Doji for a strategy question

[NinjaTrader_JessicaP]

I see what you mean.

I am going to attempt a pseudo-code version of what I posted. I hope this helps but please don't hesitate to let me know if I can take another approach. I believe this pseudo code will be useful if we do need to take another approach as a reference.

[FONT=Courier New]          make a container to store the current bar's minimum[/FONT]
[FONT=Courier New][FONT=Courier New]          make a container to store the current bar's minimum[/FONT]

          Run a sub-program called OnBarUpdate. Everything
          that is created inside this subprogram is destroyed.
          This is why we made places to store stuff earlier.
          Sub-program contents :
              If this is the first tick of the bar, reset
              our stored current minimum and maximum

              If the current ask price is lower than the lowest
              price in this bar so far, the current ask price
              is now the current bar's minimum price

              If the current bid price is higher than the
              highest price in this bar so far, the current
              bid price is now the current bar's maximum
              price

              Now you can write your own code here, and now
              you know what the bid, ask, max, and min prices
              are
[/FONT]

[outsource]

I see what you mean.

I am going to attempt a pseudo-code version of what I posted. I hope this helps but please don't hesitate to let me know if I can take another approach. I believe this pseudo code will be useful if we do need to take another approach as a reference.

[FONT=Courier New]          make a container to store the current bar's minimum[/FONT]
[FONT=Courier New][FONT=Courier New]          make a container to store the current bar's minimum[/FONT]

          Run a sub-program called OnBarUpdate. Everything
          that is created inside this subprogram is destroyed.
          This is why we made places to store stuff earlier.
          Sub-program contents :
              If this is the first tick of the bar, reset
              our stored current minimum and maximum

              If the current ask price is lower than the lowest
              price in this bar so far, the current ask price
              is now the current bar's minimum price

              If the current bid price is higher than the
              highest price in this bar so far, the current
              bid price is now the current bar's maximum
              price

              Now you can write your own code here, and now
              you know what the bid, ask, max, and min prices
              are
[/FONT]

Thank you,Jessica,

but whta is my entry logic after all that for this one?:

if (High[0] < High[1] && Low[0] > Low[1])
&& (High[0] > High[1] && Low[0] < Low[1])

[NinjaTrader_JessicaP]

Hello Outsource,

So your entry logic in pseudo code is likely not what you would like it to be. That looks like this :

[FONT=Courier New]if ...
[/FONT][FONT=Courier New]...the most recent high is less than the second most recent high...[/FONT]
[FONT=Courier New]...the most recent low is greater than the second most recent low...
[/FONT][FONT=Courier New]...the most recent high is greater than the second most recent high...
[/FONT][FONT=Courier New]...the most recent low is less than the second most recent low...
[/FONT]

Because it is impossible for the high one bar ago to be both greater than and lower than the high two bars ago, and the same is true for the low, this code block can be replaced by, simply (in real code and not pseudo code)

[FONT=Courier New]if (false)[/FONT]

A lot of programmers, in fact, use <> to mean "not equal to" for this reason.

You probably meant to split this into two code blocks, e.g.

[FONT=Courier New]if ( (High[0] < High[1]) && (Low[0] > Low[1]) )
{
    // do stuff here
}[/FONT]
[FONT=Courier New]if ( (High[0] > High[1]) && (Low[0] < Low[1]) )
{
    // do other stuff here
}[/FONT]

Otherwise, if you simply wanted to do things when the High and Low prices are not equal, that would look like this :

[FONT=Courier New]if ( (High[0] != High[1]) && (Low[0] != Low[1]) )
{
    // do stuff here
}[/FONT]

If neither of these is the case, if you could let me know what your intentions are, I can look deeper into this.

[outsource]

Hello Outsource,

So your entry logic in pseudo code is likely not what you would like it to be. That looks like this :

[FONT=Courier New]if ...
[/FONT][FONT=Courier New]...the most recent high is less than the second most recent high...[/FONT]
[FONT=Courier New]...the most recent low is greater than the second most recent low...
[/FONT][FONT=Courier New]...the most recent high is greater than the second most recent high...
[/FONT][FONT=Courier New]...the most recent low is less than the second most recent low...
[/FONT]

Because it is impossible for the high one bar ago to be both greater than and lower than the high two bars ago, and the same is true for the low, this code block can be replaced by, simply (in real code and not pseudo code)

[FONT=Courier New]if (false)[/FONT]

A lot of programmers, in fact, use <> to mean "not equal to" for this reason.

You probably meant to split this into two code blocks, e.g.

[FONT=Courier New]if ( (High[0] < High[1]) && (Low[0] > Low[1]) )
{
    // do stuff here
}[/FONT]
[FONT=Courier New]if ( (High[0] > High[1]) && (Low[0] < Low[1]) )
{
    // do other stuff here
}[/FONT]

Otherwise, if you simply wanted to do things when the High and Low prices are not equal, that would look like this :

[FONT=Courier New]if ( (High[0] != High[1]) && (Low[0] != Low[1]) )
{
    // do stuff here
}[/FONT]

If neither of these is the case, if you could let me know what your intentions are, I can look deeper into this.

