MACRo 2015- 5th International Conference on Recent Achievements in
Mechatronics, Automation, Computer Science and Robotics
Random Process Simulation Using Petri Nets
László SZILÁGYI 1, 2, 3
1 Department of Control Engineering and Information Technology
Budapest University of Technology and Economics, Hungary, e-mail: [email protected]
2 Department of Electrical Engineering, Faculty of Technical and Human Sciences,
Sapientia University, Tg. Mureş, e-mail: [email protected]
3 Hand in Scan Ltd., Budapest, Hungary, e-mail: [email protected]
Manuscript received January 12, 2015, revised February 9, 2015.
Abstract: This paper introduces a Petri net designed for the simulation of the
ancient game called Rock-Paper-Scissors or Roshambo. The network enables us to
simulate the behavior of machine players and allows us to design and evaluate strategies
against weighted random machine opponents. The paper also presents a theoretical
calculus on winning chances. Simulations carried out using the software package Pipe
(version 4.3.0) fully confirm the theoretical considerations.
Keywords: Petri nets, Roshambo, random processes, strategy.
1. Introduction
The notion of Petri network was introduced over half a century ago, but its
theory and its application spectra are still evolving quickly [1]. Although it was
primarily designed to model automata, nowadays its applications cover the
modeling of communication protocols [2], manufacturing systems [3], road
traffic [4], cyber-intrusion [5], farm processes [6], and games [7, 8, 9], and
controlling of discrete event systems [10], track-driven traffic [11], robot
movements [12]. Recently, Petri nets are also used as bricks at building
machine learning systems [13].
In this paper we employ a discrete hybrid Petri net, containing both
immediate firing and timed transitions, to implement a simulation framework of
the roshambo or Rock-Papers-Scissors (RPS) game. This game has an ancient
Asian origin [14], and currently is very popular worldwide. It is frequently used
as a decider, for example, who should execute a free kick on the soccer field?
The two players count up to three or recite the names of the game Rock-PaperScissors, and at the third they suddenly bring their hidden hands in front,
showing one of the three allowed gestures that correspond to rock, paper and
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L. Szilágyi
scissors. If the two gestures are the same, then the game is a draw. Otherwise
there is a winner: rock destroys the sharpness of the scissors, scissors cut the
paper, while paper covers the rock. Our main goal is to study the basic
strategies that can be simply implemented into random generator driven
machine players.
Figure 1: Petri network model of the RPS game
2. Methods
The basic RPS model is presented in Fig. 1. It is a Petri net consisted of 13
places, and 17 transitions. Ten transitions are immediately firing transitions,
while seven are timed ones. In the initial state there is only one token in place
P0, indicating that there is no game in progress. When the transition T ST fires,
the game begins and places PST1 and PST2 get a token each, meaning that both
players can come up with their bet. The token in PST1 enables three timed
transitions, TR1, TP1, TS1, corresponding to the three choices: rock, paper, and
scissors, respectively. When one of these three transitions fires, the
corresponding place out of PR1, PP1, and PS1 will receive one token, showing the
choice of the first player. Simultaneously PDN1 also receives a token, showing
that the first player is done with their choice. The same process happens in
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Random process simulation using Petri nets
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parallel on the right side, modeling the choice of the second player. When both
players are done with their choice, the result of the game is evaluated. This is
performed by nine immediately firing transitions, out of which one and only one
is enabled. One of the transitions TRR, TPP, and TSS will fire in case of a draw,
when both players had the same choice. In any such case, two tokens disappear
and no place receives any token. Transitions denoted by TRS, TPR, and TSP fire in
the cases when the first player is the winner, and thus the place PW1 receives a
token, indicating that player one got one point. On the other hand, transitions
denoted by TSR, TRP, and TPS fire in the cases when the second player is the
winner, and thus the place PW2 receives a token, indicating that player two got
one point. At this moment, the next transition to fire is surely T FI, action
corresponding to the final of the game. The network regains its initial state, with
the possible difference of an extra token in either PW1 or PW2. In each cycle,
exactly five firings take place, and the number of tokens in PW1 and PW2 indicate
the current score in any moment.
If we execute a random firing sequence of length 5n, it is equivalent with a
sequence of n games. The first emerging tricky question is: who will win more
games? The fastest and mistaken answer is that a draw is very likely. The truth
is, it depends on the firing rates of the random choice transitions T R1, TP1, TS1,
TR2, TP2, and TS2, denoted by rR1, rP1, rS1, rR2, rP2, and rS2, respectively. If we
denote by pRk, pPk, and pSk the probabilities of Rock, Paper, and Scissors at
player k (k is 1 or 2), then we will have:
 p Rk  rRk /  k

