Certainty from uncertainty: the probabilistic method

Transcription

1 Certainty from uncertainty: the probabilistic method Lily Silverstein UC Davis Nov 9, 2017 Seattle University 0 / 17

2 Ramsey numbers 1 / 17

3 Ramsey numbers A graph is a set of vertices {1,..., n} and edges of the form {i, j}, i, j V / 17

4 Ramsey numbers A graph is a set of vertices {1,..., n} and edges of the form {i, j}, i, j V The complete graph on n vertices, written K n, has all ( n 2) edges K 6 K 3 1 / 17

5 Ramsey numbers A red-and-blue coloring of K n assigns a color to each edge {i, j} Two colorings of K 6. 2 / 17

6 Ramsey numbers A red-and-blue coloring of K n assigns a color to each edge {i, j} Two colorings of K 6. Claim: every red-and-blue coloring of K 6 contains a red K 3 or a blue K 3. 2 / 17

7 Ramsey numbers The Ramsey number R(k, k) is the smallest number of vertices n such that every red-and-blue coloring of K n has a monochromatic (all red or all blue) K k. 3 / 17

8 Ramsey numbers The Ramsey number R(k, k) is the smallest number of vertices n such that every red-and-blue coloring of K n has a monochromatic (all red or all blue) K k. We just saw that every coloring of K 6 has a monochromatic K 3. 3 / 17

9 Ramsey numbers The Ramsey number R(k, k) is the smallest number of vertices n such that every red-and-blue coloring of K n has a monochromatic (all red or all blue) K k. We just saw that every coloring of K 6 has a monochromatic K 3. On the other hand, not every coloring of K 5 has a monochromatic K 3 : Therefore R(3, 3) = 6. 3 / 17

10 Which Ramsey numbers do we know?. 4 / 17

11 R(3, 3) = 6 Which Ramsey numbers do we know?. 4 / 17

12 R(3, 3) = 6 R(4, 4) = Which Ramsey numbers do we know?. 4 / 17

13 R(3, 3) = 6 R(4, 4) = 18 Which Ramsey numbers do we know?. 4 / 17

14 R(3, 3) = 6 R(4, 4) = 18 R(5, 5) = Which Ramsey numbers do we know?. 4 / 17

15 R(3, 3) = 6 R(4, 4) = 18 R(5, 5) =? Which Ramsey numbers do we know?. 4 / 17

16 R(3, 3) = 6 R(4, 4) = 18 R(5, 5) =? R(6, 6) =? R(7, 7) =? R(8, 8) =? Which Ramsey numbers do we know?. 4 / 17

17 Which Ramsey numbers do we know? R(3, 3) = 6 R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) R(8, 8) = ( ). 4 / 17

18 R(3, 3) = 6 Which Ramsey numbers do we know? R(4, 4) = 18 R(5, 5) = (43 48) How hard would it be to use brute force, and simply check all colorings of K 43 for monochromatic K 5 s? R(6, 6) = ( ) R(7, 7) = ( ) R(8, 8) = ( ). 4 / 17

19 R(3, 3) = 6 Which Ramsey numbers do we know? R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) How hard would it be to use brute force, and simply check all colorings of K 43 for monochromatic K 5 s? 2 (43 2 ) colorings, each with ( 43 5 ) sets of 5 vertices to check. R(8, 8) = ( ). 4 / 17

20 R(3, 3) = 6 Which Ramsey numbers do we know? R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) R(8, 8) = ( ) How hard would it be to use brute force, and simply check all colorings of K 43 for monochromatic K 5 s? 2 (43 2 ) ( colorings, each with 43 ) 5 sets of 5 vertices to check. ( ) 43 2 (43 2 ) / 17

21 R(3, 3) = 6 Which Ramsey numbers do we know? R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) R(8, 8) = ( ). How hard would it be to use brute force, and simply check all colorings of K 43 for monochromatic K 5 s? 2 (43 2 ) ( colorings, each with 43 ) 5 sets of 5 vertices to check. ( ) 43 2 (43 2 ) Number of atoms in the observable universe: / 17

22 Which Ramsey numbers do we know? R(3, 3) = 6 R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) R(8, 8) = ( ). Suppose aliens invade the earth and threaten to obliterate it in a year s time unless human beings can find the Ramsey number for red five and blue five. We could marshal the world s best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded the Ramsey number for red six and blue six, however, we would have no choice but to launch a preemptive attack. Paul Erdős 4 / 17

23 Which Ramsey numbers do we know? R(3, 3) = 6 R(4, 4) = 18 R(5, 5) = (43 48) R(6, 6) = ( ) R(7, 7) = ( ) Before Erdős, we didn t know any bounds on Ramsey numbers (although Ramsey did prove they are finite). Let s look at how Erdős found a lower bound using the probabilistic method. R(8, 8) = ( ). 4 / 17

