Rectangle Transformation Problem

Shaojiang Wang, Kun He, Yicheng Pan, Mingji Xia

Abstract: In this paper, we propose the rectangle transformation problem (RTP) and its variants. RTP asks for rectangle partitions on two rectangles of the same area which produce two identical sets of pieces. We are interested in the minimum RTP which requires to minimize the partition size. This initiates the algorithmic study of dissection problems in module number optimization, particularly in the category of rectangle partition. We mainly focus on the strict rectangle transformation problem (SRTP) in which rotation is not allowed during the transformation. It has been shown that SRTP has no finite solution if the ratio of the two parallel side lengths of input rectangles is irrational. So we turn to its complemental case, SRTP with integral input, denoted by SIRTP, in which case both side lengths are assumed integral. We give a polynomial time algorithm $$\text {ALGSIRTP}$$ ALGSIRTP which gives a solution at most $$q/p+7\log _2 p$$ q / p + 7 log 2 p to $$\text {SIRTP}(p,q)$$ SIRTP ( p , q ) ( $$q\ge p$$ q ≥ p ), where p and q are two integral side lengths of input rectangles $$p\times q$$ p × q and $$q\times p$$ q × p . Note that q / p is an intrinsic lower bound for $$\text {SIRTP}(p,q)$$ SIRTP ( p , q ) . So $$\text {ALGSIRTP}$$ ALGSIRTP is a $$(7\log p)$$ ( 7 log p ) -approximation algorithm for minimum $$\text {SIRTP}(p,q)$$ SIRTP ( p , q ) . On the other hand, we show that for any $$\varepsilon >0$$ ε > 0 and any constant range $$(1,1+\delta )$$ ( 1 , 1 + δ ) , there are integers p and q ( $$q>p$$ q > p ) of ratio q / p in this range, such that there is no solution less than $$\max \{q/p,\log _2^{1-\varepsilon } q\}$$ max { q / p , log 2 1 - ε q } to $$\text {SIRTP}(p,q)$$ SIRTP ( p , q ) . This is an almost tight bound since the algorithm $$\text {ALGSIRTP}$$ ALGSIRTP gives an upper bound $$7\log _2 p+O(1)$$ 7 log 2 p + O ( 1 ) in this case. We also raise a long series of open questions for further research along this line.