Bezier curve, Newton-Raphelson method, formula problem, please help! [modified] - Algorithms Discussion Boards

It's your lucky day Roger.
I have decided to grace your question with my massive intelect.

Roger Wright wrote:
The cable types are well characterized, with weights per foot, modulus of elasticity, and tensile strength published.

Ok,you can go a s far as the "Sagsec" type of solution quoted elsewhere in this thread, but I believe you would be better off looking for a good Rule of Thumb, that can be applied to other areas and applications.

Ok weights per foot, modulus of elasticity and tensile strength mumbo jumbo are irrelevant. You must ascertain a maximum usable length per strand and be done with that. The other factors are only of real use when discussing the virtues of a tether for an orbitting satellite. You and your decent Injun bretheren have not ventured into this field yet I pressume.

I percieve the savings from using the best solution for this on a new power route to be substantial so working on a solution is admirable.

Ok lets say we have Pole1 at height h1 and Pole2 at height h2 and a hypoteneuse of distance dh.

What we now need to know for in the field rules of thumb are
i.) The lowest point of the power line (clue the lesser of Pole1 and Pole2 is the maximum boundary.)
ii.) The distance the lowest point is reached from either Pole.
iii.) A formula to calculate the lines height at any other distance from a Pole.
note this is easiest to do once kwowledge of i and ii are achieved.

The guts of this is that gravity on earth is generally consistent from one location to another, so a section if not all when Pole1 and Pole2 heights are equal will be symmetrical.
To calculate the new shape and dimesions of the line for the remainder of the line will use the already calculated figures to calculate the rest.
So guess what you now need to use interpolation.

Regardz
Colin J Davies

The most LinkedIn CPian (that I know of anyhow) Smile | :)

I need an algorithm for combining values from a variable number of variable length categories.

For example, I might have an array of the following values with their associated categories:

value1 - category1
value2 - category1
value3 - category2
value4 - category2
value5 - category2

The result should be sets containing every possible combination of 1 value from each category:

{value1, value3}
{value1, value4}
{value1, value5}
{value2, value3}
{value2, value4}
{value2, value5}

I can't use C(n,k) because k is variable. Can anyone help? My brain has short-circuited on this one.

Thanks!

you may try something like this

 static void Main(string[] args)<br />
        {<br />
            int[] nItems = { 2, 3 };<br />
            int[] number = new int[nItems.Length];<br />
            Generate(nItems, number);<br />
        }<br />
<br />
        static void Generate(int[] CategoryItemCount, int[] currentCombination)<br />
        {<br />
            for (int i = 0; i < CategoryItemCount[0]; i++)<br />
            {<br />
                currentCombination[CategoryItemCount.Length - 1] = i;<br />
<br />
                if (CategoryItemCount.Length != 1)<br />
                {<br />
                    Generate(new List<int>(CategoryItemCount).GetRange(1, CategoryItemCount.Length - 1).ToArray(), currentCombination);<br />
                }<br />
                else<br />
                {<br />
                    // store currentCombination<br />
                }<br />
            }<br />
        }<br />

regards

Thanks! This helps.

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Yes your test passed.

As of how to accomplish that have you ever tried Google?

Failing that try Badger | [badger,badger,badger,badger...]

.

Hi Guys,

I’m creating an animation program and have built a Bezier spline motion editor. The user-controllable curve plots time(x) against value(y). So my challenge is to retrieve the 'y' value from a given 'x' on the curve. However, all the articles I've been able to find/understand only give a formula for calculating an x or y from 't' - the percentage along the given cubic bezier curve.

As 't' is more concentrated around areas which heavily curve (to make a smooth curve), it's not possible to calculate y from x. So I'm using what seems to be a recognised work around and iteratively guessing at which 't' value produces x, continually reducing the difference until I find the correct 't' which produces my required x. I then use this 't' to calculate y.

This all works perfectly well. However, this is an animation program so I need to optimize this process. My current technique requires up to 15 iterations before it finds an accurate value for 't'. I found an article online by Don Lanaster, “Some more cubic spline math BEZMATH.PS”. He mentions using the Newton-Raphelson method to reduce this process down to 3 iterations. Apparently this technique uses the slope of the curve to make the approximation of 't' more accurate. However, the formula he uses doesn't match the formula I currently have working to calculate a value on the curve. I'm sure I'm misunderstanding it!

At this point I have to confess I am a Math dummy. I can work my way around practical mathematical issues as long as I don’t have to resort to long formulas full of ancient Greek!

Here's an extract from Don's article:

--- Quote ----
Let's use a better ploy to get our approximation to close quickly. It is called the NEWTON-RAPHELSON method, but is much simpler than it sounds. Say we get an error of x - x1. x1 is the current x for our current guess. At x1, our spline curve has a slope. Find the slope.

