Mapping the Discrete Logarithm: Talk 2

Transcription

1 Mapping the Discrete Logarithm: Talk 2 Joshua Holden Joint work with Daniel Cloutier, Nathan Lindle (Senior Theses), Max Brugger, Christina Frederick, Andrew Hoffman, and Marcus Mace (RHIT REU) Rose-Hulman Institute of Technology Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 1 / 45

2 The Question Definition A functional graph is a directed graph such that each vertex must have exactly one edge directed out from it. Definition An m-ary functional graph is a functional graph where each node has in-degree of exactly zero or m. We are investigating the functional graph induced by the discrete exponentiation map x g x mod p. Question How much does the discrete exponentiation functional graph (DEFG) look like a random graph? (By random graph we mean a randomly chosen graph of a specified type on a specified number of nodes.) Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 2 / 45

3 Example: x 3 x mod 23 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 3 / 45

4 Some Measurements of Interest Number of connected components Number of cyclic nodes Number of image nodes Average component size (as seen from a node) Average cycle length (as seen from a node) Average tail length (as seen from a node) Maximum cycle length Maximum tail length Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 4 / 45

5 Strategy 1 Figure out what a random graph looks like. (Combinatorics exponential generating functions, Analysis) 2 Figure out what a discrete exponentiation functional graph (DEFG) looks like. (Number Theory, Computation) 3 Compare. (Statistics) Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 5 / 45

6 First Example of Exponential Generating Functions A permutation consists of a set of components, each of which is a (nonempty) cycle of nodes. The number of ways to make nodes into a (nonempty) cycle is counted by x + 2! x 2 2 2! + 3! 3 x 3 3! + = x + x x 3 ( = ln The number of ways to make a permutation is counted by e ln( 1 1 x ) = 1 1 x = 1 + 1!x + 2!x 2 So there are n! permutations on n nodes. 1 x 2! + 3!x 3 3! +. ). Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 6 / 45

7 Second Example of Exponential Generating Functions A component of a functional graph is a cycle of rooted trees (connected by their roots). The number of ways to make a rooted tree is counted by t(x) such that t(x) = xe t(x), so t(x) = W ( x). The number of ways to make nodes into these components is counted by ( ) ( ) c(x) = ln = ln. 1 1 t(x) W ( x) The number of ways to make nodes into a functional graph is counted by f (x) = e c(x) = n=0 n n x n n!. So there are n n functional graphs on n nodes. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 7 / 45

8 Third Example of Exponential Generating Functions A component of a binary functional graph (bfg) is a cycle of nodes, each of which is connected to a binary tree. The number of ways to make a binary tree is counted by b(x) such that b(x) = x + x b(x)2 1, so b(x) = 2 1 2x 2. x The number of ways to make nodes into these components is counted by ( ) ( ) 1 1 c 2 (x) = ln = ln. 1 xb(x) 1 2x 2 The number of ways to make nodes into a bfg is counted by ( ) 2n (2n)! f 2 (x) = e c2(x) x 2n = n 2 n (2n)!. n=0 So there are ( ) 2n (2n)! n 2 binary functional graphs on 2n nodes. n Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 8 / 45

9 Bivariate Egf s Now we can count a total number of functional graphs. How can we find out what a random one looks like? Theorem Let f n be the number of ways to carry out a task on {1,..., n}, n 1. Let g n be the number of ways to carry out a second task on {1,..., n}. Let h k,n be the number of ways to split {1,..., n} into k disjoint nonempty subsets, carry out the first task on each subset, and carry out the second task on the set of subsets. x i Let F(x) = f i i!, G(x) = x i g i i!, and H(x, u) = h k,i u k x i i!. i=1 Then H(x, u) = G(uF(x)). i=0 k,i=0 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 9 / 45

10 Examples of Bivariate Egf s The number of ways to make n nodes into a functional graph with k components is counted by ( ) ξ(x, u) = e uc(x) = e u ln 1 1+W ( x) = 1 + 1ux + (3u + 1u 2 ) x 2 2! + (17u + 9u2 + 1u 3 ) x 3 3! + The number of ways to make n nodes into a bfg with k components is counted by ξ 2 (x, u) = e uc 2(x) = e u ln = 1 + 2u x 2 ( ) 1 1 2x 2 2! + (24u + 12u2 ) x 4 4! + Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 10 / 45

11 Weighted Totals Now we can get total measurements. Theorem (Differentiation Formula) Let f n be the number of ways to carry out a task on {1,..., n}, n 1. Let g n be the number of ways to carry out a second task on {1,..., n}. Let h n be the number of ways to split {1,..., n} into zero or more disjoint nonempty subsets, carry out the first task on each subset, and carry out the second task on the set of subsets, with each way weighted by the number of subsets. x i Let F(x) = f i i!, G(x) = x i g i i!, and H(x) = i=1[ ] i=0 i=0 Then H(x) = u G(uF(x)). u=1 h i x i i!. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 11 / 45

