C3 = 5, R5C4 = 7, R6C2 = 2, R3C2 = 8 (prelim b), R3C3 = 7, R5C23 = [68], R7C2 = 4 (prelim f), R12C2 = [51], R12C1 = [69], R1C3 = 4, R89C2 = [73], R9C4 = 8
31b. Naked triple {789} in R1C789, locked for N3

32. R3C7 = 1 (hidden single in C7), R3C9 = 2, R6C4 = 4, R6C78 = [51], R6C9 = 6, R4C9 = 3, R4C456 = [216], R5C6 = 3, R7C37 = [93], R6C6 = 8, R6C5 = 9, R9C56 = [79], R2C56 = [87], R2C7 = 4 (step 1b)

and the rest is naked singles.

Thinking further about that I'd like to qualify my acceptance slightly.

If the uniqueness is as a result of the complete rotational symmetry then I think I would find this technique acceptable although I'm not sure whether I'd use it. My A123 walkthrough in the previous message didn't use it. However if the uniqueness is only because a software solver has checked and said that a puzzle has a unique solution, then I would consider it to be like a UR and definitely wouldn't use it.

Can udosuk, or anyone else, please tell us whether the uniqueness of A123 and the other puzzles in this thread is as a direct result of their rotational symmetry?

Nah. You can create puzzles which are not unique, which have a rotational cage pattern.