## Parallel resistance in electric circuits

21

1

### 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.

6

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.

– FryAmTheEggman – 2019-09-11T15:17:17.173

13

# 05AB1E, 5 3 bytes

zOz


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### Explanation

z                     # compute 1/x for each x in input
O                    # sum input
z                   # compute 1/sum


4Excepting built-ins, this is probably as low as we can go! – None – 2019-09-11T14:57:57.707

9

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


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3It looks beautiful. – Eric Duminil – 2019-09-12T11:52:48.753

Solution along the OP's recursive lines would be 22 chars: foldr1(\r s->r*s/(r+s)). – ceased to turn counterclockwis – 2019-09-12T15:53:33.343

9

# 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!

4This is horrible and beautiful simultaneously! – ceased to turn counterclockwis – 2019-09-12T15:45:56.537

1Thanks, these are the best compliments:) – flawr – 2019-09-12T16:21:28.797

7

# 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)


4I'm assuming İ is pronounced the same way i is pronounced in list. Is this a way of saying the challenge was easy? – Stewie Griffin – 2019-09-13T06:31:14.713

4

# 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... – Giuseppe – 2019-09-13T13:59:34.347

4

# 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

4

# APL (Dyalog Unicode), 4 bytes

÷1⊥÷


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1APL is the original golfing language! – None – 2019-09-12T13:35:45.333

@YiminRong It's not a golfing language... :P – Erik the Outgolfer – 2019-09-12T15:11:57.803

I know, but its byte count is on par with modern golfing languages! – None – 2019-09-12T15:33:26.163

-1 byte: ÷1⊥÷ Try it online!

@Adám Oh duh of course 1∘⊥ is the same as +/ for vectors... – Erik the Outgolfer – 2019-09-15T11:36:51.967

Heh, it is even on APLcart.

@Adám Hm... somewhat obscure entry... :P – Erik the Outgolfer – 2019-09-15T11:42:48.437

3

# R, 15 bytes

1/sum(1/scan())


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Follows the same Harmonic Mean principle seen in other answers.

3

# JavaScript, 28 bytes

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


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3

# Perl 6, 14 bytes

1/*.sum o 1/**


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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 – 2019-09-12T00:17:20.230

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 – 2019-09-12T00:28:42.520

3

$_=reduce{$a*$b/($a+$b)}@F  Try it online! 20 bytes – Nahuel Fouilleul – 2019-09-11T18:45:08.513 @NahuelFouilleul 17 bytes, no -M – Grimmy – 2019-09-12T00:26:11.127 3 # bash + coreutils, 25 bytes bc -l<<<"1/(0${@/#/+1/})"


TIO

3

# Wolfram Language (Mathematica), 10 bytes

1/Tr[1/#]&


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Is there no harmonic mean builtin, or is it longer to type? – Eric Duminil – 2019-09-12T11:53:30.593

@Eric AFAIK it's intuitively named HarmonicMean and is longer. – my pronoun is monicareinstate – 2019-09-12T15:23:42.497

3

# 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.

2

# JavaScript (ES6), 29 bytes

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


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or:

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


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

2

# MATL, 5 bytes

,1w/s


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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.

