Introduction The matrices A m satisfy a lovely recurrence: � 1 � 1 A 1 = 1 0 1 1 1 . A 2 = 1 0 0 1 0 0 and � A k − 1 � Ak − 2 A k = Ak − 2 Neil Calkin Counting Kings
Introduction For large m the matrix A m has fractal-like nature Neil Calkin Counting Kings
Introduction For large m the matrix A m has fractal-like nature Neil Calkin Counting Kings
Introduction (Curiously, this picture looks quite different to me when rotated!) Neil Calkin Counting Kings
Introduction In work with an REU some years ago, we computed the values of η m up to m = 34. Neil Calkin Counting Kings
Introduction In work with an REU some years ago, we computed the values of η m up to m = 34. These computations are not for the faint of heart: A 34 is a 14930352 × 14930352 matrix! Neil Calkin Counting Kings
Introduction In work with an REU some years ago, we computed the values of η m up to m = 34. These computations are not for the faint of heart: A 34 is a 14930352 × 14930352 matrix! 1 k λ k λ k k 31 10607.913998964 1.348525092499 32 14242.651559646 1.348340917465 33 19122.809968143 1.348167927490 34 25675.125129685 1.348005133658 Neil Calkin Counting Kings
Introduction In work with an REU some years ago, we computed the values of η m up to m = 34. These computations are not for the faint of heart: A 34 is a 14930352 × 14930352 matrix! 1 k λ k λ k k 31 10607.913998964 1.348525092499 32 14242.651559646 1.348340917465 33 19122.809968143 1.348167927490 34 25675.125129685 1.348005133658 These values, along with some other computations, allowed us to bound the value of η between 1 . 3426439 < η < 1 . 3426444 Neil Calkin Counting Kings
Introduction The dominant eigenvector How does the dominant eigenvector v m behave? Can we see fractal-like behaviour? Neil Calkin Counting Kings
Introduction The dominant eigenvector How does the dominant eigenvector v m behave? Can we see fractal-like behaviour? Pictures help give insight Neil Calkin Counting Kings
Introduction The dominant eigenvector How does the dominant eigenvector v m behave? Can we see fractal-like behaviour? Pictures help give insight Plot the coordinates of v m as a list of points. Neil Calkin Counting Kings
Introduction v 8 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 Neil Calkin Counting Kings
Introduction | bmv 9 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 10 20 30 40 50 60 70 80 90 Neil Calkin Counting Kings
Introduction v 10 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 50 100 150 Neil Calkin Counting Kings
Introduction v 11 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0 50 100 150 200 250 Neil Calkin Counting Kings
Introduction Currently, an undergrad, Nick Cohen, and I are investigating whether we can approximate v m by dominant eigenvectors for smaller A k . Neil Calkin Counting Kings
Introduction Currently, an undergrad, Nick Cohen, and I are investigating whether we can approximate v m by dominant eigenvectors for smaller A k . A big problem here seems to be that we don’t have the right notation for the problem, Neil Calkin Counting Kings
Introduction Currently, an undergrad, Nick Cohen, and I are investigating whether we can approximate v m by dominant eigenvectors for smaller A k . A big problem here seems to be that we don’t have the right notation for the problem, and possibly we don’t know even which space we should be working in! Neil Calkin Counting Kings
Introduction “What if” questions Last summer, working with REU students, we were discussing f ( m , n ) and various generating functions in particular, letting the variable x mark the number of kings on a board. Neil Calkin Counting Kings
Introduction “What if” questions Last summer, working with REU students, we were discussing f ( m , n ) and various generating functions in particular, letting the variable x mark the number of kings on a board. Let � f ( m , n , k ) x k F m , n ( x ) = k so that F m , n ( x ) is the generating function (a polynomial) counting all boards with k kings. One of the students said Neil Calkin Counting Kings
Introduction “What if” questions Last summer, working with REU students, we were discussing f ( m , n ) and various generating functions in particular, letting the variable x mark the number of kings on a board. Let � f ( m , n , k ) x k F m , n ( x ) = k so that F m , n ( x ) is the generating function (a polynomial) counting all boards with k kings. One of the students said F m , n ( 1 ) = f ( m , n ) and so n F m , n ( 1 ) 1 / n = η. lim What can we say about n F m , n ( x ) 1 / n ? lim Neil Calkin Counting Kings
Introduction Aside: Neil Calkin Counting Kings
Introduction Aside: What are the most ubiquitous counting sequences? Neil Calkin Counting Kings
Introduction Aside: What are the most ubiquitous counting sequences? Powers of 2? Neil Calkin Counting Kings
Introduction Aside: What are the most ubiquitous counting sequences? Powers of 2? Fibonacci numbers? Neil Calkin Counting Kings
Introduction Aside: What are the most ubiquitous counting sequences? Powers of 2? Fibonacci numbers? Catalan numbers? Neil Calkin Counting Kings
Introduction Aside: What are the most ubiquitous counting sequences? Powers of 2? Fibonacci numbers? Catalan numbers? The students discovered (interpreting things appropriately) Theorem n = 1 + x − x 2 + 2 x 3 − 5 x 4 + 14 x 5 − 42 x 6 + . . . 1 lim n F m , n ( x ) √ = 1 + 1 + 4 x . 2 Neil Calkin Counting Kings
Introduction Even though the series diverges at x = 1 (way before then), the function it represents still gives the correct value for η 1 . Neil Calkin Counting Kings
Introduction Even though the series diverges at x = 1 (way before then), the function it represents still gives the correct value for η 1 . The students proved analogous theorems for m = 2 and m = 3, and we are working on determining the coeffcients of 1 mn . lim m , n F m , n ( x ) Neil Calkin Counting Kings
Introduction Higher Dimensions The technique of transfer matrices doesn’t work very well for higher dimensions Neil Calkin Counting Kings
