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Trurl

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  1. Here are my equations to review. There are quite possibly hundreds of variations. This is what I meant by isolating x and y in the equations. I mean isolated p and q separately, knowing only N. There are 3 equations. I am just getting used to Latex so I hope this is readable. If necessary I will post corrections. But it is these 3 separate equations I want you to test. p^3 - (p^3*N^2) / (N^2 + p) \[p^3 - \frac {N^2 p^3} {N^2 + p}\] ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ _ The above equation should be close to zero, but sometimes the error is closer to 1 (p^3 *N) / (N^2 + p) = fraction fraction/ q = fraction / (N /p) Sqrt[fraction / (N /p)] * N = p^2 \[\frac{N p^3}{N^2+p}=\text{fraction}\] ___________________________________________________________________ The above equations are the second example. q = Sqrt[(N*q^2 + q)/ N] q^2 - (N*q^2 + q)/ N is approximately 0 In the case where N = 85 and q = 17, q^2 - (N*q^2 + q)/ N is 1/x or 1/5 or 0.2 0.2 * 85 = x or 5 in this example \[q^2 - \frac {0\text {approximately}\t ext {is}\left (N q^2 + q \right)} {N}\] Final equations are a proposed method for finding q, knowing only N I have been working with the first equation for awhile. The other 2 are based on the first, but still need tried. There is nothing fancy here. I am just comparing how numbers divide.
  2. The new "y isolated" equation does not work. It was based on the "x isolated" equation, but does not work. Does anyone see a way to isolate y knowing only pnp?
  3. \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} Yes that is exactly it. But does it work? I think it works, but does it prove useful? I call "N": pnp and "p and q": "x and y" Given N (I call pnp) we estimate x. And in the latest post I am trying to find y, the larger factor. \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} \frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \text {pnp}} {\text {pnp} y^3 + y^2} \[\frac {\text {pnp}^3 y^3 - \text {pnp} y^2 + \t ext {pnp}} {\text {pnp} y^3 + y^2}\]
  4. (y^3 * pnp^3 - y^2 * pnp + pnp) / (y^3 * pnp + y^2) = pnp^2 y is now isolated as x was. Ready to simplify. No, you uncovered a weakness of the PC method. That is that the error varies between 0 and 1. The first SemiPrime was close to 94. I factored the second to be sure. But by working with the equation for years that it could be a smaller Prime factor. The trouble is without knowing the true error testing numbers less than 94 in the example, narrows the selection but not more useful than division. We need exact results and not an estimate. But does that exist? 3 possible solutions: Isolate y and compare to x Investigate the curve of the graph of the equation between 0 and 1 Find a pattern in the error. An error not spanning such a large range. 3^4/(8637^2+3)=0.000001732 3 < 94 but the number of importance is 0.00001732. It causes 3 to be hidden when starting at 1 and not zero. Questions are good. It is quite possible I am wrong. But I see something or I wouldn’t have put so much effort. From Wikipedia: Semiprimes are highly useful in the area of cryptography and number theory, most notably in public key cryptography, where they are used by RSA and pseudorandom number generators such as Blum Blum Shub. These methods rely on the fact that finding two large primes and multiplying them together (resulting in a semiprime) is computationally simple, whereas finding the original factors appears to be difficult. In the RSA Factoring Challenge, RSA Security offered prizes for the factoring of specific large semiprimes and several prizes were awarded. The original RSA Factoring Challenge was issued in 1991, and was replaced in 2001 by the New RSA Factoring Challenge, which was later withdrawn in 2007.[7] https://en.m.wikipedia.org/wiki/Semiprime
  5. Ok. I'll try two examples: 8633, 8637. Both are semi primes. NSolve [( x^4/((8637)^2 + x)) == 1] x≈-92.935 x≈92.935 NSolve [( x^4/((8633)^2 + x)) == 1] x≈-92.914 x≈92.914 @Trurl Is this helpful for someone looking for the prime factors of the numbers 8633 and 8637? If so, how? Yes the numbers are too close to estimate. But I still have faith in the Pappy Craylar method. After all 3 < 92 and 3^4/(8637^2+3)=0.000001732 Which Is close to zero where the error is supposed to be near anyway. Remember when I said we needed to test the logarithmic curve of 3*5; 3*7; 3*11 and so forth to find the error change as 3*Prime number increases? I know your smart and are looking for holes in the PC method. That is what you are supposed to do. But I know you are smart enough to understand it. A fix to to great number of possible estimates is to isolate y as x is isolated. Then we will see a better picture of what is happening when SemiPrimes are factored.
  6. y^2 = ((pnp*y^2+y)/pnp) This is one of many isolated y’s. I have much more to share. But I want readers of this to see how simple the equation is. Remember that this thread is called Simple yet Interesting. 😒
  7. I will post here how the equations were derived if anyone is interested. My question to you is: Do you agree the Pappy Craylar method is significant to factorization? My second question to you is there a list of all the ciphers that use SemiPrimes? The hash ciphers the message but to have public keys you need a one-way function. But do one-way functions exist? I read on the web the question of whether math is created or discovered. Something that hasn’t been solved is impossible to it’s discovered it is not. Was it your understanding that was discovered? I cannot factor without error. So large SemiPrime increase in range between zero and one. But if I can post here with the derivation of x, maybe a similar equation can be found for y, the larger factor.
  8. tar

