Eight Water Lilies Buy Three Days, and Dividing Says Twenty-Six

A single water lily floats on a pond. Every day its area doubles, and on day thirty it has covered the whole surface. Now run the experiment again with eight lilies of the starting size instead of one, all planted on the same morning. Which day does the pond disappear?

Eight plants, so an eighth of the wait: $30/8 = 3.75$ days. That is the answer most people give, and it is not slightly wrong. The pond is covered on day twenty-seven.

Why thirty over eight looks right

The division is not a random slip. It is the correct answer to a neighbouring puzzle. Suppose the lily added a fixed patch of area every day rather than doubling. Covering the pond in thirty days would mean it adds one thirtieth of the pond per day, eight of them would add eight thirtieths, and the pond would vanish after $30/8 = 3.75$ days. Division is the right operation whenever the growth rate does not depend on how much is already there.

Doubling is the other kind of process. Its rate is proportional to the current area, so the eight lilies grow eight times as fast as the single one on every single day, and the gap between the two schedules never closes in the way the division assumes. Nothing about the arithmetic $30/8$ is sloppy. A model was swapped out quietly, and the answer belongs to the model that got substituted in.

It is worth measuring how far off the substitution lands. Take areas in units of one starting lily, so the single lily has area $2^t$ on day $t$ and the pond has area $2^{30}$. At the division's 3.75 days the eight lilies cover

of the surface. One ten-millionth. The pond at that moment looks exactly as empty as it did on the first morning.

Eight lilies are one lily, three days older

Eight is $2^3$, and that is the whole problem. The total area of the eight lilies on day $t$ is

which is the area a single lily has on day $t+3$. So the eight lilies are not a new problem. They are the old problem with the calendar rewritten. Setting $2^{t+3} = 2^{30}$ gives the answer in one line:

Every milestone, which is the part worth holding on to. The single lily covers half the pond on day twenty-nine, because $2^{29}/2^{30} = 1/2$. The eight lilies therefore cover half the pond on day twenty-six, and a quarter on day twenty-five, and so on all the way back. The two schedules are one curve seen through two different calendars.

The general rule, and the floor hiding in it

Start with $k$ lilies instead of eight. The pond is covered once $k \cdot 2^{t} \ge 2^{30}$, which rearranges to $t \ge 30 - \log_2 k$. The pond is only inspected once a day, so the first day of coverage is the smallest integer above that threshold:

The days saved are $\lfloor \log_2 k \rfloor$, and when $k = 2^{j}$ the floor does nothing and the saving is exactly $j$. Notice which number is absent from that expression: thirty. How long the original lily takes is irrelevant to the saving. Give the pond three hundred days and eight lilies still buy exactly three of them.

Five lilies, and where the clean rule stops

The floor function in equation (4) is not decoration. With five lilies, $\log_2 5 = 2.3219\ldots$, so the exact crossing happens at $t = 27.678$ and the first evening on which the pond is covered is day twenty-eight. The saving is two days, not 2.3219 days. A fractional head start is thrown away by a process that is only ever measured in whole days.

The same rounding makes seven lilies no better than four, and three lilies no better than two. On the powers of two the rule is exact and on everything else it is a staircase.

The assumption doing the work

Equation (2) treats eight lilies as one lily of eight times the area, and that step needs the eight patches to stay out of each other's way while they grow. If they overlap, the covered area is strictly less than $8 \cdot 2^t$ and the pond survives past day twenty-seven. Non-overlap is the intended reading of the puzzle, and it is the only place where the answer could honestly be argued.

There is a second ambiguity, and it turns out not to matter. The phrase covers the pond in thirty days can mean the pond has exactly the area $2^{30}$, or merely that day thirty is the first day it is covered, which places the pond anywhere in $(2^{29}, 2^{30}]$. Every pond size in that interval gives twenty-seven, because the eight-lily schedule crosses the same window three days earlier and the window is one doubling wide. Under continuous rather than nightly doubling the crossing falls exactly on $t = 27$, so the discrete and continuous readings agree too.

Where the schedule stops sliding

The time-shift argument is more general than the exponential that motivated it. Suppose the area obeys $A' = F(A)$ for some growth law depending on the current area and not on the date. Then the eight-lily trajectory is the one-lily trajectory started further along, which means it is the same curve read $\tau$ days early, where $\tau$ is the time a single lily needs to reach eight units. Exponential growth is what pins $\tau$ at exactly three. A law that grows more slowly than exponential over that stretch, a logistic curve for instance, makes $\tau$ larger and clears the pond before day twenty-seven, because near the end an advantage in area is worth more days than it was at the start.

Two things do break it. A growth law that depends on the calendar rather than on the current area is no longer time-translation invariant, and then eight lilies are genuinely a new problem. And a different growth factor moves the answer even with the head start unchanged: a lily that triples daily and covers the pond in thirty days gives eight lilies a saving of $\log_3 8 = 1.8928\ldots$ days, which the floor cuts to one, so the pond goes on day twenty-nine. The tidy answer of three days is an arithmetic coincidence between the head start and the doubling, and nothing forces the two to line up.

The division, meanwhile, reads as a saving of more than twenty-six days off a thirty-day schedule. The real saving is three. The ratio between the two candidate answers is $27 / 3.75 = 7.2$, and the ratio between the two pictures of the pond at 3.75 days is a factor of ten million.

Sources and further reading

• The growth law behind equation (2) and its time-translation property Exponential growth
• Doubling period as the natural clock of the problem Doubling time
• What replaces the exponential once the pond runs out of room Logistic function
• The rounding in equation (4) Floor and ceiling functions

Every number here was checked twice: symbolically, and by a day-by-day simulation in exact integer arithmetic that uses no logarithm and no exponent shift. It reads day thirty for one lily, day twenty-seven for eight, day twenty-eight for five, and it finds a saving of exactly $j$ days for $2^{j}$ lilies at every $j$ up to twenty.