Generating random math expression

Question

I have this idea running around in my head, to generate and evaluate random mathematical expressions. So, I decided to give it a shot and elaborate an algorithm, before coding it to test it.

Example:

Here are some example expressions I want to generate randomly:

4 + 2                           [easy]
3 * 6 - 7 + 2                   [medium]
6 * 2 + (5 - 3) * 3 - 8         [hard]
(3 + 4) + 7 * 2 - 1 - 9         [hard]
5 - 2 + 4 * (8 - (5 + 1)) + 9   [harder]
(8 - 1 + 3) * 6 - ((3 + 7) * 2) [harder]

The easy and medium ones are pretty straight-forward. Random ints separated by random operators, nothing crazy here. But I'm having some trouble getting started with something that could create one of the hard and harder examples. I'm not even sure a single algorithm could give me the last two.

What I am considering:

I can't say I tried those ideas, because I didn't really want to waste much time going in a direction that had no chance of working in the first place. But still, I thought of a couple solutions:

• Using trees
• Using regular expressions
• Using a crazy "for-type" loop (surely the worst)

What I'm looking for:

I'd like to know which way you believe is the best to go, between the solutions I considered, and your own ideas.

If you see a good way to start, I'd appreciate a lead in the right direction, e.g. with the beginning of the algorithm, or a general structure of it.

Also note that I will have to evaluate those expressions. This can be done either after the expression is generated, or during its creation. If you take that in consideration in your answer, that's great.

I'm not looking for anything language-related, but for the record, I'm thinking of implementing it in Objective-C, as that's the language I'm most working with recently.

Those examples did not include the : operator, as I only want to manipulate ints, and this operator adds many verifications. If your answer gives a solution handling this one, that's great.

If my question needs any clarification, please ask in the comments. Thanks for your help.

Answer 1

Here's a theoretic interpretation of your problem.

You are looking to randomly generate words (algebraic expression) from a given language (the infinite set of all syntactically correct algebraic expressions). Here's a formal description of a simplified algebraic grammar supporting only addition and multiplication:

E -> I 
E -> (E '+' E)
E -> (E '*' E)

Here, E is an expression (i.e., a word of your language) and I is a terminal symbol (i.e., it's not expanded any further) representing an integer. The above definition for E has three production rules. Based on this definition, we can randomly build a valid arithmetic as follows:

1. Start with E as the single symbol of the output word.
2. Choose uniformly at random one of the non-terminal symbols.
3. Choose uniformly at random one of the production rules for that symbol, and apply it.
4. Repeat steps 2 - 4 until only terminal symbols are left.
5. Replace all terminal symbols I by random integers.

Here's an example of the application of this algorithms:

E
(E + E)
(E + (E * E))
(E + (I * E))
((E + E) + (I * E))
((I + E) + (I * E))
((I + E) + (I * I))
((I + (E * E)) + (I * I))
((I + (E * I)) + (I * I))
((I + (I * I)) + (I * I))
((2 + (5 * 1)) + (7 * 4))

I assume you would choose to represent an expression with an interface Expression which is implemented by classes IntExpression, AddExpression and MultiplyExpression. The latter two then would have a leftExpression and rightExpression. All Expression subclasses are required to implement an evaluate method, which works recursively on the tree structure defined by these objects and effectively implements the composite pattern.

Note that for the above grammar and algorithm, the probability of expanding an expression E into a terminal symbol I is only p = 1/3, while the probability to expand an expression into two further expressions is 1-p = 2/3. Therefore, the expected number of integers in a formula produced by the above algorithm is actually infinite. The expected length of an expression is subject to the recurrence relation

l(0) = 1
l(n) = p * l(n-1) + (1-p) * (l(n-1) + 1)
     = l(n-1) + (1-p)

where l(n) denotes the expected length of the arithmetic expression after n applications of production rules. I therefore suggest that you assign a rather high probability p to the rule E -> I such that you end up with a fairly small expression with high probability.

EDIT: If you're worried that the above grammar produces too many parenthesis, look at Sebastian Negraszus' answer, whose grammar avoids this problem very elegantly.

Answer 2

first of all I'd actually generate the expression in postfix notation, you can easily convert to infix or evaluate after generating your random expression, but doing it in postfix means you don't need to worry about parenthesis or precedence.

I'd also keep a running total of the number of terms available to the next operator in your expression (assuming you want to avoid generating expressions that are malformed) i.e. somthing like this:

string postfixExpression =""
int termsCount = 0;
while(weWantMoreTerms)
{
    if (termsCount>= 2)
    {
         var next = RandomNumberOrOperator();
         postfixExpression.Append(next);
         if(IsNumber(next)) { termsCount++;}
         else { termsCount--;}
    }
    else
    {
       postfixExpression.Append(RandomNumber);
       termsCount++;
     }
}

obviously this is pseudo code so is not tested/may contain mistakes and you probably wouldn't use a string but a stack of some discriminated union like type

Answer 3

blubb's answer is a good start, but his formal grammar creates too many parantheses.

Here's my take on it:

E -> I
E -> M '*' M
E -> E '+' E
M -> I
M -> M '*' M
M -> '(' E '+' E ')'

E is an expression, I an integer and M is an expression that is an argument for a multiplication operation.

Answer 4

The parentheses in the "hard" expression represent order of evaluation. Rather than trying to generate the displayed form directly, just come up with a list of operators in random order, and derive the display form of the expression from that.

Numbers: 1 3 3 9 7 2

Operators: + * / + *

Result: ((1 + 3) * 3 / 9 + 7) * 2

Deriving the display form is a relatively simple recursive algorithm.

