# Winnie-the-Pooh and the 27 honey pots

Winnie-the-Pooh keeps his $27$ honey pots in the larder. Each pot contains up to $1$ kilogram of honey, and different pots contain different quantities of honey. All $27$ pots together contain $17$ kilogram of honey.

Every day, Winnie-the-Pooh selects $7$ pots, picks a real number $x$, and then eats exactly $x$ kilogram of honey from each of the selected pots.

Question: Is it always (independently of the initial distribution of honey) possible for Winnie to empty all $27$ honey pots in a finite number of days?

• Does he select 7 pots which have honey in them or it does not matter?
– dmg
Commented Sep 25, 2015 at 8:42
• @Gamow So he can select the same empty pot every day? Forever? Poor Winnie...
– dmg
Commented Sep 25, 2015 at 9:12
• I have a feeling that when you pick a random distribution the chance is actually pretty slim that it can be done. But don't know how to proof it
– Ivo
Commented Sep 25, 2015 at 11:04
• Is Pooh picking his x at random, or may we assume that he follows an optimal strategy, never picking an x that would put the puzzle into an unsolvable state? Commented Sep 25, 2015 at 15:54
• I believe WtP keeps his honey pots in the larder. A minor point, I know.
– A E
Commented Sep 25, 2015 at 15:59

Here is a rigorous proof that Pooh can always succeed.

Let $P_i$ be the amount of honey in the $i^{th}$ pot when sorted in decreasing honey contents, so $1\ge P_1\ge P_2\ge \dots\ge P_{27}\ge0$ and $P_1+\dots+P_{27}=17$.

Using the beginning of Sleafar's solution, we now need only show how to equalize the seven heaviest pots. At any point, let $m$ be the number of pots which have the same amount of honey as the heaviest, and let $n$ be the number of nonempty pots (so initially, $m=1$ and $n=27$). We show that, as long as $m<7$ and $n\ge 14$, it is possible to either increase $m$ or decrease $n$. We then argue why $n$ can't drop below $14$ before $m$ reaches $7$, so that eventually we equalize the top $7$ pots.

Here is the method. Let $x$ be the gap in honey contents between the heaviest $m$ pots and the next strictly lighter pot. Each day for the next $n-7$ days, Pooh eats $\frac{x}{n-7}$ kg from each of the $m$ heaviest pots. Each of these days, he must eat the same amount from some $7-m$ other pots. He does this in such a way that the $n-7$ pots $P_8,P_9,\dots,P_n$ are eaten from equally over this $n-7$ day period (for example, by arranging the nonempty pots in a circle, and eating from each consecutive segment of length $7-m$).

As long as each of the nonempty pots contain at least $\frac{x}{n-7}\cdot (7-m)$ kg of honey, then this will work, causing the $m$ heaviest pots to decrease to the next highest weight, thus increasing the number $m$. If the nonempty pots do not contain enough honey to do this, then adjust $x$ so that the smallest pot contains exactly $\frac{x}{n-7}\cdot (7-m)$. Then doing the procedure will empty that pot, thus decreasing $n$.

Now, why can't $n$ drop below $14$ before $m$ reaches $7$? First off, the amount of honey that needs to be removed from pots 8 through $n$ in order to equalize pots 1 through 7 is equal to $$6(P_1-P_2)+5(P_2-P_3)+4(P_3-P_4)+3(P_4-P_5)+2(P_5-P_6)+(P_6-P_7)$$ This is because, during the phase when pots $P_1,\dots,P_m$ are equal, we have to remove a total of $P_{m}-P_{m+1}$ from each of the first $m$ pots, and $(7-m)$ times that from the last pots 8 through $n$. A different way of writing this is $$6P_1-P_2-P_3-\dots -P_7\le 6 - 6P_7< 3$$ The last inequality follows since $P_7> \frac12$. Why is this true? If not, then the first 6 pots would initially contain at most $6$ kg, and the last 21 would contain at most $\frac12$, so that the total amount of honey was at most $6+\frac12\cdot 21=16.5$, which is impossible since we started with $17$ kg of honey.

