In this paper, we combine the notions of completing and avoiding partial latin squares. Let P be ... more In this paper, we combine the notions of completing and avoiding partial latin squares. Let P be a partial latin square of order n and let Q be the set of partial latin squares of order n that avoid P. We say that P is Q-completable if P can be completed to a latin square that avoids Q ∈ Q. We prove that if P has order 4t and contains at most t − 1 entries, then P is Q-completable for each Q ∈ Q when t ≥ 9.
We show that any partial 3r ×3r Latin square whose filled cells lie in two disjoint r ×r sub-squa... more We show that any partial 3r ×3r Latin square whose filled cells lie in two disjoint r ×r sub-squares can be completed. We do this by proving the more general result that any partial 3r by 3r Latin square, with filled cells in the top left 2r × 2r square, for which there is a pairing of the columns so that in each row there is a filled cell in at most one of each matched pair of columns, can be completed if and only if there is some way to fill the cells of the top left 2r × 2r square.
... FINAL TECHNICAL REPORT GRANT #: N00014-01-1-0917 PRINCIPAL INVESTIGATORS Tristan Denley, Talm... more ... FINAL TECHNICAL REPORT GRANT #: N00014-01-1-0917 PRINCIPAL INVESTIGATORS Tristan Denley, Talmage James Reid, Haidong Wu (e-mail: [email protected]. ... Thomassen gave a simple proof of this fact and showed that contractible edges have many applications ...
This article explores the challenges that students face in navigating the curricular structure of... more This article explores the challenges that students face in navigating the curricular structure of post-secondary degree programs, and how predictive analytics and choice architecture can play a role. It examines Degree Compass, a course recommendation system that successfully pairs current students with the courses that best fit their talents and program of study for upcoming semesters. Data are presented to demonstrate the impact that this system has had on student success. In particular the data will show that by closing the information gap, this system is able to close the educational achievement gap for low-income and minority students.
It is well known that all n×n partial Latin squares with at most n−1 entries are completable. Our... more It is well known that all n×n partial Latin squares with at most n−1 entries are completable. Our intent is to extend this well known statement to partial Latin cubes. We show that if an n×n×n partial Latin cube contains at most n − 1 entries, no two of which occupy the same row, then the partial Latin cube is completable. Also included in this paper is the problem of completing 2×n×n partial Latin boxes with at most n − 1 entries. Given certain sufficient conditions, we show when such partial Latin boxes are completable and then extendable to a deeper Latin box.
We show that for any positive integer k 4, if R is a (2k − 1) × (2k − 1) partial Latin square, th... more We show that for any positive integer k 4, if R is a (2k − 1) × (2k − 1) partial Latin square, then R is avoidable given that R contains an empty row, thus extending a theorem of Chetwynd and Rhodes. We also present the idea of avoidability in the setting of partial r-multi Latin squares, and give some partial fillings which are avoidable. In particular, we show that if R contains at most nr/2 symbols and if there is an n × n Latin square L such that n of the symbols in L cover the filled cells in R where 0 < < 1, then R is avoidable provided r is large enough.
In the spirit of Ryser's theorem, we prove sufficient conditions on k, ℓ and m so that k × ℓ × m ... more In the spirit of Ryser's theorem, we prove sufficient conditions on k, ℓ and m so that k × ℓ × m Latin boxes, i.e. partial Latin cubes whose filled cells form a k × ℓ × m rectangular box, can be extended to a k × n × m latin box, and also to a k × n × n latin box, where n is the number of symbols used, and likewise the order of the Latin cube. We also prove a partial Evans type result for Latin cubes, namely that any partial Latin cube of order n with at most n − 1 filled cells is completable, given certain conditions on the spatial distribution of the filled cells.
As usual, for simple graphs G and H, let the Ramsey number r(G,H) be defined as the least number ... more As usual, for simple graphs G and H, let the Ramsey number r(G,H) be defined as the least number n such that for any graph K of order n, either G is a subgraph of K or H is a subgraph of/(. We shall establish the values of r(aC~,bCs) and r(aCv, bC7) almost precisely (where nG is the graph consisting of n vertex disjoint copies of G) extending the work of Mizuno and Sato, who proved similar results about r(aC4, bC4). Our technique also allows us to find a general upper bound for the Ramsey number r(aC
Let G be an r-regular graph of order n and independence number α(G). We show that if G has odd gi... more Let G be an r-regular graph of order n and independence number α(G). We show that if G has odd girth 2k + 3 then α(G) ≥ n 1−1/k r 1/k. We also prove similar results for graphs which are not regular. Using these results we improve on the lower bound of Monien and Speckenmeyer, for the independence number of a graph of order n and odd girth 2k + 3.
