Overview
--------
- Solve abstractly in A-I
- Resolve values using kropke dots
- Clean up

Abstractly
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In box 5 we know that:
col 5 is one of [147], [258], or [369] in some order
row 5 is one of [123], [456], or [789] in some order
Either way, let's call column 5 [ABC], specifically B-A-C and row 5 [ADG], specifically G-A-D.
Box 5 completes [EF], [FI], [HI], [EH] clockwise.

Before you've started looking at the kropkes, *always* use the inherrant symmetry - if you deduce something about D in a column, then there's guaranteed to be an equivalent deduction about C in a row rotated by 90 degrees to the right, and about B in a row rotated 90 degrees to the left.

Letter the modular line [ADG]-[BEH]-[CFI]-... in sequence
Letter the entropic line [ABC]-[DEF]-[GHI]-... in sequence
r5c2 must be [ABC] and [BEH] so is B
r5c8 must be [ABC] and [CFI] so is C
r7c5 must be [EH] and [DEF] so is E
r2c7 must be [DG] and [DEF] so is D
r5c3 must be [CFI] and [DEF] so is F
r8c5 must be [DG] and [GHI] so is G
r5c7 must be [BEH] and [GHI] so is H
r3c5 must be [FI] and [GHI] so is I

Deductions from that make r4c7 E, r7c6 F, r6c3 I, and r3c4 H.
Deductions from that make r6c4 E, r4c4 F, r4c6 I, and r6c6 H.
Deductions from that make r5c1 E, r1c5 F, r5c9 I, and r9c5 H.

See what I mean about the symmetry?

Kropkes
-------
We can deduce r1c4 and r9c6 are [ABC], and that r4c9 and r6c1 are [ADG].
However, we can only limit r1c6 to [ABCEG], r4c1 to [ADGCH], r9c4 to [ABCDI] and r6c9 to [ADGBF]

So our four kropkes are
i) F x [ABCEG]
ii) E x [ACDGH]
iii) H x [ABC]
iv) noise x noise

As F and E both must participate, they can't be [789].
So [DEF] is [123] (any order) or [456] (D=5)

As H and E both participate they could be [14] (pairing with [28]), or [28] (pairing with [14]), but not [36], as H would only ever pair with E, breaking (iii).
Therefore [HE] is either [14] or [28]
Non-[369]s pair only with unlike residues, so remove H from (ii) r4c1 and B from (iii) r9c6.

If [DEF] is [456], D would have to be 5 and E would have to be 4, making F 6.
F=6 only pairs with [CI], and only C's available, so [ABC]'s low, and H is 8 - (iii) breaks.
Therefore [DEF] is [123].

We can now immediately reject H=6, as it needs to pair with [AC], but the 3 is [DEF].
If H was 4, E=1, pairing with D=2, making F=3 pairing with 6, which would be I, which isn't available.
Therefore H is 8, making [AC]=4, so B=5, and E=2.
We can remove all [579]=[BGI] from all kropkes, and H=8 where it's too high against E.

To be honest, it's not worth the effort changing BEHs into numbers, just continue in letters.

Our kropkes now look like:
i) F=[13] x [26]=[ACE]
ii) E=2 x [14]=[DAC] - but see below
iii) H=8 x 4=[AC]
iv) [ADF] x [ACDH]

This pruning places box 2's G in r2c6, and box 4's H in r4c3, and box 6's B in r6c8
=> r7c3 is B, r8c3 is E, and r3c3 is D

However, 8x4x2, so Hx[AC]xE, i.e. ExD is impossible.
=> box 4 row 4 has an [AC] pair, making a [DG] pair, and fixing r6c7=F and r6c9=A
=> r3c7 is C, and r2c7 is I, r7c7 is G, r2c2 is H, and r2c7 is E
=> Box 9's E is in r9c9

iv) A x [CDH] could be 4 x 2=?D, 4 x 8==H, or 6 x 3=?D, so isn't A x C
If it was A=6 x 3=D, then E=2 x 4=C, but r7c2 becomes A, preventing that.
Therefore it's A=4 x H=8., and F=3 x 6=C up top.

And that's huge. Speedrun!


Proof of uniqueness:
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Entropic%7ER5C2%7ER5C3%7ER4C3%7ER4C2%7ER3C2%7ER2C2%7ER2C3%7ER3C3%7ER3C4%7ER2C4%7ER2C5%7ER3C5%7ER4C5%7ER4C4%7ER5C4%7ER5C5%7ER5C6%7ER6C6%7ER6C5%7ER7C5%7ER8C5%7ER8C6%7ER7C6%7ER7C7%7ER8C7%7ER8C8%7ER7C8%7ER6C8%7ER6C7%7ER5C7%7ER5C8.Modular%7E3%7ER2C5%7ER3C5%7ER3C6%7ER2C6%7ER2C7%7ER2C8%7ER3C8%7ER3C7%7ER4C7%7ER4C8%7ER5C8%7ER5C7%7ER5C6%7ER4C6%7ER4C5%7ER5C5%7ER6C5%7ER6C4%7ER5C4%7ER5C3%7ER5C2%7ER6C2%7ER6C3%7ER7C3%7ER7C2%7ER8C2%7ER8C3%7ER8C4%7ER7C4%7ER7C5%7ER8C5.BlackDot%7ER9C5%7ER9C6.BlackDot%7ER1C6%7ER1C5.BlackDot%7ER4C1%7ER5C1.BlackDot%7ER6C9%7ER7C9