So, the procedure will remove less than $3$ kg from the last 20 pots. However, the first 14 pots contain at most 14 kg, meaning the lightest pots together contain at least $17-14=3$ kg. Since our procedure eliminates the lightest pots first, the first 14 pots will not be emptied, as claimed.

• Great approach and easy to follow (though it still has to sink in a little). I suggested an edit that among other things changed some instances of 27 into n. Most of my mind is convinced now that irrationality doesn't matter. Shakes up my mathematical intuitions a bit :-) Commented Sep 28, 2015 at 20:21
• Profoundly good answer! I wonder if the estimates being so tight was 'coincidental' (ie, it's something special about the choice of numbers). Commented Jan 7, 2016 at 6:39

# Update:

I made a big blunder in the previous version, this should be fixed now.

## Part 1: It is always possible to remove amount $x$ from any number $n \ge 7$ of pots if all pots have at least amount $x$

Let $p$ be the number of permutations with 7 pots containing a specific pot. Cycle all permutations and remove amount $\frac xp$ from each pot. As each pot occurs $p$ times the total removed amount from each pot is $x$.

## Part 2: A combination of pots where the 7 pots containing the most amounts are equal is know to work regardless of the amounts in other pots

Consider the following image:

The bars show the amounts in the pots, ordered by the amount. The amounts in the first 7 pots are equal. Start with the smallest pot $J$ and remove the amount of $J$ from all pots (this works as seen in part 1). Repeat with pot $I$ and then with pot $H$. At the end we have the top 7 pots containing only the black marked area, which can be emptied in the next step.

## Part 3: Removing superfluous amount in biggest pot requires 6 times the same amount 20 smallest pots

Consider the following image:

We must remove the red part in pot $A$ to reach the situation from part 2. It is easy to see that we can do this easily if we have the same 6 parts in pots $H-M$.

This is the place where I made a blunder in the previous version, assuming I could trade the red part in $A$ with a combination of parts of the same size in pots $>G$, instead of 6 times the size.

What can be said about the distribution of the amounts beyond pot $G$? Actually it doesn't matter, if we use a trick. Consider the following image.

We want to remove the red part in pot $A$. The orange part in pot $H$ has 6 times the size of the red part. We can remove the red part and the orange part of pot $H$ while also removing the orange parts in pots $A-G$. To do this we generate all permutations of 6 pots from the range $B-H$:

$\binom{7}{6}$ = 7

The number each of the pots $B-H$ appears in this permutations is:

$\binom{6}{5}$ = 6

Now we add pot $A$ to all these permutations and remove the red part from each permutation. We remove the red part 7 times from pot $A$ and 6 times from pots $B-H$ which is exactly what we see in the image above. This can be easily extended to any number of pots, as long as they contain 6 times the size of the red part.

The other conclusion is, no matter what we do, we can't remove the red part if the size of the pots beyond $G$ is less than 6 times the size of the red part.

## Part 4: Removing superfluous amount in 2 biggest pot requires 2.5 times the same amount 20 smallest pots

Consider the following image:

Similar to part 3, we now want to remove the 2 red parts in pots $A-B$. It is again easy to see that we can do this easily if we have the same 5 parts in pots $H-L$.

Using the same reasoning as in part 3, it can be shown, that the amount distribution beyond pot $G$ doesn't matter.

It can also be shown that similar applies to superfluous amounts in pots $C-F$.

## Part 5: The general case

The following image shows the general case which is guaranteed to be solvable based on the parts above:

As we have seen in part 3, the distribution in pots beyond $G$ doesn't matter. Also as seen in part 2 adding amounts in pots beyond $G$ is solvable too. What is left is to show which amounts can be solved in any case.

## Part 6: The edge case

The above image shows the edge case, where the red area is maximized (if we limit ourselves to 13 pots). The height of pots $H-M$ must be the same as the height of the red part of pot $A$. Pots $B-G$ must have also the same size. As $A$ is limited to $1$, the height of pots $B-M$ must be $\frac{1}{2}$. The size of this case is:

$1 + 12 * \frac{1}{2} = 7$

This means that 7kg honey is enough to guarantee, that the puzzle is solvable, if the pot size is limited to 1kg.

