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Latest comment: 58 minutes ago by 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D in topic Is this a possible Groth16/ZkSnark verifier‑side simplification ?
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September 16

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Geographical almost-centres and Croatia

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Given the country's unusual shape, Croatia's geographical centre is rather awkwardly placed; if you Google <croatia geographical centre> you get lots of forums and similar content discussing the idea that its centre is actually outside the country, in western Bosnia and Herzegovina. From this I'm left wondering: (1) Are there any reliable sources for this claim? I couldn't find any. If so, it would be a good addition to geography of Croatia. (2) The geographical centre article discusses different methods of calculating the geographical centre of a region and the potential problems resulting therefrom. If Croatia's centre really can be defined to be in Bosnia, do all definitions put it there, or do some definitions put it in Croatia? (3) Is there any concept of most-centred-within-boundary? [This is the biggest reason I came to the Maths desk; I'm wondering if topologists would care about it?] Let us assume that Croatia's centre is outside Croatia: is there a term that refers to the Croatian location that is least-off-centre? It wouldn't necessarily be the Croatian location closest to the centre (imagine a narrow salient that would be closest but severely off-centre), but I suppose it could. I'm thinking of the point where, if you balanced a flat map of Croatia on it, the map would topple most slowly. Obviously some points are better than others — a point at the country's southern tip would be worse than places farther northwest — so I wonder if it's reasonable to define the best place when no place is ideal. Nyttend (talk) 07:51, 16 September 2024 (UTC)Reply

PS, imagine Croatia like a balancing bird toy. If you broke off the bird's head, you probably couldn't balance it at all, but you'd do a lot better balancing on the body (or even the tail) than at the wingtips. Croatia lacks the "bird head", but you're probably better-off balancing in the northwest than anywhere else. Nyttend (talk) 07:55, 16 September 2024 (UTC)Reply

There was a discussion not that long ago here: Wikipedia:Reference desk/Archives/Miscellaneous/2024 July 18 § Lake Lats and Longs. It contains, at least implicitly, answers to some of the questions. Since the notion of centre is not well-defined, neither is that of "least off-centre". The location nearest to a given point outside the area is on its boundary. The interior point furthest from the boundary works for most actual country shapes, including Croatia, for which this is a point roughly 20 km east of zagreb. — Preceding unsigned comment added by Lambiam (talkcontribs) 16:06, 16 September 2024 (UTC)Reply
The visual center and algorithms to approximate the pole of inaccessibility are discussed in this 2016 Mapbox post. I thought that if you'd include just a spherical geometry of the Earth you'd get even more interesting questions, but it appears at a glance (? not sure?) the algorithm in the link already generalizes nicely to noneuclidean geometries and higher dimensions. SamuelRiv (talk) 17:46, 16 September 2024 (UTC)Reply
Through any point on Earth there is (by the mean value theorem) a great circle that bisects the population of Croatia. Among such circles, consider the segments that cut the territory of Croatia into exactly two pieces. I propose the midpoint of the shortest such segment as a centre. (shamelessly OR) —Tamfang (talk) 01:29, 18 September 2024 (UTC)Reply
If the border of Croatia is sufficiently fractal-like, there may be no great circle that cuts the area into just two connected pieces.  --Lambiam 09:19, 18 September 2024 (UTC)Reply
Doesn't need to be fractal; a C-shape with overlapping ends (ie a spiral with just over a full turn) can't be halved (by area) into only two pieces by an infinite straight line. There may be a coral atoll somewhere that approaches this? 213.143.143.69 (talk) 12:37, 18 September 2024 (UTC)Reply
There's quite a few simply connected shapes that create uncomfortable solutions for this algorithm (so far splitting 3+ pieces, you can also just have a wide lobe with two long thin tails side-by-side; the shortest area or even-population bisector straight line would have to cut through both tails; the longer unbroken line that cuts the lobe can be stopped by making the lobe an S-turn; which can only be resolved (to retain contiguity) by removing the requirement of a great circle.
(The spiral atoll example need not be a problem, since the great circle need not be required to cut through the entire atoll, but just one segment.)
It's a good idea, but it needs a bit more refinement to get a sensible-and-unique definition in all cases. SamuelRiv (talk) 16:32, 18 September 2024 (UTC)Reply
Seems to me your solution to the spiral problem (for which thanks) applies to any simply connected shape. —Tamfang (talk) 05:46, 23 September 2024 (UTC)Reply
Could it be meaningful to measure the density of such circles passing through a given neighborhood? I think I see an approach or two, but it needs more thought. The maximum by this measure is my new favorite center. —Tamfang (talk) 05:43, 23 September 2024 (UTC)Reply
Consider any point within the boundary. Calculate the maximum distance from that point to any other point within the boundary. Then choose the point where that maximum distance is least. This will give a 'centre' that is in some way the closest possible to the rest of the area. For convex shapes it is equivalent to the circumcentre (centre of the bounding circumcircle). Would this be unique for concave shapes? -- Verbarson  talkedits 11:38, 19 September 2024 (UTC)Reply
Not necessarily; a symmetric C-shape where the ends curl back nearly to the centre would have two possible solutions, one on each tip. -- Verbarson  talkedits 13:27, 19 September 2024 (UTC)Reply
1) For a practical problem like this, a finite set of 'diverse' solutions (or a continuous and symmetric set) is not really an issue -- just choose any one point; 2) you can also include the shortest path between two points within the concave polygon. A central consideration above is the calculation time.
Another possibility is to get the approximate convex skull, the largest enclosed convex polygon, which reduces you to an easy incenter calculation and you're done. SamuelRiv (talk) 18:24, 19 September 2024 (UTC)Reply
Also, for cases like long bulbous C-shapes, we can adapt if we prefer the narrow midway 'neck' as opposed to the widest of the 'bulbs'. (To illustrate the scenario: consider North and South American -- what the cartographer calls the natural center of the supercontinent is either around the Isthmus of Panama, or it's at the geographic center of their choice of the gigantic North or South continent (or they can put the same label on both). One could find all the local extrema of the largest enclosed convex polygon at any point (assuming they are discrete sets, so the shape is not a perfect circular atoll for example, and if the approximate algorithm for the convex skull can be so adapted) and then choose the median polygon incenter (however you want to resolve it for even numbers). SamuelRiv (talk) 18:40, 19 September 2024 (UTC)Reply
I assume the distance is measured on an interior path. —Tamfang (talk) 05:28, 23 September 2024 (UTC)Reply
Either interior or straight-line. It may depend on whether you want to site a power station or an airport? -- Verbarson  talkedits 07:47, 23 September 2024 (UTC)Reply

