Subfield: Combinatorics


Question: 
Let $n$ be a nonnegative integer. A grasshopper jumps along the real axis. He starts at point 0 and makes $n+1$ jumps to the right with pairwise different positive integral lengths $a_{1}, a_{2}, \ldots, a_{n+1}$ in an arbitrary order. Let $M$ be a set of $n$ positive integers in the interval $(0, s)$, where $s=a_{1}+a_{2}+\cdots+a_{n+1}$. Prove that the grasshopper can arrange his jumps in such a way that he never lands on a point from $M$.

Solution:
First of all we remark that the statement in the problem implies a strengthening of itself: Instead of $|M|=n$ it is sufficient to suppose that $|M \cap(0, s-\bar{a}]| \leq n$, where $\bar{a}=\min \left\{a_{1}, a_{2}, \ldots, a_{n+1}\right\}$. This fact will be used in the proof.

We prove the statement by induction on $n$. The case $n=0$ is obvious. Let $n>0$ and let the assertion be true for all nonnegative integers less than $n$. Moreover let $a_{1}, a_{2}, \ldots, a_{n+1}, s$ and $M$ be given as in the problem. Without loss of generality we may assume that $a_{n+1}<a_{n}<$ $\cdots<a_{2}<a_{1}$. Set

$$
T_{k}=\sum_{i=1}^{k} a_{i} \quad \text { for } k=0,1, \ldots, n+1
$$

Note that $0=T_{0}<T_{1}<\cdots<T_{n+1}=s$. We will make use of the induction hypothesis as follows:

Claim 1. It suffices to show that for some $m \in\{1,2, \ldots, n+1\}$ the grasshopper is able to do at least $m$ jumps without landing on a point of $M$ and, in addition, after these $m$ jumps he has jumped over at least $m$ points of $M$.

Proof. Note that $m=n+1$ is impossible by $|M|=n$. Now set $n^{\prime}=n-m$. Then $0 \leq n^{\prime}<n$. The remaining $n^{\prime}+1$ jumps can be carried out without landing on one of the remaining at most $n^{\prime}$ forbidden points by the induction hypothesis together with a shift of the origin. This proves the claim.

An integer $k \in\{1,2, \ldots, n+1\}$ is called smooth, if the grasshopper is able to do $k$ jumps with the lengths $a_{1}, a_{2}, \ldots, a_{k}$ in such a way that he never lands on a point of $M$ except for the very last jump, when he may land on a point of $M$.

Obviously, 1 is smooth. Thus there is a largest number $k^{*}$, such that all the numbers $1,2, \ldots, k^{*}$ are smooth. If $k^{*}=n+1$, the proof is complete. In the following let $k^{*} \leq n$.

Claim 2. We have

$$
T_{k^{*}} \in M \quad \text { and } \quad\left|M \cap\left(0, T_{k^{*}}\right)\right| \geq k^{*} \text {. }
$$

Proof. In the case $T_{k^{*}} \notin M$ any sequence of jumps that verifies the smoothness of $k^{*}$ can be extended by appending $a_{k^{*}+1}$, which is a contradiction to the maximality of $k^{*}$. Therefore we have $T_{k^{*}} \in M$. If $\left|M \cap\left(0, T_{k^{*}}\right)\right|<k^{*}$, there exists an $l \in\left\{1,2, \ldots, k^{*}\right\}$ with $T_{k^{*}+1}-a_{l} \notin M$. By the induction hypothesis with $k^{*}-1$ instead of $n$, the grasshopper is able to reach $T_{k^{*}+1}-a_{l}$



with $k^{*}$ jumps of lengths from $\left\{a_{1}, a_{2}, \ldots, a_{k^{*}+1}\right\} \backslash\left\{a_{l}\right\}$ without landing on any point of $M$. Therefore $k^{*}+1$ is also smooth, which is a contradiction to the maximality of $k^{*}$. Thus Claim 2 is proved.

Now, by Claim 2 , there exists a smallest integer $\bar{k} \in\left\{1,2, \ldots, k^{*}\right\}$ with

$$
T_{\bar{k}} \in M \quad \text { and } \quad\left|M \cap\left(0, T_{\bar{k}}\right)\right| \geq \bar{k} \text {. }
$$

Claim 3. It is sufficient to consider the case

$$
\left|M \cap\left(0, T_{\bar{k}-1}\right]\right| \leq \bar{k}-1
$$

Proof. If $\bar{k}=1$, then (4) is clearly satisfied. In the following let $\bar{k}>1$. If $T_{\bar{k}-1} \in M$, then (4) follows immediately by the minimality of $\bar{k}$. If $T_{\bar{k}-1} \notin M$, by the smoothness of $\bar{k}-1$, we obtain a situation as in Claim 1 with $m=\bar{k}-1$ provided that $\left|M \cap\left(0, T_{\bar{k}-1}\right]\right| \geq \bar{k}-1$. Hence, we may even restrict ourselves to $\left|M \cap\left(0, T_{\bar{k}-1}\right]\right| \leq \bar{k}-2$ in this case and Claim 3 is proved.

Choose an integer $v \geq 0$ with $\left|M \cap\left(0, T_{\bar{k}}\right)\right|=\bar{k}+v$. Let $r_{1}>r_{2}>\cdots>r_{l}$ be exactly those indices $r$ from $\{\bar{k}+1, \bar{k}+2, \ldots, n+1\}$ for which $T_{\bar{k}}+a_{r} \notin M$. Then

$$
n=|M|=\left|M \cap\left(0, T_{\bar{k}}\right)\right|+1+\left|M \cap\left(T_{\bar{k}}, s\right)\right| \geq \bar{k}+v+1+(n+1-\bar{k}-l)
$$

and consequently $l \geq v+2$. Note that

$T_{\bar{k}}+a_{r_{1}}-a_{1}<T_{\bar{k}}+a_{r_{1}}-a_{2}<\cdots<T_{\bar{k}}+a_{r_{1}}-a_{\bar{k}}<T_{\bar{k}}+a_{r_{2}}-a_{\bar{k}}<\cdots<T_{\bar{k}}+a_{r_{v+2}}-a_{\bar{k}}<T_{\bar{k}}$

and that this are $\bar{k}+v+1$ numbers from $\left(0, T_{\bar{k}}\right)$. Therefore we find some $r \in\{\bar{k}+1, \bar{k}+$ $2, \ldots, n+1\}$ and some $s \in\{1,2, \ldots, \bar{k}\}$ with $T_{\bar{k}}+a_{r} \notin M$ and $T_{\bar{k}}+a_{r}-a_{s} \notin M$. Consider the set of jump lengths $B=\left\{a_{1}, a_{2}, \ldots, a_{\bar{k}}, a_{r}\right\} \backslash\left\{a_{s}\right\}$. We have

$$
\sum_{x \in B} x=T_{\bar{k}}+a_{r}-a_{s}
$$

and

$$
T_{\bar{k}}+a_{r}-a_{s}-\min (B)=T_{\bar{k}}-a_{s} \leq T_{\bar{k}-1} .
$$

By (4) and the strengthening, mentioned at the very beginning with $\bar{k}-1$ instead of $n$, the grasshopper is able to reach $T_{\bar{k}}+a_{r}-a_{s}$ by $\bar{k}$ jumps with lengths from $B$ without landing on any point of $M$. From there he is able to jump to $T_{\bar{k}}+a_{r}$ and therefore we reach a situation as in Claim 1 with $m=\bar{k}+1$, which completes the proof.