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,
, Appendix: proofs of theorems and properties XXI Lemma 69. Let ? be an altitude ordering for w and let H be a hierarchy on V such that ?(H) is one-side increasing for ?
, As ?(H) is one-side increasing for ?, by the condition 1 of Definition 18, we can affirm that ? = 0 + 1 + · · · + n ? 1. In order to prove that (V, E ? ) is a MST of (G, ?(H)), we will prove that, for any MST G of (G, ?(H)), the sum of the weight of the edges in G is greater than or equal to ?. Let G be a MST of (G, ?(H)). As G is a MST of (G, ?(H)), by the condition 1 of Lemma 83, we have that G and G have the same quasi-flat zones hierarchy, map ?(H): ? = e?E? ?(H)(e)
, Since the partitions P i and P i?1 are distinct, then there exists a region in P i which is not in P i?1 . Therefore, there is a region X of P i which is composed of a several regions {R 1, n ? 1} is a subset of the range of ?(H)
, the lowest j such that x and y belong to the same region of P j is i. Thus, there exists an edge u = {x
, by Definition 21, there is a hierarchical watershed H w of (G, w) such that H is a flattening of H w . By Theorem 19, there is an altitude ordering ? for w such that ?(H w ) is one-side increasing for ?. Let ? be the altitude ordering for w such that ?(H w ) is one-side increasing for ?, By Lemma, vol.84
, Then, by Lemma 71, there is an edge u = {x, y} such that u is in E \ E(G ) and such that ?(H)(u) is different from the greatest weight among the edges in the path ? from x to y in (G , ?(H)). Let v be an edge of greatest weight in ?. As H is equal to QFZ(G, ?(H)), we may affirm that ?(H)(u) is lower than ?(H)(v) because, otherwise, the vertices x and y would be connected in the ?-level set of (G, ?(H)) for a ? lower than ?(H)(u), which contradicts the fact that ?(H) is a saliency map. Hence, we have ?(H)(u) < ?(H)(v). Then, by Lemma 73, as H is a flattening of H w
, As H w is one-side increasing for ?, by the second condition of Definition 18, for any watershed-cut edge u = {x, y} for ?, we have ?(H w )(u) = 0. Then, for any partition P of H w , x and y belong to the same region of P. Therefore, as any partition of H is a partition of H w , we can say that, for any partition P of H, x and y belong to the same region of P. Hence, the lowest ? such
, We will now prove the third condition for H to be a flattened hierarchical watershed of (G, w), XXIV Appendix: proofs of theorems and properties
, Let u be an edge in E ? and let R be the child of R u such that ?(H w )(u) ? ?{?(H w )(v) | R v ? R}, there exists a child R of R u such that ?(H w )(u) ? ?{?(H w )(v)
, Lemma 75. Let H be a hierarchy on V and let ? be an altitude ordering for w such that: 1. (V, E ? ) is a MST of
, Then H is a flattened hierarchical watershed of (G, w)
, In order to prove Lemma 75, we first state two auxiliary lemmas. From Property 10 of [27], we can deduce the following lemma linking binary partition hierarchies and MSTs
, Let B be a binary partition hierarchy of (G, w), Lemma, vol.76
, By Property 12 of [27] linking hierarchical watersheds and hierarchies induced by an altitude ordering and a sequence of minima, and by Lemma 83, we infer the following lemma
, Let G be a MST of (G, w) and let H be a hierarchical watershed of (G , w), Lemma
, Then H is also a hierarchical watershed of (G, w)
, ?(H)); and 2. for edge u in E ? , if u is not a watershed-cut edge for ?, then ?(H)(u) = 0; and 1. By the definition of f , as there are n?1 watershed-cut edges for ?, we can say that, for any i in {1, of Lemma 75. Let H be a hierarchy on V such that there is an altitude ordering ? for w such that: 1. (V, E ? ) is a MST of
, by the definition of f , f (u) is non-zero if and only if u is not a watershed-cut edge for ?, so the statement 2 of Definition 18 holds true for f . XXVI Appendix: proofs of theorems and properties
, no minimum of w included in X. Hence, none of the building edges of the descendants of X is a watershed-cut edge for ?. By the definition of f , we have f (u) = 0 and, for any edge v such that R v ? X, we have f (v) = 0. Hence, f (u) ? ?{f (v) such that R v is included in X}. Otherwise, let us assume that u is a watershed-cut edge for ?. Then there is at least one minimum of w included in each child of R u . By the hypothesis 3, there is a child X of R u such that ?(H)(u) ? ?{?(H)(v) such that R v is included in X}
, Therefore, f (u) ? ?{f (v) such that R v is included in X}. Then, the third condition of Definition 18 holds true for f
, QFZ(G , f ) is a hierarchical watershed of (G , w) (resp. (G, w)). Now, we only need to prove that any partition of H is a partition of QFZ(G , f ), Hence, f is one-side increasing for ? and, as stated previously
