Back to the chemistry class

Question

Introduction

One of the first things I learned in chemistry, was giving the formula for the combustion of alkanes. The most basic example was: 2CH4 + 4O2 -> 4H2O + 2CO2. Note that the equation could have been simplified by dividing every coefficient by 2, we would get CH4 + 2O2 -> 2H2O + CO2. To make things easier, we are going to ignore this to not complicate things.

You can see that with the combustion of any alkane, oxygen is used. After the reaction, only CO2 and H2O are made.

The task

The input will always be in the form C(n)H(2n+2), with n>0. These are some example inputs:

CH4
C2H6
C3H8
C4H10
C5H12
C6H14

Useful tip:

Every alkane has a standard pattern in the combustion reaction:

2C(n)H(2n+2) + (3n+1)O2 -> (2n+2)H2O + (2n)CO2

e.g.

C4H10 gives the following equation: 2C(4)H(2*4+2) + (3*4+1)O2 -> (2*4+2)H2O + (2*4)CO2. After calculating everything, we get this final equation 2C4H10 + 13O2 -> 10H2O + 8CO2

Testcases

CH4
2CH4 + 4O2 -> 4H2O + 2CO2

C3H8
2C3H8 + 10O2 -> 8H2O + 6CO2

C4H10
2C4H10 + 13O2 -> 10H2O + 8CO2

C12H26
2C12H26 + 37O2 -> 26H2O + 24CO2

Rules

• Note that I used spaces for the examples to increase readability, these are not necessary. 2CH4 + 4O2 -> 4H2O + 2CO2 and 2CH4+4O2->4H2O+2CO2 are both valid outputs. However, if you consistently use spaces for the output, you get a -10% bonus
• This is code-golf, so the program with the least amount of bytes wins!

Answer 1

Python 3, 86 * 0.9 = 77.4 bytes

s=input()
N=int(s.split('H')[1])
print("2%s + %dO2 > %dH20 + %dCO2"%(s,N*1.5-2,N,N-2))

Try it online!

Extracts the number of H's rather than the number of C's from the input. This avoid special-casing CH4 and simplifies the output expressions in terms of N=2n+2.

The output has parameters plugged in via string formatting. The first summand is just the input string, and the rest have calculated numbers plugged in. Note that N*1.5-2 (same as N*3/2-2) gives a float, but the string formatting converts it to an int.

Answer 2

Java, 202 * 0.9 = 181.8 bytes

Sometimes, I wonder if I'm just hurting myself with Java.

Thanks to @TNT and @TFeld for shaving off a good 20 bytes!

class A{public static void main(String[]a){String s=a[0].substring(1,a[0].indexOf("H"));long n=Long.parseLong((s.length()>0)?s:"1");System.out.printf("2%s + %dO2 > %dH2O + %dCO2",a[0],3*n+1,2*n+2,2*n);}}

Pretty simple. Basically, I cut the input from C to H, and get that substring. If it's nothing, I set n to one. Otherwise, I set it to the number between C and H. The code following just prints it out and puts it into proper notation.

Ungolfed

class A{
    public static void main(String[]a) {
         String s=a[0].substring(1,a[0].indexOf("H"));
         long n=Long.parseLong((s.length()>0)?s:"1");
         System.out.printf("2%s + %dO2 > %dH2O + %dCO2",a[0],3*n+1,2*n+2,2*n);
    }
}

Answer 3

Python 2, 122 91 * 0.9 = 81.9 bytes

i=input()
n=2*int(i[1:i.find('H')]or 1)
print'2%s + %dO2 > %dH2O + %dCO2'%(i,n*3/2+1,n+2,n)

Try it online!

Answer 4

JavaScript ES6, 63 * 0.9 = 56.7 bytes

_=>`2${_} + ${$=_.split`H`[1],$*1.5-2}O2 > ${$}H2O + ${$-2}CO2`

Similar to my ESMin answer

Answer 5

Pyth, 53*0.9 = 48 bytes

Previous byte count:

• 69 * 0.9 = 62

• 63 * 0.9 = 57

• 56 * 0.9 = 50

Js>z+xz\H1s[\2z%" + %dO2 > %dH2O + %dCO2"[-*1.5J2J-J2

It is just like xnor's answer, where I use the H values. A variable J is used to store the amount of H's in the molecular formula of the molecule.

Try it here

Answer 6

Javascript ES6, 96*0.9 = 86.4

f=s=>`2C${(n=(s.match(/\d+(?!.*\d)/)[0]-2))>2?n/2:''}H${n+2} + ${1.5*n+1}O2 > ${n+2}H2O + ${n}CO2`

Answer 7

CJam, 45 44 bytes

2r_'H/1=[i_((_2/1$+)@@]"+%dO2>%dH2O+%dCO2"e%

Try it online

To avoid special casing CH4, I don't use the first number at all, and only extract the number after the H. This gives the value of 2n+2. The remaining values are then calculated based on that, and formatted.

Explanation:

2       Push leading 2 for output.
r       Get input.
_'H/    Copy and split at 'H.
1=      Take second part after split, this is 2n+2.
[       Start list.
  i       Convert string value to integer.
  _((     Copy and decrement twice, giving 2n.
  _2/     Copy and divide by 2, giving n.
  1$      Copy the 2n to the top.
  +       Add, to get 3n.
  )       Increment, giving 3n+1. We now have 2n+2, 2n, 3n+1 on stack.
  @@      Rotate top 3 entries twice, to get 3n+1, 2n+2, 2n.
]       Close list.
"+%dO2>%dH2O+%dCO2"
        Format string for output.
e%      "printf" operator.

Answer 8

Perl, (84 + 1) * 0.9 = 76.5

(+1 char for running with the -n flag)

My first Perl golf!

@x=(1,m/C(\d)/g);$n=$x[$#x];say"2$_ + ".(3*$n+1)."O2 > ".(($n*=2)+2)."H20 + ${n}CO2"

It's important that STDIN not contain a trailing newline. Example usage:

llama@llama:...code/perl/ppcg64412chemistry$ printf CH4 | perl -n chemistry.pl
2CH4 + 4O2 > 4H20 + 2CO2

Ungolfed-ish:

#!/usr/bin/perl
use 5.10.0;

$_ = 'C3H8';
my @x = (1, m/C(\d)/g);
my $n = $x[$#x];
say "2$_ + ".(3*$n+1)."O2 > ".(($n*=2)+2)."H20 + ${n}CO2";

The lines

my @x = (1, m/C(\d)/g);
my $n = $x[$#x];

are fairly interesting here. $#x represents the "last populated index" of @x (and SE's syntax highlighting thinks it's a comment because it's dumb), so $x[$#x] will select the captured part of the C(\d) regex if it exists, or 1 otherwise. (Perl doesn't care that it'll be a string in all other cases except 1; you can use numerical operators on strings just fine in Perl.)

Answer 9

Python, 148 * 0.9 = 133 bytes

Prev byte counts

• 0.9*195 = 175
• 0.9*190 = 171

i=raw_input()
O=3*int(i[1:i.find('H')]or 1)+1;C=2*int(i[1:i.find('H')]or 1);print"2"+i+" + "+`O`+"O2"+" > "+i[i.find('H')+1:]+"H2O"+ " + "+`C`+"CO2"

Try it online!

Answer 10

𝔼𝕊𝕄𝕚𝕟, 64 * 0.9 = 57.6 bytes

a=ïČ`H”[1],`2⦃ï} + ⦃a*1.5-2}O2 > ⦃a}H2O + ⦃a-2}CO2`

Try it here (Firefox only)

Explanation

a=ïČ`H”[1],      // split input along H to get 2n+2
`  2⦃ï}          // 2C(n)H(2n+2) – this is the same as 2[input alkane's formula]
   + ⦃a*1.5-2}O2 // + 2(3n+1)O2
   > ⦃a}H2O      // > (2n+2)H2O
   + ⦃a-2}CO2    // + (2n)O2
`                // implicitly output the template string above

Answer 11

Pyth, 40 bytes

s[2z\+h*3Jv|hctz\H\1"O2>"yhJ"H2O+"yJ"CO2

Try it online. Test suite.

Answer 12

JS, 118 (106) bytes

x=prompt();y=x.replace("CH","C1H").match(/\d{1,}/g)[0];alert(2+x+" + "+(3*y+1)+"O2 > "+(2*y+2)+"H2O + "+(2*y)+"CO2");

Answer 13

MATLAB, 96 * 0.9 = 86.4 bytes

s=input('');i=[sscanf(s,'C%dH'),1];n=i(1);fprintf('%s + %dO2 > %dH2O + %dCO2',s,3*n+1,2*n+2,2*n)

Fairly self explanatory. Takes an input string (need to wrap it in quote marks so MATLAB doesn't try to execute it!). Then converts the number between the C and the H in the input to decimal. The value 1 is appended as an array to the output of sscanf so that in the case CH4 when we extract the first index in the array into n we get a 1 if there was no number between the C and H. Then just prints out with the spaces following the formula in the question.

This should also work with Octave using the online interpreter here.

Answer 14

Clojure/ClojureScript, 98*0.9 = 88.2 bytes

#(let[n(max 1(int((re-find #"C(.*)H"%)1)))m(* 2 n)](str "2"%" + "(inc(* 3 n))"O2 > "(+ 2 m)"H2O + "m"CO2"))

Creates an anonymous function. Try it by going here and entering (def f #(let...)), then (f "C3H8").

Answer 15

Excel, 123 * 0.9 = 110.7 bytes

=="2"&A1&" + "&(3*MID(A1,FIND("H",A1)+1,9)/2-2)&"O2 > "&MID(A1,FIND("H",A1)+1,9)&"H2O + "&(MID(A1,FIND("H",A1)+1,9)-2)&"CO2"

If we could take CH4 input as C1H4, can be reduced to 122 * 0.9 = 109.8 bytes

="2"&A1&" + "&3*MID(A1,2,FIND("H",A1)-2)+1&"O2"&" > "&MID(A1,FIND("H",A1)+1,9)&"H2O + "&(MID(A1,FIND("H",A1)+1,9)-2)&"CO2"

Answer 16

Jelly, 33 bytes

ṣ”HṪVµ2³”+2⁸_©+H“O2>”⁸“H2O+”®“CO2

Try it online!

The bonus doesn't help here.

Answer 17

C++ (clang), 160 bytes

#include<iostream>
int main(){int n=1,t;std::cin.get();std::cin>>n;t=2*n;printf("2C");n-1&&printf("%i",n);printf("H%i + %iO2 > %iH2O + %iCO2",t+2,t+n+1,t+2,t);}

Try it online!

A bit more that I expected. Reads first char and discards it, then reads int and outputs the result. Problem is with n being 1. I can't think of shorter way to output it.

Ungolfed

#include <iostream>
int main()
{
    int n = 1, t;
    std::cin.get();
    std::cin >> n;
    t = 2 * n;
    printf("2C");
    n - 1 && printf("%i", n);
    printf("H%i + %iO2 > %iH2O + %iCO2", t + 2, t + n + 1, t + 2, t);
}

Answer 18

F# (.NET Core), 113 bytes

let s=stdin.ReadLine()
float s.[1+s.IndexOf 'H'..]|>fun f->printf"2%s + %gO2 > %gH2O + %gCO2"s<|f*1.5-2.<|f<|f-2.

Try it online!