Update: here is an algorithm in Perl to generate the display form. Because + and * are distributive, it randomizes the order of the terms for those operators. That helps keep the parentheses from all building up on one side.

use warnings;
use strict;

sub build_expression
{
    my ($num,$op) = @_;

    #Start with the final term.
    my $last_num = pop @$num; 
    my $last_op = pop @$op;

    #Base case: return the number if there is just a number 
    return $last_num unless defined $last_op;

    #Recursively call for the expression minus the final term.
    my $rest = build_expression($num,$op); 

    #Add parentheses if there is a bare + or - and this term is * or /
    $rest = "($rest)" if ($rest =~ /[+-][^)]+$|^[^)]+[+-]/ and $last_op !~ /[+-]/);

    #Return the two components in a random order for + or *.
    return $last_op =~ m|[-/]| || rand(2) >= 1 ? 
        "$rest $last_op $last_num" : "$last_num $last_op $rest";        
}

my @numbers   = qw/1 3 4 3 9 7 2 1 10/;
my @operators = qw|+ + * / + * * +|;

print build_expression([@numbers],[@operators]) , "\n";

Answer 5

To expand on the tree approach, let's say each node is either a leaf or a binary expression:

Node := Leaf | Node Operator Node

Note that a leaf is just a randomly-generated integer here.

Now, we can randomly generate a tree. Deciding the probability of each node being a leaf allows us to control the expected depth, although you might want an absolute max depth as well:

Node random_tree(leaf_prob, max_depth)
    if (max_depth == 0 || random() > leaf_prob)
        return random_leaf()

    LHS = random_tree(leaf_prob, max_depth-1)
    RHS = random_tree(leaf_prob, max_depth-1)
    return Node(LHS, RHS, random_operator())

Then, the simplest rule for printing the tree is to wrap () around each non-leaf expression and avoid worrying about operator precedence.

---

For example, if I parenthesize your last sample expression:

(8 - 1 + 3) * 6 - ((3 + 7) * 2)
((((8 - 1) + 3) * 6) - ((3 + 7) * 2))

you can read off the tree that would generate it:

                    SUB
                  /      \
               MUL        MUL
             /     6     /   2
          ADD          ADD
         /   3        3   7
       SUB
      8   1

Answer 6

I would use trees. They can give you great control on the generation of the expressions. E.g. you can limit the depth per branch and width of each level separately. Tree based generation also gives the answer already during the generation, which is useful if you want to ensure that also the result (and subresults) are hard enough and/or not too hard to solve. Especially if you add division operator at some point, you can generate expressions that evaluate to whole numbers.

Answer 7

Here's a slightly different take on Blubb's excellent answer:

What you're trying to build here is essentially a parser that operates in reverse. What your problem and a parser have in common is a context-free grammar, this one in Backus-Naur form:

digit ::= '0' | '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9'
number ::= <digit> | <digit> <number>
op ::= '+' | '-' | '*' | '/'
expr ::= <number> <op> <number> | '(' <expr> ')' | '(' <expr> <op> <expr> ')'

Parsers start with a stream of terminals (literal tokens like 5 or *) and try to assemble them into nonterminals (things composed of terminals and other nonterminals, such as number or op). Your problem starts with nonterminals and works in reverse, picking anything between the "or" (pipe) symbols at random when one is encountered and recursively repeating the process until reaching a terminal.

A couple of the other answers have suggested that this is a tree problem, which it is for a certain narrow class of cases where there are no nonterminals that reference themselves directly or indirectly through another nonterminal. Since grammars allow that, this problem is really a directed graph. (Indirect references through another nonterminals count toward this as well.)

There was a program called Spew published on Usenet in the late 1980s which was originally designed to generate random tabloid headlines and also happens to be a great vehicle for experimenting with these "reverse grammars." It operates by reading a template that directs the production of a random stream of terminals. Beyond its amusement value (headlines, country songs, pronounceable English gibberish), I've written numerous templates that are useful for generating test data that's ranged from plain text to XML to syntactically-correct-but-uncompilable C. Despite being 26 years old and written in K&R C and having an ugly template format, it compiles just fine and works as advertised. I whipped up a template that solves your problem and posted it on pastebin since adding that much text here doesn't seem appropriate.

Answer 8

Here is a simple recursive method in javascript:

var genExpression = function(h){
    var n = max(1,h + random(-2,0));
    var s = "" + floor(random(1,11));
    var a = ["+","-","*","/","^"];
    var u = ["√","sin"];
    for(var i = 0; i < n; i ++){
        s += (" " + a[floor(random(0,a.length))]);
        if(random() < 0.25 && h > 4){
            s += (" " + u[floor(random(0,u.length))]);
            s += ("(" + genExpression(h - 4) + ")");
            s += (" " + a[floor(random(0,a.length))]);
        }
        if(random() < 0.5 && h > 3){
            s += (" (" + genExpression(h - 2) + ")");
        }
        else{
            s += (" " + floor(random(1,11)));
        }
    }
    return s;
};

So genExpression(4) might return something like 2 ^ (8 / 10 - 1 - 1) / 2 + 5 + (3 / 9 / 7) + √(2 ^ 3) ^ 5.

Answer 9

My manydl.c program on github is generating randomly some C code as generated plugins. The generated C code contains random expressions.

In practice, the generated C code is compiled by GCC as a plugin, then dlopen(3)-ed, then dlsym(3)-ed.

Generating C or C++ code was and is used in GCC MELT and in RefPerSys.

A book explaining why is it useful is: Jacques Pitrat's Artificial Beings, the conscience of a conscious machine

I have been able to generate half a million of more or less random C files, compile each generated file with gcc -O -fPIC -shared -Wall into a plugin, and dlopen each plugin.

Experimentally the program is never looping, but I cannot prove that.

NB: feel free to email me.