Is it possible to programmatically find ALL horizontal channels/floats at a certain time interval or at least formulate the ToR correctly? - page 2

 
so how do you define it? ))
 
sanyooooook:
you don't have a clear channel detection algorithm, in fact the programmer would have to build that algorithm himself, for 50, nah (only this made me smile nothing whiter)

or rather not even that, not defining it, but finding it...
 
sever29:

or rather, not definitions, but finding...
To find you must first recognise, to recognise you must compare it to something, and to compare it to something you must set a benchmark.
 

To begin with, I think you need to draw all the channels you see by hand... Then you can outline the rules and principles on the basis of which they were drawn. Then it is up to the programmer to make the terms of reference based on the formalisation.

 
sever29:

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sever29 31.08.2010 01:20 am | delete

I'll call it a royal, armoured, cyclic avalanche-matrix with trap elements...


that was my idea too. i just didn't get it from the side...

well, I couldn't do it.

on the renko charts. the brick levels show nice corridors and consolidation zones.

i'm currently working on my reneko levels and calculations. the thing is getting weird.

If you can't see the Renco levels, it's a very powerful normaliser.

 
Any channel, whatever it is, shifts when the trend changes, so automating such a process is rather cumbersome to carry out, but only a local channel definition is possible.
 
sever29:

Here's the problem, I know what should work, but in what way, I can't formulate it.

At a certain time interval, e.g. a week, I need to programmatically find ALL horizontal channels with a width of, for example, 50 pips.

Additional attributes of the beginning and end of the channels existence:

1. The price alternately crosses the boundaries of the channel (turns inside it twice), the trajectory of which, inside the channel, looks like v or ^ (two reversals inside the channel, alternately testing its boundaries).

2. If the price leaves the borders of the channel at a distance equal to its width (this can be variable, e.g. twice the channel width or three times, etc.), we consider the channel to be non-existent:).

We read some information from all the channels found (flat).

Help.


Channels are rarely horizontal, more often oblique. After testing the boundaries twice, what is the probability of repeating a third time?

I think where it's less than 100 pips, it's only chaos.

 
sanyooooook:
If you want to find it, you need to recognise it, if you want to recognise it, you need to compare it with something, if you want to compare it with something, you need to set a benchmark.


I have just now encountered a problem with my vocabulary...

recognize... aren't the attributes in my post enough? A) the date from which we start to identify B) the fixed width of the channel (the range of the flat) C) the trajectory of the price movement inside the borders of the channel, D) alternate testing of its borders from inside E) identification of the channel, i.e. the flat when the price turns in the channel twice. F) end of a flat when the price goes out of the channel to the outside at a distance that is equal to the channel width or double channel width, triple channel width, etc. etc.

What is missing for the benchmark definition of a flat with a given characteristic?

 
gss:

To begin with, I think you need to draw all the channels you see by hand... Then you can outline the rules and principles on the basis of which they were drawn. Then it is up to the programmer to make the terms of reference based on the formalisation.


You cannot draw them by hand, there may be hundreds and hundreds of them in a week. Do you understand what I mean? Move any channel up/down by pips and find another channel...
 
neama:


1. such an idea has crossed my mind too. however, it didn't work out from the side...

I couldn't do it.

2. on the Renco charts. the brick levels show nice corridors and consolidation zones.

I am now busy adjusting my reneko levels and calculating them.

3. cast your eyes in the direction of the reno - it's a very powerful normalizer.


1. Which one? Also to poke fun at the name of the avalanche? :)

2......

3. i don't know if you'll get it right... I'm not interested in what others are interested in and vice versa. I see my way on the fore...

Reason: