## The 2 Ohm Problem

```The most common electrical resistors commercially available are a series of
twelve basic resistance values, multiplied by arbitrary powers of ten. This
is called the "E12 series", consisting of resistors with the following values
(ohms):

10   12   15   18   22   27   33   39   47   56   68   82
100  120  150  180  220  270  330  390  470  560  680  820
1000 1200 1500 1800 2200 2700 3300 3900 4700 5600 6800 8200
etc....

The E12 series extends down into fractional values, i.e., 8.2, 6.8, 5.6, and
so on, but we will not consider these. There also exists an "E24 series" that
provides 24 resistances per decade, but those are less widely available.

Noting that the total resistance of a set of resistors R1, R2, ..., Rn
combined in parallel is

1        1      1             1
----  =  ---- + ---- + ...  + ----
Rtot      R1     R2            Rn

we might ask which integer-valued resistances can be produced by combining
distinct integer-valued resistances in parallel from the standard E12 series.

Of course, if 1/R equals a sum of unit fractions then R must divide the
product of the denominators, which implies that R cannot be synthesized by
means of 2 or more distinct E12 resistors in parallel if R is divisible by
any prime other than those that appear in the E12 sequence, namely, 2, 3, 5,
7, 11, 13, 17, 41, and 43.  It's interesting that this covers the first
seven primes, even though it's doubtful that the E12 values were chosen with
prime factors in mind.  According to the commercial literature, the E24
values were defined as the closest integers to 10r^k, k=0,1,2,..,23, where
r is the 24th root of 10, although this doesn't match all the values.

Even if R is entirely composed of "E12 primes" it may not be possible to
synthesize it by just TWO resistors in parallel.  It may require three or
more.  The only "fundamental" integer values of R that are expressible as a
sum of just two unit fractions with E12 denominators are those shown below:

1/R = 1/A + 1/B

R         A       B
-----    ------  -------
6         10      15
20         22     220
30         33     330
72        120     180
88        120     330
99        180     220
102        120     680
108        180     270
132        220     330
975       1000   39000
1125       1200   18000
1476       1800    8200
1485       2700    3300
6875      10000   22000
9375      10000  150000
16875      18000  270000
17875      33000   39000
103125     150000  330000
193875     330000  470000

I call these "fundamental" because we can multiply any of them by any
power of 10.  For example, since 1/99 = 1/180 + 1/220 it's obvious that
1/990 = 1/1800 + 1/2200, and so on.  However, although the above
solutions are fundamental in this sense, they are not all "primitive",
because they may share common E12 factors.  For example, the solution
[103125, 150000, 330000] is just 15 times the solution [6875, 10000, 22000].

To create other fundamental solutions we must allow more than two resistors
in parallel.  Here's a summary of the "most economical" solutions I've found
for the first several values of R:

R                           denominators
---  -------------------------------------------------------------
3   10  12  15   22   33   220   330
4   10  12  15
5   10  12 100  150
6   10  15
7   10  56 100  120  150
8   10 100 120  150
9   18  22 180  220
10   10
11   12 220 330
12   12
13   15 100 3900
14   33 100 120 150 180 220 330 1000 1500 1800 2200 5600 12000 18000
15   15
16   18 180 1200 1800
17   33  68 120  330  680 1200 6800
18   18
19   impossible
20   22 220
21   33 100 330  560 1000 1200 1500
22   22
23   impossible
24   33 120 330
25   33 150 330
26   39 100 390 3900
27   27
etc.

This raises some interesting questions.  For example, is the above expression
for 14 ohms optimal, or can it be done with fewer than 14 resistors?  In general,
is it possible to express 1/R as a sum of inverses of distinct "E12 integers"
for any R (greater than 1) that is composed entirely of E12 primes?

It's easy to see that we can't synthesize 1 ohm from any number of distinct E12
resistors in parallel, because the infinite sum of the inverses of all these
resistors implies that the least resistance that can be produced by such a
network is

31873589748
-----------  ohms    =    1.764  ohms
18062619415

On the other hand, this doesn't rule out the possibility that 2 ohms could
be produced. This is the "2 Ohm Problem", i.e., find a combination of distinct
E12 resistors that give exactly 2 ohms when connected in parallel.