, Let G ?,?(H) be the ?-level set of (G , ?(H)). Let ? be the greatest value in {f (u) | u ? E(G ?,?(H) )}. We will prove that the ?-level set of (G , f ) is equal to the ?-level set of (G , ?(H)), }: {?(H)(u)
, ?(H)). Then, ?(H)(u) > ? and, for any edge v in the ?-level set of (G , ?(H)), we have ?(H)(u) > ?(H)(v)
, if v is a watershed-cut edge for ?, then v ? 2 u and f (u) > f (v). Otherwise, if v is not a watershed-cut edge for ?, by the definition of f , we have f (v) = 0 and f (u) > f (v). Thus, for any edge v in the ?-level set of (G , ?(H))
, Therefore, we can conclude that the ?-level set of (G , f ) is equal to the ?-level set 1. { (R) | R is a region of B ? } = {0
, any two distinct minima M 1 and M 2 of w, we have (M 1 ) = (M 2 ); and 3. for any region R of B ? , we have that (R) is equal to ?{ (M ) such that M is a minimum of w
, We will prove that the statements 1, 2 and 3 hold true for . To prove that the statement 1 holds true, we will first prove that { (R) | R is a region of B ? } ? {0, Let be an extinction map for ?. Then, by the definition of extinction maps, there is a sequence S = (M 1
,
, any two distinct minima M 1 and M 2 of w, we have (M 1 ) = (M 2 ); and 3. for any region R of B ? , we have that (R) is equal to ?{ (M ) such that M is a minimum of w
, M n ) of minima of w such that, for any region R of B ? , the value (R) is the extinction value of R for
, M n ) be a sequence of minima of w ordered in non-decreasing order for , i.e., for any two distinct minima M i and M j , with i, j in {1, n}, if (M i ) < (M j ) then i < j. 1. (V, E ? ) is a MST of
, Let B be a binary partition hierarchy of (G, w). Then, any minimum of w is a region of B, Lemma, vol.80
, M n ) be a sequence of minima of w and let ? be the persistence map for (S, ?). The range of ? is {0, Lemma 81. Let ? be an altitude ordering on the edges of G for w, let S = (M 1, p.1
, We will prove that (1) for any building edge u for ?, ?(u) is in {0, Proof. Let be the extinction map for
, is a MST of (G, ?(H)); and 2
, By Lemma 83 (statement 1), as G is a MST of (G, ?(H)), we have that G and G have the same quasi-flat zones hierarchies
, By the definition of persistence values, we can affirm that
, Hence, by Lemma 83, G and G have the same quasi-flat zones hierarchies (for ?(H )): QFZ(G , ?(H )) = QFZ(G, ?(H )), G is a MST of, vol.84
, n ? 1} and, by Definition 18 (statement 2), only the weight of the watershed-cut edges for ? are strictly greater than zero. Then, {f (e) | e is a watershed ? cut edge f or ?} = {1, n ? 1}. Hence, for any i in {1, vol.18
, Let ? be an altitude ordering for w, let f be a map from E into R such that f is one-side increasing for ?, and let ? be the approximated extinction map for (f, ?). The range of ? is {0, Lemma, vol.87
, We will prove that: (1) for any i in {0, . . . , n}, there is a region R of B ? such that ?(R) = i; and (2) for any region R of B ? , we have ?(R) in {0
, We start by proving that there is a region R of B ? such that ?(R) = n. Let R be the set V of vertices of G. Then, by Definition 27 (statement 1), we have ?(R) = k + 1, We first prove statement
, Otherwise, let v be the building edge of the parent of R. By Definition 27, the value ? f (R) is either f (v) or ?(parent(R)). Hence, either ? f (R) is equal to f (v) for a building edge v for ?, or ? f (R) is equal to ?(V ) = n. It is enough to prove that n and f (v) are in {0, . . . , n}. As f is one-side increasing for ?, Let R be a region of B ? . If R = V , then ?(R) = n, as established in the proof of statement
, Let ? be an altitude ordering for w and let f be a map from E into R such that f is one-side increasing for ?. Let ? be the approximated extinction map for (f, ?), For any two minima M 1 and M 2 of w, if ?(M 1 ) = ?(M 2 )
, Y )) = f (u X ); and 3. there is a minimum of w included in Y
, We consider two cases: (1) sibling(Z) is a leaf-region of B ? ; and (2) sibling(Z) is a non-leaf region of B ? . (1) If sibling(Z) is a leaf-region of B ? , then, by Definition 26, Z is a dominant region for (f, ?) and sibling(Z) is not a dominant region for (f, ?). Hence, by Definition, Proof. Let X be a region such that there is a minimum M of w such that M ? X. Then, there is a child Z of X such that there is a minimum M such that M ? Z. Let Z be a child X such that there is a minimum M such that M ? Z, vol.27
, Let us now assume that sibling(Z) is a non-leaf region of B ? . Since X is not a minimum of w and since there is a minimum of w included in Z, we can conclude that there is a minimum of w included in sibling(Z) as well. Let be the non-leaf ordering for (f, ?)
