Passenger

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~~~ least squares fitting example ~~~
For example:
One phenomenon related to the time was that at time 1 the value 14 + 8j + 11l  3n was measured. At time 2, a value of 3 + 10j + 15l  7n was measured. It was found that the causeeffect of the phenomenon was approximately in accordance with the values:
x0 = 1, y0 = 14 + 8j + 11l  3n x1 = 2, y1 = 3 + 10j + 15l  7n x2 = 3, y2 = 12 + 14j + 20l  11n x3 = 4, y3 = 20 + 17j + 23l  15n x4 = 5, y4 = 25 + 19j + 24l  17n x5 = 6, y5 = 28 + 22j + 24l  21n x6 = 7, y6 = 27 + 24j + 23l  24n x7 = 8, y7 = 25 + 27j + 20l  27n x8 = 9, y8 = 23 + 27j + 17l  28n x9 = 10, y9 = 18 + 27j + 13l  32n x10 = 11, y10 = 13 + 28j + 11l  33n x11 = 12, y11 = 4 + 28j + 6l  35n x12 = 13, y12 = 1 + 28j + 2l  35n x13 = 14, y13 = 9 + 27j  3l  35n x14 = 15, y14 = 15 + 26j  6l  35n x15 = 16, y15 = 22 + 24j  10l  37n x16 = 17, y16 = 26 + 21j  13l  37n x17 = 18, y17 = 30 + 21j  17l  36n x18 = 19, y18 = 33 + 17j  18l  35n x19 = 20, y19 = 34 + 13j  19l  33n x20 = 21, y20 = 32 + 11j  17l  32n x21 = 22, y21 = 28 + 7j  15l  30n x22 = 23, y22 = 22 + 2j  11l  29n x23 = 24, y23 = 12  2j  7l  26n x24 = 25, y24 = 1  8j + 0l  22n
Based on the graphical analysis, it was decided to fit a cube parable to the phenomenon. Placing the measured response values in the least squares formula gives the cubic parabola coefficients:
f(x) = k0 + k1x^1 + k2x^2 + k3x^3, where k0 = 31.479 + 2.882j + 2.508l + 1.642n k1 = 20.364 + 4.297j + 8.735l  4.578n k2 = 2.068  0.179j  1.029l + 0.124n k3 = 0.052 + 0.000j + 0.027l + 0.001n
the recomplex cube parabola describes the phenomenon well. At initial values of x, the function values are:
f (recomplex(1)) = 13.131 + 7.000j + 10.241l  2.810n f (recomplex(2)) = 1.394 + 10.758j + 16.080l  7.009n f (recomplex(3)) = 12.409 + 14.152j + 20.185l  10.950n f (recomplex(4)) = 20.227 + 17.181j + 22.720l  14.628n f (recomplex(5)) = 25.163 + 19.842j + 23.846l  18.038n f (recomplex(6)) = 27.528 + 22.134j + 23.725l  21.175n f (recomplex(7)) = 27.638 + 24.053j + 22.520l  24.035n f (recomplex(8)) = 25.805 + 25.598j + 20.393l  26.612n f (recomplex(9)) = 22.342 + 26.766j + 17.506l  28.903n f (recomplex(10)) = 17.563 + 27.554j + 14.021l  30.901n f (recomplex(11)) = 11.782 + 27.961j + 10.100l  32.602n f (recomplex(12)) = 5.311 + 27.984j + 5.906l  34.002n f (recomplex(13)) = 1.536 + 27.621j + 1.600l  35.095n f (recomplex(14)) = 8.445 + 26.870j  2.655l  35.876n f (recomplex(15)) = 15.103 + 25.728j  6.697l  36.342n f (recomplex(16)) = 21.197 + 24.192j  10.364l  36.487n f (recomplex(17)) = 26.413 + 22.261j  13.493l  36.305n f (recomplex(18)) = 30.438 + 19.932j  15.924l  35.794n f (recomplex(19)) = 32.959 + 17.203j  17.493l  34.946n f (recomplex(20)) = 33.661 + 14.072j  18.038l  33.759n f (recomplex(21)) = 32.232 + 10.535j  17.397l  32.226n f (recomplex(22)) = 28.359 + 6.592j  15.408l  30.343n f (recomplex(23)) = 21.727 + 2.239j  11.909l  28.106n f (recomplex(24)) = 12.024  2.525j  6.737l  25.509n f (recomplex(25)) = 1.064  7.704j + 0.269l  22.547n
Algebra does not take a stand on the quality symbolically agreed for each axis. The unit vector l may represent, for example, snowfall as moles.
