22
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Introduction:

Two resistors, R1 and R2, in parallel (denoted R1 || R2) have a combined resistance Rp given as:

$$R_{P_2} = \frac{R_1\cdot R_2}{R_1+R_2}$$ or as suggested in comments:

$$R_{P_2} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$

Three resistors, R1, R2 and R3 in parallel (R1 || R2 || R3) have a combined resistance (R1 || R2) || R3 = Rp || R3 :

$$R_{P_3} = \frac{\frac{R_1\cdot R_2}{R_1+R_2}\cdot R_3}{\frac{R_1\cdot R_2}{R_1+R_2}+R_3}$$

or, again as suggested in comments:

$$R_{P_3} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}+ \frac{1}{R_3}}$$

These formulas can of course be extended to an indefinite number of resistors.


Challenge:

Take a list of positive resistor values as input, and output the combined resistance if they were placed in parallel in an electric circuit. You may not assume a maximum number of resistors (except that your computer can handle it of course).

Test cases:

1, 1
0.5

1, 1, 1
0.3333333

4, 6, 3
1.3333333

20, 14, 18, 8, 2, 12
1.1295

10, 10, 20, 30, 40, 50, 60, 70, 80, 90
2.6117  

Shortest code in each language wins. Explanations are highly encouraged.

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1
  • 6
    \$\begingroup\$ There are a few other challenges that refer to the harmonic mean (1 2 3) but I don't think there is a duplicate. In line with what flawr suggested, I think this challenge body should have that phrase listed somewhere so we can close a future dupe more easily. \$\endgroup\$ Sep 11 '19 at 15:17

44 Answers 44

13
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05AB1E, 5 3 bytes

zOz

Try it online!


Explanation

z                     # compute 1/x for each x in input 
 O                    # sum input 
  z                   # compute 1/sum
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1
  • 4
    \$\begingroup\$ Excepting built-ins, this is probably as low as we can go! \$\endgroup\$
    – user15259
    Sep 11 '19 at 14:57
11
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Haskell, 18 16 bytes

(1/).sum.map(1/)

Try it online!

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2
  • 3
    \$\begingroup\$ It looks beautiful. \$\endgroup\$ Sep 12 '19 at 11:52
  • \$\begingroup\$ Solution along the OP's recursive lines would be 22 chars: foldr1(\r s->r*s/(r+s)). \$\endgroup\$ Sep 12 '19 at 15:53
10
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MATLAB, 14 bytes

In MATLAB norm(...,p) computes the p-norm of a vector. This is usually defined for \$p \geqslant 1\$ as

$$\Vert v \Vert_p = \left( \sum_i \vert v_i \vert^p \right)^{\frac{1}{p}}.$$

But luckily for us, it also happens to work for \$p=-1\$. (Note that it does not work in Octave.)

@(x)norm(x,-1)

Don't try it online!

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2
  • 4
    \$\begingroup\$ This is horrible and beautiful simultaneously! \$\endgroup\$ Sep 12 '19 at 15:45
  • 1
    \$\begingroup\$ Thanks, these are the best compliments:) \$\endgroup\$
    – flawr
    Sep 12 '19 at 16:21
7
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Jelly,  5  3 bytes

İSİ

Try it online!

How?

Initially I forgot this form from my electronic engineering days ...how easily we forget.

İSİ - Link: list of numbers, R   e.g. [r1, r2, ..., rn]
İ   - inverse (vectorises)            [1/r1, 1/r2, ..., 1/rn]
 S  - sum                             1/r1 + 1/r2 + ... + 1/rn
  İ - inverse                         1/(1/r1 + 1/r2 + ... + 1/rn)
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1
  • 4
    \$\begingroup\$ I'm assuming İ is pronounced the same way i is pronounced in list. Is this a way of saying the challenge was easy? \$\endgroup\$ Sep 13 '19 at 6:31
5
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Octave, 15 bytes

@(x)1/sum(1./x)

Try it online!

