Checking to see if @Min is either 1 more or 1 less then Mod(x, 5) = 0

[nitekram]

I am looking to create a check to see if the @Min is plus or minus 1 from Mod($x, 5) = 0. So if the @Min was say 14, then I want to know that passed, and if the @Min was 15 or 16 that would also pass. So for every multilple of 5, I want to check if it is 1 minute before, right at a multiple of 5, or 1 minute after.

This works - I assume for 5 minutes after the hour, but how do I do it for every 5 mins?

If Mod(@MIN, 5) = '0' Or Mod(@MIN, 6) = '0' Or Mod(@MIN, 4) = '0' Then
        If Mod(@MIN, 4) = '0' Then $iMin = @MIN + 1 ; @min = 4
    If Mod(@MIN, 5) = '0' Then $iMin = @MIN
    If Mod(@MIN, 6) = '0' Then $iMin = @MIN - 1 ; @min = 6
EndIf

.

[BrewManNH]

Mod doesn't work the way you're using it.

Mod divides the first number by the second number and tells you if there's a remainder, if the return from the function is 0 it's telling you that the first number is evenly divisible by the second number. If the result is anything other than 0 then it's telling you that there's a remainder left over. It does not tell you that the number ends in 5, 6, or 4. Mod(@Min, 5) would tell you if @Min is a multiple of 5 or not, and will work for every 5 minutes in an hour. It will also work for 00 minutes, but every other number used in the second parameter will also return 0 so that's not going to help much.

[nitekram]

Mod doesn't work the way you're using it.

Mod divides the first number by the second number and tells you if there's a remainder, if the return from the function is 0 it's telling you that the first number is evenly divisible by the second number. If the result is anything other than 0 then it's telling you that there's a remainder left over. It does not tell you that the number ends in 5, 6, or 4. Mod(@Min, 5) would tell you if @Min is a multiple of 5 or not, and will work for every 5 minutes in an hour. It will also work for 00 minutes, but every other number used in the second parameter will also return 0 so that's not going to help much.

Understand, and if I put all conditions in for Mod(), meanging 4, 5, 6, 9, 10, 11, 14, 15, 16,ect. then it would work, but I would hope that there is an easier way?

EDIT, or I guess I could just not use Mod() and put conditions for each of the @Min I am looking for, I will look at doing that and see if it works for my issue.

Edited February 19, 2015 by nitekram

[SadBunny]

Something like this?

For $i = 0 To 59

    ConsoleWrite($i & " : " & _checker($i) & @CRLF)
Next

Func _checker($nr)

    $x = Mod($nr, 5)
    Return ($x == 0 Or $x == 1 Or $x == 4)
EndFunc   ;==>_checker

[BrewManNH]

I came up with something almost the same SadBunny.

For $I = 0 To 59
    $Min = Mod($I, 5)
    If $Min = 1 Or $Min = 4 Or $Min = 0 Then
        ConsoleWrite("Min is correct " & $Min & @TAB & $I & @CRLF)
    Else
        ConsoleWrite("Min isn't correct " & @CRLF)
    EndIf
Next

[SadBunny]

Note that <any integer> mod <n> is anywhere between 0 .. (n-1). So another way to put it is:

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)

    $modulo = Mod($value, $n)
    Return ($modulo <= $range Or $modulo >= $n-$range)
EndFunc   ;==>_checker

With this function you could do your task by saying _is...(@Min, 1, 5). But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).

[nitekram]

Note that <any integer> mod <n> is anywhere between 0 .. (n-1). So another way to put it is:

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)

    $modulo = Mod($value, $n)
    Return ($modulo <= $range Or $modulo >= $n-$range)
EndFunc   ;==>_checker

With this function you could do your task by saying _is...(@Min, 1, 5). But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).

I like this one, as I have no clue what it does, but it does work...thanks to all.

[nitekram]

One question, I am going to return array of the true, as well as the value. How can I return the multiple number, meaning if the value was 6, how can I return 5?