The real axis, i and its companion features have remarkable automation where different qualities can communicate with each other.
Multidimensional algebra is recursively generated from zero.
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That is, you cannot talk about any technology at code level. A new way to generate pseudorandom numbers was in the next and rndfunctions.
But it doesn't hurt if the forum is forbidden to display the code.
At least I enjoy reading other people's codes, and at best I try the other person's code myself.
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// 64bit random number generator. // CODE DELETED }
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Multidimensional algebra
The capabilities of Algebra also include a tool that can concatenate and approximate parallel cause and effect functions into compact functions. By drawing an approximately smooth graph on the screen, the algebra succeeds in approximating a given set of points to a short function. Algebra's skill is in its ability to describe phenomena and nature.
What about drawing different colored curves on the screen. The white set of points is temperature, the yellow wind direction, the orange season, the blue one represents the humidity, etc. Each set of points can be described as a single phenomenon, but what about the function that alone describes the causeeffect relationship between the whole set?
The relation of the function can also be asymmetric. The weather function can have nexcitations as the inputt, and only one relatively fuzzy parameter between 01 whether or not to go fishing.
***
You can program your computer to study math. With robust algebra axioms and inconsistent series development, bit fear further delves into geometry and symmetry. In the final solution, the computer had to partly abandon the symbolic concepts of positive, negative, real and imaginary. Algebra does not need to take a stand on how the monitor is placed on the table.
Algebra's algorithms are standard, and its objects can describe anything. One of the most beautiful algebra formulas is the least squares polynomial fit:
k_{0}∑x_{j}^{0} + k_{1}∑x_{j}^{1} + k_{2}∑x_{j}^{2} +, ..., + k_{n}∑x_{j}^{n+0} = ∑y_{j}x_{j}^{0}
k_{0}∑x_{j}^{1} + k_{1}∑x_{j}^{2} + k_{2}∑x_{j}^{3} +, ..., + k_{n}∑x_{j}^{n+1} = ∑y_{j}x_{j}^{1}
k_{0}∑x_{j}^{2} + k_{1}∑x_{j}^{3} + k_{2}∑x_{j}^{4} +, ..., + k_{n}∑x_{j}^{n+2} = ∑y_{j}x_{j}^{2}
...
k_{0}∑x_{j}^{n} + k_{1}∑x_{j}^{n+1} + k_{2}∑x_{j}^{n+2} +, ..., + k_{n}∑x_{j}^{n+n} = ∑y_{j}x_{j}^{n}If the object to be placed in the formula is valid and in equilibrium with itself, the algebra processes the substance with reasonable results, otherwise it does not. A good example is quaternions. Algebra does not comment on the quaternion internal calculation logic.
***
i^{2} is 1, but what is the object raised to another to give i? i's companions have an interesting connection:
e^{xi + yj + zk + ...} = 1
The following is a simple computation of the recomplex number with C ++ examples.
~ recomplex
~ order
~ class definition
~ addition
~ subtraction
~ multiplication
~ income of parent vectors
~ dividing
~ abs(recomplex)
~ pow(recomplex, int)
~ code notes***
~~~ recomplex ~~~
A complex number is a real subset of a recomplex number. Real numbers and imaginary numbers, in turn, are real subsets of a complex number. The prefix "re" refers to feedback, a repetition of the same basic logic in the following scale.
~~~ order ~~~
The order of magnitude is the power of two, 1, 2, 4, ..., 2^{n}. The real numbers are tuned by an order of 1. The complex numbers take effect by an order of 2. In the order of 4, the complex numbers get a parallel logic object. Their combination allows, for example, simple function mappings for parallel causeandeffect phenomena.
~~~ Class Definition ~~~
C ++ carried out the load on the operators, so that the computation of numerical structures, especially on a computer, did not deviate from the usual straightforwardness.
class recomplex { public: /************************************ * The dimension should be 2 ^ n. * * DIMENSION 1 > real numbers * * DIMENSION 2 > complex numbers * * DIMENSION 4 > 4D complex numbers * * DIMENSION 8 > 8D complex numbers * ************************************/ #define DIMENSION 4 double e[DIMENSION]; recomplex(void); ~recomplex(void); recomplex(char*, ...); recomplex(int); recomplex(int, int); recomplex(int, int, int, int); friend void print(recomplex); friend double abs(recomplex); friend recomplex operator(recomplex); friend recomplex operator+(recomplex, recomplex); friend recomplex operator(recomplex, recomplex); friend recomplex operator*(recomplex, recomplex); friend recomplex operator/(recomplex, recomplex); friend recomplex operator*(recomplex, double); friend recomplex operator/(recomplex, double); friend recomplex operator*(double, recomplex); friend recomplex operator/(double, recomplex); friend recomplex pow(recomplex, int); };
The recomplex definition includes basic calculations, a prime number and an absolute value. In addition to this, there is a value printing function as well as value generators.
