a non-geometric switch toggling game
DESCRIPTION
A Non-geometric Switch Toggling Game. Megan Duke Muskingum University. Lights Out by Tiger. Relies on the position of the chosen switch. An example of a system of 7 switches with a 5-toggle transition rule applied. Graph of a system of 7 switches with a 5-toggle transition rule applied. - PowerPoint PPT PresentationTRANSCRIPT
A Non-geometric Switch Toggling Game
Megan DukeMuskingum University
Lights Out by
TigerRelies on the position of the chosen switch
An example of a system of 7 switches with a 5-toggle transition rule applied
(7,0 )𝜏5,0
(2,5 )
(5,2 )
(0,7 )𝜏5,0
𝜏1,4
0 , 7 1 , 6 2 , 5 3 , 4
4 , 3 5 , 2 6 , 1 7 , 0
𝜏5,0𝜏5,0
𝜏1,4
Graph of a system of 7switches with a 5-toggle transition ruleapplied
0 , 9 1 , 8
2 , 7 3 , 6 4 , 5
5 , 4 6 , 3
7 , 2 8 , 1 9 , 0
Graph of a system of 9switches with a 4-toggle transition ruleapplied
States and are in different components of the graph.
Achieving the state depends on the parity of , the number of switches in the system and , the number of switches being toggled.If is odd, a system of switches can be transitioned from to .
If is even, a system of switches can be transitioned from to only when is even.
When is odd and
𝑘=2𝑚+1
decreases the number of switches on by
𝑛=2𝑘+𝑟Method:1. Apply to to get
2. Apply to times to get
3. Apply to to get
𝒏=𝟏𝟎 ,𝒌=𝟑
0 , 1 0 1 , 9
2 , 8 3 , 7 4 , 6 5 , 5 6 , 4
7 , 3 8 , 2 9 , 1 1 0 , 0
(7,3 )(6,4 )(5,5 )(4,6 )(3,7 )
} 𝜏2 ,14 times
(10,0 )𝜏3,0
(0,10 )𝜏3,0
When is odd and
This case is always done in steps.
𝑛−𝑘=2𝑎
Method:1. Apply to to get
2. Apply to to get
3. Apply to to get
Case: is odd
0 , 1 0 1 , 9 2 , 8 3 , 7 4 , 6 5 , 5
6 , 4 7 , 3 8 , 2 9 , 1 1 0 , 0
0 , 8 1 , 7 2 , 6 3 , 5
4 , 4 5 , 3
6 , 2 7 , 1 8 , 0 A system of switches with a -toggle transition rule applied
A system of switches with a -toggle transition rule applied
When is even and is odd (𝑥 , 𝑦 )𝜏𝑘−𝑤 ,𝑤
→(𝑥−𝑘+2𝑤 , 𝑦+𝑘−2𝑤 )
Given is even, is even.
From an initial state , any sequence of transition rules yielding the state will have .
𝑛≢0mod 2There are no transition rules to go from to .
When is even and
𝑘=2𝑚decreases the number of switches on by
𝑛=2𝑘+2𝑟
Method:1. Apply to to get
2. Apply to times to get
3. Apply to to get
Consider only when is even
0 , 1 4 1 , 1 3 2 , 1 2 3 , 1 1 4 , 1 0 5 , 9 6 , 8
7 , 7 8 , 6
9 , 5 1 0 , 4 1 1 , 3 1 2 , 2 1 3 , 1 1 4 , 0 (10 ,4 )(8 ,6 )(6 ,8 )(4,10 )
} 𝜏3 ,13 times
(14 ,0 )𝜏4 ,0
(0,1 4 )𝜏4 ,0
𝒏=𝟏𝟒 ,𝒌=𝟒
Transitioning from a given initial state to a specified terminal state depends on the parity of .
If is odd, a system of switches can be transitioned from to.
If is even, a system of switches can be transitioned from toonly when .
References