Literally every time I'm serving some soup I'm thinking of this little mathematical problem I devised.

Imagine you have a very large (= infinite, for the purposes of the actual problem) bowl of soup and want to divide it into two halves using a (finite) ladle. The trouble is that some of the good stuff floated to the surface and every scoop takes a part $p$ of it, $p < \frac12$ (so its amount goes down geometrically with a quotient of $1-p$). The other good stuff is evenly dissolved in the volume. The protocol is that you always take a full scoop in this manner, so the only freedom is in what order do you empty the ladle into $A$ or $B$'s plate. What order would you take so that, asymptotically, both parties get fair shares in both respects?

Obviously the answer can't be $ABABAB\ldots$, because $A$ would always get $1/(1-p)$-times more than $B$. $\overline{ABBA}$ does not work either because $1+(1-p)^{-3}$ is always greater than $(1-p)+(1-p)^2$. In fact (except for a countable number of special values of $p$ (*)) it can't be any periodic infinite word of $A$ and $B$ for a slight variant of the same reason. One more thing I found is that the greedy algorithm (go $A$ until it has more than $B$, then swap) won't work either because while it would always satisfy splitting of the geometrical part, it would break the linear one (number of scoops). That's all I know at the moment.

Rephrased: for a given $1-p =: q \in (\frac12, 1)$, find a subset $M ∈ \mathbb{N}$ such that both

  • $\displaystyle\lim_{N→∞} \frac{\#\left( M \cap \{1,\ldots,N\} \right)}{N} = \frac12$,
  • $\displaystyle\sum_{n=1}^∞ q^n \chi_M(n) = \frac12 \sum_{n=1}^∞ q^n$.

I'm pretty sure the answer will lie somewhere in the theory of non-integer base systems or combinatorics on words or both, but I know little of those.

(*) For example, for $q = φ - 1$ the equality $q + q^2 + q^3 = 1 + q^4 + q^5$ holds, so for $p = 2 - φ \approx 0.381966$ the sequence $\overline{ABBBAA}$ would work.

Alex Ravsky
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The Vee
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  • If, say, $p=1/3$, then the infinite series adds up to $1/2$, so you want each plate to get $1/4$, but the first ladle has more than that, as it has $1/3$, so there's no way to do what you want. A little arithmetic shows that this argument goes through for all $p<1/2$. – Gerry Myerson Jan 21 '21 at 07:52
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    I think you can show that as $p \to 0$ the answer converges to the Thue-Morse sequence: https://en.wikipedia.org/wiki/Thue%E2%80%93Morse_sequence#Equitable_sequencing – Qiaochu Yuan Jan 21 '21 at 07:57
  • @GerryMyerson Fixed, my apologies for the seriously broken first version. – The Vee Jan 21 '21 at 08:04
  • @QiaochuYuan That's true, and certainly a great observation for where to start looking, but unfortunately it won't work outside of that limit, for a fixed $p$. – The Vee Jan 21 '21 at 08:24
  • I've tried to think about this first as a finite bowl of soup, intending to consider the limit as the bowl gets larger. My problem is that in this situation a geometric model is not realistic, because the soup should start to get more concentrated and the chance of scooping up some good stuff should start to increase. In particular, the last scoop takes all of the remaining good stuff. – tomi Jan 21 '21 at 08:35
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    Lol i don't know how to solve this but a very nice question! – dezdichado Jan 24 '21 at 19:50
  • I am writing a paper and going to give a talk on my partial solutions of this problem at [Western Ukrainian online mathematical seminar](http://www.math.lviv.ua/seminar/index.php) at Monday, February 1, at 15:05 [GMT+2](https://time.is/en/GMT+2). I am going to provide a link to the seminar later. – Alex Ravsky Jan 29 '21 at 12:48
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    @AlexRavsky That's amazing!! I'm honoured and I'd love to see the seminar. Only there's a realistic possibility I will be offline that day :-( I suppose you wouldn't consider taking a recording? – The Vee Jan 29 '21 at 18:36
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    I am going to ask my colleague to record my talk and then to put it at Youtube. – Alex Ravsky Jan 29 '21 at 22:25
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    The paper is [here](https://mega.nz/file/M1RWERwL#-xleLdm6gU_YZ00r6k7BgK3sjPsfbhdGxCFP-3KkBgg). – Alex Ravsky Jan 30 '21 at 21:15
  • Your comments, remarks, suggestions etc. are welcome. – Alex Ravsky Jan 31 '21 at 05:12

