The other three switch puzzle: how to wire them all to a singe light

dave y. wrote:


Which is the same as
(A AND B AND C) OR (A AND NOT B AND NOT C) OR (NOT A AND B AND NOT C) OR (NOT A AND NOT B AND C)

Which is not three 3-way switches.

Yet, there is another claim of solution also with a hint of non-standard wiring.

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snipped-for-privacy@aol.com wrote:

More than a claim. The solution was posted earlier in this thread. (I didn't find it withe a quick look, so I can't credit the author. Here it is again: A B o-------o o------o / \ / power-----o / o------load-----ground C / \ D / o-------o o------o
There are other configurations, but this one meets code. The center switch is a "four-way", for its four terminals. It can be made by connecting the NO of each pole of a DPDT with to the NC of the other. Depending on its position, wires A - B and C - D are connected, or A - D and C - B. As many additional four-ways can be cascaded as needed.
The end switches are three-way -- SPST -- and pretty self explanatory.
Jerry
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Jerry Avins wrote:

The "claim" I was referring to was that of a solution using three 3-way switches, which is not what you show below.
And although a three 3-way solution has been claimed, no such solution has yet been posted.

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snipped-for-privacy@aol.com wrote:

...

We agree. I misunderstood.
Jerry
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A1 A2 A3 Y3 L1 C1 C2 C3 D3 L2 B1 B2 B3 Z3
L1 and L2 are the lamp terminals The Cs and Ds are common, the A,B,Y,Zs are common. The 3 switch is DPDT.
HOT-C1 A1-A3-Z3 B1-B3-Y3 C3-A2 D3-B2 C2-L1 NEU-L2
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On Thu, 02 Nov 2006 14:57:44 -0600, snipped-for-privacy@grace.speakeasy.net (Matthew Russotto) proclaimed to the world:

The question asked was how to do this with three SPDT switches. There are several ways of doing it with different switches easily found in lots of different places, but you have to have at least one double pole switch.
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(UTC):

Sure, that's easy.
The 4 devices have a total of 11 connections, meaning there's a maximum of 11 nodes in any wiring.
Without loss of generality, you can assign "ground" to node 1 and "power" to node 2. At this point there are 11^11 possible hookups.
Then you can prune away:
* any hookup which has a disconnected partition
* any hookup which skips a node number (which means you'll never use a node beyond 9)
I'm going to go on record saying I think the solution must involve either a DPDT "4-way" switch or some kind of switching device (e.g. diode or relay).
--Keith Lewis klewis mitre.org The above may not (yet) represent the opinions of my employer.
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Keith A. Lewis wrote:
...

Solutions have been posted using two SPDTs and one DPDT. (Those are standard.) The problem as posed requires three SPDTs. Either there's a solution or there isn't, but a soluution can't "require" something not allowed.
Jerry
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Can I be the first arrogant person to say that it cannot be achieved with three single pole changeover switches. Even with polarity swapping tricks it doesn't seem feasible. Perhaps the original puzzle was not remembered correctly.
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Clive Mitchell wrote:

I'm inclined to agree with you, but 70+ tears have taught me to be very careful of claiming that if I can't do a thing, nobody can. :-)
Jerry
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Jerry Avins wrote:

I don't cry much. That should be "years".
Jerry
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proclaimed to the world:

Post reading again when you should have proof read instead ;-)
You are making me feel better about my own posts. People claim that I misspell but really, I never miss spelling.
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It seems that there is no solution. On sci.physics, Jim Black wrote a program that searched for solutions, and it found none.
Heinz
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Clive Mitchell wrote...

I haven't yet managed to achieve the clarity of thought to formulate a formal proof, but I think that it's self-evidently impossible. For every switch to be effective each node must connect one switch terminal to one and only one other (including the connections via the lamp and supply). It's obvious that every switch must be fully connected and multiple connections are not allowed, since this would lead to more than one switch controlling the same element of the current path. Hence, because terminals must be connected in pairs, the total number of terminals must be an even number. Thus 2xSPDT switches are allowed (6 terminals) and 2xSPDT plus any number (n) of DPDT changeover switches (4n+6 terminals) but 3xSPDT switches cannot work (9 terminals), since there will always be one terminal that remains unconnected however the switches are arranged.
Clearly, whilst this is a necessary condition, it is not sufficient to define a multiple switching network, since every possible current path must also pass through every switch. I believe that this additional condition will forbid networks containing more than one pair of SPDT switches, but I cannot prove that assertion.
Comments?
David
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message

with
tricks
remembered
and
cannot
must
Again, not rigorous, but to continue your train of thought, an SPDT switch permits two possible current paths. To permit either of these two paths to be usable, both paths must pass through all other switches.
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I spent a while trying to work it out with three-phase AC, but I couldn't get that to work either.
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On Fri, 03 Nov 2006 13:52:57 -0600, snipped-for-privacy@grace.speakeasy.net (Matthew Russotto) proclaimed to the world:

Try to find a solution using the same hardware plus as many diodes as you like and use DC.
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On Fri, 03 Nov 2006 17:41:30 -0500, Paul M <PaulMatWiredogdotcom> wrote:

If you can throw in additional components, you can get at least a partial solution fairly easily - even for AC. But I'll explain it for DC to avoid pondian issues over the names for the wires.
You wire two of the switches as the normal arrangement to switch positive on or off. The output of this goes to one end of the bulb, and to negative via a resistor. The other switch connects the other end of the bulb to either positive or negative.
Drawbacks: a) half the time when the bulb is off there is current going through the resistor b) the bulb is only partly lit in half the cases when it is on as the resistor is in series. c) the other half the time the resistor is in parallel with the bulb, also wasting current.
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Nick Atty wrote:

Can you try an ascii drawing? The way I understand this is not ok, because when the last switch is connected to plus, there is no way to switch the lamp on with the other two switches.
Heinz
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wrote:

I've not got time to try the ascii at the moment, but let me try to explain it.
You can treat the other two switches as a single on/off connected to plus. When they are "on" then both sides of the lamp will be plus, so the lamp will be off. When they are "on", then that side of the lamp will be connected to negative by the resistor, so current will flow through the lamp and resistor in series.
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