Series/Parallel of 10 resistors question

The total resistance is 7/5 (if R = 1). But, I don't know how to get there using on series/parallel connections.

You win.

John S

only

Yup. Same value I got, 1.4 ohms.

:) I wasn't trying to win, just point out that it takes a little more analysis that was shown by Martin on the Wolfram website.

These problems are indeed fun, even if as Jim says impractical. Stretches the mind. No idea what practical thing can happen. But John Conway hoped nothing practical would ever be done with his work on 26 dimensions -- until someone found something despite his hopes. So you never know where something like this takes you until you go for it.

Jon

There is a mapping from n to R that gives the result.

F_i(A,B) = i*(A + B) + (1 - i)*A*B/(A + B)

is one such mapping,

F_a_k(....F_a_0(R,R),...,R)

where a_k is the binary sequence for an integer m

Zero and open are useful values. If they're free, why not?

You're no better at keeping technical discussions technical than Slowman. Gotta be a prick.

I had arrived at it from: (1) hard work as a teenager when I saw the problem and didn't know there was an easy approach such as Norton and Thevenin eqivalents and hadn't come up with branch-current, mesh, and nodal analyses yet on my own (I _much_ prefer nodal analysis, by the way, as it 'sings in my mind' like a beautiful song whereas the others are tinny and clunky by comparison.) (2) Reading others' works saying so. (3) Trying my imagination at it from time to time afterwards, just to make sure.

But it is provable, I think. Lay out all possible nodes that reach from the starting point to the ending point without visiting any node twice. For example, in the above unbalanced wheatstone bridge, with 0 the starting point and 5 the ending point: 0 / \ / R R 1 / R / \ 2-----R-----3 \ / R / 4 R R / \ / 5

We get:

0 2 4 5 0 2 3 5 0 1 3 5 0 1 3 2 4 5

You might notice above that there is a "2 3" in one path and a "3 2" in another, which implies a reversal. Not sure if that means anything yet. But there it is.

Let's take another 7 resistor version, but only with series parallel connections:

0--R--1--R--2--R--3--R--4--R--5 | | | | '-----R-----+ '---R-------------'

In this case, we get:

0 1 2 3 4 5 0 1 4 5 0 1 2 4 5

No reversals. (Remember that we aren't allowed to visit a node more than once.)

Let's try this one:

,---R-------, | | 0--R--1--R--2--R--3--R--4--R--5 | | '-----R-----+

0 1 3 4 5 0 1 2 4 5 0 1 2 3 4 5 0 1 3 2 4 5

Hmm. There is that pesky reversal, again. A "2 3" and then a "3 2" appears.

Might be a clue here somewhere about defining what is series parallel only and what might be excluded from that privileged set.

But I'm just shooting in the dark.

Jon

Hi,

The last statement is wrong. I just realized that there are circuits that are valid in the sense of the problem definition which can not be reduced to a single resistor by those two rules, such as:

|---R----R---| | | |

• --| R |-- - | | | |---R----R---|

None of the resistors are in parallel or in series - so none of the above rules can be applied.

Regards, Timo

Be sure to wear steel-toed shoes ;-) ...Jim Thompson

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I love to cook with wine.     Sometimes I even put it in the food.

Oh crap. That's 2 _more_ things I missed. I thought the op wanted simple series/parallel and practical solutions, not complex arrays & theoretical solutions. Grumble. Displeased with myself, but glad you posted.

Ed

Yes. I agree. I was slightly pleased to see it and the way I read it confirmed that my construction was sufficient. I did say that I wasn't sure if some networks of deltas might generate some new values.

I agree with Jon (this was the sort of net that I was worried about).

But this particular one is too symmetrical and does not contribute any new values to the sequence. A basic symmetry argument shows that the two nodes are at V/3 and 2V/3 and the overall impedance is R. If you draw it right the equipotential surfaces become clear.

I think the first one that could introduce a novel value requires one extra resistor so one leg has 3R in it. Can anyone find or derive the full sequence for every possible network from identical resistors?

I'd also be interested in the number of unique values.

The two that my construction does not generate for N=4 are

R||R+R||R and (R+R)||(R+R)

I think that is all of them

I haven't so far been able to sketch all the missing N=5 networks. It is obviously at least 4.

It seems clear that for any composite number PQ less than the total number of resistors N you can take them in groups of P or Q and apply the series parallel rule to the clumps which definitely generates new networks. 6 is the first composite with distinct factors.

--
Regards,
Martin Brown

Bitrex nailed a good paper here:

Take a look. It's all you could hope to want, I think, if you include the references as well.

Jon

Thanks! Not only that but some of the results are applicable to software testability (a subject I am very interested in).

--
Regards,
Martin Brown

--
Here's a tabulation of all of the 55 possible combinations of from one
to ten simple series, parallel and series-parallel connected equal
value resistors normalized to one ohm:

news:n7u2a75jqn978kb45guihdtf2qv4l9qnbs@4ax.com

Please share! What parts relate to that (what do you _see_ in your mind about this?) I'm very curious.