, by the definition of dominant regions (Definition 26), we have that either Z or sibling(Z) is a dominant region for (f, ?)
, Since both Z and sibling(Z) include at least one minimum of w, we may say that there is a child Y of X for which the hypothesis 1, 2 and 3 hold true, Otherwise, if sibling(Z) is a dominant region for (f, ?)
, Let ? be an altitude ordering for w and let f be a map from E into R such that f is one-side increasing for ?. Let ? be the approximated extinction map for (f, ?), Lemma 90
, If Y 1 is a minimum of w, then the property holds true. Otherwise, if Y 1 is not a minimum of w, it means that there is a minimum M of w such that M ? Y 1 . By Lemma 89, there is a child Y 2 of Y 1 such that ?(Y 2 ) = ?(Y 1 ) = f (u) and such that there is a minimum of w included in Y 2 . Again, if Y 2 is a minimum of w, then the property holds true. Otherwise, we can apply this same reasoning indefinitely. We can define a sequence
, Let ? be an altitude ordering for w and let f be a map from E into R such that f is one-side increasing for ?, Lemma 91
, Otherwise, by Lemma 89, there in Y 2 . Again, if Y 2 is a minimum of w, then the property holds true. Otherwise, we can apply this same reasoning indefinitely, X is a minimum of w, then it is trivial
, Proof of Lemma 88, order to prove that (1) for any two minima M 1 and M 2 of w, if ?(M 1 ) = ?(M 2 ), then M 1 = M 2 , we will prove that
, Appendix: proofs of theorems and properties XXXVII
, By Lemma 86, for any i in {1, . . . , n ? 1}, there is a watershed-cut edge such that f (u) = i. Then, for any i in {1, As w has n minima, it suffices to prove that, for any i in {1, . . . , n}, there is a minimum M of w such that ?(M ) = i. By Lemma 90, for any watershed-cut edge u for B ? , there is a minimum M such that ?(M ) = f (u)
, Since w has n minima, it implies that the values ?(M 1 ) and ?(M 2 ) are distinct for any pair (M 1 , M 2 ) of distinct minima of w. Hence, for any two minima M 1 and M 2 of w, if ?(M 1 ) = ?(M 2 ), then M 1 = M 2
, Let ? be an altitude ordering for w and let f be a map from E into R such that f is one-side increasing for ?. Let ? be the approximated extinction map for (f, ?), Lemma 92
, Let ? be an altitude ordering for w and let f be a map from E into R such that f is one-side increasing for ?, Lemma 93
, Let X be a region of B ? . Then ?(X) is greater than or equal to the supremum descendant value of X (for
, X) = ?(V ) = k + 1, where k is the supremum descendant value of X for (f, ?) (first case of Definition 27). Then, ?(X) is clearly than the supremum descendant value of X
, not a dominant region for (f, ?), then ?(X) = f (u) (third case of Definition 27)
, then there is no descendant of X whose building edge is a watershed-cut edge for ?. Hence, for any edge v such that R v ? X, u is not a watershed-cut edge for ? and, since f is one-side increasing for ?, we have f (v) = 0 Definition 18 (statement 2). Therefore, the supremum descendant value of X is zero. By Definition 18 (statement 1), we have {f (v), XXXVIII Appendix: proofs of theorems and properties (a) If there is no minimum M of w such that M ? X
, If X sibling(X), then, by the definition of non-leaf ordering
, Since f is one-side increasing for ?, by the statement 3 of Definition 18, there is a child Y of parent(X) such that f (u) ? ?{f (v), Thus, we have (X) ? (sibling(X))
, Then, we have f (u) ? (X) or f (u) ? (sibling(X)). In the case where f (u) ? (sibling(X)), this also implies that f (u) ? (X) because (X) ? (sibling(X)). Therefore, ?(X)
