New results for the Mondrian art problem
The Mondrian problem consists of dissecting a square of side length n ∈ N into non-congruent rectangles with natural length sides such that the difference d(n) between the largest and the smallest areas of the rectangles partitioning the square is minimum. In this paper, we compute some bounds on d(...
| Authors: | , , |
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| Format: | article |
| Publication Date: | 2021 |
| Country: | España |
| Institution: | Universitat de Lleida (UdL) |
| Repository: | Repositori Obert UdL |
| OAI Identifier: | oai:repositori.udl.cat:10459.1/70756 |
| Online Access: | https://doi.org/10.1016/j.dam.2021.01.016 http://hdl.handle.net/10459.1/70756 |
| Access Level: | Open access |
| Keyword: | Non-congruent rectangles Partition Mondrian problem |
| Summary: | The Mondrian problem consists of dissecting a square of side length n ∈ N into non-congruent rectangles with natural length sides such that the difference d(n) between the largest and the smallest areas of the rectangles partitioning the square is minimum. In this paper, we compute some bounds on d(n) in terms of the number of rectangles of the square partition. These bounds provide us optimal partitions for some values of n ∈ N. We provide a sequence of square partitions such that d(n)/n2 tends to zero for n large enough. For the case of 'perfect' partitions, that is, with d(n) = 0, we show that, for any fixed powers s1, . . . , sm, a square with side length n = p s1 1 · · · p smm , can have a perfect Mondrian partition only if p1 satisfies a given lower bound. Moreover, if n(x) is the number of side lengths x (with n ≤ x) of squares not having a perfect partition, we prove that its 'density' n(x) x is asymptotic to (log(log(x))2 2 log x , which improves previous results. |
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