Bernd is a high school student who has some problems in chemistry. In class he has to design chemical equations for some experiments they are doing, such as the combustion of heptane C7H16:

C7H16 + 11O2 → 7CO2 + 8H2O

Since mathematics isn't exactly Bernd's strongest subject, he often has a hard time finding the exact ratios between the pro- and educts of the reaction. Since you are Bernd's tutor, it is your job to help him! Write a program, that calculates the amount of each substance needed to get a valid chemical equation.


The input is a chemical equation without amounts. In order to make this possible in pure ASCII, we write any subscriptions as ordinary numbers. Element names always start with a capital letter and may be followed by a minuscule. The molecules are separated with + signs, an ASCII-art arrow -> is inserted between both sides of the equation:


The input is terminated with a newline and won't contain any spaces. If the input is invalid, your program may do whatever you like.

You may assume, that the input is never longer than 1024 characters. Your program may either read the input from standard input, from the first argument or in an implementation defined way at runtime if neither is possible.


The output of your program is the input equation augmented with extra numbers. The number of atoms for each element must be the same on both sides of the arrow. For the example above, a valid output is:


If the number for a molecule is 1, drop it. A number must always be a positive integer. Your program must yield numbers such that their sum is minimal. For instance, the following is illegal:


If there is no solution, print


instead. A sample input that has no solution is







This is , so shortest code wins!

Your program must terminate in reasonable time for all reasonable inputs

  • 2
    \$\begingroup\$ I could be wrong, but this seems like a natural candidate for a programming challenge rather code golf. \$\endgroup\$
    – DavidC
    Commented Oct 16, 2012 at 2:06
  • 2
    \$\begingroup\$ I once wrote a chemical equation solver on my TI-89 graphing calculator, using the built-in solve( function and eval( to interpret the input :) \$\endgroup\$
    – mellamokb
    Commented Oct 16, 2012 at 12:48
  • 3
    \$\begingroup\$ @mellamokb why don't you post it, you'll get an upvote from me for originality \$\endgroup\$ Commented Oct 16, 2012 at 15:57
  • 7
    \$\begingroup\$ "Since you are Bernds tutor, it is your job to help him!" - I would have thought a tutor should be teaching Bernd to think for himself, rather than write software for him so he doesn't have to :P \$\endgroup\$
    – naught101
    Commented Jan 30, 2015 at 4:44
  • 1
    \$\begingroup\$ @KuilinLi It is not wrong, just different. \$\endgroup\$
    – FUZxxl
    Commented Dec 25, 2016 at 9:37

6 Answers 6


C (gcc), 442 bytes

b(k){if(k<0)for(n=j=0;!n&&j<u;j++)for(i=0;i<=c;i++)n+=t[i]*v[i][j];else for(t[k]=0;n&&t[k]++<30;)b(k-1);}
main(int r,char**a){for(s=m[0]=a[1];*s;){if(*s==45)r=0,s++;if(*s<65)m[++c]=++s;j=*s++;if(*s>96)j=*s+++j<<8;for(i=0,t[u]=j;t[i]-j;i++);u+=i==u;for(n=0;*s>>4==3;)n=n*10+*s++-48;n+=!n;v[c][i]=r?n:-n;}b(c);for(i=0,s=n?"Nope!":a[1];*s;putchar(*s++))s==m[i]&&t[i++]>1?printf("%d",t[i-1]):0;putchar(10);}

Try it online!

Ungolfed version

// element use table, then once parsed reused as molecule weights

// molecules
char*s,*m[99]; // name and following separator
c,v[99][99]; // count-1, element vector


// brute force solver, n==0 upon solution - assume at most 30 of each molecule
    if(k<0)for(n=j=0;!n&&j<u;j++)for(i=0;i<=c;i++)n+=t[i]*v[i][j]; // check if sums to zero
    else for(t[k]=0;n&&t[k]++<30;)b(k-1); // loop through all combos of weights

main(int r,char**a){
    // parse
        // parse separator, advance next molecule
        // parse element
        // lookup element index
        // parse amount
        // store element count in molecule vector, flip sign for other side of '->'
    // solve
    // output
  • \$\begingroup\$ +1 this is much more respectable than the pres. debate \$\endgroup\$
    – ardnew
    Commented Oct 17, 2012 at 2:36
  • 2
    \$\begingroup\$ Try using commas as statement separators to avoid curly braces. Also try replacing if-then-else-constructs with ternary operators to shorten the code. t[i]>1?printf("%s",t[i]):0; is one byte shorter. Also: m[0] is the same as *m. \$\endgroup\$
    – FUZxxl
    Commented Oct 17, 2012 at 9:32

Wolfram Language (Mathematica), 510 bytes

I employed the augmented chemical composition matrix approach described in

L.R.Thorne, An innovative approach to balancing chemical - reaction equations : a simplified matrix - inverse technique for determining the matrix null space. Chem.Educator, 2010, 15, 304 - 308.

