Newton raphson method codings
i need generalized coding for load flow analysis using newton raphson method in power system analysis..........
What does that have to do with J2me?
Similar Messages
-
Square root of an interval using Newton's method
Hello!
I am trying to create a method that calculates the square root of an interval, and I am having trouble with both the actual calculation part, as well as the base case for the recursion. I implemented a simple counter for the recursion, but was not seeing any kind of pattern for the values. (I am pretty sure the "better" values should converge to 0).
I was wondering if anybody wanted to take a swing at it and help me out. :)
Here is the code for my program, followed by the code for Newton's method for calculating square roots of doubles. I am supposed to use it as a reference.
I made the simple arithmetic methods with the help of http://en.wikipedia.org/wiki/Interval_arithmetic . They seem to work fine, so I am having issues with troubleshooting!
Thanks!
public class Interval {
double x1;
double x2;
public Interval(double newx1, double newx2){
x1 = newx1;
x2 = newx2;
public String toString(){
return "[" + this.x1 + ", " + this.x2 + "]";
//Add an interval to the current one.
public Interval add(Interval j){
double tempx1 = this.x1 + j.x1;
double tempx2 = this.x2 + j.x2;
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Subtract an interval from the current one.
public Interval sub(Interval j){
double tempx1 = this.x1 - j.x2;
double tempx2 = this.x2 - j.x1;
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Multiply an interval with the current one.
public Interval mul(Interval j){
double minx1 = Math.min(this.x1*j.x1, this.x2*j.x2);
double minx2 = Math.min(this.x1*j.x2, this.x2*j.x1);
double maxx1 = Math.max(this.x1*j.x1, this.x2*j.x2);
double maxx2 = Math.max(this.x1*j.x2, this.x2*j.x1);
double tempx1 = Math.min(minx1, minx2);
double tempx2 = Math.max(maxx1, maxx2);
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
//Divide the current interval by a new one.
public Interval div(Interval j){
double minx1 = Math.min(this.x1/j.x1, this.x2/j.x2);
double minx2 = Math.min(this.x1/j.x2, this.x2/j.x1);
double maxx1 = Math.max(this.x1/j.x1, this.x2/j.x2);
double maxx2 = Math.max(this.x1/j.x2, this.x2/j.x1);
double tempx1 = Math.min(minx1, minx2);
double tempx2 = Math.max(maxx1, maxx2);
Interval tempInterval = new Interval(tempx1, tempx2);
return tempInterval;
static Interval step(Interval x, Interval y) {
// Compute a "better" guess than x for the square root of y:
// Code for doubles: Interval better = x - (x*x - y)/(2*x);
Interval two = new Interval(2.0, 2.0);
Interval better = x.sub( ( (x.mul(x)).sub(y) ).div(two.mul(x)) );
// For doubles:
if ( Math.abs(better.x2 - better.x1) < 0.001 ) { // base case
System.out.println(better.toString());
return better;
else {
return step(better, y); // try to get even better...
static Interval sqrt(Interval y) {
return step(y, y); //: start guessing at the square root
public static void main(String args[]){
Interval i = new Interval(4.0, 8.0);
Interval j = new Interval(4.0, 8.0);
Interval addij = i.add(j);
Interval subij = i.sub(j);
Interval mulij = i.mul(j);
Interval divij = i.div(j);
Interval sqrtj = i.sqrt(j);
System.out.println("Intervals:");
System.out.println(i.toString());
System.out.println(j.toString());
System.out.println("Add: " + addij.toString());
System.out.println("Sub: " + subij.toString());
System.out.println("Mul: " + mulij.toString());
System.out.println("Div: " + divij.toString());
System.out.println("Sqrt: " + sqrtj.toString());
}and newton's root finder for doubles:
public class SquareRoot {
static final double ALLOWED_ERROR = 0.001;
* Newton's method for finding square roots.
static double step(double x, double y) {
// Compute a "better" guess than x for the square root of y:
double better = x - (x*x - y)/(2*x);
// Are we close enough?
if ( Math.abs(x - better) < ALLOWED_ERROR ) { // => stop: base case
return better;
else {
return step(better, y); // try to get even better...
static double sqrt(double y) {
return step(y, y); //: start guessing at the square root
public static void main(String[] args) {
System.out.println(Math.sqrt(1234));
System.out.println(sqrt(1234));
// NOTE: you may need to adjust the error bound for these two to agree
}Nathron wrote:
Here is the code for my program, followed by the code for Newton's method for calculating square roots of doubles. I am supposed to use it as a reference.The only thing I can see that looks suspicious is the call to step(better, y) in your reference code.
Are you sure it shouldn't be step(y, better) or step(better, x))? Newton-Rhapson is supposed to be a progressive method, but as far as I can see the value of y can never change with the way you've got it. And if you've copied that to your new code, it might explain the problem.
Winston -
Method that fines zeros of a function.
I've a method:
double y (double x)
I need a method:
find_zeroes(range_min, range_max )
that gives me back an array with the passages on zero of the "y" function in the range between range_min and range_max
I mean that if at [x], [y=-0.001] and at [x+0.001], [y=+0.001] then probably at [x] there will be a zero of the function so, it will be returned as an element of the array.
is there any code around that does this work?
thanksIs this homework?
Look up Newton-Raphson method. There are other methods. -
Problem:
Using the Newton-Raphson method, find the square root of an input positive integer number (ensure that the input is positive). Compare the result with Math.sqrt (n).
Hints:
* Our function is f(x) = x^2 - n where n is the input number. Then, Newton-Raphson method sas that xk+1 is a better approximation for the square root of n than xk where
xk+1 = (xk + n / xk) ---------->all the k are below the x.
* You can take x0 as n.
* Stop when |xk+1 - xk| < e where e is a very small number (take as 1.0E-09).
Sample execution:
Type the integer whose square root you would like to compute: 2
Square root of 2 is 1.414213562373095 by Newton-Raphson method and
1.4142135623730951 by Math class.
I didnt understand what is formula of Newton-Raphson method exactly.Therefore, i couldnt apply the method into java.It is my lab assignment please help me until friday morning.Thanks.umutcan55 wrote:
I didnt understand what is formula of Newton-Raphson method exactly.You could [start with the Wikipedia article|http://en.wikipedia.org/wiki/Newton's_method] for example.
Therefore, i couldnt apply the method into java.So you don't actually have a Java question, right? So this is not the correct forum.
It is my lab assignment please help me until friday morning.That's irrelevant. -
"Eval Parsed Formula Node VI" does not return outputs in predefined order
I make a data analysis program, where the data consists of some million events and each event has e.g. 4 channels and 1-5 hits on each channel.
I would like the user to select different expressions of these channels to give coordinates to plot in a 2D histogram (increment a bin in Intensity Graph), e.g. for some experiment you want to show x=ch1-ch2; y=ch1+ch2+ch3+ch4; while in another experiment you want x=ch1-123; y=123-ch2;
There are other VIs that use static LabView-code for the normal things, but now after a few years of adding to this program I find that it would be quite good with a general plotter and let the user specify simple expressions like this. Also with the "normal" static plots, there is a need to filter out bad data and then it would be much simpler both to program and use if you could write text expressions with boolean logic to combine multiple filters. Making a LabView code and GUI that allows AND/OR/parenthesis combinations of expressions will be quite an effort and not very reusable.
So, with the above motivation, I hope you see that it would make sense to have a useable string formula evaluator in LabView. I find some info about MathScript's user-defineable functions, but haven't managed to get MathScript working in LV2010 yet (maybe some licensing or installation issues, I think I had it in 8.6). But I think it would be possible to do most of what I want for the display-part (not the filtering) with the simpler Eval/Parse Formula Node VIs and suitable use of the limited variable name space. So I started testing, and found a quite annoying issue with how the evaulator works.
To the parser, you are expected to send an array of output variable names. But then it ignores this array, and returns one output per assignment/semicolon in the formula, in the order of the formula text. Since the static parts of my program need to know what the output values mean (which of them is x and which is y), I would have to rely on the user not using any intermediate variable and defining x before y. The attached screenshot demonstrates the problem, and also that it has been solved by NI statff in the "Eval Formula Node VI" which does the appropriate array-searching to extract only the pre-defined outputs, in their expected order. But using that VI is about 100 times as slow, I need to pre-compile the formula and then only use the evaulator in the loop that runs over a million events.
I don't know if I'll take the time to make my own tweks to the parsing stage (e.g. preparation of array-mapping to not have to repeat the search or maybe hacking the output list generated by the parser) or if I'll have to make it in a static Formula Node in the block-diagram (which supports more functions), so that the user has to run with development environment and stop the program to change the plotting function. But I wanted to share this trouble with you, in hope of improvments in future LabView versions or ideas from other people on how I could accomplish my aim. I have MATLAB-formula node possibility too, but is far as I have seen the only place the user could update the formula would then be in a separate .m file, which is re-read only when typing "clear functions" in the Matlab console window. (Having this window is also an annoyance, and perhaps the performance of calling Matlab in every iteration is not great.)
Besides this issue, it also seems very strange there is virtually no support for conditional expressions or operators in Formula Node evaulated formulas (called Mathematics VIs in the documentation). Maybe using (1+sign(a-b))/2 you can build something that is 0 when a<b and 1 when a>b, but it is not user friendly! Would it really be diffcult to add a function like iif(condition, return_value_if_true, return_value_if_false) to replace the unsupported "condition ? if_true : if_false" operator? Would it really be difficult to add support for the < <= >= > == || && operators? Although compiled to an assemply language, this can't exactly be difficult for a CPU.
Attachments:
LV script test.png 62 KB
LV script test.vi 18 KB(1) You can put any kind of code inside an event structure with the limitation that it should be able to complete quickly and without user interaction. For example a while loop for a newton-raphson method is fine, but an interactive while loop where the stop condition depends on user iteraction is not recommended. By default, event structures lock the front panel until the event completes, so you would not even be able to stop it.
(2) Yes, you can do all that. LabVIEW has no limitation as a programming language. Use shift registers to hold data and state information, the manipulate the contents as needed.
(3) I would recommend to use plain LabVIEW primitives instead of formula nodes. Show us your code so we can better see what it's all about. Obviously you have a mismatch betweeen scalars and arrays. It is impossible from the given information where the problem is.
(4) Yes, look inside the nonlinear curve fit VI (you can open it an inspect the code!). One of the subVIs does exactly that. Just supply your model VI.
LabVIEW Champion . Do more with less code and in less time . -
Besides using the math class, may I know if there is another way to find the square root of an integer?
thank you.Besides using the math class, may I know if there is
another way to find the square root of an integer? There are several numerical methods available. The most common is built on Newton-Raphson.
The idea is very simple. Say you want to calculate the squareroot of N. One can note that this squareroot will be somewhere between N and 1/N. So an approximate value will be in the middle like
N1 = (N + 1/N)/2.
Now you're closer and can get even closer by repeating this again with
N2 = (N1 + 1/N1)/2.
This is repeated again and again until Nx*Nx (where x are the numbers 1,2,3,4,5,6 etcetera) is sufficiently close to N. When it is Nx is the squareroot of N. -
Hi everyone
I wrote an APR calculator for an ericsson T610 (don't ask why), everything works fine except it takes a really long time to run through the algorithm. (which I'm about to believe is too powerful for the handset)
I don't really know how to optimize my code to reduce the number of attempts before hitting the correct interest rate(required value)
(for those who have worked with the Bisection and Newton methods for calculating APR will know the required value progresses between a min and a max range until its at the correct value, every time a possible value is determined its used to calculate the sum of present values which in turn must equal the original advance otherwise it continues up and down the range until the correct value is found)
What I want to know really, is this type of calculation really fit for the handset, should it be able to handle the calculation, and I need to change my code?
Any ideas, direction to other forums or direction to some where use full will be greatly appreciated.
Cheerio.Hi sabre150
I will gladly post it, just to let you know I have solved the equation (using bisection method not newton), my problem is with the speed of the process on the handset. anyway here goes...
The manual given to me can be found at
http://www.oft.gov.uk/NR/rdonlyres/19342C6E-BB3E-41E9-AA15-29054CD31D77/0/oft144.pdf
the formula is on page (pdf) 14 (in the book page 9)
on page (pdf) 54 you will find psudo code for the bisection method (in the book page 49)
on page (pdf) 57 you will find psudo code for the newton method (in the book page 52)
to save anyone from attempting alot of reading
'A' represents (future value) total sum of installments
'pV' represents (present value)
' t ' represents the term (in months)
' i ' represents the APR in bisection method
' x ' represents apr in newton method
'dV' represents derivative value
for example :
advance = 91500
term = 300 (months)
rate = 9.5%
to calculate installments use: advance * rate / ( 1 - ( 1 + rate ) ^ ( -NoOfInstallments ) )
in this case the installments are 799.4324446897131 per month
bisection method:
A = pV ( 1 + i ) ^ ( t )
Pv = A / ( 1 + i ) ^ ( t / 12)
total_pVs_for_advance = 91500 (there is just one)
using the correct value for i the total_sum_Of_pVs_for_installments will equal total_pVs_for_advance
value i will then be the apr.
the information above can be used with the psudo code from the bisection method pointed out in page (pdf) 54
the result should read 10.48......
in newton's method: unfortunatly they dont explain alot regarding the newton method
as explained in the book when using newton method it is more convenient to have the present value formula as
Ax^t . i would imagine this is read as A * ( 1 / ( 1+x ) ^ t (please correct me if i'm wrong)
x is the intermediate value and has the the value
1 / ( 1 + x )
also this method requires the calculation of a derivative value dV
which has the formula
tAx^t -1. i would imagine this is read as t * A * ( 1 / ( 1+ x )) ^ t-1 (please correct me if i'm wrong)
using the information above and the psudo code for the newton method the answer reads
10.48....... -
Hi,
I've been asked to see if I can fix a C program so that it will run on a mac. It currently compiles and runs perfectly on a linux machine, but as soon as you run it on a mac, it will compile, but quit with a bus error very early on. The full code is here:
// Program for the construction of cartograms (density-equalizing map
// projections) using the Gastner-Newman technique. The program needs polygon
// coordinates and a count of cases in each region as input, calculates the new
// coordinates, and writes these to a file. Postscript images of the original
// map and the cartogram are created.
// WRITTEN BY MICHAEL GASTNER, September 20, 2004.
// If you use output created by this program please acknowledge the use of this
// code and its first publication in:
// "Generating population density-equalizing maps", Michael T. Gastner and
// M. E. J. Newman, Proceedings of the National Academy of Sciences of the
// United States of America, vol. 101, pp. 7499-7504, 2004.
// The input coordinates in MAPGENFILE must be in ArcInfo "generate" format of
// the type:
// 1
// 0.3248690E+06 0.8558454E+06
// 0.3248376E+06 0.8557575E+06
// 0.3248171E+06 0.8556783E+06
// 0.3247582E+06 0.8556348E+06
// 0.3246944E+06 0.8555792E+06
// 0.3246167E+06 0.8555253E+06
// 0.3250221E+06 0.8557324E+06
// 0.3249436E+06 0.8557322E+06
// 0.3248690E+06 0.8558454E+06
// END
// 1
// 0.3248690E+06 0.8558454E+06
// 0.3249651E+06 0.8558901E+06
// 0.3250519E+06 0.8559769E+06
// 0.3250691E+06 0.8561246E+06
// 0.3249678E+06 0.8560541E+06
// 0.3249003E+06 0.8560088E+06
// 0.3076424E+06 0.8477603E+06
// 0.3075691E+06 0.8477461E+06
// 0.3075595E+06 0.8476719E+06
// END
// 2
// 0.6233591E+06 0.6056502E+06
// 0.6235193E+06 0.6056467E+06
// 0.6235054E+06 0.6055372E+06
// 0.7384296E+06 0.2334260E+06
// 0.7383532E+06 0.2345770E+06
// END
// END
// The number of cases in CENSUSFILE must be given in the form
// region #cases (optional comment), e. g:
// 43 0.9 Alabama
// 51 0.3 Alaska
// 37 0.8 Arizona
// 47 0.6 Arkansas
// 25 5.4 California
// 32 0.8 Colorado
// 19 0.8 Connecticut
// 29 0.3 Delaware
// 28 0.3 District of Columbia
// 49 2.5 Florida
// 45 1.3 Georgia
// 1 0.4 Hawaii
// The output coordinates are written to CARTGENFILE in ArcInfo "generate"
// format. A postscript image of the original input is prepared as MAP2PS,
// an image of the cartogram as CART2PS.
// Modified on Oct 29, 2004. The number of divisions in each dimension lx, ly
// will now be determined from the input map. Also fixed array bound
// violations in intpol and newt2, and centered the ps-files.
// Modified on Dec 3, 2004. Introduced pointers bbmaxx,bbmaxy,bbminx,bbminy -
// the bounding box coordinates for each polygon - to reduce calculations in
// crnmbr. Thanks to Chris Brunsdon for the suggestion.
// Program now terminates with error message if there is no density specified
// in CENSUSFILE for a region on the map.
// "Donut" polygons are now handled properly. Polygons oriented clockwise will
// be considered exterior boundaries. If they are oriented anti-clockwise they
// are holes inside an enclosing polygon. (Unconventional for mathematicians,
// but this is ArcGIS standard.) The identifier is either that of the enclosing
// polygon, or -99999 in which case it is a hole in the preceding polygon.
// The population can now be a floating-point number.
// If a region contains exactly zero population, it will be replaced by
// MINPOPFAC times the smallest positive population in any region.
// Obviously, MINPOPFAC should be <1, I will set it to 0.1 by default. This
// should reduce the value of sigma necessary for the integrator to finish
// properly.
// Modified on March 31, 2005. Initialized maxchange in nonlinvoltra() as
// INFTY. Replaced crnmbr() by a similar, but faster routine interior().
// Many thanks to Stuart Anderson for pointing out this shortcut.
// TO DO LIST:
// - Do Gaussian blur within nonlinvoltra as a simple call to calcv and
// eliminate gaussianblur().
// - Modify the NR code for the FFTs to deal more naturally with the fencepost
// issue. No more copying of arrays in sinft and cosft.
// - Write truly two-dimensional versions of coscosft, cossinft, and sincosft.
// See Chan and Ho: A new two-dimensional fast cosine transform algorithm,
// IEEE Transactions on Signal Processing, 39, 481ff. (1991).
// - Read polygon data directly from shapefile instead of generate file.
// - Reorganize data structures. It is currently quite a mess to read.
// Inclusions
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
// Definitions
#define CART2PS "./cart.ps","w" // Cartogram image.
#define CARTGENFILE "./cartogram.gen" // Cartogram generate file.
#define CENSUSFILE "./census.dat","r" // Input.
#define CONVERGENCE 1e-100 // Convergence criterion for integrator.
// #define DISPLFILE "./displ.dat","w"
#define FALSE 0
#define HINITIAL 1e-4 // Initial time step size in nonlinvoltra.
#define IMAX 50 // Maximum number of iterations in Newton-Raphson routine.
#define INFTY 1e100
#define LINELENGTH 1000
#define MAP2PS "map.ps","w" // Map image.
#define MAPGENFILE "./map.gen","r" // Input coordinates.
#define MAX(a,b) ((b>a)?(b):(a))
#define MAXINTSTEPS 3000 // Maximum number of time steps in nonlinvoltra.
#define MAXNSQLOG 18 // The number of sample points for rho_0 is <~2^MAXNSQLOG.
#define MIN(a,b) ((b<a)?(b):(a))
#define MINH 1e-5 // Smallest permitted time step in the integrator.
#define MINPOPFAC 0.1 // Replace 0 population by a fraction of the minimum.
#define NR_END 1
#define NSUBDIV 1 // Number of linear subdivisions for digitizing the density.
#define PADDING 1.5 // Determines space between map and boundary.
#define PI 3.141592653589793
#define SIGMA 0.1 // Initial width of Gaussian blur.
#define SIGMAFAC 1.2 // Increase sigma by this factor. Must be > 1.
#define SWAP(a,b) tempr=(a);(a)=(b);(b)=tempr;
#define TIMELIMIT 1e8 // Maximum time allowed in integrator.
#define TOLF 1e-3 // Sensitivity w. r. t. function value in newt2.
#define TOLINT 1e-3 // Sensitivity of the integrator.
#define TOLX 1e-3 // Sensitivity w. r. t. independent variables in newt2.
#define TRUE !FALSE
// Types
typedef int BOOLEAN;
typedef struct
float x;
float y;
} POINT;
// Globals
float bbmaxx,*bbmaxy,*bbminx,*bbminy,**gridvx,*gridvy,maxx,maxy,minpop,
minx,miny,polymaxx,polymaxy,polyminx,polyminy,*rho,**rho_0,**vx,**vy,*x,
*xappr,xstepsize,**y,*yappr,ystepsize;
int lx,ly,maxid,nblurs=0,npoly,npolycorn,*npolyinreg,nregcorn,nregion,
polygonid,**polyinreg,*regionid,*regionidinv,*within;
POINT *polycorn,*regcorn;
// Function prototypes
int readline(char line[],FILE *infile);
void countpoly(FILE *infile);
void countcorn(FILE *infile);
void readcorn(FILE *infile);
void makeregion(void);
void pspicture(FILE *outfile);
void bboxes(void);
void interior(void);
double regionarea(int ncrns,POINT *polygon);
void digdens(void);
void four1(float data[],unsigned long nn,int isign);
void realft(float data[],unsigned long n,int isign);
void cosft(float z[],unsigned long n,int isign);
void sinft(float z[],unsigned long n,int isign);
void coscosft(float **y,int isign1,int isign2);
void cossinft(float **y,int isign1,int isign2);
void sincosft(float **y,int isign1,int isign2);
float **dmatrix(long nrl,long nrh,long ncl,long nch);
float *d3tensor(long nrl, long nrh, long ncl, long nch, long ndl, long ndh);
void free_matrix(float **m,long nrl,long nrh,long ncl,long nch);
void free_f3tensor(float *t,long nrl,long nrh,long ncl,long nch,long ndl,
long ndh);
void fourn(float data[],unsigned long nn[],int ndim,int isign);
void rlft3(float *data,float **speq,unsigned long nn1,unsigned long nn2,
unsigned long nn3,int isign);
void gaussianblur(void);
void initcond(void);
void calcv(float t);
float intpol(float **arr,float x,float y);
BOOLEAN newt2(float h,float *xappr,float xguess,float *yappr,float yguess,
int j,int k);
BOOLEAN nonlinvoltra(void);
POINT transf(POINT p);
void cartogram(void);
// Function to read one line.
int readline(char line[],FILE *infile)
if (fgets(line,LINELENGTH,infile)==NULL) return 1;
return 0;
// Function to count the number of polygons.
void countpoly(FILE *infile)
char line[LINELENGTH];
npoly = 0;
while (!readline(line,infile)) if (line[0] == 'E') npoly++;
npoly--; // The .gen file ends with two consecutive "END"s.
// Function to count polygon corners. Also determines minimum/maximum x-/y-
// coordinate.
void countcorn(FILE *infile)
BOOLEAN bnd_ok = FALSE;
char line[LINELENGTH];
float x,y;
int polyctr=0,ratiolog;
npolycorn = (int*)calloc(npoly,sizeof(int));
polycorn = (POINT*)malloc(npoly*sizeof(POINT));
readline(line,infile); // Skip first line.
readline(line,infile);
sscanf(line,"%f %f",&x,&y);
polyminx = x; polymaxx = x; polyminy = y; polymaxy = y;
npolycorn[0] = 1;
while (!readline(line,infile))
if (line[0] != 'E')
sscanf(line,"%f %f",&x,&y);
if (x < polyminx) polyminx = x;
if (x > polymaxx) polymaxx = x;
if (y < polyminy) polyminy = y;
if (y > polymaxy) polymaxy = y;
npolycorn[polyctr]++;
else
readline(line,infile);
polycorn[polyctr] = (POINT)malloc(npolycorn[polyctr]sizeof(POINT));
polyctr++;
if (ceil(log((polymaxx-polyminx)/(polymaxy-polyminy))/log(2))+
floor(log((polymaxx-polyminx)/(polymaxy-polyminy))/log(2))>
2*log((polymaxx-polyminx)/(polymaxy-polyminy))/log(2))
ratiolog = (int)floor(log((polymaxx-polyminx)/(polymaxy-polyminy))/log(2));
else
ratiolog = (int)ceil(log((polymaxx-polyminx)/(polymaxy-polyminy))/log(2));
lx = (int)pow(2,(int)(0.5*(ratiolog+MAXNSQLOG)));
ly = (int)pow(2,(int)(0.5*(MAXNSQLOG-ratiolog)));
if ((polymaxx-polyminx)/lx > (polymaxy-polyminy)/ly)
maxx = 0.5((1PADDING)*polymaxx(1-PADDING)polyminx);
minx = 0.5((1-PADDING)*polymaxx(1PADDING)polyminx);
maxy = 0.5(polymaxypolyminy(maxx-minx)ly/lx);
miny = 0.5(polymaxy+polyminy-(maxx-minx)ly/lx);
else
maxy = 0.5((1PADDING)*polymaxy(1-PADDING)polyminy);
miny = 0.5((1-PADDING)*polymaxy(1PADDING)polyminy);
maxx = 0.5(polymaxxpolyminx(maxy-miny)lx/ly);
minx = 0.5(polymaxx+polyminx-(maxy-miny)lx/ly);
// Uncomment the next lines for interactive choice of boundary conditions.