The slope is expressed as rise/run. Now, on any triangle...

rise = run x (rise/run)

This gives us a very good improvement for our next approximation. It turns out that the "adjust for slope" method converges very rapidly. Three passes are usually good enough. If our curve has an equation of...

x = At^3 + Bt^2 + Ct + D

...its slope will be...

x' = 3At^2 +2Bt + C

And the dt/dx slope will be its inverse or 1/(3At^2 + 2Bt +C). This is easily calculated. The next guess will be...

nextguess = currentt + (curentx - x)(currentslope)

----- end quote ----

So here's the function I call iteratively with varying values for 't' and it returns 'x' for me to compare with what I want:

public float CalcBezierValue(float t, float A, float B, float C, float D)
{
t = 1 - t; // Reverse the normalised percentage

float F1 = t * t * t; // These vars used purely for visual clarity
float F2 = 3 * t * t * (1 - t);
float F3 = 3 * t * (1 - t) * (1 - t);
float F4 = (1 - t) * (1 - t) * (1 - t);

return A * F1 + B * F2 + C * F3 + D * F4;
}

As you can see the base formula which produces the return value is quite different from the one in the article (x = At^3 + Bt^2 + Ct + D). Consequently I’m failing to comprehend how to calculate the slope value or its inverse which seem to be required to reduce the number of calls to this function.

Can anyone help me?

Many thanks for your time,

Simon

-- modified at 12:01 Wednesday 13th September, 2006

srev wrote:
I’m creating an animation program and have built a Bezier spline motion editor. The user-controllable curve plots time(x) against value(y). So my challenge is to retrieve the 'y' value from a given 'x' on the curve. However, all the articles I've been able to find/understand only give a formula for calculating an x or y from 't' - the percentage along the given cubic bezier curve.

This is one of many, many reasons why I won't use a Bezier spline for camera or motion control. A) its always more difficult B) it is always more time-consuming and cpu-hogging.

I tend to use Catmull subdivision surface variants. Catmull-Rom for splines although I have my own variation of it. The original code was from a motion spline in AI Game Programming Wisdom[^] I modified it for uneven weighted distances and some other actions. If you are trying to do smooth motion cameras without a huge CPU overhead, I highly recommend the book. I use it for just about everything within, it's a great book all around, but in related to this there is a chapter on least-acceleration motion curves, PID controllers and subdivision splines, all slightly different approaches to motion control.

If you don't want the book, take a peek at these:
http://www.mvps.org/directx/articles/catmull/[^]
http://www.lighthouse3d.com/opengl/maths/index.php?catmullrom[^]
http://www.cubic.org/docs/hermite.htm[^]

_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)

Jeffry thanks for these links, they look really usefull. And Amazon has my money for the book you recommended.

Thanks again!

Hi,

If this is your function, y = CalcBezierValue(t, A, B, C, D), the slope function may be (I have not tested it at all)

public float CalcBezierSlope(float t, float A, float B, float C, float D)
{
t = 1 - t; // Reverse the normalised percentage

float F1 = 3 * t * t; // These vars used purely for visual clarity
float F2 = 6 * t - 9 * t * t;
float F3 = 3 * (1 - t) * (1 - t) - 6 * t * (1 - t);
float F4 = -3 * (1 - t) * (1 - t);

// I am not sure about the minus, but since there's function composition t = 1 - t ...
return ( -1 ) * A * F1 + B * F2 + C * F3 + D * F4;
}

and your iteration function may look like

public float CalcBezierNextGuess(float x0, float t, float A, float B, float C, float D)
{
// I am not sure about the minus sign, may be a plus
return t - (CalcBezierValue(t,A,B,C,D) - x0) / CalcBezierSlope(t, A, B, C, D);
}

I might be completely wrong, but I have no time to test this.

Regards,

Guys,

Thanks so much for everyone's feedback. From all the feedback on this and other forums I have the answer.

The function CalcBezierValue() computes the cubic Bézier curve as the polynomial:
x = A*t^3 + 3*B*(t^2)*(1-t) + 3*C*t*(1-t)^2 + D*(1-t)^3

To use the technique in the article I need the derivative:
x' = 3*A*t^2 + 6*B*t*(1-t) - 3*B*t^2 + 3*C*(1-t)^2 - 6*C*t*(1-t) - 3*D*(1-t)^2

The final code looks like this:

// Calc the derivative value of the curve at the specified percentage (rather than the polynomial)
public float CalcBezierDerivative(float P, float A, float B, float C, float D)
{
P = 1 - P; // Reverse the normalised percentage

float tR = 3 * A * (P * P);
float tS = 6 * B * P * (1 - P);
float tT = 3 * B * (P * P);
float tU = 3 * C * ((1 - P) * (1 - P));
float tV = 6 * C * P * (1 - P);
float tW = 3 * D * ((1 - P) * (1 - P));

return tR + tS - tT + tU - tV - tW;
}

Many thanks for all your responses.

Simon

2^32,582,657-1[^]

"In this house, we obey the laws of thermodynamics!" - Homer Simpson

Web - Blog - RSS - Math - LinkedIn

My friend posed the following question to me, that a friend of his was given in her math class:

Suppose in a simplified game of (American) football that the only possible amount of points one can score at a time are 3 or 7. What is the highest score that is impossible to achieve?

For example, 17 is a possible score (7 + 7 + 3) but 11 is not (there is no combination of 3's and 7's you can add together to get 11). Of course, since this is given in the context of a game, there can be no negative scores, so (7 + 7 - 3) doesn't work.

