5 Bit Barker Code

[jafal]

hi all

I want to perform multiplication and addition for this array so that the result is as follows
thanks in advance

int [] intArray = new  int[]  {1, 1, 1,-1, 1};
println("Original Array:");
for (int i=0; i<intArray.length; i++)
print(intArray[i] + "  ");
println();
println(" reverse order:");
for (int i=intArray.length-1; i>=0; i--)
print(intArray[i] + "  ");

------------------------------------------------------
i want to reach this result 

{1, 1, 1,-1, 1} orginal * reverse  {1,-1, 1, 1, 1}

 1 1  1 -1 1
  -1 -1 -1 1 -1
      1  1 1 -1  1
         1 1  1 -1  1
           1  1  1 -1  1
            
 ------------------------------------------------------           
 1 0 1 0  5   0  1  0  1
  

[matheplica]

You can change your array to a IntList how have reverse method implement

[jafal]

thank you for this hint

[jafal]

hi all

i made a code but its very long how i can use for loop to make it shorter

int[] numbers = new int[5];

void setup() {
  size(200, 200);
  numbers[0] = 1;  
  numbers[1] = 1; 
  numbers[2] = 1;  
  numbers[3] = -1; 
  numbers[4] = 1;  
  int a = numbers[4] *numbers[0]; 
  int b = numbers[4] * numbers[1]; 
  int c = numbers[4] * numbers[2]; 
  int d = numbers[4] * numbers[3]; 
  int e = numbers[4] * numbers[4]; 
  println(a, b, c, d, e);

  int a1 = numbers[3] *numbers[0]; 
  int b1 = numbers[3] * numbers[1]; 
  int c1 = numbers[3] * numbers[2]; 
  int d1 = numbers[3] * numbers[3]; 
  int e1= numbers[3] * numbers[4];  
  println(a1, b1, c1, d1, e1);

  int a2 = numbers[2] *numbers[0]; 
  int b2 = numbers[2] * numbers[1]; 
  int c2 = numbers[2] * numbers[2]; 
  int d2 = numbers[2] * numbers[3]; 
  int e2= numbers[2] * numbers[4];  
  println(a2, b2, c2, d2, e2);
  int a3 = numbers[1] *numbers[0]; 
  int b3 = numbers[1] * numbers[1]; 
  int c3 = numbers[1] * numbers[2]; 
  int d3 = numbers[1] * numbers[3]; 
  int e3= numbers[1] * numbers[4];
  println(a3, b3, c3, d3, e3);
  int a4 = numbers[0] *numbers[0]; 
  int b4 = numbers[0] * numbers[1]; 
  int c4 = numbers[0] * numbers[2]; 
  int d4 = numbers[0] * numbers[3]; 
  int e4= numbers[0] * numbers[4];
  println(a4, b4, c4, d4, e4);
  int x1 = a;
  int x2 = b+a1;
  int x3 = c+b1+a2;
  int x4 = d+c1+b2+a3;
  int x5 = e+d1+c2+e3+a4;
  int x6 = e1+d2+c3+b4;
  int x7 = e2+d3+c4;
  int x8 = e3+d4;
  int x9 = e4;
  println(x1, x2, x3, x4, x5, x6, x7, x8, x9);
}

void draw() {
}

[Flolo]

Hey
you can do the upper half with a nested for loop but I don’t know if you can do the lower half with a for loop. Unfortunately, I didn’t manage to do it

I hope I could help you

Here is my code:

int[] numbers = new int[]{1, 1, 1, -1, 1};

ArrayList<Integer> abc;

void setup() {
  size(200, 200);

  surface.setVisible(false);//so that the canvas is not displayed

  abc = new ArrayList<Integer>();

  //index 0  = 4 * 0
  //index 1  = 4 * 1
  //index 2  = 4 * 2
  //index 3  = 4 * 3
  //index 4  = 4 * 4

  //index 5  = 3 * 0
  //index 6  = 3 * 1
  //index 7  = 3 * 2
  //index 8  = 3 * 3
  //index 9  = 3 * 4

  //index 10 = 2 * 0
  //index 11 = 2 * 1
  //index 12 = 2 * 2
  //index 13 = 2 * 3
  //index 14 = 2 * 4

  //index 15 = 1 * 0
  //index 16 = 1 * 1
  //index 17 = 1 * 2
  //index 18 = 1 * 3
  //index 19 = 1 * 4

  //index 20 = 0 * 0
  //index 21 = 0 * 1
  //index 22 = 0 * 2
  //index 23 = 0 * 3
  //index 24 = 0 * 4

  //int a = 0; 
  //int b = 1; 
  //int c = 2; 
  //int d = 3; 
  //int e = 4; 
  //println(a, b, c, d, e);

  //int a1 = 5; 
  //int b1 = 6; 
  //int c1 = 7; 
  //int d1 = 8; 
  //int e1 = 9;  
  //println(a1, b1, c1, d1, e1);

  //int a2 = 10; 
  //int b2 = 11; 
  //int c2 = 12; 
  //int d2 = 13; 
  //int e2 = 14;  
  //println(a2, b2, c2, d2, e2);