24 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

25 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

26 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

27 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

28 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

29 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

30 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

31 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent / 17

32 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent. 6 5 Let G(n) denote a random coloring of the complete graph on n vertices / 17

33 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent. 6 5 Let G(n) denote a random coloring of the complete graph on n vertices. 7 4 What s the probability that G(n) has a monochromatic K k? / 17

34 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent. 6 5 Let G(n) denote a random coloring of the complete graph on n vertices What s the probability that G(n) has a monochromatic K k? P [G(6) has a monochromatic K 3 ] = 1 P [G(5) has a monochromatic K 3 ] < / 17

35 Randomly colored graphs Color each of the ( n 2) edges of Kn either red or blue, with probability 1 2 each. The choices are independent. 6 5 Let G(n) denote a random coloring of the complete graph on n vertices What s the probability that G(n) has a monochromatic K k? P [G(6) has a monochromatic K 3 ] = 1 P [G(5) has a monochromatic K 3 ] < 1 Idea: if for any k and n we have 9 2 P [G(n) has a monochromatic K k ] < 1, 1 then R(k, k) > n. 5 / 17

36 All the probability you need to know (for the next proof) Independence If A and B are independent events, then P [A B] = P [A] P [B]. Subadditivity If A 1,..., A n are (not necessarily independent) events, then n P [A 1 A 2 A n ] P [A i ]. i=1 6 / 17

37 Theorem (Erdős, 1947) For all k 3, R(k, k) > 2 k/2. Proof. Recall G(n) is our random blue-and-red coloring of K n. For a fixed set of k vertices, S, let X S denote the event: the set of edges {{i, j} : i, j S} is monochromatic. Outline of proof: 1. P [X S ] = 2 1 (k 2). [ ] 2. P S =k X S ( n k) 2 1 ( k 2). 3. If n = 2 k/2, then ( n k) 2 1 ( k 2) < 1. Conclusion: if n = 2 k/2, then G(n) has positive probability of no monochromatic K k s, hence R(k, k) > n. 7 / 17

38 A bit more probability Expectation If X is a discrete random variable that takes values a 1, a 2,..., a n, then n E [X ] := a i P [X = a i ]. Linearity of expectation If X = n i=1 X i, then i=1 E [X ] = n E [X i ]. i=1 Useful observation about expectation If E [X ] = b, then there is some a b such that P [X = a] > 0, and there is some c b such that P [X = c] > 0 8 / 17

39 A different lower bound Theorem For any integer n, R(k, k) > n ( n k) 2 1 ( k 2). Proof. Outline: 1. Let X = S =k X S. Then E [X ] = ( n k) 2 1 ( k 2) = m. 2. There must be a blue-and-red coloring with no more than m monochromatic K k. 3. Discard one vertex from each monochromatic K k. We now have a coloring of the complete graph on (at least) n m vertices that has no monochromatic K k s. Hence R(k, k) > n m. 9 / 17

40 Why I care 10 / 17

41 Why I care Finding the exact values of R(k, k) is a very hard problem that no one knows how to solve efficiently. 10 / 17

42 Why I care Finding the exact values of R(k, k) is a very hard problem that no one knows how to solve efficiently. The probabilistic approach is easy computationally, and gives us at least a partial answer. 10 / 17

43 Dimension of a solution space Let M be a collection of monomials in n variables.. M = {x 3 1 x 2 2 x 3, x 4 1, x 2 3 x 2 4, x 4 x 3 5 } 11 / 17

44 Dimension of a solution space Let M be a collection of monomials in n variables. Fix some field k (e.g., real numbers R, complex numbers C). Let V (M) be the set V (M) = {(a 1, a 2,..., a n ) k n : f (a) = 0 for all f M} M = {x 3 1 x 2 2 x 3, x 4 1, x 2 3 x 2 4, x 4 x 3 5 }. V (M) = all vectors in R 5 that simultaneously satisfy x 3 1 x 2 2 x 3 = 0 x 4 1 = 0 x 2 3 x 2 4 = 0 x 4 x 3 5 = 0 11 / 17

45 Dimension of a solution space Let M be a collection of monomials in n variables. Fix some field k (e.g., real numbers R, complex numbers C). Let V (M) be the set V (M) = {(a 1, a 2,..., a n ) k n : f (a) = 0 for all f M} M = {x 3 1 x 2 2 x 3, x 4 1, x 2 3 x 2 4, x 4 x 3 5 }. V (M) = all vectors in R 5 that simultaneously satisfy x 3 1 x 2 2 x 3 = 0 x 4 1 = 0 x 2 3 x 2 4 = 0 x 4 x 3 5 = 0 V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} 11 / 17

