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Simulated Annealing - When brute-force takes too long

RTFC

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RTFC

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Simulated Annealing (SA) is a simple technique for finding an acceptable solution (but not necessarily always the absolute best one that exists!) to very hard combinatorial problems, that is, ones for which a brute-force approach of cycling through all possible alternatives to find the global optimum just takes too darn long. Typically, one would be seeking some specific sequence or permutation of a (sub)set, and the number of possibilities is astronomically large. In addition, for SA to be applicable, you'll need to be able to quantify in some way how good any particular trial solution is, or how far distant from the ideal result. The real power of SA lies in these so-called cost functions; you can define as many as you like, with different weights if you like, and these conditions are even allowed to be conflicting. User-defined settings define how tenaciously it should be exploring solution space before homing in on a region with desirable properties, and finding its minimum. You can think of it as a learning algorithm that is allowed to make lots of mistakes in the beginning, before gradually avoiding ever more potentially bad decisions.

A simple Analogy:

You're standing in the middle of an extensive, rough terrain with hills, ridges, bumps, valleys, and tiny, deep depressions. You've got a GPS altimeter to tell you how high above sea-level you currently are, but this area is perpetually shrouded in thick fog. Now find the lowest point. And quickly please. Now what? There's no time to systematically grid and traverse the entire area, and you cannot see more than a few feet ahead. Following the local gradient down-slope will likely get you stuck at a very local minimum, whereas a much lower point may be just beyond the hext hill.

Luckily, you brought your magic boots (the red ones with the twinkling silver stars). These allow you to make huge leaps through the fog, landing safely somewhere else. And although you don't know in advance where you'll end up, the boots magically remember their last-previous departure point, so if you don't like your new surroundings, you can go back one jump (but never more than that). Now to find the lowest point in the landscape, you keep tracking your altimeter changes with every jump. Some jumps will get you to higher ground, others might land you in a deep valley. The trick is not to settle for ever-lower heights immediately, but to allow going back up ever so often, so you can get beyond some high ridge behind which you might find a much lower depression than your best-previous result. Crucially, to decide whether or not to gain height in the misty terrain, you roll some dice you've got in your pocket, and the more jumps you've made, the lower you make the upper bound below which you would still go back uphill. So in the beginning you'll be jumping all over the place (allowing you to sample the terrain extensively), but eventually you'll be limiting yourself to some deep valley you've found, which might be the deepest one all round, but even if not, it will still be a pretty good guess. And you will have found this deep valley in a tiny fraction of the time it would take to do a full land survey.

Simulated annealing needs to be defined in terms of the specific problem you're trying to solve. So it's not possible to provide you with a generic UDF that'll figure out in advance what your ideal solution would look like (for example, in the analogy above, we might be looking for the highest point instead of the lowest one). You need to define that ideal in terms of one or more cost functions that SA will then attempt to minimise. What can be done is to provide you with specific examples.

Example 1.

From a list of user-defined pre-supplied values, select the fewest terms that sum to (or approximate) a predefined target total. This solution was written in response to this thread. Current cost updates are periodically written to console. This example attempts to satisfy two conditions simultaneously: getting a sum that matches the target, and using the smallest number of terms. 

; Simulated Annealing example (combinatorial minimisation), by RTFC (22 Feb 2016)

; Note that this algorithm converges on A *local* minimum (in terms of the
; user-defined cost-function(s)), which is not necessarily THE *global* minimum.
; Note also that the search path, duration, and final result may differ from run to run.
; Several parameters can be tweaked to adjust this.

#include <Array.au3>
#include <Math.au3>

Global $temperat,$path,$kk,$nswap,$nswapstep,$cost,$altcost,$tempstep
Global $costadjust,$ttlsites,$totalcost,$factor,$maxsumlength,$maxsize
Global $site1,$site2,$index1,$index2,$weight_sum,$weight_length
Global $options,$sumlength,$prevlength

; initialise
SRandom(@SEC+@AutoItPID)    ; initialise radnomising seed
$verbose=True           ; T: write regular progress updates to console
$factor=1               ; optional Oracle-response adjustment (not used here)
$prevlength=$sumlength  ; to enable reverting to previous state after _TryChange
$minsumlength=0         ; if nothing else is know, we'll start with a single array entry (base-0)

$options=10             ; larger value = larger likelihood of swapping vs changing sumlength
if $options<3 then Exit ; minimum size for this set-up (see Func _TrySwap)

; adjust the balance of conditions here (see _Cost function)
$weight_sum=1           ; relative importance of matching condition 1 (sum = target)
$weight_length=1        ; relative importance of matching condition 2 (lowest number of terms)

; summation results buffer
Global $bestsum[10]
Global $bestsumlength=UBound($bestsum)

; define the summation result we wish to achieve (try different values here!)
$target = 27
; Note: if you change this value, you may have to adjust $minsumlength below as well!

