Readme.md

@@ -2,7 +2,7 @@
 
 ## Tooling
 
-- Python 3.6
+- Python 3.5
 - [MyPy](http://www.mypy-lang.org/ ) für statische Typchecks
 - [Pandoc](https://pandoc.org/ ) für die Dokumentation
 - Python Module: siehe [requirements.txt](https://pip.pypa.io/en/latest/user_guide/#requirements-files )

Readme.txt

@@ -1,23 +0,0 @@
-README
------
-
-Für die Ausführung des Algorithmus wird Python 3 (empfohlene Version: 3.6.1) benötigt.
-Die Packages, die zusätzlich gebraucht werden, können der requirements.txt entnommen werden.
-(Installation kann hier einzeln oder über den Befehl: python -m pip install -r requirements.txt)
-
-Zur Ausführung bitte im Terminal in den Ordner src gehen und dort das Skript main.py starten.
-Parameter, die hierbei möglich sind:
-    -h zeigt alle Optionen an
-    -p aktiviert die Ausgabe über den Plotter als Diagramm
-    -l wird benötigt falls die Eingabe eine Liste von Problemen ist (d.h. für jobshop1.txt)
-    -i Index des Problems in der Liste (nur relevant bei -l)
-    -t setzt die Starttemperatur des Simulated Annealings
-    -s setzt die maximalen Umformungsschritte pro Generierung einer neuen Lösung
-    -a setzt die Wahrscheinlichkeit, pro Umformungsschritt auch eine Lösung zu akzeptieren, obwohl
-        noch nicht die maximalen Umformungsschritte erreicht sind
-
--t -s und -a müssen nicht alle gesetzt sein, dann wird der jeweilige Defaultwert verwendet
-Defaultwerte: max_temp = 300, max_steps = 250, accept_prob = 0.01
-
-Beispielaufruf: 
-    python .\main.py -p -l -i 2 -t 50 ..\inputdata\jobshop1.txt

doc.md

@@ -1,36 +0,0 @@
-## scheduling problem defined by:
-  1. $m$ specialized machines
-  2. tasks $\tau$ of the form $(e, i)$ with $t \in \mathbb{N}$ the execution time and $i \in \{1,2,\dots,m\}$ the machine the task has to run on
-  3. $n$ jobs $T_k$ with $\forall T_k:$ linear order of tasks, with $k \in \{1,2,\dots,n\}$
-  4. Additionally, a multiset $\Omega$ of arbitrary but fixed size that contains wait states $\omega := (1, i)$ with $i \in \{1,2,\dots,m\}$ the blocked machine.
-
-The goal is to find the fastest feasible schedule $\sigma_{min}$.
-
-## evaluative function
-  - minimize the execution time of $\sigma$
-  - upper bound: largest processing time first
-  - lower bound: max sum of execution times on one machine
-
-## solutions
-  - list of tuples $(t, \tau)$ with $t \in \mathbb{N}$ the scheduled begin of $\tau$
-
-## operations
-  - $\operatorname{ins}(\omega, t)$: block a machine at time $t$ for $w$ time steps.
-  - $\operatorname{xchg}(\tau_1,\tau_2)$: exchange the position of two tasks.
-  
-Both operations require that the start times are recomputed.
-
-## neighbourhood of solution
-
-  - $\operatorname{neighbours}(\sigma) = \{x \in \Sigma | \delta(\sigma, x) \leq n\}$ with $\Sigma$ the set of all feasible schedules.
-  - $\delta$: $\delta ( \sigma )=0$, $\delta ( \operatorname{op}(x)) = \delta (x) + 1$ (ass. ins has the same penalty xchg has), $x$ either op($y$) or $\sigma$ 
-
-## constraints
-  - only schedule new $\tau$ if another $\tau$ is finished
-  - only schedule $\tau \in T_k$ that has no unscheduled predecessor in $T_k$
-  - only one task on a machine any given time
-
-## implementation in Python
-  - translate problem into list of jobs, jobs into lists of tasks, ie problem = [$T_0, T_1,\dots,T_{k-1}$], $T_i$ = [$\tau_1,\tau_2,\dots$]
-  - address tasks based on their indices, ie [0][1] is the second task of the first job.
-  - compute only one possible next solution, rate, drop/accept. $\delta$ is computed iteratively during generation