, In the base step, we consider that parent(X) is V . In the inductive step, we show that, if the property holds true for parent(X), then it also holds true for X. Please note that, if parent(X) is not a dominant region for (f, ?), the property holds for parent(X) as proven in the previous case. (a) Base step: if parent(X) is V , then ?(X) = ?(V ) = (V ) + 1 (first case of Definition 27). We can see that (V ) ? (X) because
, Since ?(X) = ?(parent(X)), we have ?(X) ? (parent(X)). We can affirm that, for any edge v in E ? such that R v ? X, we also have R v ? parent(X). Hence, (parent(X)) ? (X). Therefore, ?(X)
, Appendix: proofs of theorems and properties XXXIX Proof of Lemma 92. We will prove that, for any region X of B ?
, The edge u is not a watershed-cut edge for ? because the child X of R u does not include any minimum of w. Hence, since f is one-side increasing for ?, by the statement 2 of Definition 18, X, then X is not a dominant region for
, X is a minimum of w, then ?(X) = ?{? f (M ) such that M is a minimum of w included in X} = ?{? f (X)}
, To prove that ?(X) ? ?{?(Y ) | Y ? X}, it is enough to demonstrate that, for any region Z of B ? , we have ?(Z) ? ?{?(Y ) | Y is a child of Z}. Let Z be a region of B ? . If Z is a leaf region of B ? , then ?(Z) ? ?{?(Y ) | Y is a child of Z} = ?{} = 0 because, by Lemma 87, ?(Z) is in {0, . . . , n}. Let us now assume that Z is not a leaf region of B ? and let Y be a child of Z. If Y is a dominant region for (f, ?), then ?(Y ) = ?(Z) and, consequently, ?(Z) ? ?(Y ). Otherwise, if Y is not a dominant region for (f, ?), then ?(Y ) = f (v), where v is the building edge of, Z. By Lemma, vol.93
, We can now prove that ?(X) = ?{? f (M ) such that M is a minimum of w included in X} in the case where X is not a minimum of w. By Lemma 91, there is a minimum M of w such that M ? X and such that ?(M ) = ?(X). Let M be the minimum of w such that ?(M ) = ?(X). Since ?(X) ? ?{?(Y ) | Y ? X}, we can say that ?(X) = ?{? f (M )
, Let ? be an altitude ordering for w and let f be a map from E into R such that the approximated extinction map for (f, ?) is an extinction map for ?. Then, f is one-side increasing for ?. XL Appendix: proofs of theorems and properties = {0, Lemma 94, p.1
, ? 1} and let R be a region of B ? such that ?(R) = i. If R is not a dominant region for (f, ?), then ?(R) = f (u), where u is the building edge of the parent of R and, then, we can affirm that there exists an edge in E ? whose weight for f is i. Otherwise, if R is a dominant region for (f, ?), then ?(R) = ?(parent(R)). If parent(R) is not a dominant region for (f, ?), then ?(parent(R)) = ?(v), where v is the building edge of the parent of parent(R) and we have our property. Otherwise, if parent(R) is a dominant region for (f, ?), then ?(parent(R)) = ?(parent(parent(R))). We can see that, for any leaf region R of B ? , we have ?(R) = ?{?(M ) such that M is a minimum of w included in R} = 0
, Let ? be an altitude ordering for w and let f be a map from E into R. Let ? be the approximated extinction map for (f, ?) such that ? is an extinction map. Then
, at least one minimum of w included in each child of R u . Hence, by Property 23, the value of each child of R u in ? is greater than zero. As there is at most one child of R u that is a dominant region for (f, ?), then there is a child of R u that is not a dominant region for (f, ?)