Jon

How many ways can you hook up one resistor? ONE. How many unique ways can you add a resistor to make a composite resistor? TWO, series or parallel. For each of the two composite resistors, how can you add a resistor to make a new composite resistor? TWO, series or parallel. For each of the 4 possible composite resistors, how can you add a resistor to make a new composite resistor? repeat 'till you run out of resistors...add 'em up.

Is it not that simple?

For unequal resistors, gets much more complicated, but the same concept should work.

This method leaves out the combinations where the new composite is built of composite blocks, as you're adding one resistor at a time. For instance, it does not generate a series connection of two parallel-connected blocks.

--

Tauno Voipio

Read the thread and concentrate on those posts by Martin Brown. Perhaps they will answer your question.

On Wed, 19 Oct 2011 22:28:55 +0100, "TTman" wrote:

The correct answer is:

18362390002030400342502466214352424101003520520206203414135725725624524 13011204205245248562476247624924924682346724875913948275247646824359245 12342385245684287524858456745672713841344756456252345922592472456467285 8272772752757257278572572752475723457257275177312347246534

Which is:

One novemoctogintillion, eight hundred thirty-six octooctogintillion, two hundred thirty-nine septoctogintillion, two hundred three quinoctogintillion, forty quattuoroctogintillion, thirty-four treoctogintillion, two hundred fifty duooctogintillion, two hundred forty-six unoctogintillion, six hundred twenty-one octogintillion, four hundred thirty-five novemseptuagintillion, two hundred forty-two octoseptuagintillion, four hundred ten septseptuagintillion, one hundred sexseptuagintillion, three hundred fifty-two quinseptuagintillion, fifty-two quattuorseptuagintillion, twenty treseptuagintillion, six hundred twenty duoseptuagintillion, three hundred forty-one unseptuagintillion, four hundred thirteen septuagintillion, five hundred seventy-two novemsexagintillion, five hundred seventy-two octosexagintillion, five hundred sixty-two septsexagintillion, four hundred fifty-two sexsexagintillion, four hundred thirteen quinsexagintillion, eleven quattuorsexagintillion, two hundred four tresexagintillion, two hundred five duosexagintillion, two hundred forty-five unsexagintillion, two hundred forty-eight sexagintillion, five hundred sixty-two novemquinquagintillion, four hundred seventy-six octoquinquagintillion, two hundred forty-seven septenquinquagintillion, six hundred twenty-four sexquinquagintillion, nine hundred twenty-four quinquinquagintillion, nine hundred twenty-four quattuorquinquagintillion, six hundred eighty-two trequinquagintillion, three hundred forty-six duoquinquagintillion, seven hundred twenty-four unquinquagintillion, eight hundred seventy-five quinquagintillion, nine hundred thirteen novemquadragintillion, nine hundred forty-eight octoquadragintillion, two hundred seventy-five septenquadragintillion, two hundred forty-seven sexquadragintillion, six hundred forty-six quinquadragintillion, eight hundred twenty-four quattuorquadragintillion, three hundred fifty-nine trequadragintillion, two hundred forty-five duoquadragintillion, one hundred twenty-three unquadragintillion, four hundred twenty-three quadragintillion, eight hundred fifty-two novemtrigintillion, four hundred fifty-six octotrigintillion, eight hundred forty-two septentrigintillion, eight hundred seventy-five sextrigintillion, two hundred forty-eight quintrigintillion, five hundred eighty-four quattuortrigintillion, five hundred sixty-seven tretrigintillion, four hundred fifty-six duotrigintillion, seven hundred twenty-seven untrigintillion, one hundred thirty-eight trigintillion, four hundred thirteen novemvigintillion, four hundred forty-seven octovigintillion, five hundred sixty-four septenvigintillion, five hundred sixty-two sexvigintillion, five hundred twenty-three quinvigintillion, four hundred fifty-nine quattuorvigintillion, two hundred twenty-five trevigintillion, nine hundred twenty-four duovigintillion, seven hundred twenty-four unvigintillion, five hundred sixty-four vigintillion, six hundred seventy-two novemdecillion, eight hundred fifty-eight octodecillion, two hundred seventy-two septendecillion, seven hundred seventy-two sexdecillion, seven hundred fifty-two quindecillion, seven hundred fifty-seven quattuordecillion, two hundred fifty-seven tredecillion, two hundred seventy-eight duodecillion, five hundred seventy-two undecillion, five hundred seventy-two decillion, seven hundred fifty-two nonillion, four hundred seventy-five octillion, seven hundred twenty-three septillion, four hundred fifty-seven sextillion, two hundred fifty-seven quintillion, two hundred seventy-five quadrillion, one hundred seventy-seven trillion, three hundred twelve billion, three hundred forty-seven million, two hundred forty-six thousand, five hundred thirty-four

The permutations in some of the equations derived there are closely related to the path testing coverage problem for branch decision points in software if you do a little bit of relabelling of the entities.

The traditional metric is cyclomatic complexity index first derived by McCabe - it sort of works (it is correct for what it sets out to do). What it is also good for is spotting latent maintenance traps in inherited code. If the complexity is beyond a certain limit there is a very good chance that it will contain bugs.

--
Regards,
Martin Brown