2

# 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 https://wiki.php.net/rfc/arrow_functions – Ismael Miguel – 2019-09-13T16:41:05.617 I think you're right! But nowhere to demo? – None – 2019-09-13T18:28:58.147 You have to download it yourself. However, since PHP 7.4.0RC1 was released on the 5th of this month (https://www.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. – Ismael Miguel – 2019-09-13T21:41:08.510 2 # Python 3, 30 bytes lambda r:1/sum(1/v for v in r)  Try it online! 2 # Perl 5 (-p), 17 bytes $a+=1/$_}{$_=1/$a  Try it online! 2 # 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)! – None – 2019-09-12T13:39:08.157 Actually, 2^61 is only 2.305843e+18 and the observable universe is 8.8 × 10^26 m in diameter. – me' – 2019-09-14T09:22:52.117 Yeah, serious overestimation! Actual magnitude would be around the size and mass of Deimos, smaller moon of Mars. – None – 2019-09-14T14:48:36.923 2 # 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  2 # 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]. 1 # 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  1 # 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 1 # 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. 1 # 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  1 # J, 6 bytes 1%1#.%  Try it online! 2it's a pity "sum under reciprocal" is the same number of bytes: +/&.:% – ngn – 2019-09-13T16:53:45.197 @ngn Yes, but your solution looks more idiomatic to J. – Galen Ivanov – 2019-09-13T17:55:31.300 1 # Ruby, 26 22 bytes -4 bytes thanks to @ValueInk. ->x{1/x.sum{|i|1.0/i}}  Try it online! @ValueInk: Doh! Thanks a lot – Eric Duminil – 2019-09-13T20:38:40.647 1 # [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. 1 # 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  1 ## C#, 16 bytes 1/l.Sum(n=>1f/n)  ## Try It 1 Since the Sum() function can also accept a selector, you can cut 9 bytes by ommiting the .Select() part, like so – pappbence96 – 2019-09-13T13:21:24.190 1Just realized, one more byte can be saved by writing 1f/n instead of 1.0/n as both methods will avoid integer division. – pappbence96 – 2019-09-13T13:29:47.583 nice catch mate! – koviroli – 2019-09-13T13:33:09.080 1 # expl3 (LaTeX3 programming layer), 65 bytes The following defines a function that prints the result to the terminal (unfortunately expl3 has very verbose function names): \def\1#1{\fp_show:n{1/(\clist_map_function:nN{#1}\2)}}\def\2{+1/}  A complete script which can be run from the terminal including all the test cases as well as the setup to enter expl3: \RequirePackage{expl3}\ExplSyntaxOn \def\1#1{\fp_show:n{1/(\clist_map_function:nN{#1}\2)}}\def\2{+1/} \1{1, 1} \1{1, 1, 1} \1{4, 6, 3} \1{20, 14, 18, 8, 2, 12} \1{10, 10, 20, 30, 40, 50, 60, 70, 80, 90} \stop  If run with pdflatex <filename> the following is the console output: This is pdfTeX, Version 3.14159265-2.6-1.40.20 (TeX Live 2019) (preloaded format=pdflatex) restricted \write18 enabled. entering extended mode (./cg_resistance.tex LaTeX2e <2018-12-01> (/usr/local/texlive/2019/texmf-dist/tex/latex/unravel/unravel.sty (/usr/local/texlive/2019/texmf-dist/tex/latex/l3kernel/expl3.sty (/usr/local/texlive/2019/texmf-dist/tex/latex/l3kernel/expl3-code.tex) (/usr/local/texlive/2019/texmf-dist/tex/latex/l3backend/l3backend-pdfmode.def)) (/usr/local/texlive/2019/texmf-dist/tex/latex/l3packages/xparse/xparse.sty) (/usr/local/texlive/2019/texmf-dist/tex/generic/gtl/gtl.sty)) > 1/(\clist_map_function:nN {1,1}\2)=0.5. <recently read> } l.3 \1{1, 1} ? > 1/(\clist_map_function:nN {1,1,1}\2)=0.3333333333333333. <recently read> } l.4 \1{1, 1, 1} ? > 1/(\clist_map_function:nN {4,6,3}\2)=1.333333333333333. <recently read> } l.5 \1{4, 6, 3} ? > 1/(\clist_map_function:nN {20,14,18,8,2,12}\2)=1.129538323621694. <recently read> } l.6 \1{20, 14, 18, 8, 2, 12} ? > 1/(\clist_map_function:nN {10,10,20,30,40,50,60,70,80,90}\2)=2.611669603067675. <recently read> } l.7 \1{10, 10, 20, 30, 40, 50, 60, 70, 80, 90} ? ) No pages of output. Transcript written on cg_resistance.log.  ## Explanation \fp_show:n : evaluates its argument as a floating point expression and prints the result on the terminal, every expandable macro is expanded during that process. \clist_map_function:nN : takes two arguments, a comma separated list and a function/macro, if called like \clist_map_function:nN { l1, l2, l3 } \foo it expands to something like \foo{l1}\foo{l2}\foo{l3}. In our case instead of \foo the macro \2 is used, which expands to +1/ so that the expression expands to +1/{l1}+1/{l2}+1/{l3} 1 # SimpleTemplate, 72 bytes Okay, this was VERY hard! Specifically with a language that has VERY poor math support. {@fnx A,S}{@eachA}{@set/_ 1 _}{@set+S S,_}{@/}{@set/S 1 S}{@returnS}{@/}  Creates a function x that must be called by passing an array as the first and only argument. Ungolfed: {@fn calc values, sum} {@each values as value} {@set/ value 1 value} {@set+ sum sum, value} {@/} {@set/ sum 1 sum} {@return sum} {@/}  Explanation: • {@fn calc values, sum} - Creates the function calc, which has 2 variables (values and sum) taking the values of the first arguments (defaults to null) • {@each values as value} - Loops over each value in values • {@set/ value 1 value} - Divides 1 by value, storing it into value • {@set+ sum sum, value} - Adds sum to value, storing it into sum • {@/} - Closes the {@each} (usually optional) • {@set/ sum 1 sum} - Divides 1 by sum, storing it into sum • {@return sum} - Returns sum (duh) • {@/} - Closes the {@fn} (usually optional, even if it shouldn't be for functions) To use the function, you can use this code: {@set values 10, 10, 20, 30, 40, 50, 60, 70, 80, 90} {@//array of values} {@call calc into result values} {@echo result} {@//2.6116696030677}  If you are curious, the ungolfed example compiles to this PHP code: // {@fn calc values, sum}$DATA['calc'] = function()use(&$FN, &$_){
$DATA = array( 'argv' => func_get_args(), 'argc' => func_num_args(), 'VERSION' => '0.62', 'EOL' => PHP_EOL, 'PARENT' => &$_
);
$_ = &$DATA;
$DATA["values"] = &$DATA["argv"][0];
$DATA["sum"] = &$DATA["argv"][1];

