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Excercise07 Updates to all subtitle files #18

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8 changes: 4 additions & 4 deletions Uebungen/Blatt07/Exercise07-Task1-de.sbv
Original file line number Diff line number Diff line change
Expand Up @@ -178,8 +178,8 @@ Es stand ja auch in der Aufgabe, wenn ein
Kunde seine Waren aufs Band legt, belegen

0:04:01.970,0:04:20.850
diese zunächst
Platz 1.
diese zunächst Platz 1. Das bedeutet, wenn der Kunde
etwas auf Platz 1 legt, wird dieser von "leer" zu "voll".

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Es steht oben, nach erfolgreichem Ablegen
Expand All @@ -189,7 +189,7 @@ wechselt der Kunde (bzw. Kassierer) in den
Zustand beschäftigt.

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Das bedeutet, der Kunde wechselt von warten,
Das bedeutet, der Kunde wechselt von wartend,
während er die Waren ablegt oder sie abgelegt

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Expand All @@ -201,7 +201,7 @@ Jetzt ist der Kunde beschäftigt und der Platz

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Im Text stand, dass, wenn der Kunde beschäftigt
war, das nach kurzer Zeit wieder wartend wird.
war, dass er nach kurzer Zeit wieder wartend wird.

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Wir müssen also von beschäftigt wieder eine
Expand Down
20 changes: 10 additions & 10 deletions Uebungen/Blatt07/Exercise07-Task1-en.sbv
Original file line number Diff line number Diff line change
Expand Up @@ -97,15 +97,15 @@ empty or full.

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So, in this case, space two would be empty
and in this case, space one would be full.
and space one would be full.

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So that we have both states for space one
and for space two.

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And these places represent states, which were
always/ either of them has to be selected
And these places represent states, where
one of them has to be selected at all times,

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because a space is either empty or full and
Expand Down Expand Up @@ -211,7 +211,7 @@ and space two should be full afterwards.

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And then again, we can think about the process
of how the space two becomes empty.
of how space two becomes empty.

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So, we have a representation for getting it
Expand All @@ -232,7 +232,7 @@ belt is not empty.

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So this means this transaction should probably
represent these states, so the cashier is
represent this state, so the cashier is

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waiting and space two is full.
Expand Down Expand Up @@ -303,7 +303,7 @@ would be going out of waiting and empty but
we would then have tokens in busy and full.

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And then again, we could, when we have a token
And then again, when we have a token
here, then it could go via this transition

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Expand Down Expand Up @@ -333,7 +333,7 @@ And again, after doing that, he could again
fire this transition, which we already checked

0:11:12.430,0:11:19.060
that it's still is just one token in each
is still just one token in each
column.

0:11:19.060,0:11:24.310
Expand All @@ -344,7 +344,7 @@ So we have just one token here and only one.

0:11:27.280,0:11:40.800
So there's not zero tokens, it's exact exactly
one token in each/ yeah, call them here, so.
one token in each column here.

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And then finally this transition can only
Expand Down Expand Up @@ -379,13 +379,13 @@ working as expected, in the sense that no

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token is additionally added in while executing
the graph, or that token
the graph, or that a token

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gets completely removed.

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He could check further interactions between
We could check further interactions between
those but I think you get the picture of what's

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Expand Down
14 changes: 7 additions & 7 deletions Uebungen/Blatt07/Exercise07-Task2-de.sbv
Original file line number Diff line number Diff line change
Expand Up @@ -204,11 +204,11 @@ die Nachbedingung ist, dass eine Marke wieder
rauskommt, also eine Marke auftaucht.

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Jetzt sind wir schon an einem Punkt, ich nenne
ihn a., wo keine andere Markierung mehr möglich
Jetzt sind wir schon an einem Punkt,
wo keine andere Markierung mehr möglich

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ist.
ist. (a.)

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Es hätte aber auch die Möglichkeit gegeben,
Expand Down Expand Up @@ -277,7 +277,7 @@ Dann wären wir jetzt an diesem Punkt, dass
die Marke hier liegt.

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Und dann könnte wieder t3 nur schalten.
Und dann könnte wieder nur t3 schalten.

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Dann würde die Marke verschwinden als Vorbedingung
Expand Down Expand Up @@ -367,11 +367,11 @@ aufgeschrieben.
Ich hoffe, dass die Lösung trotzdem klar
daraus hervorgeht.

0:08:49.610,0:08:58.970
0:08:49.610,0:09:07.500
Ich schreibe mal noch dazu, Stellenreihenfolge
(s1,s2,s3,s4).

0:08:58.970,0:09:10.830
0:09:07.500,0:09:10.830
Kommen wir jetzt zum allerletzten Petrinetz.

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Expand Down Expand Up @@ -491,7 +491,7 @@ Und jetzt sind wir quasi wieder im Startzustand.

0:11:16.980,0:11:24.680
Mit anderen Worten, dieser Schritt, dass t3
und t2 abläuft, kann auch je immer und immer
und t2 abläuft, kann jetzt auch immer und immer

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und immer wieder im Kreis laufen.
Expand Down
40 changes: 20 additions & 20 deletions Uebungen/Blatt07/Exercise07-Task2-en.sbv
Original file line number Diff line number Diff line change
@@ -1,6 +1,6 @@
0:00:02.190,0:00:12.000
Within this task, we shall give a marking
that has no successful marking and in each
that has no successor marking and in each

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case, we shall also say which transitions
Expand Down Expand Up @@ -29,7 +29,7 @@ So, anyways, it has to start with T2.

