All Hail Hypnorenban

Renbans, entropies, and German whispers can be solved only up to sum-to-10 parity, only the black kropkes can distinguish high from low, so it makes most sense to solve for A-I rather than 1-9 until parity is known. However, it also makes sense to colour for entropy with 3 colours.

The 3 renbans on row 1 must be [ABC]/[DEF]/[GHI] in some order.
The rightmost segment can't be [DEF] because the kropke demands a [456] participant, label it [GHI].
If box 1's component was [DEF], its other whole renban would be [123] or [789], but whatever it is, you can't complete the kropke with a number from the other set.
=> Top row is [ABC]/[DEF]/[GHI] triples, with some extra restrictions on the dotted cells

Column 1's whole renbans must include the two of [ABC] that aren't in r1c1, and obviously can't be [ABC], so one of them must be [BCD] and r1c1 is A.
The other one must contain a G, so the dogleg renban fully in box 1 can have an [EF] or [HI] pair.
If it's [HI], it must be completed with a G, either 7 or 3, so must be 3 paired with 6, D, which must be on a [DEF] renban that goes into box 2, impossible.
=> it's an [EF] pair, with its renban completed with a D, that's 4 or 6, so pairing with 2,8, or 3, or in letters B,H, or G on the kropke, but clearly not B.
With D and another from [EF] taken on row 2, one of the other renbans must be [123], and it can't be the one partly in box 1.
This forces the renban mostly in box 2 to be [GHI], and the straggler in column 9 is left as [EF].
The other dogleg renban in box 1 must be [GHI], but the I is in box 1.


Column 1's other 2 renbans must now be [BCD] above [GHI]
The entropy line in column 3 must coexists with the [BCD] renban, so must contain A
The entropy line in column 2 must contain an A, B, or C, which must be external to box 4.

Row 9 and column 9 are very similar in composition, and should be considered together
If they start with a G or H, then the renbans must be [ABC] and [DEF], leaving [GI] or [HI] in box 9
If they start with an I, they could be [ABC] with [DEF] leaving [GH], [ABC] with [FGH] leaving [DE], or [CDE] with [FGH] leaving [AB].
However, if one of them leaves [AB], or [DE], the other must too, and the dogleg renban in box 9 contains a duplicate.
Therefore at least one starts with a G or H, and the other is a different member of [GHI], and both have [ABC] and [DEF] as their renbans.

Row 9 has the [ABC] before the [DEF] in order to mate at the kropke with [GHI]
Column 9 already has an [EF] in r2c9, so the order's defined.
A black kropke of [DEF]o[ABC] can be DoB (4o2), FoC (6o3), or FoB (4o8)
A black kropke of [DEF]o[GHI] can be FoH (4o8), DoG (6o3), or DoH (4o8)
These push I into column 9 of box 9 an dout of r1c9
They also discover Es on r1c5 and r9c6

Box 5, being 3 renbans, must be [ABC]/[DEF]/[GHI], but row 3 rejects [GHI] and [DEF], and column 6 rejects [DEF], so can be fully coloured.
This places D in r4c1 and r2c2, F in r4c9, E in r2c9, D in r3c9, F in r2c1, and E in r3c1

The ABC triple in box 5 and D in r4c1 prevent the renban in column 7 from being ABC or BCD, but it could be C-E-D.
However, hypothecating a C into r3c7 forces F into r9c7. The dogleg renban partly in column 7 needs to contain only values from [GHI], but there's no room in box 9 for those.
=> r3c7 is not C, it's F
=> r9c7 is D, and r9c5 is F
=> r5c5 is D
=> r6c8 is naked D
=> r1c4 is D and r1c6 is F
=> r1c3 is B to be on the kropke with D, and r1c2 is C, and that's huge!
=> A r9c3 is C by sudoku

However, what's huge is that [BD] kropke pair has resolved the letters into numbers: A=1
The other kropkes have a bunch of implications from their resolution from simple scanning.

The 4 and [56] pair in column 4 prevent the renban in row 7 from continuing in the [456] range, and there's no room for 2 more [123]s in the row, so they're a [78] pair, and r7c3 must be 6
=> r7c1 is 9, r9c1 is 7, and r8c1 is 8
=> r9c9 is 9, r8c9 is 7, and r1c9 is 8
=> r8c2 is 5, and r8c3 is 4
=> r8c4 is 3, as it can't be 6
=> r8c8 is a naked 6, r7c6 is a naked 4, and r6c8 is a naked 4

By sudoku, r7c7 and r7c8 become a [35] pair, and row 8 needs a [129] triple, however, r8c7 and r8c5 are restricted to [12], placing 9 in r8c6
=> r6c5 is 9, resolving the [79] pair in box 2, which resolves the [78] pair in box 8.

The renban on r7c6 can't continue [56] by 6 pressure, nor [23] by [123] presure, so continues [35] in known order.
=> r6c4 is 6 and r5c4 is 5, forcing 5 into r4c3 and 6 into r5c2.
=> r6c2 is 7, making r5c3 9, resolving the [79] pair in box 1, and the [78] pair in box 5.
=> the straight renban in column 7 uniquely continues 7 and 8, which forces 9 into r4c8 and resolves the [79] pair in box 3.

The 1 in r6c3 resolves the [12]s practically everywhere, and the rest of the empty cells drop in instantly, needing just a little help from the modular line in box 2.

Proof of uniqueness:
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Renban%7ER1C1%7ER1C2%7ER1C3.Renban%7ER5C9%7ER6C9%7ER7C9.Renban%7ER7C1%7ER8C1%7ER9C1.Renban%7ER2C6%7ER2C7%7ER2C8.Renban%7ER8C4%7ER8C3%7ER8C2.Renban%7ER3C7%7ER4C7%7ER5C7.Renban%7ER4C4%7ER4C5%7ER4C6.Renban%7ER1C7%7ER1C8%7ER1C9.Renban%7ER2C2%7ER2C1%7ER3C1.Renban%7ER9C7%7ER9C6%7ER9C5.Renban%7ER6C8%7ER7C8%7ER8C8.Renban%7ER4C2%7ER3C2%7ER3C3.Renban%7ER7C5%7ER7C4%7ER7C3.Renban%7ER6C4%7ER5C4%7ER5C5.Renban%7ER1C4%7ER1C5%7ER1C6.Renban%7ER2C9%7ER3C9%7ER4C9.Renban%7ER8C9%7ER9C9%7ER9C8.Renban%7ER9C4%7ER9C3%7ER9C2.Renban%7ER6C1%7ER5C1%7ER4C1.Renban%7ER2C3%7ER2C4%7ER2C5.Renban%7ER6C7%7ER7C7%7ER7C6.Renban%7ER5C6%7ER6C6%7ER6C5.Entropic%7ER5C2%7ER6C2%7ER7C2.Entropic%7ER4C3%7ER5C3%7ER6C3.Modular%7E3%7ER3C4%7ER3C5%7ER3C6.Modular%7E3%7ER3C8%7ER4C8%7ER5C8.Modular%7E3%7ER8C7%7ER8C6%7ER8C5.BlackDot%7ER1C3%7ER1C4.BlackDot%7ER2C2%7ER2C3.BlackDot%7ER7C5%7ER7C6.BlackDot%7ER9C7%7ER9C8.BlackDot%7ER5C9%7ER4C9.WhiteDot%7ER9C6%7ER9C7