~~~ addition ~~~
recomplex operator+(recomplex a, recomplex b) { for (int n=0; n<DIMENSION; n++) a.e[n]+=b.e[n]; return a; }
If C is a stranger, a few words about it. C is based on saving sheets. All unnecessary and unnecessary deliberation has been removed from the syntax. The operators of C are very efficient. For example, +=, *=, /=, ... will first execute this. mission, followed by placement. a.e [n] + = b.e [n] thus means the same as a.e[n] = a.e[n]+b.e[n]. Because the latter style consumes more paper, C uses the shorter notation a.e[n] += b.e[n] for the same thing.
For example:
recomplex A(2, 3, 5, 7); recomplex B(7, 5, 3, 2); print(A + B);
Prints a value of 9 + 8i  8j + 9k.~~~ subtraction ~~~recomplex operator(recomplex a, recomplex b) { for (int n=0; n<DIMENSION; n++) a.e[n]=b.e[n]; return a; }
For exsample:recomplex A(2, 3, 5, 7); recomplex B(7, 5, 3, 2); print(AB);
The difference value is 5  2i + 2j + 5k.~~~ multiplication ~~~recomplex operator*(recomplex a, recomplex b) { int *ConstantIncome(void); int *R=ConstantIncome(); int i, j, n=DIMENSION*2; double x[DIMENSION*2]; double y[DIMENSION*2]; double t[DIMENSION*2]; for (i=j=0; i<DIMENSION; i++, j+=2) x[j+1]=y[j+1]=t[j]=t[j+1]=0.0, x[j]=a.e[i], y[j]=b.e[i]; for (i=0; i<n; i++) for (j=0; j<n; j++) t[R[i*n+j]]+=x[i]*y[j]; for (i=j=0x00; i<DIMENSION; i++, j+=2) t[i]=(double)(t[j]  t[j+1]); return *(recomplex*) t; }
The multiplication first requires an input definition for the base vectors. Input looping does not take a stand on the formal definitions of positive, negative, real, imaginary, etc. The result of the multiplication is returned to the sum form only in the last loop.
For exsample:
recomplex A(2, 3, 5, 7); recomplex B(7, 5, 3, 2); print(A * B);
The result is 32 + 32i + 87k. Multiplication is an exchange. If A and B are complex, the result is complex:
recomplex A(2, 3, 0, 0); recomplex B(7, 5, 0, 0); print(A * B);
Prints 1 + 31i.
~~~ basic vectors revenue ~~~
The product of the base vectors is generated by a recursive function. The table summarizes all income for the parent vectors related to the order 4:
# 1  1  i  i  j  j  k  k  ############################################ 1 # 1  1  i  i  j  j  k  k  # 1 # 1  1  i  i  j  j  k  k  # i # i  i  1  1  k  k  j  j  # i # i  i  1  1  k  k  j  j  # j # j  j  k  k  i  i  1  1  # j # j  j  k  k  i  i  1  1  # k # k  k  j  j  1  1  i  i  # k # k  k  j  j  1  1  i  i  #
For example, ijk = i, and especially k^{2} = i. The trade remains because:
ikj = jik = jki = kij = kji = i
void GenerateBaseVectorsInputs(int *R, int n) { int I=0, J=n1, i; int X=n, Y=J+n, j; int k=DIMENSION*2; for (i=I; i<=J; i++) for (j=I; j<=J; j++) { R[i*k+X+j]=R[i*k+j]+X; R[(X+j)*k+i]=R[i*k+j]+X; R[(Yi)*k+Yj]=JR[i*k+j]; } if (n+n < DIMENSION*2) { GenerateBaseVectorsInputs(R, n+n); } } int* ConstantIncome(void) { static int R[DIMENSION*DIMENSION*4]={1}; if (R[0] == 1) { int FirstThereWasZero=0; R[0]=(int)FirstThereWasZero; GenerateBaseVectorsInputs(R, 1); } return R; }
~~~ dividing ~~~
recomplex operator/(recomplex x, recomplex y) { recomplex z; for (int i, n=DIMENSION; n>=2; n/=2) { for (z=y, i=n/2; i<n; i++) z.e[i] = z.e[i]; x=x*z; y=y*z; } return x/y.e[0]; }
In a divisor, a divisor is produced for the divisor, multiplying by the divisor and divisor. At the end of each loop step, the divider axis values are less than half (the rest add up to zero). When the value of the divisor is real, the value divides the elements of the divisor. For example, 2 + 3i + 5j + 7k has a compound number of 2 + 3i  5j  7k.