September 19

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Property of a matrix

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If I raise a permutation matrix to a high power it is still a permutation matrix (all values 0 or 1). If I raise a stochastic matrix to a high power it is still a stochastic matrix (all values <=1). But there are a lot of matrices raised to a high power have elements with very large values. Is there a name for the property of a matrix that when raised to a high power remain reasonably valued? RJFJR (talk) 22:12, 19 September 2024 (UTC)Reply

The spectrum of such a matrix is inside the closed unit disk (a necessary but not sufficient condition). A matrix whose spectrum is in the open unit disk always satisfies this property. I'm not sure there is a general name for this property though. Tito Omburo (talk) 23:07, 19 September 2024 (UTC)Reply
That's it. Thank you! RJFJR (talk) 02:08, 20 September 2024 (UTC)Reply
Note that when the spectral radius is strictly less than   the powers of the matrix tend to the null matrix, which is not very exciting. I also vaguely remember seeing a theorem that states or implies that when the spectral radius is equal to   the growth of the entries is polynomial in the value of the exponent.  --Lambiam 21:55, 20 September 2024 (UTC)Reply

September 20

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Adef
=
B+C

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1. Since the expression in the title is a definition, we can conclude, that A exists if and only if both B and C exist as well.

2. However, we can't conclude that if B or C exist then A exists. Check: A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product: if only the price   of the first product exists, we still can't conclude - that the price   of the second product exists - nor that the total price   exists. The defintion   only tells us (besides the relation  ), that the total price   exists if and only if both - the price   of the first product exists - and the price   of the second product exists.

