# Parallel resistance in electric circuits

### 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.

• 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. Sep 11, 2019 at 15:17

# 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

• Excepting built-ins, this is probably as low as we can go!
– user15259
Sep 11, 2019 at 14:57

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


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• It looks beautiful. Sep 12, 2019 at 11:52
• Solution along the OP's recursive lines would be 22 chars: foldr1(\r s->r*s/(r+s)). Sep 12, 2019 at 15:53

# 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!

• This is horrible and beautiful simultaneously! Sep 12, 2019 at 15:45
• Thanks, these are the best compliments:) Sep 12, 2019 at 16:21

# Jelly,  5  3 bytes

İSİ


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### 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)

• I'm assuming İ is pronounced the same way i is pronounced in list. Is this a way of saying the challenge was easy? Sep 13, 2019 at 6:31

# Octave, 15 bytes

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


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Harmonic mean, divided by n. Easy peasy.

• @tsh you know, I don't think I ever noticed that. I guess it's almost the harmonic mean... Sep 13, 2019 at 13:59

# PowerShell, 22 bytes

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


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

# APL (Dyalog Unicode), 4 bytes

÷1⊥÷


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• APL is the original golfing language!
– user15259
Sep 12, 2019 at 13:35
• @YiminRong It's not a golfing language... :P Sep 12, 2019 at 15:11
• I know, but its byte count is on par with modern golfing languages!
– user15259
Sep 12, 2019 at 15:33
• -1 byte: ÷1⊥÷ Try it online!
Sep 15, 2019 at 9:27
• @Adám Oh duh of course 1∘⊥ is the same as +/ for vectors... Sep 15, 2019 at 11:36

# R, 15 bytes

1/sum(1/scan())


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Follows the same Harmonic Mean principle seen in other answers.

# JavaScript, 28 bytes

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


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$_=reduce{$a*$b/($a+$b)}@F  Try it online! • 20 bytes Sep 11, 2019 at 18:45 • @NahuelFouilleul 17 bytes, no -M Sep 12, 2019 at 0:26 # Perl 6, 14 bytes 1/*.sum o 1/**  Try it online! 1 / ** is an anonymous function that returns a list of the reciprocals of its arguments. 1 / *.sum is another anonymous function that returns the reciprocal of the sum of the elements of its list argument. The o operator composes those two functions. • Very nice. I don't see HyperWhatevers used often enough in golfing since they can't be used in more complex expressions. If they were closer to normal Whatevers, I'd expect sumething like this to work, but alas... – Jo King Sep 12, 2019 at 0:17 • Yeah, this is probably the first time I've even thought about using one for golfing, and I was disappointed to discover its limitations. – Sean Sep 12, 2019 at 0:28 # bash + coreutils, 25 bytes bc -l<<<"1/(0${@/#/+1/})"


TIO

# Wolfram Language (Mathematica), 10 bytes

1/Tr[1/#]&


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• Is there no harmonic mean builtin, or is it longer to type? Sep 12, 2019 at 11:53
• @Eric AFAIK it's intuitively named HarmonicMean and is longer. Sep 12, 2019 at 15:23

# 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.

# 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


# PHP, 51 bytes

Reciprocal of sum of reciprocals. Input is $a. 1/array_reduce($a,function($c,$i){return$c+1/$i;});


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• 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 Sep 13, 2019 at 16:41 • I think you're right! But nowhere to demo? – user15259 Sep 13, 2019 at 18:28 • 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. Sep 13, 2019 at 21:41 # 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. # Python 3, 30 bytes lambda r:1/sum(1/v for v in r)  Try it online! # Perl 5 (-p), 17 bytes $a+=1/$_}{$_=1/$a  Try it online! # 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. • “Only 2⁶¹ resistors” would probably fill the observable universe (many times over)! – user15259 Sep 12, 2019 at 13:39 • Actually, 2^61 is only 2.305843e+18 and the observable universe is 8.8 × 10^26 m in diameter. – me' Sep 14, 2019 at 9:22 • Yeah, serious overestimation! Actual magnitude would be around the size and mass of Deimos, smaller moon of Mars. – user15259 Sep 14, 2019 at 14:48 # 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. # 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]. # Rattle, 24 bytes |>II^[=1+#ge_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  # 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 # 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.

# 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


# 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


# J, 6 bytes

1%1#.%


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• it's a pity "sum under reciprocal" is the same number of bytes: +/&.:%
– ngn
Sep 13, 2019 at 16:53
• @ngn Yes, but your solution looks more idiomatic to J. Sep 13, 2019 at 17:55

# [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.

# Forth (gforth), 49 bytes

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


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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