4 Answers4


Proposition. Let $(a_n)$ be an absolutely convergent series of real numbers such that for any natural $k$ we have $$|a_{2k-1}-a_{2k}|\le \sum_{n=k+1}^\infty |a_{2n-1}-a_{2n}|.\label{1}\tag{1}$$ Then there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} a_n=\tfrac 12\sum_{n=1}^\infty a_n$.

Proof. For each natural $k$ put $b_k=|a_{2k-1}-a_{2k}|$. It suffices to show that we can consecutively choose signs for $b_k$, providing $\sum_{k=1}^\infty \pm b_k=0$. At the beginning choose a sign “$+$” for $b_1$ and for each $k>1$ choose for $b_k$ a sign “$+$”, if $\sum_{i=1}^{k-1} \pm b_i\ge 0$ and “$-$”, otherwise. It is easy to check that Condition \eqref{1} provides $\sum_{k=1}^\infty \pm b_k=0$. $\square$

Corollary. For each $q\in (1/\sqrt{2},1)$ there exists a subset $M$ of natural numbers such that $|M\cap \{2k-1,2k\}|=1$ for each natural $k$ and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. It suffices to check that for any any natural $k$ we have $$q^{2k-1}-q^{2k}\le \sum_{n=k+1}^\infty q^{2n-1}-q^{2n}=\frac{q^{2k+1}}{1+q}. \square$$

Alex Ravsky
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  • Very nice!! Do you think this idea could be extended to $q \in (1/2,1/\sqrt2]$, too, perhaps by taking larger chunks than pairs? – The Vee Jan 25 '21 at 10:07
  • @TheVee I tried to do this (basically for fours), but a straightforward extension does not work and my tries suggest to look for other ideas. – Alex Ravsky Jan 25 '21 at 11:09
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    I would echo @Alex Ravsky. I have spent my time trying the blocks but the 'remaining terms' became 'small' too quickly. However, I was concentrating on getting formulae using GPs which was perhaps not a good idea although it worked well for larger $q$s. –  Jan 25 '21 at 11:25

I wrote a short paper devoted to your problem.

Solving it, I introduced the following notion, considered in a separate question. A number $q\in (1/2, 1)$ is approximating, if there exist non-negative numbers $A$ and $N$ such that for each $x_0\in [0,A]$ there exist $n\le N$ and a polynomial $P(x)$ of the form $\sum_{i=1}^n \pm x^i$ such that $P(1)=0$ and $|x_0-P(q)|\le Aq^n$.

Proposition. For each approximating number $q$ there exists a subset $M$ of natural numbers such that $\lim_{N\to\infty} |M \cap \{1,\ldots, N\} |/N = 1/2$, and $\sum_{n\in M} q^n=\tfrac 12\sum_{n=1}^\infty q^n$.

Proof. Let the number $q$ is approximating with the constants $A$ and $N$. Put $k_0=0$. Then we can inductively build an increasing sequence $(k_n)$ of natural numbers such that $k_{n+1}-k_n\le N$ for each $n$, assigning signs to numbers $q^i$ for $k_{n-1}<i\le k_n$ (a half of the assigned signs are “$+$” and the other half are “$-$”) assuring $\sum_{i=1}^{k_n} \pm q^i\le Aq^{k_n}$. Let $M$ be the set of natural $n$ such that $q^n$ has “$+$” sign. $\square$

In the paper is shown that each $q\in (q_\infty,1)$ is approximating, where $q_\infty=0.5845751\dots$ is a unique positive root of a polynomial $Q_\infty(x)=x^4+x^3+2x^2-1$.