, Since ? is an extinction map and since there is no minimum of w
, Let ? be an altitude ordering for w and let f be a map from E into R. Let be a map from each region of B ? into its supremum descendant value (for
, Let ? be the approximated extinction map for (f, ?) such that ? is an extinction map. Then
, Let ? be the approximated extinction map for (f, ?) such that ? is an extinction map. Then, for any region R of B ? , ?(R) ? ?{?(X) | X ? R}. dominant region for (f, ?), Property 98. Let ? be an altitude ordering for w and let f be a map from E into R
, At most one of the children of R u is a dominant region for (f, ?), by Property 99, we have ?(R u ) ? {f (v) | v is the building edge of a region X ? R}
, Let f be a map from E into R, let ? be a lexicographic ordering for (w, f ), and let ? be the approximated extinction map for (f, ?), vol.28
, Thus, by Lemma 85, ? is an extinction map for ?. Backward implication: Let ? be an extinction map for ?. Then, by Property 94, f is one-side increasing for ?. Hence, Theorem 20, f is one side increasing for the lexicographic ordering ? for (w, f )
, If f is the saliency map of a hierarchical watershed of (G, w), then f is the saliency map of a hierarchical watershed of (G, w) for S. To prove Property 30
, Let f be a map from E into R and let ? be an altitude ordering for w. Let ? be the approximated extinction map for (f, ?), and let S be the estimated sequence of minima for (f, ?)
, Therefore, S is a sequence of minima ordered according to their extinction values in ?. Therefore, S corresponds to the estimated sequence of minima for, Proof. Let ? be an extinction map for ?. Then, there exists a sequence S = (M 1
, Proof of Property 30. Let f be the saliency map of a hierarchical watershed of (G, w)
, Hence, for any u in E ? , we have f (u) = min{?(R u ), f (u)}. By the definition of approximated extinction maps (Definition 27), we can conclude that, for any child X of R(u), we have either ?(X) = ?(R u ) or ?(X) = f (u). Hence, f (u) = min{?(R) | R is a child of R u }. By Property 100, ? is the extinction map for S and ?. Hence, f maps any building edge u for ? into its persistence value for, Then, by Theorem 28, ? is an extinction map. By Property 99, for any u in E ? , we have ?(R u ) ? f (u)
, The following statements hold true: 1. The hierarchy H is a hierarchical watershed of (G, w) if and only if the watersheding ?(f ) of f (for ?) is equal to f
, The watersheding ?(f ) of f is the saliency map of a hierarchical watershed of (G, w)
, The watersheding ?(?(f )) of ?(f ) is equal to ?(f )
, To prove Theorem 33, we first present the following lemma
, We will prove that f is also the saliency map of the hierarchy induced by (S, ?). By Property 30, the map f is the saliency map of a hierarchical watershed of (G, w) for S. Therefore, by Lemma 8.2.5, the map f is the saliency map of the hierarchy induced by (S, ?). Thus, the watersheding ?(f ) of f is equal to f . Backward implication: Let ?(f ) be equal to f . Let S be the estimated sequence of minima for (f, ?). By Theorem 32, the map ?(f ) is the saliency map of a hierarchical watershed of (G, w) for S. Hence, S be a sequence of minima of w, and let f be the saliency map of a hierarchical watershed of (G, w) for S. Let ? be a lexicographic ordering for
, Statement 2: Let S be the estimated sequence of minima for f and ?. By Theorem 32, the watersheding ?(f ) of f is the saliency map of a hierarchical watershed of (G, w) for S. Hence, ?(f ) is the saliency map of a hierarchical
, By the statement 2, the watersheding ?(f ) of f is the saliency map of a hierarchical watershed of (G, w). Hence, by the first statement, the watersheding ?(?(f )) of (?(f ), vol.3
, Let (G, w) be a weighted graph and let ? be an altitude ordering for w. map for (S, ?), Lemma, vol.101
, By Property 6, since f is the saliency map of the hierarchical watershed of (G, w) for S, then f is the saliency map of the hierarchy induced by S and by an altitude ordering for w. Since ? is the unique altitude ordering for w, then f is the saliency map of the hierarchy
, Therefore, the value f (u) is min{ (X), (Y )}, where X and Y are the children of R u