; define our array of summation terms to select from
$dim=9                  ; this is just the way this problem was presented
$ttlsites=$dim*$dim
$maxsize=$ttlsites-1
$maxsumlength=$ttlsites-2   ; need at least one tail slot for swapping

; enable this for the predefined problem...
$doIntegerTest=True ; False
if $doIntegerTest Then
    Global $aArray = [9,2,2,3,1,1,6,3,4, _
                    4,2,3,4,5,6,7,8,7, _
                    7,2,3,4,5,1,7,2,2, _
                    2,2,3,1,5,5,7,1,4, _
                    3,2,1,2,3,6,6,6,4, _
                    3,2,3,4,5,6,7,8,3, _
                    2,2,3,8,1,4,7,1,2, _
                    1,7,3,5,5,6,7,1,2, _
                    7,2,3,7,5,1,7,8,9]

    ; in this specific case, we already know that we'll need at least 4 terms
    ; because a) 9=max value in array, b) there are only 2 nines in the array, and c) target =3x9
    $minsumlength=3     ; base-0, so 4 entries altogether

Else
    ; OR use this for random floats in range 1-$dim (just as an example)
    Global $aArray[$ttlsites]
    For $cc=0 to $maxsize
        $aArray[$cc]=random(1,$dim,0)   ; non-integer values used here!
    Next

    ; no clue about this constraint in this case
    $minsumlength=0 ; one entry (base-0)
EndIf

;______START OF ANNEALING ROUTINE____________

$nover =1000            ; maximum number of changes at any temperature (for more complicated problems, set this several orders of magnitude higher)
$nlimit=Int($nover/4)   ; maximum number of successful changes before continuing
$nwrite=Int($nover/5)   ; default status update interval if verbose=.t.
$tempsteps=100          ; number of temperature steps to try
$tfactor=0.95           ; annealing schedule: temperature is reduced by this factor after each step

While True
    $temperat=0.5       ; initial temperature; smaller = more aggressive + more myopic search
    $absimp=0           ; counter
    $nswapstepzero=0    ; counter
    $sumlength=$minsumlength        ; base-0

    ; prep the cost vars
    $totalcost=_Cost()
    $cost=$totalcost
    $lowestcost=$totalcost
    $initcost=$totalcost

    ; main loop starts here
    For $tempstep=1 to $tempsteps       ; try up to N temperature steps
        $nswap=0
        $nswapstep=0
        For $kk=1 to $nover
            _TrySwap()      ;   swap and determine cost adjustment

            Switch _AskOracle()     ; Feel the Force, Luke.
                Case True
                    $nswap+=1
                    $totalcost+=$costadjust
                    $cost=$altcost

                    If $lowestcost>$totalcost Then
                        $nswapstep+=1
                        $absimp+=1
                        $lowestcost=$totalcost

                        ; ensure results buffer is sufficiently large
                        If $bestsumlength<=$sumlength Then
                            $bestsumlength+=5
                            ReDim $bestsum[$bestsumlength]
                        EndIf

                        ; flush current-best summation
                        For $bc=0 to $sumlength-1
                            $bestsum[$bc]=$aArray[$bc]
                        Next

                        ; pad tail with zeroes
                        For $bc=$sumlength to $bestsumlength-1
                            $bestsum[$bc]=0
                        Next

                        _ScreenOut()
                        If $totalcost<=0 Then ExitLoop
                    Endif

                Case Else   ; restore the previous state
                    $sumlength=$prevlength
                    $aArray[$index1]=$site1
                    $aArray[$index2]=$site2
            EndSwitch

            ; show we're still alive
            If $verbose And mod($kk,$nwrite)=0 Then _ScreenOut()
            If $nswap>=$nlimit Or $lowestcost<=0 then ExitLoop
        Next

        ; optional early-out scenario (disable for a more thorough search)
        If $nswapstep=0 then $nswapstepzero+=1
        If $nswapstepzero=10 then ExitLoop      ; no more improvements in the last N temperature steps

        ; reduce temperature = likelihood of following a trajectory away from the nearest LOCAL optimum (in the hope of getting nearer to the GLOBAL optimum)
        $temperat*=$tfactor
    Next

    ; present final result
    _Arraysort($bestsum)    ; just for clarity

    $summation="Best result so far (at a cost of " & $lowestcost & ") is: " & @CRLF
    $terms=0
    $result=0
    For $cc=0 to $sumlength
        If $aArray[$cc]>0 then
            $summation&=$aArray[$cc] & "+"
            $result+=$aArray[$cc]
            $terms+=1
        Endif
    Next
    $summation=StringTrimRight($summation,1) & " = " & $result & " (target = " & $target & ")" & @CRLF
    $summation&="Number of summation terms: " & $terms & @CR & "Temperature steps: " & $tempstep & @CR & @CR & "Press <Ok> to try again, <Cancel> to Quit"

    if Msgbox($MB_OKCANCEL,"Simulated Annealing Test Result",$summation)=$IDCANCEL then Exit

    ; shuffle entries a bit for variety
    For $cc=1 to $maxsize*10
        $index1=random(0,$maxsize,1)
        $index2=random(0,$maxsize,1)
        $tmp=$aArray[$index1]
        $aArray[$index1]=$aArray[$index2]
        $aArray[$index2]=$tmp
    Next
WEnd

Exit


Func _AskOracle()

    If $costadjust<0 Then
        Return True
    Else        ; this is where all the magic happens!
        Return (random()<Exp(-($costadjust*$factor)/$temperat))
    Endif

EndFunc


Func _TrySwap()

    $index1=0       ; these vars are all Globals
    $index2=0
    $altcost=0
    $prevlength=$sumlength

    ; decide whether to reduce/increase number of terms, or swap an existing term
    Switch Random(1,$options,1)
        Case 1  ; crop
            $sumlength=_Max($minsumlength,$sumlength-1)

        Case 2  ; extend
            $sumlength=_Min($maxsumlength,$sumlength+1)

        Case Else   ; this likelhood is determined by the value of $options (>=3)
            $index1=random(0,$sumlength,1)
            $index2=random($sumlength+1,$maxsize,1)
    EndSwitch

    ; store current contents, in case we decide later that this was a bad idea
    $site1=$aArray[$index1]
    $site2=$aArray[$index2]