12 Example of Average Measurements The total number of components in all functional graphs with n nodes is counted by [ ] ( ) 1 Ξ(x) = u euc(x) = u=1 1 + W ( x) ln W ( x) = 1x + 5 x ! + 38x 3! + 390x 4! +. So the average number of components in a functional graph with n nodes is 1, 5 2 2, , ,.... Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 12 / 45

13 Example of Variance Measurements The sum of squares of the number of components in all functional graphs with n nodes is counted by [ ( )] Ξ (x) = u u u euc(x) u=1 ( ( ) ( )) 1 1 = ln + ln W ( x) 1 + W ( x) 1 + W ( x) = 1x + 7 x 2 2! + 62x 3 3! + 700x 4 4! +. ( N ) Then we use the formula Var(z) = 1 N i=1 z 2 i z 2. So the variance in the number of components in a functional graph with (e.g.) 2 nodes is ( ) = Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 13 / 45

14 Example of Cumulative Measurements We want to find the average cycle length as seen from a node over all functional graphs. Every node in a component sees the cycle length of that component. So each component should be weighted by the number of nodes times the number of cyclic nodes. This is counted by [ 2 Ξ(x) = u w ec(x) = 1x + 10 x 2 ] 1 = 1 ut(wx) u=1,w=1 2! + 117x 3 3! x 4 4! +. W ( x) (1 + W ( x)) 4 Then we divide by the number of graphs and the number of nodes in each graph. So the average cycle length as seen from a node is 1, , , 1648,... for n = 1, 2, 3, 4,... nodes Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 14 / 45

15 Extremal Measurements Theorem Let f n be the number of ways to carry out a task on {1,..., n}, n 1. Let g n be the number of ways to carry out a second task on {1,..., n}. Let h n be the number of ways to split {1,..., n} into zero or more disjoint nonempty subsets, carry out the first task on each subset, and carry out the second task on the set of subsets, with each way weighted by the size of the largest subset. x i k Let F(x) = f i i!, F [k] (x) = i=1 i=1 x i G(x) = g i i!, and H(x) = i=0 Then H(x) = k=0 i=0 f i x i i!, h i x i i!. [ ] G(F(x)) G(F [k] (x)). Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 15 / 45

16 Explanation of the Extremal Measurements Formula G(F(x)) counts the number of ways to carry out both tasks. G(F [k] (x)) counts the number of ways to carry out both tasks if the largest subset has size at most k. So [ G(F(x)) G(F [k] (x)) ] counts the number of ways to carry out both tasks if the largest subset has size at least k + 1. Let H i (x) count the number of ways to carry out both tasks if the largest subset has size exactly i. Then H(x) = = = ih i (x) = i=1 k=1 k=0 i=1 k=1 i H i (x) = [ G(F(x)) G(F [k 1] (x)) [ ] G(F(x)) G(F [k] (x)) k=1 i=k ] H i (x) Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 16 / 45

17 Example of Extremal Measurements Recall that a permutation consists of a set of components, each of which is a (nonempty) cycle of nodes. The number of ways to make nodes into a (nonempty) cycle is counted by x + 2! x 2 2 2! + 3! 3 x 3 3! + = x + x x 3 ( = ln 1 x The number of ways to make a permutation weighted by the size of the largest cycle is counted by [ e ln( ] 1 1 x ) k e i=1 xi i = x + 3 x ! + 13x 3! + 67x 4! +. k=0 So the average maximum cycle size in a permutation with n nodes is 1, 3 2!, 13 3!, 67,... for n = 1, 2, 3, 4,... nodes. 4! ). Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 17 / 45

18 Random Graph Measurements (Standard Results from Literature) The expected mean values for the measurements of interest in a random functional graph of size n, as n, are: Number of components ln (2n) + γ 2 Number of cyclic nodes πn/2 1 3 Number of image nodes (1 e 1 )n Notation h(n) k(n) will mean h(n) = k(n) + lower order terms Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 18 / 45

19 Random Graph Measurements (II) Average cycle length Average tail length Maximum cycle length πn/8 πn/8 c 1 πn n c 1 = 0 [ ( 1 exp v e u du )] u dv Maximum tail length 2πn ln n Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 19 / 45

20 First data set p = = (2)(50021) + 1 ( safe prime ) p = = (2 3 )(3)(11)(379) + 1 p = = (2 2 )(3)(5)(7)(11)(23) + 1 For each g = 1, 2,..., p 1 we build the graph. Then we compute measurements over all graphs. The original code was written in C++ by Daniel Cloutier (2006), using the RHIT parallel cluster. There were later revisions by Nathan Lindle (2008) and by Andrew Hoffman and Marc Mace (2009). Result Arbitrary random graphs aren t a good model. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 20 / 45

