someone sent me a difficult 6x6 from Simon Tatham's website, and it _is_ difficult
(my solver rates it slightly above the regular difficult 6x6 on the site (!))

I started solving it, but am momentarily stuck (would like to find a solving path not
involving trial-and-error )

Anyone have any ideas?

Hi, Patrick.
Step 1: to determine that "4-" is [1,5] but not [2,6].

Applying the parity rule, since "30x" is even and "6x" is odd we arrive to the conclusion that the sum of "2:" and "2:" in the right hand corner (columns e and f) must be odd, which means that one of them must be even [2,4] and the other odd ([1,2] or [3,6]).

The sum of "30x" and "6x" is always 17 ([2,3,5] plus [1,6] or [1,5,6] plus [2,3]).

Now, suppose "4-" is [2,6], the two division cages must be [2,4] and [3,6]. As a consequence, the three cages have a sum of 8 + 6 + 9 = 23. Now we have: 18 + 17 + 23 = 58, so "1-" in f2-f3 = 5, that is, [2,3], which is impossible since no more 2's can be located in columns e or f. Then, "4-" = [1,5].

The two division cages, "2:" in the right corner must be [2,4] and [3,6] with a sum of 15. And then, 18 + 17 + 6 + 15 = 56, thus "1-" = [3,4]. Clearly the division cages could not be [2,4] and [1,2] because 18 + 17 + 6 + 9 = 50 and "1-" would sum 13, that is, [6,7] (out of range).

We find very interesting conclusions from here, for instance, "a3" = 5; "7+" can't be [2,5] and "60x" must be [1,2,5,6] or [1,3,4,5] since [2,2,3,5] is now impossible. The 5 of "30x" is in d5, otherwise b5 = 5, but [1,2,6] or [1,3,4] would inhibit at the same time [2,4] or [3,6] for cage "2:" in e6-f6. So, d5 = 5 and, since there is a 5 in either b1 or c1 (another 5 is in b6 or c6), f1 = 1, f4 = 5 and e4 = 1.

Let's continue with the analytical solution. The 5 in column e must be in e2 (since there is a 5 in b1 or c1). This means that d1 + e1 + d2 = 12. The possibilities, without any 5, are [3,3,6] or [2,4,6] for those positions, but the first one is not possible because it would inhibit "30x"; thus [2,4,6].

The cage "24x" must be [2,3,4] or [1,4,6] since [2,2,6] is not possible now. Consequently, this cage has a 4 and, since there is another 4 in [2,4,6], "1-" in b1-c1 must be [5,6] and not [4,5].

The rest is as follows: d2 = 6, "30x" = [2,3,5], d3 = 1 and e3 = 6, d1 = 4 and e1 = 2.
Then a1 = 3, a4 = 4, a5 = 1, a2 = 2, b2 = 4 and a6 = 6.

b2 = 4 >>> f2 = 3, f3 = 4, c2 = 1, c3 = 2, b3 = 3 and b4 = 6. Thus b1 = 5 and c1 = 6. "7+" is now [3,4] so c4 = 3, c5 = 4 and, obviously, c6 = 5. Also, d4 = 2 and d6 = 3.

Finally, "2:" in e5-f5 = [3,6] with e5 = 3, f5 = 6.
And "2:" in e6-f6 = [2,4] with e6 = 4 and f6 = 2. And we complete the puzzle with b5 = 2 and b6 =1.

The full analytical solution we have arrived is:

356421
241653
532164
463215
124536
615342

Comment: The puzzle is quite difficult. I believe this is because of the presence of four division cages "2:". In general, the division cages, as the subtraction cages, increase the level of difficulty.
Merry Christmas and a Happy New 2024 for everyone.

Thanks clm and wishing all the same as well!

Hi everyone.
I also examined the problem before seeing the other solution.
Here is how I solved it:

I don't know where to draw the line as to where a solution is considered to be T&E,
but here is a logical solution:

C1~1 (examine column C)
D1,D2~5

If we look at the cages in columns DEF all but the two 2: cages have a fixed parity.
Taking parity into consideration, one of the cages E56, F56 is {2,4} and the other {1,2} or {3,6}.

Now examine the possible values for the cage EF6. They must be {1,2}, {2,4}, {3,6}.
These interact strongly with the 60* cage at B5.
If EF6 is {1,2} then EF5={2,4}, D6=6, B5=1, D56={1,5}, D45={1,5} which is impossible.
If EF6 is {3,6} then D6=2, B5=3, EF5={2,4}, D5=5, D4=3, C45={1,6}, C23={2,4}, C6=5, C1=3.
This is also impossible because B1 is the only 5 remaining in Col B and B1,C1 must differ by 1.

This establishes that EF6={2,4}. Either D6=6 && B5=4 or D6=3 && B5=2.

The next step was to check the first possibility D6=6 && B5=4.
Note that since A6 is now odd and A3+A4+A5=10 that A1 and A2 have the same parity.
Thay obviously can't be {1,3} as that would require B2=8.
Checking possible values for B2:
B2=6 -> A12={1,4}, A6=21 - 5 - 10 = 6 which is impossible.
B2=4 implies that one of A12 is odd which is impossible.
B2=3 -> A12={2,4}, A6=5, B6=1, B34={2,6} which is impossible.
B2=2 -> A1=2, A2=6, A6=3, B34={3,6}, B1=5, B6=1, C3=2 leaving no place for a 2 in Column D.

Hence we have established that D6=3 and B5=2.
The rest of the solution is routine.
B34={3,6}, D5=5, D4=2
There is an Xwing on value 4 in rows 1,2 columns BD. So A1, A2 C1,C2, E1, E2, F1, F4 are all ~4.
ETC.

Thanks clm and crabmouse,
these analyses are actually very useful for improving my solver