Quote Originally Posted by Lvl 2 Expert View Post
As elegant as Warty's method is, it seems you need less questions if you use a larger numerical system. That's because for every time you ask "is the next digit 1?" you also have to ask "is there a digit after this?" you're distinguishing between three states for each digit, 1, 0 and not there, lowering efficiency.

If you use for instance hexadecimal notation things get better. A hexadecimal digit represents 4 bits, and you can pin down the value of a hexadecimal digit in 4 questions (1-8, 1-4, 1-2, 1), but you only need a single question after that to ask about the next digit.

This means it becomes even more efficient when you take things a full byte at a time, or some even larger number. You could also try working with a notation system one short of the round numbers. "In a pentadecimal system, is the next digit between 8 and 15 or non-existent?" (If we start counting question length for efficiency this is going to kill it.)

But I feel like there should be a cleverer way to get rid of those extra questions...
In Hex, the 4 questions you need to ask are basically equivalent to 4 "is the next bit" questions (except in reverse in your example). So it's basically the same as only asking every 4 bits. Which does make a lot of sense (I'd make it probabilistic, you only ask after an appropriate number of zeros).

The Penta-decimal system is interesting.
Alternatively once we stop to ask the question, we could also pick up on other short cuts. Finding the number of consecutive 1s and 0s will be quicker in some cases.

*For non-existent read are all digits above this point zero.

Regardless at worst for the integers (if they plan to use the number to full accuracy in any way), we can find it almost as easily as they can use it for size comparisons.
Which I think counts for finding a number on a plane.