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July 15

What’s the complexity of solving the ᴇᴄᴅʟᴘ using this minors based las vegas algorithm ?

I was reading this paper. The almost principal minor (ᴀᴘᴍ) case sounds to yield a subexponential algorithm for solving the ᴇᴄᴅʟᴘ both in number of matrices rows and kernel counts, yet they talk about using a supercomputer for achieving the result of their paper where an older version of their source code can be found here.

So given finite field size, what’s the estimated complexity of the whole algorithm in the best design scenario ? 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 11:48, 15 July 2025 (UTC)

Histograms

Can bins in histograms have different widths? In different-width histograms, values are signified by area of bins rather than height, width of bins signify range of values belonging to each bin (the larger the range, the wider the bin), and height of bins signify value divided by range. None of histogram makers I have found on the web have an option to make histograms with different-width bins. --40bus (talk) 12:07, 15 July 2025 (UTC)

Google "histogram unequal class width" or "histogram uneven bin width" for loads of options. Whether that's a good idea depends on what you're trying to do. For most variables it would make the visual interpretation of the histogram harder than necessary. For age groups (e.g. [0,18), [18, 65], (65, ∞)) it would be quite useful (but then one wouldn't normally plot that on a linear age axis, so not a good example..). --Wrongfilter (talk) 12:26, 15 July 2025 (UTC)

A search for "frequency density" gives several tutorial examples of different-width histograms ([1], [2], [3], [4]), including one by age groups ([5]).  ​‑‑Lambiam 04:56, 16 July 2025 (UTC)

If a ≠ b, a^b and b^a are both algebraic numbers, must a and b be algebraic numbers?

If a ≠ b, a^b and b^a are both algebraic numbers, must a and b be algebraic numbers? 59.126.168.120 (talk) 19:21, 15 July 2025 (UTC)

Algebraic number _{Inserted at start for those who don't know what it is.} SGBailey (talk) 05:20, 26 July 2025 (UTC)

The numbers $0^{\pi }$ and $\pi ^{0}$ are both algebraic. Excluding such trivial counterexamples, consider the system of equations $x^{y}=2,y^{x}=3.$ It is solved by $x=1.36280\,{\cdot \cdot \cdot }\,,y=2.23925\,{\cdot \cdot \cdot }\,.$ It will be astounding, should these values prove to be algebraic.  ​‑‑Lambiam 06:07, 16 July 2025 (UTC)

By the Gelfond-Schneider theorem, if a and b are both algebraic numbers other than 0 or 1, and b is irrational, then $a^{b}$ is transcendental. Therefore, should $a^{b},b^{a},a,b$ be all algebraic and a and b be both other than 0 or 1, then a and b must both be rational. Conversely, if a and b are both rational, then clearly both $a^{b}$ and $b^{a}$ are algebraic. I conjecture this is the only possibility. Duckmather (talk) 18:24, 17 July 2025 (UTC)

I doubt that the system $x^{y}=2,y^{x}=3.$ has a solution in the rationals and even suspect that this is not hard to prove, which will definitely settle the question in the heading in the negative.  ​‑‑Lambiam 12:41, 19 July 2025 (UTC)

Neither of the individual equations has a rational solution beyond the trivial $x=2^{k},y={\frac {1}{k}}$ with $k\in \mathbb {Z} \backslash \{0\}$, and the similar $y=3^{k},x={\frac {1}{k}}$ with $k\in \mathbb {Z} \backslash \{0\}$. This is clear simply from the fact that $x=2^{1/y}$, or $y=3^{1/x}$, can only be rational when ${\frac {1}{y}}$ or ${\frac {1}{x}}$ respectively is an integer. Consequently there are no rational solutions to the combination of the two equations, and thus the solution to $x^{y}=2,y^{x}=3$ is composed of $x,y$ at least one of which is transcendental. GalacticShoe (talk) 19:38, 19 July 2025 (UTC)

July 17

Fractions of metric units

Is there any application that uses thirds, sixths and twelfths of metric units? Most metric units are in base 10, and do not divide evenly by these numbers. Are there any metric units (other than units of time) that are not in base 10, and divide evenly by these numbers? --40bus (talk) 11:26, 17 July 2025 (UTC)

Units of measure, including the metric units, are not numbers and are not "in a base". Numbers come into play when reporting the value of a physical quantity. Such values are presented in two parts: a numerical value together with a unit of measurement. The numerical value is customarily presented using decimal numbers (and the SI system prescribes their use), but this is independent of the system of units used.  ​‑‑Lambiam 20:21, 17 July 2025 (UTC)

Since altbn254 is using an isomorphism, how to map points from $\mathbb {F} _{p}^{12}$ back to $\mathbb {F} _{p}^{2}$ using SageMath provided they have a suitable order ?

Given the curve’s specifications, and the following code :

#set up the F_p^2 = F_p[i] / (i^2 + 1) field
p=21888242871839275222246405745257275088696311157297823662689037894645226208583
F2.<z2> = GF(p^2,modulus=x^2+1)
ec = EllipticCurve([F2(0),F2(3/(9+z2))]) # set up the curve over F_p^2 = F_p[i] / (i^2 + 1)​
X,Y = ( # set up the generator
21280594949518992153305586783242820682644996932183186320680800072133486887432 * z2 +
150879136433974552800030963899771162647715069685890547489132178314736470662,
1081836006956609894549771334721413187913047383331561601606260283167615953295 * z2 +
11434086686358152335540554643130007307617078324975981257823476472104616196090
)
pt = ec([X,Y,1])
#declare the isomorphsim to map the point to F_p¹²
G2.<w2> = GF(p^2,modulus=x^2-18*x+82)
F12.<z12> = GF(p^12,modulus=x^12-18*x^6+82)
X1=(X[0]-X[1]*9) + (X[1])*w2
Y1=(Y[0]-Y[1]*9) + (Y[1])*w2
print(Y1^2-X1^3-3/w2)
X2=X1[0] + X1[1]*z12^6
Y2=Y1[0] + Y1[1]*z12^6
print(Y2^2-X2^3-3/z12^6)
X3 = X2*z12^2
Y3 = Y2*z12^3
ec12 = EllipticCurve([F12(0),F12(3)]) # declare the F_p¹² curve linked to the isomorphism.
P0=ec12(X3,Y3) # X3 and Y3 represent the converted generator coordinates. See https://github.com/ethereum/py_pairing/blob/dd5ede17919c2afd042e4fbb7fda06d250df09be/py_ecc/bn128/bn128_curve.py#L100

but this is for mapping a point to the $y=x^{3}+3$ curve defined over $\mathbb {F} _{p}^{12}$. How to perform the reverse by going from $\mathbb {F} _{p}^{12}$ to the curve defined as ${\frac {Y^{2}=X^{3}+3}{i+9}}$ over the field $\mathbb {F} _{p}^{2}={\frac {F_{p}[i]}{i^{2}+1}}$ 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 15:54, 17 July 2025 (UTC)

Where are you getting this 254 and 12th power from? Neither of those are described above; they all mention altbn128, not 254, and the base field and its quadratic extension, not a 12th power extension Sesquilinear (talk) 18:28, 17 July 2025 (UTC)

$\mathbb {F} _{p}^{12}$ is the field’s for pairings because of the embedding degree while the curve over $\mathbb {F} _{p}^{2}={\frac {F_{p}[i]}{i^{2}+1}}$ is used as notation for point compression. There’re different bn128 curves wearing the same name, hence the one used with cryptocurrencies is sometimes labelled as altbn254. Everything is written in the specifications here. According to the implementation at https://github.com/ethereum/py_pairing/blob/dd5ede17919c2afd042e4fbb7fda06d250df09be/py_ecc/bn128/bn128_curve.py#L100 the mapping although only implemented for going from $\mathbb {F} _{p}^{12}$ to the curve defined as $\mathbb {F} _{p}^{2}={\frac {F_{p}[i]}{i^{2}+1}}$ is an isomorphism which suggest that doing in the other direction is possible and my question is to do it in SageMath. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 20:00, 17 July 2025 (UTC)

Okay, I think I see the confusion. The Ethereum spec there is quite terse and jumps through the intermediate steps assuming you already know them.

My exercise for you is: describe as many groups relevant to the pairing and its implementation as you can, and tell me how many you think are isomorphic to each other, as well as which ones will be strict subgroups of each other; this last part is the most important part and I don't think you'll be able to understand anything in elliptic curve cryptography without being able to tell me that last part. Sesquilinear (talk) 20:52, 17 July 2025 (UTC)

This is about maths : as shown by https://github.com/ethereum/py_pairing/blob/dd5ede17919c2afd042e4fbb7fda06d250df09be/py_ecc/bn128/bn128_curve.py#L100, you need to map points to embedding field in order to perform the pairings.

As for the isomorphism : as you said the specification don’t define it so I fail to understand it fully… I can’t explain it to you as a result. I’m asking the result in SageMath, but I would be unable to explain it mathematically too. The isomorphism is between ec12 = EllipticCurve([F12(0),F12(3)]) and EllipticCurve([F2(0),F2(3/(9+z2))]). This is shown by the code I wrote : but how it works ? I don’t understand fully (again). 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 22:04, 17 July 2025 (UTC)

Explain, in your own words, what the groups involved are. Sesquilinear (talk) 22:10, 17 July 2025 (UTC)

