Generalized Splines. Madeline Handschy, Julie Melnick, Stephanie Reinders. Smith College. April 1, 2013
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1 Smith College April 1, 213
2
3 What is a Spline?
4 What is a Spline? are used in engineering to represent objects.
5 What is a Spline? are used in engineering to represent objects.
6 What is a Spline? are used in engineering to represent objects. We will use splines to graphically represent systems of congruences.
7 What is a System of Congruences?
8 What is a System of Congruences? We say x y mod a if x y is a multiple of n. Example 3 13 mod 5 because 13 3 = 1 = 2 5
9 What is a System of Congruences? We say x y mod a if x y is a multiple of n. Example 3 13 mod 5 because 13 3 = 1 = 2 5 A spline is a graphical representation of a system of congruences. We label the nodes of the graph with the variables and the edges with the moduli. Below is the spline for the above congruence. y 13 x a 3 5
10 Paths with 3 Vertices System of Congruences x y mod a 1 y z mod z y x a 1
11 Why is this useful? System of Congruences x 1 x 2 mod a 1 x 2 x 3 mod x 3 x 4 mod a 3 x 4 x 5 mod a 4 x 5 x 6 mod a 5 x 6 x 1 mod a 6 x 5 x 1 mod a 7 x 5 x 2 mod a 8 x 5 x 3 mod a 9 x 4 x 2 mod a 1 x 4 x 1 mod a 11 x 4 x 6 mod a 12
12 Why is this useful? System of Congruences x 1 x 2 mod a 1 x 2 x 3 mod x 3 x 4 mod a 3 x 4 x 5 mod a 4 x 5 x 6 mod a 5 x 6 x 1 mod a 6 x 5 x 1 mod a 7 x 5 x 2 mod a 8 x 5 x 3 mod a 9 x 4 x 2 mod a 1 x 4 x 1 mod a 11 x 4 x 6 mod a 12 x 5 x 4 x 6 x 3 x 1 x 2
13 Main Questions Can we list all possible splines of a certain shape? Can we find a basis for the set of triangular splines? What can we find out about splines other than cycles?
14 General Triangle System of Congruences x y mod a 1 y z mod z x mod a 3 a 3 z x a 1 y
15 Triangle Solutions System of Congruences x y mod a 1 y z mod z x mod a 3 1? lcm(, a 3 ) a 3 1 a 3? a 3 a 1 a 1 a 1 1
16 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3 z a 3 y Theorem There is a minimal value for y such that a spline of this form exists. a 1
17 Existence of Solution as Application of CRT System of Congruences y mod 3 y z mod 4 z mod 5 z y
18 Chinese Remainder Theorem System of Congruences x a 1 mod n 1 x mod n 2 There exists a solution x if the integers n 1, n 2 are coprime. If gcd(n 1, n 2 ) 1 a solution x exists if and only if a 1 mod gcd(n 1, n 2 ).
19 Existence of Solution as Application of CRT System of Congruences y mod a 1 z y mod z mod a 3
20 Existence of Solution as Application of CRT System of Congruences y mod a 1 z y mod z mod a 3 By the Chinese Remainder Theorem, this system of congruences will have a solution if and only if y mod gcd(, a 3 )
21 Existence of Solution as Application of CRT System of Congruences y mod a 1 z y mod z mod a 3 By the Chinese Remainder Theorem, this system of congruences will have a solution if and only if y mod gcd(, a 3 ) Thus y = k gcd(, a 3 ) for some k Z.
22 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3
23 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3 Thus y = la 1 for some l Z.
24 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3 Thus y = la 1 for some l Z. Recall y = k gcd(, a 3 ) for some k Z.
25 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3 Thus y = la 1 for some l Z. Recall y = k gcd(, a 3 ) for some k Z. We want to minimize y...
26 Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3 Thus y = la 1 for some l Z. Recall y = k gcd(, a 3 ) for some k Z. We want to minimize y... Let y = lcm(gcd(, a 3 ), a 1 ).
