Mixing
There are three lights which can be in one of three states. Can we get the system of lights into a specific state?
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There are three lights in a line. Each light can be in one of three states: off, light red, and dark red. There is a cycle of states: OFF, then LIGHT RED, then DARK RED, then back to OFF.
There are three switches which control the lights like so:
Switch A - advances the cycle for the first two lights
Switch B - advances the cycle for the all three lights
Switch C - advances the cycle for the last two lights.
If we start with all three lights in the off state can the switches be pushed in some order so that the three lights in the line are in: OFF-LIGHT RED-DARK RED?
I'm trying to model this with linear algebra. Where A,B,C are the lights in a row and we push A x times, B y times, and C z times. Of course the numbers are mod 3 because after 3 pushes we wrap back to the off state.
Any suggestions?
linear-algebra modular-arithmetic
asked 1 hour ago
Idle Math Guy
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up vote
1
down vote
favorite
If there is someone who can come up with a better title for this please edit the title.
There are three lights in a line. Each light can be in one of three states: off, light red, and dark red. There is a cycle of states: OFF, then LIGHT RED, then DARK RED, then back to OFF.
There are three switches which control the lights like so:
Switch A - advances the cycle for the first two lights
Switch B - advances the cycle for the all three lights
Switch C - advances the cycle for the last two lights.
If we start with all three lights in the off state can the switches be pushed in some order so that the three lights in the line are in: OFF-LIGHT RED-DARK RED?
I'm trying to model this with linear algebra. Where A,B,C are the lights in a row and we push A x times, B y times, and C z times. Of course the numbers are mod 3 because after 3 pushes we wrap back to the off state.
Any suggestions?
linear-algebra modular-arithmetic
asked 1 hour ago
Idle Math Guy
340114
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If there is someone who can come up with a better title for this please edit the title.
There are three lights in a line. Each light can be in one of three states: off, light red, and dark red. There is a cycle of states: OFF, then LIGHT RED, then DARK RED, then back to OFF.
There are three switches which control the lights like so:
Switch A - advances the cycle for the first two lights
Switch B - advances the cycle for the all three lights
Switch C - advances the cycle for the last two lights.
If we start with all three lights in the off state can the switches be pushed in some order so that the three lights in the line are in: OFF-LIGHT RED-DARK RED?
I'm trying to model this with linear algebra. Where A,B,C are the lights in a row and we push A x times, B y times, and C z times. Of course the numbers are mod 3 because after 3 pushes we wrap back to the off state.
Any suggestions?
linear-algebra modular-arithmetic
asked 1 hour ago
Idle Math Guy
340114
If there is someone who can come up with a better title for this please edit the title.
There are three lights in a line. Each light can be in one of three states: off, light red, and dark red. There is a cycle of states: OFF, then LIGHT RED, then DARK RED, then back to OFF.
There are three switches which control the lights like so:
Switch A - advances the cycle for the first two lights
Switch B - advances the cycle for the all three lights
Switch C - advances the cycle for the last two lights.
If we start with all three lights in the off state can the switches be pushed in some order so that the three lights in the line are in: OFF-LIGHT RED-DARK RED?
I'm trying to model this with linear algebra. Where A,B,C are the lights in a row and we push A x times, B y times, and C z times. Of course the numbers are mod 3 because after 3 pushes we wrap back to the off state.
Any suggestions?
linear-algebra modular-arithmetic
linear-algebra modular-arithmetic
asked 1 hour ago
Idle Math Guy
340114
asked 1 hour ago
Idle Math Guy
340114
asked 1 hour ago
Idle Math Guy
340114
asked 1 hour ago
Idle Math Guy
340114
asked 1 hour ago
Idle Math Guy
340114
340114
add a comment |Â
add a comment |Â
2 Answers
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We can represent the states of the lights with $0=Off, 1=Light Red, 2=Dark Red$. Then we want $$x+yequiv 0\ x+y+zequiv 1\y+zequiv 2$$
where all the equivalences are $mod 3$ The first two tell us $z=1$. The last two tell us $x=2$. Then from the first $y=1$. So we use A twice and the others once.
answered 1 hour ago
Ross Millikan
283k23191359
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
add a comment |Â
up vote
3
down vote
Note that we can switch the first, second and third light independently, so any state can be reached:
- Switching $A$ twice and then $B$ is the same as switching the last light.