46 Dimension of a solution space Let M be a collection of monomials in n variables. Fix some field k (e.g., real numbers R, complex numbers C). Let V (M) be the set V (M) = {(a 1, a 2,..., a n ) k n : f (a) = 0 for all f M} In general, V (M) is a union of subspaces of k n. The dimension of V (M) is the maximum dimension of one of these subspaces. M = {x 3 1 x 2 2 x 3, x 4 1, x 2 3 x 2 4, x 4 x 3 5 }. V (M) = all vectors in R 5 that simultaneously satisfy V (M) = x 3 1 x 2 2 x 3 = 0 x 4 1 = 0 x 2 3 x 2 4 = 0 x 4 x 3 5 = 0 {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} 11 / 17

47 Dimension of a solution space Let M be a collection of monomials in n variables. Fix some field k (e.g., real numbers R, complex numbers C). Let V (M) be the set V (M) = {(a 1, a 2,..., a n ) k n : f (a) = 0 for all f M} In general, V (M) is a union of subspaces of k n. The dimension of V (M) is the maximum dimension of one of these subspaces. M = {x 3 1 x 2 2 x 3, x 4 1, x 2 3 x 2 4, x 4 x 3 5 }. V (M) = all vectors in R 5 that simultaneously satisfy V (M) = x 3 1 x 2 2 x 3 = 0 x 4 1 = 0 x 2 3 x 2 4 = 0 x 4 x 3 5 = 0 {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} dim V (M) = 3 11 / 17

48 Vertex covers M = {x1 3 x2 2 x 3, x1 4, x3 2 x4 2, x 4 x5 3 }. V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} dim V (M) = 3 12 / 17

49 Vertex covers From our set of monomials M we define a hypergraph H where the vertices of H are {x 1,..., x n} for each f M we include the edge {x i : x i divides f } M = {x1 3 x2 2 x 3, x1 4, x3 2 x4 2, x 4 x5 3 }. V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} x 5 x 4 x 3 dim V (M) = 3 x 1 x 2 12 / 17

50 Vertex covers From our set of monomials M we define a hypergraph H where the vertices of H are {x 1,..., x n} for each f M we include the edge {x i : x i divides f } M = {x1 3 x2 2 x 3, x1 4, x3 2 x4 2, x 4 x5 3 }. V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} x 5 x 4 x 3 dim V (M) = 3 Minimal vertex covers of H: {x 1, x 4 }, {x 1, x 3, x 5 } x 1 x 2 A vertex cover of H is a set S of vertices, such that every edge of H contains at least one member of S. A vertex cover is minimal if no subset of it is a vertex cover. 12 / 17

51 Vertex covers From our set of monomials M we define a hypergraph H where the vertices of H are {x 1,..., x n} for each f M we include the edge {x i : x i divides f } M = {x1 3 x2 2 x 3, x1 4, x3 2 x4 2, x 4 x5 3 }. V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} x 5 x 4 x 3 x 1 x 2 dim V (M) = 3 Minimal vertex covers of H: {x 1, x 4 }, {x 1, x 3, x 5 }...variables set to zero in V (M) subspaces! A vertex cover of H is a set S of vertices, such that every edge of H contains at least one member of S. A vertex cover is minimal if no subset of it is a vertex cover. 12 / 17

52 Vertex covers From our set of monomials M we define a hypergraph H where the vertices of H are {x 1,..., x n} for each f M we include the edge {x i : x i divides f } M = {x1 3 x2 2 x 3, x1 4, x3 2 x4 2, x 4 x5 3 }. V (M) = {(0, a 2, a 3, 0, a 5 ) a 2, a 3, a 5 R} {(0, a 2, 0, a 4, 0) a 2, a 4 R} x 5 x 4 x 3 x 1 x 2 dim V (M) = 3 Minimal vertex covers of H: {x 1, x 4 }, {x 1, x 3, x 5 }...variables set to zero in V (M) subspaces! A vertex cover of H is a set S of vertices, such that every edge of H contains at least one member of S. A vertex cover is minimal if no subset of it is a vertex cover. dim(v(m)) = n minimum size of a vertex cover 12 / 17

53 Applications to optimization & engineering Facility location Each vertex is the location of a possible bus stop. Each edge is a neighborhood; the vertices it contains are the bus stops accessible from that neighborhood. Minimum vertex cover = fewest bus stops to build to provide service to everyone. Electrical circuit design Each vertex is a component on an integrated chip. Each edge is a set of components that form a feedback loop. Minimum vertex cover = smallest set of component changes to break all cycles. Constraint satisfaction Each vertex is a true/false statement. Each edge is a set of propositions that can t simultaneously be true. Minimum vertex cover = smallest set of propositions that need to be removed to make the system satisfiable. x 5 x 4 x 3 x 1 x 2 13 / 17