One slight tweak was added: I divided the transpose of the null-space vector by the greatest common divisor of the elements to ensure integer values in any solutions. My implementation does not yet handle cases where there is more than one solution to balancing the equation.

b@t_ :=Quiet@Check[Module[{s = StringSplit[t, "+" | "->"], g = StringCases, k = Length, 
  e, v, f, z, r},
e = Union@Flatten[g[#, _?UpperCaseQ ~~ ___?LowerCaseQ] & /@ s];v = k@e;
s_~f~e_ := If[g[s, e] == {}, 0, If[(r = g[s, e ~~ p__?DigitQ :> p]) == {}, 1, 
   r /. {{x_} :> ToExpression@x}]];z = k@s - v;
r = #/(GCD @@ #) &[Inverse[Join[SparseArray[{{i_, j_} :> f[s[[j]], e[[i]]]}, k /@ {e, s}], 
Table[Join[ConstantArray[0, {z, v}][[i]], #[[i]]], {i, k[#]}]]][[All, -1]] &
Row@Flatten[ReplacePart[Riffle[Partition[Riffle[Abs@r, s], 2], " + "],2 Count[r,_?Negative]->" -> "]]],"Nope!"]

Try it online!


It works by setting up the following chemical composition table, consisting of chemical species by elements, to which an addition nullity vector is added (becoming the augmented chemical composition table:

chemical composition table

The inner cells are removed as a matrix and inverted, yielding.


The right-most column is extracted, yielding:

{-(1/8), -(11/8), 7/8, 1}

Each element in the vector is divided by the gcd of the elements (1/8), giving:

{-1, -11, 7, 8}

where the negative values will be placed on the left side of the arrow. The absolute values of these are the numbers needed to balance the original equation:


  • \$\begingroup\$ don't forget to add the exclamation point! \$\endgroup\$
    – ardnew
    Commented Oct 18, 2012 at 17:45
  • \$\begingroup\$ :} ok, and I upped the character count \$\endgroup\$
    – DavidC
    Commented Oct 18, 2012 at 18:52
  • \$\begingroup\$ I think you mean the right-hand column, not the left-hand one. I appreciate the explanation (+1) but I do wonder: if it weren't the case that the number of molecules is one more than the number of elements, how do you pad? Off to read the paper now. \$\endgroup\$ Commented Oct 19, 2012 at 20:51
  • \$\begingroup\$ For some reason, I only came across your comment today. Yes, I did mean "right-hand column". So much time has passed since I worked on this that I cannot see (or remember) where padding is used. Sorry. \$\endgroup\$
    – DavidC
    Commented Jun 26, 2015 at 16:00

Python 2, 879 bytes

import sys,re
from sympy.solvers import solve
from sympy import Symbol
from fractions import gcd
from collections import defaultdict

for p in eq.split('->'):
 for k in p.split('+'):
  c = [Ls.pop(0), 1]
  for e,m in re.findall('([A-Z][a-z]?)([0-9]*)',k):
   m=1 if m=='' else int(m)
Ys=dict((s,eval('Symbol("'+s+'")')) for s in Os if s not in Ls)
Qs=[eval('+'.join('%d*%s'%(c[1],c[0]) for c in Ss[s]),{},Ys) for s in Ss]+[Ys['a']-a]
if k:
 N=[k[Ys[s]] for s in sorted(Ys)]
 for a1, a2 in zip(N[0::2],N[1::2]):g=gcd(g,a2)
 N=[i/g for i in N]
 pM=lambda c: str(c) if c!=1 else ''
 print '->'.join('+'.join(pM(N.pop(0))+str(t) for t in p.split('+')) for p in eq.split('->'))
else:print 'Nope!'

Try it online!

Could be much less than 880, but my eyes are killing me already...


Python 2, 640 bytes

previous byte counts: 794, 776, 774, 765, 759, 747, 735, 734, 720, 683, 658, 655, 654, 653, 651, 640

The second indentation level is only a tab, the third is a tab then a space.