21 Next Step So what is a better model? We will use some number theory(!) to understand why DEFG s aren t quite as random as the graphs we have analyzed so far. We need to account for a little bit more structure. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 21 / 45

22 m-ary Functional Graphs Let φ(k) = # {1 c k gcd(c, k) = 1} be the Euler phi function. Proposition If m (p 1) then there are φ( p 1 m ) m-ary functional graphs produced by varying g for a given p. Furthermore, the values of g that produce an m-ary graph are precisely those for which g is a strictly perfect mth power modulo p. Proof (sketch). Fix a primitive root r. Then g is a strictly perfect mth power modulo p iff g = r a and m = gcd(a, p 1). If y = r b, then g x = r ax y = r b mod p iff ax b mod p 1. This has m solutions if m b and no solutions otherwise, so the graph is m-ary. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 22 / 45

23 Image Nodes Proposition The number of image nodes in an m-ary functional graph with p 1 nodes is exactly p 1 m. Proof. In any finite directed graph, the sum of the in-degrees equals the sum of the out-degrees. In a functional graph, the sum of the out-degrees equals the number of nodes. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 23 / 45

24 Permutation Measurements: Means (Standard Results from Literature) The expected mean values for the measurements of interest in a random unary functional graph (permutation) of size n are: Number of components = n i=1 1 i = Ψ(n + 1) + γ ln n + γ Average cycle length = n Maximum cycle length c 2 n n c 2 = 0 [ ( 1 exp v e u du )] u dv where Ψ(x) = d dx ln(γ(x)). Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 24 / 45

25 Example: x 5 x mod 23 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 25 / 45

26 Permutation Measurements: Variances (Standard Results from Literature) The expected variances for the measurements of interest in a random unary functional graph (permutation) of size n are: Number of components = Ψ(n + 1) + Ψ (n + 1) + γ π2 6 ln n + γ π2 6 Average cycle length = n Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 26 / 45

27 Binary Functional Graphs For binary functional graphs, not all of the results we need seem to be in the literature. A general technique was given by Flajolet and Odlyzko: Explicitly define the structure. Convert to exponential generating functions. Mark the structures of interest. Compute expected value generating functions. Perform automatic singularity analysis to get asymptotic form of coefficients. Normalize. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 27 / 45

28 Example: x 3 x mod 23 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 28 / 45

29 Singularity Analysis How do we get information about the coefficients of generating functions? One way is to use complex analysis, which leads to theorems such as: Notation If f (x) is a generating function, [x n ]f (x) is the coefficient of x n in f (x). Theorem (Example of a Transfer Theorem ) Let α R \ {0, 1, 2, 3, }. Then [x n ](1 x) α n α 1 Γ( α). We did automatic analysis using a Maple package. (Flajolet, Salvy, Zimmerman, et al.) It applies transfer theorems to convert the type of singularity to the asymptotics of the coefficients. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 29 / 45

30 Example of Singularity Analysis Let f (x) = 1 1 2x 2 = (1 2x 2 ) 1/2. Let z = 2x 2, so [z n ](1 z) 1/2 = [2 n x 2n ](1 2x 2 ) 1/2. Now [z n ](1 z) 1/2 n 1/2 Γ(1/2) = 1 πn, so [x 2n ](1 2x 2 ) 1/2 2n πn, and the number of bfg s with n nodes is [ ] x n 2 n/2 n! n even f (x) πn/2 n! 0 n odd. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 30 / 45

31 Alternative Technique In some cases a more precise technique was also used: Explicitly define the structure. Convert to exponential generating functions. Mark the structures of interest. Compute expected value generating functions. Obtain DEs whose solutions are the generating functions. Use the DEs to find recurrence relations. Solve the recurrence relations. Normalize. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 31 / 45

32 Example of Converting to a Recurrence Relation Let f (x) = 1 1 2x 2 = (1 2x 2 ) 1/2 again. so If f (x) = f (x) = 2 2x (1 2x 2 = ) 3/2 1 2x 2 f (x), (1 2x 2 )f (x) = 2xf (x). b i x i, then f (x) = ib i x i 1, so i=0 i=1 2 jb j 1 x j = j=1 b j+1 x j, j=0 and we can extract the recurrence relation: b n+2 = 2n + 2 n + 2 b n, b 0 = 1, b 1 = 0. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 32 / 45

33 Example of Converting to a Recurrence Relation, cont. The solution to is b n+2 = 2n + 2 n + 2 b n, b 0 = 1, b 1 = 0 n! ( [x n n ]f (x) = b n = 2)! 2 2 n/2 n even 0 n odd. Again, Maple can do all of this automatically. The two methods agree: The total number of (labeled) binary functional graphs of size n is ( n 2 n! ) 2! 2 2 n/2 n! 2 n/2. πn/2 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 33 / 45