EDIT So something like this:

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    Local $aAnswer[2]
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n)
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        $aAnswer[0] = True
        $aAnswer[1] = $value
        Return $aAnswer
    Else
        Return ($modulo <= $range Or $modulo >= $n - $range)
    EndIf

EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

.

Edited February 19, 2015 by nitekram

[SadBunny]

Well if you're writing it like that, you can change the

Else
      Return ($modulo <= $range Or $modulo >= $n - $range)
 EndIf

to:

EndIf
    Return False

... as your current code reaches the else block if and only if that same truth formula already returned False. No need to do it again.

[SadBunny]

But... Instead of returning an array, which you then have to read out again in your calling code, why not just have the function return -1 (or any other negative value) in case of failure? Like this...

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    Local $modulo = Mod($value, $n)
    Return (($modulo <= $range Or $modulo >= $n - $range) ? $value : -1)
EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

[nitekram]

I think this will work? It might not be the cleanest, but I am doing a search on those minutes being multiple of 5 with the variance.

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    Local $aAnswer[2]
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n)
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        $aAnswer[0] = True
        If $modulo <= $range Then

            IniWrite('p:\uip\testingtime.ini', @MDAY, $modulo, $range)
            If Mod($value, 5) = 0 Then $aAnswer[1] = $value
            If Mod($value + 1, 5) = 0 Then $aAnswer[1] = $value + 1
            If Mod($value - 1, 5) = 0 Then $aAnswer[1] = $value - 1
            ConsoleWrite('TRUE1 ' & $aAnswer[1] & @CRLF)

        ElseIf $modulo >= $n - $range Then

            IniWrite('p:\uip\testingtime.ini', @MDAY, $modulo, $n - $range)
            If Mod($value, 5) = 0 Then $aAnswer[1] = $value
            If Mod($value + 1, 5) = 0 Then $aAnswer[1] = $value + 1
            If Mod($value - 1, 5) = 0 Then $aAnswer[1] = $value - 1
            ConsoleWrite('TRUE2 ' & $aAnswer[1] & @CRLF)
        EndIf

        Return $aAnswer
    Else
        Return False
    EndIf

EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

.

[kylomas]

nitekram,

Nice, but should you not be setting $aAnswer[0] to false at some point...

expandcollapse popup

consoleWrite('!Truth table' & @CRLF)

for $1 = 1 to 60
    ConsoleWrite(stringformat('%-10i %-10i %-10i %-10i %-10s', $1, 2, 5, _
        _isValueWithinRangeFromMultipleOfN($1, 2, 5)[1], _
        _isValueWithinRangeFromMultipleOfN($1, 2, 5)[0]) & @CRLF)
next

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    Local $aAnswer[2]
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n)
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        $aAnswer[0] = True
        If $modulo <= $range Then

            IniWrite('p:\uip\testingtime.ini', @MDAY, $modulo, $range)
            If Mod($value, 5) = 0 Then $aAnswer[1] = $value
            If Mod($value + 1, 5) = 0 Then $aAnswer[1] = $value + 1
            If Mod($value - 1, 5) = 0 Then $aAnswer[1] = $value - 1
            ;ConsoleWrite('TRUE1 ' & $aAnswer[1] & @CRLF)

        ElseIf $modulo >= $n - $range Then

            IniWrite('p:\uip\testingtime.ini', @MDAY, $modulo, $n - $range)
            If Mod($value, 5) = 0 Then $aAnswer[1] = $value
            If Mod($value + 1, 5) = 0 Then $aAnswer[1] = $value + 1
            If Mod($value - 1, 5) = 0 Then $aAnswer[1] = $value - 1
            ;ConsoleWrite('TRUE2 ' & $aAnswer[1] & @CRLF)
        EndIf

        Return $aAnswer
    Else
        Return False
    EndIf

EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

kylomas

edit: your math might be off by 1 also...

Edited February 19, 2015 by kylomas

[SadBunny]

I have no idea why you make that function so complex and redundant while it's actually simple enough to use an almost-oneliner 

- Why are you set on working with an array as a return value while it's possible with an int?

- Why are you duplicating the exact same code in an if-else for cases $modulo >= $n-$range and $modulo <= $range, while the work is the exact same in both cases and you already know that at least one of them is true (because you're inside the first If)?