For exsample:
recomplex A(2, 3, 5, 7); recomplex B(7, 5, 3, 2); print(A / B);
The result of the split calculation is 0.5488136  0.1575801i + 0.7181670j + 0.3977540k. If A and B are complex, the result is complex:
recomplex A(2, 3, 0, 0); recomplex B(7, 5, 0, 0); print(A / B);
Returns 0.3918919 + 0.1486486i.
~~~ abs(recomplex) ~~~
double abs(recomplex x) { recomplex z; double r, c; int i, j=1, n; for (n=DIMENSION; n>=2; n/=2, j+=j) { for (z=x, i=n/2; i<n; i++) z.e[i]=z.e[i]; x=x*z; } r=fabs(x.e[0]); c=1.0/(double)j; return pow(r, c); }
In the eigenvalue, x is multiplied by its compound number until its value is completely real. The value then returns the junction power collected by j.
For exsample:
recomplex A(2, 3, 5, 7); recomplex B(7, 5, 3, 2); printf("%25.20f \n", abs(A / B));
C double is accurate because the absolute value of the result is exactly 1.0000000000000000000000
~~~ pow(recomplex, int) ~~~
recomplex pow(recomplex x, int n) { if (n) { recomplex t=x; int i=n<0? n: n; for (i; i; i) t=t*x; return n>0? t: recomplex(1) / t; } else { return recomplex(1); } }
The limit for x^{0} is 1 as x approaches zero. Calculators have varying interpretations of whether 0^{0} is approximately 1 or whether the calculation is interrupted and an error condition is printed. In the least squares fit, x^{0} must also be 1 for x. The exponent n can be a positive or a negative integer.
~~~ code notes ~~~
Recomplex figures form the municipality. All municipal billing rules apply.
The code below can be used to test how recomplex numbers behave.
#include <memory.h> #include <stdlib.h> #include <stdio.h> #include <math.h> class recomplex { public: /************************************ * The dimension should be 2 ^ n. * * DIMENSION 1 > real numbers * * DIMENSION 2 > complex numbers * * DIMENSION 4 > 4D complex numbers * * DIMENSION 8 > 8D complex numbers * ************************************/ #define DIMENSION 4 double e[DIMENSION]; recomplex(void); ~recomplex(void); recomplex(char*, ...); recomplex(int); recomplex(int, int); recomplex(int, int, int, int); friend void print(recomplex); friend double abs(recomplex); friend recomplex operator(recomplex); friend recomplex operator+(recomplex, recomplex); friend recomplex operator(recomplex, recomplex); friend recomplex operator*(recomplex, recomplex); friend recomplex operator/(recomplex, recomplex); friend recomplex operator*(recomplex, double); friend recomplex operator/(recomplex, double); friend recomplex operator*(double, recomplex); friend recomplex operator/(double, recomplex); friend recomplex pow(recomplex, int); }; /************************************************************************ * The recomplex(void) generator initializes the values to zero. * * ************************************************************************/ recomplex::recomplex(void) { memset(this, 0, sizeof(recomplex)); } recomplex::~recomplex(void) { } /************************************************************************ * Generator recomplex(char *, ...) * * * * For example: * * * * recomplex x("2d", 2, 3), x gets a value of 2.0000 + 3.0000i * * recomplex y("2f", 1.2345, 6.7898), y gets a value of 1.2345 + 6.7898i * * * * The format first gives the recomplex of chapters, for example 4. * *  'd' means that the numbers are given as integers. * *  'f' means that the numbers are given in floating point form. * ************************************************************************/ recomplex::recomplex(char *f, ...) { char format='d'; int i, kpl=atoi(f); int step=(int)sizeof(int); memset(this, 0x00, sizeof(recomplex)); unsigned long P=(unsigned long)&f+sizeof(char*); for (i=0; f[i]; i++) { if (f[i] == 