One proposal for redistricting reform is the concept of "compactness". A "compact" congressional district was defined by Ballotpedia as "[a district where] the constituents residing within an electoral district should live as near to one another as possible". A metric is to measure "[the] ratio of the circumference of a district and its total area".

Is it possible to draw a "compact" congressional map that is also a gerrymander, and if so, is it computationally feasible often enough to be used in real life?

  • 2
    "Prevent" is the wrong metric. Nothing will absolutely prevent gerrymandering. The goal is to make gerrymandering more difficult / less extreme. Does compactness make gerrymandering more difficult? Almost certainly. Feb 24, 2022 at 1:11
  • Could you provide a working definition of gerrymandering? Google's says "manipulate the boundaries so as to favor one party or class". Compactness isn't the only factor - just look at Michigan's fairly compact map, which is in litigation right now. A previous (dismissed) suit decried how it splits up majority-black districts in Detroit, which absolutely affects the playing field - hey look, two Dems got consolidated, creating a new R+8 district with no incumbent. That's certainly a desired result for the GOP Feb 24, 2022 at 1:17
  • Depending on the extent to which it's weighted vs other metrics, I suspect emphasizing compactness favors the GOP, since large cities (> 1 representative equivalent) are overwhelmingly blue and most smaller cities (< 1/2 representative equivalent) are more of a mixed bag. See Georgia's fairly compact map, and compare the partisan leans around Atlanta (D +16, +50, +52, +60) to the state's other semi-major cities' districts (Augusta R+17, Savannah R+20; Columbus is split so it's not a great analogy) Feb 24, 2022 at 1:24
  • @Punintended - Compactness demonstrably makes gerrymandering harder by disallowing a number of maps that would produce outcomes disproportionate to the overall population, and the Republican Party employs much more gerrymandering, so I think the most it could do is not disfavor the Republican Party as much as some other measures
    – Obie 2.0
    Feb 24, 2022 at 1:29
  • @Obie2.0 I don't disagree, but I was trying to frame things neutrally in lieu of pointing fingers about current gerrymandering practicees. One method the GOP uses is siloing of like voters, which can be prevented with compactness (eg, fixing the Alabama map) or not (see Georgia above). Using this tool, Dems go from 168 to 151, while GOP goes from 195-180. The difference is similar but the advantage persists Feb 24, 2022 at 1:36

2 Answers 2


No prevention, but some inhibition.

Gerrymandering relies on clever mappings to maximize the number of voting districts with small but statistically significant majorities for your own party (which as a corollary gives to other parties preferentially districts with huge majorities from them). It's only important for non-proportional representative voting systems though, like for example in the US or UK. Indeed gerrymandered voting maps tend to be highly non-compact, sometimes not even connected voting districts while voting district maps drawn by neutral commissions tend to be quite compact.

Restricting the freedom to choose arbitrarily shaped district maps will restrict the ability to maximize gerrymandering. One can expect that with your compactness requirement the maximal possible amount of gerrymandering cannot be achieved.

However, since voters of the same political orientations seem to be concentrated geographically (city, countryside, ...) even compact voting district maps can be gerrymandered a lot. Therefore it doesn't prevent it.


Suppose there is a state with only three voting districts and two regions of voters (one leaning for one party and another leaning for another party). You can surely cut the regions so that either party gets two out of the three voting districts but still have fairly compact regions. Therefore compactness is not sufficient to ensure fairness of elections in non-proportional voting systems.

By the way, Wikipedia lists a few more options to achieve more competitive voting maps without much human intervention: Minimum district to convex polygon ratio, Shortest splitline algorithm, Minimum isoperimetric quotient or Efficiency gap calculation. Or simply change the voting system to something that is proportional.

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    "Or simply change the voting system to something that is proportional." Note that this simply causes a different problem. In any case I am for convex polygons as the solution.
    – Joshua
    Jun 22, 2023 at 21:02

Effectiveness of the proposal aside, there is an inherent problem with the premise of using perimeter and area as a ratio when using (e.g.) geographical areas. In general, the more vertices a shape has (with a fixed area) the larger the perimeter. In other words, depending on the unit of measurement the perimeter can be any value (with a reasonable lower bound and infinity as the upper bound). This is quite famously known as the Coastline Paradox, where depending on what units you use, you can basically get any measure you want for the perimeter of the coastline, meaning that this measure (ratio of area and circumference) can be gamed, obviously, in the same way as current political gerrymandering. To directly answer the question, it is possible to draw a compact district (because you can essentially get any circumference you want in practice). The computational feasibility is NP-Hard (i.e. most difficult) since I believe districts would still be computed based on a minimal cover algorithm so a heuristic would probably be used anyway.

  • The measure given in the question is not a measure of compactness for precisely this reason. Popular measures of compactness measure the ratio of the area of a shape to the area of some other shape whose length scale is related to the perimeter of the shape in some fashion, such as a circle whose circumference is equal to its perimeter. If you double the perimeter of the shape while making it "wrinkly" such that its area remains unchanged, you only decrease such a measure of compactness.
    – Obie 2.0
    Feb 24, 2022 at 0:58
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    For another measure of compactness, the Reock score, the ratio used is that of the area of the shape to that of the minimum bounding circle. If a shape is made more wrinkly without substantially extending it in some direction, this measure of compactness will stay almost unchanged.
    – Obie 2.0
    Feb 24, 2022 at 1:04
  • We also don't know for sure that an NP-hard problem is the "most difficult", because it has not been proven that exist no polynomial time algorithms for NP-hard problems, even if it does seem likely.
    – Obie 2.0
    Feb 24, 2022 at 1:15
  • @Obie2.0 I'm not sure that uberhaxed meant NP-hard problems are "the most difficult"; they only said "most difficult". Without "the" it could just mean "very difficult". Regardless, you don't need to speculate about P = NP to argue that NP-hard problems aren't "the most difficult"; that would be blatantly and obviously false (which is why I'm a little doubtful that uberhaxed would have meant it that way, for you to correct), because there are classes like NEXPTIME (and indeed an infinite series of ever-harder classes we haven't named).
    – Ben
    Feb 24, 2022 at 2:32
  • @Ben - Well, true.
    – Obie 2.0
    Feb 24, 2022 at 3:17

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