each one of the 4 types has distinct color:
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Be yourself
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or instead of last 29 orders executed, choose last 29 errors:
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- Post #4,623
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{quote} If(cond1 && (cond3 && cond4 || !cond4 ) ) || (!cond2 && (cond3 && cond4 || !cond4 )) we can simplify it sure cond1 = 1cond2 so if cond1 is flase then cond2 is true , if cond1 is true then cond1 is false so the final answer is (cond3 && cond4 || !cond4 ) proof (cond1 && (cond3 && cond4 || !cond4 ) ) || (!cond2 && (cond3 && cond4 || !cond4 )) = (cond1 || !cond2 ) && (cond3 && cond4 || !cond4 ) as you said ( common factor ) = True && (cond3 && cond4 || !cond4 ) = (cond3 && cond4 || !cond4 ) Note : I didn't realize from the begining that All...
you build summary on my wrong description
let me calrify it more:
if cond2 is TRUE then 100% cond1 is False
but cond2 has 52
different values from A to Z +# + a to zso our statement must be only:
1) If(cond1 && (cond3 && cond4 || !cond4 ) ) || ( cond2 && (cond3 && cond4 || !cond4 ))
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I guess that
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- Jan 21, 2021 5:15pm Jan 21, 2021 5:15pm
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omar--This is the solution
Thank you very much for great assistance.
could it be summarized assuming cond1 and cond2 not related to each other.
the common term: (cond3 && cond4 || !cond4 ). could be extracted as a common factor or not in this case
Thank you very much for great assistance.
could it be summarized assuming cond1 and cond2 not related to each other.
the common term: (cond3 && cond4 || !cond4 ). could be extracted as a common factor or not in this case
Be yourself
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and the above condition will be applied for all expert-scripts.
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- Post #4,627
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- Jan 21, 2021 7:00pm Jan 21, 2021 7:00pm
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I POLOGIZE FOR LAST HOURS VERSIONS
C/D or CLS or DEL will not work at all for ONE-TIME or REPEAT
yesterday when i make small adjusttmnts, I also forget some parenthesis
it will be deleted now, and uploded again.
Be yourself
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Be yourself
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- Jan 21, 2021 7:46pm Jan 21, 2021 7:46pm
{file} I POLOGIZE FOR LAST HOURS VERSIONS C/D or CLS or DEL will not work at all for ONE-TIME or REPEAT yesterday when i make small adjusttmnts, I also forget some parenthesis it will be deleted now, and uploded again.
Maestro you made a masterpiece!
day by day... you are drawing a wonderfull rainbow... that we can see in this dark days...
Thank you again!
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- Jan 21, 2021 8:53pm Jan 21, 2021 8:53pm
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I will redesign the dashboard
you know. best single idea come after practicing all stupid ideas.
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you know. best single idea come after practicing all stupid ideas.
f
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- Jan 21, 2021 11:24pm Jan 21, 2021 11:24pm
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ShortQuiz:
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{quote} Shows chart window open in the same session. Like BTCUSD, HK50m etc.
and exact answer beacuse of this---answer on chart.
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- Jan 22, 2021 3:03am Jan 22, 2021 3:03am
I will redesign the dashboard you know. best single idea come after practicing all stupid ideas. {image} f
- Post #4,635
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- Edited 5:03am Jan 22, 2021 4:36am | Edited 5:03am
omar--This is the solution Thank you very much for great assistance. {image} could it be summarized assuming cond1 and cond2 not related to each other. the common term: (cond3 && cond4 || !cond4 ). could be extracted as a common factor or not in this case
As a theory and by logic if cond1 and cond2 are not related yes it could be simplified and common factor can be taken.
BUT
in your case , and after my understanding , here cond2 is subset of cond1
which means
Records ( orders ) selected based on cond2 "j" are part of the records selected by "All" so you need to filter .
if you follow the logic to simplify the statement then what happens
the result will be
( Cond1 || Cond2 ) && ... the rest of the term
( cond1 || cond2 ) will be always True statement and in this case
All records will be deleted in all cases ( only the magic_status and magic number filter will work in this case ).
So following logic in this case is against your purpose , since we start with wrong assumption that ( they are not related ).
I think this is clear for both of us now.
to make it short : we can't simplify it more in this case
Regards,
