, q 1 ? L c (Lemma 8). Hence, the number of solutions counted with multiplicity is at least three which contradicts Assumption A 3 . Hence, Assumption ? 3 is satisfied
, Ball(P ) is a regular square system, its solution set is a zero-dimensional manifold in the compact set B Ball(P ) (regular value theorem). Hence, Ball(P ) has a finite number of solutions in B Ball . Since ? P (Definition 39) is surjective (Lemma 40), the set L (Definition 37) is also finite. Hence, the set L c ? L n is finite (since L c ? L n is the image of L under the surjective function (q 1 , q 2 ) ? q 1 ). Moreover, by Assumption ? 2 , the set L n ? L c does not intersects the boundary of B. Hence, Assumption A 2 and A 4 are satisfied in B. To prove that Assumption A 3 is satisfied, let p = (?, ?) ? ? C (C) and |? ?1 (p)| 3. For pairwise distinct points q 1 , q 2 , q 3 ? ? ?1 (p), by Lemma 38
, C (p) consists of at most two distinct points. We consider two cases: (a) ? ?1
, By Lemma 38, the pair (q 1 , q 2 ) is in L n , and hence, there exists a solution X = (?, ?, y, r, t) ? R × R × R n?2 × R n?2 × R of Ball(P ), C (p) has two distinct elements, say q 1
, Since X is a regular solution (Assumption ? 1 ), by Lemma 54 we have that none of q 1 , q 2 is in L c
, Lemma 8, the multiplicity of {P
, y) = 0 ? R n?1 , x 1 ? ? = x 2 ? ? = 0} at q. If m = 1, then we are done. If m > 1, then by Lemma 8 we have that q ? L c . Hence, there exists a solution of Ball(P ) of the form X = (?, ?, y, r, 0) ? R × R × R n?2 × R n?2 × R such that ? P (X) = (q, q) (Lemma 40). Since X is regular (Assumption ? 1 ), C (p) has a unique point q. Let m denote the multiplicity of the system {P (x 1
, Assumption A 3 is satisfied. Now, Since Assumptions A 1 ,A 2 , A 3 and A 4 are satisfied and since all solutions of Ball(P ) are regular
, Using Lemma 58, we are ready to check Assumptions A 2 , A 3 , A 4 and A 5 using ? 1 , ? 2 and ? 3, Since Lemma, vol.58
, B and a function P from B to R n?1 . Output: True if and only if P satisfies Assumption A 1 in B. 1: L := {B} 2: while L = ? do 3: B := pop(L) 4: if 0 ? P (B) and rank
, Subdivide U and add its children to L. 11: return Solutions Remark 61. Notice that the output of Semi-algorithm 2, as described in the algorithmic part, may not be a set of pairwise disjoint boxes. More precisely, two boxes of the output may intersect in their boundaries if a solution of the Ball system is on or near their common boundary. To solve this issue, one could use the so-called ?-inflation (see for instance, U := pop(L). 5: if 0 ? Ball(P )(U) or ( f Ball (U)) ? (B × B) = ? then 6: Continue. 7: if rank, vol.9
, Under Assumption A 1 in B, if Semi-algorithm 2 stops, it returns a finite set Solutions of pairwise disjoint (2n ? 1)-boxes such that every solution of Ball(P ) in B Ball is contained in a box of this set and each box contains at most one solution. Moreover, Semi-algorithm 2 stops if and only if Ball(P ) satisfies Assumptions ? 1
, Assume that Semi-algorithm 2 stops and Solutions is the output set. Then, by Remark 61, we have that the boxes in Solutions are pairwise disjoint. Moreover, B Ball is covered by a set of boxes such that every box of this set satisfies: (a) one of the conditions in Step (5) which implies that no interesting solutions (i.e., that characterize singular points in ? C (C)) are in this box, or (b) the conditions in Step (7) which guarantee that if a solution X exists in this box
, Thus, X satisfies Assumptions ? 1 and ? 2
, Hence, every solution of Ball(P ) in B Ball is regular and contained in a box of Solutions. The condition rank
, According to the correctness proof, every solution X of Ball(P ) in B Ball is regular and satisfies ? P (X) ? B × B. Thus, Assumptions ? 1 and ? 2 are satisfied in B Ball . On the other hand, assume that ? 1 and ? 2 hold. We prove that Semi-Algorithm 2 stops in two steps: Step 1: By Assumption ? 1 all solutions of the square system Ball(P ) are regular. Hence, they form a zero dimen
, Step 2: We prove that for any box U ? L with a small enough width, one of the conditions in Step (5) or the conditions in Steps (7) are satisfied. Thus, in both cases U will be removed from L, and hence
, be a chain of those boxes. Since J Ball(P ) is a box function, we have that lim i?? J Ball(P ) (U i ) = J Ball(P ) (X)
, Consider the box function det( J Ball(P ) ). Notice that det( J Ball(P ) (U i )) converges to det(J Ball(P ) (X)). However, 0 ? det( J Ball(P ) (U i )) for all i ? N * but det
, For any box in L with a small enough width one of the conditions of Step (5) is satisfied or all of the conditions in Step (7) are satisfied which proves the lemma
, Semi-algorithm 2 requires a closed (2n ? 1)-box U 0 that contains B Ball . For instance the following set could be used: {(q, r, t) ? i ? {3, vol.63, p.0
, Finally, using Lemma, vol.58
, A 4 and A 5 in B. 1: Check Assumption A 1 (Semi-algorithm 1). 2: Compute a closed (2n ? 1)-box U 0 that contains B Ball (Remark 63). 3: L := the output of Semi-algorithm 2. 4: for all distinct U, U ? L do 5: Keep refining U, U until their plane projections are disjoint (ignoring the twin solutions) or it is guaranteed that one of them has no solution of Ball(P ). Ball(P ) in U (resp. U ) exists, Output: True, if and only if P satisfies Assumption A 1, vol.3
, If Semi-algorithm 3 stops, L is a set of pairwise disjoint boxes each of which containing at most one solution of
, We propose a regular square system that encodes the singularities of the plane projection of a generic curve in
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