Harmonic mean, divided by n. Easy peasy.

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1
  • \$\begingroup\$ @tsh you know, I don't think I ever noticed that. I guess it's almost the harmonic mean... \$\endgroup\$
    – Giuseppe
    Sep 13 '19 at 13:59
4
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PowerShell, 22 bytes

$args|%{$y+=1/$_};1/$y

Try it online!

Takes input via splatting and uses the same 1/sum of inverse trick as many of the others are doing

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4
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APL (Dyalog Unicode), 4 bytes

÷1⊥÷

Try it online!

-1 thanks to Adám.

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7
  • 1
    \$\begingroup\$ APL is the original golfing language! \$\endgroup\$
    – user15259
    Sep 12 '19 at 13:35
  • \$\begingroup\$ @YiminRong It's not a golfing language... :P \$\endgroup\$ Sep 12 '19 at 15:11
  • \$\begingroup\$ I know, but its byte count is on par with modern golfing languages! \$\endgroup\$
    – user15259
    Sep 12 '19 at 15:33
  • \$\begingroup\$ -1 byte: ÷1⊥÷ Try it online! \$\endgroup\$
    – Adám
    Sep 15 '19 at 9:27
  • \$\begingroup\$ @Adám Oh duh of course 1∘⊥ is the same as +/ for vectors... \$\endgroup\$ Sep 15 '19 at 11:36
3
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R, 15 bytes

1/sum(1/scan())

Try it online!

Follows the same Harmonic Mean principle seen in other answers.

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3
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JavaScript, 28 bytes

a=>a.map(y=>x+=1/y,x=0)&&1/x

Try it Online!

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3
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Perl 5 -pa -MList::Util=reduce, 26 bytes

$_=reduce{$a*$b/($a+$b)}@F

Try it online!

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2
3
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Perl 6, 14 bytes

1/*.sum o 1/**

Try it online!

1 / ** is an anonymous function that returns a list of the reciprocals of its arguments. 1 / *.sum is another anonymous function that returns the reciprocal of the sum of the elements of its list argument. The o operator composes those two functions.

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2
  • \$\begingroup\$ Very nice. I don't see HyperWhatevers used often enough in golfing since they can't be used in more complex expressions. If they were closer to normal Whatevers, I'd expect sumething like this to work, but alas... \$\endgroup\$
    – Jo King
    Sep 12 '19 at 0:17
  • \$\begingroup\$ Yeah, this is probably the first time I've even thought about using one for golfing, and I was disappointed to discover its limitations. \$\endgroup\$
    – Sean
    Sep 12 '19 at 0:28
3
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bash + coreutils, 25 bytes

bc -l<<<"1/(0${@/#/+1/})"

TIO

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3
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Wolfram Language (Mathematica), 10 bytes

1/Tr[1/#]&

Try it online!

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2
  • \$\begingroup\$ Is there no harmonic mean builtin, or is it longer to type? \$\endgroup\$ Sep 12 '19 at 11:53
  • \$\begingroup\$ @Eric AFAIK it's intuitively named HarmonicMean and is longer. \$\endgroup\$ Sep 12 '19 at 15:23
3
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MathGolf, 3 bytes

∩Σ∩

The same as other answers, using the builtins (\$\frac{1}{n}\$) and Σ (sum): $$M(x_1,...,x_n)=\frac{1}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}}$$

Try it online.

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3
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Java 8, 24 bytes

a->1/a.map(d->1/d).sum()

I noticed there wasn't a Java answer yet, so figured I'd add one.

Try it online.

Explanation:

Uses the same Harmonic Mean approach as other answers:

$$M(x_1,...,x_n)=\frac{1}{\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}}$$

a->                       // Method with DoubleStream parameter and double return-type
     a.map(d->1/d)        //  Calculate 1/d for each value `d` in the input-stream
                  .sum()  //  Then take the sum of the mapped list
   1/                     //  And return 1/sum as result
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2
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PHP, 51 bytes

Reciprocal of sum of reciprocals. Input is $a.

