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Let's say I have an array of any fixed length i.e. it could be 3, 5, 7 ... upto 30.

How can I generate an algorithm such that if I pick an item from the array, every single day, it should not pick the same item index on any two consecutive days.

Note: I do not have the ability to store, what was picked the previous day.

For example, if my array is ["a", "b", "c", "d", "e", "f"]

valid result would be ["a", "b", "c", "d", "e", "f", "d", "a", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f"]

Invalid result would be

["a", "b", "c", "d", "e", "f", "a", "b", "b", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f"]

Note the consecutive b in above

An ideal result would be: where the whole array is perfectly rotated after the end of length is reached.

["a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f", "a", "b", "c", "d", "e", "f"]

Madhur Ahuja
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  • If you want it to give you the same result all day, independent of how often it's called, then I would suggest using the day of the month as an indexer of some kind. Boil that down to fit your array and you'll have a daily unique and incrementing result by default. – Antiga Apr 18 '16 at 19:18
  • Do you have access to day number? You could just use day number modulo the number of elements in the array as the index of an element to be picked. Otherwise, you could swap the picked element with the element at index `0` and always pick elements from range `[1, n - 1]`, so that you never pick the previous element as it is on the position 0 – Alexey Subach Apr 18 '16 at 19:18
  • so you want a (pseudo-) random sequence and no index following itself. right? picked one (and the same) index each day. How long should this sequence go (how many days/months/years)? and do you have to be able to pick values from the past? – Thomas Apr 18 '16 at 20:39
  • Sry, didn't see the part about the perfectly rotated. Easiest solution: `arr[ Math.floor(Date.now()/86400000) % arr.length ]` – Thomas Apr 18 '16 at 20:47

5 Answers5

1
var d = new Date();
var n = d.getDay();
var chr = String.fromCharCode(97 + n);

Run this every day to generate the next value in your sequence.

So your sequence will be:

a, b, c, d, e, f, g

Daniel Paczuski Bak
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1

Try using mod (%) that way the number can never exceed the array length. Demo: 

var array = ["a", "b", "c", "d", "e", "f"];
var output = [];
for (var day = 1; day < 365; day++) {
    output.push(array[day % array.length])
}
console.log(output);
Caleb O'Leary
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1

This is very easy to accomplish using getDate which returns the current day of the month (1 - 31 depending on the month). Combining this with the modulus operator ensures that you will always get a valid item in non-consecutive order.

function log(msg) {
  document.querySelector('pre').innerHTML += msg + '\n';
}

function logItem(items, date) {
  var index = (date.getDate() - 1) % items.length;
  var item = items[index];
  log(date.toDateString() + ' : ' + item);
}

function addDay(date) {
  date.setDate(date.getDate() + 1);
}
var items = ['a', 'b', 'c', 'd', 'e', 'f'];
var day = new Date(2016, 0, 1);
for (var i = 0; i < items.length * 3; i++) {
  logItem(items, day);
  addDay(day);
}
<pre></pre>

There is one problem with this exact implementation: it is possible for the end of the month to have the same relative value as the first of the month.

function log(msg) {
  document.querySelector('pre').innerHTML += msg + '\n';
}

function logItem(items, date) {
  var index = (date.getDate() - 1) % items.length;
  var item = items[index];
  log(date.toDateString() + ' : ' + item);
}

function addDay(date) {
  date.setDate(date.getDate() + 1);
}
var items = ['a', 'b', 'c', 'd', 'e', 'f'];
var day = new Date(2016, 0, 30);
for (var i = 0; i < 4; i++) {
  logItem(items, day);
  addDay(day);
}
<pre></pre>

To handle this, you can determine how many days are in the month, determine if your start and end indices are the same and change one if they are. How you pick an alternative value is up to you.

function log(msg) {
  document.querySelector('pre').innerHTML += msg + '\n';
}

function logItem(items, date) {
  // ---
  var daysInMonth = getDays(date);
  var day = date.getDate();
  var index = (day - 1) % items.length;
  if (day === daysInMonth && (day - 1) % items.length === 0) {
    // Change to a different item
    index = Math.floor(Math.random() * (items.length - 3)) + 2;
  }
  // ---
  var item = items[index];
  log(date.toDateString() + ' : ' + item);
}

function getDays(date) {
    return new Date(date.getYear(), date.getMonth() + 1, 0).getDate();
}

function addDay(date) {
  date.setDate(date.getDate() + 1);
}
var items = ['a', 'b', 'c', 'd', 'e', 'f'];
var day = new Date(2016, 0, 30);
for (var i = 0; i < 4; i++) {
  logItem(items, day);
  addDay(day);
}
<pre></pre>
Community
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Mike Cluck
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1

I made a solution using the modulo operater: %. Not sure if this is exactly what you want, but it seems to work properly.

var items = ["a","b","c","d","e","f"];
var day = new Date().getDay();

console.log(items[day % items.length]);

You could presumably put that into a function, or display it however you'd like.

Jon Laing
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0

Try something like this 

var a = ["a", "b", "c", "d", "e", "f"],
  b = [],
  random;

var i = function(val) {
  random = val[Math.floor(Math.random() * val.length)];
  (random != b[b.length - 1]) ? b.push(random): b.push('duplicate');
  console.log(b);
}

setInterval(() => {
  i(a)
}, 1000);
Nenad Vracar
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  • This doesn't take the day into account at all. If this algorithm was done during different runs of the program on consecutive days, it could produce the same output on both days. – Mike Cluck Apr 18 '16 at 19:38
  • @Mike C you are right, i used `setInterval` just for demo purpose – Nenad Vracar Apr 18 '16 at 19:40