Smith College April 1, 213
What is a Spline?
What is a Spline? are used in engineering to represent objects.
What is a Spline? are used in engineering to represent objects.
What is a Spline? are used in engineering to represent objects. We will use splines to graphically represent systems of congruences.
What is a System of Congruences?
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
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
Paths with 3 Vertices System of Congruences x y mod a 1 y z mod z y x a 1
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
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
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?
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
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
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
Existence of Solution as Application of CRT System of Congruences y mod 3 y z mod 4 z mod 5 z 15 2 5 4 y 5 4 3 5 4 12 3 3 3
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 ).
Existence of Solution as Application of CRT System of Congruences y mod a 1 z y mod z mod a 3
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 )
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.
Existence of Solution as Application of CRT System of Congruences y mod a 1 y z mod z mod a 3
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.
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.
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...
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 ).
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 ).
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
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
Square Solutions System of Congruences mod a 1 z mod z w mod a 3 w mod a 4 Rewrite this as...
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
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
Bases: Can we find a basis for the set of triangular splines? Can we find a basis for the set of triangular splines?
Bases: What Solutions Have We Mentioned So Far? a 3 1 1 a 3 z y lcm(, a 3 ) a 3 a 1 a 1 a 1 1
Bases: What Solutions Have We Mentioned So Far? a 3 1 1 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
Bases: What Solutions Have We Mentioned So Far? a 3 1 1 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
Bases: Do these vectors form a basis? 1 1, 1 z y, Do these vectors form a basis? lcm(, a 3 )
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.
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.
Goal 1: Show that every linear combination of these vectors a solution 1 α 1 1
Goal 1: Show that every linear combination of these vectors a solution 1 z α 1 + β y 1
Goal 1: Show that every linear combination of these vectors a solution 1 z lcm(, a 3 ) α 1 + β y + γ 1
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
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
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
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 ))
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 ))
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
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
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.
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.
Goal 1: Show that every linear combination of these vectors a solution We showed that the first edge, a 1, is satisfied.
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.
Bases: Work in Progress Goal 1 Complete: We showed every linear combination of the vectors is a solution.
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.
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.
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 ).
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.
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.
Thanks! Thank you to Our advisor Professor Julianna Tymoczko The Smith College Department of Mathematics and Statistics The National Science Foundation, Grant DMS-1143716