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November 15

Are there morphisms when enlarging a prime field sharing a common suborder/subgroup ?

Simple question : I have a prime field having modulus $p$ where p−1 contains $O$ as prime factor, and I have a larger prime field $q$ also having $O$ as it’s suborder/subgroup. Are there special cases where it’s possible to lift 2 $p$’s elements to modulus $q$ while keeping their discrete logarithm if those 2 elements lies only within the $O$’s subgroup ? Without solving the discrete logarithm of course ! 82.66.26.199 (talk) 11:36, 15 November 2024 (UTC)

Clearly it is possible, since any two groups of order o are isomorphic. Existence of a general algorithm, however, is equivalent to solving the discrete log problem (consider the problem of determining a non-trivial character). Tito Omburo (talk) 11:40, 15 November 2024 (UTC)

So how to do it without solving the discrete logarithm ? Because of course, I was meaning without solving the discrete logarithm. 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 12:51, 15 November 2024 (UTC)

It can't. You're basically asking if there is some canonical isomorphism between two groups of order O, and there just isn't one. Tito Omburo (talk) 15:00, 15 November 2024 (UTC)

Even if it’s about enlarging instead of shrinking ? Is in theory impossible to build a relation/map or is that no such relation exists yet ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 08:48, 16 November 2024 (UTC)

At least into the group of complex roots of unity, where a logarithm is known, it is easily seen to be equivalent to discrete logarithm. In general, there is no relation between the groups of units in GF(p) and GF(q) for p and q distinct primes. Any accidental isomorphisms between subgroups are not canonical. Tito Omburo (talk) 15:02, 16 November 2024 (UTC)

November 16

What’s the secp256k1 elliptic curve’s rank ?

Simple question : what’s the rank of secp256k1 ?
I failed to find how compute the rank of an elliptic curve using the version of online tools like SageMath or Pari/gp since it’s the only thing I have access to… 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 15:44, 16 November 2024 (UTC)

I don't know a clear answer but a related question is discussed here. 2601:644:8581:75B0:0:0:0:2CDE (talk) 01:57, 17 November 2024 (UTC)

Although I know it doesn t normally apply to this curvd, I was reading this paper https://pdfupload.io/docs/4ef85049. As a result, I am very curious about knowing the rank of secp256k1 which is why I asked it especially if it allows me know how to compute them on ordinary curves. 2A01:E0A:401:A7C0:417A:1147:400C:C498 (talk) 11:01, 17 November 2024 (UTC)

Maybe by some chance, this might have the answer. ExclusiveEditor _{Notify Me!} 19:20, 17 November 2024 (UTC)

Same question by same questioner, so not by chance.  --Lambiam 06:51, 18 November 2024 (UTC)

Yes, It’s me who asked the question. He didn’t replied to my last comment about the elliptic curve prime case. I’m meaning the paper 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 07:08, 18 November 2024 (UTC)

November 17

Final four vote probability

In a social deduction game at the final four where nobody is immune and each of the four gets one vote what is the probability of a 1–1–1–1 vote? (78.18.160.168 (talk) 22:26, 17 November 2024 (UTC))

Social deduction games exist in many different versions, with different rules. Can you provide (a link to) a description of the precise rules of the version of the game you want us to consider?

Moreover, if the players can follow different strategies, or can follow their intuitions instead of rolling the dice and using the outcome according to the fixed strategy, the situation cannot be viewed as a probability problem. Can we assume that the players play with the same given independent and identically random strategy?  --Lambiam 06:47, 18 November 2024 (UTC)

I was thinking of The Traitors, but it could also be applied to Survivor: Pearl Islands. There are no dice. In The Traitors before the final four banishment vote, there is a vote on whether to end game or banish again. If everyone votes to end the game the game ends but if one or more people votes to banish again, the game continues. I jumped ahead to the banishment vote because I have not seen a season where all four people vote to end the game. PS my IP address has changed. (78.16.255.186 (talk) 20:24, 18 November 2024 (UTC))

I don't understand the rules from the description in The Traitors and don't know what a "1" vote signifies, but in any case, this does not look like it can be modelled as a mathematical probability problem, for a host of reasons. The outcome of a vote will generally depend on the dispositions of the participants (are they more rational or more likely to choose on a whim; are they good in interpreting the behaviour of others) as well on their past behaviours. It is not possible to assign probabilities to such factors, and there is no mathematical model for how such factors influence the voting.  --Lambiam 03:58, 19 November 2024 (UTC)

If you simplify much further to just "if you have four people, and each one randomly chooses someone (that is not the person themself), what's the probability that each person gets chosen once", then we can generalize this to some arbitrary $n$ people.