/* printf("For the %i polygon(s) under consideration:\n",npoly);
printf("minimum x = %f\tmaximum x = %f\n",polyminx,polymaxx);
printf("minimum y = %f\tmaximum y = %f\n",polyminy,polymaxy);
while (!bnd_ok)
printf("Type in your choice of boundaries.\nminx = ");
scanf("%f",&minx);
printf("maxx = ");
scanf("%f",&maxx);
printf("miny = ");
scanf("%f",&miny);
printf("maxy = ");
scanf("%f",&maxy);
if (minx>polyminx || maxx<polymaxx || miny>polyminy || maxy<polymaxy)
printf("Invalid choice, does not enclose the polygon(s).\n");
else bnd_ok = TRUE;
// Function to read polygon corners. The first and last vertex of each polygon
// must be identical.
void readcorn(FILE *infile)
char line[LINELENGTH];
float xcoord,ycoord;
int i,id,polyctr=0;
polygonid = (int)malloc(npolysizeof(int));
xstepsize = (maxx-minx)/lx;
ystepsize = (maxy-miny)/ly;
if (fabs((xstepsize/ystepsize)-1)>1e-3)
fprintf(stderr,"WARNING: Area elements are not square: %f : %f\n",
xstepsize,ystepsize);
readline(line,infile);
sscanf(line,"%i",&id);
polygonid[polyctr] = id;
i = 0;
while (!readline(line,infile))
if (line[0] != 'E')
sscanf(line,"%f %f",&xcoord,&ycoord);
polycorn[polyctr].x = (xcoord-minx)/xstepsize;
polycorn[polyctr][i++].y = (ycoord-miny)/ystepsize;
else
// Is first and last vertex the same?
if (fabs(polycorn[polyctr][0].x-
polycorn[polyctr][npolycorn[polyctr]-1].x)+
fabs(polycorn[polyctr][0].y-
polycorn[polyctr][npolycorn[polyctr]-1].y)>1e-12)
fprintf(stderr,
"ERROR: %i-th polygon does not close upon itself.\n",
polyctr+1);
fprintf(stderr,"Identifier %i, first point %f %f.\n",
polygonid[polyctr],polycorn[polyctr][0].x*xstepsize+minx,
polycorn[polyctr][0].y*ystepsize+miny);
exit(1);
readline(line,infile);
sscanf(line,"%i",&id);
i = 0;
polyctr++;
if (polyctr<npoly) polygonid[polyctr] = id;
polyminx = (polyminx-minx)/xstepsize;
polyminy = (polyminy-miny)/ystepsize;
polymaxx = (polymaxx-minx)/xstepsize;
polymaxy = (polymaxy-miny)/ystepsize;
// Function to make regions from polygons.
void makeregion(void)
BOOLEAN repeat;
int i,lastid,minid,polyctr,ptctr,regctr;
// Count the number of regions.
nregion = 0;
maxid = minid = polygonid[0];
for (polyctr=0; polyctr<npoly; polyctr++)
if (polygonid[polyctr] == -99999) continue;
if (polygonid[polyctr]>maxid || polygonid[polyctr]<minid) nregion++;
else
repeat = FALSE;
for (i=0; i<polyctr; i++)
if (polygonid[polyctr]==polygonid)
repeat = TRUE;
break;
if (!repeat) nregion++;
if (polygonid[polyctr]>maxid) maxid = polygonid[polyctr];
if (polygonid[polyctr]<minid) minid = polygonid[polyctr];
if (minid < 0)
fprintf(stderr,
"ERROR: Negative region identifier %i.\n",minid);
exit(1);
// Match region identifiers.
regionid = (int)malloc(nregionsizeof(int));
nregion = 0;
maxid = minid = polygonid[0];
for (polyctr=0; polyctr<npoly; polyctr++)
if (polygonid[polyctr] == -99999) continue;
if (polygonid[polyctr]>maxid || polygonid[polyctr]<minid)
regionid[nregion++] = polygonid[polyctr];
else
repeat = FALSE;
for (i=0; i<polyctr; i++)
if (polygonid[polyctr]==polygonid)
repeat = TRUE;
break;
if (!repeat) regionid[nregion++] = polygonid[polyctr];
if (polygonid[polyctr]>maxid) maxid = polygonid[polyctr];
if (polygonid[polyctr]<minid) minid = polygonid[polyctr];
regionidinv = (int)malloc((maxid+1)sizeof(int));
// Negative number for unused identifiers.
for (i=0; i<=maxid; i++) regionidinv = -1;
for (regctr=0; regctr<nregion; regctr++)
regionidinv[regionid[regctr]] = regctr;
// Which polygons contribute to which regions?
npolyinreg = (int*)calloc(nregion,sizeof(int));
polyinreg = (int*)malloc(nregion*sizeof(int));
lastid = polygonid[0];
for (polyctr=0; polyctr<npoly; polyctr++)
if (polygonid[polyctr] != -99999)
npolyinreg[regionidinv[polygonid[polyctr]]]++;
lastid = polygonid[polyctr];
else npolyinreg[regionidinv[lastid]]++;
for (regctr=0; regctr<nregion; regctr++)
polyinreg[regctr] = (int)malloc(npolyinreg[regctr]sizeof(int));
for (regctr=0; regctr<nregion; regctr++) npolyinreg[regctr] = 0;
lastid = polygonid[0];
for (polyctr=0; polyctr<npoly; polyctr++)
if (polygonid[polyctr] != -99999)
polyinreg[regionidinv[polygonid[polyctr]]]
[npolyinreg[regionidinv[polygonid[polyctr]]]++] = polyctr;
lastid = polygonid[polyctr];
else polyinreg[regionidinv[lastid]][npolyinreg[regionidinv[lastid]]++]
= polyctr;
// Make regions from polygons. Start and end each polygon at (0,0).
nregcorn = (int*)calloc(nregion,sizeof(int));
regcorn = (POINT*)malloc(nregion*sizeof(POINT));
for (regctr=0; regctr<nregion; regctr++)
for (i=0; i<npolyinreg[regctr]; i++)
nregcorn[regctr] += npolycorn[polyinreg[regctr]]+1;
nregcorn[regctr]++;
for (regctr=0; regctr<nregion; regctr++)
regcorn[regctr] = (POINT)malloc(nregcorn[regctr]sizeof(POINT));
ptctr = 0;
regcorn[regctr][ptctr].x = regcorn[regctr][ptctr++].y = 0.0;
for (polyctr=0; polyctr<npolyinreg[regctr]; polyctr++)
for (i=0; i<npolycorn[polyinreg[regctr][polyctr]]; i++)
regcorn[regctr][ptctr++] = polycorn[polyinreg[regctr][polyctr]];
regcorn[regctr][ptctr].x = regcorn[regctr][ptctr++].y = 0.0;
// Function to prepare a map in postscript standard letter format.
void pspicture(FILE *outfile)
char line[LINELENGTH];
float addx,addy,b,conv,g,r;
int ptctr,regctr;
if (11*lx > 8.5*ly)
conv = (float)8.5*72/lx;
addx = 0;
addy = 1136-8.5*36ly/lx;
else
conv = (float)11*72/ly;
addx = 8.536-11*36lx/ly;
addy = 0;
fprintf(outfile,"0.5 setlinewidth\n");
for (regctr=0; regctr<nregion; regctr++)
fprintf(outfile,"newpath\n");
fprintf(outfile,"%f %f moveto\n",
regcorn[regctr][1].x*conv+addx,
regcorn[regctr][1].y*conv+addy);
for (ptctr=2; ptctr<nregcorn[regctr]; ptctr++)
if (fabs(regcorn[regctr][ptctr].x)+
fabs(regcorn[regctr][ptctr].y)>1e-12)
fprintf(outfile,"%f %f lineto\n",
regcorn[regctr][ptctr].x*conv+addx,
regcorn[regctr][ptctr].y*conv+addy);
else
fprintf(outfile,"closepath\n");
if (ptctr<nregcorn[regctr]-1)
ptctr++;
fprintf(outfile,"%f %f moveto\n",
regcorn[regctr][ptctr].x*conv+addx,
regcorn[regctr][ptctr].y*conv+addy);
// Determine colors for map (without better knowledge I will do it
// arbitrarily).
if (regctr%3 == 0)
r = (float)regctr/nregion;
g = 1-(float)regctr/nregion;
b = fabs(1-2*(float)regctr/nregion);
else if (regctr%3 == 1)
b = (float)regctr/nregion;
r = 1-(float)regctr/nregion;
g = fabs(1-2*(float)regctr/nregion);
else
g = (float)regctr/nregion;
b = 1-(float)regctr/nregion;
r = fabs(1-2*(float)regctr/nregion);
fprintf(outfile,"%f %f %f setrgbcolor\ngsave\nfill\n",r,g,b);
fprintf(outfile,"grestore\n0 setgray stroke\n");
fprintf(outfile,"showpage\n");
// Function to find the bounding box for each polygon.
void bboxes(void)
int i,j;
float maxx, minx, maxy, miny;
bbmaxx = (float)malloc(npolysizeof(float));
bbmaxy = (float)malloc(npolysizeof(float));
bbminx = (float)malloc(npolysizeof(float));
bbminy = (float)malloc(npolysizeof(float));
for (i = 0; i < npoly; i++)
maxx = polycorn[0].x;
maxy = polycorn[0].y;
minx = maxx;
miny = maxy;
for (j = 0; j < npolycorn; j++)
if (polycorn[j].x > maxx) maxx = polycorn[j].x;
if (polycorn[j].x < minx) minx = polycorn[j].x;
if (polycorn[j].y < miny) miny = polycorn[j].y;
if (polycorn[j].y > maxy) maxy = polycorn[j].y;
bbmaxx=maxx;
bbmaxy=maxy;
bbminx=minx;
bbminy=miny;
void interior(void)
int i,inhowmanyregions,inregion[2],j,k,l,m,n,regctr;
// Initialize within[][]. -1 means outside all regions.
for (i=0; i<=lx; i++) for (j=0; j<=ly; j++) within[j] = -1;
// Fill within[][].
for (i=0; i<nregion; i++) for (j=0; j<npolyinreg; j++)
for (k=0, n=npolycorn[polyinreg[j]]-1;
k<npolycorn[polyinreg[j]]; n=k++)
for (l=(int)ceil(MIN(polycorn[polyinreg[j]][k-1].y,
polycorn[polyinreg[j]][k].y));
l<MAX(polycorn[polyinreg[j]][k-1].y,
polycorn[polyinreg[j]][k].y); l++)
for (m=(int)floor(bbminx[polyinreg[j]]);
m<(polycorn[polyinreg[j]][n].x-
polycorn[polyinreg[j]][k].x)*
(l-polycorn[polyinreg[j]][k].y)/
(polycorn[polyinreg[j]][n].y-
polycorn[polyinreg[j]][k].y)+
polycorn[polyinreg[j]][k].x;
m++)
within[m][l] = i-within[m][l]-1;
// Function to determine polygon area. This is needed to determine the average
// population.
// The problem in short is to find the area of a polygon whose vertices are
// given. Recall Stokes' theorem in 3d for a vector field v:
// integral[around closed curve dA]v(x,y,z).ds =
// integral[over area A]curl(v).dA.
// Now let v(x,y,z) = (0,Q(x,y),0) and dA = (0,0,dx*dy). Then
// integral[around closed curve dA]Q(x,y)dy = integral[over area A]dQ/dxdxdy.
// If Q = x:
// A = integral[over area A]dx*dy = integral[around closed curve dA]x dy.
// For every edge from (x,y) to (x[i1],y[i1]) there is a
// parametrization
// (x(t),y(t)) = ((1-t)xtx[i+1],(1-t)y+ty[i1]), 0<t<1
// so that the path integral along this edge is
// int[from 0 to 1]{(1-t)xt*x[i+1]}(y[i1]-y)dt =
// 0.5(y[i1]-y)(x+x[i1]).
// Summing over all edges yields:
// Area = 0.5*[(x[0]+x[1])(y[1]-y[0]) + (x[1]+x[2])(y[2]-y[1]) + ...
// ...(x[n-1]+x[n])(y[n]-y[n-1])+(x[n]x[0])(y[0]-y[n])]
// ArcGIS treats a clockwise direction as positive, so there is a minus sign.
double regionarea(int ncrns,POINT *polygon)
double area=0;
int i;
for (i=0; i<ncrns-1; i++)
area -=
0.5(polygon.xpolygon[i+1].x)(polygon[i1].y-polygon.y);
return area -= 0.5(polygon[ncrns-1].x+polygon[0].x)
(polygon[0].y-polygon[ncrns-1].y);
// Function to digitize density.
void digdens(void)
char line[LINELENGTH];
double area,avgdens,*cases,dens,minpop=INFTY,ncases,totarea=0.0,totpop=0.0;
FILE* infile;
int i,id,ii,inhowmanyregions,inregion[2],j,jj,regctr;
if ((infile=fopen(CENSUSFILE))==NULL)
fprintf(stderr,"ERROR: Cannot find CENSUSFILE.\n");
exit(1);
// Find the minimum positive number of cases.
while (!readline(line,infile))
sscanf(line,"%i %lf",&id,&ncases);
if (ncases<minpop && ncases>1e-12) minpop = ncases;
fclose(infile);
// Store the number of cases in an array.
cases = (double)malloc(nregionsizeof(double));
for (i=0; i<nregion; i++) cases = -1.0;
infile = fopen(CENSUSFILE);
while (!readline(line,infile))
sscanf(line,"%i %lf",&id,&ncases);
if (id>maxid || regionidinv[id]<0)
fprintf(stderr,"ERROR: Identifier %i in CENSUSFILE does not\n",id);
fprintf(stderr,"match any identifier in generate file.\n");
exit(1);
if (ncases>1e-12) totpop += (cases[regionidinv[id]] = ncases);
else totpop += (cases[regionidinv[id]] = MINPOPFAC*minpop);
for (regctr=0; regctr<nregion; regctr++) if (cases[regctr] < 0.0)
fprintf(stderr,"ERROR: No density for region %i?\n",regionid[regctr]);
fprintf(stderr,"cases = %f\n",cases[regctr]);
exit(1);
fclose(infile);
// Calculate regions' areas, total area to be mapped, regional and average
// densities.
area = (double)malloc(nregionsizeof(double));
for (regctr=0; regctr<nregion; regctr++)
totarea += (area[regctr] = regionarea(nregcorn[regctr],regcorn[regctr]));
dens = (double)malloc(nregionsizeof(double));
for (regctr=0; regctr<nregion; regctr++)
dens[regctr] = cases[regctr]/area[regctr];
avgdens = totpop/totarea;
// Digitize density.
for (i=0; i<=lx; i++) for (j=0; j<=ly; j++) rho_0[j] = 0; // Initialize.
printf("digitizing density ...\n");
for (i=0; i<lx; i++) for (j=0; j<ly; j++)
if (within[j]==-1) rho_0[j] = avgdens;
else rho_0[j] = dens[within[j]];
// Fill the edges correctly.
rho_0[0][0] += rho_0[0][ly] + rho_0[lx][0] + rho_0[lx][ly];
for (i=1; i<lx; i++) rho_0[0] += rho_0[ly];
for (j=1; j<ly; j++) rho_0[0][j] += rho_0[lx][j];
for (i=0; i<lx; i++) rho_0[ly] = rho_0[0];
for (j=0; j<=ly; j++) rho_0[lx][j] = rho_0[0][j];
// Replace rho_0 by Fourier transform
coscosft(rho_0,1,1);
free(area);
free(cases);
for (i=0; i<npoly; i++) free(polycorn);
free(polycorn);
for (i=0; i<nregion; i++) free(regcorn);
free(regcorn);
free(dens);
free(npolycorn);
free(nregcorn);
free(polygonid);
free(regionid);
free(regionidinv);
free(bbmaxx);
free(bbmaxy);
free(bbminx);
free(bbminy);
for (i=0; i<nregion; i++) free(polyinreg);
free(polyinreg);
free(npolyinreg);
// Function to replace data[1...2*nn] by its discrete Fourier transform, if
// isign is input as 1; or replaces data[1...2*nn] by nn times its inverse
// discrete Fourier transform, if isign is input as -1. data is a complex array
// of length nn or, equivalently, a real array of length 2*nn. nn MUST be an
// integer power of 2 (this is not checked for!).