To be honest, this is jadkik94's answer, but so many bytes were shaved, I had to do it. Tell me if I can shave any bytes off!

from sympy import*
import sys,re
from sympy.solvers import*
from collections import*
for p in P(q,v):
 for k in P(p,u):
  for e,m in re.findall('([A-Z][a-z]*)(\d*)',k):
   m=int(m or 1)
Y=dict((s,Symbol(s))for s in set(O)-set(L))
Q=[eval(w('%d*%s'%(c[1],c[0])for c in S[s]),{},Y)for s in S]+[Y['a']-a]
if k:
 N=[k[Y[s]]for s in sorted(Y)]
 print v.join(w((lambda c:str(c)*(c!=1))(N.pop(0)/g)+str(t)for t in P(p,u))for p in P(q,v))

Try it online!

  • \$\begingroup\$ save three bytes: ''.join(map(chr,range(97,122))) :D \$\endgroup\$
    – aliqandil
    Commented Jan 3, 2017 at 14:53
  • \$\begingroup\$ :(, that doesn't work. However, map(chr,range(97,123)) works for 12 bytes saved. \$\endgroup\$
    – Adalynn
    Commented Jan 3, 2017 at 15:11
  • \$\begingroup\$ oh right! it's python 2! \$\endgroup\$
    – aliqandil
    Commented Jan 11, 2017 at 23:25

JavaScript, 682 bytes

x=>{m=1;x.split(/\D+/g).map(i=>i?m*=i:0);e=new Set(x.replace(/\d+|\+|->/g,"").match(/([A-Z][a-z]*)/g));e.delete``;A=[];for(let z of e){t=x.split`->`;u=[];for(c=1;Q=t.shift();c=-1)Q.split`+`.map(p=>u.push(c*((i=p.indexOf(z))==-1?0:(N=p.substring(i+z.length).match(/^\d+/g))?N[0]:1)));A.push(u)}J=A.length;for(P=0;P<J;P++){for(i=P;!A[i][P];i++);W=A.splice(i,1)[0];W=W.map(t=>t*m/W[P]);A=A.map(r=>r[P]?r.map((t,j)=>t-W[j]*r[P]/m):r);A.splice(P,0,W)}f=e.size;if(!A[0][f])return"Nope!";g=m=-m;_=(a,b)=>b?_(b,a%b):a;c=[];A.map(p=>c.push(t=p.pop())&(g=_(g,t)));c.push(m);j=x.match(/[^+>]+/g);return c.map(k=>k/g).map(t=>(t^1?t:"")+(z=j.shift())+(z.endsWith`-`?">":"+")).join``.slice(0,-1);}

This is a much more golfed (decades of characters!) of Kuilin's answer. Might be noncompeting because certain JS features postdate the challenge.


Javascript, 705 bytes

(non-competing, some features postdate the challenge)

Other solutions all had elements of brute-forcing. I tried for a more deterministic approach by representing the chemical equation as a set of linear equations, and then solving using the Gauss-Jordan algorithm to take the reduced row-echelon form of that matrix. In order to isolate out the trivial case where everything is zero, I assume that one of the elements is a constant number - and that number is determined by just all the numbers multiplied together, in order to not have fractions. Then as a final step we'll divide each by the gcd to satisfy the last condition.