, Let (G, w) be a tree with a unique altitude ordering and let ? be the altitude ordering for w. Let S be a sequence of minima of w and let be the extinction map for (S, ?), Lemma, vol.103
, Let f be the saliency map of the hierarchical watershed of (G, w) for S. Let u be a watershed-cut edge for ? and let M be a minimum of w. If M is not the minimum of greatest extinction value among the minima, Lemma, vol.104
, By Lemma 103, the weight of each watershed-cut edge v such that R v ? R u is the extinction value of a minimum included in R u . Moreover, by Lemma 19, f is one-side increasing for ? and, consequently, the watershed-cut edges have pairwise distinct edge weights in f . Therefore, for ? 1 minima included in R u , there is a watershed-cut edge v such that R v ? R u and such that f (u) is the extinction value of this minimum. Hence, there is only one minimum M included in R u such that there is no watershed-cut edge v such that R v ? R u and f (u) = (M )
, Let f be the saliency map of the hierarchical watershed of (G, w) for S. Let u be a watershed-cut edge for ?. Let M be the minimum of maximum extinction value among the minima included in R u . Then, for any watershed, Lemma 105. Let (G, w) be a tree with a unique altitude ordering and let ? be the altitude ordering for w
, By Lemma 103, the value f (v) is the extinction value of a minimum included in R u . By Lemma 104, f (u) is different from the extinction value of M . As the minima of w have pairwise distinct extinction values and since the extinction value of M is maximal among the minima
, Let M x be the minimum included in X such that (X) = (M x ) and let M y be the minimum included in Y such that (Y ) = (Y x ). The region R u is a maximal region of B ? for f if and only if: 1. either M x is the only minimum included in X; or 2, Lemma 106. Let (G, w) be a weighted graph and let ? be the unique altitude ordering for w
, We first prove the forward implication. We consider the conditions 1 and 2 separately
, a maximal region of B ? for f , we will first prove that f (u) > {f (v) | R v ? X} and, then, we will prove that f (u) > {f (v) | R v ? Y }. Let r be the building edge of X. By Definition 2, r is not a watershed-cut edge for ?. Hence, since f is one-side increasing for ? by Lemma 19, we can say that f (r) = 0 by the second statement of Definition 18. Therefore, by the third statement of Definition 18, for any edge r such that R r ? R v , we have f (r ) = 0. Since u is a watershed-cut edge for ?, we have f (u) > 0 by the second statement of Definition 18, By Lemma, vol.102
, In this case, for any region Z such that Z ? R v , we have (Z) = (Z) because the position of the minima included in R u are the same in the sequences S and S . Since f (v) is defined by the extinction value of the children of R v
, ) is min{ (X), (Y )}. Let M x be the minimum of second greatest extinction value among the minima included in X. By Property 106, since (M x ) > (M y ), we have (M x ) < (M y ). Therefore, we can say that M x (resp. M y ) is still the minimum of w of greatest extinction value for among the minima included in X (resp. Y ). Hence, we have that (X) = (M x ) = (Y ) and that (Y ) = (M y ) = (X). Hence, f (u) = min{ (X)
, The only minima that had their extinction values changed in with respect to were two minima M x and M y included in R u . Hence, the greatest extinction value among the minima included in R u is still the same even though the minimum carrying this extinction value has changed. Therefore, the extinction value of both children of R v has not been
, We will consider the two following cases: (a) R v ? X; and (b) R v ? Y . Appendix: proofs of theorems and properties XLIX
, Otherwise, let us assume that M x is included in R v . If M x = R v , then v is not a watershed-cut edge for ? and we have f (u) = f (u) = 0 by the second statement of Definition 18. Otherwise, let R be the child of R v that includes M x . By our assumption, M x is the minimum of w of greatest extinction value among the minima included in X. Therefore, (R ) = (M x ). Moreover, f (v), being min{ (R ), (sibling(R ))}, is equal to (sibling(R )). Let M x be the minimum of w of second greatest extinction value among the minima included in X, If M x is not included in R v , we can conclude that the extinction value for of all regions included in R v have not been changed with respect to , which implies that f (u) = f (u), vol.106
