How to get random math operator for a quiz qustion in javascript

Question

I want to know how can we get a random math operator to use and process in a math quiz question.

A random question is given to a user to solve, question has only two number to solve for any of the addition, subtraction, multiplication or division.

I have this code to generate two random number.

HTML

    <div id="num1"></div><!--end of num1-->
    <div id="num2"></div><!--end of num2-->
    <div id="operator"></div><!--end of operator-->  
    <div id="answer"></div><!--end of answer-->
    <button onclick="New()">New</button>

Javascript

function New(){
num1=document.getElementById("num1");
num2=document.getElementById("num2");   
rnum1 = Math.floor((Math.random()*100)+1);
rnum2 = Math.floor((Math.random()*100)+1);
num1.innerHTML=rnum1
num2.innerHTML=rnum2    
}

How can I generate random operator from +-*/ to use and process in code like

operator.innerHTML = '+';
answer.innerHTML = rnum1 + rnum2;

score 7 · Accepted Answer · answered Jan 20 '14 at 05:52

You can do this:

var operators = [{
        sign: "+",
        method: function(a,b){ return a + b; }
    },{
        sign: "-",
        method: function(a,b){ return a - b; }
    }];

var selectedOperator = Math.floor(Math.random()*operators.length);

operators[selectedOperator].sign                  //this will give you the sign
operators[selectedOperator].method(rnum1, rnum2)  //this will give you the answer

score 3 · Answer 2 · answered Apr 12 '16 at 04:43

I know I'm a little late to the party, but I thought I'd share a nice little hash map I wrote that takes advantage of the ES6 arrow functions for clarity.

const OPMAP = {
  '*': (n1, n2) => n1 * n2,
  '/': (n1, n2) => n1 / n2,
  '+': (n1, n2) => n1 + n2,
  '-': (n1, n2) => n1 - n2
}

Simply call OPMAP[operator] to get your anonymous function, and OPMAP[operator](n1, n2) to get your result.

score 2 · Answer 3 · answered Jan 20 '14 at 05:53

2

Define an array of operators:

var ops=['+','-','*','/'];

//random operator

var opindex = Math.random()*4; //good that your rnum2 cannot be zero
var operator = ops[opindex];

//calculate the expected result:

var res;
switch (opindex){
  case 0: res=rnum1+rnum2; break;
  case 1: res=rnum1-rnum2; break;
  case 2: res=rnum1*rnum2; break;
  case 3: res=rnum1/rnum2; break;
}

answered Jan 20 '14 at 05:53

Schien

3,855
1
16
29

+1 for nice solution but as Derek answered 1st I am accepting that answer. I appreciate your time and solution. – Jan 20 '14 at 06:13

score 0 · Answer 4 · answered Aug 07 '18 at 13:45

I would like to share my approach for generating random basic operations. I want it to have a bit of control in regards to the operations/operators being selected thus, selection of these elements is done by range, e.g. Math.random() returns a number between 0..1, thus if the value is lower than 0.5 it uses a given operator, in other words, the given operator has 50% of changes of being selected. Here is the full implementation, (for now it also supports adding one variable X in the equation):

function _getRandomNumber() {
    var randomPercentage = Math.random();
    var cardinality = (randomPercentage <= 0.20) ? -1.0 : 1.0;
    randomPercentage = Math.random();
    var randomNumber = Math.round((randomPercentage * 1000000) * 100) / 100;
    return (cardinality * randomNumber);
}

function _getOperator() {
    var randomPercentage = Math.random();
    if (randomPercentage < 0.10) {
        //  Nested expression
        return `(${_getExpression()})`;
    } else if (randomPercentage < 0.20) {
        // a header (variable)
        return `x`;
    } else {
        return _getRandomNumber();
    }
}

function _getBinaryOperator() {
    var randomPercentage = Math.random();
    if (randomPercentage < 0.25) {
        return `+`;
    } else if (randomPercentage < 0.50) {
        return `-`;
    } else if (randomPercentage < 0.75) {
        return `*`;
    } else {
        return `/`;
    }
}

function _getBinaryExpression() {
    return `${_getOperator()} ${_getBinaryOperator()} ${_getOperator()}`;
}

function _getExpression() {
    return _getBinaryExpression();
}



module.exports = {
    "getExpression": _getExpression
};

console.log(_getExpression());

Cheers, Herb

score -1 · Answer 5 · answered Jan 20 '14 at 05:54

-1

Example:

var operators = ['+','-','*','/'];

function New(){
    num1=document.getElementById("num1");
    num2=document.getElementById("num2");
    oper=document.getElementById("operator");
    answer=document.getElementById("answer");
    rnum1 = Math.floor((Math.random()*100)+1);
    rnum2 = Math.floor((Math.random()*100)+1);
    op = operators[Math.floor(Math.random()*4)];
    num1.innerHTML=rnum1;
    num2.innerHTML=rnum2;
    oper.innerHTML=op;
    answer.innerHTML = eval(rnum1 + op + rnum2);
}

answered Jan 20 '14 at 05:54

Maurice Perry

32,610
9
70
97

You can't calculate the answer by doing it this way. – Derek 朕會功夫 Jan 20 '14 at 05:56
You *can* use `eval`, but it isn't considered a safe practice. – Derek 朕會功夫 Jan 20 '14 at 05:57
It's safe in this case – Maurice Perry Jan 20 '14 at 05:59
rnum1 is a number between 1 and 100 – Maurice Perry Jan 20 '14 at 06:00
Well I still consider it to be [a bad practice](http://stackoverflow.com/questions/86513/why-is-using-the-javascript-eval-function-a-bad-idea). – Derek 朕會功夫 Jan 20 '14 at 06:06
+1, Your solution is nice and short, I like Derek answer more safe to use. I appreciate your time and help. – Jan 20 '14 at 06:16

How to get random math operator for a quiz qustion in javascript

5 Answers5