    ; swap contents for now
    $aArray[$index1]=$site2
    $aArray[$index2]=$site1

    ; compute the new sum (as either length or content has changed)
    $altcost=_Cost()

    ; performance difference between original and new state
    $costadjust=$altcost-$cost  ; $cost is already filled in previous pass

EndFunc


Func _Cost()

    Local $cc,$result=0
    For $cc=0 to $sumlength
        $result+=$aArray[$cc]
    Next
    Return (Abs($result-$target)*$weight_sum) + (($sumlength-$minsumlength)*$weight_length)

EndFunc


Func _ScreenOut()

    ConsoleWrite("Simulated Annealing. Initial total cost: " & $initcost & @CRLF)
    ConsoleWrite("Step: " & $tempstep & " of " & $tempsteps & "; Temperature: " & $temperat & @CRLF)
    ConsoleWrite("Executed Swaps: " & $nswap & "; Lowest Cost so far: " & $lowestcost & @CRLF)
    ConsoleWrite("Total Improvements: " & $absimp & "; Improvements this step: " & $nswapstep & @CRLF  & @CRLF)

EndFunc

 

Example 2. The Travelling Salesman problem (TSP)

This is a classic combinatorial minimisation problem, and relevant to real-world logistics: to find the shortest route for visiting all cities exactly once, before returning to the original starting point. As it is quite entertaining to see how the algorithm gradually solves this brain teaser,  I've added a simple GUI that visualises the cities (red circles) and the changing routes between them (blue). The problem becomes exponentially harder to solve when the number of cities is increased. This example (adapted from Press et al., Numerical recipes, 2nd ed., pp. 438-443) employs a single cost function of the total route distance.

TSP.au3

It's important to stress that simulated annealing cannot guarantee that the global optimum will always be found, only that it will likely come up with a fairly good solution, and much faster than brute force ever could. If that's good enough for you, then those red, silver-starred boots might fit you too.:D

 

 

Edited by RTFC
example added

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Melba23 Universalist

Melba23

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RTFC,

Wow, I expected something complex but this is going to take a fair while to understand. I am now reading this to try and get some idea of what is going on in there.

M23

Public_Domain.png.2d871819fcb9957cf44f4514551a2935.png Any of my own code posted anywhere on the forum is available for use by others without any restriction of any kind

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RTFC Universalist

RTFC

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@JohnOne: Yeah, best not play too much with it, or you might go blind.;)

@M23: It really works mostly like I describe in the analogy. Rolling the dice = _AskOracle(), using an exponential distribution. We always accept a lower cost, but if it's higher, then we only accept the jump if we sample below our annealing temperature (well technically, cost scaled by temperature), and the temperature itself is gradually lowered. And the name annealing is apt, as the metal is cooled slowly to give the atoms the opportunity to settle into the crystal lattice.

Edited by RTFC

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RTFC Universalist

RTFC

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Added a 3rd example in the first post, a Sudoku Generator & Solver.

Paraphrasing the script remarks section:

This example illustrates how some types of problem can cause Simulated Annealing to get stuck in a local optimum other than the global one. To get around this, we can apply a thorough reshuffle of all non-fixed parameters, and try again. The harder the problem is, the larger the average number of required reshuffles to find the full solution (see listed examples in script).

Sudoku puzzles with very few given clues (or none) are easy to solve, because many paths exist that lead to full solutions (non-unique for number of clues < 17). Sudoku's with many clues are also easy to solve, because there are only relatively few paths left, many of which yield the full solution.

Sudoku's with (or close to) the minimum number of clues that identify a unique solution are the hardest, because many paths do exist, but most lead to a sub-optimal, incomplete solution.:think: The location of the clues also becomes increasingly important the closer we get to this minimum. Reshuffling allows us to explore this landscape from different starting points.

The new example script (#3) displays a temporary result (with timing) each time it gets stuck; when the true solution is found, it plays a sound before exiting. Note that this may take a long time.:yawn:

NB This is obviously not the fastest way to solve a Sudoku; the point of this example is to show that SimAnn can (eventually) find it, without knowing how to solve it, just by getting feedback on its current attempts, and despite the solution space itself being rather large. Furthermore, this example does not imply that all intractable problems can be solved by repeated reshuffle + retry. For example, it would be useless to attempt to quickly generate bitcoins this way.

 

EDIT: after fixing a bug in updating $totalcost, it turns out sudoku isn't a particularly good example of simulated annealing after all (it just takes too long). Until I find a better way to implement it in this context I've removed this example.

Edited by RTFC

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Acanis Wayfarer

Acanis

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Hey, thanks for providing examples :) I would really like to understand, how TSP works (in AutoIt). 

 

I wanted to try to adapt it to a script of mine, but I find it difficult to follow the example and replace individual parts. Since the cities are global and everything is too connected and the variable names are also a bit hard to read for me. (I am not a very experienced hobby developer^^)

I created a script with a more complicated coordinate system using ChatGPT and a lot of time and implemented a "nearest neighbor" algo. To make it a bit better with my limited knowledge, I tried to use it at least twice, hoping that the script wouldn't do something really stupid at least at the first step.

My hope was somewhat that the "understanding" of the coordinate system is mainly in the calculation of the distance and then I can simply adjust the input (the cities) and replace the function for calculating the distance and then I would be able to progress step by step somehow. But that doesn't work with the structure of the example and since my understanding is running close to the limit, I can't really make any progress or even get started. 

Could someone perhaps help me? I think my example code and the visualization could be used to make the TSP example more comprehensible (using visible coordinates and with this kind of visualisation)... I would be really grateful ^^!