, By our assumption, M y is the minimum of w of greatest extinction value among the minima included in Y . Hence, f (v), being min{ (R ), (sibling(R ))}, is equal to (sibling(R )). By our hypothesis, If M y is not included in R v , we can conclude that the extinction value for of all regions included in R v have not been changed with respect to , which implies that f (u) = f (u)
, Hence, there is no minimum included in R v whose extinction value for S is equal to (M x ). Therefore, by Lemma 105, f (v) < (M x ) and, consequently, f (v) = f (v), which contradicts our assumption. Hence, M x is the minimum of greatest extinction value among the minima included in X. The same argument can be used to L Appendix: proofs of theorems and properties prove that M y is the minimum of greatest extinction value among the minima, We now prove the backward implication. Let f and f be equal. Using Lemma 106, we will prove by contradiction that R u is a maximal region of B ? for f and that M x (resp. M y ) is the minimum of w of greatest extinction value among the minima included in X (resp. Y )
, Let M x be the minimum included in X with the second greatest extinction value among all minima included in X. As R u is not a maximal region, Lemma 106, we have that (M x ) ? (Y ). We know that (M x ) = (M y ) = (Y ) and that (M y ) = (M x ) = (X). Since (M x ) ? (Y ) and since (M x ) = (Y ), then M x became the minimum of greatest extinction value for S among the minima included in X. However, (M x ) is still less than (M y ) = (M x ). In the end, we will have (X) = (M x ) and (Y ) = (M x ). Since f (u) = min{ (X), (Y )}, we have that f (u) = (M x ). By our assumption that (M x ) > (M y )
, | w) is equal to |S H | |Mw| . Hence, we need to prove that |S H | is equal to 2 m . To this end, we will prove that, given any sequence S in S w (H), we can obtain another sequence in S H only by, for each maximal region R of B ? for f , swapping the order of two minima included in R. By Lemma 108, for each maximal region R of B ? for f , there is only one pair of minima of (G, w) and let ? be an altitude ordering for (G, w) such that both ?(H 1 ) and ?(H 2 ) are one-side increasing for ?, Proof of Property 37. Let f be the saliency map of H. By Property, vol.35
, Appendix: proofs of theorems and properties LI
, ?(H 2 )). We will prove that the hierarchy QFZ(G, f 3 ) is a flattened hierarchical watershed of (G, w), Let H 1 and H 2 be two hierarchical watersheds of (G, w) and let ? be an altitude ordering for w such that both ?(H 1 ) and ?(H 2 ) are one-side increasing for ?. Let f 3 denote the map (?(H 1 )
, The following Lemmas 109, 112 and 113 prove respectively that the conditions 1, 2 and 3 for QFZ
, Let f 1 and f 2 be two maps from E into R and let G be a subgraph of G such that G is a MST of both (G, f 1 ) and (G, f 2 ). Then G is also a MST of (G
, the context of graphs and we state two well-known properties of spanning trees in Lemmas 110 and 111. Let x and y be two vertices in V and let ? = (x 0
,
, As G is a spanning tree, by Lemma 110, the graph (V, E(G ) ? {u}) contains a cycle ? which includes the edge u. Since G is a MST for (G, f 1 ) and for (G, f 2 ), by the forward implication of Lemma 111, Lemma 111. Let G be a spanning tree of a weighted graph
, (v), f 2 (v)) ? min(f 1 (u), f 2 (u)) and, consequently, f 3 (v) ? f 3 (u). Hence, for any edge v in ?, the cycle ?, we have min
, Thus, by the backward implication of Lemma 111, G is a MST of
, The following lemma proves that the condition 2 for QFZ(G, f 3 ) to be a flattened hierarchical watershed hold true. LII Appendix: proofs of theorems and properties
, The following lemma proves that the condition 3 for QFZ(G, f 3 ) to be a flattened hierarchical watershed holds true
, Let f 1 and f 2 be two maps from E into R and let B be a binary partition hierarchy of (G, w) such that f 1 and f 2 are one-side increasing for ?