    Trurl,

    I like your electric fields idea regarding the applications of the twelve sections of the sphere but since I last posted on that thread I have run into a major roadblock to all the applications I was considering.  The subdivisions do NOT appear to be of equal area as I had figured earlier in the thread, using a spherical calculator.  I made some incorrect assumptions about angles and unfortunately it appears the subdivisions near the center of each diamond are  of greater area then nearer the corners.    So most applications are thwarted until I come up with a way to have each designated subsection span an equal area.

    There are still some applications that will work if exactness is not required.  If you quarter a section each of the quarters is of equal area, so if you want to divide the sphere into 48 equal sections for your application the scheme is still good.

    One such idea is to for instance mount 48 cameras on a tower or aircraft each pointing in the direction of the center of a quartered section of each diamond.  Then the output of the cameras could be sent in  a one to one way to 48 screens positioned in the same pattern around a viewer.  Thus placing the viewer at the top of the tower or in the craft.

    1. Show previous comments  1 more
    2. Trurl

      Trurl

      So the 48 cameras are some sort of virtual reality?

      Application: What if you divide the Earth into sections to map satellites?

      What about navigation, especially in Space?

      What about video game 3D world positioning?

      The changing shape and propagation of the radio wave?

      When you draw you start with geometry primitives. Why not create a CAD script?

      It is hard for me to think of something mathematical that doesn’t have an application no matter how small.

      I know my application suggestions are not descriptive and it would take much work to complete them, but I envision a 3D model manipulated by change, shape, and division of spheres. Instead of editing polygons you’d edit spheres. If you tried 3D printing you know the inside of the model is structured like a bridge trusses. Those supports could be your spheres divisions.

    3. tar

      tar

      Nice,i

      I do need to learn some CAD and Rendering programs.

       There are many applications I think for the general Sperical Rhombic Dodecahedral structure.

      And interesting aspect is it filled with symmetries and dual figures. And it is the basis for  a dense packing scheme that puts 12 spheres around a center sphere.  This pattern is a cuboctrahedral pattern that when built out, putting always 12 spheres around each sphere in the same pattern it looks more and more like a cuboctahedron as you build out, and contains 3 intersectinng square planes and 4 intersecting hexagonal planes.

      The math is already there, the switch is to think of a cube with the corners cut off to the center of the edges.  The center of each edge is the center of a 1/12th section of the sphere.

      I am working on a scheme to each sphere in the scheme.  The scheme fills space and is completely scalable and very symmetric, having direct analogs to the cube, the sphere, the tetrahedron and the cuboctahedron.

      I like your ideas.  Please use the system freely. It is a mathematical system so I can not claim any ownership.  What I will claim is the discovery of the way to break the sphere into TAR radians using the four axis of the cube that extend through the corners, assigning a color to each 360 group of great circles and using the intersection of the two colors in each of the diamonds to name each degree sized area.  I recently found out that all the degree sized area are not equal in area, so there is as of yet not a clean mathematical way to use the system.

      Where I see some possibility however is in labeling the spheres in the dense packing situation.  There is a relationship between the spheres in the dense packing system and the TAR radian system but I have not discovered it yet.

