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Hm. This doesn't seem like it does much if there's a sufficiently high concentration of cracked districts, relative to packed districts. I gerrymandered during
by vitus 3y ago
Hm. This doesn't seem like it does much if there's a sufficiently high concentration of cracked districts, relative to packed districts.
I gerrymandered during the define phase using classic packing / cracking strategies, such that I had 8 majority-B districts (2:3) and 2 majority-A districts (4:1 and 5:0), and unsurprisingly, the only districts I was able to combine that were majority-A were those that included the packed districts.
If the overall split was, say, 27:23 instead of 25:25 such that we could define 9 majority-B districts in the define phase, then I would only have been able to define a single majority-A district in the combine phase.
(And yes, all of these gerrymandered districts would be considered safe B seats, as one would expect with a 20% margin)
There are also potentially issues if the packed districts are geographically clustered -- we see this a lot in states with a single predominant urban center (e.g. Kansas, Minnesota, Kentucky). In those cases, you might be forced to combine multiple packed districts due to pathological maps. For instance, consider a map where a Democratic bastion is districted into concentric rings -- that satisfies the contiguity requirement, yet only the outermost district abuts any Democratic-minority districts.
- maxpalmer 3y agoSuppose you're defining as Party B, and you draw 8 majority-B districts (2:3) and 2 majority-A districts. Then, when Party A is combining, they would pair each of the majority-A districts with a majority-B district with a smaller margin, resulting in 2 A districts and 3 B districts. This is an improvement compared to if B drew 5 districts unilaterally, where it could draw 4 majority-B districts.
- vitus 3y agoYes, it is an improvement in this specific scenario, since there's limited granularity in how we can draw the districts. If instead we had 9 population nodes we could assign to each Define district, then we might be able to draw nine 4:5 districts in favor of B, and a single 9:0 district in favor of A. In that case, A cannot recover any of the unfairness that B introduced.