How Close Can You Get to Pi Ohms Using Ten Resistors?

[ceptimus]

Tommy has just ten resistors, a 1-ohm, a 2-ohm, a 3-ohm and so on up to a 10-ohm.

He wants to wire them up in a combination that gives a total resistance as near as possible to π ohms, that is to say 3.14159265(etc.) ohms.

He doesn't have to use all the resistors - if he can get nearer to π ohms by only using (say) eight of the resistors, that's fine.

What's the best he can do?

 

[NateTG]

There's a lot of possible arrangements, and they're hard to describe.

For example: getting $$\frac{22}{7}$$ is possible by using:

a 3 ohm and 4 ohm resistor in parrelel to get a $$\frac{12}{7}$$ Ohm resistor
and a 2 ohm and 5 ohm resistor in parralel to get a $$\frac{10}{7}$$ Ohm resistor
and then putting the two in series. That's a pretty good set-up.
$$\frac{22}{7}=3.\bar{1}\bar{4}\bar{2}\bar{8}\bar{5}\bar{7}$$
so the error is less than .002.

That leaves 6 resistors unused.

 

[AKG]

Tommy has just ten resistors, a 1-ohm, a 2-ohm, a 3-ohm and so on up to a 10-ohm.

He wants to wire them up in a combination that gives a total resistance as near as possible to π ohms, that is to say 3.14159265(etc.) ohms.

He doesn't have to use all the resistors - if he can get nearer to π ohms by only using (say) eight of the resistors, that's fine.

What's the best he can do?

Quick guess, I'm sure you can get closer, but of all the possible combinations with only 2 resistors, putting 5 and 8 in parallel gives:

3.076923077

 

[NateTG]

Here's a slightly better answer:
5 2 Ohm resistors in parralel for a $$\frac{2}{5}$$ Ohm resistor
put it in series with a 10 Ohm resistance for a $$\frac{52}{5}$$ Ohm resistor
Put it in parralel with two 9 ohm resistors.
That makes for a $$\frac{468}{149} \approx 3.14093$$

This is a pretty ugly problem...

 

[Gokul43201]

Any reasonable way to make 355/113 ~3.1415929 ?

 

[NateTG]

Any reasonable way to make 355/113 ~3.1415929 ?

Code tags used for fixed spacing.
Numbers represent resistors of the appropriate value.

   |-1-|
 |-|   |---------|
 | |-2-|         |
-|               |-2--
 | |-4-| |-2-|   |
 |-|   |-|   |-1-|
   |-5-| |-7-|

Please double check my math.

 

[ceptimus]

Code tags used for fixed spacing.
Numbers represent resistors of the appropriate value.

   |-1-|
 |-|   |---------|
 | |-2-|         |
-|               |-2--
 | |-4-| |-2-|   |
 |-|   |-|   |-1-|
   |-5-| |-7-|

Please double check my math.

You used the two-ohm and one-ohm resistors more than once.

 

[NateTG]

You used the two-ohm and one-ohm resistors more than once.

Oh. I thought it was any 10 resistors with values 1-10...

 

[verty]

Somebody should write a quick proggy to brute force it. I will see if I can do that.

 

[Gokul43201]

I doubt you can do better than 22/7 without repeats...

 

[The Bob]

I am assuming that we are using the proper way of calculating electronic resistance otherwise I cannot do this with as much ease.

The Bob (2004 ©)

 

[The Bob]

Code tags used for fixed spacing.
Numbers represent resistors of the appropriate value.

    |-1-|
  |-|   |---------|
  | |-2-|         |
--|               |-2-|--
  | |-4-| |-2-|   |
  |-|   |-|   |-1-|
    |-5-| |-7-|

Please double check my math.

NateTG. I just want to double check my double checking of your calculations. I believe your resistance would be 3.709302326 etc. Ω. Here are my calculations of your answer:

1Ω and 2Ω in Parellel = 1/1 + 1/2 = 2/2 + 1/2 = 3/2. Take Reciprocal = 2/3
4Ω and 5Ω in Parellel = 1/4 + 1/5 = 5/20 + 4/20 = 9/20. Take Reciprocal = 20/9
2Ω and 7 Ω in Parellel = 1/2 + 1/7 = 7/14 + 2/14 = 9/14. Take Reciprocal 14/9

Then 20/9 + 14/9 + 1 = 20/9 + 14/9 + 9/9 = 43/9

Then the parellel resistance of the two sides (e.g. the 2/3 and the 43/9):
1/(43/9) + 1/(2/3) = 1.709302326 etc.

