How to make the first two numbers of a sequence Math.Andom are NOT EQUAL

Question

I made a lottery system in order to learn, I will not post the entire code here just the part I’m stuck.

The PC draw is a sequence of 6 numbers that I present in the DOM (that doesn’t matter), so I want the first two numbers to NEVER be equal.

I tried it with While but it’s not working, look:

    //gera o sorteio de 6 numeros de 1 a 20
    var rn1 = 1 + (Math.round(Math.random() * 19));
    var rn2 = 1 + (Math.round(Math.random() * 19));
    var rn3 = 1 + (Math.round(Math.random() * 19));
    var rn4 = 1 + (Math.round(Math.random() * 19));
    var rn5 = 1 + (Math.round(Math.random() * 19));
    var rn6 = 1 + (Math.round(Math.random() * 19));

    //eis oq eu tentei
    if(rn2 == rn1){ // se rn1 = rn2
        while(rn2 == rn1){ // comece o loop
            gerarnew1 = 1 + (Math.round(Math.random() * 19)); // gera um novo valor de 1 a 20 e roda até que...
                if(gerarnew1 != rn1){ // ...não seja mais = rn1
                    rn2 = gerarnew1; // entao atribuo esse novo numero a rn2
                    break; // paro o loop e, pela logica rn2 agora é != rn1
                }
        }
    }
    //entao sao armazenados aqui e apresentados dps no resto do codigo
    var rsorteio = new Array(rn1,rn2,rn3,rn4,rn5,rn6);

But as I’m posting here it’s certainly because it didn’t work, I run the algorithm several times and one hour it appears with the first two numbers equal.

What am I doing wrong?

Suggest I post the entire code?

Answer 1

Follows a solution using Fisher-Yates shuffle.

This algorithm has the characteristic of not repeating the data, because it simply changes the position of the members.

The algorithm is a very simple loop, it starts from the first item, and swaps with some of the following randomly (even with "himself") and can be considered "shuffled" already. This is done with the next item, drawing from it even to the last one (no longer deals with the first one, only from then on). The same is done with the third item (no longer deals with the previous two) and so on.

Follows the code:

var max = 20;
var set = [];
var i, j, t;
for( var i = 1; i <= max; i++ ) set.push( i );

for( i = 1; i < ( max - 1 ); i++ ) {
  j = i + ( Math.round( Math.random() * ( max - i - 1 ) ) );
  t = set[j];
  set[j] = set[i - 1];
  set[i - 1] = t;
}


// Demonstrando o resultado pegando 6 números
for( i = 0; i < 6; i++ ) document.body.innerHTML += set[i]+'<br>';
document.body.innerHTML += 'Clique novamente em "executar" para outro sorteio';

Modern version (1964 :P) of the algorithm:

https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle#The_modern_algorithm

Answer 2

Here is a suggestion more DRY. Ie without repeating code.

var numeros = [];

function novoNumero(index) {
    var sugestao = 1 + (Math.round(Math.random() * 19));
    if (index == 1 && sugestao == numeros[0]) sugestao = novoNumero(index);
    return sugestao;
}
for (var i = 0; i < 6; i++) {
    var numero = novoNumero(i);
    numeros.push(numero);
}

console.log(numeros.join(' - '));
// dá por exemplo: 16 - 12 - 17 - 2 - 19 - 19 

Here’s a test running this code 10,000 times.
Take a look at the console: http://jsfiddle.net/6vvyh0a3/

In this code I put the function novoNumero accepts as parameter the index, ie the position of the number in the array. The cycle for 0 to 5 and every iteration calls the function.

The function creates a new number and, if the value that variable index received be 1, then compares with the first number already generated in the previous iteration (ie numeros[0]). If they are equal, the function calls itself and generates a new number that will again compare.

After that, sure not to have repeated numbers, push (insert) this number in the array.

Answer 3

I believe that the most appropriate way for this problem (not to repeat elements) is to use the set-type javascript because it does not allow duplicated elements. This simplifies both the number generation and the control/end of the loop to generate the card and the game.

Basically it would be something like:

var cartela = new Set();
while(cartela.size < 6) cartela.add(número aleatório)

I separated the code into three functions one that generates the chart, another for the soterio and one that counts the correct.

function gerarSorteio(){
    var sorteio = new Set();  
    while(sorteio.size < 6){
        sorteio.add(Math.round(Math.random() * 20));
    }
    return sorteio;
}

function marcarDezenas(){
    var cartela = new Set();
    while(cartela.size < 6){
        cartela.add(parseInt(prompt("Informe a Dezena:")));
    }
    return cartela;
}

function resultado(sorteio, cartela){
    var acertos = [];
    for (let item of sorteio){
        if(cartela.has(item)){
            acertos.push(item);
        }
    }
    return acertos;
}

var sorteio = gerarSorteio();
var cartela = marcarDezenas();
var acertos = resultado(sorteio, cartela);

console.log(sorteio);
console.log(cartela);
console.log(acertos);

Example - replit