 

My code: 

#include <Array.au3>
#include <GUIConstantsEx.au3>
#include <WindowsConstants.au3>
#include <GDIPlus.au3>
#include <EditConstants.au3>

Global $iMapWidth = 60, $iMapHeight = 60, $iMapRoot = 3

; Beispielkoordinaten
Local $aInputArray = ["2:50:9", "2:50:2", "2:50:4", "1:50:3", "1:50:8", "1:49:8", "3:49:4", "3:51:3", "3:51:8", "2:51:7", "1:51:4", "2:48:8", "2:48:4", "1:48:3", "1:48:2", "1:48:1", "3:48:4", "4:49:4", "3:50:6", "4:50:4", "4:50:5", "5:51:1"]
Local $sStartCoordinate = "3:51:1"
Local $sHomeCoordinate = "2:50:8"


;~ Local $aBestRoute = getBestRoute($sStartCoordinate, $aInputArray, 5)
Local $aBestRoute = getBestRoute($sStartCoordinate, $aInputArray, 5, $sHomeCoordinate)

DrawMap($aInputArray, $sStartCoordinate, $aBestRoute, $aBestRoute[UBound($aBestRoute) - 1])

Func getBestRoute($sStartCoordinate, $aTargets, $iNumberOfTargets, $sHomeCoordinate = False)
    Local $aBestRoutes[10], $aBestRoute[0]
    Local $sEndCoordinate = $sHomeCoordinate ? $sHomeCoordinate : $sStartCoordinate
    Local $aPossibleFirstTargets = getNearestTargets($sStartCoordinate, $aTargets)
    Local $aTargetsWithoutStartC = $aTargets
    Local $iIdStartCoordInTargets = _ArraySearch($aTargets, $sStartCoordinate) ; Falls ein Ziel auch der Startpunkt ist

    If $iIdStartCoordInTargets >= 0 Then _ArrayDelete($aTargetsWithoutStartC, $iIdStartCoordInTargets) ; Wenn Ziel auch Startpunkt, darf das nicht für die Suche des weiteren Weges genutzt werden

    ; Finde mit den neuen Startkoordinaten die besten Wege
    For $i = 0 To UBound($aPossibleFirstTargets) - 1
        $aBestRoutes[$i] = getShortRoute($aPossibleFirstTargets[$i], $aTargetsWithoutStartC, $iNumberOfTargets - 1, $sEndCoordinate)
    Next

    ; Lösche leere Einträge, falls es weniger als 10 Ziele gab
    For $i = UBound($aBestRoutes) - 1 To 0 Step -1
        If $aBestRoutes[$i] = "" Then
            _ArrayDelete($aBestRoutes, $i)
        EndIf
    Next

    ; Finde die beste Route
    Local $iShortestDistance = -1
    For $i = 0 To UBound($aBestRoutes) - 1
        Local $aTemp = $aBestRoutes[$i]
        $aTemp[1] += getDistance($sStartCoordinate, $aTemp[2])
        Local $iTotalDistance = $aTemp[1]

        ; Vergleiche die Gesamtentfernung mit der bisher kürzesten Route
        If $iShortestDistance = -1 Or $iTotalDistance < $iShortestDistance Then
            $iShortestDistance = $iTotalDistance
            $aBestRoute = $aTemp ; Aktuelle Route als beste Route speichern
        EndIf
    Next

    _ArrayInsert($aBestRoute, 2, $sStartCoordinate)

    Return $aBestRoute ; Die kürzeste Route wird zurückgegeben
EndFunc

; Finde die bis zu 10 nächsten Koordinaten, die alle potenziell gute Start-Koordinaten für die weitere Suche wären
Func getNearestTargets($sStartCoordinate, $aTargets)
    Local $aNearestTargets[0]
    Local $iMaxTargets = UBound($aTargets) > 10 ? 10 : UBound($aTargets)

    For $i = 1 To $iMaxTargets
        Local $iShortestDistance = -1
        Local $sNearestTarget = ""

        For $j = 0 To UBound($aTargets) - 1
            Local $iDistance = getDistance($sStartCoordinate, $aTargets[$j])

            If $iShortestDistance = -1 Or $iDistance < $iShortestDistance Then
                $iShortestDistance = $iDistance
                $sNearestTarget = $aTargets[$j]
            EndIf
        Next

        If $sNearestTarget <> "" Then
            _ArrayAdd($aNearestTargets, $sNearestTarget)
            _ArrayDelete($aTargets, _ArraySearch($aTargets, $sNearestTarget))
        EndIf
    Next

    Return $aNearestTargets
EndFunc

Func getShortRoute($sStartCoordinate, $aInputArray, $iNumberOfTargets, $sEndCoordinate)
    Local $aVisited[0] ; Leeres Array für besuchte Ziele
    Local $iTotalDistance = 0
    Local $sCurrentCoordinate = $sStartCoordinate
    Local $iVisitedTargets = 0

    _ArrayAdd($aVisited, $sStartCoordinate) ; Ist visited, weil das hier ja mit NearestTargets arbeitet, also schon das erste Ziel nach der echten Startkoordinate ist

    Local $iIdOfStartCoordinateInInput = _ArraySearch($aInputArray, $sStartCoordinate)
    If $iIdOfStartCoordinateInInput >= 0 Then _ArrayDelete($aInputArray, $iIdOfStartCoordinateInInput) ; Startkoordinate aus Array für mögliche Ziele entfernen, weil hier dann schon abgelaufen als wirklich erstes Ziel!