1.709302326 + 2 = 3.709302326Ω

Hope I am right in you being right.

The Bob (2004 ©)

 

[The Bob]

Out of interest ceptimus, had NateTG not got it in post 2#?

The Bob (2004 ©)

 

[NateTG]

I knew there was an error.
It should have been more like:

   |-1-|
 |-|   |-1-------|
 | |-1-|         |
-|               |-2--
 | |-4-| |-2-|   |
 |-|   |-|   |-1-|
   |-5-| |-7-|

Then it would be:
$$\frac{1}{\frac{3}{2}}+\frac{1}{\frac{43}{9}}=\frac{2}{3}+\frac{9}{43}=\frac{43*2+9*3}{43*3}=\frac{113}{129}$$
So the resistance is the reciprocal:
$$\frac{129}{113}$$
Then the total resistance is:
$$\frac{129}{113}+\frac{226}{113}=\frac{355}{113}$$

Unfortunately the condidition was that only one resistor of each value could be used, so the solution wasn't legal.

I can replace one of the 2 Ohm resistors with a 6 and a 3 Ohm resistor in parralel. That leaves the 8, 9 and 10 Ohm resistors, but even putting the lot of them in parralel gives a resistance that's too large

 

[ceptimus]

Out of interest ceptimus, had NateTG not got it in post 2#?

The Bob (2004 ©)

No. You can do much better. Here are my best efforts so far using 3, 4, and 5 resistors. By using 6, 7, 8, 9 or all 10 of the resistors, you can get much closer to PI.

3 resistors: 3.158  (error: 1.6 * 10^-2)
-+--7-----8--+- 
 |           |
 +--4--------+

4 resistors: 3.14102  (error: 5.67 * 10^-4)
-+--1--+--3--+- 
 |     |     |
 +--5--+-10--+

5 resistors: 3.141553  (error: 4.01 * 10^-5)
-+--7-----9--+- 
 |           |
 +--1--+--3--+
 |     |  
 +-10--+

 

[Gokul43201]

hummm...nice.

 

[NateTG]

Brute force or Brain Teaser?

Is there an elegant solution, or should I just dust off my C skills?

 

[ceptimus]

Is there an elegant solution, or should I just dust off my C skills?

It's brute force, so far as I know. But optimizing the program so that you're not looking at years of run time, is a nice Brain Teaser.

 

[The Bob]

Can I ask that we post the resistors like NateTG in post 6# pleaes because I just got confused (at a glance) with what ceptimus put in the last few posts (with the code diagram).

The Bob (2004 ©)

 

[ceptimus]

How are the programs coming along?

Here is a 6-resistor version:

6 resistors: 3.14159292  (Error: 2.67 * 10^-7)
-+-------10--------+-
 |                 |
 +--5--+-----+--2--+
 |     |     |
 +--8--+     |
 |           |
 +--7-----9--+

With 9 resistors, I've found 105768/33667

I'll keep my 10-resistor solution secret for now.

 

[ceptimus]

OK, as everyone seems to have given up, here are my best answers:

8 resistors: 3.141592564  (Error: 8.93 * 10^-8)

@-+--------7--------+-@
  |                 |
  +--1--+--9--------+
  |     |           |
  +--3--+--2--+-10--+
  |     |     |
  +--4--+--8--+

9 resistors: 3.1415926575 (105768/33667) (Error: 3.91 * 10^-9) 

  +--5--+--8-----+
  |     |        |
  +-10--+--4--+  |
  |           |  |
@-+-----9-----+  +-@
  |           |  | 
  +--6--+--1--+  |
  |     |        |
  +--7--+--2-----+

10 resistors: 3.1415926555 (524935/167092) (Error: 1.95 * 10^-9)

  +-10--+-----5-----+--9--+
  |     |           |     |
  +--6--+--4--+--2--+     |
  |           |     |     |
@-+           +--7--+     +-@
  |           |     |     |
  |           +--8--+     |
  |           |           |
  +-----1-----+-----3-----+

 

[check]

Very good ceptimus. I tried writing a 'brute force' program to figure it out... I couldn't wrap my head around how to do it though. Oh well. lol
But great brain teaser.