1/array_reduce($a,function($c,$i){return$c+1/$i;});

Try it online!

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3
  • \$\begingroup\$ With PHP7.4, I think you can do this: 1/array_reduce($a,fn($c,$i)=>$c+1/$i); (38 bytes). Read more in wiki.php.net/rfc/arrow_functions \$\endgroup\$ Sep 13 '19 at 16:41
  • \$\begingroup\$ I think you're right! But nowhere to demo? \$\endgroup\$
    – user15259
    Sep 13 '19 at 18:28
  • \$\begingroup\$ You have to download it yourself. However, since PHP 7.4.0RC1 was released on the 5th of this month (php.net/archive/2019.php#2019-09-05-1), you probably are safe using it. If you have doubts, you can ask in the meta. \$\endgroup\$ Sep 13 '19 at 21:41
2
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JavaScript (ES6), 29 bytes

a=>a.reduce((p,c)=>p*c/(p+c))

Try it online!

or:

a=>1/a.reduce((p,c)=>p+1/c,0)

Try it online!

But with this approach, using map() (as Shaggy did) is 1 byte shorter.

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2
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Python 3, 30 bytes

lambda r:1/sum(1/v for v in r)

Try it online!

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2
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Perl 5 (-p), 17 bytes

$a+=1/$_}{$_=1/$a

Try it online!

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2
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x86-64 Machine code - 20 18 bytes

0F 57 C0             xorps       xmm0,xmm0  
loopHead
F3 0F 53 4C 8A FC    rcpss       xmm1,dword ptr [rdx+rcx*4-4]
0F 58 C1             addps       xmm0,xmm1  
E2 F6                loop        loopHead
0F 53 C0             rcpps       xmm0,xmm0  
C3                   ret  

Input - Windows calling convention. First parameter is the number of resistors in RCX. A pointer to the resistors is in RDX. *ps instructions are used since they are one byte smaller. Technically, you can only have around 2^61 resistors but you will be out of RAM long before then. The precision isn't great either, since we are using rcpps.

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3
  • \$\begingroup\$ “Only 2⁶¹ resistors” would probably fill the observable universe (many times over)! \$\endgroup\$
    – user15259
    Sep 12 '19 at 13:39
  • \$\begingroup\$ Actually, 2^61 is only 2.305843e+18 and the observable universe is 8.8 × 10^26 m in diameter. \$\endgroup\$
    – me'
    Sep 14 '19 at 9:22
  • \$\begingroup\$ Yeah, serious overestimation! Actual magnitude would be around the size and mass of Deimos, smaller moon of Mars. \$\endgroup\$
    – user15259
    Sep 14 '19 at 14:48
2
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MATL, 5 bytes

,1w/s

Try it online!

I'm not sure if "do twice" (,) counts as a loop, but this is just the harmonic mean, divided by n.

Alternately, ,-1^s is five bytes as well.

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2
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Intel 8087 FPU machine code, 19 bytes

 D9 E8      FLD1                    ; push 1 for top numerator on stack
 D9 EE      FLDZ                    ; push 0 for running sum 
        R_LOOP: 
 D9 E8      FLD1                    ; push 1 numerator for resistor
 DF 04      FILD WORD PTR[SI]       ; push resistor value onto stack 
 DE F9      FDIV                    ; divide 1 / value 
 DE C1      FADD                    ; add to running sum 
 AD         LODSW                   ; increment SI by 2 bytes 
 E2 F4      LOOP R_LOOP             ; keep looping 
 DE F9      FDIV                    ; divide 1 / result                  
 D9 1D      FSTP WORD PTR[DI]       ; store result as float in [DI]

This uses the stack-based floating point instructions in the original IBM PC's 8087 FPU.

Input is pointer to resistor values in [SI], number of resistors in CX. Output is to a single precision (DD) value at [DI].

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2
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Rattle, 24 bytes

|>II^[=1+#g`e_1P+~s]`e_1

Try it Online!