'f') { step=sizeof(double); format='f'; } } for (i=0; i<kpl; i++, P+=step) e[i] = (char)format=='d'? *(int*)P: *(double*)P; } /************************************************************************ * Somewhere around 20 elements when the parent vector ID is larger * * like 'z', the base vector ID may be a letter abnormal character. * * In fact, the calculation of the base vectors no comment is made on * * symbolic symbols. * ************************************************************************/ void print(recomplex a) { printf("%0.9f ", a.e[0]); for (int n=1; n<DIMENSION; n++) printf("%c %0.9f%c ", a.e[n]<0? '': '+', fabs(a.e[n]), 'h'+n); printf("\n"); } recomplex::recomplex(int a) { memset(this, 0, sizeof(recomplex)); e[0]=a; } recomplex::recomplex(int a, int b) { memset(this, 0, sizeof(recomplex)); e[0]=a; e[1]=b; } recomplex::recomplex(int a, int b, int c, int d) { memset(this, 0, sizeof(recomplex)); e[0]=a; e[1]=b; e[2]=c; e[3]=d; } recomplex operator(recomplex a) { for (int n=0; n<DIMENSION; n++) a.e[n]=a.e[n]; return a; } recomplex operator+(recomplex a, recomplex b) { for (int n=0; n<DIMENSION; n++) a.e[n]+=b.e[n]; return a; } recomplex operator(recomplex a, recomplex b) { for (int n=0; n<DIMENSION; n++) a.e[n]=b.e[n]; return a; } recomplex operator*(recomplex a, recomplex b) { int *ConstantIncome(void); int *R=ConstantIncome(); int i, j, n=DIMENSION*2; double x[DIMENSION*2]; double y[DIMENSION*2]; double t[DIMENSION*2]; for (i=j=0x00; i<DIMENSION; i++, j+=2) x[j+1]=y[j+1]=t[j]=t[j+1]=0.0, x[j]=a.e[i], y[j]=b.e[i]; for (i=0; i<n; i++) for (j=0; j<n; j++) t[R[i*n+j]]+=x[i]*y[j]; for (i=j=0x00; i<DIMENSION; i++, j+=2) t[i]=(double)(t[j]  t[j+1]); return *(recomplex*) t; } void GenerateBaseVectorsInputs(int *R, int n) { int I=0, J=n1, i; int X=n, Y=J+n, j; int k=DIMENSION*2; for (i=I; i<=J; i++) for (j=I; j<=J; j++) { R[i*k+X+j]=R[i*k+j]+X; R[(X+j)*k+i]=R[i*k+j]+X; R[(Yi)*k+Yj]=JR[i*k+j]; } if (n+n < DIMENSION*2) { GenerateBaseVectorsInputs(R, n+n); } } int* ConstantIncome(void) { static int R[DIMENSION*DIMENSION*4]={1}; if (R[0] == 1) { int FirstThereWasZero=0; R[0]=(int)FirstThereWasZero; GenerateBaseVectorsInputs(R, 1); } return R; } recomplex operator/(recomplex x, recomplex y) { recomplex z; for (int i, n=DIMENSION; n>=2; n/=2) { for (z=y, i=n/2; i<n; i++) z.e[i] = z.e[i]; x=x*z; y=y*z; } return x/y.e[0]; } recomplex operator/(recomplex x, double k) { for (int i=0; i<DIMENSION; i++) x.e[i]/=(double)k; return x; } recomplex operator*(double k, recomplex x) { for (int i=0; i<DIMENSION; i++) x.e[i]*=(double)k; return x; } recomplex operator*(recomplex x, double k) { return k*x; } recomplex operator/(double k, recomplex x) { recomplex y("1f", k); return (recomplex)y/x; } double abs(recomplex x) { recomplex z; double r, c; int i, j=1, n; for (n=DIMENSION; n>=2; n/=2, j+=j) { for (z=x, i=n/2; i<n; i++) z.e[i]=z.e[i]; x=x*z; } r=fabs(x.e[0]); c=1.0/(double)j; return pow(r, c); } recomplex pow(recomplex x, int n) { if (n) { recomplex t=x; int i=n<0? n: n; for (i; i; i) t=t*x; return n>0? t: recomplex(1) / t; } else { return recomplex(1); } } /************************************************************************ * the mainfunction tests the income swap law. * ************************************************************************/ void main(void) { recomplex x("4f", 0.12, 0.34, 0.56, 0.78); recomplex y("4f", 0.98, 0.76, 0.54, 0.32); printf("x = "); print(x); printf("y = "); print(y); printf("\n"); printf("x*y = "); print(x*y); printf("y*x = "); print(y*x); printf("\n"); }
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By eliminating symbolic signs for multiplication, complex spaces are possible.