/edit - Why are you forcing (potentially, i.e. if the minute is not in the accepted ranges) 4 addition and 6 modulo operations while 1 is enough?

- Why are you iniwriting anyway from within a function that has no other work than performing a simple check? That's questionable code flow

 

I guess it all comes down to taste, but my suggestion would be to rethink the whole thing (and then accept my suggestions )

 

At the end of the day it doesn't matter, to each is own If it works for you it works for you. Have a good one!

Edited February 19, 2015 by SadBunny

[SadBunny]

nitekram,

Nice, but should you not be setting $aAnswer[0] to false at some point...

 

Not necessarily, his code now either returns

either an array with the first element set to True and the second element set to $value

or False.

It could work, provided that you know that really well from the calling code. Problem is you can't treat the function as a boolean any more, as the truth value of a filled array is also False (which I did not expect but just found out; if you dimension and fill an array and then write "If $array then <do affirmative>" it will still fail )

[kylomas]

nitekram,

The modulo routine needs to account for op1 being < op2 like...

ConsoleWrite('!Truth table' & @CRLF)

for $1 = 1 to 60
    ConsoleWrite(stringformat(' $value = %-10i $result = %-10i', $1, _isValueWithinRangeFromMultipleOfN($1, 2, 10)) & @CRLF)
next

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)

    ; account for value less than iteration number
    if $value < $n then return ((($n - $value) > $range) ? -1 : $value)

    Local $modulo = Mod($value, $n)
    Return ($modulo <= $range or $n - $modulo <= $range) ? $value : -1

EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

[SadBunny]

 

nitekram,

The modulo routine needs to account for op1 being < op2 like...

ConsoleWrite('!Truth table' & @CRLF)

for $1 = 1 to 60
    ConsoleWrite(stringformat(' $value = %-10i $result = %-10i', $1, _isValueWithinRangeFromMultipleOfN($1, 2, 10)) & @CRLF)
next

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)

    ; account for value less than iteration number
    if $value < $n then return ((($n - $value) > $range) ? -1 : $value)

    Local $modulo = Mod($value, $n)
    Return ($modulo <= $range or $n - $modulo <= $range) ? $value : -1

EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

 

Why would you break the algorithm for the integers smaller than $n? (In this case, the first few minutes of the hour? Not to mention ALL negative numbers?) /me puzzled. 

Btw, minutes go from 0 to 59. Luckily, with the modulo technique, it also works for all representable integers

This is becoming a weird thread. Don't know if it's me.

Edited February 19, 2015 by SadBunny

[kylomas]

@SadBunny - I'm not, it is for values smaller than the interval.  Run my code with the 0 test commented out and you will see it.

[SadBunny]

@SadBunny - I'm not, it is for values smaller than the interval.  Run my code with the 0 test commented out and you will see it.

 

I have run it with your addition commented out, I wrote the code  (Edit: also I ran it from 0 to 59 instead of from 1 to 60 ) What I don't understand is why would you want to exclude all numbers smaller than $n * 1? OP wants to get a True if the current minute is divisible by 5, with a tolerance of 1 minute. Your addition will break that for every whole hour and the first and fourth minute past the whole hour. (Or, in terms of my algorithm, all minutes of the hour between 0 and the first instance of the chosen interval, so with your example minutes 0 to 9, while minute 0, 1, 2, 8 and 9 should clearly also trigger.)

Why would you want to do that?

P.S. Remember that 0 is just as much a multiple of N as any other multiple of N.

Edited February 19, 2015 by SadBunny

[kylomas]

 I wrote the code

 

Not quite, my formula is slightly different but the net result is the same.

Why would you want to do that?

 

'cause I'm cross-eyed...You're right, the formula stands on it's own, my mistake...

[SadBunny]

Not quite, my formula is slightly different but the net result is the same.

 

'cause I'm cross-eyed...You're right, the formula stands on it's own, my mistake...

 

I am too, I didn't see that you slightly rewrote my thingy. Ok, mystery solved