27 Existence of Solution as Application of CRT a 3 z a 1 lcm(gcd(, a 3 ), a 1 )) Next we check that a spline exists: We must have a solution to the system of congruences z lcm(gcd(, a 3 ), a 1 )) mod z mod a 3 By the Chinese Remainder Theorem, we have a solution z if and only if lcm(gcd(, a 3 ), a 1 )) mod gcd(, a 3 ).
28 General Square System of Congruences x y mod a 1 y z mod z w mod a 3 w x mod a 4 w a 3 z a 4 x a 1 y
29 Square Solutions System of Congruences x y mod a 1 y z mod z w mod a 3 w x mod a 4 1 a 3 1? a 3?? a 3? lcm(a 3, a 4 ) a 3 a 4 1 a 1 1 a 4 a 1? a 4 a 1 a 4 a 1
30 Square Solutions System of Congruences mod a 1 z mod z w mod a 3 w mod a 4 Rewrite this as...
31 Square Solutions System of Congruences z mod z w mod a 3 w mod a 4 This is essentially the same as a triangle. a 4 w a 3 z a 1 a 4 w a 3 z
32 Square Solutions System of Congruences y mod a 1 y z mod z w mod a 3 w mod a 4 w a 3 z a 4 a 1 y
33 Bases: Can we find a basis for the set of triangular splines? Can we find a basis for the set of triangular splines?
34 Bases: What Solutions Have We Mentioned So Far? a a 3 z y lcm(, a 3 ) a 3 a 1 a 1 a 1 1
35 Bases: What Solutions Have We Mentioned So Far? a a 3 z y lcm(, a 3 ) a 3 1 a 1 In the center triangle, y is the minimal solution stated in the theorem we mentioned earlier. y = lcm(a 1, gcd(, a 3 )) a 1 a 1
36 Bases: What Solutions Have We Mentioned So Far? a a 3 z y lcm(, a 3 ) a 3 1 a 1 In the center triangle, y is the minimal solution stated in the theorem we mentioned earlier. y = lcm(a 1, gcd(, a 3 )) We can write each of these solutions as a vector: 1 1, 1 z y, a 1 lcm(, a 3 ) a 1
37 Bases: Do these vectors form a basis? 1 1, 1 z y, Do these vectors form a basis? lcm(, a 3 )
38 Bases: Do these vectors form a basis? 1 1, 1 z y, Do these vectors form a basis? lcm(, a 3 ) Goal 1: Show that every linear combination of these vectors is a solution.
39 Bases: Do these vectors form a basis? 1 1, 1 z y, Do these vectors form a basis? lcm(, a 3 ) Goal 1: Show that every linear combination of these vectors is a solution. Goal 2: Show that every solution can be written as a linear combination of these vectors.