One approach to finding such an expression is to first consider each individual
basic resistor, multiplied by powers of 10, focusing on eliminating the large
primes from the denominators of these individual sums. For the basic resistors
22, 33, 39, 47, 58, 68, and 82 the smallest sums that eliminate the primes
greater than 5 from the denominators are

1    1       1
-- + ---  =  --
22   220     20

1    1       1
-- + ---  =  --
33   330     30

1       1        77
-- + + -----  =  ----
39     39000     3000

1     1        1        213
-- + ----- + ------  =  -----
47   47000   470000     10000

1     1        143
-- + -----  =  ----
56   56000     8000

1    1      1       1        653
-- + --- + ---- + ------  =  -----
68   680   6800   680000     40000

1    1      1      1        1        271
-- + --- + ---- + ----- + ------  =  -----
82   820   8200   82000   820000     20000

The sum of the seven series listed above is 3561/20000, which differs from 1/2 by
6439/20000.  Thus we seek sums k1 to k5 of reciprocal powers of 10 such that

/ 1 \       / 1 \       / 1 \       / 1 \       / 1 \       6439
( --- )k1 + ( --- )k2 + ( --- )k3 + ( --- )k4 + ( --- )k5 = -----
\ 10/       \ 12/       \ 18/       \ 15/       \ 27/      20000

Since there are only 32 distinct sums of reciprocal powers of 10 up to 10^4, it
is feasible to make an exhaustive search. The first solution we find is

1      1      1                      1      1                 1
k1 = 1 + ---- + ---- + ----     k2 = k3 = 1 + ---- + ----     k4 = 1 + ----    k5 = 0
10^2   10^3   10^4                   10^1   10^2              10^4

Thus we can produce 2 ohms by combining the following thirty-two integer resistances
from the E12 series:

10       1000  10000  100000
12  120  1200
15                    150000
18  180  1800
22  220
33  330
39             39000
47             47000  470000
56             56000
68  680  6800         680000
82  820  8200  82000  820000

This solution makes no use of any resistors from the "27" sequence.  There are, of
course, many other solutions. Consider, for example, the expression

1.1111   1.1101   1.001   1.0011   1.001   1.1   1.1   1.0111   1.0111   114653
------ + ------ + ----- + ------ + ----- + --- + --- + ------ + ------ = ------
82       68       56      47      39      33    22     18       12     360000

To complete a "2 ohm" solution based on this, we need to find series using the
resistances 10, 15, and 27 that sum to Q = 65347/360000.  Each of the following
expressions equals this quantity, so each of them (along with the expression above)
gives a distinct solution of the 2 ohm problem:

Q  =  1.10001/10 + 1.01111/15 + 0.11100/27

=  1.10101/10 + 1.01011/15 + 0.11010/27

=  1.10111/10 + 1.01001/15 + 0.11001/27

=  1.11101/10 + 1.00011/15 + 0.10110/27

=  1.11111/10 + 1.00001/15 + 0.10101/27

One peculiar feature of these solutions can be seen if we interpret the numerators
of these terms as binary numbers. For example, the numerator of 10 in the first of
these expressions is 1.10001, which we interpret as the binary number 110001 = 49.
Similarly we determine the base 2 corresponding to each of the other numerators,
adding a zero to each numerator of 27 so that it has the same number of digits
as the other numerators. This gives the results

49    47    56
53    43    52
55    41    50
61    35    44
63    33    42

The differences between consecutive numbers in the first column are 4, 2, 6, 2,
and these are the negatives of the differences of consecutive numbers in the
second and third columns. If we reverse the digits of the numerators, but still
interpret them as binary numbers, we have the values

35    61    14
43    53    22
59    37    38
47    49    26
63    33    42

Now the differences between consecutive numbers in the first column are 8, 16, -12,
16, and these are the same as the differences of the third column, and the negatives
of the differences of the second column.

We can simplify the situation by removing all the terms from the numerators
that are common to all rows, leading to the denominators for five equal
expressions (swapping the third and fourth sets to put them in numerical order)

.0000   .1110   .1100
.0100   .1010   .1010
.0110   .1000   .1001
.1100   .0010   .0110
.1110   .0000   .0101

Taking these as binary digits in forward order or reverse order, we get the
sequences

forward             reverse

0  14  12            0  7   3
4  10  10            2  5   5
6   8   9            6  1   9
12   2   6            3  4   6
14   0   5            7  0  10

In the forward case, the differences between consequtive numbers in the first
column are equal and opposite the differences in the second, which are twice
the differences of the third column, whereas in the reverse case the differences
in the first and third columns are equal and opposite the differences in the
second column.

The forward case occurs because the component-wise difference between any two
sets of numerators, interpreted as decimal numbers, is proportional to 1, -1,
-0.9, which implies that each set gives the same quantity when divided by their
respective denominators, because of the basic equality

1     1     / 1     1 \           1     1     0.9
--- - --- - ( --- - --- )    =    --- - --- -  ---    =     0
10    15    \ 27   270/           10    15     27

The decimal differences between the numerators of consecutive sets (multiplied
by 10000) are shown below.

.0000     .1110     .1100
100      -100       -90
.0100     .1010     .1010
10       -10        -9
.0110     .1000     .1001
990      -990      -891
.1100     .0010     .0110
10       -10        -9
.1110     .0000     .0101

On the other hand, when we interpret the numbers in binary, the decimal
difference given by 1 - .1 = .9 becomes 1 - .1 = .1 (which is 1/2), so the
differences are proportional to 1, -1, -0.5 instead of 1, -1, -0.9. It would
be interesting to know if there exist larger sets of numbers with this same
property, i.e., the component-wise differences are in fixed proportions in
two different bases.
```