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So, we can start with T2 and when we process
this transaction, when we fire it then there
this transition, when we fire it then there

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will be markings in S2 and S3 because of these
Expand All @@ -50,8 +50,8 @@ So, when we do that, we remove one token from
S2 and add a token in S1.

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And now, what is another reachable marking
is we can still fire T3.
And now, there is still another reachable marking
as we can still fire T3.

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And when we do this, so let me properly remove
Expand All @@ -63,7 +63,7 @@ this token here as well.

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And then we are in this marking here now,
we can find no more transition.
we can fire no more transition.

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So, we reached a final marking.
Expand All @@ -73,7 +73,7 @@ And now we can name actually this marking
as we know which marking it is.

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And it is in S1 there are two tokens and in
In S1 there are two tokens and in
S2, there are zero tokens.

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Expand Down Expand Up @@ -136,7 +136,7 @@ quite promising so we could choose T2 to fire

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and in this case we would get to this point
here where/ so, well this was the sequence,
here where... So, well this was the sequence.

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I'll write it down here below.
Expand All @@ -147,7 +147,7 @@ in the final marking because there is no transition

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that we can fire now because T2 requires a
token in S1 as well does T1, and T3 requires
token in S1 and so does T1, and T3 requires

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a token in S2.
Expand Down Expand Up @@ -201,15 +201,15 @@ one token in S3.
And now we’ve reached a different marking.

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So, in this case, so this one should be the
So, in this case, this one should be the
same, but S4 would be one in this case.

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And let me write this arrow again.
Let me write this arrow again.

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So, S4 would be one and in order to reach
this sequence, we execute it first of all
this sequence, we executed first of all

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T1 then T3, and then T2.
Expand All @@ -226,8 +226,8 @@ So now in this state, we could once more fire
T1.

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So, by firing T1 at this stage we get to token
into S2 and removing it from S1.
So, by firing T1 at this stage we get the token
into S2 and remove it from S1.

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And now we can fire only one transition again
Expand All @@ -241,11 +241,11 @@ in S4 and again a token in S1 and the token
in S2 is removed.

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So (?roughly) we can fire T3 here once more
So we can fire T3 here once more
and then again we can either fire T2 which

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we get a token here and we have reached a
will get the token here and we have reached a
final marking as we can fire no more transition

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Expand All @@ -270,7 +270,7 @@ finally fire T2 but there are no further options
but cycling in this case.

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Now let's consider the last Petri Net in three.
Now let's consider the last Petri Net N3.

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And in this case, there is only T3 active,
Expand Down Expand Up @@ -306,7 +306,7 @@ So I can put this token back in here and we
add a token in S4.

0:10:59.200,0:11:19.230
And this was call it T3 and we could now either
And this, i'll call it T3, and we could now either
fire T1 or T2 because both just require a

0:11:19.230,0:11:27.180
Expand All @@ -321,11 +321,11 @@ We remove a token from S4 and get a token
into S1.

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So now this is marking without successor because
So now this is a marking without successor because
now we can't fire any transition anymore.

0:11:51.991,0:11:59.200
T1, and T2, require both a token in S4, and
T1, and T2, both require a token in S4, and
T3 requires a token in S3.

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Expand Down Expand Up @@ -370,7 +370,7 @@ Or we could again fire T3 and then T2, so
either of these sequences would be okay.

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So, for instance, T2/T3/T1 would be another
So, for instance, T3/T2/T3/T1 would be another
option of a sequence that will lead to the

0:13:49.990,0:13:55.699
Expand Down
12 changes: 6 additions & 6 deletions Uebungen/Blatt07/Exercise07-Task3-de.sbv
Original file line number Diff line number Diff line change
Expand Up @@ -36,7 +36,7 @@ da noch ein Pfeil drauf zeigen.

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Von der Startmarkierung s1 gehen zwei Pfeile
aus, da können wir gucken, in welche Transaktion.
aus, da können wir gucken, in welche Transitionen.

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Einmal hier hin und einmal hier hin.
Expand Down Expand Up @@ -68,10 +68,10 @@ Dann schreiben wir das auf.

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Wir haben jetzt also eine Marke in s1, eine
Marke in s2, und null Marken in s3 und in
Marke in s2, und null Marken in s3 und s4

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s4. t1 kann nochmal feuern, warum nicht?
t1 kann nochmal feuern, warum nicht?

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Die Vorbedingung ist eine Marke, eine haben
Expand Down Expand Up @@ -157,7 +157,7 @@ an Stelle s3 und null an Stelle s4.

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Wir zeichnen wieder einen Pfeil dahin und
schreiben dran, welch Transaktion, nämlich
schreiben dran, welche Transaktion, nämlich

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t2, geschaltet hat.
Expand All @@ -182,7 +182,7 @@ lösche ich die.

0:04:25.510,0:04:39.789
Und es kommt als Nachbedingung eine in s4
und als Nachbedingung eine in s2 hinein.
und eine in s2 hinein.

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Also sind wir jetzt wieder in dem Zustand,
Expand Down Expand Up @@ -233,7 +233,7 @@ Was kann hier schalten?

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Hier könnte zum Beispiel t4 schalten, weil
das ja eine Vorbedingung ist, wenn t4 schaltet.
s4 ja eine Vorbedingung ist.

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Dann geht die Marke hier raus und da rein.
Expand Down
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