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November 20

Sequences: Is there a name for a sequence, all of whose members are different from each other?

2A06:C701:7455:4600:C907:E8C0:F042:F072 (talk) 09:07, 20 November 2024 (UTC)

A term used in the literature: injective sequence.^[1]  --Lambiam 13:18, 20 November 2024 (UTC)

November 21

Is it possible to adapt Nigel’s Smart algorithm for establshing an isomorphism when the curve is only partially anomalous ?

An anomalous elliptic curve is a curve for which $\#E(\mathbf {F} _{q})=q$. But in my case, the curve has order j×q and the underlying field has order i×q. In the situation I’m thinking about, I do have 2 points such as both G∈q and P∈q subgroup and where P=s×G.

So since the scalar $s$ lies in a common part of the additive group from both the curve along it’s underlying base field, is it possible to transfer the discrete logarithm to the underlying finite field ? Or does anomalous curves requires the whole embedding field’s order to match the one of the curve even if the discrete logarithm solution lies into a common smaller group ?

If yes, how to adapt the Nigel’s smart algorithm used for solving the discrete logarithm inside anomalous curves ? The aim is to etablish an isomorphism between the common subgroup generated by E and $F_{p}$ 82.66.26.199 (talk) 19:47, 21 November 2024 (UTC)

November 22

Dihedral primes in base 36

Fourteen-segment display (alphanumeric display) can be used in base 36 (the largest case-insensitive alphanumeric numeral system using ASCII characters), thus we can use fourteen-segment display to define dihedral primes in base 36 (with A=10, B=11, C=12, …, Z=35), just like seven-segment display to define dihedral primes in base 10. If we use fourteen-segment display to define dihedral primes in base 36 (with A=10, B=11, C=12, …, Z=35), which numbers will be the dihedral primes in base 36 with <= 6 digits? 218.187.66.155 (talk) 19:14, 22 November 2024 (UTC)

It depends on how you encode each symbol on a fourteen-segment display (in particular, the number 0 and the letter O will need to be distinguished). If we go by File:Arabic number on a 14 segement display.gif and File:Latin alphabet on a 14 segement display.gif, then there are ten valid inversions, which are as follows: 0 <-> 0, 2 <-> 5, 8 <-> 8, H (17) <-> H, I (18) <-> I, M (22) <-> W (32), N (23) <-> N, O (24) <-> O, X (33) <-> X, and Z (35) <-> Z. Of these, only 5, H, N, and Z are coprime to 36, so any dihedral prime must necessarily end with one of these. Duckmather (talk) 04:02, 25 November 2024 (UTC)

We can use an encoding that the inversions not only include the ones which you listed, but also include 1 <-> 1, 3 <-> E (14), 6 <-> 9, 7 <-> L (21), and S (28) <-> S, if so, then which numbers will be the dihedral primes in base 36 with <= 6 digits? (Also, why 2 <-> 5? They are not rotated 180 degrees) 210.243.207.143 (talk) 20:31, 26 November 2024 (UTC)

We can also consider “horizontal surface” “vertical surface”, and “rotate 180 degrees”, separately, and consider normal glyphs and fourteen-segment display glyphs separately (see Strobogrammatic number, we can also find the strobogrammatic numbers (as well as the strobogrammatic primes) in base 36):

Horizontal surface:

0 <-> 0 (only normal glyph)

1 <-> 1

2 <-> 5 (only fourteen-segment display glyph)

3 <-> 3

7 <-> J (19) (only fourteen-segment display glyph)

8 <-> 8

B (11) <-> B

C (12) <-> C

D (13) <-> D

E (14) <-> E

H (17) <-> H

I (18) <-> I

K (20) <-> K

M (22) <-> W (32)

O (24) <-> O

X (33) <-> X

Vertical surface:

0 <-> 0 (only normal glyph)

1 <-> 1

2 <-> 5 (only fourteen-segment display glyph)