    For $i = 1 To $iNumberOfTargets
        Local $iShortestDistance = -1
        Local $sNextCoordinate = ""
        Local $iNextIndex = -1

        For $j = 0 To UBound($aInputArray) - 1
            If Not _ArraySearch($aVisited, $aInputArray[$j]) >= 0 Then
                Local $iDistance = getDistance($sCurrentCoordinate, $aInputArray[$j])

                If $iShortestDistance == -1 Or $iDistance < $iShortestDistance Then
                    $iShortestDistance = $iDistance
                    $sNextCoordinate = $aInputArray[$j]
                    $iNextIndex = $j
                EndIf
            EndIf
        Next

        If $iNextIndex <> -1 Then
            $iTotalDistance += $iShortestDistance
            $sCurrentCoordinate = $sNextCoordinate

            _ArrayAdd($aVisited, $sNextCoordinate)
            _ArrayDelete($aInputArray, $iNextIndex)
            $iVisitedTargets += 1
        Else
            ; Kein weiteres Ziel gefunden, brechen Sie die Schleife ab
            ExitLoop
        EndIf
    Next

    $iTotalDistance += getDistance($aVisited[UBound($aVisited) -1], $sEndCoordinate)

    ; Erstellen Sie das Ergebnisarray
    Local $aResult[2] = [$iVisitedTargets + 1, $iTotalDistance] ; +1, da die Startkoordiante nicht als besuchtes Ziel gilt
    For $i = 0 To $iVisitedTargets
        _ArrayAdd($aResult, $aVisited[$i])
    Next

    _ArrayAdd($aResult, $sEndCoordinate)

    Return $aResult
EndFunc

Func DrawMap($aInputArray, $sStartCoordinate, $aBestRoute, $sEndCoordinate)
    Local $iGridSize = 20
    Local $iGuiWidth = 900
    Local $iGuiHeight = 900
    Local $iLineSpacing = $iGuiWidth / $iGridSize ; Berechnen des Abstands zwischen den Linien

    ; GDI+ initialisieren
    _GDIPlus_Startup()

    ; GUI erstellen
    Local $hGUI = GUICreate("Map Visualization", $iGuiWidth, $iGuiHeight)
    GUISetState(@SW_SHOW)

    ; Grafikobjekt erstellen
    Local $hGraphic = _GDIPlus_GraphicsCreateFromHWND($hGUI)
    Local $hBrushOrange = _GDIPlus_BrushCreateSolid(0xFFFFA500) ; Orange
    Local $hBrushRed = _GDIPlus_BrushCreateSolid(0xFFFF0000) ; Rot
    Local $hBrushBlack = _GDIPlus_BrushCreateSolid(0xFF000000) ; Schwarz
    Local $hBrushGrey = _GDIPlus_BrushCreateSolid(0xFF008000) ; Grau
    Local $hPen = _GDIPlus_PenCreate(0xFF000000, 1) ; Schwarzer Stift mit einer Breite von 1 für das Gitter

    ; Arrays mit 2D-Koordinaten erstellen
    Local $aInputArray2D[0]
    Local $aBestRoute2D[0]

    For $i = 0 To UBound($aInputArray) - 1
        _ArrayAdd($aInputArray2D, convertCoordTo2D($aInputArray[$i]))
    Next

    For $i = 2 To UBound($aBestRoute) - 1
        _ArrayAdd($aBestRoute2D, convertCoordTo2D($aBestRoute[$i]))
    Next

    Local $sMiddle2D = $iGridSize / 2
    Local $aStartCoordinate2D = StringSplit(convertCoordTo2D($sStartCoordinate), ':', 2)
    Local $aEndCoordinate2D = StringSplit(convertCoordTo2D($sEndCoordinate), ':', 2)
    Local $iXDifference = $aStartCoordinate2D[0] - $sMiddle2D
    Local $iYDifference = $aStartCoordinate2D[1] - $sMiddle2D

    For $i = 1 To $iMapWidth * $iMapRoot
        For $j = 1 To $iMapHeight * $iMapRoot
            ; Überprüfen Sie, ob die Koordinate in der besten Route ist
            If _ArraySearch($aBestRoute2D, $i & ":" & $j) >= 0 Then
                Local $iRouteIndex = _ArraySearch($aBestRoute2D, $i & ":" & $j)
                If $iRouteIndex >= 1 Then ; Beachten Sie den Index 1, um den ersten Zielpunkt einzubeziehen
                    _GDIPlus_GraphicsFillRect($hGraphic, ($i - 1 - $iXDifference) * $iLineSpacing, ($j - 1 - $iYDifference) * $iLineSpacing, $iLineSpacing, $iLineSpacing, $hBrushRed)
                    ; Beschriftung mit der Reihenfolge hinzufügen
                    _GDIPlus_GraphicsDrawString($hGraphic, $iRouteIndex & "", ($i - 1 - $iXDifference) * $iLineSpacing + 5, ($j - 1 - $iYDifference) * $iLineSpacing + 5, "Arial", 10, 0, 0xFFFFFFFF)
                EndIf
            ElseIf _ArraySearch($aInputArray2D, $i & ":" & $j) >= 0 Then
;~              ConsoleWrite("Zeichnen aus InputArray:" & @CRLF & $i & ":" & $j & " | " & $aInputArray[_ArraySearch($aInputArray2D, $i & ":" & $j)] & @CRLF)
                ; Koordinate im InputArray, aber nicht in der besten Route
                _GDIPlus_GraphicsFillRect($hGraphic, ($i - 1 - $iXDifference) * $iLineSpacing, ($j - 1 - $iYDifference) * $iLineSpacing, $iLineSpacing, $iLineSpacing, $hBrushOrange)
            EndIf