Introduction Higher Dimensions The technique of transfer matrices doesn’t work very well for higher dimensions the matrices get way too big, way too quickly, and don’t satisfy the same sorts of nice recurrences). Neil Calkin Counting Kings
Introduction Higher Dimensions The technique of transfer matrices doesn’t work very well for higher dimensions the matrices get way too big, way too quickly, and don’t satisfy the same sorts of nice recurrences). Are there other techniques that could work in 2 dimensions which might generalize better to higher dimensions? Neil Calkin Counting Kings
Introduction A statistical approach via a beautiful result of Knuth: Neil Calkin Counting Kings
Introduction A statistical approach via a beautiful result of Knuth: Theorem Given a rooted tree, take a random walk from the root to a leaf, choosing uniformly from among the children at each node. Let X j be the number of children seen at the j th vertex. Let X = X 1 X 2 ... X l be the product of these values: then E ( X ) = # leaves in the tree Neil Calkin Counting Kings
Introduction A statistical approach via a beautiful result of Knuth: Theorem Given a rooted tree, take a random walk from the root to a leaf, choosing uniformly from among the children at each node. Let X j be the number of children seen at the j th vertex. Let X = X 1 X 2 ... X l be the product of these values: then E ( X ) = # leaves in the tree This is a surprising, surprisingly trivial, surprisingly powerful theorem! Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns at depth i in the tree we will have constructed an m × i board. Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns at depth i in the tree we will have constructed an m × i board. Let X i be the number of columns of height m which can be placed adjacent to the current column. Choose one of them uniformly, to create an m × ( i + 1 ) board. Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns at depth i in the tree we will have constructed an m × i board. Let X i be the number of columns of height m which can be placed adjacent to the current column. Choose one of them uniformly, to create an m × ( i + 1 ) board. Let X = X 1 X 2 . . . X n : then E ( X ) = f ( m , n ) . Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns at depth i in the tree we will have constructed an m × i board. Let X i be the number of columns of height m which can be placed adjacent to the current column. Choose one of them uniformly, to create an m × ( i + 1 ) board. Let X = X 1 X 2 . . . X n : then E ( X ) = f ( m , n ) . This is an exact formula! Neil Calkin Counting Kings
Introduction Construct a tree corresponding to m × n configurations of kings: start with a a column of non-attacking kings, chosen uniformly from the set of f ( m , 1 ) permissible columns at depth i in the tree we will have constructed an m × i board. Let X i be the number of columns of height m which can be placed adjacent to the current column. Choose one of them uniformly, to create an m × ( i + 1 ) board. Let X = X 1 X 2 . . . X n : then E ( X ) = f ( m , n ) . This is an exact formula! Unfortunately, computing E ( X ) is not easy: but we can sample. Neil Calkin Counting Kings
Introduction The Distribution of X How should X be distributed? Neil Calkin Counting Kings
Introduction The Distribution of X How should X be distributed? Consider log X = � n i = 1 log X i . Neil Calkin Counting Kings
Introduction The Distribution of X How should X be distributed? Consider log X = � n i = 1 log X i . There are f ( m , 1 ) permissible columns that could be placed at step i − 1, and the degree X i depends exactly upon which column we place. Neil Calkin Counting Kings
Introduction The Distribution of X How should X be distributed? Consider log X = � n i = 1 log X i . There are f ( m , 1 ) permissible columns that could be placed at step i − 1, and the degree X i depends exactly upon which column we place. Let d j be the degree corresponding to the j th permissible column, and Y j ( n ) be the number of copies of the j th column which we have seen. Then f ( m , 1 ) � log X = Y j log d j j = 1 Neil Calkin Counting Kings
Introduction Consequently, log X is asymptotically normal: Neil Calkin Counting Kings
Introduction Consequently, log X is asymptotically normal: It is relatively easy to determine E ( log X ) Neil Calkin Counting Kings
Introduction Consequently, log X is asymptotically normal: It is relatively easy to determine E ( log X ) If we can estimate Var ( log X ) either exactly or through sampling, we can estimate E ( X ) . Neil Calkin Counting Kings
Introduction Higher dimensions: Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that building each board a column at a time. For an m × n × p board, the tree will have depth np . Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that building each board a column at a time. For an m × n × p board, the tree will have depth np . Still very easy to compute X i at a given depth in the tree. Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that building each board a column at a time. For an m × n × p board, the tree will have depth np . Still very easy to compute X i at a given depth in the tree. Knuth’s theorem still holds. Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that building each board a column at a time. For an m × n × p board, the tree will have depth np . Still very easy to compute X i at a given depth in the tree. Knuth’s theorem still holds. How well can we estimate the variance of log X ? Neil Calkin Counting Kings
Introduction Higher dimensions: Build a 3 dimensional m × n × p board by building first an m × n board then build a board on top of that, then a board on top of that building each board a column at a time. For an m × n × p board, the tree will have depth np . Still very easy to compute X i at a given depth in the tree. Knuth’s theorem still holds. How well can we estimate the variance of log X ? Can a better understanding of how fast the two dimensional version converges give us information on how many times we have to sample the three dimensional version? Neil Calkin Counting Kings
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