, we have that, for any building edge u of B, f 1 (u) ? ?{f 1 (v) | R v ? X} (resp. f 2 (u) ? ?{f 2 (v) | R v ? X}) for a child X of R u . We need to prove that, Proof. Since f 1 (resp. f 2 ) is one-side increasing for ?, vol.18
, Let X and Y be the children of R u . If f 1 (u) ? ?{f 1 (v) | R v ? X} (resp. f 1 (u) ? ?{f 1 (v) | R v ? Y }), we can affirm that f 3 (u) ? ?{f 1 (v) | R v ? X} (resp. f 3 (u) ? ?{f 1 (v) | R v ? Y }) as well, Since f 3 (e) = min(f 1 (e), f 2 (e))
, By Lemma 69, we can affirm that (V, E ? ) is a MST of both (G, ?(H 1 )) and (G, ?(H 2 )). Let f 3 denote the map (?(H 1 ), ?(H 2 )). By Lemma 109, (V, E ? ) is a MST of (G, f 3 ) as well, vol.46
, C(b, a); and 2. if min(a, b) < min(c, d), then C(a, b) < C(c, d); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d)
, Let H 1 and H 2 be two hierarchical watersheds of (G, w) and let ? be an altitude ordering for w such that both ?(H 1 ) and ?(H 2 ) are one-side increasing for ?. Let C be a positive function from R 2 into R such that
, C(0, 0) = 0; and
, C(b, a); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d) then C(a, b) ? C(c, d)
, By Property 22, we need to prove that there exists a binary partition hierarchy B of (G, w) such that the following statements hold true: 1. (V, E(B )) is a MST of (G, f 3 ); and 2
, Let C be a function from R 2 into R such that, for any two real values x and y, we have C(x, y) = C(y, x)
, As min(a, b) = min(c, d) and max(a, b) = max(c, d), then either we have (i) a = c and b = d which implies that C(a, b) = C(c, d); or (ii) c = b and d = a which implies that C(c, d) = C(b, a), which, by our hypothesis on C
, The following three lemmas prove that the conditions 1, 2 and 3 for QFZ(G, f 3 ) to be a flattened hierarchical watershed of (G, w) hold true
, Lemma 115. Let C be a positive function such that
, C(0, 0) = 0; and
, C(b, a); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d) then C(a, b) ? C(c, d)
, Let f 1 and f 2 be the saliency maps of two hierarchies on V and let G be a subgraph of G such that G is a MST of both (G, f 1 ) and (G, f 2 ). Then G is also a MST of
, Let f 3 denote the map C(f 1 , f 2 )
, Lemma 111, for any edge v in the cycle ?, we have f 1 (v) ? f 1 (u) and f 2 (v) ? f 2 (u). Therefore, for any edge v in the cycle ?, we have min(f 1 (v), f 2 (v)) ? min(f 1 (u), f 2 (u)) and max(f 1 (v), f 2 (v)) ? max
, Then, we should consider the three following cases: 1. If min(f 1 (v), f 2 (v)) < min
, (v)) = min(f 1 (u), f 2 (u)) and max
, C(0, 0) = 0; and
, C(b, a); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d) then C(a, b) ? C(c, d)
, Let f 1 and f 2 be the saliency maps of two hierarchies on V and let B be a binary partition hierarchy of (G, w) such that both f 1 and f 2 are one-side increasing for ?
, Lemma 117. Let C be a positive function such that
, C(0, 0) = 0; and
, C(b, a); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d) then C(a, b) ? C(c, d)
, Let f 1 and f 2 be the saliency maps of two hierarchies on V and let B be a binary partition hierarchy of (G, w) such that both f 1 and f 2 are one-side increasing for ?. Let f 3 denote the map C(f 1 , f 2 ). Then, for any
, We need to prove that, for any building edge u of B, there is a child X of R u such that f 3 (u) ? ?{f 3 (v) | R v ? X}. Let u be a building edge of B and let X and Y be the children of R u . We should consider the following four cases: 1. If f 1 (u) ? ?{f 1 (v) | R v ? X} and f 2 (u) ? ?{f 2 (v) | R v ? X}, then, for any building edge e such that R e ? X, we have f 1 (u) ? f 1 (e) and f 2 (u) ? f 2 (e), ? min(f 1 (u), f 2 (u)). If min(f 1 (e), f 2 (e)) < min, vol.2
, Thus in all cases we have C(f 1 (u), f 2 (u)) ? C(f, p.1
, ) ? ?{f 1 (v) | R v ? X} and f 2 (u) ? ?{f 2 (v)
, Let v be an edge such that R v ? X. By our assumption, we have f 1 (u) ? f 1 (v). Indeed, since f is a one-side increasing map, we can say that either f 1 (u) = f 1 (v) = 0 or f 1 (u) > f 1 (v) because only the watershed-cut edges for ? have non-zero and pairwise distinct weights. If f 1 (u) = f 1 (v) = 0, this implies that neither u nor v are watershed-cut edges for ? and therefore f 2 (u) = f 2 (v) = 0, which implies that f 3 (u) = 0 ? f 3 (v) = 0. Otherwise
, Therefore, we have f 3 (u) ? ?{f, vol.3
, C(0, 0) = 0; and
, C(b, a); and 3. if min(a, b) = min(c, d) and max(a, b) < max(c, d) then C(a, b) ? C(c, d)