    4. tar

      tar

      a scheme to label each sphere in the scheme

  9. Thanks @Sensei and @Ghideon The error from the equation to x^4 changes as the 2 Prime products change in the distance between the factors. But much of the usefulness of the Pappy Craylar method relies on the X factor being exact. I prose isolating y (the larger factor) as x has been isolated. The distance between x and y should prove useful. Again thanks for the help with large digits. Now we have a meaningful test of the PC method.
  10. PNP = 85 NSolve [( x^4/(85^2 + x)) == 1] 85 {{x -> 9.22249}, {x -> -0.00294118 + 9.21954 I}, {x -> -0.00294118 - 9.21954 I}, {x -> -9.2166}} Clear[x] NSolve [( x^4/(35794234179725868774991807832568455403003778024228226193532908\ 190484670252364677411513516111204504060317568667^2 + x)) == 1] Clear[x] NSolve [( x^4/(35794234179725868774991807832568455403003778024228226193532908\ 190484670252364677411513516111204504060317568667^2 + x)) == 1] Clear[x] NSolve [( x^4/((6863*7759)^2 + x)) == 1] {{x -> 7297.26}, {x -> 0. + 7297.26 I}, {x -> 0. - 7297.26 I}, {x -> -7297.26}} Clear[x] NSolve [( x^4/((6863*7759)^2 + x)) == 0.8] {{x -> 6901.32}, {x -> 0. + 6901.32 I}, {x -> 0. - 6901.32 I}, {x -> -6901.32}} I don't know why but Mathematica won't solve my large RSA number. Above are 3 different attempts. PNP=85 at 1, you have a value of 9.22249. This means the unknown x is less that 9.22249. Ignore the imaginary numbers because you should be only concerned with the real numbers. You can be sure that those weird values (often imaginary) are not the Prime factors. Thanks for the reply, Ghideon. I think it is clear it works with small values. But as PNP gets larger, the error does not increase. The errors are the same. The error differs as the distance from x to PNP changes. I don't yet know the error or the precision of float points. Mathematica should be able to crunch it. Look at the 3rd program paragraph. I tested an error of 0.8 from 0 to the computed value of the equation. I have never programmed hundred digit+ numbers. I think Mathematica will handle the calculation but as I show in the 2 program paragraph above it just refuses to compute. I thought that maybe someone on this forum would know how to program a hundred digit number and not have a error in floats. I know it is possible, but to use floats without rounding requires a program. I have looked for such programs in C++ but it was so complex. I chose Mathematica to have a ready made program, but it must not see the 100 digit number as a number. But the problem is logic and not number crunching. If I am correct and the equation: x^4/(85^2+x) error is always below 1 then we could pinpoint x.
  11. I too have had such writing problems. You are writing for a message board now and it is informal. For essays you have to outline your ideas. Write a draft. You think faster than you write that is why you outline with brief descriptions. Easy to understand points are a must. If you jump between topics which make sense to you and not the reader you will never get a point across. Simplify. It is like I know how it works but how do I explain it. You are just trying to form ideas and put them into grammatically correct sentences at the same time. Slow you pace and write ideas on scratch paper. I was taking a master’s class in adult education. They told me I was writing musings for the internet. I thought the ideas behind the writing was the most important. But presentation of the idea is a tricky art. Knowing science is one thing but teaching it is it’s own art. Same with writing. Clarity and explaining your point is more important than trying to explain everything at once. Too many ideas to ideas is usually good. You just have to record them before you lose them. Document with a journal.
  12. Your work on dividing the sphere is interesting. As a graphic artist I appreciate all the lines drawn on the volleyball. What are your applications? The one I think of is antenna signal propagation. The shape of impedance and conductance on the antenna would be the division of a sphere that changes size and shape. If you use the division of the sphere as a reference, you have the 3D way to explain electricity, magnetism, and waves; like a sine curve is a reference to the 2D.
  13. Clear [x, pnp, f]; pnp = 85 For [x = 3, (x < (pnp/2)), (x = x + 2) ; f[x] = x^4/(pnp^2 + x) If[f[x] < 1, Print[x]] ] 85 $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of If[f[5]<1,Print[x]]. $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of If[f[7]<1,Print[x]]. $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of If[f[9]<1,Print[x]]. General::stop: Further output of $RecursionLimit::reclim2 will be suppressed during this calculation. Clear [x, pnp, y]; pnp = 85 For [x = 3, (x < (pnp/2)), (x = x + 2) ; y = x^4/(pnp^2 + x) Print[y]] 85 y (125 Null)/1446 (2401 Null)/7232 (6561 Null)/7234 (14641 Null)/7236 (28561 Null)/7238 (10125 Null)/1448 (4913 Null)/426 (130321 Null)/7244 (194481 Null)/7246 (279841 Null)/7248 (3125 Null)/58 (531441 Null)/7252 (707281 Null)/7254 (923521 Null)/7256 (1185921 Null)/7258 (300125 Null)/1452 (1874161 Null)/7262 (2313441 Null)/7264 (2825761 Null)/7266 Well I think the syntax is correct now. Mathematica does not want to do what I tell it. I showed the Null because if you divide those values without the Null you get the answer. I am only looking for x values that f[x] <1 (near zero). Again if anyone can tell me how to perform math on hundred-digit numbers let me know. Mathematica would handle it, but the recursion error is in my way. I'm so close. I want to be able to just cut in past pnp and find the zero values. Also note the equation is simplified. That is thanks to an unanimous post. Mathematica simplified in one keyword what I was unable to do on paper. I think I'm going to turn to free Linux opensource math programs. Thanks Sensei for promptly answering my post. I think the problem is that Mathematica expects an integer value, but I have seen it use decimals many times before. It is just frustrating when the computer language doesn't do what you want. But I don't mind if someone programs it before I do, because that may lead to the Pappy Craylar Method being proven true.
  14. All those thoughts are a blessing. Not racing thoughts but the potential ideas. If you are control of those thoughts you can focus them. Too many ideas is better than no ideas. That is one heck of a brainstorming session. You should not write in proper pose. Instead write down ideas and create outlines. I listened to an Audible book on how to create things by writing. It can help you write a book but it helps in all areas of creativity. The book is called Accidental Genius by Mark Levy.
  15. Thanks. The code is that simple. Put it seem to Print f of x instead of x. In[63]:= Clear x Clear pnp pnp = 85 f[x] = ((x ^ 3 * pnp^2) / (pnp^2 + x) - x^3) x = 3; While [f[x_] < 1, Print[x]; (x + 2)] Out[63]= 3 Clear Out[64]= 85 Clear Out[65]= 85 Out[66]= -(81/7228)
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