Any questions from newcomers on MQL4 and MQL5, help and discussion on algorithms and codes - page 1760

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Nikolay Ivanov #:

1) It's not rounding it's a margin of error... and a very big one... in my example it's 10, but on some instruments it could be 100, 1000, 10000 etc... a nice rounding...

2) So you will tell the customer that it is possible and then make his processor 24 by 7 check all the icons on all the charts every second to see if the circle size can change there... + He'll need to make sure your code is always on. I think it's sabotage :)

Bottom line... At first you said the task can be solved, but then it turned out that the values will be approximate and have a large margin of error, and the watchdog code must be running to fix and adjust the values by hovering the processor...

And you think this is called solving a problem? It's called grasping, then convincing/proving/questioning that it's right...

I have a rating of 4.87. You're almost always happy with everything. Considering there are inadequates among the customers, I think it's not bad.

Sorry, but the level of bullshit is off the charts. I'm not gonna answer you again.

 
Nikolay Ivanov #:

1)Look for fractures in all lines.

2) From those where there is a break, choose the one with 0 bar extremum...

Would such an algorithm work?

Yes, didn't put the question correctly, sorry!

We don't touch the bars/curves, only the indicator buffers

Buff[1] < Buff[0], одна из двух высоких= Buff[1] > Buff[0]
 
Andrey Sokolov #:

I have a rating of 4.87. You're almost always happy with everything. Considering there are inadequates among the customers, I think that's pretty good.

Sorry, but the level of bullshit is off the charts. I'm not answering you again.

Save your stubbornness for arbitration... tell disgruntled customers the nonsense you're talking about... Sorry, here's a thermometer with an accuracy of 10 degrees instead of the 0.1 :) enjoy it :)

 
Vitaly Muzichenko #:

At zero.

One of the two low lines= Bar[1] <Bar[0], one of the two high lines=Bar[1] >Bar[0].

Searching along the top 2 lines and the bottom 2 lines

At any given time there could be a break on the second highest and first lowest, or on the lowest and highest, or on the first highest and second lowest. Or on both at once.

How do you determine the position of the lines?

Maybe this will help.

//+------------------------------------------------------------------+
bool BufferSide(double &buff_up,double &buff_dn)
  {
   bool sig=false;
   if(buff_up[0]<buff_up[1] && buff_dn[0]>buff_dn[1]) sig=true;
   return(sig); 
  }
 
Vitaly Muzichenko #:

Yes, didn't put the question right, sorry!

Do not touch the bars/candles, only the indicator buffers

I meant "bars" of indicator buffers :)

would the algorithm I suggested not work?

 
Nikolay Ivanov #:

I was referring to the "bars" of the indicator buffers :)

Wouldn't the algorithm I suggested work?

Seems good, hadn't thought of it.

Main problem: How to optimally calculate the 2 upper and lower lines without a cycle?

Alternatively, when calculating these 4 lines, score the last 2 prices in an array to compare them later.

 
Vitaly Muzichenko #:

Main problem: How to optimally calculate the top and bottom 2 lines without a cycle?

I don't quite get the point, what if I put all the prices into an array and sort in ascending order, then take from the array the extreme ones at both ends?

ArraySort();
 
Nikolay Ivanov #:

I don't quite get the point, what if I put all the prices into an array and sort them in ascending order, then take from the array the outermost prices at both ends?

Right, I'll try, I need 4 prices, the first 2 and the last 2. It's easy to do from an array.

 
Vitaly Muzichenko #:

Right, I'll try it, I need 4 prices, the first 2 and the last 2. It's easy to do from an array.

But to determine the line number from the value found - is it necessary?

 
Nikolay Ivanov #:

but do you need to determine the line number from the value found?

I don't think so, I'll make a structure array, write symbol, price[1] and price[0] there.

That's quite enough. Also, lines/buffers there can be from 2 to 11

The only thing would be to sort it by price[1]
1...175317541755175617571758175917601761176217631764176517661767...1965
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