Actually, I'll be more explicit on one point. The G1 and G2 they imply an isomorphism between aren't the full groups of elements of the elliptic curves over the given fields. Indeed, with Hasse's theorem on elliptic curves you can show that's impossible for a curve over $F_{p}$ and a curve over $F_{p^{2}}$ when $p$ is big enough. Sesquilinear (talk) 23:13, 17 July 2025 (UTC)

The curve over contains $\mathbb {F} _{p}^{12}$ contains 21888242871839275222246405745257275088548364400416034343698204186575808495617 has a subgroup/suborder. By reverse mapping, I’m also talking about points being in such correct subgroup. For the other finite fields, the specification the order is 21888242871839275222246405745257275088548364400416034343698204186575808495617 directly. I’m not wanting to map between $\mathbb {G} _{1}$ and $\mathbb {G} _{2}$. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 06:44, 18 July 2025 (UTC)

Have all of your elliptic curve related questions been an attempt to solve the discrete logarithm problem for this zero knowledge proof system? Sesquilinear (talk) 06:33, 18 July 2025 (UTC)

In the case of this question, this is for solving a class of diffie Hellman. Although looking unlikely : Satoh’s latest Miller’s inversion algorithm seems to work for weil_pairing inversion (resulting root of unity of finite field elements in my case do satisfy the required unprobable criteria for the algorithm). The algorithm outputs a $\mathbb {G} _{2}$ point having the 21888242871839275222246405745257275088548364400416034343698204186575808495617 order that need to be mapped back from the ec12 curve to the ec curve of the SageMath code. 2A01:E0A:ACF:90B0:0:0:A03F:E788 (talk) 07:01, 18 July 2025 (UTC)

July 18

Tetration

Are there any web programs that can calculate last 100 or more digits of tetrational numbers, even power towers of 100 or more numbers? --40bus (talk) 04:31, 18 July 2025 (UTC)

July 26

Aurifeuillian factors

If k is a positive integer which is not perfect power, then k^core(k)-(-1)^floor(core(k)/2) (where core(k) is the squarefree part of k, i.e. OEIS: A007913) always have Aurifeuillian factors, and for squarefree k == 1 mod 4, the left Aurifeuillian factors are OEIS: A230376 and the right Aurifeuillian factors are OEIS: A230378, and for squarefree k == 2, 3 mod 4, the left Aurifeuillian factors are OEIS: A230377 and the right Aurifeuillian factors are OEIS: A230379, but there is still no OEIS sequence for all positive integers k which are not perfect powers, so

1. Please give the (left and right) Aurifeuillian factors of k^core(k)-(-1)^floor(core(k)/2) (where core(k) is the squarefree part of k, i.e. OEIS: A007913) for all positive integers k<=200 which are not perfect powers
2. If possible, please give a PARI/GP program that can compute the (left and right) Aurifeuillian factors of k^core(k)-(-1)^floor(core(k)/2) (where core(k) is the squarefree part of k, i.e. OEIS: A007913) for given positive integer k which is not perfect power
3. For which positive integers k which are not perfect powers, the left Aurifeuillian factors of k^core(k)-(-1)^floor(core(k)/2) (where core(k) is the squarefree part of k, i.e. OEIS: A007913) is prime? Should there be infinitely many such positive integers k which are not perfect powers?
4. For which positive integers k which are not perfect powers, the right Aurifeuillian factors of k^core(k)-(-1)^floor(core(k)/2) (where core(k) is the squarefree part of k, i.e. OEIS: A007913) is prime? Should there be infinitely many such positive integers k which are not perfect powers?

61.229.98.43 (talk) 19:52, 26 July 2025 (UTC)

Divisibility of units

Are there any SI units that divide evenly by 3 (and any of its multiples)? The only such metric units that come to mind are second, whose superunits are 60 times the previous unit (and thus divide evenly by 3), but second itself and its subunits do not, as well as degree of angle, but it is not an SI unit, the corresponding SI unit being radian, which does not divide evenly by 3. And of units that do not divide evenly, are thirds of these units ever used? --40bus (talk) 21:34, 26 July 2025 (UTC)

According to our article, minutes are non-SI units, so even the example you gave isn't valid. A minute is said to be "accepted for use" though, and if you look at the corresponding table it says minutes, hours and days as time units, and degrees and arcminutes as angle units, have subdivisions divisible by 3. That's 24 in the case of a day, and 60 for everything else. Astronomers also use light-hours, light-minutes, and light-seconds on occasion, but I gather these aren't strictly SI units. A parsec is based on an arcsecond, but it doesn't scale; 60 parsecs do not make an parmin, and if anything it would be the other way around. (Astronomy seems to be the outlier here since it still uses a plethora of non-metric units while the rest of the scientific community has (mostly) converted to SI units.) Really, the whole point of SI units is to use precise decimals instead of less precise fractions, and to reduce the number of units so there is less need for awkward conversion factors. --RDBury (talk) 23:03, 26 July 2025 (UTC)