# Question about logic riddle (patch of lily pads that doubles every day)

I wasn't sure if the Math SE is the right SE to post this question, but after reading this question at the Meta SE, I think this SE is the right place.

In a video on YouTube about logic riddles, one of the riddles is:

In a lake, there is a patch lily pads that doubles in size every day. It takes the patch 48 days to cover the entire lake. How long does it take for the patch to cover half of the lake?

The answer is 47 days, because if the lake is half full at day 47 and the patch doubles every day, then on day 48 the lake must be entire full. More formally the part of the lake covered with the patch $c$ is given by $$c(t) = 2\times c(t-1)$$ There are many variations of variations of this riddle on the Internet, but the essence is the same.

However I think that day 45 is more likely. Let me explain why.

Firstly the function of $c$ is really given by $$c(t) = \min\{2\times c(t-1), 1\}$$ since the lake can't be covered for more than $100\%$.

Say the initial part of the lake covered is $1/2^k = (1/2)^k$ with $k \in \mathbb{R}$, so $c(0) = (1/2)^k$. In this case the lake would be covered in $k$ days, since $$\frac{1}{2^k} \times 2^k = 1$$

But for any other $k \in \mathbb{R}^+ - \mathbb{N}$ (so for every postive real non intiger value of $k$), which is far more likely since $\mathbb{N} \subset \mathbb{R}^+$, you would need $k + 1$ days to cover the entire lake.

A concrete example: say $0.4\text{ }(=40\%)$ of the lake is covered at day the beginning of day $x - 2$, so $c(x-2) = 0.4$. This would mean $c(x-1) = 2\times c(x-2) = 0.8$ and $$c(x) = \min\{c(x-1),1\} = \min\{1.6,1\} = 1$$ thus the lake is entirely covered at day $x$.

So the lake is entirely covered at the beginning of day $x$, even since day $x-1$ begins with $0.8 = 80%$ of coverage. If we take $x = 48$, the lake is covered at day 48 like in the original riddle, but the lake isn't half covered at day $47 = x - 1$. So it must be covered half in the course of day $x - 2$.

Am I missing something or is the lake indeed almost always half covered two days instead of one day before it is fully covered?

Disclaimer: This is not meant as critic on the maker of the video. Her videos are awesome!

Edit: Yes, I'm probably overthinking this, act as if I'm not. :)

• I feel this is taking the scenario too literally - this sort of riddle seems more of an abstract puzzle. It's a logic puzzle, so it isn't intended to make sense without the assumption that it takes an integer number of days for the lake to be fully covered. – Uzai Aug 14 '16 at 21:23
• I think you're overthinking the puzzle, but let's go along with it. Either they are reporting the exact time the lake is covered (so in your $k\notin\mathbb N$ case they would have said something like "it takes 47.5276 days for the lake to be covered"), or they are rounding up to the end of the day (so they say 48 days if the lake is covered at some point during the course of the 48th day). In either case, if you apply the same logic to the lake being half covered, you get 47 days. – Rahul Aug 14 '16 at 21:27
• If you take "it takes 48 days to cover the lake" to mean "we walked away and came back 48 days later and saw the lake was covered" then the answer can be anything from a nanosecond to 47 days. If you take it to mean "sometime during day 48 (or 47) it became full" then sometime during day 47 (or 46) it became half full. If it means, as was obviously intended "at exactly 48 days the lake became full" then at exactly 47 days the lake became half full. I don't see many other options. I don't get 45 at all. – fleablood Aug 15 '16 at 0:31
• Or maybe we can think of it as the lilies stay exactly the same and then at the stroke of midnight it spontaneously doubles. So maybe it was 99.9% covered on day 47. Then it was 49.95% covered on day 46. So it still wasn't half covered until day 47. – fleablood Aug 15 '16 at 0:37

Suppose we think of the growth as a continuous process. If $c_0$ is the initial coverage of the lake at the start of day 1, then the coverage of the lake at the end of day $t$ is $c(t)=c_02^t$. Now $c(t)=1$ when $c_02^t=1$ i.e at $$t_1=-\frac{\ln c_0}{\ln 2}$$
while $c(t)=1/2$ when $c_02^t=1/2$, i.e. at $$t_{1/2}=-\frac{\ln c_0}{\ln 2}-1$$
So, as was obvious without any calculation, $t_{1/2}=t_1-1$. If, for example $t_1$ turns out to be $47.879$, then $t_{1/2}=46.879$. That is if the lake is covered during the 48th day, then it is half covered during the 47th.
• Thanks for your response. If $c(t_1) = 1$ (with the independent variable in days), then $c(t_1+1) = 1$ too, since the lake can't be covered more than 100%. Say $\tau = t_1 +1$. Now $c(\tau) = 1$ but also $c(\tau - 1) = 1$. We could have started with $c(\tau) = 1$ instead of $c(t_1)$, this must not make any difference I think. So if $c(\tau)$ or $c(t)$ equals 1 at a certain day, it can just as good be equal to 1 the day before, right? – Kevin Aug 14 '16 at 21:43
• Perhaps I should have been explicit, but $t_1$ is the smallest $t$ such that the lake is covered. (I did not define the function $c$ so that $c(t)$ is equal to the minimum of $c_02^t$ and $1$, because we can ignore the constraint that the coverage cannot be more than $1$; all we are interested in is the first point in time in which the lake is covered. ) – smcc Aug 14 '16 at 21:50