3. For concluding, that if any value of the above three values A,B,C exists then all of them exist, it's sufficient to write down three definitions:

 
 
 

4. Question: as I've indicated, it's sufficient (to write down all three defintions), but is there any shorter notation, to make sure that if any value of the above three values A,B,C exists then - all of them exist - and satisfy  

HOTmag (talk) 13:26, 20 September 2024 (UTC)Reply

Are you asking about a shorter notation? I may be looking mighty foolish, but isn't it circular and therefore meaningless to write the system of three definitions like you have above? Or rather, wouldn't writing down the first be exactly as meaningful as writing down all three, since both situations only relate A, B, C to one another? I get they're definitions and not merely statements, but I'm not really seeing the difference here. You'd have to define B, C in terms other than A to reach a more meaningful domain. Again, I could look totally foolish right now, since I've never answered a question here before. Remsense ‥  13:34, 20 September 2024 (UTC)Reply
It seems you haven't read 2#. HOTmag (talk) 13:40, 20 September 2024 (UTC)Reply
I don't really understand it, no. You're defining A, B, C as the prices of products, but then you're writing abstract definitions of them in terms of each other. Am I missing something? Oh, did you mean to say A is the total? It makes more sense to me that way. Remsense ‥  13:43, 20 September 2024 (UTC)Reply
In any case, I still don't understand what the extra two definitions achieve: either B and C are defined outside of A or they're not, right? If we wanted to know them in terms of A, we already got that in the first definition. Remsense ‥  13:45, 20 September 2024 (UTC)Reply
I've just added an addition to 2#, to make it clear. HOTmag (talk) 13:49, 20 September 2024 (UTC)Reply
I return to my initial question then, are you just looking for a shorter notation for this? Remsense ‥  13:52, 20 September 2024 (UTC)Reply
Yep. I've just made it clear in 4# (thanks to your question). HOTmag (talk) 13:55, 20 September 2024 (UTC)Reply
Sorry for being slow on the uptake. I'm not sure if this needs to fit into any particular system or paradigm: anything wrong with  . Sorry if that hurts anyone to see, haven't flexed these muscles in a while Remsense ‥  14:31, 20 September 2024 (UTC)Reply
Not to bug you, but only since I'm relatively unsure of myself in this area—was this answer something like what you were looking for? Remsense ‥  20:10, 20 September 2024 (UTC)Reply
You limit the set of the Bs and the Cs to be the positive integers, but my question is general, without limiting anything. HOTmag (talk) 02:05, 22 September 2024 (UTC)Reply
You gave the example of prices, so that's what I picked the positive reals* based on. Clearly, you can replace the set with whatever you want. Remsense ‥  02:08, 22 September 2024 (UTC)Reply
If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does your definition let us deduce the other definition:   which I would like to deduce, bearing in mind that not all products have a price? Note: Since the latter is a defintion of C, then the existence of A and of B must be derived from the very existence of C. HOTmag (talk) 02:21, 22 September 2024 (UTC)Reply
It seems the bit about "deducing a definition" articulates a fundamental confusion you have about what you're trying to accomplish. Definitions are stated, not deduced. Remsense ‥  05:34, 22 September 2024 (UTC)Reply
By asking whether a new definition can be deduced from an old definition, I mean whether a new claim, that was presented before as an additional definition, can be derived as consequence, from a given assumption that was presened before as an old definition.
In our case, the given assumption, was presented before as an old definition: Adef
=
B+C
. The new claim, was presened before as an additional definition:   The question is, whether we can deduce the latter from the former, i.e. whether we can deduce the new claim from the given assumption. HOTmag (talk) 13:08, 22 September 2024 (UTC)Reply
If the values we assign to B and C are logically independent from one another, then no. That's what logical independence means.Remsense ‥  13:17, 22 September 2024 (UTC)Reply
This is totally confused. The addition operation   is special in that it is operand-wise strictly monotonic and therefore has, at least in the integers and reals (but not in the natural numbers) an operand-wise inverse, which we can denote using the subtraction operation  . In general, this is not possible.
Defining some quantity   by an equation of the form   only makes sense if all terms in the right-hand side are defined. It is not just that this fails to define   if   or   is not defined. It just does not make sense. And if   is not defined, writing something like   only increases the confusion.
There is another issue in which the definedness of a defined term depends on the definedness of another term, namely when defining a function. Suppose we define a new function   using existing known functions with a limited domain. For example, we may define real-valued function   on the real numbers by the equation
 
For   to be defined by this equation for some given value of   it is necessary that both   and   are defined. This is more a matter of common sense than anything else, and I see no need for some notational device to express this dependency.  --Lambiam 22:29, 20 September 2024 (UTC)Reply
If A denotes the total price, B denotes the price of the first product, and C denotes the price of the second product, does the definition Adef
=
B+C
make senee? If it does, can you deduce the definition:   HOTmag (talk) 02:13, 22 September 2024 (UTC)Reply
You can't define something that already has a meaning. So if A, B and C all have independent meanings, as shown by the use of "denote", then none of them can be defined in terms of the other two. It may be true that A = B + C based on these meanings, but that's not a definition. In any case, you can't deduce a definition. If A has no independent meaning then you can define it to be anything, B + C, B - C, or B * C. --RDBury (talk) 05:29, 22 September 2024 (UTC)Reply
Well, I'm presenting an analogous question (taken from a discipline very close to arithmetic), in my following thread. I hope my new question explains also my old question, but if my old question is still not clear, you can ignore it, and focus on my new question. HOTmag (talk) 13:38, 22 September 2024 (UTC)Reply

September 22

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How can we briefly characterize a given set of vectors, as "linearly dependent - every proper sub set being linearly independent", while we only refer to the vectors rather than to their set?

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For example: S is a set of the following vectors:

A=(1,1,0),
B=(1,0,0),
C=(0,1,0).

Note: A=B+C, and B=A-C, and C=A-B, so the set S is linearly dependent.

Using A,B,C only, i.e without using S, what's the shortest description, claiming that the set S is linearly dependent but every proper sub set of S is linearly independent? HOTmag (talk) 13:30, 22 September 2024 (UTC)Reply

@HOTmag: I'd just say 'S has k linearly independent elements' (in the give example k=2). --CiaPan (talk) 14:46, 22 September 2024 (UTC)Reply
Is your response a suggestion of rephrasing my question?
If it's intended to be an answer, then please note: My condition requires to be "using A,B,C only, i.e without using S". Additionally, where does your description claim, that S is linearly dependent? HOTmag (talk) 14:56, 22 September 2024 (UTC)Reply
One way to characterise the set is "a set of vectors, any one of which can be written in terms of the others in a unique way". The set is just A, B and C, i.e. any property of them is a property of the set of them. --2A04:4A43:900F:F4C3:49F4:4EFB:C442:608F (talk) 15:15, 22 September 2024 (UTC)Reply
Since my condition requires to be "using A,B,C only, i.e without using S", so I guess you mean the following: "Each vector, can be written as a unique linear combination of the other vectors". Thanks. HOTmag (talk) 15:52, 22 September 2024 (UTC)Reply
Assuming the vectors are  , form the matrix   whose columns are  . The stated condition is then:
  1. the matrix of   minors of   is zero, and
  2. each of the columns of the matrix of   minors of V has a non-zero entry.
- Tito Omburo (talk) 17:36, 22 September 2024 (UTC)Reply
Your description, both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:41, 23 September 2024 (UTC)Reply
You can say, "each of the sets {A,B}, {A,C} and {B,C} is linearly independent".  --Lambiam 21:05, 22 September 2024 (UTC)Reply
You also have to add that the set {A,B,C} is linearly dependent, but then the description - both uses sets, and also becomes longer than the original one indicated in the title. HOTmag (talk) 08:40, 23 September 2024 (UTC)Reply
There is a unique linear combination generating 0, and in this linear combination all coefficients are nonzero.2404:2000:2000:8:FDE8:8311:95E3:654D (talk) 00:00, 23 September 2024 (UTC)Reply
Yes, and by symbolic notation the description even becomes shorter. Thanks. HOTmag (talk) 08:42, 23 September 2024 (UTC)Reply

September 23

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Converting map scale into height and width

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How can I calculate the height and width in meters of a future geographic map of the world based on its desired scale in cm and km? 212.180.235.46 (talk) 15:35, 23 September 2024 (UTC)Reply