// From "Numerical Recipes in C".
void four1(float data[],unsigned long nn,int isign)
double theta,wi,wpi,wpr,wr,wtemp;
float tempi,tempr;
unsigned long i,istep,j,m,mmax,n;
n=nn<<1;
j=1;
for (i=1; i<n; i+=2)
if (j>i)
// This is the bit-reversal section of the routine.
SWAP(data[j],data);
SWAP(data[j1],data[i1]); // Exchange the two complex numbers.
m=n>>1;
while (m>=2 && j>m)
j -= m;
m>>=1;
j += m;
// Here begins the Danielson-Lanczos section of the routine.
mmax=2;
while (n>mmax) // Outer loop executed log_2 nn times.
istep = mmax<<1;
// Initialize the trigonometric recurrence.
theta = isign*(6.28318530717959/mmax);
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
wr = 1.0;
wi = 0.0;
for (m=1; m<mmax; m+=2) // Here are the two nested inner loops.
for (i=m; i<=n; i+=istep)
j=i+mmax; // This is the Danielson-Lanczos formula
tempr=wrdata[j]-widata[j+1];
tempi=wrdata[j1]widata[j];
data[j]=data-tempr;
data[j1]=data[i1]-tempi;
data += tempr;
data[i+1] += tempi;
wr = (wtemp=wr)wpr-wiwpi+wr; // Trigonometric recurrence.
wi = wiwprwtempwpiwi;
mmax=istep;
// Function to calculate the Fourier Transform of a set of n real-valued data
// points. It replaces this data (which is stored in array data[1...n]) by the
// positive frequency half of its complex Fourier Transform. The real-valued
// first and last components of the complex transform are returned as elements
// data[1] and data[2] respectively. n must be a power of 2. This routine also
// calculates the inverse transform of a complex data array if it is the
// transform of real data. (Result in this case must be multiplied by 2/n).
// From "Numerical Recipes in C".
void realft(float data[],unsigned long n,int isign)
double theta,wi,wpi,wpr,wr,wtemp;
float c1=0.5,c2,h1i,h1r,h2i,h2r;
unsigned long i,i1,i2,i3,i4,np3;
theta = 3.141592653589793/(double) (n>>1); // Initialize the recurrence
if (isign == 1)
c2 = -0.5;
four1(data,n>>1,1); // The forward transform is here.
else // Otherwise set up for an inverse transform.
c2 = 0.5;
theta = -theta;
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
wr = 1.0+wpr;
wi = wpi;
np3 = n+3;
for (i=2; i<=(n>>2); i++) // Case i=1 done separately below.
i4 = 1(i3=np3-(i2=1+(i1=ii-1)));
// The two separate transforms are separated out of data.
h1r = c1*(data[i1]+data[i3]);
h1i = c1*(data[i2]-data[i4]);
h2r = -c2*(data[i2]+data[i4]);
h2i = c2*(data[i1]-data[i3]);
// Here they are recombined to form the true transform of the original
// data.
data[i1] = h1r+wrh2r-wih2i;
data[i2] = h1iwrh2iwih2r;
data[i3] = h1r-wrh2r+wih2i;
data[i4] = -h1iwrh2iwih2r;
wr = (wtemp=wr)wpr-wiwpi+wr; // The recurrence.
wi = wiwprwtempwpiwi;
if (isign == 1)
data[1] = (h1r=data[1])+data[2]; // Squeeze the first and last data
// together to get them all within the original array.
data[2] = h1r-data[2];
else
data[1] = c1*((h1r=data[1])+data[2]);
data[2] = c1*(h1r-data[2]);
// This is the inverse transform for the case isign = -1.
four1(data,n>>1,-1);
// Function to calculate the cosine transform of a set z[0...n] of real-valued
// data points. The transformed data replace the original data in array z. n
// must be a power of 2. For forward transform set isign=1, for back transform
// isign = -1. (Note: The factor 2/n has been taken care of.)
// From "Numerical Recipes in C".
void cosft(float z[],unsigned long n,int isign)
double theta,wi=0.0,wpi,wpr,wr=1.0,wtemp;
float *a,sum,y1,y2;
int j,n2;
// Numerical Recipes starts counting at 1 which is rather confusing. I will
// count from 0.
a = (float)malloc((n+2)sizeof(float));
for (j=1; j<=n+1; j++) a[j] = z[j-1];
// Here is the Numerical Recipes code.
theta=PI/n; //Initialize the recurrence.
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
sum = 0.5*(a[1]-a[n+1]);
a[1] = 0.5*(a[1]a[n1]);
n2 = n+2;
for (j=2; j<=(n>>1); j++)
wr = (wtemp=wr)wpr-wiwpi+wr;
wi = wiwprwtempwpiwi;
y1 = 0.5*(a[j]+a[n2-j]);
y2 = (a[j]-a[n2-j]);
a[j] = y1-wi*y2;
a[n2-j] = y1+wi*y2;
sum += wr*y2;
realft(a,n,1);
a[n+1] = a[2];
a[2] = sum;
for (j=4; j<=n; j+=2)
sum += a[j];
a[j] = sum;
// Finally I revert to my counting method.
if (isign == 1) for (j=1; j<=n+1; j++) z[j-1] = a[j];
else if (isign == -1) for (j=1; j<=n+1; j++) z[j-1] = 2.0*a[j]/n;
free(a);
// Function to calculate the sine transform of a set of n real-valued data
// points stored in array z[0..n]. The number n must be a power of 2. On exit
// z is replaced by its transform. For forward transform set isign=1, for back
// transform isign = -1.
void sinft(float z[],unsigned long n,int isign)
double theta,wi=0.0,wpi,wpr,wr=1.0,wtemp;
float *a,sum,y1,y2;
int j;
unsigned long n2=n+2;
// See my comment about Numerical Recipe's counting above. Note that the last
// component plays a completely passive role and does not need to be stored.
a = (float*) malloc((n+1)*sizeof(float));
for (j=1; j<=n; j++) a[j] = z[j-1];
// Here is the Numerical Recipes code.
theta = PI/(double)n; // Initialize the recurrence.
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
a[1] = 0.0;
for (j=2; j<=(n>>1)+1; j++)
// Calculate the sine for the auxiliary array.
wr = (wtemp=wr)wpr-wiwpi+wr;
// The cosine is needed to continue the recurrence.
wi = wiwprwtempwpiwi;
// Construct the auxiliary array.
y1 = wi*(a[j]+a[n2-j]);
y2 = 0.5*(a[j]-a[n2-j]);
// Terms j and N-j are related.
a[j] = y1+y2;
a[n2-j] = y1-y2;
// Transform the auxiliary array.
realft(a,n,1);
// Initialize the sum used for odd terms below.
a[1] *= 0.5;
sum = a[2] = 0.0;
// Even terms determined directly. Odd terms determined by running sum.
for (j=1; j<=n-1; j+=2)
sum += a[j];
a[j] = a[j+1];
a[j+1] = sum;
// Change the indices.
if (isign == 1) for (j=1; j<=n; j++) z[j-1] = a[j];
else if (isign == -1) for (j=1; j<=n; j++) z[j-1] = 2.0*a[j]/n;
z[n] = 0.0;
free(a);
// Function to calculate a two-dimensional cosine Fourier transform. Forward/
// backward transform in x: isign1 = +/-1, in y: isign2 = +/-1.
void coscosft(float **y,int isign1,int isign2)
float temp[lx+1];
unsigned long i,j;
for (i=0; i<=lx; i++)
cosft(y,ly,isign2);
for (j=0; j<=ly; j++)
for (i=0; i<=lx; i++) temp=y[j];
cosft(temp,lx,isign1);
for (i=0; i<=lx; i++) y[j]=temp;
// Function to calculate a cosine Fourier transform in x and a sine transform
// in y. Forward/backward transform in x: isign1 = +/-1, in y: isign2 = +/-1.
void cossinft(float **y,int isign1,int isign2)
float temp[lx+1];
unsigned long i,j;
for (i=0; i<=lx; i++)
sinft(y,ly,isign2);
for (j=0; j<=ly; j++)
for (i=0; i<=lx; i++) temp=y[j];
cosft(temp,lx,isign1);
for (i=0; i<=lx; i++) y[j]=temp;
// Function to calculate a sine Fourier transform in x and a cosine transform
// in y. Forward/backward transform in x: isign1 = +/-1, in y: isign2 = +/-1.
void sincosft(float **y,int isign1,int isign2)
float temp[lx+1];
unsigned long i,j;
for (i=0; i<=lx; i++)
cosft(y,ly,isign2);
for (j=0; j<=ly; j++)
for (i=0; i<=lx; i++) temp=y[j];
sinft(temp,lx,isign1);
for (i=0; i<=lx; i++) y[j]=temp;
// Function to allocate a float matrix with subscript range
// m[nrl..nrh][ncl..nch]. From "Numerical Recipes in C".
float **dmatrix(long nrl,long nrh,long ncl,long nch)
long i, nrow=nrh-nrl1,ncol=nch-ncl1;
float **m;
/* allocate pointers to rows */
m=(float **) malloc((unsigned int)((nrow+NR_END)sizeof(float)));
if (!m)
fprintf(stderr,"allocation failure 1 in matrix()\n");
exit(1);
m += NR_END;
m -= nrl;
/* allocate rows and set pointers to them */
m[nrl]=(float *) malloc((unsigned int)((nrowncol+NR_END)sizeof(float)));
if (!m[nrl])
fprintf(stderr,"allocation failure 2 in matrix()\n");
exit(1);
m[nrl] += NR_END;
m[nrl] -= ncl;
for(i=nrl1;i<=nrh;i+) m=m[i-1]+ncol;
/* return pointer to array of pointers to rows */
return m;
// Function to allocate a float 3tensor with range
// t[nrl..nrh][ncl..nch][ndl..ndh]. From "Numerical Recipes in C".
float *d3tensor(long nrl, long nrh, long ncl, long nch, long ndl, long ndh)
long i,j,nrow=nrh-nrl1,ncol=nch-ncl+1,ndep=ndh-ndl1;
float *t;
/* allocate pointers to pointers to rows */
t=(float *) malloc((sizet)((nrow+NREND)sizeof(float*)));
if (!t)
fprintf(stderr,"allocation failure 1 in f3tensor()\n");
exit(1);
t += NR_END;
t -= nrl;
/* allocate pointers to rows and set pointers to them */
t[nrl]=(float **) malloc((sizet)((nrowncol+NREND)*sizeof(float)));
if (!t[nrl])
fprintf(stderr,"allocation failure 2 in f3tensor()\n");
exit(1);
t[nrl] += NR_END;
t[nrl] -= ncl;
/* allocate rows and set pointers to them */
t[nrl][ncl]=(float *) malloc((sizet)((nrowncol*ndep+NREND)sizeof(float)));
if (!t[nrl][ncl])
fprintf(stderr,"allocation failure 3 in f3tensor()\n");
exit(1);
t[nrl][ncl] += NR_END;
t[nrl][ncl] -= ndl;
for(j=ncl1;j<=nch;j+) t[nrl][j]=t[nrl][j-1]+ndep;
for(i=nrl1;i<=nrh;i+) {
t=t[i-1]+ncol;
t[ncl]=t[i-1][ncl]+ncol*ndep;
for(j=ncl1;j<=nch;j+) t[j]=t[j-1]+ndep;
/* return pointer to array of pointers to rows */
return t;
void free_matrix(float **m,long nrl,long nrh,long ncl,long nch)
free((char*) (m[nrl]+ncl-1));
free((char*) (m+nrl-1));
void free_f3tensor(float *t,long nrl,long nrh,long ncl,long nch,long ndl,
long ndh)
free((char*) (t[nrl][ncl]+ndl-1));
free((char*) (t[nrl]+ncl-1));
free((char*) (t+nrl-1));
// Function to replace data by its ndim-dimensional discrete Fourier transform,
// if isign is input as 1. nn[1..ndim] is an integer array containing the
// lengths of each dimension (number of complex values), which MUST be all
// powers of 2. data is a real array of length twice the product of these
// lengths, in which the data are stored as in a multidimensional complex
// array: real and imaginary parts of each element are in consecutive
// locations, and the rightmost index of the array increases most rapidly as
// one proceeds along data. For a two-dimensional array, this is equivalent to
// storing the arrays by rows. If isign is input as -1, data is replaced by its
// inverse transform times the product of the lengths of all dimensions.