I wrote a small program to brute-force the solution, and searching up to 1000 I found the maximum impossible score is 11. However, I am not able to prove it. This problem reminds me of some things we did in Number Theory class but it's been a while and I've hence forgotten a lot of it.

Any pointers?

--
Marcus Kwok

It's trivial to show that the score can never be 11, so you now just have to prove that all numbers greater than 11 can be scored. All numbers after 11 fall into one of 3 series:

12 + 3n + 0 = {12, 15, 18, ...}
12 + 3n + 1 = 13 + 3n = {13, 16, 19, ...}
12 + 3n + 2 = 14 + 3n = {14, 17, 20, ...}

So, now just show that 12 (4 field goals), 13 (2 field goals and a touchdown), and 14 (2 touchdowns) can be scored, then all other numbers can be scored by tacking on more field goals (3n).

Fantastic! Thanks. Of course, now that I know the answer, it seems so obvious Badger | [badger,badger,badger,badger...]

--
Marcus Kwok

Anyone know of a way to take an inverse power to get the number a number was raised by?

x = 2^(20 - a) or x = 2^a

I need to determine a, but my math is way rusty. I think I need the nth root of x, but don't know how to write it, and actually need to find out what n is.

Do I need to do x^(1/a) = 2? I get stuck after that. Probably basic math, but it's slipped my brain. Smile | :)

Any help would be greatly appreciated.

Found this mildly funny: http://mathworld.wolfram.com/Power.html[^]

A number to the first power is, by definition, equal to itself, i.e., z^1==z.

It is therefore impressive that Captain Kirk (William Shatner) is able to detect one more heartbeat aboard the starship Enterprise than can be accounted for by amplifying an auditory sensor intensified by a factor of "1 to the fourth power" in the Season 1 Star Trek episode "Court Martial" (1967).

- S
50 cups of coffee and you know it's on!

Hi,

x = 2^(20 - a) or x = 2^a

Use log

log x = (20 - a) * log 2 or log x = a * log 2

Regards

Thanks!

- S
50 cups of coffee and you know it's on!

you can do this:

if x=n^a

then a=log(x)/log(n)

Log is the logarithm ( base 10 or naturals, this works whith any base )

Greetings

Hi, probably a better place for this, hopefully someone can help.

For one of my programs i'm trying to impliment my own version of the RSA algorithm, the encryption works fine, however the decrption does not, even though it uses basically the same algorithm. The code is as follows:

<br />
<br />
//		g++ main.c -o main<br />
<br />
#include <stdio.h><br />
#include <stdlib.h><br />
#include "math.h"<br />
<br />
<br />
int do_crypto(int M, int e, int N);<br />
<br />
int main()<br />
{<br />
	<br />
	printf("...\n");<br />
	<br />
	//	TWO PRIMES FOR OUR KEY<br />
	int p = 17;<br />
	int q = 11;<br />
	//	THIRD PART OF OUR KEY (PUBLIC)<br />
	int e = 7;<br />
	//	CALCULATE THE OTHER PART OF THE PUBLIC KEY<br />
	int N = p * q;	//	= 187<br />
	<br />
	//	THE CHARACTER TO ENCODE AS ASCII<br />
	int M = 88;<br />
	//int M = "M";<br />
	<br />
	//	ENCRYPT A CHARACTER<br />
	//int C = int ( M * exp(e) ) % N;<br />
	int C = do_crypto(M, e, N);		//	int M, int e, int N)<br />
	printf("C = %d \n", C);<br />
	<br />
	<br />
	//	CALCULATE THE DECRYPT KEY<br />
	//int d = ( 1 % ( (p-1) * (q-1) ) ) / e;<br />
	int d =  ( ( (p-1) * (q-1) ) / e );<br />
	printf("d = %d \n", d);<br />
	<br />
	<br />
	// DECRYPT THE CHARACTER<br />
	int m = do_crypto(11, 23, 187);		//	int C, int d, int N)<br />
	printf("m = %d \n", m);<br />
	<br />
	<br />
	return 0;<br />
}<br />
<br />
int do_crypto(int M, int e, int N)<br />
{<br />
	int iret = int ( M * exp(e) ) % N;<br />
	<br />
	return iret;<br />
}<br />
<br />

When decrypting I have put in actual values for the keys etc, and the result should be 88.
The actual formulas are:
ENCRYPT: C = Me (Mod N)
DECRYPT: M = Cd (Mod N)
*** Please note the 'e' and 'd' are supposed to be superscript, e.g. raised to the power of...

Have you figured this out or still having trouble with it?

"Any sort of work in VB6 is bound to provide several WTF moments." - Christian Graus

i have a map that contain (22 * 18 ) images i don't want to add all of the images
i want have a 3 * 3 table
the problem is i want when user drag the map the images is load in the imagesboxes without user understand that all of the images dosn't load

How Can I do it(!!!!!) WTF | :WTF:

MHF

Write a C program that accepts as input any of the following:

>> a listing of edges of a graph given as pairs of positive integers

>> the adjacency matrix

>> the incidence matrix

and outputs the other two...