For a map of the world, you will have to specify which map projection it will use, and then to which part(s) of the projection the scale is to apply (and for some projections, how much of the globe to include). Mapping a spherical surface onto a plane surface cannot be done with a constant scale factor. Eg: you might specify a Mercator projection, between latitudes +/-85deg, with a scale of 1:10,000,000 at the equator, which would give you a map about 4m wide along the equator, and (judging by the illustration for Mercator) slightly more than that high. Other projections will vary considerably. -- Verbarson  talkedits 18:07, 23 September 2024 (UTC)Reply
My own map with van der Grinten projection and equatorial scale of 1:20 000 000 for example is 1.18 m high and 1.94 m wide. Out of curiosity, I started to think about dimensions of imaginative humongous world maps. With that as benchmark, if my calculations are correct, a 1:160 000 scale world map, for example, would be 147 m long and 242 m wide. 212.180.235.46 (talk) 20:35, 23 September 2024 (UTC)Reply
The scale of a map is the ratio between the distance between two points on the map and that between the corresponding points on the ground. Any map projection necessarily distorts the distances, so the scale you get if you take New York and San Francisco as the two points is generally substantially different from what you get if you take Berlin and Moscow. For maps of the whole Earth it is non-trivial to decide on a "nominal scale". In spite of the name, the van der Grinten projection is not a geometric projection in which points on the map are found by following rays projected along a straight line out of a shrunk globe. The equatorial and central-meridional scales are well defined, though. The length ℓeq of the equator is approximately 40,000 km. If the equatorial scale is chosen to be 1 : Seq, the width of the map is ℓeq / Seq. So for Seq = 160,000 this comes out as about 250 m. The length ℓmer of a meridian, measured from one pole to the other, is about 40,000 km. Letting Scm stand for the central-meridional scale factor, the height of the map will be ℓmer / Scm. If these scale factors are chosen to be equal, the map will have a height of half its width. For a circular map as shown in the article, Scm = 2 Seq.  --Lambiam 10:14, 24 September 2024 (UTC)Reply


September 26

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Given the optimal ate pairing e(A,B)=y is to possible to determine I and J such as e(I,J)=2y or even e(I,J)=3y?

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Simple question : let’s say I have a pairing friendly curve having a very large trace, and that I have a pairing with points A∈  and B∈  such as the optimal ate pairing e(A,B)=y, then is it possible to fully determine 2 point I and J such as e(I,J) equals a target multiple of the finite’s field element y ?

Or does it requires full pairing or full generalized Miller’s inversion and thus would be impossible in practice on a curve like bn254 ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 15:51, 26 September 2024 (UTC)Reply

September 27

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Uses of Pascal's tetrahedron

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The following is a use of Pascal's triangle:

To find how many ways there are to make a total of   circles all black or red, the formula is just  . For example, there are   ways to make a group of 6 circles, all black or red, classified by whether each circle is black or red. An example is black-red-black-red-black-red.

But how about finding the number of ways to make a group of 6 circles, all black or red, classified by how many are black or red. To find out how many ways there are to make a total of   circles all black or red classified by how many are black and how many are red, you use the n+1th row of Pascal's triangle. For  , this means we use the seventh row, which is  . This means that there is one way to color 6 circles where all of them are black, 6 where 5 are black and one is red, 15 where 4 are black and 2 are red, 20 where 3 are black and 3 are red, 15 where 2 are black and 4 are red, 6 where one is black and 5 are red, and one where all 6 are red.

How about a similar application for Pascal's tetrahedron?? Here is the seventh layer of the tetrahedron:

1
6 6
15 30 15
20 60 60 20
15 60 90 60 15
6 30 60 60 30 6
1 6 15 20 15 6 1

Just as the seventh row of Pascal's triangle can be used for the classification of ways to make 6 circles all of which are black or red classified by how many are black and how many are red, it is likewise true that the seventh layer of Pascal's tetrahedron can be used for... Georgia guy (talk) 14:06, 27 September 2024 (UTC)Reply