void fourn(float data[],unsigned long nn[],int ndim,int isign)
int idim;
unsigned long i1,i2,i3,i2rev,i3rev,ip1,ip2,ip3,ifp1,ifp2;
unsigned long ibit,k1,k2,n,nprev,nrem,ntot;
double tempi,tempr;
float theta,wi,wpi,wpr,wr,wtemp;
for (ntot=1, idim=1; idim<=ndim; idim++)
ntot *= nn[idim];
nprev = 1;
for (idim=ndim; idim>=1; idim--)
n = nn[idim];
nrem = ntot/(n*nprev);
ip1=nprev << 1;
ip2 = ip1*n;
ip3 = ip2*nrem;
i2rev = 1;
for (i2=1; i2<=ip2; i2+=ip1)
if (i2 < i2rev)
for (i1=i2; i1<=i2+ip1-2; i1+=2)
for (i3=i1; i3<=ip3; i3+=ip2)
i3rev = i2rev+i3-i2;
SWAP(data[i3],data[i3rev]);
SWAP(data[i31],data[i3rev1]);
ibit = ip2>>1;
while (ibit>=ip1 && i2rev>ibit)
i2rev -= ibit;
ibit >>= 1;
i2rev += ibit;
ifp1 = ip1;
while (ifp1 < ip2)
ifp2 = ifp1 << 1;
theta = 2isignPI/(ifp2/ip1);
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
wr = 1.0;
wi = 0.0;
for (i3=1; i3<=ifp1; i3+=ip1)
for (i1=i3; i1<=i3+ip1-2; i1+=2)
for (i2=i1; i2<=ip3; i2+=ifp2)
k1 = i2;
k2 = k1+ifp1;
tempr = (float)wrdata[k2]-(float)widata[k2+1];
tempi = (float)wrdata[k21](float)widata[k2];
data[k2] = data[k1]-tempr;
data[k2+1] = data[k1+1]-tempi;
data[k1] += tempr;
data[k1+1] += tempi;
wr = (wtemp=wr)wpr-wiwpi+wr;
wi = wiwprwtempwpiwi;
ifp1 = ifp2;
nprev *= n;
// Function to calculate a three-dimensional Fourier transform of
// data[1..nn1][1..nn2][1..nn3] (where nn1=1 for the case of a logically two-
// dimensional array). This routine returns (for isign=1) the complex fast
// Fourier transform as two complex arrays: On output, data contains the zero
// and positive frequency values of the third frequency component, while
// speq[1..nn1][1..2*nn2] contains the Nyquist critical frequency values of the
// third frequency component. First (and second) frequency components are
// stored for zero, positive, and negative frequencies, in standard wrap-around
// order. See Numerical Recipes for description of how complex values are
// arranged. For isign=-1, the inverse transform (times nn1nn2nn3/2 as a
// constant multiplicative factor) is performed, with output data (viewed as
// real array) deriving from input data (viewed as complex) and speq. For
// inverse transforms on data not generated first by a forward transform, make
// sure the complex input data array satisfies property 12.5.2 from NR. The
// dimensions nn1, nn2, nn3 must always be integer powers of 2.
void rlft3(float *data,float **speq,unsigned long nn1,unsigned long nn2,
unsigned long nn3,int isign)
double theta,wi,wpi,wpr,wr,wtemp;
float c1,c2,h1r,h1i,h2r,h2i;
unsigned long i1,i2,i3,j1,j2,j3,nn[4],ii3;
if (1+&data[nn1][nn2][nn3]-&data[1][1][1] != nn1nn2nn3)
fprintf(stderr,
"rlft3: problem with dimensions or contiguity of data array\n");
exit(1);
c1 = 0.5;
c2 = -0.5*isign;
theta = 2isign(PI/nn3);
wtemp = sin(0.5*theta);
wpr = -2.0wtempwtemp;
wpi = sin(theta);
nn[1] = nn1;
nn[2] = nn2;
nn[3] = nn3 >> 1;
// Case of forward transform. Here is where most all of the compute time is
// spent. Extend data periodically into speq.
if (isign == 1)
fourn(&data[1][1][1]-1,nn,3,isign);
for (i1=1; i1<=nn1; i1++)
for (i2=1, j2=0; i2<=nn2; i2++)
speq[i1][++j2] = data[i1][i2][1];
speq[i1][++j2] = data[i1][i2][2];
for (i1=1; i1<=nn1; i1++)
// Zero frequency is its own reflection; otherwise locate corresponding
// negative frequency in wrap-around order.
j1 = (i1 != 1 ? nn1-i1+2 : 1);
// Initialize trigonometric recurrence.
wr = 1.0;
wi = 0.0;
for (ii3=1, i3=1; i3<=(nn3>>2)+1; i3+,ii3=2)
for (i2=1; i2<=nn2; i2++)
if (i3 == 1)
j2 = (i2 != 1 ? ((nn2-i2)<<1)+3 : 1);
h1r = c1*(data[i1][i2][1]+speq[j1][j2]);
h1i = c1*(data[i1][i2][2]-speq[j1][j2+1]);
h2i = c2*(data[i1][i2][1]-speq[j1][j2]);
h2r = -c2*(data[i1][i2][2]speq[j1][j21]);
data[i1][i2][1] = h1r+h2r;
data[i1][i2][2] = h1i+h2i;
speq[j1][j2] = h1r-h2r;
speq[j1][j2+1] = h2i-h1i;
else
j2 = (i2 != 1 ? nn2-i2+2 : 1);
j3 = nn3+3-(i3<<1);
h1r = c1*(data[i1][i2][ii3]+data[j1][j2][j3]);
h1i = c1*(data[i1][i2][ii31]-data[j1][j2][j31]);
h2i = c2*(data[i1][i2][ii3]-data[j1][j2][j3]);
h2r = -c2*(data[i1][i2][ii31]+data[j1][j2][j31]);
data[i1][i2][ii3] = h1r+wrh2r-wih2i;
data[i1][i2][ii3+1] = h1iwrh2iwih2r;
data[j1][j2][j3] = h1r-wrh2r+wih2i;
data[j1][j2][j3+1] = -h1iwrh2iwih2r;
// Do the recurrence.
wr = (wtemp=wr)wpr-wiwpi+wr;
wi = wiwprwtempwpiwi;
// Case of reverse transform.
if (isign == -1)
fourn(&data[1][1][1]-1,nn,3,isign);
// Function to perform Gaussian blur.
void gaussianblur(void)
float **blur,***conv,***pop,**speqblur,**speqconv,*speqpop;
int i,j,p,q;
blur = d3tensor(1,1,1,lx,1,ly);
conv = d3tensor(1,1,1,lx,1,ly);
pop = d3tensor(1,1,1,lx,1,ly);
speqblur = dmatrix(1,1,1,2*lx);
speqconv = dmatrix(1,1,1,2*lx);
speqpop = dmatrix(1,1,1,2*lx);
// Fill population and convolution matrix.
for (i=1; i<=lx; i++) for (j=1; j<=ly; j++)
if (i > lx/2) p = i-1-lx;
else p = i-1;
if (j > ly/2) q = j-1-ly;
else q = j-1;
pop[1][j] = rho_0[i-1][j-1];
conv[1][j] = 0.5*
(erf((p+0.5)/(sqrt(2.0)(SIGMApow(SIGMAFAC,nblurs))))-
erf((p-0.5)/(sqrt(2.0)(SIGMA*pow(SIGMAFAC,nblurs)))))
(erf((q+0.5)/(sqrt(2.0)(SIGMApow(SIGMAFAC,nblurs))))-
erf((q-0.5)/(sqrt(2.0)(SIGMA*pow(SIGMAFAC,nblurs)))))/(lxly);
// Fourier transform.
rlft3(pop,speqpop,1,lx,ly,1);
rlft3(conv,speqconv,1,lx,ly,1);
// Multiply pointwise.
for (i=1; i<=lx; i++)
for (j=1; j<=ly/2; j++)
blur[1][2*j-1] =
pop[1][2j-1]*conv[1][2j-1]-
pop[1][2j]*conv[1][2j];
blur[1][2*j] =
pop[1][2j]*conv[1][2j-1]+
pop[1][2j-1]*conv[1][2j];
for (i=1; i<=lx; i++)
speqblur[1][2*i-1] =
speqpop[1][2i-1]*speqconv[1][2i-1]-
speqpop[1][2i]*speqconv[1][2i];
speqblur[1][2*i] =
speqpop[1][2i]*speqconv[1][2i-1]+
speqpop[1][2i-1]*speqconv[1][2i];
// Backtransform.
rlft3(blur,speqblur,1,lx,ly,-1);
// Write to rho_0.
for (i=1; i<=lx; i++) for (j=1; j<=ly; j++) rho_0[i-1][j-1] = blur[1][j];
free_f3tensor(blur,1,1,1,lx,1,ly);
free_f3tensor(conv,1,1,1,lx,1,ly);
free_f3tensor(pop,1,1,1,lx,1,ly);
free_matrix(speqblur,1,1,1,2*lx);
free_matrix(speqconv,1,1,1,2*lx);
free_matrix(speqpop,1,1,1,2*lx);
// Function to initialize rho_0. The original density is blurred with width
// SIGMA*pow(SIGMAFAC,nblurs).
void initcond(void)
float maxpop;
int i,j;
// Reconstruct population density.
coscosft(rho_0,-1,-1);
// There must not be negative densities.
for (i=0; i<lx; i++) for (j=0; j<ly; j++) if (rho_0[j]<-1e10)
fprintf(stderr,"ERROR: Negative density in DENSITYFILE.\n");
exit(1);
// Perform Gaussian blur.
printf("Gaussian blur ...\n");
gaussianblur();
// Find the mimimum density. If it is very small suggest an increase in
// SIGMA.
minpop = rho_0[0][0];
maxpop = rho_0[0][0];
for (i=0; i<lx; i++) for (j=0; j<ly; j++) if (rho_0[j]<minpop)
minpop = rho_0[j];
for (i=0; i<lx; i++) for (j=0; j<ly; j++) if (rho_0[j]>maxpop)
maxpop = rho_0[j];
if (0<minpop && minpop<1e-8*maxpop)
fprintf(stderr,"Minimimum population very small (%f). Integrator\n",
minpop);
fprintf(stderr,
"will probably converge very slowly. You can speed up the\n");
fprintf(stderr,"process by increasing SIGMA to a value > %f.\n",
SIGMA*pow(SIGMAFAC,nblurs));
// Replace rho_0 by cosine Fourier transform in both variables.
coscosft(rho_0,1,1);
// Function to calculate the velocity field
void calcv(float t)
int j,k;
// Fill rho with Fourier coefficients.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
rho[j][k] = exp(-((PIj/lx)*(PI*j/lx)+(PI*k/ly)*(PI*k/ly))*t)rho_0[j][k];
// Calculate the Fourier coefficients for the partial derivative of rho.
// Store temporary results in arrays gridvx, gridvy.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
gridvx[j][k] = -(PIj/lx)rho[j][k];
gridvy[j][k] = -(PIk/ly)rho[j][k];
// Replace rho by cosine Fourier backtransform in both variables.
coscosft(rho,-1,-1);
// Replace vx by sine Fourier backtransform in the first and cosine Fourier
// backtransform in the second variable.
sincosft(gridvx,-1,-1);
// Replace vy by cosine Fourier backtransform in the first and sine Fourier
// backtransform in the second variable.
cossinft(gridvy,-1,-1);
// Calculate the velocity field.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
gridvx[j][k] = -gridvx[j][k]/rho[j][k];
gridvy[j][k] = -gridvy[j][k]/rho[j][k];
// Function to bilinearly interpolate a two-dimensional array. For higher
// accuracy one could consider higher order interpolation schemes.
float intpol(float **arr,float x,float y)
int gaussx,gaussy;
float deltax,deltay;
// Decompose x and y into an integer part and a decimal.
gaussx = (int)x;
gaussy = (int)y;
deltax = x-gaussx;
deltay = y-gaussy;
// Interpolate.
if (gaussx==lx && gaussy==ly)
return arr[gaussx][gaussy];
if (gaussx==lx)
return (1-deltay)arr[gaussx][gaussy]deltayarr[gaussx][gaussy1];
if (gaussy==ly)
return (1-deltax)arr[gaussx][gaussy]deltaxarr[gaussx1][gaussy];
return (1-deltax)(1-deltay)arr[gaussx][gaussy]+
(1-deltax)deltayarr[gaussx][gaussy1]
deltax(1-deltay)arr[gaussx1][gaussy]
deltaxdeltayarr[gaussx1][gaussy1];
// Function to find the root of the system of equations
// xappr-0.5h*v_x(t+h,xappr,yappr)-x[j][k]-0.5*hvx[j][k]=0,
// yappr-0.5h*v_y(t+h,xappr,yappr)-y[j][k]-0.5*hvy[j][k]=0
// with Newton-Raphson. Returns TRUE after sufficient convergence.
BOOLEAN newt2(float h,float *xappr,float xguess,float *yappr,float yguess,
int j,int k)
float deltax,deltay,dfxdx,dfxdy,dfydx,dfydy,fx,fy;
int gaussx,gaussxplus,gaussy,gaussyplus,i;
// Initial guess.
