taylansan 8 Posted January 19 Share Posted January 19 Hello, I'm checking a formula as the following: (x / (x+y)) / ((x-1) / (x-1+y)) With this calculation, I'm trying to get exactly 0.5 (not 4.9999, not 5.0000001). So, I have my code as: ;Local $x = 3, $y = 1 ;Local $x = 15, $y = 6 Local $x = 85, $y = 35 $result = ($x / ($x + $y)) * (($x - 1) / ($x - 1 + $y)) ConsoleWrite("Output will be: " & $x & " " & $y & " --> " & $result & @CRLF) If $result = 0.5 Then ConsoleWrite("YES, we're inside IF" & @CRLF) Else ConsoleWrite("Noooo" & @CRLF) EndIf If I use the combination of (3,1) and (15,6) the result is exactly 0.5 and I can see the console "YES, we're inside IF". But the combination of (85,35) returns "Noooo". We can also see that all these three combinations are exactly 0.5: If I change the condition as below, then yeah, it will return. But I'm looking for the exact 0.5 value. ;If $result = 0.5 Then If $result > 0.4999999999 Then So, my question is: Why the combination of (85,35) didn't return "YES, we're inside IF" even the output shows as it's exactly 0.5? Notes / assumptions: We only support up to 15 decimals. These rational numbers are cut at 15th digit, so we can never calculate the exact value. This assumption can be wrong. Because: (3,1)'s 2nd multiplier (2/3) will return 0.666666666666667; but we're inside that if. (15,6)'s 2nd multiplier (14/20) will return 0.714285714285714; but we're inside that if again. (85,35)'s 1st multiplier (85/120) will return 0.708333333333333 and 2nd multiplier (84/119) will return 0.705882352941177; but we're not inside that if anymore. If I add a rounding with 15 digits, it will work: $result = Round(($x / ($x + $y)) * (($x - 1) / ($x - 1 + $y)), 15) TY. Link to post Share on other sites

AspirinJunkie 82 Posted January 19 Share Posted January 19 5 minutes ago, taylansan said: Why the combination of (85,35) didn't return "YES, we're inside IF" even the output shows as it's exactly 0.5? Because it is not exactly 0.5. It is only rounded on output, so you don't see this. Consolewrite basically makes an implicit string() around the number variable to be output. And this has the effect of rounding if the difference occurs after the 14th decimal place. Modify your output line as follows to avoid this implicit rounding: ConsoleWrite(StringFormat("Output will be: %d %d --> %.17f\n", $x, $y, $result)) taylansan 1 Link to post Share on other sites

Nine 1,606 Posted January 19 Share Posted January 19 No computer can calculate exactly 2/3 (as your first example). A true mathematician would create a computer formula that fits the limitations of a machine. “They did not know it was impossible, so they did it” ― Mark Twain Spoiler Block all input without UAC Save/Retrieve Images to/from Text Monitor Management (VCP commands) Tool to search in text (au3) files Date Range Picker Virtual Desktop Manager Sudoku Game 2020 Overlapped Named Pipe IPC HotString 2.0 - Hot keys with string x64 Bitwise Operations Multi-keyboards HotKeySet Recursive Array Display Fast and simple WCD IPC Multiple Folders Selector GIF Animation (cached) Screen Scraping Link to post Share on other sites

pixelsearch 470 Posted January 20 Share Posted January 20 Just a thought : when using OP's formula, we got 2 divisions first, then 1 final multiplication, which increases the resulting decimal issue : $result = ($x / ($x + $y)) * (($x - 1) / ($x - 1 + $y)) Why not multiplying twice in the first place, then dividing only once ? $result = ($x * ($x - 1)) / (($x + $y) * ($x - 1 + $y)) Link to post Share on other sites

seadoggie01 301 Posted January 20 Share Posted January 20 Pixelsearch's answer is interesting and might be good enough. If you do need to verify that a number is exactly 1/2 though, you could do something like this... ; Top half of the division $numerator = ($x * ($x - 1)) ; Bottom half of the division $denomenator = (($x + $y) * ($x - 1 + $y)) ; Get the factors of the numerator and denomenator Local $aNumFactors = Factor($numerator) ; Factor isn't a function, left to you to figure out Local $aDenFactors = Factor($denomenator) ; Returns an array of numbers ; Now, compare the arrays and remove non-unique elements (I'm not doing this, it'd be non-trivial to write. Check on the _Array* functions?) ; Then, check to see if you have nothing in $aNumFacors (it's a 1) ; and if you have a 2 remaining in $aDenFacors All my code provided is Public Domain... but it may not work. Use it, change it, break it, whatever you want. Spoiler My Humble Contributions:Personal Function Documentation - A personal HelpFile for your functionsAcro.au3 UDF - Automating Acrobat ProToDo Finder - Find #ToDo: lines in your scriptsUI-SimpleWrappers UDF - Use UI Automation more Simply-erKeePass UDF - Automate KeePass, a password manager Link to post Share on other sites

jchd 1,810 Posted January 21 Share Posted January 21 (edited) On 1/19/2023 at 8:59 PM, taylansan said: (x / (x+y)) / ((x-1) / (x-1+y)) This is not what you use subsequently. You use (x/(x + y))*((x - 1)/(x + y - 1)) As @pixelsearchadvised, compute separately numerator and denominator, and only then perform the final division. Note that this is essentially what you show in your examples in the picture joined. Since you only expect a result exactly = 1/2, you should make your life easier by computing the inverse fraction ((x + y)/x)*((x + y - 1)/(x - 1)) in the form ((x + y)*(x + y - 1))/(x*(x - 1)) and comparing to 2. Your computation remains then in ℤ unless the result != 2 For $x = 1 To 1000 For $y = 1 To 1000 If (($x + $y)*($x + $y - 1))/($x*($x - 1)) = 2 Then ConsoleWrite("(" & $x & ", " & $y & ") verifies the condition" & @LF) EndIf Next Next (3, 1) verifies the condition (15, 6) verifies the condition (85, 35) verifies the condition (493, 204) verifies the condition Note also you can spare the division: For $x = 1 To 1000 For $y = 1 To 1000 If (($x + $y)*($x + $y - 1)) = 2*($x*($x - 1)) Then ConsoleWrite("(" & $x & ", " & $y & ") verifies the condition" & @LF) EndIf Next Next Edited January 21 by jchd Danyfirex 1 This wonderful site allows debugging and testing regular expressions (many flavors available). An absolute must have in your bookmarks.Another excellent RegExp tutorial. Don't forget downloading your copy of up-to-date pcretest.exe and pcregrep.exe hereRegExp tutorial: enough to get startedPCRE v8.33 regexp documentation latest available release and currently implemented in AutoIt beta. SQLitespeed is another feature-rich premier SQLite manager (includes import/export). Well worth a try.SQLite Expert (freeware Personal Edition or payware Pro version) is a very useful SQLite database manager.An excellent eBook covering almost every aspect of SQLite3: a must-read for anyone doing serious work.SQL tutorial (covers "generic" SQL, but most of it applies to SQLite as well)A work-in-progress SQLite3 tutorial. Don't miss other LxyzTHW pages!SQLite official website with full documentation (may be newer than the SQLite library that comes standard with AutoIt) Link to post Share on other sites

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