It counts how many ways there are to make 6 circles all of which are black, red, or green, classified by how many are black, how many are red, and how many are green. For example, if you want there to be 2 of each colour, you get   ways, which is exactly the middle entry of this layer. Double sharp (talk) 14:55, 27 September 2024 (UTC)Reply
I would call 1, 6, 15, ... the sixth row, and the top row, with a single 1, the zeroth row. That's what Wikipedia's does in the articles I've seen at least. Anyway, this should be in the article Pascal's pyramid, and can be further extended to Pascal's simplex. The article on the Multinomial theorem is relevant here as well. --RDBury (talk) 17:25, 27 September 2024 (UTC)Reply
RDBury, is the top row "1" not really a row?? Georgia guy (talk) 21:02, 27 September 2024 (UTC)Reply
Well, you can call it whatever you want and index it as you like; I'm not going to get into the philosophy of rows. But the formulas are simpler if you start with the top "1" as row 0. --RDBury (talk) 00:55, 28 September 2024 (UTC)Reply
RDBury, does that statement parallel the statement that trigonometry is simpler if you use radians as opposed to degrees?? Georgia guy (talk) 01:04, 28 September 2024 (UTC)Reply
The OP was calling what most people call row 6 "the seventh row" and I thought that was worth pointing out for future reference. Getting caught up in what counts as a "row" and whether statements are parallel is a matter for philosophy and linguistics. --RDBury (talk) 02:20, 28 September 2024 (UTC)Reply

September 28

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Bitcoin price rigging

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I have been day trading bitcoin for almost a year now. I read articles from bitcoin news to try and guess when the dips and dives will happen and so far it has helped. I just read and article about bitcoin possibly losing freedom to governments and large institutions being able to rig the price. This has happened recently with the German government of Saxony selling off a hoard of seized coin, The Mt. Gox dispersal, and the US dept of Justice mysteriously moving 30/230k of its seized hoard two days after Trumps bitcoin speech in Nashville raised the price. https://mpost.io/u-s-unloads-2-billion-in-bitcoin-from-silk-road-seizure/

Is it possible for a whale (large bitcoin hodler) to make smaller transactions in a short time to move the price for a larger transaction at a later time? On the upswing this is called 'pump and dump' and 'poop and scoop' on the down swing. Both are illegal in most market trades.

This post may fit better in Humanities where finance is listed or IT where crypto may belong. 2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 08:38, 28 September 2024 (UTC)Reply


I'm pretty sure that this isn't a math question. But the article on Pump and dump does explicitly mention cryptocurrencies. --RDBury (talk) 13:22, 28 September 2024 (UTC)Reply
The reason I posted in math is because I am wondering if a small buy or sell in a quick time frame will actually move the price enough for traders like me who watch the minute candle scale. I use MetaTrader 4 where the price moves in microseconds every time a buy/sell happens. Is there a formula for volume needed to move the price in a small time frame or article about time/volume/price/ ratio calculations?
Pump and dump, Bear raid, Short and distort, and Uptick rule, all help to explain how it is possible for whales to control the price. Fear_of_missing_out#Investing is the main cause of up-spikes since I have been investing and most are followed by dives. https://www.weforum.org/agenda/2024/08/explainer-carry-trades-and-how-they-impact-global-markets/ The carry trade crash the 1st week of August caused another huge dive. Foreign_exchange_market#Carry_trade2604:3D08:5E7A:6A00:D94:3638:168B:18A0 (talk) 17:48, 28 September 2024 (UTC)Reply
The time needed to move the needle with high-frequency trading is a combination of the latency of the connection between the computers of the flash traders and those of the exchanges, which depends on the current state of the networking technology and the physical distance between the traders and the exchange, as well as any regulations enforcing a time lag. It then will take time before small-time traders can see the needle having moved. A mathematical model will require too many parameters to be of practical use.  --Lambiam 08:51, 29 September 2024 (UTC)Reply

September 29

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Is this a possible Groth16/ZkSnark verifier‑side simplification ?

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Hello,

The verification algorithm is already simple but I was thinking about some costly environments like blockchains having low block limits. This might be naïve thinking but I was wondering at possibility : normally the prover gives 3 elliptic curves points to the verifier A ; B ; C When public inputs are used C/the inputs vector is split.

But as a simplification part, why not completely ditch the C parts of the proof when public inputs are used ? That way, the verifier would have to compute 1 pairing in less for verifying the proof. I’m meaning e(C,verifying_key_part). It seems to me the requirement to pair with public inputs would still ensure the security of the system… Is it because skipping that pairing would allow to forge public inputs ? As far I understand, a malicious prover would still have to satisfy all constraints of the quadratic arithmetic program and thus would have to use public inputs satisfying constraints. Or is it because it would be impossible to rework the protocol to have the prover being able to produce proofs that verify ?

Or even maybe both of the assumptions above ? 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 11:51, 29 September 2024 (UTC)Reply