- Switching $C$ twice and then $B$ is the same as switching the first light.
- Switching $A$ once, switching $C$ once and switching $B$ twice is
the same as switching the middle light.
answered 1 hour ago
Servaes
18.9k33684
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We can represent the states of the lights with $0=Off, 1=Light Red, 2=Dark Red$. Then we want $$x+yequiv 0\ x+y+zequiv 1\y+zequiv 2$$
where all the equivalences are $mod 3$ The first two tell us $z=1$. The last two tell us $x=2$. Then from the first $y=1$. So we use A twice and the others once.
answered 1 hour ago
Ross Millikan
283k23191359
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
add a comment |Â
up vote
3
down vote
accepted
We can represent the states of the lights with $0=Off, 1=Light Red, 2=Dark Red$. Then we want $$x+yequiv 0\ x+y+zequiv 1\y+zequiv 2$$
where all the equivalences are $mod 3$ The first two tell us $z=1$. The last two tell us $x=2$. Then from the first $y=1$. So we use A twice and the others once.
answered 1 hour ago
Ross Millikan
283k23191359
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We can represent the states of the lights with $0=Off, 1=Light Red, 2=Dark Red$. Then we want $$x+yequiv 0\ x+y+zequiv 1\y+zequiv 2$$
where all the equivalences are $mod 3$ The first two tell us $z=1$. The last two tell us $x=2$. Then from the first $y=1$. So we use A twice and the others once.
answered 1 hour ago
Ross Millikan
283k23191359
We can represent the states of the lights with $0=Off, 1=Light Red, 2=Dark Red$. Then we want $$x+yequiv 0\ x+y+zequiv 1\y+zequiv 2$$
where all the equivalences are $mod 3$ The first two tell us $z=1$. The last two tell us $x=2$. Then from the first $y=1$. So we use A twice and the others once.
answered 1 hour ago
Ross Millikan
283k23191359
edited 52 mins ago
answered 1 hour ago
Ross Millikan
283k23191359
answered 1 hour ago
Ross Millikan
283k23191359
answered 1 hour ago
Ross Millikan
283k23191359
283k23191359
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
add a comment |Â
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
1
1
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
Why does the last one fail? What is wrong with $(x,y,z)=(2,1,1)$?
– Servaes
1 hour ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
@Servaes: I confused myself. That does work. Fixed.
– Ross Millikan
53 mins ago
add a comment |Â
up vote
3
down vote
Note that we can switch the first, second and third light independently, so any state can be reached:
- Switching $A$ twice and then $B$ is the same as switching the last light.
- Switching $C$ twice and then $B$ is the same as switching the first light.
- Switching $A$ once, switching $C$ once and switching $B$ twice is
the same as switching the middle light.
answered 1 hour ago
Servaes
18.9k33684
add a comment |Â
up vote
3
down vote
Note that we can switch the first, second and third light independently, so any state can be reached:
- Switching $A$ twice and then $B$ is the same as switching the last light.
- Switching $C$ twice and then $B$ is the same as switching the first light.
- Switching $A$ once, switching $C$ once and switching $B$ twice is
the same as switching the middle light.
answered 1 hour ago
Servaes
18.9k33684
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Note that we can switch the first, second and third light independently, so any state can be reached:
- Switching $A$ twice and then $B$ is the same as switching the last light.
- Switching $C$ twice and then $B$ is the same as switching the first light.
- Switching $A$ once, switching $C$ once and switching $B$ twice is
the same as switching the middle light.
answered 1 hour ago
Servaes
18.9k33684
Note that we can switch the first, second and third light independently, so any state can be reached:
- Switching $A$ twice and then $B$ is the same as switching the last light.
- Switching $C$ twice and then $B$ is the same as switching the first light.
- Switching $A$ once, switching $C$ once and switching $B$ twice is
the same as switching the middle light.
answered 1 hour ago
Servaes
18.9k33684
answered 1 hour ago
Servaes
18.9k33684
answered 1 hour ago
Servaes
18.9k33684
answered 1 hour ago
Servaes
18.9k33684
18.9k33684
add a comment |Â
add a comment |Â
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