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

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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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You and me both!

Jerry

In message , Jerry Avins writes

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.

In message , Paul M writes

Yeah, I could only make it work with two 2-way SPST and 1 double pole DPDT changeover switch (crossover).

Yeah, I found that one as well. :)

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

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

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.

I just noticed, a 4-way switch has only 4 terminals, so it isn't simply DPDT.

It's a DPDT with the NO of each pole internally wires to the NC of the other.

Jerry

I spent a while trying to work it out with three-phase AC, but I couldn't get that to work either.

Here's your DPDT:

W X C D Y Z

Wire X to Y, and W to Z, and you have a four-way. XY and WZ are the inputs, C and D are the outputs.

I wonder if they sell regular DPDT wall switches. Would they be called

There's a "Two circuit switch", but that's not it. That's equivalent to a DPDT with two contacts missing:

W C D Z

A google search reveals that you can get DPDT wall switches, called exactly that. I found one from Hubbell Wiring Devices for $75 which seems a bit excessive. It also appears you can get some called "Two Circuit Three Way", which should be the same thing.

In message , Doghouse writes

Yeah, but by bridging the two diagonally opposing pairs of contacts on a typical DPDT switch it becomes a crossover switch.

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.

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.

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.

Arrggh!

s/on/off/

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