Cal Aldred retweetledi
@BEBischof @CalAldred I literally generated the first hundred or so, if anyone is curious. This is for both even and odd fiber degree: jacobianfun.org/counterexamples
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Cal Aldred
510 posts
Cal Aldred
@CalAldred
Baseball Data Scientist. Opinions my own.
Wrigleyville Katılım Haziran 2013
653 Takip Edilen521 Takipçiler
Coming soon from the machine gods: the infinite ladder is just one row of a two-parameter factory of Jacobian counterexamples. The original map sits at (1,1), and every mapping degree from 3 onward appears.
Cal Aldred@CalAldred
Came up with a way to generate an infinite family of counterexamples, via Fable + GPT 5.6 Sol
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Came up with a way to generate an infinite family of counterexamples, via Fable + GPT 5.6 Sol
levent@__alpoge__
hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)
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There's an infinite ladder of this geometric construction too, per GPT 5.6 Sol and Fable 5! Integrate D^{m−1} for a quadratic D, mark a root of the primitive, forget the mark. Generically 2m−1 sheets, so no two rungs are polynomially equivalent.
Andy Jiang@davikrehalt
@littmath @__alpoge__ GPT: Take π: P¹ × Sym²(P¹) → Sym³(P¹), (p, {q,r}) ↦ {p,q,r}. R be its ramification divisor; H ⊂ Sym³(P¹) ≅ P³ be hyperplane tangent but not osculating to the small diagonal; X := (P¹ × Sym²(P¹)) \ (R ∪ π⁻¹(H)) ≅ A³; Y := Sym³(P¹) \ H ≅ A³. π|X: X → Y is counterexample
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@__alpoge__ Used Fable + GPT 5.6 Sol to generate an infinite family of counterexamples, not sure how useful this is
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hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final
((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)
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As @cwhite23_ says:
"The goal isn't always to have the highest r-squared and lowest RMSE value."
rat king fan Robert Orr@NotTheBobbyOrr
a great example of how heuristics can occasionally be more practical and useful than overly complicated ML algos that have more predictive power
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Cal Aldred retweetledi
you are supposed to be fascinated by trees
Sarah 📸@swrightCO
You are supposed to be fascinated by trees
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@bubbleboi I find it’s quite important to tune gamma for xgboost models
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Cycle check, my friend who works at OpenAI just texted me this.
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Someone needs to make a benchmark for the benchmarks that benchmarks how hard it is for an AI agent to tell that the benchmark is in fact a benchmark
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@MLBRandomStats @ESPNLunardi A 40 point win against a top 10 team?
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@ESPNLunardi What puts Michigan’s resume ahead of Purdue’s to you?
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Random thoughts from a rare Saturday at home:
• Michigan will not lose its 1-seed, but cede No. 1 overall to Arizona (if the Wildcats complete the win at TCU)
• Wisconsin may just sneak into the next projection
• Houston is the most forgotten excellent team in a long time
• Georgia will look back on today as a very big win
• Same for Florida, same for Virginia, same for Kansas (oops on that last one...)
• Most miserable loss of the day games may have been Yale at Princeton (likely putting Dartmouth in the next bracket!)
Finally, the most important game underway at the moment is Oklahoma at Texas A&M (with Cal at VaTech a close second).
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@alz_zyd_ It drives down the value of homes, but still gets passed on as a cost to renters, right? Seems fairly straightforward
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@alz_zyd_ Seems simple, from symmetry you know each player must have 1/3 chance to win no matter the number they choose and that all 3 players must have the same strategy. You can chain the results up from the choose 1 case which is (1-p1)^2 + p1^2 = 1/3, p1 = 0.5
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Game theory puzzle:
There are 3 people. Each person announces an integer. The smallest unique integer wins: e.g. if your opponents both pick 1, you win with any number. If all 3 pick the same number, the winner is picked randomly
Question: what's the Nash equilibrium?
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Whatever is in the air for the airlines needs to chill tf out. My goodness pls it’s Christmas
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@AriannaBSophia Even after it’s filled up with bags? TIL, never got in trouble for that before
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