3 <-> E (14) (only fourteen-segment display glyph)

8 <-> 8

A (10) <-> A

H (17) <-> H

I (18) <-> I

J (19) <-> L (21) (only fourteen-segment display glyph)

M (22) <-> M

O (24) <-> O

T (29) <-> T

U (30) <-> U

V (31) <-> V (only normal glyph)

W (32) <-> W

X (33) <-> X

Y (34) <-> Y

Rotate 180 degrees:

0 <-> 0

1 <-> 1

2 <-> 2 (only fourteen-segment display glyph)

3 <-> E (14) (only fourteen-segment display glyph)

5 <-> 5 (only fourteen-segment display glyph)

6 <-> 9

7 <-> L (21) (only fourteen-segment display glyph)

8 <-> 8

H (17) <-> H

I (18) <-> I

M (22) <-> W (32)

N (23) <-> N

O (24) <-> O

S (28) <-> S (only normal glyph)

X (33) <-> X

Z (35) <-> Z 218.187.66.221 (talk) 18:45, 27 November 2024 (UTC)

November 23

radial distance between a circle and another enclosing circle

On an x-y plane, draw a circle, radius r1 centered on the origin, 0,0. Draw a second circle centered on some offset value -x, y = 0, radius r2 which greater than r1+x so that the second circle completely encloses the first and does not touch it. Draw a line at angle a beginning at the origin and crossing both circles. How do I calculate the distance along this line between the two circles? ```` Dionne Court (talk) 06:07, 23 November 2024 (UTC)

Given:

• inner circle: centre at $(0,0),$ radius $r_{1},$ equation $x^{2}+y^{2}=r_{1}^{2};$
• outer circle: centre at $(u,0),$ radius $r_{2}>r_{1}+u,$ equation $(x-u)^{2}+y^{2}=r_{2}^{2}.$
• line through origin at angle $\alpha ,$ parametric equation $(x,y)=(\lambda \cos \alpha ,\lambda \sin \alpha ).$

The line crosses the inner circle at $(x,y)=(\pm r_{1}\cos \alpha ,\pm r_{1}\sin \alpha ),$ both obviously at distance $r_{1}$ from the origin.

To find its crossings with the outer circle, we substitute the rhs of the line's equation for $(x,y)$ into the equation of the outer circle, giving $(\lambda \cos \alpha -u)^{2}+(\lambda \sin \alpha )^{2}=r_{2}^{2}.$ We need to solve this for the unknown $\lambda$. This is a quadratic equation; call its roots $\lambda _{1}$ and $\lambda _{2}.$ The corresponding points are at distances $|\lambda _{1}|$ and $|\lambda _{2}|$ from the origin.

The crossing distances are then $|\lambda _{1}|-r_{1}$ and $|\lambda _{2}|-r_{1}.$

If you use $\left|(|\lambda _{1}|-r_{1})\right|$ and $\left|(|\lambda _{1}|-r_{1})\right|,$ this will work for any second circle, also of it intersects the origin-centred circle or is wholly inside, provided the quadratic equation has real-valued roots.  --Lambiam 08:46, 23 November 2024 (UTC)

November 27

How much did UPS pay in workers comp claims for heat-related incidents last year? On the flip side, how much would it cost to air-condition their package vans and warehouses?

Did they also pay hazard bonuses for working in the heat?

Is it cheaper for UPS to just air condition their warehouses and package vans?

After paying the initial installation fees for the new HVAC systems, how much will it cost for UPS to run air conditioning and maintain their HVAC systems for one year (at least only when the weather is hot?)

And how much did they pay out in heat-related workers comp claims for one year?

