The boundary of a shape and its classification

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Figure 5. A shape x, its boundary b(x) and two possible neighborhoods ∆(b(x)) coequal to x. Figure 6 illustrates this provision for a line shape to be a subshape of the boundary of a plane segment. Neighborhoods ∆ 1 (l) and ∆ 2 (l) are necessarily disjoint, since ∆1 ( l ) ≤ x ∧ ∆ 2 ( l ) ⋅ x = 0 ⇒ ∆1 ( l ) ⋅ ∆ 2 ( l ) = 0. We note that every shape is a neighborhood of its boundary. We can now state the condition that suffices for a shape to be a subshape of the boundary of a segment: (4) Any subshape of the boundary of a segment is also a subshape of the boundary of any other segment that contains the given segment provided there is a neighborhood coequal and disjoint from the latter segment. ∀ , y ∈ Σ , x ≤ y : ∀l ≤ b( x ) : l ≤ b( y ) ⇔ ∃∆( l ) : ∆( l ) ⋅ y = 0. x seg Figure 6. A subshape l of the boundary b(x) of a segment x and two neighborhoods ∆ 1 (l) and ∆ 2 (l) coequal to x. By comparing a neighborhood of a segment with any shape coequal with this neighborhood it follows that: (5) A segment and each of its neighborhoods can always be partitioned, with respect to any other shape coequal to the neighborhood, into two shapes and corresponding neighborhoods one contained in the second shape and the other disjoint from it. ∀l ∈ Σseg , ∀∆( l ), ∀x ∈ Σ : c ( x ) = c ( ∆( l )) ⇒ ∃∆( m), ∆( l − m) : ∆( l ) = ∆( m) + ∆( l − m) ∧ ∆( m ) ≤ x ∧ ∆( l − m) ⋅ x = 0. Figure 7 illustrates this proposition for a pair of plane segments.
Figure 7. Two plane segments x and y and subshape l of the boundary of x, with the three possible partitions of l and x with respect to y: (i) m = 0, (ii) m ≠ 0 ∧ m ≠ l, (iii) m = l. Consider a shape x {x 1 , …, x n }. When n = 1, b(x) = b(x 1 ). When n = 2, consider a segment l ≤ ( b( x 1 ) − b( x 2 )) ⇒ l ≤ b( x 1 ) . There are two coequal disjoint neighborhoods ∆ 1 (l) and ∆ 2 (l) with ∆ 1 (l) ≤ x 1 ≤ x = x 1 + x 2 and ∆ 2 (l) · x 1 = 0. By (5), partition ∆ 2 (l) with respect to x 2 into neighborhoods ∆ 2 (m) and ∆ 2 (l – m) with m ≤ l, ∆ 2 (m) ≤ x 2 and ∆ 2 (l – m) · x 2 = 0. ∆ 2 (l – m) ≤ ∆ 2 (l) and ∆ 2 (l – m) · x 1 = 0 imply that ∆ 2 (l – m) · x = 0. Further, since ∆ 2 (m) ≤ ∆ 2 (l), then ∆ 2 (m) · x 1 = 0. Given m · b(x 2 ) = 0 and ∆ 2 (m) ≤ x 2 , consider a shape neighborhood ∆*(m) with ∆*(m) · ∆ 2 (m) = 0 and ∆*(m) ≤ x 2 . Provision (2) specifies that ∆*(m) · x 1 ≠ 0, which implies that x 1 · x 2 ≠ 0, in turn, contradicts the assumption that x 1 and x 2 are disjoint. Therefore, m = 0 and ∆ 2 (l – m) is a neighborhood of l, or ∆ 2 (l – m) = ∆(l). Given ∆ 1 (l) ≤ x and ∆(l) · x = 0, it follows by provision (3) that l ≤ b(x). Likewise, for a segment l ≤ ( b( x 2 ) − b( x 1 )) ⇒ l ≤ b ( x 2 ) . Whence, (b(x 1 ) – b(x 2 )) + (b(x 2 ) – b(x 1 )) = b(x 1 ) ⊕ b(x 2 ) ≤ b(x). Consider a segment l ≤ b(x). There are two coequal disjoint neighborhoods ∆ 1 (l) and ∆ 2 (l) with ∆ 1 (l) ≤ x and ∆ 2 (l) · x = 0. By (5), partition ∆ 1 (l) with respect to x 1 into neighborhoods ∆ 1 (m) and ∆ 1 (l – m) with m ≤ l, ∆ 1 (m) ≤ x 1 and ∆ 1 (l – m) · x 1 = 0 and, therefore, ∆ 1 (l – m) ≤ x 2 . Consider ∆ 2 (m) ≤ ∆ 2 (l); since ∆ 2 (l) · x = 0 it follows that ∆ 2 (m) · x 1 = 0 and, by (3), m ≤ b(x 1 ). Consider ∆ 2 (l – m) ≤ ∆ 2 (l); since ∆ 2 (l) · x = 0 it follows that ∆ 2 (l – m) · x 2 = 0 and, by (3), l – m ≤ b(x 2 ). Thus, b(x) ≤ b(x 1 ) + b(x 2 ). Lastly, consider a segment l ≤ ( b( x 1) ⋅ b ( x 2 )) ⇒ l ≤ b( x 1) , There are two coequal disjoint neighborhoods ∆ 1 (l) and ∆ 2 (l) with ∆ 1 (l) ≤ x 1 and ∆ 2 (l) · x 1 = 0. By (5), partition ∆ 2 (l) with respect to x 2 into ∆ 2 (m) and ∆ 2 (l – m), with m ≤ l, ∆ 2 (m) ≤ x 2 and ∆ 2 (l – m) · x 2 = 0. Consider ∆ 1 (m) ≤ ∆ 1 (l) ≤ x 1 ; since ∆ 2 (m) ≤ x 2 and x 1 · x 2 = 0, it follows from the negation of (2) and (3) that ∆ 1 (m) ≤ x, ∆ 2 (m) ≤ x and ∆ 1 (m) · ∆ 2 (m) = 0 and, therefore, m · b(x) = 0. Since ∆ 1 (l) ≤ x 1 and x 1 · x 2 = 0, it follows that ∆ 1 (l) · x 2 = 0. Consider ∆ 1 (l – m) ≤ ∆ 1 (l) ≤ x 1 ; then, ∆ 1 (l – m) · x 2 = 0, ∆ 2 (l – m) · x 2 = 0 and ∆ 1 (l – m) · ∆ 2 (l – m) = 0. Again, from the negation of (2) and (3) it follows that l − m ≤/ b( x 2 ) , which contradicts l ≤ (b(x 1 ) · b(x 2 )). Thus, m = l and l · b(x) = 0 and, therefore, (b(x 1 ) · b(x 2 )) · b(x) = 0. Whence, b(x) ≤ b(x 1 ) + b(x 2 ) – (b(x 1 ) · b(x 2 )) = b(x 1 ) ⊕ b(x 2 ) and, thus, b(x) = b(x 1 ) ⊕ b(x 2 ).
Page 1 and 2: The boundary of a shape and its cla
Page 3 and 4: such that y ≤ x and b(y) ≤ b(x)
Page 5: within the proximity of the segment
Page 9 and 10: Consider a segment l ≤ b(x · y).
Page 11 and 12: which contradicts the definition of
Page 13 and 14: Let x denote the boundary of a segm
Page 15 and 16: 0, or ∆ 2 (l - m) · Γ(s ) i = 0
Page 17 and 18: way if and only if it is shared bet
Page 19 and 20: Figure 18. The boundary segments of
Page 21 and 22: Figure 20. Inner and outer neighbor
Page 23 and 24: Figure 21. Examples showing that ea
Page 25 and 26: SUM (x, y) 1 (I x , I y , M, N, O x
Page 27 and 28: Acknowledgement The authors would l

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