notes.md

@@ -26,7 +26,7 @@
     - $S = \left\{(o_j,t) | o_j \in O \cup \left\{w_n | n \in \mathbb{N} \wedge w_n \text{ v.d.F. } (1, m) \right\} \wedge o_j \text{ v.d.F. } (d, m, j) \wedge t \in T \forall o \in O : \exists (o,t) \in S\right\}$
     - indirekt lässt sich durch laufende Operation und Zeitpunkt auch Belegung einer Maschine zu einem Zeitpunkt ermitteln
     - Optimierung: sparse speichern
-1. Liste von (T, $o_j$) mit $T \in \mathbb{N}$ (Time), $o_j \in O$ (Tasks), j bezeichnet den Job
+1. Liste von (T, $o_j$) mit $T \in \mathbb{N}$ (Time), $o_j \in O$ (Tasks)
     - Operationen:
         - Vertauschen von 2 Jobs auf einer Maschine, selbstinvers
         - Verzögern von Operationen (keine expliziten Wartezustände nötig)

requirements.txt

@@ -1,5 +1,3 @@
 mypy
-arpeggio
+Arpeggio
 matplotlib
-numpy
-tkinter

shell.nix

@@ -1,3 +0,0 @@
-with import <nixpkgs> {};
-
-  (python3.withPackages (ps: [ps.numpy (ps.matplotlib.override {enableQt=true;}) ps.mypy ps.arpeggio])).env

src/Generator/generator.py

@@ -38,8 +38,8 @@ def accept(solution):
     Maybe skip this during the first step to generate a more
     random solution.
     """
-    return tighten(solution)
-    #return solution
+    #return tighten(solution)
+    return solution
 
 
 def tighten(solution):

src/Output/output.py

@@ -11,9 +11,7 @@ def create_plot(problem, solution):
 
     with plt.xkcd():
         fig,ax = plt.subplots()
-        col = colors.XKCD_COLORS
-        del col['xkcd:white']
-        colorlist = list(col.values())
+        colorlist = list(colors.XKCD_COLORS.values())
         random.shuffle(colorlist)
         for m in range(0, problem.machines):
             mach_ops = [ x for x in solution if problem.problem_data[x[1][0]][x[1][1]][1] == m ]

src/Parser/__init__.py

@@ -6,33 +6,33 @@ from collections.abc import Mapping
 __all__ = ["js1_style", "js2_style"]
 
 grammar = """
-        // starting point for jobshop1 input file
+        # starting point for jobshop1 input file
         job_shop1       = skip_preface
-        // eat away lines of preface, until first problem_instance is
-        // encountered; then the list of instances start
+        # eat away lines of preface, until first problem_instance is
+        # encountered; then the list of instances start
         skip_preface    = (!problem_instance r"[^\n]+" skip_preface) / (eol skip_preface) / instance_list
         instance_list   = problem_instance (sep_line trim_ws eol problem_instance eol?)* eof_sep
         problem_instance = trim_ws "instance" ' ' instance_name trim_ws eol trim_ws eol sep_line description eol problem_data
         description     = r"[^\n]*"
         instance_name   = r"\w+"
         sep_line        = trim_ws plus_line trim_ws eol
-        // lines out of multiple + signs
+        # lines out of multiple + signs
         plus_line       = r"\+\+\++"
-        // EOF is a builtin rule matching end of file
+        # EOF is a builtin rule matching end of file
         eof_sep         = trim_ws plus_line " EOF " plus_line trim_ws eol* EOF
-        // entry point for jobshop2 input files
+        # entry point for jobshop2 input files
         job_shop2       = problem_data EOF
         problem_data    = trim_ws num_jobs ' ' num_machines eol job_data+
-        // used for skipping arbitrary number of non-breaking whitespace
+        # used for skipping arbitrary number of non-breaking whitespace
         trim_ws         = r'[ \t]*'
-        // git may change line-endings on windows, so we have to match on both
+        # git may change line-endings on windows, so we have to match on both
         eol             = "\n" / "\r\n"
         nonneg_num      = r'\d+'
         num_jobs        = nonneg_num
         num_machines    = nonneg_num
         machine         = nonneg_num
         duration        = nonneg_num
-        // task data for 1 job
+        # task data for 1 job
         job_data        = ' '* machine ' '+ duration (' '+ machine ' '+ duration)* trim_ws eol
         """
 

