Solve any equation with this 175 Byte function

[minxomat]

Hi.
 
It is winter outside, irish coffee is warming my heart and crazy things happen at the german forum at 1am.
 
So i came up with this function which is basically the whole newton equation solving algorithm in 287 (see #4 for 175 version) Bytes (using 6 vars) total. I know, it is highly unoptimized (For-Loops cause math errors and a huge performance drop, so does Dim) and totally against any rules which apply to good coding standards. But it was fun.
 
Example: 11^x+12^x  = 242/x+(144*(Cos(0)*2))/x
 
Next you want to choose a start value. 1 should do it in this case, and this is also the default used in the function. The complete Example:
 

ConsoleWrite(n("11^x+12^x", "242/x+(144*(Cos(0)*2))/x") & @LF)


; Function.........: n
; Does.............: solve equations
; Size.............: 278 Bytes
; Syntax...........: n(lterm, rterm, startval)
; Author...........: minx

Func n($0,$1,$2=1)
Dim $3=$0&"-("&$1&")",$4,$5=1/10^9
For $c=0 To 10^4
$6=$2+$5
$z=Execute(StringReplace($3,"x","$2"))
$2-=$z/((Execute(StringReplace($3,"x","$6"))-$z)/$5)
If IsInt($c/1000)+IsInt($c/2000)=1 Then $4=$2
If Abs($2-$4)+($4=$2)=1 Then Return $2
Next
EndFunc

cheers!

Edited December 14, 2013 by minx

[Marsi]

Tried to shorten it a bit

Func n($0,$1,$2=1)
$0&='-('&$1&')'
For $3=0 To 10^4
$4=Execute(StringReplace($0,'x',$2))
$2-=$4/(Execute(StringReplace($0,'x',$2+1/10^9))-$4)/10^9
If IsInt($3/1000) And IsInt($3/2000)=0 Then $1=$2
If Abs($2-$1+$1=$2)=1 Then Return $2
Next
EndFunc ; 254 Byte

And for the crazy ones:

Func n($0,$1,$2=1,$3=-1,$4=0,$5=Assign,$6=StringReplace)
    Return ($3=-1)?n($0&'-('&$1&')',$1,$2,0):0*$5('4' _
    ,Execute($6($0,'x',$2)))*$5('2',$2-$4/(Execute($6 _
    ($0,'x',$2+1/10^9))-$4)/10^9)*((IsInt($3/1000)And _
    IsInt($3/2000)=0)?$5('1',$2):0)+Abs($2-$1+$1=$2)= _
    1?$2:n($0,$1,$2,$3+1)
EndFunc ; 285 Bytes (without _)

M

[minxomat]

Shortened the shortened version again, 250 Bytes now:
 

Func n($0,$1,$2=1)
$0&='-('&$1&')'
For $3=0 To 10^4
$4=Execute(StringReplace($0,'x',$2))
$2-=$4/(Execute(StringReplace($0,'x',$2+1/10^9))-$4)/10^9
If IsInt($3/1000)+IsInt($3/2000)=1 Then $1=$2
If Abs($2-$1+$1=$2)=1 Then Return $2
Next
EndFunc

[Marsi]

Func n($0,$1,$2=1)
$0&='-('&$1&')'
For $3=0 To 999
$4=Execute(StringReplace($0,'x',$2))
$2-=$4/(Execute(StringReplace($0,'x',$2+1/10^9))-$4)/10^9
Next
Return $2
EndFunc

175

At 2:00 noone sees, that Abs($2-$1+$1=$2)=1 -> $1=$2

M

Edit:

Func n($0,$1,$2=1,$3=Assign,$4=StringReplace)
Return IsString($1)?n($0&'-('&$1&')',0,$2):0*$3(6, _
Execute($4($0,'x',$2)))*$3(2,$2-$6/(Execute($4($0, _
'x',$2+1/10^9))-$6)/10^9)+$1=999?$2:n($0,$1+1,$2)
EndFunc

Edited December 15, 2013 by Marsi

[minxomat]

Only possibility to optimize this further is to reduce the step depth (999 in the 175B version). In the most cases, 9 steps are sufficient for a 4bit comma precision, so that would be 173 Bytes