,
,
,
112 1. The function ReduceRectangle(R) returns a straight rectangle Rec, such that Rec ? R, and either R \ ,
, The agent traveled a distance of at most 21(P erimeter(R) ? P erimeter(Rec)) during the execution of ReduceRectangle(R)
, All other necessary objects will be defined within the proof. For the notation, refer to Fig. 6.2. Consider the execution of function ReduceRectangle(R) starting at the center p of R
The variable N ewRectangle is set to a straight rectangle R such that R ? R , and either R \ R does not contain ? k (z) or the agent has seen the treasure ,
The inequality P erimeter(R ) ? P erimeter(R ) ? 2 holds ,
The agent traveled a distance of at most 21(P erimeter(R) ? P erimeter(R )) during the execution of function ReduceRectangle(R) ,
, Note that the points defined in lines 4 to 6 (in particular the points A,B, d and e) exist and ABde is a straight rectangle such that ABde ? R in view of the fact that (R , (L 1 , x 1 )) is a basic configuration. Moreover, since z ? (L 1 , x 1 ), we have ? k (z) ? (L 1 , x 1 ). However, edCD ? (L 1 , x 1 ) ? [de] and R \ ABde = edCD, So we have (R \ ABde) ? (L 1 , x 1 ) = ? and Property P1 is satisfied. Property P2 also holds because
, For every integer i ? 3 and for every triangle T of Slicing(i), there is at least one tile t of T iling(4?(i) ? 2), such that t ? T and one side of t is included in a side of S. For the base case i = 3
, included in t , for all integers r < r . Hence H 3 is true. Assume that H j is true for some integer j ? 3 and let us prove that H j+1 is also true. included in L. By the inductive hypothesis, there exists a tile t of T iling(4?(j) ? 2) such that t ? T and one side of t is included in L. For any integers r < r , every tile of T iling(r) contains side length of a tile of T iling(4?(j + 1) ? 2), ? 2 ? 10. Moreover, each side of every tile t of T iling(r) contains at least one side of a tile of T iling(r )
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