Any other amounts which we would try to add, don't change anything. Amounts added to pots beyond $G$ are OK as shown in part 2. Amounts added to pots $B-G$ reduce the required amount in pots beyond $G$ as can be seen in the following image:

One more thing we could do, is to use the other 14 pots we didn't use yet to further maximize the red area, like in the following image:

We used here one more pot. We increased the size of the red area above $A$ by some amount $x$. We also increased the size of the red area beyond pot $G$ by $6x$. But at the sime time we decreased the size of the black area by $7x$ which in the end gives a total change of $0$.

The pot size which would still work for 17kg is (based on the first edge case):

$x + 12 * \frac{x}{2} = 17$

$x = \frac{17}{7} = 4,28...$

Also, this solutions works with any number of pots $\ge 13$.

Alternatively we can make Winnie-the-Pooh really happy and tell him, he can eat from 17 pots each day (which would still work with 17kg and 1kg max per pot).

• great post, I was just wondering what your thoughts were on the situation where a pot contains an irrational amount of honey? (Say root(2)/2?) Irrationality messes with my head, and I'm not sure if it changes anything Commented Sep 28, 2015 at 11:44
• the finite number of days part worries me, though maybe I'm just not fully understanding your solution Commented Sep 28, 2015 at 11:49
• @user1868155 Hehe, irrational numbers were also one of my first ideas, but they simply doesn't matter as you found out yourself. Even if the total amount would be irrational it would still work, think of any of the colored boxes above as an irrational number. Apart from that, the algorithm I described uses a finite number of steps (which can still be optimized by summing up the same permutations used in different steps). Commented Sep 28, 2015 at 16:44

YES, HE CAN!

Rigorous proof and exact bounds: If the starting amount of honey in each pot is less or equal to $17/7$ kilograms, then Pooh can always do this. If the starting amount of honey in each pot can be more than $17/7$ kilograms, then Pooh can not always do this.

More generally: If the total amount of honey is $A$, the number of pots is $X$, each of the pots contains at most $P$ kilograms of honey and Pooh chooses every day $r \leq X$ pots, then he can always empty them if and only if $P \leq A/r$. (I have provided computations for $A=17, X=27, r=7$, but the same approach works for all triplets $(A, X, r)$.

Proof:

PART 1.

The idea is to arrange the pots by amount of honey in decreasing order - #1, #2, #3, etc. and then take honey in such way so that we get more and more pots with the same amount of honey as pot #7.

Arrange the pots by amount of honey:

$x_1\geq x_2\geq x_3\geq x_4\geq...\geq x_{27}$.

Assume that we have:

$x_1\geq x_2\geq ...\geq x_k>x_{k+1}=x_{k+2}=...=x_l>x_{l+1}\geq x_{l+2}\geq ...\geq x_{27}$,

where $k<7$ and $l\geq 7$ (this means that all pots with numbers from $k$ up to $l$ have the same amount of honey as pot #7). Now Pooh takes honey from the first $l$ pots, such that the amount of honey taken from each of pots $k+1, ... , l$ is $(7-k)/(l-k)$ of the amount of honey taken from each of pots $1,2,...,k$ and one of these two conditions is satisfied:

1. the amount of honey in pot $k$ becomes equal to the amount of honey in pots $k+1,k+2,...,l$;
2. the amount of honey in pot $l+1$ becomes equal to the amount of honey in pots $k+1, k+2,...,l$.

It is easy to see that Pooh can complete this task in $l$ days - just take honey from pots $1$, $2$, ... ,$k$ + some honey from $7-k$ pots among $k+1$,$k+2$,...$l$ in cyclic manner. After he completes this step, he repeats the process until all pots get equal amount of honey in them and eventually become empty.

Now we prove that eventually we will get $x_1=x_2=...=x_{27}=0$. Indeed, if we assume that Pooh gets stuck, then all of the pots except for at most $6$ should become empty. Let the number of non-empty pots be $m \leq 6$ and the total amount of honey remaining is $H \in [0,\min(mP,17))$. Clearly, the amount of honey taken from these $m$ pots is exactly $m(17-H)/7$ (because every time we take out honey, $m/7$ of it is taken out of them). Since each pot contains in the beginning at most $P$ kilograms of honey, we have

$$\frac{(17-H)m}{7}+H\leq mP.$$

The inequality above just states that the total amount of honey remaining in the last $m$ pots + the amount of honey taken out of them is no more than the maximum amount of honey there was in the pots initially.

