USD Contest - Fall 1


2017-09-14 00:00 CEST

USD Contest - Fall 1


2017-09-21 00:00 CEST
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Problem B
Best Compromise

Photo by Andy Nelson

There is a saying “a good compromise makes no one happy”. As distressing as that may be, I guess most people still think it is a good idea to seek the best agreement. The problem though, is to define the measure in which the word “best” gets a meaning. Let us consider $n$ people, each with a firm belief on $m$ different yes/no issues. We may think of the people’s beliefs as binary strings $b_ i$ of length $m$ for $0<i\leq n$, i.e. one string per person, with entries reflecting a person’s belief in each of the $m$ issues. An agreement is also a binary string $a$ of length $m$, indicating the agreed outcome of each of the $m$ issues. Some would argue that the agreement $a$ that minimises the maximum of $H(b_ i-a)$ over all $i$, where $H(x)$ is the Hamming function counting the number of non-zero entries of the vector $x$, is the best compromise. Unfortunately, it is widely believed that finding this agreement is computationally infeasible for $n$ and $m$ large. Another reasonable suggestion is the agreement $a$ that minimises $\left(\sum _{i=1}^ n H(b_ i-a)^ p\right)^{1/p}$ for some positive integer $p$. Note in particular that when $p\rightarrow \infty $, this measure coincides with the former. I for one am most content with the choice of $p=1$ though, don’t you agree?


On the first line of input is a single positive integer $t$, telling the number of test scenarios to follow. Each test scenario is described by two positive integers $n\leq 100$ and $m\leq 100$, on a line of their own. Then follow $n$ lines, each containing a binary string $b_ i$ on $m$ ‘0’ and ‘1’ characters.


For each test scenario, output the assignment $a$ that minimises the measure above with $p = 1$, on a row of its own. If there are several solutions, output anyone.

Sample Input 1 Sample Output 1
5 5
1 7