src/example.py

@@ -1,10 +1,10 @@
-import Parser.js1_style as p
-#import Parser.js2_style as p
+#import Parser.js1_style as p
+import Parser.js2_style as p
 from SchedulingAlgorithms import simanneal as sim
 from Output import output as o
 
-problem = p.parse_file("../inputdata/jobshop1.txt")[0]
-#problem = p.parse_file("../inputdata/sample")
+#problem = p.parse_file("../inputdata/jobshop1.txt")[0]
+problem = p.parse_file("../inputdata/sample")
 sim.init(problem)
 solution = sim.anneal()
 o.create_plot(problem, solution)

src/main.py

@@ -13,9 +13,6 @@ Command line options:
     -p      activate pretty output (requires tkinter)
     -l      assume that a file contains multiple problems, default is only 1
     -i      index of the problem you want solved. has no effect without l
-    -t      set parameter max_temp of simulated annealing
-    -s      set parameter max_steps of simulated annealing
-    -a      set parameter accept_prob of simulated annealing
 
 Invocation:
     python [-hlp] file
@@ -27,8 +24,8 @@ def main():
     js1 = False
     plot = False
     try:
-        opts, args = getopt.getopt(sys.argv[1:], 'hpli:t:s:a:')
-    except getopt.GetoptError as err:
+        opts, args = getopt.getopt(sys.argv[1:], 'hpli:')
+    except getoptGetoptError as err:
         print(err)
         sys.exit()
     if ('-h', '') in opts:
@@ -41,12 +38,6 @@ def main():
         js1 = True
     idx = [int(x[1]) for x in opts if x[0]=='-i']   
     idx = idx[0] if idx else -1
-    max_temp = [int(x[1]) for x in opts if x[0]=='-t']   
-    max_temp = max_temp[0] if max_temp else -1
-    max_steps = [int(x[1]) for x in opts if x[0]=='-s']   
-    max_steps = max_steps[0] if max_steps else -1
-    accept_prob = [int(x[1]) for x in opts if x[0]=='-a']   
-    accept_prob = accept_prob[0] if accept_prob else -1
     if not args:
         print("No file given.")
         sys.exit()            
@@ -65,28 +56,7 @@ def main():
         problem = problem[idx]
     print(problem)
     sim.init(problem)
-    if not max_temp == -1:
-        if not max_steps == -1:
-            if not accept_prob == -1:
-                solution = sim.anneal(max_temp, max_steps, accept_prob)
-            else:
-                solution = sim.anneal(max_temp = max_temp, max_steps = max_steps)
-        else:
-            if not accept_prob == -1:
-                solution = sim.anneal(max_temp = max_temp, accept_prob = accept_prob)
-            else:
-                solution = sim.anneal(max_temp = max_temp)
-    else:
-        if not max_steps == -1:
-            if not accept_prob == -1:
-                solution = sim.anneal(max_steps = max_steps, accept_prob = accept_prob)
-            else:
-                solution = sim.anneal(max_steps = max_steps)
-        else:
-            if not accept_prob == -1:
-                solution = sim.anneal(accept_prob = accept_prob)
-            else:
-                solution = sim.anneal()
+    solution = sim.anneal()
     print(solution)
     print(sim.rate(solution))
     if plot: