I use conditional formatting to color cells according to the probability that I can lift them—if I lifted 50kg for 10 reps then I can definitely do 50kg for 9 reps, so that cell is green. But if e1RM(50,10) > e1RM(40,15) then I can probably do that too so it's light green. The visualization naturally becomes Pareto-like.
If I'm feeling strong I can aim for higher weight, lower reps. Or if I'm feeling weak I can close out a (weight, reps) that's below my current e1RM but I haven't accomplished yet. The end result is that I'm always "accomplishing" some sort of PR no matter how I feel.
I call this e1RM Bingo.
My main finding for “pick whatever weight you want today” was that picking a lot of different weights made the curve less identifiable, so my latest iteration encourages you to pick a ladder for a few sentinel exercises per mesocycle in order to improve the statistical power. In addition, strength improves more quickly at >80% of 1RM, and hypertrophy depends on proximity to failure, so if you pick a lower weight, you really need to go to failure, which burns you out for the rest of your session, where leaving 1-2 reps in reserve is probably sufficient for hypertrophy and leaves a lot more gas in the tank for the rest of the session. Definitely open to suggestion/discussion here.
https://curvefit.app (it runs on Cloudflare free tier, so I won’t have to start running ads or charging until I hit a couple thousand users)
There are weight x rep combinations that have a e1rm of 80kg, 85kg, 90kg, and so on.
These are just equal elevation contours through the e1rm(x, y) function.
The Pareto concept doesn't require that we calculate a function of all the dimensions and find contours; that sort of thing is not involved.
But we could apply it here like this. Suppose we conduct a weight lifting contest as follows: contestants can lift any weight any number of times, and record the weight and reps.
Then, how do we rank the results to find a winner, or winners? We have multiple dimensions, not a single dimension like "seconds to run 10 km".
We can find the Pareto front set of the performances by eliminating all that have been dominated. A lift is dominated if another lift is no worse (no less weight, and no fewer reps), and strictly better: eight the weight is higher, or there are more reps, or both.
We then end up with undominated winners, e.g. there could be three like this: { (100kg, 1), (80kg, 2), (70kg, 5) } but (70kg, 4) would not belong, due to being dominated by the third one, and (90kg, 1) would not due to being dominated by the first. The middle one is not dominated by either: though it's less weight than the 100kg, it is more reps, and though it is fewer reps than the 70kg, it is more weight.
Given the Pareto front set, if we want to determine a single winner, we need a function to reduce the parameters to a single value. (The function should be such that if we included the eliminated losers under that function, none of them would emerge winner over the Pareto front set). This e1rm function looks like it fits the bill.
If we have this function, we don't need the Pareto concept; we just run all the results through the function and pick the contestant(s) that maximize it.
But they go even a step further, they extend into 3 dimensions to also add body weight as a variable. So your graph would really have to be a 3D volume. Because different levels of body weight have different capabilities.
Please don't make an app based on this.
Maybe in vein but did anyone already figure this one out? The closest I got was PT sans, open-licensed commissioned by the Russian ministry for communication (I found it surprising that a country that doesn't use Latin script made the best font!), but it's not widely shipped so you need to figure out how to include font files whenever you want to use it
"The Pareto Front today claimed responsiblity for...."
Anyways I'll namedrop Iosevka as perfect monospace font for working on 13" laptop
When explaining it to some coworkers, I stumbled on a fairly intuitive explanation: "I've run farther before, and I've run faster before, but I've never run _this_ far, _this fast."
There was some pushback about why not just call it a PR (personal record), but I would only use that term for fixed distances (1mi, 5k, 10k, etc.) or a consistent route that I've run many times before. Nobody would say "I set my 7.40 mile PR today." More importantly, it misses the comparison to all farther (and faster) runs—it's not exciting to set a 5k PR just because you've barely run that distance before, and the pace is actually slower that a 10k you've done.
(Had a Pareto run of 7.40 miles @ 6:28/mi last week!)
A point is Pareto so long as it is non-dominated—that is, you're not looking for dominating points, you're looking for points that are "no worse" than all others, in all criteria, when you consider that point as a reference.
So your Pareto Runs are indeed Pareto points. However, your run with your fastest possible speed, even if your distance was really bad, is also still a Pareto efficient point.
(Taking >= as more efficient here) By definition, the point A is Pareto if there is no point B such that in all criteria, B >= A, and for at least one criteria B > A. Take the run with the best speed. It is Pareto because we cannot find a single point B that satisfies both of these conditions. Your "Pareto Run" doesn't satisfy this set of conditions because it is worse in terms of speed, even if it has better distance than the max speed point.
The only way your Pareto runs would be the only Pareto points in your record is if they simultaneously hit maxima for distance and speed when compared to all other points. So, for them to be the sole Pareto point, the clause ""I've run farther before, and I've run faster before..." would have to be false! The point would have to break both your all time records to be the solitary Pareto point. With running, because of how speed and distance are related this will basically never happen.
The definition of Pareto efficiency is essentially negative in nature--it's not about finding specific dominating points, it's about finding points that are not dominated by any others on any criterion, period. All criteria are weighted equally in the search for Pareto points. It doesn't build in any weighting like considering maximum across criteria as "better" than points that only maximize one criteria. For a "biobjective" problem like your runs, the Pareto set will always contain the points (MAX, -) and (-, MAX)--they may not be unique over the criteria but there will always be at least one representative for each, I believe.
ChatGPT 5.6 Luna on the right (cheaper) cover most of the frontier, with a point for Deepseek flash, and higher performance overlapping heavily between 5.6 Sol and Fable.
That DeepSeek point will probably move back towards Luna as deepseek announced a "significant" price increase coming to their API [1], which kind of demonstrates that beating the Pareto frontier is where the difficulty actually is).
[1] https://www.bloomberg.com/news/articles/2026-08-06/deepseek-...
I think it's great and hope the price can stay the same.
As the number of objectives (dimensions) increases, the number of samples you need to cover the frontier increases exponentially. You will very rarely find solutions that actually dominate other solutions in many practical optimization scenarios. With 2 dimensions you have a 25% chance of domination. With 10 dimensions it's a .098% chance.
The most useful cases I've seen tend to occur where we just optimize for two things at once. The chances of domination are high, it's easy to visualize and very efficient to implement. As we get into higher dimensional spaces, things get weird really fast.
The geometric problem of computing a d-dimensional Pareto set of cardinality n
https://en.wikipedia.org/wiki/Maxima_of_a_point_set
has a truly weird property not covered by the computational complexity discussion on that page. It says there's an algorithm achieving O(n log(n)^(d-3) log log n), which is true and also a lie. The algorithm that achieves that asymptotic form is a galactic algorithm; and not an ordinary one in the sense of "has a large constant multiplicative factor", but one with this property (I've never found any other algorithm which exhibits it):
The runtime is within a bounded constant factor of n^2, for all n up to some critical N whose size is exponential in d (I think it was exactly 2^d or something).
I.e. the runtime has "two shapes": it's purely quadratic up to a galactically-large constant, and thereafter has a transition into to a slower function. The asymptotic version in the textbooks isn't achievable in the real world (for all but very small dimension).
There's an elementary proof using generating functions.
edit to add: If anyone's curious about it, a simplified version of the recurrence relation that's enough to exhibit this behavior (you can instantly see it if you graph this numerically) is
f(n,d=0) = 1
f(n=1,d) = 1
f(n,d) = n + 2f(⌊n/2⌋, d) + 2f(⌊n/2⌋, d-1)I've built large, deep product evaluation frameworks, and it is 100% of the time a running argument with stakeholders, inside and out, "well you should have measured it this way" or "I think we should be targeting X not Y" or "why didn't you consider Z in the metric??"
The Pareto Front in practice is squishy, fuzzy, and often quite moist and moldy.
As you say, the most useful things happen in low-dimensional spaces.
The 80/20 “rule,” as far as I know, is meant to be descriptive after the fact. It can’t be used as a planning assumption. To be fair to those managers, they don’t really mean to be rigorous. They are just trying to justify cutting scope.
If one option is at least as good on every relevant dimension and better on one, just pick it. That's not really a trade-off, and it shouldn't need escalation. Eg, if two SaaS tools cost the same and have similar support, but one fits your use case better, you choose that one. Otherwise, you just suck at your job!
The interesting decisions only start once you're already on the frontier, where getting more of one thing means giving up something else. If the better tool costs 50% more, now you're trading capability against cost, and that may need sign-off.
Basically, everyone should be able to get to the frontier on their own. Coordination and arbitration at higher levels of the org / between different departments should happen on the frontier, where the trade-offs involve several people or teams.
The objectives are matching arguments to parameters.
A set of functions is identified among the candidates: those that are possible for the call at all, like having a compatible number of parameters.
Essentially, the overload rule says that the Pareto front set of candidates must contain one member, otherwise the call is considered ambiguous, and diagnosable rule violation.
The objectives being optimized are individual parameter positions, each in the dimension of suitability: being a better match.
One candidate is better than another if it is no worse a type match in every parameter, and strictly better in at least one parameter.
Is that trying to say:
"for every solution not in the set, there exists at least one objective such that at least one solution in the Pareto set beats that solution in that objective" i.e. every non-Pareto-front solution is beaten in some objective(s) by a Pareto-front solution, however it may be unbeaten in other objectives.
Or is it:
"for every objective in the system, every solution that is not in the set is beaten in that objective by one or more Pareto-set solutions."
Or is it:
"For every solution not in the set, there exists at least one Pareto solution which beats it in every objective."
Matthias Ehrgott's books on multicriteria optimization explain Pareto efficiency very well without sacrificing rigor. I think they do a better job than this article.
Example: Which LLM gives me the best ELI5 explanations for a given price. https://evalry.com/benchmarks/explain-like-i-m-5-321
https://github.com/PatMyron/cloud#compute--memory-unit-price...