We can rewrite the inequality as: $$0<H(7-m)\leq m(7P-17) \rightarrow P>17/7$$.

Therefore we can conclude that if $P\leq 17/7$, then Pooh can always do this.

PART 2

The idea is to fill up the first pot entirely.

Now we prove the inverse. Let the first pot has $P$ kilograms of honey and the others have $(17-P)/26$ each. Now every time we take $x$ kilograms of honey from the first pot, we take at least $6x$ kilograms of honey from the remaining pots. In order to empty the first pot, we have to take out $P$ kilograms of honey from it and therefore we have to take out $6P$ kilograms of honey from the other pots. However, $6P>17-P$, which is a contradiction.

Remark: If the total amount of honey is $A$ and the number of pots Pooh can use on every day is $r$, then using the same calculations it is easy to see that the optimal $P=A/r$. The total number of pots does not matter at all, it could be anywhere from $7$ to infinity.

• I'm trying to work through your procedure, but I run into trouble for the initial situation. Here we have either k=7 or l=7, or in other words, the assumption as stated doesn't hold. This is probably minor but I don't yet have the insight to see what the fix should be. Commented Sep 28, 2015 at 8:24
• Thanks for the remark @Oliphaunt, it was a typo - l>=7. Commented Sep 28, 2015 at 15:16
• 17/6 kg can't be right. If you put 17/6 kg in the first 6 pots, you used up all 17kg and poor Winnie can't eat any honey at all. In my updated answer it's 17/7 (off-by-one-error somewhere?). Commented Sep 28, 2015 at 16:51
• Yes, messed up a bit the calculations. Will fix it now, thanks for the remark! Commented Sep 28, 2015 at 17:05
• @Oliphaunt The distinct restriction only adds an epsilon which can be as small as required. And Winnie could then eat 21*epsilon=0,0000... Commented Sep 28, 2015 at 17:11

It is not always possible.

First note that, suppose that it would be possible then it could always be finished in $\binom{27}{7}$ days, because this is the number of possible ways to pick 7 pots, and eating from the same 7 pots on multiple days can always be combined to eating that amount in 1 day.

In all those combinations every pot occurs the same amount of time. So let's say it was possible to distribute the 17kg evenly (which isn't allowed by the rules). Then there is a solution in exactly $\binom{27}{7}$ days in which every day the same exact amount is eaten.

This means that given a distribution you are always allowed to subtract $x$ from every pot and try to solve that distribution instead.

So this is what we do:
put 1 gram in the first pot, put 1/10 gram in the second, 1/100 in the third and so on. Whatever is left from the 17kg we distribute evenly among all pots.

According to what I said before, we can neglect that final part so we just see if we can solve it with the 1, 1/10, 1/100 distribution. And this is simple: we can't solve it because the honey in pot one is even more than all the honey of the other pots combined.

• There is one thing that makes me doubt my solution: Just because we can subtract any $x$ amount from all pots doesn't mean that we should do it, I think.
– Ivo
Commented Sep 25, 2015 at 12:38
• I agree. Suppose there are 3 pots and we have to eat 2 at a time. Then 2,2,4 is actually solvable while 0,0,2 isn't. Commented Sep 25, 2015 at 13:44
• @Anachor you're right. thanks for the example. Guess my solution is not sufficient
– Ivo
Commented Sep 25, 2015 at 13:46
• I think you have the right idea though. If one pot is bigger than the rest combined, it can't be done. Perhaps distribute the remainder by multiplication instead of addition? Commented Sep 25, 2015 at 14:53
• I like you insight about the even distribution. Imagine the top seven pots all had the same amount of honey. Then by reducing the top seven, you can get the top eight to all have the same amount of honey. Then reduce the top eight evenly until the top 9 have the same amount, etc. In this way, as long as the top 7 have the same amount, you can eat all the honey. Commented Sep 25, 2015 at 18:06

We can formulate the problem as follows. This is not an answer but perhaps it leads to one.