Jessica,i don`t know how programmatically,but can you imagine that the bar can do many things as it developes intrabar,can you imagine all the people?.....The code that i posted just illustrates how an inside bar becomes an outside bar - intrabar.the code that does it for me exposes the patterns via int series.So they are ID`ed as,lets say:

the inside bar == 1

the oustside bar == 2

so itry to get the logic like

if this.pattern[0] == 1
&& this.pattern[0] == 2

{entry}

but it doesn`t work so far,but you are amazing in your assistance!

[NinjaTrader_JessicaP]

Hello Outsource,

I appreciate your patience, but in your simple example, you have "this.pattern" taking on multiple values in a single tick, which it can not do.

Are you asking how to tell if something has taken on the first pattern, and then if it has, if it has taken on the second pattern, within the same bar?

If this is the case, you will need to refer to the 3rd example in the first post I sent to you. If I can clarify that post, or provide any other assistance, please let me know.

[outsource]

Are you asking how to tell if something has taken on the first pattern, and then if it has, if it has taken on the second pattern, within the same bar?

If this is the case, you will need to refer to the 3rd example in the first post I sent to you. If I can clarify that post, or provide any other assistance, please let me know.

Yes,because this one always shows up first(when and if it does):

if (High[0] < High[1]) && (Low[0] > Low[1])

after a while this one shows up during the same bar:

if (High[0] > High[1]) && (Low[0] < Low[1])

Our quest so far is how to tell the strategy to b able to recognize both and open a position.The strategy calls an indicator where those two gus are ID`ed.

[NinjaTrader_JessicaP]

Hello outsource,

I would like at this time to return to a piece of code I posted earlier, and reproduce it in pseudo-code, so that it is easier to understand what this does.

[FONT=Courier New]private bool tradedYet = false;
protected override void OnBarUpdate()
{

    // ...

    if(FirstTickOfBar)
    {
        tradedyet = false;
    }

    if(! tradedYet)
    {        [/FONT][FONT=Courier New] if([/FONT][FONT=Courier New]Position.MarketPosition != MarketPosition.Long && Low[0] < Open[0] && Close[0] > Open[0] )
[/FONT][FONT=Courier New]         {
            [/FONT][FONT=Courier New]EnterLong();[/FONT][FONT=Courier New]
            tradedYet = true;[/FONT][FONT=Courier New]
        [/FONT][FONT=Courier New]}
        [/FONT][FONT=Courier New]else if([/FONT][FONT=Courier New]Position.MarketPosition != MarketPosition.Short && High[0] > Open[0] && Close[0] < Open[0] )
        [/FONT][FONT=Courier New]{
            [/FONT][FONT=Courier New]EnterShort();[/FONT][FONT=Courier New][FONT=Courier New]
            tradedYet = true;[/FONT]        [/FONT][FONT=Courier New]}
    }

    // ...

} [/FONT]

Here is the same thing in pseudo-code

[FONT=Courier New]keep track of our state between ticks
run a sub program called OnBarUpdate
    if this is the first tick in the bar
        reset our state to "not yet traded"
    if we are in a "not yet traded state"
        enter long or short side of the market depending
        set our state to "we have performed a trade"
[/FONT]

So this accomplishes something very close to what you were asking just now. Let's take what you just asked, and re-write it in pseudo code.

I will refer to this,

[FONT=Courier New]( (High[0] < High[1]) && (Low[0] > Low[1]) )[/FONT]

as your "thin" condition, since you start off with a wide range and end up with a short range, and

[FONT=Courier New]( (High[0] > High[1]) && (Low[0] < Low[1]) )[/FONT]

as your "thick" condition, since you start off with a thin range and end up with a wide range. Our pseudo code for what you just asked for now looks like,

[FONT=Courier New]keep track of our state between ticks
run a sub program called OnBarUpdate
    if this is the first tick in the bar
        reset our state to "[B]thick condition not yet met[/B]"
    if we are in a "[B]thick condition not yet met[/B]" state
        set our state to "[B]thick condition has been met[/B]"
    [B]otherwise
        if our "thin condition" has been met
            do something
[/B][/FONT]

That looks like, in regular code,

[FONT=Courier New]private bool [B]thickMet[/B] = false;
protected override void OnBarUpdate()
{

    // ...