function solve(x) {
	//firstly we find bigNumber, which will be all numbers multiplied together, in order to assume the last element is a constant amount of that
	bigNumber = 1;
	arrayOfNumbers = new Set(x.split(/\D+/g));
	for (let i of arrayOfNumbers) bigNumber *= parseInt(i);
	//first actual step, we split into left hand side and right hand side, and then into separate molecules
	//number of molecules is number of variables, number of elements is number of equations, variables refer to the coefficients of the chemical equation
	//note, the structure of this is changed a lot in the golfed version since right is the same as negative left
	left = x.split("->")[0].split("+");
	righ = x.split("->")[1].split("+");
	molecules = left.length + righ.length;
	//then let's find what elements there are - this will also become how many equations we have, or the columns of our matrix minus one
	//we replace all the non-element characters, and then split based on the uppercase characters
	//this also sometimes adds a "" to the array, we don't need that so we just delete it
	//turn into a set in order to remove repeats
	elems = new Set(x.replace(/\d+|\+|->/g,"").match(/([A-Z][a-z]*)/g));
	rrefArray = [];//first index is rows, second index columns - each row is an equation x*(A11)+y*(A21)+z*(A31)=A41 etc etc, to solve for xyz as coefficients
	//loop thru the elements, since for each element we'll have an equation, or a row in the array
	for (let elem of elems) {
		buildArr = [];
		//loop thru the sides
		for (let molecule of left) {
			//let's see how many of element elem are in molecule molecule
			//ASSUMPTION: each element happens only once per molecule (no shenanigans like CH3COOH)
			index = molecule.indexOf(elem);
			if (index == -1) buildArr.push(0);
			else {
				index += elem.length;
				numberAfterElement = molecule.substring(index).match(/^\d+/g);
				if (numberAfterElement == null) buildArr.push(1);
				else buildArr.push(parseInt(numberAfterElement));
		//same for right, except each item is negative
		for (let molecule of righ) {
			index = molecule.indexOf(elem);
			if (index == -1) buildArr.push(0);
			else {
				index += elem.length;
				numberAfterElement = molecule.substring(index).match(/^\d+/g);
				if (numberAfterElement == null) buildArr.push(-1);
				else buildArr.push(parseInt(numberAfterElement)*(-1));
	//Gauss-Jordan algorithm starts here, on rrefArray
	for (pivot=0;pivot<Math.min(molecules, elems.size);pivot++) {
		//for each pivot element, first we search for a row in which the pivot is nonzero
		//this is guaranteed to exist because there are no empty molecules
		for (i=pivot;i<rrefArray.length;i++) {
			row = rrefArray[i];
			if (row[pivot] != 0) {
				workingOnThisRow = rrefArray.splice(rrefArray.indexOf(row), 1)[0];
		//then multiply elements so the pivot element of workingOnThisRow is equal to bigNumber we determined above, this is all to keep everything in integer-space
		multiplyWhat = bigNumber / workingOnThisRow[pivot]
		for (i=0;i<workingOnThisRow.length;i++) workingOnThisRow[i] *= multiplyWhat
		//then we make sure the other rows don't have this column as a number, the other rows have to be zero, if not we can normalize to bigNumber and subtract
		for (let i in rrefArray) {
			row = rrefArray[i];
			if (row[pivot] != 0) {
				multiplyWhat = bigNumber / row[pivot]
				for (j=0;j<row.length;j++) {
					row[j] *= multiplyWhat;
					row[j] -= workingOnThisRow[j];
					row[j] /= multiplyWhat;
		//finally we put the row back
		rrefArray.splice(pivot, 0, workingOnThisRow);
	//and finally we're done!
	//sanity check to make sure it succeeded, if not then the matrix is insolvable
	if (rrefArray[0][elems.size] == 0 || rrefArray[0][elems.size] == undefined) return "Nope!";
	//last step - get the results of the rref, which will be the coefficients of em except for the last one, which would be bigNumber (1 with typical implementation of the algorithm)
	bigNumber *= -1;
	gcd_calc = function(a, b) {
		if (!b) return a;
		return gcd_calc(b, a%b);
	coEffs = [];
	gcd = bigNumber;
	for (i=0;i<rrefArray.length;i++) {
		num = rrefArray[i][molecules-1];
		gcd = gcd_calc(gcd, num)
	for (i=0;i<coEffs.length;i++) coEffs[i] /= gcd;
	//now we make it human readable
	//we have left and right from before, let's not forget those!
	out = "";
	for (i=0;i<coEffs.length;i++) {
		coEff = coEffs[i];
		if (coEff != 1) out += coEff;
		out += left.shift();
		if (left.length == 0 && righ.length != 0) {
			out += "->";
			left = righ;
		} else if (i != coEffs.length-1) out += "+";
	return out;


s=x=>{m=1;x.split(/\D+/g).map(i=>i!=""?m*=i:0);e=(new Set(x.replace(/\d+|\+|->/g,"").match(/([A-Z][a-z]*)/g)));e.delete("");A=[];for(let z of e){t=x.split("->");u=[];for(c=1;Q=t.shift();c=-1)Q.split("+").map(p=>u.push(c*((i=p.indexOf(z))==-1?0:(N=p.substring(i+z.length).match(/^\d+/g))?N[0]:1)));A.push(u)}J=A.length;for(P=0;P<J;P++){for(i=P;!A[i][P];i++);W=A.splice(i,1)[0];W=W.map(t=>t*m/W[P]);A=A.map(r=>!r[P]?r:r.map((t,j)=>t-W[j]*r[P]/m));A.splice(P,0,W)}f=e.size;if (!A[0][f])return "Nope!";g=m=-m;_=(a,b)=>b?_(b,a%b):a;c=[];A.map(p=>c.push(t=p.pop())&(g=_(g,t)));c.push(m);j=x.match(/[^+>]+/g);return c.map(k=>k/g).map(t=>(t==1?"":t)+(z=j.shift())+(z.endsWith("-")?">":"+")).join("").slice(0,-1);}


  • 1
    \$\begingroup\$ Noncompeting, as some features postdate the challenge. \$\endgroup\$
    – Adalynn
    Commented Dec 31, 2016 at 0:55
  • \$\begingroup\$ Oh wow I didn't notice how old this was. Thanks! \$\endgroup\$
    – kuilin
    Commented Jan 1, 2017 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.