Alex Ravsky
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    Here's the bounty as this is the closest we could get to this point, great job and thanks for the follow-up! I'll leave accepting an answer a few more days' time. – The Vee Jan 29 '21 at 19:03
  • @TheVee I suggest not to accept my answers in order to indicate that despite partial progress the problem is still open. – Alex Ravsky Jan 30 '21 at 11:00

There is an uncountable set of $q\in (1/2,1)$ for which there are only two sets $M$ (which are complements of each other) such that $\sum_{m\in M}q^m=\frac{1}2\cdot \frac{1}{1-q}$ - and uncountably many of these forced sets are not of asymptotic density $1/2$.

To prove this, let us define, for any subset $M$ of the natural numbers, the summing function $$f_M(q)=\frac{\sum_{m\in M}q^m}{\sum_{n\in\mathbb N}q^n}=(1-q)\cdot \sum_{m\in M}q^m$$ Note that for any $M$, the function $f_M$ is continuous where $|q|<1$ - and, in fact, the family of such functions is equicontinuous on this range.

Let's also define the left shift operator $$L(M)=\{m' \in \mathbb N : m'+1\in M\}$$ and note that $$f_{M}(q) = (1-q)\cdot \chi_{M}(0) + qf_{L(M)}(q)$$ where $\chi_M$ is the indicator function of $M$.

Let's say a set $M$ has $n$-bounded runs if there are no values $k$ such that $k\in M$ but $k+1,k+2,\ldots,k+n$ are all not in $M$ or such that $k\not\in M$ and the following $n$ integers are all in $M$.

We can now state a lemma:

Lemma: For any $n$, there is some $\varepsilon > 0$ such that if $M$ has $n$-bounded runs, then for all $q\in[1/2,1/2+\varepsilon)$ and all $M'\subseteq \mathbb N$ we have $f_M(q)=f_{M'}(q)$ implies $M=M'$.

Proof. First, we need to choose $\varepsilon$ so that $\left|f_{M}(q) - \frac{1}2\right| > \varepsilon$ for all sets $M$ with $n$-bounded runs and all $q\in [1/2,1/2+\varepsilon)$ - this is possible since $|f_M(1/2)-\frac{1}2|$ can never be less than $\frac{1}{2^{n+1}}$ for such a set, considering binary representations, and the family of functions $f$ is equicontinuous, so the values $|f_M(q)-\frac{1}2|$ must remain bounded away from zero for all $q$ close enough to $\frac{1}2$ - which allows us to choose a $\varepsilon$ satisfying our condition. Note that this, more helpfully, ensures that $f_{M}(q)$ cannot ever be in $[1-q, q]$.

One can now fix any $M$ with $n$-bounded runs and any $M'$ with $f_M(q)=f_{M'}(q)$ for some $q\in [1/2,1/2+\varepsilon)$ and prove inductively that $M=M'$. Consider the equation: $$f_M(q)=f_{M'}(q)$$ $$(1-q)\chi_{M}(0)+qf_{L(M)}(q) = (1-q)\chi_{M'}(0)+qf_{L(M')}(q)$$ Note, however, that this expression is either less than $1-q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=0$) or greater than $q$ (in which case it must be that $\chi_{M}(0)=\chi_{M'}(0)=1$) by choice of $\varepsilon$. Thus, $0$ is either in both or neither of the sets. However, this also immediately then simplifies as: $$f_{L(M)}(q)=f_{L(M')}(q)$$ which then allows us to repeat this argument inductively to see that $M=M'$, proving the lemma.