            ; Endpunkt zeichnen, wenn er nicht die Start-Koordinate ist
            If Not _2DCoordArraysEqual($aStartCoordinate2D, $aEndCoordinate2D) And _2DCoordStringsEqual($i & ":" & $j, convertCoordTo2D($sEndCoordinate)) Then
                _GDIPlus_GraphicsFillRect($hGraphic, ($i - 1 - $iXDifference) * $iLineSpacing, ($j - 1 - $iYDifference) * $iLineSpacing, $iLineSpacing, $iLineSpacing, $hBrushGrey)
                    ; "FIN" in die Koordinaten schreiben
                _GDIPlus_GraphicsDrawString($hGraphic, "FIN", ($i - 1 - $iXDifference) * $iLineSpacing + 5, ($j - 1 - $iYDifference) * $iLineSpacing + 5, "Arial", 10, 0, 0xFFFFFFFF)
            EndIf
        Next
    Next

    ; Startpunkt zeichnen und die Länge des Lösung angebe
    _GDIPlus_GraphicsFillRect($hGraphic, ($sMiddle2D - 1) * $iLineSpacing, ($sMiddle2D - 1) * $iLineSpacing, $iLineSpacing, $iLineSpacing, $hBrushBlack)
    _GDIPlus_GraphicsDrawString($hGraphic, $aBestRoute[1], ($sMiddle2D - 1) * $iLineSpacing + 10, ($sMiddle2D - 1) * $iLineSpacing + 15, "Arial", 12, 0, 0xFFFFFFFF)

    ; Gitter zeichnen
    For $i = 1 To $iGridSize
        For $j = 1 To $iGridSize
            _GDIPlus_GraphicsDrawRect($hGraphic, ($i - 1) * $iLineSpacing, ($j - 1) * $iLineSpacing, $iLineSpacing, $iLineSpacing, $hPen)
        Next
    Next

    ; GUI anzeigen
    GUISetState(@SW_SHOW, $hGUI)

    ; Loop bis der Benutzer die Anwendung schließt.
    Do
    Until GUIGetMsg() = $GUI_EVENT_CLOSE

    ; Ressourcen aufräumen
    CleanupResources($hBrushOrange, $hBrushRed, $hBrushBlack, $hPen, $hGraphic, $hGUI)
EndFunc

Func CleanupResources($hBrushOrange, $hBrushRed, $hBrushBlack, $hPen, $hGraphic, $hGUI)
    _GDIPlus_BrushDispose($hBrushOrange)
    _GDIPlus_BrushDispose($hBrushRed)
    _GDIPlus_BrushDispose($hBrushBlack)
    _GDIPlus_PenDispose($hPen)
    _GDIPlus_GraphicsDispose($hGraphic)
    _GDIPlus_Shutdown()
    GUIDelete($hGUI)
EndFunc

Func _2DCoordArraysEqual($coord1, $coord2)
    ; Überprüfen, ob beide Parameter Arrays sind
    If IsArray($coord1) And IsArray($coord2) Then
        ; Überprüfen Sie die Anzahl der Elemente in beiden Arrays
        If UBound($coord1) <> UBound($coord2) Then
            Return False
        EndIf

        ; Vergleichen Sie die Werte in den Arrays
        For $i = 0 To UBound($coord1) - 1
            If $coord1[$i] <> $coord2[$i] Then
                Return False
            EndIf
        Next

        ; Wenn alle Vergleiche erfolgreich sind, geben Sie True zurück
        Return True
    EndIf

    ; Wenn keiner der obigen Fälle zutrifft, geben Sie False zurück
    Return False
EndFunc

Func _2DCoordStringsEqual($coord1, $coord2)
    ; Zerlegen Sie die Koordinaten in Arrays
    Local $aCoord1 = StringSplit($coord1, ':')
    Local $aCoord2 = StringSplit($coord2, ':')

    ; Überprüfen Sie die Anzahl der Elemente in beiden Arrays
    If UBound($aCoord1) <> UBound($aCoord2) Then
        Return False
    EndIf

    ; Vergleichen Sie die Werte in den Arrays
    For $i = 1 To UBound($aCoord1) - 1
        If $aCoord1[$i] <> $aCoord2[$i] Then
            Return False
        EndIf
    Next

    ; Wenn alle Vergleiche erfolgreich sind, geben Sie True zurück
    Return True
EndFunc

Func convertCoordTo2D($sCoord)
    $aCoord = getArrayFromCoordString($sCoord)

    Local $ax = ($aCoord[0][0] - 1) * $iMapRoot + Mod($aCoord[0][2] - 1, $iMapRoot) + 1
    Local $ay = ($aCoord[0][1] - 1) * $iMapRoot + Int(($aCoord[0][2] - 1)/$iMapRoot) + 1

    Return $ax & ":" & $ay
EndFunc

Func convert2DTo3D($sCoord)
    Local $aCoord = StringSplit($sCoord, ':', 2)
    Local $ax = $aCoord[0]
    Local $ay = $aCoord[1]
    Local $ar = Mod(($ax - 1), $iMapRoot) + 1 + Mod(($ay - 1), $iMapRoot) * $iMapRoot
    Local $axx = Ceiling($ax / $iMapRoot)
    Local $ayy = Ceiling($ay / $iMapRoot)
    Return $axx & ":" & $ayy & ":" & $ar
EndFunc