40 Goal 1: Show that every linear combination of these vectors a solution 1 α 1 1
41 Goal 1: Show that every linear combination of these vectors a solution 1 z α 1 + β y 1
42 Goal 1: Show that every linear combination of these vectors a solution 1 z lcm(, a 3 ) α 1 + β y + γ 1
43 Goal 1: Show that every linear combination of these vectors a solution 1 z lcm(, a 3 ) α 1 + β y + γ 1 α + βz + γ lcm(, a 3 ) = α + βy α
44 Goal 1: Show that every linear combination of these vectors a solution 1 z lcm(, a 3 ) α 1 + β y + γ 1 Is this a solution? α + βz + γ lcm(, a 3 ) = α + βy α
45 Goal 1: Show that every linear combination of these vectors a solution Is this a solution? α + βz + γ lcm(, a 3 ) a 3 α a 1 α + βy
46 Goal 1: Show that every linear combination of these vectors a solution Is this a solution? α + βz + γ lcm(, a 3 ) a 3 α + βy α a 1 Remember: y = lcm(a 1, gcd(, a 3 ))
47 Goal 1: Show that every linear combination of these vectors a solution Check edge a 1 : α + βz + γ lcm(, a 3 ) a 3 α a 1 α + β lcm(a 1, gcd(, a 3 ))
48 Goal 1: Show that every linear combination of these vectors a solution Check edge a 1 : α + βz + γ lcm(, a 3 ) a 3 α a 1 α + β lcm(a 1, gcd(, a 3 )) Need to show: α + β lcm(gcd(, a 3 ), a 1 ) α mod a 1
49 Goal 1: Show that every linear combination of these vectors a solution Check edge a 1 : α + βz + γ lcm(, a 3 ) a 3 α a 1 α + β lcm(a 1, gcd(, a 3 )) Need to show: α + β lcm(gcd(, a 3 ), a 1 ) α mod a 1 We can write lcm(gcd(, a 3 ), a 1 ) as some multiple k of a 1. Then we have α + βka 1
50 Goal 1: Show that every linear combination of these vectors a solution Check edge a 1 : α + βz + γ lcm(, a 3 ) a 3 α a 1 α + β lcm(a 1, gcd(, a 3 )) Need to show: α + β lcm(gcd(, a 3 ), a 1 ) α mod a 1 We can write lcm(gcd(, a 3 ), a 1 ) as some multiple k of a 1. Then we have α + βka 1 = α + na 1 for some integer n.
51 Goal 1: Show that every linear combination of these vectors a solution Check edge a 1 : α + βz + γ lcm(, a 3 ) a 3 α a 1 α + β lcm(a 1, gcd(, a 3 )) Need to show: α + β lcm(gcd(, a 3 ), a 1 ) α mod a 1 We can write lcm(gcd(, a 3 ), a 1 ) as some multiple k of a 1. Then we have α + βka 1 = α + na 1 for some integer n. Thus α + β lcm(gcd(, a 3 ), a 1 ) α mod a 1.
52 Goal 1: Show that every linear combination of these vectors a solution We showed that the first edge, a 1, is satisfied.
53 Goal 1: Show that every linear combination of these vectors a solution We showed that the first edge, a 1, is satisfied. In a similar fashion we can show that edges and a 3 are also satisfied. Theorem Every linear combination of 1 z 1, y, 1 lcm(, a 3 ) is a solution to a triangular spline.
54 Bases: Work in Progress Goal 1 Complete: We showed every linear combination of the vectors is a solution.
55 Bases: Work in Progress Goal 1 Complete: We showed every linear combination of the vectors is a solution. Goal 2: Every solution can be written as a linear combination of the vectors. We are still working to prove this.
56 Notation Let S n = the star graph with n vertices of degree 1 K n = the complete graph on n vertices W n = the wheel graph on n vertices C n = the cycle graph on n vertices.
57 Chinese Remainder Theorem: Star Graphs System of Congruences x a 1 mod n 1 x mod n 2 There exists a solution x if the integers n 1, n 2 are coprime. If gcd(n 1, n 2 ) 1 a solution x exists if and only if a 1 mod gcd(n 1, n 2 ).
58 Wheels: Relationship to star graphs W n = C n 1 + S n 1 Corollary: A sufficient condition for W n to have non-trivial solutions is for the labels of the edges adjacent to the center vertex (i.e. the vertex of degree n - 1) to be pairwise relatively prime.
59 Complete Graphs: Relationship to star graphs K n = K n 1 + S n 1 K n = C 3 + n 1 S i i=3 Corollary: A sufficient condition for K n to have non-trivial solutions is for n-3 of the vertices to have the following condition on their adjacent edges: all n-1 edges adjacent to the vertex v i have labels that are pairwise relatively prime.
60 Thanks! Thank you to Our advisor Professor Julianna Tymoczko The Smith College Department of Mathematics and Statistics The National Science Foundation, Grant DMS
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