OK, I agree that this is the standard solution. Here's my interpretation:

Start counting days d on the day the Guru speaks. Each islander knows that if the observable blue eyed islanders n are greater than d then that islander must leave at midnight. On day 1 the blue eyed islanders n value is one less than the n' for those without blue eyes.  When n' equals d then the blue eyed islanders will leave.<br>  <br>  n' = n + 1<br> <br>  <br> <br> Leave IFF d > n (supposing no one else has left yet)<br> <br>  <br> <br> If n on d_1 was more than it is on d_2 then islander knows they don't have blue eyes and won't leave because other islanders cannot leave with certainty when d < or = n. Again, leave if and only if d > n but since n' is strictly greater than n ... those w/ non-blue eyes won't leave that day which happens to be when d = n' ... And if anyone left the night before (for d>1, to be clear) then the islander may conclude that they do not have blue eyes.<br> So, on d=100, the blue eyed islanders each see n=99 each reasoning that "if I didn't have blue eyes the rest would've left yesterday". And when any leave, the remaining islanders each reason that "if they were certain, they must have observed fewer other blue eyed islanders than days counted ... I did not. So I must not have blue eyes." The floor of (n<d) is the floor of (n'-1 <d). Islanders leave as soon as they are certain and that's the floor of the function only. If any leave, they know they're operating with different values and therefore have insufficient data to ever conclude the color of their eyes; since they know the Guru will never again speak.<br> <br>  <br> <br> As mentioned by /u/Mendoza2909, knowing with certainty that your eyes aren't one color doesn't imply with necessity that they are any other specific color. Which is necessary to know in order to leave, apparently.<br> <br>