// {@each values as value}
// loop variables
$tmp_1_val = isset($DATA['values']) ? $tmp_1_val = &$DATA['values'] : null;
$tmp_1_keys = gettype($DATA['values']) == 'array'
? array_keys($DATA['values']) : array_keys(range(0, strlen($tmp_1_val = $tmp_1_val . '') - 1));$tmp_1_key_last = end($tmp_1_keys); // loop foreach($tmp_1_keys as $tmp_1_index =>$tmp_1_key){
$DATA['loop'] = array( 'index' =>$tmp_1_index,
'i' => $tmp_1_index, 'key' =>$tmp_1_key,
'k' => $tmp_1_key, 'value' =>$tmp_1_val[$tmp_1_key], 'v' =>$tmp_1_val[$tmp_1_key], 'first' =>$tmp_1_key === $tmp_1_keys[0], 'last' =>$tmp_1_key === $tmp_1_key_last, );$DATA['__'] = $tmp_1_key;$DATA['value'] = $tmp_1_val[$tmp_1_key];

// {@set/ value 1 value}
$DATA['value'] = array_map(function($value)use(&$DATA){return (1 /$value);}, $FN['array_flat']((isset($DATA['value'])?$DATA['value']:null))); // {@set+ sum sum, value}$DATA['sum'] = array_sum($FN['array_flat'](array((isset($DATA['sum'])?$DATA['sum']:null),(isset($DATA['value'])?$DATA['value']:null)))); // {@/} }; // {@set/ sum 1 sum}$DATA['sum'] = array_map(function($value)use(&$DATA){return (1 / $value);},$FN['array_flat']((isset($DATA['sum'])?$DATA['sum']:null)));

// {@return sum}
return (isset($DATA['sum'])?$DATA['sum']:null);

// {@/}
};

// {@set values 10, 10, 20, 30, 40, 50, 60, 70, 80, 90}
$DATA['values'] = array(10,10,20,30,40,50,60,70,80,90); // {@call calc into result values}$DATA['result'] = call_user_func_array(isset($DATA['calc']) && is_callable($DATA['calc'])? $DATA['calc']: (isset($FN["calc"])? $FN["calc"]: "calc"), array((isset($DATA['values'])?$DATA['values']:null))); // {@echo result} echo implode('',$FN['array_flat']((isset($DATA['result'])?$DATA['result']:null)));


0

# Pyth, 6 bytes

c1scL1


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Uses the modified formula $$\\frac{1}{R_P}=\frac{1}{R_1} + \frac{1}{R_2}\$$.

c1      # 1/
s     #   sum(                     )
cL1  #       map(lambda x: 1/x, Q)  # Q (=input) is implicit


0

# Japt v2.0a0, 7 bytes

1÷Ux!÷1


Try it

0

# OCaml, 50 bytes

fun l->1./.(List.fold_left(fun a e->a+.1./.e)0. l)


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0

# Stax, 5 bytes

{u+Fu


Run and debug it at staxlang.xyz!

{  F     For each:
u         Invert
u    Invert
Implicit print as fraction


0

# Scala, 15 bytes

1/_.map(1/).sum


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0

# Python, 30 bytes

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


-2, thanks @JoKing

Welcome to PPCG! Unfortunately, submissions should either be a full program (preface with v=input();print) or a function (preface with lambda v:) to be valid. A snippet which presumes pre-defined variables isn't allowed. – Value Ink – 2019-09-16T18:41:56.210

Welcome to the site! This is a snippet, which means that you don't output or input according to our Input/Output rules. You can correct this by wrapping the statement in print(...) and by taking v as input, either in a lambda or by using input.

– caird coinheringaahing – 2019-09-16T18:42:48.173

That's unfortunate, thanks for pointing me to the IO conventions. I have updated my answer. :) – pikaynu – 2019-09-17T09:54:17.720

0

# jq, 16 characters

1/(map(1/.)|add)


Sample run:

bash-5.0\$ jq '1/(map(1/.)|add)' <<< '[10, 10, 20, 30, 40, 50, 60, 70, 80, 90]'
2.611669603067675


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0

# C (gcc), 65 62 bytes

Thanks to ceilingcat for -3 bytes.

float a(b,c)int*b;{float d=0;for(;c--;)d+=1./b[c];return 1/d;}


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0

# C++ (clang), 78 bytes

#import<list>
void f(std::list<int>r,double&s){s=0;for(int p:r)s+=1./p;s=1/s;}


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77 bytes – ceilingcat – 2019-12-04T21:13:00.203