    if(FirstTickOfBar)
    {
        [B]thickMet[/B] = false;
    }

    if(! [B]thickMet[/B])
    {        [/FONT][FONT=Courier New] if [/FONT][B][FONT=Courier New]( (High[0] > High[1]) && (Low[0] < Low[1]) )[/FONT][/B][FONT=Courier New]
[/FONT][FONT=Courier New]         {[/FONT][FONT=Courier New]
            [B]thickMet[/B] = true;[/FONT][FONT=Courier New]
        [/FONT][FONT=Courier New]}[/FONT][FONT=Courier New]
    }
    else
    {[/FONT][FONT=Courier New][FONT=Courier New] if [/FONT][B][FONT=Courier New]( (High[0] < High[1]) && (Low[0] > Low[1]) )[/FONT][/B][FONT=Courier New]
[/FONT][FONT=Courier New]         {[/FONT][FONT=Courier New]
            [B]// do stuff[/B][/FONT][FONT=Courier New]
        [/FONT][FONT=Courier New]}
[/FONT]    }
} [/FONT]

This is the sequence of events this specifies :

1. On our first tick, we start testing to see if the thick condition was met
2. Around the 3rd tick of the bar, the thick condition is met.
   1. We now start looking for the thin condition
   
   
3. Around the 6th tick of the bar, the thin condition is met
   1. The code where I have "// do stuff" is executed
   
   

If I did not understand the goal correctly, please copy the last pseudo-code sample in this post and modify it in your reply. That will make it very easy to get you the right assistance.

[outsource]

Hi,Jessica,i think you understood correctly,but it still doesn`t work.Here is my version:

if(! thickMet)
{ if (High[0] > High[1]) && (Low[0] < Low[1])
{ thickMet = true;}
}
else
{
if (High[0] < High[1]) && (Low[0] > Low[1])
{xR.Set(0, true);}
}

then,im trying to call the sequence via strategy

if(this.pa.XR[0] == true)

{enter}

no luck

[NinjaTrader_JessicaP]

Hello outsource,

I had to take some guesses at the code outside the sample you provided, but I am providing a sample which incorporates what you gave me, which does work. Please review and modify this code sample to suit your needs, and please let me know if you have any questions.

[outsource]

Hello outsource,

I had to take some guesses at the code outside the sample you provided, but I am providing a sample which incorporates what you gave me, which does work. Please review and modify this code sample to suit your needs, and please let me know if you have any questions.

Thank you,Jessica,i`ll try it asap and get back to you!

[outsource]

Jessica,i cant only see now how i call the value via strategy?Or is it the strategy that you provided?For me it would be easier if it was an indicator.Can i just move all the parts to the new indicator?If so,what is the value to call?

Thank you

[NinjaTrader_JessicaP]

Hello outsource,

Bear in mind that you can not place trades from indicators. If you need some help getting started with strategies, you can refer to this section of the help guide

http://ninjatrader.com/support/helpGuides/nt7/running_ninjascript_strategies.htm

That said, I have just tested, and the code sample I provided will in fact compile into an Indicator. I am providing a code sample in which I have pasted it into an indicator.

Please let us know if there are any other ways we can help.

[outsource]

Hello outsource,

Bear in mind that you can not place trades from indicators. If you need some help getting started with strategies, you can refer to this section of the help guide

http://ninjatrader.com/support/helpGuides/nt7/running_ninjascript_strategies.htm

That said, I have just tested, and the code sample I provided will in fact compile into an Indicator. I am providing a code sample in which I have pasted it into an indicator.

Please let us know if there are any other ways we can help.

Thanks,Jessica!Seems to be working perfectly great!Thank you for your assistance and have a great Weekend!

[outsource]

Hello,Support,

Let`s get to the topic please,shall we?

The question about the following sequence which i need to try a little tweak on:

    if(FirstTickOfBar)
    {
        thickMet = false;
    }

    if(! thickMet)
    {         if ( (High[0] > High[1]) && (Low[0] < Low[1]) )
         {
            thickMet = true;
        }
    }
    else
    { if ( (High[0] < High[1]) && (Low[0] > Low[1]) )
         {
            // do stuff
        }
    }
}

Now,following the example above,is the tripple alternation possible?For example:

 if(FirstTickOfBar)
    {
        thickMet = false;
    }

    if(! thickMet)
    {         if ( (High[0] > High[1]) && (Low[0] < Low[1]) )
         {
            thickMet = true;
        }

         else    if ( High[0] = High[1] )
         {
            thickMet = true;
         }
    }
    else
    { if ( (High[0] < High[1]) && (Low[0] > Low[1]) )
         {
            // do stuff
        }
    }
}

Will it be working properly that way?

Thank you

[NinjaTrader_ZacharyG]

Hello outsource,

In your modified logic, your script is first going to check if High[0] > High[1] AND if Low[0] < Low[1]. If both of these conditions are true, thickMet is set to true.

Now if both of these conditions are NOT true, or if just one is NOT true, your script is going to evaluate your else if statement. If High[0] equals High[1], then thickMet is set to true.

Please note that in a conditional statement, you need to check for equality with two equal signs (==).

else if (High[0] == High[1])