Candlestick analysis. Pattern coding - page 2

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Integer:
You just need to code it first, then, increase the efficiency.

just encode in codobase was https://www.mql5.com/ru/code/291 , and it seems to me that the problem is "solved head-on" and there is no coding

The CandleCode method http://www.forextimes.ru/foreks-stati/candlecode-zhiv is exactly what I'm talking about - i.e. each candle is converted into a digital code, but I doubt that this method will allow to determine the relative position of bars

 
IgorM:

just encode in codobase was https://www.mql5.com/ru/code/291 , and it seems to me that the problem is "solved head-on" and there is no coding

The CandleCode method http://www.forextimes.ru/foreks-stati/candlecode-zhiv seems to be exactly what I'm talking about - i.e. each candle is converted into a digital code, but I doubt that this method will allow to determine the relative position of bars

According to CandleCode one candle has 256 variants. If it's two candlesticks, it is 256*256, and multiplied by the large number of candlesticks' relative positions. If it is three candlesticks then it's a problem...
 
Can draw a horizontal grid and encode by the position of bar prices in the grid. Encodes by three levels, by four levels, by five levels, etc., thus adjusting the total number of variants and accuracy.
 
Integer:
Maybe draw a horizontal grid and code by the position of bar prices in the grid. Encode by three levels, by four, by five, etc., thus adjusting the total number of variants and accuracy.

cool!

Let me try to formalise the task: develop a method for determining the mutual positioning of bars for the 5 given examples

1. red bar Open > Close

2. in example 1, red bar High and green bar Low are insignificant or may not exist (in other examples by analogy) is specified by a constant in the

1. 2. 3. 4. 5.

how to effectively code this "stuff" for the above 5 examples?

 
IgorM:

cool!

Let me try to formalise the task: develop a method for determining the mutual positioning of bars for the 5 given examples

1. red bar Open > Close

2. in example 1, the red bar's High and the green bar's Low are insignificant or may be absent (the rest of the examples by analogy) set by a constant in pp

1. 2. 3. 4. 5.

how to effectively code this "stuff" for the above 5 examples?

First, code each candle separately. And then code the cumulative one.

For comparison take the average size of a candle

 
papaklass:

IgorM, create a separate thread. This discussion is an afterthought here!

They've totally muzzled poor abolk, he's afraid to appear in a branch with his code. :)

I will not mind if moderators post messages to a separate thread or who suggested the competition between programmers will copy my posts to a new thread and add their vision of the problem
 
So how did you decide on the independent expertise?
 
Vinin:

First encode each candle individually. Then code the cumulative.

For comparison, take the average size of the candle

+++
 
IgorM:

cool!

Let me try to formalise the task: develop a method for determining the mutual positioning of bars for the 5 examples given

1. red bar Open > Close

2. in example 1, the red bar's High and the green bar's Low are insignificant or may be absent (the rest of the examples by analogy) set by a constant in pp

1. 2. 3. 4. 5.

how do I effectively code this "stuff" for the 5 examples?

if(Open[j]<Close[j])//Черная свеча
  {
  if(High[j]==Close[j]&&Open[j]==Low[j]) //Черный бар без хвостов
  if(High[j]>Close[j]&&Open[j]>Low[j])   //Черный бар хвост верх и вниз
  if(High[j]>Close[j]&&Open[j]==Low[j])  //Черный бар хвост в верх
  if(High[j]==Close[j]&&Open[j]>Low[j])  //Черный бар хвост вниз
  }
//----------
if(Open[j]>Close[j])//Белая свеча
  {
  if(High[j]==Open[j]&&Close[j]==Low[j])  //Белый бар без хвостов
  if(High[j]>Open[j]&&Close[j]>Low[j])    //Белый бар хвост верх и вниз
  if(High[j]>Open[j]&&Close[j]==Low[j])   //Белый бар хвост в верх
  if(High[j]==Open[j]&&Close[j]>Low[j])   //Белый бар хвост вниз
  }
//----------
if(Open[j]==Close[j])//Нет свечи
  {
  if(High[j]>Open[j]&&Close[j]>Low[j])     //+
  if(High[j]==Open[j]&&Close[j]==Low[j])   //-
  if(High[j]==Open[j]&&Close[j]>Low[j])    //хвост в низ
  if(High[j]>Open[j]&&Close[j]==Low[j])    //хвост в верх
  }

Decode the bar, add your condition, in the right line and get what you want.

By the way, this is one example where if is not replaceable and there are quite a few

 
Alexey: You can decode the bar, add your condition to the right string and get what you want.

By the way, this is one example where if is not replaceable and there are quite a few

This is all clear, and any "average programmer" can implement it, I ask the audience with a question how to encode, well, let it be my question on developing the alphabet for candlestick combinations: comb_A, comb_B, comb_C, comb_D, comb_E - respectively fig. №№ 1-5
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