Let us assign each person some number from $1$ to $n$, so that each choice can be thought of as a mapping from $[[1,n]]$ to itself. When each person is chosen exactly once, this corresponds to a mapping from $[[1,n]]$ to itself where no number is mapped to itself. This is a derangement, and we can see that the number of ways of tied voting is exactly the number of derangements for $n$ people. Thus, the probability for $n$ is the number of derangements divided by the number of mappings where no one votes for themselves.

The number of derangements on $n$ elements is the subfactorial of $n$, denoted $!n$. As for total number of mappings, each of the $n$ people has $n-1$ choices, so there are $(n-1)^{n}$ such mappings. This brings the probability to $!n/(n-1)^{n}$.

For $n=4$ the number of derangements is $9$, and there are $81$ mappings where no one votes for themselves, so the probability is $1/9$. More generally, $!n=round(n!/e)$, so the probability in general is ${\frac {round(n!/e)}{(n-1)^{n}}}$. Note that this tends to $0$ as $n$ increases. GalacticShoe (talk) 06:00, 19 November 2024 (UTC)

November 19

Basic equations / functions in predicting probability of success in insurgent vs. conventional military engagements in mid-to-late 20th century warfare / calculations for probability of the success of insurgent movements (esp. with consideration of intangible factors)

can someone kindly uncover casualty rolls -
I am thinking in particular about the Ukrainian Insurgent Army and the debates which went on within the American special services in the late 40s through early 50s about providing assistance to them after the breakthrough of the 'Iron company' (you can look up on ukr, pol, rus wiki about the so-called Iron company of the UPA ; Залiзна сотнья) from Transcarpathia in Communist-occupied Ukraine through Czechoslovakia through to Bavaria (where there were already in residence many leaders of the Ukrainian movement who had been interned by the Germans, most prominent among these Stepan Andriiovich, of course, working to raise the Ukrainian issue in the consciousness both of the public in Western 'free' world, and in the minds of the military-political authorities, who were still reeling from the taste in their mouths of the 'betrayal' of Poland, which Churchill railed against, closer, as he was, to the heart of the issue, if we have these figures, we can make at the very least basic calculations, and predict with a degree of accuracy, for example, based on the help that the Americans were considering to render to the Ukrainian freedom fighters, the successes which they could have achieved considering also the concurrent armed struggles in Romania, in Poland, in the Baltic states — Preceding unsigned comment added by 130.74.59.208 (talk) 15:15, 18 November 2024 (UTC)

This all seems very interesting, but I don't see it as mathematics question. I suggest you try the History Stack Exchange. --RDBury (talk) 19:19, 18 November 2024 (UTC)

i should like to refuse with one regard only the question pertains to application of mathematics and hard sciences in interpretation of historical events and possibilities 130.74.59.186 (talk) 20:02, 18 November 2024 (UTC)

Full stops were invented for a reason: they are very useful in making text understandable.  --Lambiam 04:07, 19 November 2024 (UTC)

There is no mathematical theory that can be used for determining the probabilities of the possible outcomes of a real-world conflict. It is not even clear that the notion of probability applies in such situations.  --Lambiam 04:14, 19 November 2024 (UTC)

This seems like more the province of game theory than probability. That it's modelled using probability in e.g. simulations, such as computer games or board games, is due to the limitations of their models. They can't fully model the behaviour of all actors so they add random probabilistic factors to compensate. But those actually engaged in conflict aren't going to be using randomness, just the best strategy based on what they know about the conflict, including what the other side(s) will do. That's game theory.--217.23.224.20 (talk) 15:49, 19 November 2024 (UTC)

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)