How well will UPS come out ahead from simply air conditioning all places and vehicles that need air conditioned? --2600:8803:1D13:7100:BD6D:70D0:30AC:B227 (talk) 01:13, 27 November 2024 (UTC)

This is not a mathematics question. We don’t answer requests for opinions, predictions or debate. Dolphin (t) 04:59, 27 November 2024 (UTC)

The largest prime factor found by trial division

The largest prime factor found by Lenstra elliptic-curve factorization is 16559819925107279963180573885975861071762981898238616724384425798932514688349020287 of 7³³⁷+1 (see [2]), and the largest prime factor found by Pollard's p − 1 algorithm is 672038771836751227845696565342450315062141551559473564642434674541 of 960¹¹⁹-1 (see [3]), and the largest prime factor found by Williams's p + 1 algorithm is 725516237739635905037132916171116034279215026146021770250523 of the Lucas number L₂₃₆₆ (see [4]), but what is the largest prime factor found by trial division? (For general numbers, not for special numbers, e.g. 7*2^20267500+1 divides the number 12^220267499+1 found by trial division, but 12^220267499+1 is a special number since all of its prime factors are == 1 mod 2^20267500, thus the trial division only need to test the primes == 1 mod 2^20267500, but for general numbers such as 3*2¹⁰⁰+1, all primes may be factors) 61.229.100.16 (talk) 20:51, 27 November 2024 (UTC)

I don't have an answer, and Mersenne primes have properties that reduce the number of primes that need to be searched, meaning that it doesn't technically need full trial division, but I would nevertheless like to raise two famous examples which I'm fairly sure were done through manual checking:

1. In 1903, Frank Nelson Cole showed that $2^{67}-1$ is composite by going up to a blackboard and demonstrating by hand that it equals $193707721\times 761838257287$. It took him "three years of Sundays" to do so, and I'm fairly sure he would have done it manually.
2. In 1951, Aimé Ferrier showed that ${\tfrac {2^{148}+1}{17}}$ is prime through use of a desk calculator, and I imagine a lot of handiwork.

GalacticShoe (talk) 02:29, 28 November 2024 (UTC)

November 29

In finite fields of large characteristics, what does prevent shrinking the modulus field size down to their larger order in order to solve discrete logarithms ?

In the recent years, several algorithms were proposed to leverage elliptic curves for lowering the degree of a finite field and thus allow to solve discrete logairthm modulo their largest suborder/subgroup instead of the original far larger finite field. https://arxiv.org/pdf/2206.10327 in part conduct a survey about those methods. Espescially since I don’t see why a large chararcteristics would be prone to fall in the trap being listed by the paper.

I do get the whole small characteristics alogrithms complexity makes those papers unsuitable for computing discrete logarithms in finite fields of large charateristics, but what does prevent applying the descent/degree shrinking part to large characteristics ? 2A01:E0A:401:A7C0:68A8:D520:8456:B895 (talk) 11:00, 29 November 2024 (UTC)

Try web search for Lim-Lee small subgroup attack. 2601:644:8581:75B0:0:0:0:C426 (talk) 23:09, 1 December 2024 (UTC)

First the paper apply when no other information is known beside the 2 finite field’s elements and then this is different from https://arxiv.org/pdf/2206.10327. While Pollard rho can remain more efficient, if the subgroup is too large, then it’s still not enough fast. I’m talking about shrinking modulus size directly. 2A01:E0A:401:A7C0:69D2:554C:93AF:D6AC (talk) 15:17, 2 December 2024 (UTC)

The problem with descent methods like the one described in large characteristic is that the complexity is $max(q,k)^{\log k}$. This is explained more clearly in [5]. If the characteristic is small, then the problem is O(k) bits, and the complexity id pseudo-polynomial in the number of bits. If the characteristic is large, then the compexity is $q^{\log k}$ which is exponential in the number of bits of q. Tito Omburo (talk) 16:36, 2 December 2024 (UTC)

November 30

Linear differential function

Differential 102.213.69.166 (talk) 04:35, 30 November 2024 (UTC)

What are you talking about? ‹hamster717🐉› ^{(discuss anything!🐹✈️ • my contribs🌌🌠)} 14:02, 30 November 2024 (UTC)