Func getStringFromArrayCoord($aArray)
    Return $aArray[0][0] & ":" & $aArray[0][1] & ":" & $aArray[0][2]
EndFunc

Func getArrayFromCoordString($sString)
    Local $aTmp1 = StringSplit($sString, ':', 2)
    Local $aCoord[1][3] =  [[$aTmp1[0], $aTmp1[1], $aTmp1[2]]]

    Return $aCoord
EndFunc

Func getDistance($sStart, $sTarget)
    Local $aDist[9]

    ; If the start and destination are the same, the distance is 1
    If StringCompare($sStart, $sTarget) = 0 Then Return 1

    Local $aTmp1 = getArrayFromCoordString($sStart), $aTmp2 = getArrayFromCoordString($sTarget)
    Local $x1 = Int($aTmp1[0][0]), $y1 = Int($aTmp1[0][1]), $r1 = Int($aTmp1[0][2])
    Local $x2 = Int($aTmp2[0][0]), $y2 = Int($aTmp2[0][1]), $r2 = Int($aTmp2[0][2])

    $aDist[0] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 - $iMapWidth) & ':' & ($y2 - $iMapHeight) & ':' & ($r2))
    $aDist[1] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 - $iMapWidth) & ':' & ($y2)               & ':' & ($r2))
    $aDist[2] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 - $iMapWidth) & ':' & ($y2 + $iMapHeight) & ':' & ($r2))
    $aDist[3] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2)              & ':' & ($y2 - $iMapHeight) & ':' & ($r2))
    $aDist[4] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2)              & ':' & ($y2)               & ':' & ($r2))
    $aDist[5] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2)              & ':' & ($y2 + $iMapHeight) & ':' & ($r2))
    $aDist[6] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 + $iMapWidth) & ':' & ($y2 - $iMapHeight) & ':' & ($r2))
    $aDist[7] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 + $iMapWidth) & ':' & ($y2)               & ':' & ($r2))
    $aDist[8] = getDistanceWithBorder($x1 & ':' & $y1 & ':' & $r1, ($x2 + $iMapWidth) & ':' & ($y2 + $iMapHeight) & ':' & ($r2))

    Return Min9($aDist[0], $aDist[1], $aDist[2], $aDist[3], $aDist[4], $aDist[5], $aDist[6], $aDist[7], $aDist[8])
EndFunc

Func getDistanceWithBorder($sStart, $sTarget)
    Local $aTmp1 = StringSplit($sStart, ':', 2), $aTmp2 = StringSplit($sTarget, ':', 2)

    Local $x1 = Int($aTmp1[0]), $y1 = Int($aTmp1[1]), $r1 = Int($aTmp1[2])
    Local $x2 = Int($aTmp2[0]), $y2 = Int($aTmp2[1]), $r2 = Int($aTmp2[2])

    Local $ax = ($x1 - 1) * $iMapRoot + Mod($r1 - 1, $iMapRoot)
    Local $ay = ($y1 - 1) * $iMapRoot + Int(($r1 - 1)/$iMapRoot)
    Local $bx = ($x2 - 1) * $iMapRoot + Mod($r2 - 1, $iMapRoot)
    Local $by = ($y2 - 1) * $iMapRoot + Int(($r2 - 1)/$iMapRoot)

    Local $iDist = Min2(Abs($ax - $bx), Abs($ay - $by))

    $ax += $ax > $bx ? -$iDist : $iDist
    $ay += $ay > $by ? -$iDist : $iDist

    Return $iDist + Abs($ax - $bx) + Abs($ay - $by)
EndFunc

Func Min($a, $b)
    Return $a < $b ? $a : $b
EndFunc   ;==>Min

Func Min2($a, $b)
    Return $a < $b ? $a : $b
EndFunc

Func Min3($a, $b, $c)
    Return $a < $b ? $a < $c ? $a : $c : $b < $c ? $b : $c
EndFunc

Func Min9($1, $2, $3, $4, $5, $6, $7, $8, $9)
    Return Min3(Min3($1, $2, $3), Min3($4, $5, $6), Min3($7, $8, $9))
EndFunc

Func Max($a, $b)
    Return $a > $b ? $a : $b
EndFunc

A section of the coordinate system to understand it: 

;~ ---------------------------  +  ------------------------  +  ------------------------
;~ 60:60:1 | 60:60:2 | 60:60:3  |  1:60:1 | 1:60:2 | 1:60:3  |  2:60:1 | 2:60:2 | 2:60:3
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:60:4 | 60:60:5 | 60:60:6  |  1:60:4 | 1:60:5 | 1:60:6  |  2:60:4 | 2:60:5 | 2:60:6
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:60:7 | 60:60:8 | 60:60:9  |  1:60:7 | 1:60:8 | 1:60:9  |  2:60:7 | 2:60:8 | 2:60:9
;~ ---------------------------  +  ------------------------  +  ------------------------
;~ 60:1:1  | 60:1:2  | 60:1:3   |  1:1:1  | 1:1:2  | 1:1:3   |  2:1:1  | 2:1:2  | 2:1:3
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:1:4  | 60:1:5  | 60:1:6   |  1:1:4  | 1:1:5  | 1:1:6   |  2:1:4  | 2:1:5  | 2:1:6
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:1:7  | 60:1:8  | 60:1:9   |  1:1:7  | 1:1:8  | 1:1:9   |  2:1:7  | 2:1:8  | 2:1:9
;~ ---------------------------  +  ------------------------  +  ------------------------
;~ 60:2:1  | 60:2:2  | 60:2:3   |  1:2:1  | 1:2:2  | 1:2:3   |  2:2:1  | 2:2:2  | 2:2:3
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:2:4  | 60:2:5  | 60:2:6   |  1:2:4  | 1:2:5  | 1:2:6   |  2:2:4  | 2:2:5  | 2:2:6
;~ ---------------------------  |  ------------------------  |  ------------------------
;~ 60:2:7  | 60:2:8  | 60:2:9   |  1:2:7  | 1:2:8  | 1:2:9   |  2:2:7  | 2:2:8  | 2:2:9
;~ ---------------------------  +  ------------------------  +  ------------------------