*xappr = xguess;
*yappr = yguess;
for (i=1; i<=IMAX; i++)
// fx, fy are the left-hand sides of the two equations. Find
// v_x(t+h,xappr,yappr), v_y(t+h,xappr,yappr) by interpolation.
fx = xappr-0.5*h*intpol(gridvx,*xappr,*yappr)-x[j][k]-0.5*hvx[j][k];
fy = yappr-0.5*h*intpol(gridvy,*xappr,*yappr)-y[j][k]-0.5*hvy[j][k];
// Linearly approximate the partial derivatives of fx, fy with a finite
// difference method. More elaborate techniques are possible, but this
// quick and dirty method appears to work reasonably for our purpose.
gaussx = (int)(*xappr);
gaussy = (int)(*yappr);
if (gaussx == lx) gaussxplus = 0;
else gaussxplus = gaussx+1;
if (gaussy == ly) gaussyplus = 0;
else gaussyplus = gaussy+1;
deltax = x[j][k] - gaussx;
deltay = y[j][k] - gaussy;
dfxdx = 1 - 0.5h
((1-deltay)*(gridvx[gaussxplus][gaussy]-gridvx[gaussx][gaussy])+
deltay*(gridvx[gaussxplus][gaussyplus]-gridvx[gaussx][gaussyplus]));
dfxdy = -0.5h
((1-deltax)*(gridvx[gaussx][gaussyplus]-gridvx[gaussx][gaussy])+
deltax*(gridvx[gaussxplus][gaussyplus]-gridvx[gaussxplus][gaussy]));
dfydx = -0.5h
((1-deltay)*(gridvy[gaussxplus][gaussy]-gridvy[gaussx][gaussy])+
deltay*(gridvy[gaussxplus][gaussyplus]-gridvy[gaussx][gaussyplus]));
dfydy = 1 - 0.5h
((1-deltax)*(gridvy[gaussx][gaussyplus]-gridvy[gaussx][gaussy])+
deltax*(gridvy[gaussxplus][gaussyplus]-gridvy[gaussxplus][gaussy]));
// If the current approximation is (xappr,yappr) for the zero of
// (fx(x,y),fy(x,y)) and J is the Jacobian, then we can approximate (in
// vector notation) for |delta|<<1:
// f((xappr,yappr)+delta) = f(xappr,yappr)+J*delta.
// Setting f((xappr,yappr)+delta)=0 we obtain a set of linear equations
// for the correction delta which moves f closer to zero, namely
// J*delta = -f.
// The improved approximation is then x = xappr+delta.
// The process will be iterated until convergence is reached.
if ((fx*fx + fy*fy) < TOLF) return TRUE;
deltax = (fy*dfxdy - fxdfydy)/(dfxdxdfydy - dfxdy*dfydx);
deltay = (fx*dfydx - fydfxdx)/(dfxdxdfydy - dfxdy*dfydx);
if ((deltax*deltax + deltay*deltay) < TOLX) return TRUE;
*xappr += deltax;
*yappr += deltay;
//printf("deltax %f, deltay %f\n",deltax,deltay);
fprintf(stderr,"newt2 failed, increasing sigma to %f.\n",
SIGMA*pow(SIGMAFAC,nblurs));
return FALSE;
// Function to integrate the nonlinear Volterra equation. Returns TRUE after
// the displacement field converged, after MAXINTSTEPS integration steps, or
// if the time exceeds TIMELIMIT.
BOOLEAN nonlinvoltra(void)
BOOLEAN stepsize_ok;
#ifdef DISPLFILE
FILE *displfile = fopen(DISPLFILE);
#endif
float h,maxchange=INFTY,t,vxplus,vyplus,xguess,yguess;
int i,j,k;
do
initcond();
nblurs++;
if (minpop<0.0)
fprintf(stderr,
"Minimum population negative, will increase sigma to %f\n",
SIGMA*pow(SIGMAFAC,nblurs));
while (minpop<0.0);
h = HINITIAL;
t = 0; // Start at time t=0.
// (x[j][k],y[j][k]) is the position for the element that was at position
// (j,k) at time t=0.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
x[j][k] = j;
y[j][k] = k;
calcv(0.0);
// (gridvx[j][k],gridvy[j][k]) is the velocity at position (j,k).
// (vx[j][k],vy[j][k]) is the velocity at position (x[j][k],y[j][k]).
// At t=0 they are of course identical.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
vx[j][k] = gridvx[j][k];
vy[j][k] = gridvy[j][k];
i = 1; // i counts the integration steps.
// Here is the integrator.
do
stepsize_ok = TRUE;
calcv(t+h);
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
// First take a naive integration step. The velocity at time t+h for
// the element [j][k] is approximately
// v(th,x[j][k]+hvx[j][k],y[j][k]hvy[j][k]).
// The components, call them vxplus and vyplus, are interpolated from
// gridvx and gridvy.
vxplus = intpol(gridvx,x[j][k]hvx[j][k],y[j][k]hvy[j][k]);
vyplus = intpol(gridvy,x[j][k]hvx[j][k],y[j][k]hvy[j][k]);
// Based on (vx[j][k],vy[j][k]) and (vxplus,vyplus) we expect the
// new position at time t+h to be:
xguess = x[j][k] + 0.5h(vx[j][k]+vxplus);
yguess = y[j][k] + 0.5h(vy[j][k]+vyplus);
// Then we make a better approximation by solving the two nonlinear
// equations:
// xappr[j][k]-0.5hv_x(t+h,xappr[j][k],yappr[j][k])-
// x[j][k]-0.5hvx[j][k]=0,
// yappr[j][k]-0.5hv_y(t+h,xappr[j][k],yappr[j][k])-
// y[j][k]-0.5hvy[j][k]=0
// with Newton-Raphson and (xguess,yguess) as initial guess.
// If newt2 fails to converge, exit nonlinvoltra.
if (!newt2(h,&xappr[j][k],xguess,&yappr[j][k],yguess,j,k))
return FALSE;
// If the integration step was too large reduce the step size.
if ((xguess-xappr[j][k])*(xguess-xappr[j][k])+
(yguess-yappr[j][k])*(yguess-yappr[j][k]) > TOLINT)
if (h<MINH)
fprintf(stderr,
"Time step below %f, increasing SIGMA to %f\n",
h,SIGMA*pow(SIGMAFAC,nblurs));
nblurs++;
return FALSE;
h /= 10;
stepsize_ok = FALSE;
break;
if (!stepsize_ok) continue;
else
t += h;
maxchange = 0.0; // Monitor the maximum change in positions.
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
if ((x[j][k]-xappr[j][k])*(x[j][k]-xappr[j][k])+
(y[j][k]-yappr[j][k])*(y[j][k]-yappr[j][k]) > maxchange)
maxchange =
(x[j][k]-xappr[j][k])*(x[j][k]-xappr[j][k])+
(y[j][k]-yappr[j][k])*(y[j][k]-yappr[j][k]);
x[j][k] = xappr[j][k];
y[j][k] = yappr[j][k];
vx[j][k] = intpol(gridvx,xappr[j][k],yappr[j][k]);
vy[j][k] = intpol(gridvy,xappr[j][k],yappr[j][k]);
h *= 1.2; // Make the next integration step larger.
if (i%10==0) printf("time %f\n",t);
i++;
} while (i<MAXINTSTEPS && t<TIMELIMIT && maxchange>CONVERGENCE);
if (maxchange>CONVERGENCE)
fprintf(stderr,
"WARNING: Insufficient convergence within %i steps, time %f.\n",
MAXINTSTEPS,TIMELIMIT);
#ifdef DISPLFILE
// Write displacement field to file.
fprintf(displfile,"time %f\nminx %f\nmaxx %f\nminy %f\nmaxy %f\n",
t,minx,maxx,miny,maxy);
fprintf(displfile,"sigma %f\n",SIGMA*pow(SIGMAFAC,nblurs-1));
fprintf(displfile,"background %f\nlx\nly\n\n",0,lx,ly);
for (j=0; j<=lx; j++) for (k=0; k<=ly; k++)
fprintf(displfile,"j %i, k %i, x %f, y %f\n",j,k,x[j][k],y[j][k]);
fclose(displfile);
#endif
return TRUE;
// Function to transform points according to displacement field.
POINT transf(POINT p)
float deltax,deltay,den,t,u;
int gaussx,gaussy;
POINT a,b,c,d,ptr;
p.x = (p.x-minx)*lx/(maxx-minx);
p.y = (p.y-miny)*ly/(maxy-miny);
gaussx = (int)p.x;
gaussy = (int)p.y;
if (gaussx<0 || gaussx>lx || gaussy<0 || gaussy>ly)
fprintf(stderr,"ERROR: Coordinate limits exceeded in transf.\n");
exit(1);
deltax = p.x - gaussx;
deltay = p.y - gaussy;
// The transformed point is the intersection of the lines:
// (I) connecting
// (1-deltax)(x,y[gaussx][gaussy])deltax(x,y[gaussx1][gaussy])
// and
// (1-deltax)(x,y[gaussx][gaussy1])+deltax(x,y[gaussx+1][gaussy1])
// (II) connecting
// (1-deltay)(x,y[gaussx][gaussy])deltay(x,y[gaussx][gaussy1])
// and
// (1-deltay)(x,y[gaussx1][gaussy])+deltay(x,y[gaussx+1][gaussy1]).
// Call these four points a, b, c and d.
a.x = (1-deltax)*x[gaussx][gaussy] + deltax*x[gaussx+1][gaussy];
a.y = (1-deltax)*y[gaussx][gaussy] + deltax*y[gaussx+1][gaussy];
b.x = (1-deltax)*x[gaussx][gaussy+1] + deltax*x[gaussx1][gaussy1];
b.y = (1-deltax)*y[gaussx][gaussy+1] + deltax*y[gaussx1][gaussy1];
c.x = (1-deltay)*x[gaussx][gaussy] + deltay*x[gaussx][gaussy+1];
c.y = (1-deltay)*y[gaussx][gaussy] + deltay*y[gaussx][gaussy+1];
d.x = (1-deltay)*x[gaussx+1][gaussy] + deltay*x[gaussx1][gaussy1];
d.y = (1-deltay)*y[gaussx+1][gaussy] + deltay*y[gaussx1][gaussy1];
// Solve the vector equation a+t(b-a) = c+u(d-c) for the scalars t, u.
if (fabs(den=(b.x-a.x)(c.y-d.y)+(a.y-b.y)(c.x-d.x))<1e-12)
fprintf(stderr,"ERROR: Transformed area element has parallel edges.\n");
exit(1);
t = ((c.x-a.x)(c.y-d.y)+(a.y-c.y)(c.x-d.x))/den;
u = ((b.x-a.x)(c.y-a.y)+(a.y-b.y)(c.x-a.x))/den;
if (t<-1e-3|| t>1+1e-3 || u<-1e-3 || u>1+1e-3)
fprintf(stderr,"WARNING: Transformed area element non-convex.\n");
ptr.x = (1-(a.x+t(b.x-a.x))/lx)minx + ((a.x+t(b.x-a.x))/lx)maxx;
ptr.y = (1-(a.y+t(b.y-a.y))/ly)miny + ((a.y+t(b.y-a.y))/ly)maxy;
return ptr;
// Function to read spatial features from user-specified file and map to
// cartogram.
void cartogram(void)
char id[LINELENGTH],line[LINELENGTH];
FILE infile,outfile;
float xcoord,ycoord;
POINT p;
infile = fopen(MAPGENFILE);
outfile = fopen(CARTGENFILE,"w");
while (!readline(line,infile))
if (sscanf(line,"%s %f %f",&id,&xcoord,&ycoord)==3)
p.x = xcoord;
p.y = ycoord;
p = transf(p);
fprintf(outfile,"%s %f %f\n",id,p.x,p.y);
else if (sscanf(line,"%f %f",&xcoord,&ycoord)==2)
p.x = xcoord;
p.y = ycoord;
p = transf(p);
fprintf(outfile,"%f %f\n",p.x,p.y);
else
sscanf(line,"%s",&id);
fprintf(outfile,"%s\n",id);
fclose(infile);
fclose(outfile);
main(void)
BOOLEAN n;
char c;
FILE genfile,psfile = fopen(MAP2PS);
float oldlx,oldly,oldmaxx,oldmaxy,oldminx,oldminy,totarea;
int i,polyctr,regctr;
// Read the polygon coordinates.
if ((genfile = fopen(MAPGENFILE)) == NULL)
fprintf(stderr,"ERROR: Cannot find MAPGENFILE\n");
exit(1);
countpoly(genfile);
fclose(genfile);
genfile = fopen(MAPGENFILE);
countcorn(genfile);
fclose(genfile);
genfile = fopen(MAPGENFILE);
readcorn(genfile);
fclose(genfile);
makeregion();
printf("%i polygon(s), %i region(s)\n",npoly,nregion);
printf("lx=%i, ly=%i\n",lx,ly);
// Calculate total area.
//totarea = 0.0;
//for (regctr=0; regctr<nregion; regctr++)
// printf("region %i has area %f, contains %i polygons\n",regionid[regctr],
// regionarea(nregcorn[regctr],regcorn[regctr]),npolyinreg[regctr]);
// totarea += regionarea(nregcorn[regctr],regcorn[regctr]);
//printf("totarea = %f\n",totarea);
// Make map.
pspicture(psfile);
fclose(psfile);
// Allocate memory for arrays.
gridvx = (float*)malloc((lx+1)*sizeof(float));
gridvy = (float*)malloc((lx+1)*sizeof(float));
rho = (float*)malloc((lx+1)*sizeof(float));
rho_0 = (float*)malloc((lx+1)*sizeof(float));
vx = (float*)malloc((lx+1)*sizeof(float));
vy = (float*)malloc((lx+1)*sizeof(float));
x = (float*)malloc((lx+1)*sizeof(float));
xappr = (float*)malloc((lx+1)*sizeof(float));
y = (float*)malloc((lx+1)*sizeof(float));
yappr = (float*)malloc((lx+1)*sizeof(float));
within = (int*)malloc((lx+1)*sizeof(int));
for (i=0; i<=lx; i++)
gridvx = (float)malloc((ly+1)sizeof(float));
gridvy = (float)malloc((ly+1)sizeof(float));
rho = (float)malloc((ly+1)sizeof(float));
rho_0 = (float)malloc((ly+1)sizeof(float));
vx = (float)malloc((ly+1)sizeof(float));
vy = (float)malloc((ly+1)sizeof(float));
x = (float)malloc((ly+1)sizeof(float));
xappr = (float)malloc((ly+1)sizeof(float));
y = (float)malloc((ly+1)sizeof(float));
yappr = (float)malloc((ly+1)sizeof(float));
within = (int)malloc((ly+1)sizeof(int));
// Digitize the density.
bboxes();
interior();
digdens();
// Solve the diffusion equation.
do n = nonlinvoltra(); while (!n);
// Print cartogram generate file.
cartogram();
// Read the transformed polygon coordinates.
oldlx = lx;
oldly = ly;
oldmaxx = maxx;
oldmaxy = maxy;
oldminx = minx;
oldminy = miny;
genfile = fopen(CARTGENFILE,"r");
countpoly(genfile);
fclose(genfile);
genfile = fopen(CARTGENFILE,"r");
countcorn(genfile);
fclose(genfile);
lx = oldlx;
ly = oldly;
maxx = oldmaxx;
maxy = oldmaxy;
minx = oldminx;
miny = oldminy;
genfile = fopen(CARTGENFILE,"r");
readcorn(genfile);
fclose(genfile);
makeregion();
// Make cartogram
psfile = fopen(CART2PS);
pspicture(psfile);
fclose(psfile);
The part of the code where the bus error occurs is this:
void interior(void)
int i,inhowmanyregions,inregion[2],j,k,l,m,n,regctr;
// Initialize within[][]. -1 means outside all regions.
for (i=0; i<=lx; i++) for (j=0; j<=ly; j++) within[j] = -1;
// Fill within[][].
for (i=0; i<nregion; i++) for (j=0; j<npolyinreg; j++)
for (k=0, n=npolycorn[polyinreg[j]]-1;
k<npolycorn[polyinreg[j]]; n=k++)
for (l=(int)ceil(MIN(polycorn[polyinreg[j]][k-1].y,
polycorn[polyinreg[j]][k].y));
l<MAX(polycorn[polyinreg[j]][k-1].y,
polycorn[polyinreg[j]][k].y); l++)
for (m=(int)floor(bbminx[polyinreg[j]]);
m<(polycorn[polyinreg[j]][n].x-
polycorn[polyinreg[j]][k].x)*
(l-polycorn[polyinreg[j]][k].y)/
(polycorn[polyinreg[j]][n].y-
polycorn[polyinreg[j]][k].y)+
polycorn[polyinreg[j]][k].x;
m++)
within[m][l] = i-within[m][l]-1;
I really have no idea what is going on. Can anyone help?
Thanks,
mooseguyorangekay wrote:
That is absolutely unreadable.
I agree completely. How about also providing a link to a set of input files so that we could actually run the code ourselves? Otherwise, there is no chance to debug the source code, it is just a mess.
You might have better luck with an older version of the file. To quote the comments:
// Modified on March 31, 2005. Initialized maxchange in nonlinvoltra() as
// INFTY. Replaced crnmbr() by a similar, but faster routine interior().
// Many thanks to Stuart Anderson for pointing out this shortcut.
I suspect that the "interior" function is just plain incorrect. It doesn't matter if it runs on some other OS. Something about it is wrong and the code is so cryptic that it can't be deciphered. -
I need some help! I admit, I'm not a java guru...far from it, just a college student with a book that doesn't have many examples for what I'm looking for. Any help will be greatly appreciated!!!
My problem: My java program is to find a root using the Newton-Rapson method. The input is a function, i.e. f(x) = x^2 + x + 2. The program is to prompt the user for the beginning numbers, i.e. 225 and -43, and then find the answer to the nearest hundredth.
I can do this on paper - no problem, my problem is I don't know enough about Java to program it to do the math.
Unfortunately, if there are any predefined classes or methods that will do this problem for me, I'm NOT allowed to use them. Every step has to be hard coded. :(
I'm not sure how to get Java to recognize the + sign or
the ^ sign or any other non-numeric signs. I know there is something called substring, but I can't seem to find examples where it uses an integer. All the examples I have found are chars (like reading the string "hello"). Not helpful when you are trying to mix integers with symbols.
Any help offered would be greatly appreciated, and I'm sure would in-turn help me the the bisection algorithm and all the future algorithm's I have to put into a program.
Thanks for your time.[mode=Pedantic] Its Newton-Raphson [mode]
Anyway use this as a starting point, its from a guy by the name of Neil Broadbent (see: http://www.neiljohan.com/java/ for lots of java samples)
This will calculate the square root of a number using Newton-Raphson factoring, you can use this as a starting point.
//Program to calculate the square root of a number using the Newton-Raphson.
//It stops when the difference between consecutive approximations is less than 0.00005
//by Neil Broadbent
import java.io.IOException;
import NeilClass.*;
public class NewtonRaphsonClass
private static boolean debugging = true;
/** @param tFactorial the value from which to find the square root
* precondition tFactorial>0
* @return x1 so that x1>0 */
public static double iSquareRoot (double tFactorial) throws IOException
double x1=0;
double x0=tFactorial/2;
double a=tFactorial;
boolean finished=false;
/* check pre-condition */
if (debugging && (tFactorial<=0)){
System.out.println("Pre-condition violoated");
finished=true;
while (finished==false)
x1=(x0+(a/x0))/2;
if (x1>x0){
if ((x1-x0)<0.00005){
finished=true;
else if (x0>x1){
if ((x0-x1)<0.00005){
finished=true;
x0=x1;
/* check post-condition */
if (debugging && (x1<=0)){
System.out.println("Post-condition violoated");
return x1;
} -
Hi, i have written a program for solving a power flow question,using Gaussian elimination method to solve for matrices.The program is pretty long,and i have tried breaking the program into smaller modules,but there were errors in the numeric results at the end.
I need some suggests as to how to break the program into smaller modules???
/* Example 10.6 Pg 347 from Bergen.
Demonstration of Newton-Raphson Iteration.
import java.util.*;
import java.io.*;
import java.lang.Math;
import java.lang.ArrayIndexOutOfBoundsException;
public class NewtonRaphsonIt
public static void main( String args[])throws ArrayIndexOutOfBoundsException
int Deg = 360;
int Loop = 3;
double pi = 3.141592653589793238;
double limit = Math.abs(5E-6);
int i,j,k,u;
double V1,V2,V3,P2,P3,Q3,deltaP2,deltaP3,deltaQ3;
double Theta1,Theta2,Theta3;
double P1,Q1,Q2;
double P_Loss,Q_Loss,PF;
double maxi,temp,temp2;
int tempi=0;
//values given
P2=0.6661; P3=-2.8653; Q3=-1.2244;
V1=1; V2=1.05; Theta1=0;
//values supposely to be chosen as a rough guess
Theta2=Theta3=0; V3=1;
double [][] Y_bus = {{-19.98,10,10},{10,-19.98,10},{10,10,-19.98}};
double [][] J = new double [3][3];
double [] fx = new double [3];
double [] oldx = {Theta2,Theta3,V3};
double [] x = {Theta2,Theta3,V3};
double [] V = {V1,V2,Theta1};
double [] Bus = {P2,P3,Q3};
double [] delta = new double [3];
double [] ANS = new double [3];
int f = fx.length;
int y = Y_bus.length;
int z = J.length;
int c = Bus.length;
int h = delta.length;
int e = ANS.length;
int ox= oldx.length;
//System.out.println("Arrary size for Ybus is:" +n);
//output Ybus Matrix
System.out.println(" Ybus is:");
System.out.println("---------");
for ( i=0; i<y; i++)
for ( j=0; j<y; j++)
System.out.print(+Y_bus[i][j]+" "+", ");
System.out.println();
for ( u=1; u<Loop+1; u++){
//while(delta[0] < limit){
//Jacobian matrix
//deltaP2/deltaTheta2
J[0][0] = (Math.abs(V[1]))* (Math.abs(V[0])) * Y_bus[1][0] * (Math.cos(x[0]))
+ (Math.abs(V[1]))* (Math.abs(x[2])) * Y_bus[1][2] * (Math.cos(x[0]-x[1]));
//deltaP2/deltaTheta3
J[0][1] = -(Math.abs(V[1]))* (Math.abs(x[2])) * Y_bus[1][2] * (Math.cos(x[0]-x[1]));
//deltaP2/delta|V3|
J[0][2] = (Math.abs(V[1])) * Y_bus[1][2] * (Math.sin(x[0]-x[1]));
//deltaP3/deltaTheta2
J[1][0] = -(Math.abs(V[1]))*(Math.abs(x[2])) * Y_bus[2][0] * (Math.cos(x[1]-x[0]));
//deltaP3/deltaTheta3
J[1][1] = (Math.abs(V[0])) *(Math.abs(x[2])) * Y_bus[2][0] * (Math.cos(x[1]))
+(Math.abs(V[1])) *(Math.abs(x[2])) * Y_bus[2][1] * (Math.cos(x[1]-x[0]));
//deltaP3/delta|V3|
J[1][2] = (Math.abs(V[0])) * Y_bus[2][0] * (Math.sin(x[1]))
+(Math.abs(V[1]))* Y_bus[2][1] * (Math.sin(x[1]-x[0]));
//deltaQ3/deltaTheta2
J[2][0] = - Y_bus[2][0] * (Math.abs(x[2])) * (Math.abs(V[1])) *(Math.sin(x[1]-x[0]));
//deltaQ3/deltaTheta3
J[2][1] = (Math.abs(x[2]))*(Math.abs(V[0])) * Y_bus[2][0] * (Math.sin(x[1]))
+(Math.abs(x[2]))*(Math.abs(V[1])) * Y_bus[2][1] * (Math.sin(x[1]-x[0]));
//deltaQ3/delta|V3|
J[2][2] = -((Math.abs(V[0])) * Y_bus[2][0] * (Math.cos(x[1]))
+ (Math.abs(V[1])) * Y_bus[2][1] * (Math.cos(x[1]-x[0]))
+ (Math.abs(x[2]))* 2 * Y_bus[2][2]);
//output Jacobian Matrix
System.out.println();
System.out.println(+u+")"+" Jacobian Matrix is:");
System.out.println("-----------------------");
for ( i=0; i<z; i++)
for ( j=0; j<z; j++)
System.out.print(+J[i][j]+" "+", ");
System.out.println();
// variable P2
fx[0] = (Math.abs(V[1]))*(Math.abs(V[0])) * Y_bus[1][0] * (Math.sin(x[0]-V[2]))
+ (Math.abs(V[1]))*(Math.abs(x[2])) * Y_bus[1][2] * (Math.sin(x[0]-x[1]));
//System.out.println("P2 is :"+fx[0]);
// variable P3
fx[1] = (Math.abs(x[2]))*(Math.abs(V[0])) * Y_bus[2][0] * (Math.sin(x[1]-V[2]))
+ (Math.abs(x[2]))*(Math.abs(V[1])) * Y_bus[2][1] * (Math.sin(x[1]-x[0]));
//System.out.println("P3 is :"+fx[1]);
// variable Q3
fx[2] = -((Math.abs(x[2]))*(Math.abs(V[0])) * Y_bus[2][0] * (Math.cos(x[1]-V[2]))
+ (Math.abs(x[2]))*(Math.abs(V[1])) * Y_bus[2][1] * (Math.cos(x[1]-x[0]))
+ (Math.abs(x[2]))*(Math.abs(x[2])) * Y_bus[2][2]);
//System.out.println("Q3 is :"+fx[2]);
//Output f(x)
System.out.println();
System.out.println(+u+")"+" Function f(x) is:");
System.out.println("--------------------");
for ( k=0; k<f; k++)
System.out.println(+fx[k]);
System.out.println();
//ouput P2,P3,Q3 as a matrix
System.out.println();
System.out.println(+u+")"+" Matrix [P2,P3,Q3] is:");
System.out.println("------------------------");
for ( k=0; k<c; k++)
System.out.println(+Bus[k]);
System.out.println();
//finding values of deltaP2,deltaP3,deltaQ3
System.out.println();
System.out.println(+u+")"+" Matrix [d.P2,d.P3,d.Q3] is:");
System.out.println("------------------------------");
for ( k=0; k<h; k++)
System.out.println(delta[k]= Bus[k]- fx[k]);
System.out.println();
//find delta x={deltaTheta2,deltaTheta3,delta|V3|}
//Gaussian Elimination method
//solve Ax=b
//matrix J will be destroyed,solution will be in matrix delta
if(J[0].length != z)
System.out.println("Matrix J must be a Square Matrix!");
return;
for(j=0; j<z; j++)
maxi=-1E-12;
for(i=j; i<z; i++)
if(Math.abs(J[i][j])> maxi)
tempi=i;
maxi=Math.abs(J[i][j]);
//System.out.println(+tempi);
if(maxi<1E-12)
System.out.println
("The Set either has mutltiple solutions or no unique solution!");
return;
//Permuting tempi row with the j-th row
if(tempi != j)
for(k=j; k<z; k++)
temp=J[j][k]; //store 1st row into temp space
J[j][k]=J[tempi][k]; //store last row into 1st row
J[tempi][k]=temp; //store 1st row back into last row
temp=delta[j]; //store 1st row into temp space
delta[j]=delta[tempi]; //store last row into 1st row
delta[tempi]=temp; //store 1st row back into last row
//System.out.println("deltatemp:"+delta[j]);
//Divide j-th row by J[j][j]
temp=J[j][j];
delta[j]=delta[j]/temp;
System.out.println("delta1:"+delta[j]);
for(k=j; k<z; k++)
J[j][k]=J[j][k]/temp;
for(i=0; i<z; i++)
if(i != j)
temp=J[i][j];
temp2=delta;
for(k=0; k<z; k++)
J[i][k] = J[i][k]-temp*J[j][k];
delta[i]= temp2-temp*delta[j];
System.out.println("delta:"+delta[j]);
System.out.println("J:"+J[i][k]);
// values of deltax in radians
System.out.println(+u+")"+" The solution for AX=b is:");
System.out.println("----------------------------");
for(i=0; i<z; i++)
System.out.println(+delta[i]);
x[i] = delta[i];
// values of deltax in degrees
System.out.println();
System.out.println(+u+")"+" The solution for deltaX is:");
System.out.println("------------------------------");
ANS[0] = x[0] * Deg/(2*pi);
ANS[1] = x[1] * Deg/(2*pi);
ANS[2] = x[2];
for(i=0; i<e; i++)
System.out.println(+ANS[i]);
//finding X^1=x^0 + deltax^0
System.out.println();
System.out.println(" The solution for X after iteration:"+"("+u+")");
System.out.println("---------------------------------------");
for(i=0; i<e; i++)
x[i] = oldx[i] + ANS[i];
System.out.println(+x[i]);
System.out.println();
//System.out.println("oldx:" +oldx[i]);
//System.out.println();
//update the oldx with the new values from x
for(i=0; i<ox; i++)
oldx[i] = x[i];
//System.out.println("oldx:" +oldx[i]);
//System.out.println();
//convert to radians for Math.sin/cos()
x[0] = x[0] * (2*pi)/Deg;
x[1] = x[1] * (2*pi)/Deg;
x[2] = x[2];
}//for loop
//if (delta[0] == limit)
// break;
//}//while loop
// calcuate P1,Q1,Q2
P1 = (Math.abs(V[0]))*(Math.abs(V[1])) * Y_bus[0][1]* (Math.sin(V[2]-x[0]))
+ (Math.abs(V[0]))*(Math.abs(x[2])) * Y_bus[0][2] * (Math.sin(V[2]-x[1]));
System.out.println("P1 is :"+P1);
System.out.println();
Q1 = -((Math.abs(V[0]))*(Math.abs(V[1])) * Y_bus[0][1] * (Math.cos(V[2]-x[0]))
+ (Math.abs(V[0]))*(Math.abs(x[2])) * Y_bus[0][2] * (Math.cos(V[2]-x[1]))
+ (Math.abs(V[0]))*(Math.abs(V[0])) * Y_bus[0][0]);
System.out.println("Q1 is :"+Q1);
System.out.println();
Q2 = -((Math.abs(V[1]))*(Math.abs(V[0])) * Y_bus[1][0] * (Math.cos(x[0]-V[2]))
+ (Math.abs(V[1]))*(Math.abs(x[2])) * Y_bus[1][2] * (Math.cos(x[0]-x[1]))
+ (Math.abs(V[1]))*(Math.abs(V[1])) * Y_bus[1][1]);
System.out.println("Q2 is :"+Q2);
System.out.println();
//Power Loss P_Loss = (P1+P2) - P3
P_Loss = (P1+P2) - P3;
System.out.println("Power Loss is :"+P_Loss);
System.out.println();
System.out.println("P2 is :"+P2);
System.out.println();
System.out.println("P3 is :"+P3);
System.out.println();
System.out.println("Q3 is :"+Q3);
System.out.println();
//Reactive Power Loss
Q_Loss = (Q1+Q2) - Q3;
System.out.println("Reactive Power Loss:" +Q_Loss);
System.out.println();
System.out.println("Q1 is :"+Q1);
System.out.println();
System.out.println("Q2 is :"+Q2);
System.out.println();
System.out.println("Q3 is :"+Q3);
System.out.println();
//Power factor PF = P/sqrt[(P^2 + Q^2)]
PF = P3/(Math.sqrt( (P3*P3) + (Q3*Q3)));
System.out.println("Power Factor is :"+PF);
System.out.println();Start off by creating a Matrix class, rather than doing it all with 2d arrays in the main method? (The Matrix class might do little more than be a wrapper around a 2d array, but at least it would be a bit more modular.)
-
Hello everybody,
Does anybody know if non-linear second order ODE can be solved with the Runge-Kutta solver. This appears to be the most common solver.
I am trying to solve a non-linear second order differential equation of the form x*x’’ +x’^2 =Const. I was looking for an example that I could use to start with and follow it. I found one called “SHOOTING METHOD”.
for f(x(t),y(t),y) I used (1/x)*(-y*y+Const)
Originally, I used x(0) =0.01 and x(1) = 0.01 for my boundary conditions but the solution just increases linearly with time as opposed to reaching a maximum and decreasing back to zero in time.
Any help is greatly appreciated. Thanks in advance.
RobertoA shooting method reformulates the BVP as an optimization of the initial conditions of IVP. In your 2nd order ODE that means that the initial conditions are y(0) and y'(0), and the specified boundary conditions are y(0) and y(1). The first initial condition is just the first boundary condition, but y'(0) must be determined to produce y(1). I wrote a VI that sweeps over a series of y'(0) values, solves the ODE for each set of initial conditions, and graphs y(1)-desired_boundary. If y(1)-desired_boundary=0 then the initial conditions exactly produces the desired boundary conditions. Please see 'sweep rhs1.vi' in the attached archive. Run using the defaults. Once the zero is bracketed then an automatic root finder may be employed to solve for the initial condition with more precision. The zero-finder must have a bracketed interval (change in sign of the function around the desired level). See optimize rhs1.vi for an example using a Newton-Raphson zero finder to solve for the y'(0)initial condition.
Please note that the Newton-Raphson is a 1-D implementation only.
-Jim
Attachments:
shooting method.zip 43 KB -
Schnittpunkte interpolierter Kurven
Hallo zusammen,
soweit ich weiss gibt es keinen DIAdem-Befehl mit dem sich die Schnittpunkte von per Akima-Spline Interpolierten Kurven direkt berechnen lassen, oder habe ich da was übersehen?
Ich muss die Schnittpunkte von Kurven bestimmen die aus wenigen Messpunkten bestehen und die für den Report durch Akima-Splines "verbunden" werden. Momentan berechne ich die Schnittpunkte indem ich die (interpolierten) Kanäle voneinander abziehe und mit PNo den Wert suche der Null am nächsten ist.
Für Kurven mit steil abfallenden Flanken ist das aber leider zu ungenau, es sei den ich berechne die Akima Splines mit 10000 oder mehr Stützwerten, was aber in den meisten fällen überflüssig ist. Was ich bräuchte wäre eine Funktion die die Schnittpunkte iterativ a la Newton-Raphson Verfahren oder änlich berechnet.
Habt Ihr eine Idee wie ich das am besten löse?
Danke,
Elmar
Attachments:
Schnittpunkte.pdf 8 KBHallo Walter,
ich habe das Problem jetzt vorerst so gelöst das ich in den Bereichen in denen sich die Schnittpunkte befinden die Stützstellenanzahl stark erhöhe.
Den Algorithmus in VBS nachzubilden war mir zu mühsam.
Ich fände es sehr gut wenn DIAdem zukünftig eine Funktion hätte um Schnittpunkte zu berechnen. Bei dieser Funktion sollte die Interpolations-Methode die verwendet wird wählbar sein (linear, Akima, etc.).
Das würde uns die Arbeit sehr erleichtern da wir oft Schnittpunkte zwischen zwei gemessenen und interpolierten Kurven berechnen müssen.
Gruss
Elmar -
How to build an equation with variables
I want to solve an equation shown in below by the Newton Raphson zero finder vi:
z-a*b*c*exp(-z/a),
a, b, and c are variables
I can build equation with constant a,b,c, but how can I build this one?
Many thanks!
Solved!
Go to Solution.A couple of points I usually feel compelled to make when I see the use of the NR method.
1) Not an issue in a simple case like this, but the NR routine is typically a poor choice when you have a 1D problem and you calculate the derivative numerically.
2) If I am going to hardwire the formula like the previous example, then I will just use a VI instead of a formula. I only use formulas when I want to be able to make changes on-the-fly. I am just a bit dubious of the formula parsing VIs in general.
3) There is a VI called Substitute Variables which will take a set of rules and substitute numbers for parameters in your formula. Here is a quick example of its use. Here the names and number of parameters is fixed by the cluster itself. If I need full flexibility I use a table to hold the parameters and their values, but I do not want to muddy the waters here. You can just choose your favorite UI and convert (if necessary) to the proper cluster for the substitution rules. -
I expect no replies to this thread - because there are no
answers, but I want to raise awareness of a faculty of other
languages that is missing in Flash that would really help 3D and
games to be built in Flash.
Below is an optimisation of the Quake 3 inverse square root
hack. What does it do? Well in games and 3D we use a lot of vector
math and that involves calculating normals. To calculate a normal
you divide a vector's parameters by it's length, the length you
obtain by pythagoras theorem. But of course division is slow - if
only there was a way we could get 1.0/Math.sqrt so we could just
multiply the vector and speed it up.
Which is what the code below does in Java / Processing. It
runs at the same speed as Math.sqrt, but for not having to divide,
that's still a massive speed increase.
But we can't do this in Flash because there isn't a way to
convert a Number/float into its integer-bits representation. Please
could everyone whinge at Adobe about this and give us access to a
very powerful tool. Even the guys working on Papervision are having
trouble with this issue.that's just an implementation of newton's method for finding
the zeros of a differentiable function. for a given x whose inverse
sq rt you want to find, the function is:
f(y) = 1/(y*y) - x;
1. you can find the positive zero of f using newton's method.
2. you only need to consider values of x between 1 and 10
because you can rewrite x = 10^^E * m, where 1<=m<10.
3. the inverseRt(x) = 10^^(-E/2) * inverseRt(m)
4. you don't have to divide E by 2. you can use bitwise shift
to the right by 1.
5. you don't have to multiply 10^^(-E/2) by inverseRt(m): you
can use a decimal shift of inverseRt(m);
6. your left to find the positive zero of f(y) = 1/(y*y) - m,
1<=m<10.
and at this point i realized what, i believe, is a much
faster way to find inverse roots: use a look-up table.
you only need a table of inverse roots for numbers m,
1<m<=10.
for a given x = 10^^E*m = 10^^(e/2) *10^^(E-e/2)*m, where e
is the largest even integer less than or equal to E (if E is
positive, e is the greatest even integer less than or equal to E,
if E is negative), you need to look-up, at most, two inverse roots,
perform one multiplication and one decimal shift:
inverseRt(x) = 10^^(-e) * inverseRt(10) *inverseRt(m), if
E-e/2 = 1 and
inverseRt(x) = 10^^(-e) * inverseRt(m), if E-e/2 = 0. -
Equation with dynamic variable
Is it possible to use one equation with 2 or more variables and make one of these variable as an output without changing the equation?
Take the equation below example:
R=x/(0.28706/0.46153+x)*(101.325/P)
As usual, by inserting the values of x and P, then value of R will be answered.
But if only the value of R and x can be inserted, how can the value of P to be answered without altering the equation.
The equation is being used multiple time in the same VI. One is for generating one part of a graph (Mollier Diagram) and the others are to be used to find the value at the intersection.
I hope my question is understandable.
Thank you in advance
-sikat
Solved!
Go to Solution.R = (x/(a+x)*(b/P) ... thx, this really will be helping me to make this Eq more understandable
Formula Node:
It seem that when i entered an equation(Eq), only the left side of the Eq can only be one variable thus producing a single output.
Case Structure(or SELECT function):
It changes the formula to a new form for every output. Let say i want make x an output. That means i need to create a new Eq : x=(P*R/b-1)*a, which i want to avoid if posible .
.... Does this means that a new Eq must be created/converted for each output.
Newton Raphson zero finder vi: (base on this N-R Example)
I have tried using this VI by making my Eq equivalent to zero and using SELECT function depending on situation. Somehow i think it will work, but the VI seem too bulky for one simple formula (with 2 variable and 1 output), which i want to avoid since many more formula will be used.
I ll be glad to see more suggestions.
Maybe you are looking for
-
How do I use "saved passwords" to access a site?
I have a list of saved passwords in firefox. So what steps do I use to access the site. If I click on the saved password entry nothing happens? I want to use firefox saved passwords software. I used to use Roboform where I had a list of logins and ju
-
New Apple ipad air display not working,please help?
Please can anyone offer help? Week old ipad air display has stopped working after app froze and son double tapped home button to shut app down.Thank you in advance for any offers.
-
TYPE and SIZE, 2 more columns to the variable in the Command behavior?
DW 8.02 added 2 more columns to the variable in the Command behavior, TYPE and SIZE. What do you put in here? What is the difference for Access text, Date/Time, number and Memo columns. The Dreamweaver DOCS don't cover this update and you can't OK th
-
Hi; I installed the complete CS6 on a Win 7 x64 Acer M5 laptop. Everything works except In Design. I re-installed In Design and it made no difference. The program closes before fully starting and Windows just says the program closed and has a problem
-
Small(hopefully) Problem with 5:1 headphones on audigy 2
I recently purchased a Razer Barracuda headset and there seems to be a small problem. Ive got all of the plugs in their color-coded slot but the left and right channel seem to be backwards. Any help would be appreciated.