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@itsananderson
Created July 19, 2015 06:09
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Solution for the consecutive numbers problem
// It's a bit messy, but it seems to work
function player(name) {
this.name = name;
this.questionCount = 0;
this.number;
}
player.prototype.setOther = function(other, otherNumber) {
this.other = other;
this.otherNumber = otherNumber;
this.possibleNumbers = [otherNumber - 1, otherNumber + 1];
}
player.prototype.knowsNumber = function() {
if (this.number) {
this.end();
return true;
}
// If more questions have been asked than the other person's number,
// and they still don't know what they are, we're the higher number
if (this.questionCount >= this.otherNumber) {
this.number = this.possibleNumbers[1];
setTimeout(function() {
this.other.knowsNumber();
}.bind(this));
this.end();
return true;
} else {
this.questionCount++;
setTimeout(this.ask.bind(this));
}
}
player.prototype.ask = function() {
if (this.other.knowsNumber()) {
// If they figured theirs out first, that means we're the lower number
this.number = this.possibleNumbers[0];
} else {
this.questionCount++;
}
}
player.prototype.end = function() {
console.log(this.name, this.number);
}
var i = 0;
var skew = 1;
var inter = setInterval(function() {
skew = (skew+1)%2;
if (skew === 0) {
i++;
}
if (i >= 10) {
clearInterval(inter);
}
var player1 = new player("player1");
var player2 = new player("player2");
// Associate players
player1.setOther(player2, i+1-skew);
player2.setOther(player1, i+skew);
// Kick off the questions
console.log("--");
console.log("Should be player1:" + (i+skew) + " player2:" + (i+1-skew));
player1.ask();
}, 100);
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