The $27$ pots correspond to a choice of values $p_i$ with $0 \leq i < 27$ satisfying:

1. $p_i \neq p_j$ whenever $i \neq j$;
2. $0 \leq p_i \leq 1$;
3. $\sum p_i = 17$

So each pot configuration lies in $Z'$ the intersection of a hypercube $[0,1]^{27}\subset\mathbb{R}^{27}$ with the hyperplane $\pi$ defined by $(3)$. Of course, the point must also satisfy $(1)$, so it must be in the complement of the union of another $\binom{27}{2}$ hyperplanes. Let $Z \subset \mathbb{R}^{27}$ be the set defined by these conditions.

A solution for a given configuration corresponds to a choice of values $v_j \geq 0$ with $0 \leq j < \binom{27}{7}$. Each such value is the amount of honey Pooh eats from each pot in a given combination of $7$ out of the $27$ pots, so they must satisfy $$\sum v_j = \frac{17}{7},$$and also for each $0 \leq i < 27$ $$\sum v_{j_i} = p_i ,$$ where the sum has $\binom{26}{6}$ terms and is carried over the $v_j$'s which involve pot $i$. These last $27$ equations can be written in matrix form as $M\cdot V = P$, where $V$ is the column vector ${\left(v_j\right)}$, $P$ is the column vector $\left(p_i\right)$ and $M$ is a $27 \times \binom{27}{7}$ matrix of $0$'s and $1$'s. Each line has $\binom{26}{6}$ nonzero entries and each column has $7$ nonzero entries.

So $V$ must lie in the set $S$, defined by the intersection of a hyperplane $\sigma$ in $\binom{27}{7}$-dimensional space with its first quadrant. A solution for $P \in Z$ exists if and only if $P \in M(S)$. Thus, a solution always exists if and only if $Z \subset M(S)$. It might however be easier to show that $Z' \subset M(S)$; this suffices since $Z \subset Z'$.

It's easy to see $M(\sigma) \subset \pi$, and it should not be too hard to show that's actually an equality (this is necessary). It's also easy to see that if $V$ has nonnegative coordinates then so does $M(V)$. However this still needs to be improved.

An attempt at a TLDR version of the proof using ideas from all the other solutions (especially Artur Kirkoyan's take on the problem) . Definitely not as rigorous but I hope still convincing.

Suppose Pooh follows two rules while eating honey

1. Never eat from a pot with less if you could be eating from a pot with more.
2. If you must choose one pot from several that are equal, cycle between these pots as to eat from them equally.

Suppose Pooh follows these rules but still ends up with a set $S$ of less than seven nonempty pots at the end. By following the rules, the pots in $S$ must have always been the pots with strictly the most honey, and have therefore been eaten from constantly. Since each pot in $S$ holds at most 1 kg, the constant eating means at most $7$ kg total has been eaten. But this is ridiculous, as there must be more than 11 kg outside the set $S$.

I found the following beautiful solution to this problem at this math blog, and thought it worthwhile to share here.

Divide a circle into 27 arcs whose lengths are proportional to the amounts of honey in each pot. Inscribe a regular heptagon in this circle. Rotate the heptagon clockwise. While the heptagon rotates, Pooh drinks from the pots corresponding to the seven arcs it is touching, drinking at the same rate of rotation. When the heptagon switches arcs, Pooh switches out the corresponding pot for the next. At any point, Pooh is drinking from exactly seven pots at an equal rate. By the time the heptagon has rotated 360º/7, returning to its starting position, all pots will have been emptied.

My instincts told me to consider irrationality, so irrationality I will consider.

## All Rational

Consider the case where all the pots contain a rational amount of honey. A rational number is one that can be expressed as $$\dfrac{p}{q}$$ where $$p$$ and $$q$$ are both integers. Hence, it can be seen that there exists some value $$a$$ for which every pot of honey contains an amount of honey equal to $$ba$$ where $$b$$ is some integer number. (for instance, if the honey in pot $$n$$ is $$\dfrac{p_n}{q_n}$$ I can let $$a = \dfrac{1}{q_1q_2...q_{27}}$$) In effect, all the pots of honey contain an amount of honey that is some integer number amount of a small division.

