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This is a homework Question and has to do with Pigeonhole principle. Could use a hint.

Q. The numbers ${0,1,2,.....9}$ are randomly assigned to the vertices ${x_0,x_1,...x_9}$ of a decagon. Show that there are 3 consecutive vertices whose sum is at least 14.

(hint given by Prof: Consider the numbers $S_0=x_0+x_1+x_2,$ $S_1=x_1+x_2+x_3,$ $ ...,$ $S_9=x_9+x_0+x_1$)

I would like to solve it but I don't even have a sense of direction on how to start with this.

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HINT: Following the suggestion, let $S_0=x_0+x_1+x_2$, $S_1=x_1+x_2+x_3$, and so on up through $S_9=x_9+x_0+x_1$. The numbers $S_0,\dots,S_9$ are the sums of all ten of the possible sets of three adjacent numbers. Suppose that none of them is $14$ or more, i.e, that $S_k\le 13$ for $k=0,1,\dots,9$. Then

$$S_0+S_1+S_2+\ldots+S_9\le 10\cdot13=130\;.$$

  1. What is $S_0+S_1+S_2+\ldots+S_9$ algebraically, i.e., in terms of the numbers $x_0,x_1,\dots,x_9$?
  2. What is $x_0+x_1+\ldots+x_9$? In view of (1), what does that tell you about $S_0+\ldots+S_9$?
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  • $\begingroup$ so for 1. $S_0 + S_1..S_9$ algebraically in terms of $x_0,x_1,..,x_9$ is $3(x_0+x_1+x_2...+x_9)+$ For 2) $x_0+x_1+..x_9=45$ as i added values $0+1..+9=45$ , so $S_0+S_1..S_9$ is in between $45$ and $130$. $\endgroup$ – Kj Tada Jul 5 '13 at 9:41
  • $\begingroup$ @KjTada: (1) Almost: it’s $3(x_0+x_1+\ldots+x_9)$. I suspect that you missed the $x_0$ in $S_8=x_8+x_9+x_0$. (2) You have the pieces, but you’ve not seen how they fit together. With (1) corrected, you know that $$S_0+\ldots+S_9=3(x_0+\ldots+x_9)=3\cdot45=135\;.$$ Is this possible if all of the sums $S_k$ are at most $13$? $\endgroup$ – Brian M. Scott Jul 5 '13 at 9:46
  • $\begingroup$ thanks Brian, no by Pigeon hole then there must be at least one $S_k$ whose sum is greater than 13. $\endgroup$ – Kj Tada Jul 5 '13 at 9:50
  • $\begingroup$ @KjTada: You’ve got it. You're welcome! $\endgroup$ – Brian M. Scott Jul 5 '13 at 9:51
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As an alternative answer to this, let's use Dijkstra's generalization of the pigeonhole principle: for any set or bag (multiset), $\textit{average} \le \textit{maximum}$. (See EWD980, EWD1094; see also section 16.4 in Gries & Schneider, "A Logical Approach to Discrete math".)

For the pigeonhole principle, the key choices are: what are the pigeons, and what are the holes? For Dijkstra's generalization, the key choice is: what is the bag?

In this case we need to prove that there is a "3-sum" (i.e., the sum of the numbers assigned to 3 consecutive vertices) that is at least 14, or in other words, that the maximum 3-sum is at least 14.


Proof. The above suggests to look at the bag of all 3-sums, and investigate its average.

To calculate the average we need to know the sum of all 3-sums and the number of 3-sums. Since each vertex occurs in exactly 3 3-sums, the sum of all 3-sums is 3 times the sum of all vertex numbers, or $3 \times (0 + 1 + \dots + 9)$, or 135. The number of 3-sums is obviously 10. Therefore the average 3-sum is 135/10 or 13.5.

Now by the generalized principle ($\textit{average} \le \textit{maximum}$) it follows that the maximum 3-sum is at least 13.5. And since all 3-sums are integers (since each vertex number is an integer) we can round up, and the maximum 3-sum is at least 14.

Therefore we have proven that there is a 3-sum that is at least 14.

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