Maybe this one helps either, to understand it:
  

#include <Array.au3>

$aTable = Create(60, 60, 3) ; Zum Anzeigen des ganzen Systems als Array
_ArrayDisplay($aTable)

Func Create($iTableHeight = 3, $iTableWidth = 3, $iTableElementsRoot = 3)
    Local $aRet[$iTableHeight * $iTableElementsRoot][$iTableWidth * $iTableElementsRoot]
    For $y = 0 To $iTableHeight - 1 Step 1
        For $x = 0 To $iTableWidth - 1 Step 1
            For $i = 1 To $iTableElementsRoot ^ 2 Step 1
                $aRet[$y * $iTableElementsRoot + Int(($i-1)/$iTableElementsRoot)][$x * $iTableElementsRoot + Mod($i - 1, $iTableElementsRoot)] = $i
            Next
        Next
    Next
    Return $aRet
EndFunc

 

Edited by Acanis
RTFC Universalist

RTFC

Posted
  • Author
Posted (edited)

@Acanis: As they say in German: "Da gibt's noch Luft  nach oben." :D You're making things faaaaaar too complicated (I'm not even going to go into how you're setting up/evaluating your 3D/2D grid...:huh2:). Was this ChatGPT's idea? But you earn some points for at least annotating your code. Or was that ChatGPT as well?

I edited the TSP example on the assumption that the desired number of intermediate points is fixed, and point duplication is not allowed. You can play with $tempfactor (and $maxStepsWithoutImprovement) to adjust how much exploration of the solution space is allowed. If $verbose = true (default false now), the best sequence of journey legs so far is ArrayDisplayed every time a better solution is found (press <Esc> (every time) to continue); press <Space> to terminate prematurely.

This is all I'm going to write for you here; visualisation and any changes you'll have to implement yourself.:P Hope it helps.

  Reveal hidden contents

 

Edited by RTFC
minor efficiency optimisation in code

My Contributions and Wrappers

  Reveal hidden contents
Acanis Wayfarer

Acanis

Posted
Posted (edited)
  On 3/5/2024 at 3:34 PM, RTFC said:

@Acanis: As they say in German: "Da gibt's noch Luft  nach oben." :D You're making things faaaaaar too complicated (I'm not even going to go into how you're setting up/evaluating your 3D/2D grid...:huh2:). Was this ChatGPT's idea? But you earn some points for at least annotating your code. Or was that ChatGPT as well?

I edited the TSP example on the assumption that the desired number of intermediate points is fixed, and point duplication is not allowed. You can play with $tempfactor (and $maxStepsWithoutImprovement) to adjust how much exploration of the solution space is allowed. If $verbose = true (default false now), the best sequence of journey legs so far is ArrayDisplayed every time a better solution is found (press <Esc> (every time) to continue); press <Space> to terminate prematurely.

This is all I'm going to write for you here; visualisation and any changes you'll have to implement yourself.:P Hope it helps.

Expand  

Hehe, Iam german and your right... ;) But I try to improve :D

The grid stuff is really old and I got some help of a more experienced programmer to finish it... ^^ (And I liked it, because it was there and understandable for me^^) ChatGPT did help me with the visualisation, because I never used GDI+ before ^^ I try to annotate my code, but ChatGPT did help, too :D

I planned the desired number of intermediate points to be an input to the function. But if Iam able to understand your code, Ill try to change that myself. :)

In any case, MANY THANKS in advance!

Ill add the visualisation stuff later and test it a lot to understand the code and the impact of the parameters. :) Really cool! :)

Edited by Acanis
Acanis Wayfarer

Acanis

Posted
Posted (edited)

I still dont understand everything, but working on it :) I reduced the global variables, put all the stuff into an function and renamed a lot of the variables. Maybe its worse for experienced eyes, but it helps me a lot :D

It seems that after the refactoring everything still works as before and I was able to customize it so that the drawing function can be used again without further customization. 

Ill play around with the parameters and compare the solutions with the one of my own try; but it feels pretty good for the example data :)! Ill post the updated code, so other interested people can use it :)
 

  Reveal hidden contents

 

Edited by Acanis
RTFC Universalist

RTFC

Posted
  • Author
Posted (edited)

In path searches through a subset of points more generally (for simulated annealing or any other time-consuming algorithm), it may be of significant advantage (speedwise, that is) to compute all point-to-point distances in advance, and store these in a single look-up reference (table or matrix). Here's a simple, generic example of doing just that using matrices:

  Reveal hidden contents

This example is, however, far from optimised. One improvement would be to get rid of the final column vector copy to the output matrix, and instead repeatedly re-map a column in the final output as our vector to directly collect results in.

  Reveal hidden contents

This is faster, but still evaluates each point individually. A more efficient approach would be to create separate matrices for each coordinate dimension, and perform each math operation once per dimension only, collecting final results in the first dimension's container. Note that this much faster solution requires more memory.

  Reveal hidden contents

 

Edited by RTFC
more examples added

My Contributions and Wrappers

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