Our plan now is simply to take $$a$$ amount of honey from the seven pots that contain the most honey. We can ensure that this is doable by letting $$a = \dfrac{a}{7}$$. We will only run into an issue if at one point of time there exists only six or less pots with honey in them. In the worst case scenario, this happens when we have six pots with an extreme amount of honey in them and 21 pots with barely a trickle of honey in them. Because we are dealing with real numbers, the pots can have amounts of honey really really close to $$1$$, so we can assume in the worst case scenario all six of the top-pots contain $$1$$ kg of honey. The other $$21$$ pots must then contain $$11$$ kg of honey. But we can easily see that if this were the case, every single turn you would choose the first six pots (for they contain the most honey) and a pot from the remaining $$21$$ pots, and you would effectively deplete $$11*6$$ kg of honey from the first six pots ($$66$$ kilograms) which is more than enough to balance the amount of honey in the pots. Meaning that if all the pots contain a rational amount of honey, life is good and Pooh is a happy, albeit unhealthy, bear. -I will try and word this better, currently it's a little bit badly worded.

## Some Irrational

Here comes the interesting part. Say for simplicity you have one pot with an irrational amount of honey in it. You might wish to panic, but fear not, for all the pots added up must total $$17$$kg! This means that there must be at least one other irrational pot out there! Our aim then is to convert the irrational pots into rational ones, for then our job is easy.

Consider the case where we have only two irrational pots. The two irrational pots must be a 'conjugate pair', i.e. if one of the irrational pots was $$\pi /10$$ the other must be of the form $$k - \pi/10$$ where $$k$$ is some constant greater than $$\pi$$. Another thing we must note is that we can 'truncate' an irrational number: for instance if we have the irrational number $$0.35817395185712$$ we can take $$0.00017395185712$$ away from it to rationalize it, and this way we avoid the trouble of accidentally 'grounding' a pot, or eating all of its honey. The concept of the conjugate irrational pot pairs is very important. For example, say we had 25 rational pots, and 2 irrational pots, one which contains an irrational number $$w$$, and one which contains its conjugate irrational number, $$\bar{w}$$. We can then choose to rationalize $$w$$, and choose six other pots (which don't include the $$\bar{w}$$), but in the process we create six other $$\bar{w}$$ pots. In essence, we can then choose these seven $$\bar{w}$$ pots and rationalize them. I am really bad with words, here is a diagram:

Of course, this is a really simple case. Much more complex cases exist, for instance what if 26 of the pots contained a different irrational number, and the 27th pot contained the conjugates to all the other 26 pots? (because a pot containing$$(\bar{z} + \bar{y})$$ acts as a conjugate for both $$z$$ and $$y$$)The key thing to note is that with this 1:1 conjugate ratio business, the above tactic can be applied multiple times over and over to remove each set of conjugates.

The REAL problem arises when you have a set of three complex numbers that combined create a real one. In this case there is no longer a 1:1 conjugate ratio, (i.e. $$w$$ does still have a conjugate pair $$\bar{w}$$, but it exists split into two different pots). But doesn't this remind you of something? Indeed, we can use the method shown in the image to OBTAIN $$\bar{w}$$ if it happens to be split into multiple pots.

-checking logic and rewriting, is still a work in progress-

• Is irrationality really a necessary consideration here? Every amount of honey can be represented as a rational number, since atoms are discrete entities. How would you get an irrational amount of honey? Commented Sep 28, 2015 at 16:53
• @MarkPeters as far as I know, the honey atom hasn't been isolated yet ;-) Also, the puzzle is tagged as math so irrationality should be fair game (I draw the line at complex amounts of honey though). Truth be told, I've been considering irrationality as well and I still think I may have a counterexample... Commented Sep 28, 2015 at 17:10

I'm in two minds about this. The above arguments sound convincing, but it seems that if we simply go through the $\binom{27}{7}$ possibilities taking $x_i$ for $i\in(1,\ldots,888 030)$ out of the relevant pots, we have 27 equations with 888030 unknowns. Okay, there's a constraint of $x_i\geq0$, which is concerning, but $888030\gg27$.