This answer can probably be golfed more (can anyone outgolf me in my own language?)

Explanation

|                 take input, parse automatically (Rattle's input parsing is powerful 
                   enough to recognise "1,2,3" etc. as a list)
 >                move pointer to the right (to slot 1)
  I               flatten input list and insert into consecutive memory slots
   I^             get length of this list (which is still on top of the stack)
     [ ... ]`     loop structure: loop as many times as the length of the list

=1                set top of stack to 1
  +#              increment top of stack by list's iterator 
                   (the number of times the loop has run)
    g`            get the value in storage at the calculated index (iterator + 1)
      e_1         get 1/x of this value
         P        set the pointer to slot 0
          +~      increment the current value by the stored value in slot 0
            s     save the result to slot 0
e_1                get 1/x of this value
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1
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Dart, 42 bytes

f(List<num>a)=>a.reduce((p,e)=>p*e/(p+e));

Try it online!

Having to explicitly specify the num type is kinda sucky, prevents type infering, because it would infer to (dynamic, dynamic) => dynamic which can't yield doubles for some reason

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1
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PHP, 40 bytes

for(;$n=$argv[++$i];$r+=1/$n);echo 1/$r;

Try it online!

Tests: Try it online!

Similar to Yimin Rong's solution but without built-ins and all program bytes are included in the bytes count.

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1
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Python 3, 58 44 bytes

f=lambda x,y=0,*i:f(x*y/(x+y),*i)if y else x

A recursive function. Requires arguments to be passed unpacked, like so:

i=[10, 10, 20]
f(*i)

or

f(10, 10, 20)

Explanation:

# lambda function with three arguments. *i will take any unpacked arguments past x and y,
# so a call like f(10, 20) is also valid and i will be an empty tuple
# since y has a default value, f(10) is also valid
f=lambda x,y=0,*i: \

# a if case else b
# determine parallel resistance of x and y and use it as variable x
# since i is passed unpacked, the first item in the remaining list will be y and
# the rest of the items will be stored in i
# in the case where there were no items in the list, y will have the default value of 0
f(x*y/(x+y),*i) \

# if y does not exist or is zero, return x
if y else x
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1
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Charcoal, 7 bytes

I∕¹Σ∕¹A

Try it online! Link is to verbose version of code. Works by calculating the current drawn by each resistor when 1V is applied, taking the total, and calculating the resistance that would draw that current when 1V is applied. Explanation:

      A Input array
    ∕¹  Reciprocal (vectorised)
   Σ    Sum
 ∕¹     Reciprocal
I       Cast to string for implicit print
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1
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J, 6 bytes

1%1#.%

Try it online!

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2
  • 2
    \$\begingroup\$ it's a pity "sum under reciprocal" is the same number of bytes: +/&.:% \$\endgroup\$
    – ngn
    Sep 13 '19 at 16:53
  • \$\begingroup\$ @ngn Yes, but your solution looks more idiomatic to J. \$\endgroup\$ Sep 13 '19 at 17:55
1
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[MATLAB], 15 bytes

One more byte than flawr excellent answer, but I had to use other functions so here goes:

@(x)1/sum(1./x)

It's rather explicit, it sums the inverse of the resistances, then invert the sum to output the equivalent parallel resistance.

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1
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Forth (gforth), 49 bytes

: f 0e 0 do dup i cells + @ s>f 1/f f+ loop 1/f ;

Try it online!

Input is a memory address and array length (used as an impromptu array, since Forth doesn't have a built-in array construct)

Uses the sum-of-inverse method as most other answers are

Code Explanation

: f           \ start a new word definition
  0e          \ stick an accumulator on the floating point stack
  0 do        \ start a loop from 0 to array-length -1
    dup       \ copy the array address
    i cells + \ get the address of the current array value
    @ s>f     \ get the value and convert it to a float
    1/f f+    \ invert and add to accumulator
  loop        \ end the loop definition
  1/f         \ invert the resulting sum
;             \ end the word definition
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