  //int a3 = 15; 
  //int b3 = 16; 
  //int c3 = 17; 
  //int d3 = 18; 
  //int e3 = 19;
  //println(a3, b3, c3, d3, e3);

  //int a4 = 20; 
  //int b4 = 21; 
  //int c4 = 22; 
  //int d4 = 23; 
  //int e4 = 24;
  //println(a4, b4, c4, d4, e4);

  for (int i = numbers.length-1; i >= 0; i--) {
    for (int j = 0; j < numbers.length; j++) {
      int a = numbers[i] * numbers[j];
      print(a + " ");
      abc.add(a);
    }
    println("");
  }

  int x1 = abc.get(0);
  int x2 = abc.get(1)  + abc.get(5);
  int x3 = abc.get(2)  + abc.get(6)  + abc.get(10);
  int x4 = abc.get(3)  + abc.get(7)  + abc.get(11) + abc.get(15);
  int x5 = abc.get(4)  + abc.get(8)  + abc.get(12) + abc.get(19) + abc.get(20);
  int x6 = abc.get(9) + abc.get(13) + abc.get(17) + abc.get(21);
  int x7 = abc.get(14) + abc.get(18) + abc.get(22);
  int x8 = abc.get(19) + abc.get(23);
  int x9 = abc.get(24);
  println(x1, x2, x3, x4, x5, x6, x7, x8, x9);
}

void draw() {
}

[jafal]

Flolo

hi
nice code for 5 elements but for 16 or 32 it is going to be much longer i have tried to use for loop but failed until now thanks for your time

[Chrisir]

What is the purpose of the arrays?

[jafal]

@Chrisir

thanks for response …actually it is Barker code 5 bits calculation and it might up to length codes of N < 10^22 and calculation 13 bit length as i did with the 5 bit length with out for loop is going to be very long code

this is the idea and you can see what i calculated i take length 5 which is {1, 1, 1,-1, 1};

[jafal]

@Chrisir the idea of solving is simple

ECE3084 Lecture 12.5: Autocorrelation, Pulse Compression, and Barker Codes (Signals & Systems)

[jafal]

@Chrisir

this is some information with c

Implementing iterative autocorrelation process in C++ using for loops

[Chrisir]

Sorry, I am on a journey now and can’t really help you

[jafal]

it is fine when you can i am not in hurry have fun

[Chrisir]

Okay, when you only have 16 or 32 items specifically then write a code that use println in a for loop

Let it println the lines of code you need for the main Sketch such as

@Flolo

int x7 = abc.get(14) + abc.get(18) + abc.get(22);
  int x8 = abc.get(19) + abc.get(23);

[jafal]

@Chrisir you are true but if my array like this

int [] intArray = new  int[]  {1, 1, 1,-1, 1,+1 +1 +1 +1 +1 −1 −1 +1 +1 −1 +1 −1 +1};

solving it in this way going to take very long lines

 int x1 = abc.get(0);
  int x2 = abc.get(1)  + abc.get(5);
  int x3 = abc.get(2)  + abc.get(6)  + abc.get(10);

[Chrisir]

When you write the code lines automatically by another Sketch B it doesn’t matter

In fact you can let B insert “\n” after each „+“ automatically

Just use ctrl-t in your main Sketch A after pasting the result from B into A

[Chrisir]

Here is an example for a Sketch that writes code

It generates for loop, your Sketch B generates code with + etc.

Tools for you: Search file name, Join pde-files and others

Did you ever had to write 9 nested for-loops? for (int i0 = 0; i0 < 2; i0++) { for (int i1 = 0; i1 < 2; i1++) { for (int i2 = 0; i2 < 2; i2++) { .... ...... I wrote another Sketch to write these 9 lines with for-loops for me, so I don’t have to do it. You can copy paste the auto-generated code from the direct window in processing: void setup() { size(800, 600); int numberOfDesiredForLoops = 9; for (int i = 0; i < numberOfDesiredForLoops; i++) { …

[paulgoux]

int[] a = new  int[]  {1,2,3,4,5};
int[] ar = new int [a.length];
int[][] b = new int[a.length][a.length*2-1];
int counter = 0;
int[] c = new int[a.length*2-1];
int[][]d = new int [a.length][a.length];
for (int i=0; i<a.length; i++)ar[i] = a[a.length-1-i];
for (int i=0; i<a.length; i++) {

  println("ar",ar[i],"a",a[i]);
  for (int j=0; j<a.length; j++) {
    b[i][j+i] = ar[i]*a[j];
    
    println("output",b[i][j+i],ar[i],a[j]);
  }
  println("index", i, a[i]);
}
for (int i=0; i<b.length; i++) {
  String s = "";
  for (int j=0; j<b[i].length; j++) {
    s+= b[i][j];
     
  }
  println(s);
}
for (int i=0; i<b[0].length; i++) {

  counter = 0;
  for (int j=0; j<b.length; j++) {
    int pos = j;
    counter+= b[j][i];
  }
  c[i] = counter;
  counter++;
}
println("",counter);
for (int i=0; i<c.length; i++) {
print(c[i]);
}

[jafal]

@paulgoux

thank you it is creative idea

[paulgoux]

It probably can be simplified but for loop and array logic isn’t my forte

Good little exercise though

[jafal]

@paulgoux

the way you used is very nice