In this paper, we combine the notions of completing and avoiding partial latin squares. Let P be ... more In this paper, we combine the notions of completing and avoiding partial latin squares. Let P be a partial latin square of order n and let Q be the set of partial latin squares of order n that avoid P. We say that P is Q-completable if P can be completed to a latin square that avoids Q ∈ Q. We prove that if P has order 4t and contains at most t − 1 entries, then P is Q-completable for each Q ∈ Q when t ≥ 9.
We show that any partial 3r ×3r Latin square whose filled cells lie in two disjoint r ×r sub-squa... more We show that any partial 3r ×3r Latin square whose filled cells lie in two disjoint r ×r sub-squares can be completed. We do this by proving the more general result that any partial 3r by 3r Latin square, with filled cells in the top left 2r × 2r square, for which there is a pairing of the columns so that in each row there is a filled cell in at most one of each matched pair of columns, can be completed if and only if there is some way to fill the cells of the top left 2r × 2r square.
... FINAL TECHNICAL REPORT GRANT #: N00014-01-1-0917 PRINCIPAL INVESTIGATORS Tristan Denley, Talm... more ... FINAL TECHNICAL REPORT GRANT #: N00014-01-1-0917 PRINCIPAL INVESTIGATORS Tristan Denley, Talmage James Reid, Haidong Wu (e-mail: [email protected]. ... Thomassen gave a simple proof of this fact and showed that contractible edges have many applications ...
This article explores the challenges that students face in navigating the curricular structure of... more This article explores the challenges that students face in navigating the curricular structure of post-secondary degree programs, and how predictive analytics and choice architecture can play a role. It examines Degree Compass, a course recommendation system that successfully pairs current students with the courses that best fit their talents and program of study for upcoming semesters. Data are presented to demonstrate the impact that this system has had on student success. In particular the data will show that by closing the information gap, this system is able to close the educational achievement gap for low-income and minority students.
It is well known that all n×n partial Latin squares with at most n−1 entries are completable. Our... more It is well known that all n×n partial Latin squares with at most n−1 entries are completable. Our intent is to extend this well known statement to partial Latin cubes. We show that if an n×n×n partial Latin cube contains at most n − 1 entries, no two of which occupy the same row, then the partial Latin cube is completable. Also included in this paper is the problem of completing 2×n×n partial Latin boxes with at most n − 1 entries. Given certain sufficient conditions, we show when such partial Latin boxes are completable and then extendable to a deeper Latin box.
We show that for any positive integer k 4, if R is a (2k − 1) × (2k − 1) partial Latin square, th... more We show that for any positive integer k 4, if R is a (2k − 1) × (2k − 1) partial Latin square, then R is avoidable given that R contains an empty row, thus extending a theorem of Chetwynd and Rhodes. We also present the idea of avoidability in the setting of partial r-multi Latin squares, and give some partial fillings which are avoidable. In particular, we show that if R contains at most nr/2 symbols and if there is an n × n Latin square L such that n of the symbols in L cover the filled cells in R where 0 < < 1, then R is avoidable provided r is large enough.
In the spirit of Ryser's theorem, we prove sufficient conditions on k, ℓ and m so that k × ℓ × m ... more In the spirit of Ryser's theorem, we prove sufficient conditions on k, ℓ and m so that k × ℓ × m Latin boxes, i.e. partial Latin cubes whose filled cells form a k × ℓ × m rectangular box, can be extended to a k × n × m latin box, and also to a k × n × n latin box, where n is the number of symbols used, and likewise the order of the Latin cube. We also prove a partial Evans type result for Latin cubes, namely that any partial Latin cube of order n with at most n − 1 filled cells is completable, given certain conditions on the spatial distribution of the filled cells.
As usual, for simple graphs G and H, let the Ramsey number r(G,H) be defined as the least number ... more As usual, for simple graphs G and H, let the Ramsey number r(G,H) be defined as the least number n such that for any graph K of order n, either G is a subgraph of K or H is a subgraph of/(. We shall establish the values of r(aC~,bCs) and r(aCv, bC7) almost precisely (where nG is the graph consisting of n vertex disjoint copies of G) extending the work of Mizuno and Sato, who proved similar results about r(aC4, bC4). Our technique also allows us to find a general upper bound for the Ramsey number r(aC
Let G be an r-regular graph of order n and independence number α(G). We show that if G has odd gi... more Let G be an r-regular graph of order n and independence number α(G). We show that if G has odd girth 2k + 3 then α(G) ≥ n 1−1/k r 1/k. We also prove similar results for graphs which are not regular. Using these results we improve on the lower bound of Monien and Speckenmeyer, for the independence number of a graph of order n and odd girth 2k + 3.
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Papers by Tristan Denley