(MM club member)

2006-08-26 05:47 PM

Math Help

Okay I am not the best person at math, I never got past algerbra, so I am requesting what is most likely a simple question.

I have an array with elements that contain 0 or -1, and I want to see if all elements are = -1, and I thought that the best way to do this would be a bitwise operation but not sure which one would be correct

Code:


$array = 0,0,0,-1,0,-1; example of what the array would contain.
;Option 1
For $ = 0 to 5
  $2 = $2 | $array[$]
Next

;Option 2
For $ = 0 to 5
  $2 = $2 ^ $array[$]
Next

;Option 3

For $ = 0 to 5
  $2 = $2 ~ $array[$]
Next

I just am not sure which operator I would use.

Thanks

Witto

(MM club member)

2006-08-26 06:03 PM

Re: Math Help

I do not know what your goal is, but I would loop through an array like this
Code:


Break On
$array = 0,0,0,-1,0,-1
For $n = 0 to Ubound($array)
	If $array[$n] = -1
		? 'Element ' + $n ' is -1'
	EndIf
Next

Gargoyle

(MM club member)

2006-08-26 06:16 PM

Re: Math Help

In the end I need to know if all the elements are -1 and be able to set a variable to say all = -1 (true) or all elements do not = - 1 (false)

so to continue the examples above

Code:


$array = 0,0,0,-1,0,-1; example of what the array would contain.
;Option 1
For $ = 0 to 5
  $2 = $2 | $array[$]
Next
If $2 = -1 
  $Flag = "True"
Else
  $Flag = "False"
EndIf
;Option 2
For $ = 0 to 5
  $2 = $2 ^ $array[$]
Next
If $2 = -1 
  $Flag = "True"
Else
  $Flag = "False"
EndIf
;Option 3

For $ = 0 to 5
  $2 = $2 ~ $array[$]
Next
If $2 = -1 
  $Flag = "True"
Else
  $Flag = "False"
EndIf

Witto

(MM club member)

2006-08-26 06:25 PM

Re: Math Help

Code:


Break On
$array = 0,0,0,-1,0,-1
$flag = 'true'
For $n = 0 to Ubound($array)
	If $array[$n] = -1
		? 'Element ' + $n ' is -1'
	Else
		$flag = "false"
		$n = Ubound($array)
	EndIf
Next
? "All elements were -1 = " + $flag

Gargoyle

(MM club member)

2006-08-26 06:34 PM

Re: Math Help

Or as my wife just pointed out...

Code:


If Ascan($array,"0") = -1
$Flag = True
Else 
$Flag = False
EndIf

Les

(KiX Master)

2006-08-26 06:40 PM

Re: Math Help

By assumption is easy as long as the rules don't change.

Gargoyle

(MM club member)

2006-08-26 06:45 PM

Re: Math Help

And since I am making the rules....

Witto

(MM club member)

2006-08-26 06:48 PM

Re: Math Help

Gargoyle, that is even better. Like you said, the array only contains 0 or -1

NTDOC

(KiX Master)

2006-08-26 08:01 PM

Re: Math Help

Your wife? Some type of inside humor or does she actually code too?

Gargoyle

(MM club member)

2006-08-26 08:06 PM

Re: Math Help

No actually she is an accountant and looks at the logic differently. She has some basic programming skills (as in MSBasic) but that is it.

So by talking it through with her, sometimes she sees a different way of doing it.

Lonkero

(KiX Master Guru)

2006-08-27 12:47 AM

Re: Math Help

woman had a better logic than you... scary.

DrillSergeant

(MM club member)

2006-08-27 08:57 PM

Re: Math Help

Great help for the next KiXGolf tournament

Sealeopard

(KiX Master)

2006-08-27 10:21 PM

Re: Math Help

Here's my GOLFed down version:
Code:


$a=-1,-1,-1,0,-1
if ubound($a)*3+2=len(join(split(join($a),'-')))
  ? 'Array contains only -1s'
else
  ? 'Array does not only contain -1s'
endif

for a score of 45 for the "-1" check

NTDOC

(KiX Master)

2006-08-27 10:46 PM

Re: Math Help

Not sure how you get 45 with that. Including spaces it should be 149

Including spaces this one is: 73

$=-1,-1,-1,0,-1
IIf(UBound($)*3+2=Len(Join(Split(Join($),'-'))),'T','F')

Lonkero

(KiX Master Guru)

2006-08-27 10:49 PM

Re: Math Help

Code:


$a=-1,-1,-1,0,-1
  ? 'Array does '
if ubound($a)*3+2<>len(join(split(join($a),'-')))
 'not '
endif
  'only contain -1s'

Lonkero

(KiX Master Guru)

2006-08-27 10:50 PM

Re: Math Help

actually....
Code:


$a=-1,-1,-1,0,-1
  ? 'Array does '
if 0<len(join(split(join($a),'-1'),''))
 'not '
endif
  'only contain -1s'

Sealeopard

(KiX Master)

2006-08-27 11:38 PM

Re: Math Help

Knew you'd golf it down, NTDOC thought you were already sleeping

Les

(KiX Master)

2006-08-28 01:43 AM

Re: Math Help

All's fair in a game of golf.

Code:

$=-1,-1,-1,0,-1
IIf(UBound($)*3+2=Len(Join(Split(Join($),-))),T,F)

Lonkero

(KiX Master Guru)

2006-08-28 02:06 AM

Re: Math Help

well, this should still be shorter:
Code:


$=-1,-1,-1,0,-1
IIf(len(join(split(join($a),'-1'),'')),F,T)

Lonkero

(KiX Master Guru)

2006-08-28 02:09 AM

Re: Math Help

actually, this is even shorter:
Code:


$=-1,-1,-1,0,-1
IIf(join(split(join($a),'-1'),''),F,T)

Les

(KiX Master)

2006-08-28 03:01 AM

Re: Math Help

I'm losing sight of what is supposed to be true or false. Anyway, back to the original spousal revelation golfed down a bit.

$=-1,-1,-1,0,-1
IIf(Ascan($array,0)=-1,T,F)

Lonkero

(KiX Master Guru)

2006-08-28 06:47 AM

Re: Math Help

Code:


$=-1,-1,-1,0,-1
IIf(Ascan($array,0)<0,T,F)

NTDOC

(KiX Master)

2006-08-28 10:45 AM

Re: Math Help

Code:

$=-1,-1,-1,0,-1
IIf(Ascan($,0)<0,T,F)

NTDOC

(KiX Master)

2006-08-28 10:48 AM

Re: Math Help

To fit Golf rules though it should be this

Code:

Dim $
$=-1,-1,-1,0,-1
IIf(Ascan($,0)<0,T,F)

Lonkero

(KiX Master Guru)

2006-08-28 02:40 PM

Re: Math Help

well, think it actually should be:
Code:


IIf(Ascan($,0)<0,T,F)

as the input is provided by the caller.
odd thing is, none of us saw the varname typos up there...
weird.

Gargoyle

(MM club member)

2006-08-28 05:02 PM

Re: Math Help

You mean where everyone kept assigning $= and then doing an Ascan($Array).

I figured since you had gone way beyone where I was I would just leave it alone...