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hakona
61
hakona 2014.09.01 05:01 

could any one tell me please the logic behind that pattern , how to program it ,,, just the idea not the code


whroeder1
13596
whroeder1 2014.09.01 13:00  
hakona: could any one tell me please the logic behind that pattern , how to program it ,,, just the idea not the code
The hard part is splitting the chart into the 3 parts (a, b, and c)
Not compiled, not tested.
int iHi = LocalExtremeBar(LE_Window, 0, +1),
    iLo = LocalExtremeBar(LE_Window, 0, -1);
if(iHi > iLo){ // As pictured
   int iC = iHi;
   int iB = LocalExtremeBar(LE_Window, iC, -1); // End of A/Start of B
   int iA = LocalExtremeBar(LE_Window, iB, -1); // Start of A
   if(High[iA] >  High[iC]
   &&  Low[0]  > (High[iB] + Low[iB])/2.0 ){    // Pattern confirmed
:
Not compiled, not tested.
Pre Build 600
int     LocalExtremeBar(int length, int start, double d){
   if (d > 0.0) return LocalExtremeArray(High, length, start, d);
                return LocalExtremeArray( Low, length, start, d);
}
int     LocalExtremeArray(double arr[], int length, int iStart, double d){
   while(true){
      int iPrev = iStart;     iStart = ArrayExtrema(arr, length, iStart, d);
      if (iStart == iPrev)    break;
   }
   return(iStart);
}
int     ArrayExtrema(double arr[], int length, int iStart, double d){
   // Like ArrayMaximum but with direction and duplicate control.
   int iExtreme = iStart;  double value = arr[iExtreme];
   int iLimit = MathMin(iStart + length, ArraySize(arr));
   for(iStart++; iStart<iLimit; iStart++)  if( (arr[iStart]-value) *d >= 0.){
      iExtreme = iStart;  value = arr[iExtreme];  }   // largest index (>=)
   return iExtreme;
}
Pre Build 600
hakona
61
hakona 2014.09.01 14:28  
thank u 
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