 p Pk  rPk /  k
 p  r /
Sk
k
 Sk
(1)
where  k  rRk  rPk  rSk . This way, the probability of winning for the first
player is:
p w1  p R1 p S 2  p P1 p R 2  p S1 p P 2
(2)
while the second player wins with the probability:
p w2  p R 2 p S1  p P 2 p R1  p S 2 p P1 .
(3)
There is also a probability of a draw, defined as:
pdraw  pR1 pR 2  pP1 pP 2  pS1 pS 2 .
(4)
Having this accepted, it is likely to expect a score that is proportional with
pw1 vs. pw2. The score is expected equal for example when all transfer rates are
equal. On the other hand, when rR1=3, rS2=2, and rP1=rS1=rR2=rP2=1, then
pw1=0.4 and pw2=0.3, hence the score is expected to be proportional to 4:3.
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L. Szilágyi
Now let us suppose that both players are fully random machines, making
random choices with all possible outcomes having equal probability. In this
case all three transition rates are equal for both players. For the sake of
simplicity, without loss of generalization, let us suppose rR1=rP1=rS1=1 and
rR2=rP2=rS2=1. In this case, there are equal chances for the outcome of any
game: pw1=pw2=pdraw=1/3.
But let us further suppose that the first player wants to win the game by
attacking, and the second player wants to win by defense. In this order, the first
player intensifies the frequency of rock for example, throwing more rocks
towards the second player via setting rR1=1+x, where we suppose x is a positive
number. The second player on their turn identifies the situation and comes up
with more papers to cover the rocks of the opponent (rP2=1+y, with y>0) and
with less scissors to avoid unnecessary losses (rS2=1+z, with z<0). This way we
get: σ1=3+x, σ2=3+y+z, and so
 pR1

 pP1
p
 S1
 1 x

pR 2   3  x
  1
pP 2   
3 x
pS 2   1

 3 x


1

3 y  z 
1 y 
,
3 y  z 
1 z 
3  y  z 
(5)
leading to
 3  x  y  z  xz 


 p w1   (3  x)(3  y  z ) 

  3  x  y  z  xy 
 p w2   
.
p
  (3  x)(3  y  z ) 
 draw   3  x  y  z 
 (3  x)(3  y  z ) 


(6)
From the above formula it can be obviously deduced that the second player
is winning more frequently if xz<xy. If really x and y are positive, and z
negative, the inequality will always hold, meaning that attacking on one
arbitrarily selected choice is never a good strategy. The best score achieved by
the first player will be when x=0.
On the other hand, if x is chosen negative, the second player can easily
compensate by setting a positive z, and a negative y. These strategies will be
numerically evaluated in the next section.
The implementation and simulations were carried out using the Pipe
software package [15], version 4.3.0.
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3. Results and Discussion
The RPS game model was initialized with various parameter sets (x,y,z). For
each setting, a sequence of five thousand firings was performed, equaling one
thousand games. The expected scores were computed using Eq. (6), while the
simulated score was given by the Petri network. Results are summarized in
Table 1.
The second test (second data row of the table) reveals that the best way of
avoiding the loss of the game series is to keep on generating real random
choices. But if we are able to detect the strategy of the opponent, we are able to
compensate and win games more frequently. See details in the bottom nine rows
of Table 1.
Table 1: Tested cases.
x
0
1
1
1
1
1
1
1
-0.5
-0.5
-0.5
Rate shifts
y
z
0
0
0
0
0.5
0
1
0
2
0
0
-0.5
1
-0.5
2
-0.5
-0.5
0
-0.5
1
-0.5
2
pdraw
1/3
1/3
9/28
5/16
3/10
7/20
9/28
11/36
8/25
12/35
16/45
Probabilities
pw1
1/3
1/3
9/28
5/16
3/10
6/20
8/28
10/36
8/25
10/35
12/45
pw2
1/3
1/3
10/28
6/16
4/10
7/20
11/28
15/36
9/25
13/35
17/45
Theoretical
score
333:333
333:333
321:357
313:375
300:400
300:350
286:393
278:417
320:360
286:371
267:378
Simulated
score
313:356
310:387
277:403
297:360
272:405
277:415
320:363
287:381
250:388
If the attack comes on another track, namely the first player attacks with
more frequent scissors or papers, the applicable defense strategy equally exists,
due to the symmetry of the game.
4. Conclusion
Petri networks are indeed excellent tools in modeling random processes.
Using the built model, we were able to confirm the validity of analytically
computed strategies. By employing the developed strategies, we were able to
beat a weighted random machine player.
Further work could be invested in implementing the strategy into the Petri
net, making it able to identify the parameters of the random machine opponent.
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L. Szilágyi
Acknowledgements
This work was funded by Hungarian National Research Funds (OTKA),
under grant number PD 103921.
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