I studied this problem for over a decade until it was resolved.
Complex spaces fulfill all municipal conditions.
Please find the code below to test the municipal rules.!Moderator Note
No one should have to reverse engineer your code to understand what you are talking about.
Explain what you want to say.
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#include <math.h> #include <stdio.h> typedef unsigned __int64 uint64; #define TRUE 1 #define FALSE 0 #define INC 3 #define NEW 7 #define SEQUEL 9 class rnd64 { public: rnd64(void); ~rnd64(void); uint64 rnd(void); private: int prime_number(int); uint64 next(void); int status; uint64 dx; uint64 dy; uint64 end; uint64 x, y; uint64 R16A; uint64 R16B; uint64 Z[2]; uint64 increaseA; uint64 increaseB; }; rnd64::rnd64(void) { dx=0xfedca201L; dy=0x012357bfL; R16A=(uint64)1; R16B=(uint64)2; Z[0]=Z[1]=0x00; x=5; y=3; end=2; increaseA=0x012357bfL; increaseB=0xfedca201L; } rnd64::~rnd64(void) { } int rnd64::prime_number(int st) { if (st == INC) { x+=2; y=3; if (x>=(1<<16)) x=5; end=unsigned(sqrt(double(x)+0.25)); return SEQUEL; } else { if (!(x%y)) return FALSE; if ((y+=2)>end) return TRUE; return SEQUEL; } } inline uint64 uabs(uint64 &a, uint64 &b) { return a>b? ab: ba; } inline void swap(uint64 &a, uint64 &b) { uint64 c=a; a=b; b=c; } uint64 rnd64::next(void) { status=prime_number(NEW); if (status == TRUE) { increaseB=increaseA; increaseA+=x; prime_number(INC); } else if (status == FALSE) { prime_number(INC); } R16A = increaseA; increaseA=dx; R16B += increaseB; increaseB+=dy; if (status==TRUE) swap(R16A, R16B); R16A += (R16A>>32)^(R16B<<32); R16B = (R16A<<32)^(R16B>>32); return R16A^R16B; } uint64 rnd64::rnd(void) { uint64 p1=next(); if (status==TRUE) { uint64 a[2], b[2]; uint64 p2=next(); a[0]=b[0] = Z[0]; a[1]=b[1] = Z[1]; ++a[unsigned(p1%2)]; ++b[unsigned(p2%2)]; uint64 A=uabs(a[0], a[1]); uint64 B=uabs(b[0], b[1]); if (Z[0] == Z[1]) { if (unsigned(next()%2)) { A=1; B=0; } else { A=0; B=1; } } if (A < B) { ++Z[unsigned(p1%2)]; return p1; } else { ++Z[unsigned(p2%2)]; return p2; } } else { ++Z[unsigned(p1%2)]; return p1; } } void main(void) { rnd64 R; int st=0; uint64 bit[2]; bit[0]=bit[1]=0x00; for (;;) { uint64 P=R.rnd(); ++bit[unsigned(P%2)]; if (st==0 && bit[1]>bit[0]) { printf("%d", bit[0]>bit[1]? 1: 0); st=1; } else if (st==1 && bit[0]>bit[1]) { printf("%d", bit[0]>bit[1]? 1: 0); st=0; } } }
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e^{πi }+ 1 = 0
1 
k_{0}∑x_{j}^{0} + k_{1}∑x_{j}^{1} + k_{2}∑x_{j}^{2} +,…,+ k_{n}∑x_{j}^{n+0} = ∑y_{j}x_{j}^{0}
k_{0}∑x_{j}^{1} + k_{1}∑x_{j}^{2} + k_{2}∑x_{j}^{3} +,…,+ k_{n}∑x_{j}^{n+1} = ∑y_{j}x_{j}^{1}
k_{0}∑x_{j}^{2} + k_{1}∑x_{j}^{3} + k_{2}∑x_{j}^{4} +,…,+ k_{n}∑x_{j}^{n+2} = ∑y_{j}x_{j}^{2}
..
.
k_{0}∑x_{j}^{n} + k_{1}∑x_{j}^{n+1} + k_{2}∑x_{j}^{n+2} +,…,+ k_{n}∑x_{j}^{n+n} = ∑y_{j}x_{j}^{n}One of the most beautiful algebra formulas is the least squares polynomial formula.
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Fractals
in Linear Algebra and Group Theory
Posted
I calculated this fractal with 4D complex numbers. (j and k axis fractal zoom)