Unique way to partition into two parts of equal weight - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:11:41Z http://mathoverflow.net/feeds/question/22745 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22745/unique-way-to-partition-into-two-parts-of-equal-weight Unique way to partition into two parts of equal weight Ewan Delanoy 2010-04-27T16:46:30Z 2010-04-28T11:13:57Z <p>A special case says it all ... Let $w_1 &lt; w_2 &lt; \ldots &lt; w_{12}$ be an increasing sequence of $12$ integers ("weights") such that the total weight $W=\sum_{k=1}^{12}w_k$ is even. </p> <p>Say that $I \subseteq \lbrace 1,2, \ldots ,12 \rbrace$ is an exact subset iff the sum $\sum_{k \in I}w_k$ equals $\frac{W}{2}$. My question is : is there a sequence for which $\lbrace 1,2,5,7,10,12 \rbrace$ is exact and is the only exact subset (up to complementation) ? </p> http://mathoverflow.net/questions/22745/unique-way-to-partition-into-two-parts-of-equal-weight/22764#22764 Answer by Tony Huynh for Unique way to partition into two parts of equal weight Tony Huynh 2010-04-27T19:06:27Z 2010-04-27T19:25:04Z <p>I believe the answer is <strong>yes</strong>. Consider the sequence </p> <p>100, 200, 201, 202, 500, 601, 700, 701, 801, 1000, 1194, 1200.</p> <p>It is easy to see that the set indexed by {1,2,5,7,10,12} is the unique exact subset (up to complementation).</p> http://mathoverflow.net/questions/22745/unique-way-to-partition-into-two-parts-of-equal-weight/22833#22833 Answer by damiano for Unique way to partition into two parts of equal weight damiano 2010-04-28T11:13:57Z 2010-04-28T11:13:57Z <p>Just an expansion on my comment. I will assume that exact sequences need to have half the number of indices of the whole sequence. Then a sequence is exact for some choice of weights if and only if it is exact for one weight. (This is already argued, both in my comment and in Tony Huynh's.)</p> <p>The final question to be answered is when is a subsequence exact for some choice of weights. This is again very easy. A subset <em>I</em> of <em>{1,...,2n}</em> of size <em>n</em> obviously determines an increasing bijection $j_I$ between <em>I</em> and its complement. The subset <em>I</em> is an exact subsequence for some choice of weights if and only if the function $i \mapsto j_I(i)-i$ for $i \in I$ does not have constant sign.</p>