Then we finish with another lemma:

Lemma 2: Let $M$ be a non-empty set with $\ell$-bounded runs such that $0\not\in M$. Define $M_n = \{0\} \cup (M + n + 1)$ to be a right shift of $M$ with an initial element appended. For all large enough $n$, there exists some $q\in [1/2,1]$ such that if $f_{M'}(q)=1/2$ and $0\in M'$, then $M'=M_n$.

Proof. Note that $L(M_n)$ is always a set with $\ell$-bounded runs since the initial large run of elements not in the set is not preceded by an element that is in our set, as our definition would require - and all other runs must be in $M$ itself. Moreover, note $$f_{M_n}(q)=(1-q) + q^{n+1}f_{M}(q).$$ Note, in particular, that $f_{M_n}(1/2)\geq \frac{1}2$ and that if $q^{n+1} \leq q - \frac{1}2$, then $f_{M_n}(q)\leq \frac{1}2$ - and note that, for any $\varepsilon>0$ there is some large enough $n$ so that this is true of some $q$ in $(1/2,1/2+\varepsilon)$. In particular, suppose we choose some $\varepsilon$ so that Lemma 1 applies to sets with $\ell$-bounded runs with that $\varepsilon$ and choose $n$ big enough so that $q_0^{n+1}\leq q_0 - \frac{1}2$ for some $q_0\in (1/2,1/2+\varepsilon)$. Then, by the intermediate value theorem, there has to be some $q\in [1/2,q_0]$ such that $f_{M_n}(q) = \frac{1}2$.

However, if we let $M'$ be such that $f_{M'}(q)=\frac{1}2$ and $0\in M'$, we can then realize that $$f_{L(M')}(q) = f_{L(M_n)}(q)$$ and apply Lemma 1 to see that $L(M')=L(M_n)$, which then, since both $M'$ and $M_n$ contain zero, implies that $M'=M_n$, completing Lemma 2.

To then reach our end goal: fix any pair $(q,M)$ with the property from Lemma 2 that if for any $M'$ with $0\in M'$ we have $f_{M'}(q)=\frac{1}2$ implies $M=M'$. Note that if we had some arbitrary set $M''$ with $f_{M''}(q)=\frac{1}2$, we would also have that $f(\mathbb N\setminus M'')=\frac{1}2$ - and either $M''$ or its complement must contain $0$ - thus must equal $M$.

Together, this gives us a method to construct a $q$ for which fairly splitting soup is impossible from any set with bounded runs but not with asymptotic density $1/2$. This gives an uncountable set of $q$ for which fairly splitting soup is not possible. Note that if you choose the set $M$ to be have a periodic pattern, the produced counterexamples will be algebraic.

Milo Brandt
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Let $\{a_i\}$ be a G.P. with ratio $r\ge \frac{1}{2}$and sum $S$. Then, for any number $T$ in $(0,S]$ there is a subsequence $\{b_i\}$ of $\{a_i\}$ with sum $T$.

Proof. Let $b_1=a_m$ where $m$ is minimal such that $a_m\le T$. Then $$T<a_{m-1}=\frac{a_m}{r}\le \frac{a_m}{1-r}=a_m+\frac{a_{m+1}}{1-r}.$$ Therefore the initial conditions are reproduced for $T-a_m$ and $\{a_i\}_{i>m}$ and we can now successively choose $b_2, b_3,...$

Soup for $q\ge \frac{1}{\sqrt2}$

Let $r=q^2$. From the above result we can choose a set $I$ of positive integers such that $$\sum {{q^{2i}}_{i\in I}}= \frac{1}{2(1-q^2)}.$$

Then, for the sequence of As and Bs, place the As in positions $2i-1$ if $i\in I$ and in position $2i$ otherwise.

The absolute differences between the amount of 'good' given to A and B in successive pairs are then proportional to $1,q^2,q^4, ...$ and, by the definition of $I$, the total signed difference is zero.