34 BFG Measurements: Means (Dan Cloutier, 2006 and Nathan Lindle, 2008) The expected mean values for the measurements of interest in a random binary functional graph of size n are: ( ) ( 1 n Number of components = Ψ ln (2n) + γ 2 Number of cyclic nodes = ) + πγ( n 2 + 1) Γ( n ) Γ( n ) πn/2 1 ( ) 1 γ + ln(2) 2 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 34 / 45

35 BFG Measurements: Means (II) Average cycle length = ( 1 2 ) πγ( n 2 + 1) Γ( n ) πn/8 πγ(2 + n 2 Average tail length = ) nγ( n ) Γ( n ) nγ( n ) πn/8 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 35 / 45

36 BFG Measurements: Means (III) (Cloutier, 2006) Maximum cycle length c 1 πn n c 1 = 0 [ ( 1 exp v e u du )] u dv (Holden, 2009) Maximum tail length 2πn ln 2 ln n 2 2 γ ln n.5 ln n Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 36 / 45

37 BFG Measurements: Variances (Nathan Lindle, 2008) The expected variances for the measurements of interest in a random binary functional graph of size n are: Number of components = 1 2 Ψ ( n ) + 2 ln(n) ln(2)2 2γ ln(2) γ 2 4 Number of cyclic nodes ( ) 1 γ + ln(2) γ2 + γ ln(2) + ln(2) 2 + n/2 1 = 2Γ( n )2 + 4 n 2 Γ( n )2 πγ( n 2 + 1)Γ( n ) πγ( n 2 + 1)2 Γ( n )2 ( 2 π ) 2πn n + 2 π l=0 Ψ(l ) (2l + 1) 1 ( n 4 Ψ ) 2 2 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 37 / 45

38 BFG Measurements: Variances (II) Average cycle length = 1 ( 3πΓ( n 2 + 1)2 6 πγ( n 2 + 1)Γ( n ) + 16 n 2 Γ( n )2 + 8Γ( n )2 ) 12 Γ( n )2 ( 2 3 π ) 2πn n π 16 Average tail length = 1 (18Γ( n ) + 8Γ( n ) n 2 9 n 2 6 ( n 2 )Γ( n ) πγ( n 2 + 1) 9 πγ( n 2 + 1)) ( πγ( n 2 + 1) + n 2 πγ( n 2 + 1) 2Γ( n ))2 ( n 2 )2 Γ( n )2 1 4 ( 2 3 π ) n 8 2πn π 16 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 38 / 45

39 Ternary Functional Graphs For ternary functional graphs, a slightly different technique was used: Explicitly define the structure. Convert to exponential generating functions. Mark the structures of interest. Compute expected value generating functions. Obtain DEs whose solutions are the generating functions. Use the DEs to find recurrence relations. Recursively compute expected values. Normalize. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 39 / 45

40 TFG Structure A ternary functional graph (tfg) is a set of components. Each component is a cycle of nodes, each of which is connected to two ternary trees. A ternary tree consists of either a root node or a root node and three interchangeable nonempty ternary tree branches. Shorthand: TernFunGraph = set(components) Component = cycle(node*set(terntree, cardinality=2) TernTree = Node + Node*set(TernTree, cardinality = 3) Node = Atomic Unit Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 40 / 45

41 Convert to Generating Functions f (z) = e c(z) c(z) = ln 1 1 z 2 t2 (z) (Ternary functional graphs) (Connected components) t(z) = z + 1 3! zt3 (z) (Ternary trees) Note the increasing difficulty in solving explicitly for the tree function. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 41 / 45

42 TFG Generating Functions The expected mean values for the measurements of interest in a random ternary functional graph of size n are given by the following generating functions: (Max Brugger and Christina Frederick, 2007) [ ] Number of components: u euc(z) u=1 ( ) Number of cyclic nodes: (Brugger, 2008) Average tail length: u eln [ u uzt2 (z) u=1 zut(z) ( zt2 (z) ) 2 ( uzt2 (z) ) ] u=1 Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 42 / 45

43 TFG Generating Functions (II) For the following measurements, a double marking technique was used: (Brugger and Frederick, 2007) [ ( )] 2 1 Average component size: u w ec(z) ln uwzt2 (uwz) [ ( )] 2 1 Average cycle length: u w ec(z) ln uwzt2 (wz) u=1,w=1 u=1,w=1 Generating functions for maximum measurements have not yet been derived in the ternary case. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 43 / 45

44 Next Question How do the actual discrete exponentiation functional graphs compare with this model? We will look at the collected data and some statistical analysis. Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 44 / 45

45 Thanks! Joshua Holden (RHIT) Mapping the Discrete Logarithm: Talk 2 45 / 45