54 Applications to optimization & engineering Facility location Each vertex is the location of a possible bus stop. Each edge is a neighborhood; the vertices it contains are the bus stops accessible from that neighborhood. Minimum vertex cover = fewest bus stops to build to provide service to everyone. Electrical circuit design Each vertex is a component on an integrated chip. Each edge is a set of components that form a feedback loop. Minimum vertex cover = smallest set of component changes to break all cycles. Constraint satisfaction Each vertex is a true/false statement. Each edge is a set of propositions that can t simultaneously be true. Minimum vertex cover = smallest set of propositions that need to be removed to make the system satisfiable. Unfortunately... MINIMUM VERTEX COVER is NP-hard. In practice we can only solve small instances, or ones with special structure. 13 / 17

55 Finding a minimum vertex cover is a very hard problem that no one knows how to solve efficiently. 14 / 17

56 Finding a minimum vertex cover is a very hard problem that no one knows how to solve efficiently. Can we use a probabilistic approach to get a partial answer, using a much easier computation? 14 / 17

57 A method for generating random instances of this problem: Choose our number of variables, n. Choose a maximum total degree for the monomials, D. (Total degree = sum of exponents, e.g., tdeg(x 5 1 x 2x 2 3 ) = 8.) Choose some probability 0 < p < 1. Finally, for each of the ( ) n+d n monomials in n variables of total degree up to D, we select it to be in our set with probability p. For each monomial the choice is independent. For example, if n = 5, D = 5, and p = 0.05, then / 17

58 monomials of total degree 5 in the variables x 1,..., x 5 x 1, x 2, x 3, x 4, x 5, x 2 1, x 1x 2, x 1x 3, x 1x 4, x 1x 5, x 2 2, x 2x 3, x 2x 4, x 2x 5, x 2 3, x 3x 4, x 3x 5, x 2 4, x 4x 5, x 2 5, x 3 1, x 2 1 x 2, x 2 1 x 3, x 2 1 x 4, x 2 1 x 5, x 1x 2 2, x 1x 2x 3, x 1x 2x 4, x 1x 2x 5, x 1x 2 3, x 1x 3x 4, x 1x 3x 5, x 1x 2 4, x 1x 4x 5, x 1x 2 5, x 3 2, x 2 2 x 3, x 2 2 x 4, x 2 2 x 5, x 2x 2 3, x 2x 3x 4, x 2x 3x 5, x 2x 2 4, x 2x 4x 5, x 2x 2 5, x 3 3, x 2 3 x 4, x 2 3 x 5, x 3x 2 4, x 3x 4x 5, x 3x 2 5, x 3 4, x 2 4 x 5, x 4x 2 5, x 3 5, x 4 1, x 3 1 x 2, x 3 1 x 3, x 3 1 x 4, x 3 1 x 5, x 2 1 x 2 2, x 2 1 x 2x 3, x 2 1 x 2x 4, x 2 1 x 2x 5, x 2 1 x 2 3, x 2 1 x 3x 4, x 2 1 x 3x 5, x 2 1 x 2 4, x 2 1 x 4x 5, x 2 1 x 2 5, x 1x 3 2, x 1x 2 2 x 3, x 1x 2 2 x 4, x 1x 2 2 x 5, x 1x 2x 2 3, x 1x 2x 3x 4, x 1x 2x 3x 5, x 1x 2x 2 4, x 1x 2x 4x 5, x 1x 2x 2 5, x 1x 3 3, x 1x 2 3 x 4, x 1x 2 3 x 5, x 1x 3x 2 4, x 1x 3x 4x 5, x 1x 3x 2 5, x 1x 3 4, x 1x 2 4 x 5, x 1x 4x 2 5, x 1x 3 5, x 4 2, x 3 2 x 3, x 3 2 x 4, x 3 2 x 5, x 2 2 x 2 3, x 2 2 x 3x 4, x 2 2 x 3x 5, x 2 2 x 2 4, x 2 2 x 4x 5, x 2 2 x 2 5, x 2x 3 3, x 2x 2 3 x 4, x 2x 2 3 x 5, x 2x 3x 2 4, x 2x 3x 4x 5, x 2x 3x 2 5, x 2x 3 4, x 2x 2 4 x 5, x 2x 4x 2 5, x 2x 3 5, x 4 3, x 3 3 x 4, x 3 3 x 5, x 2 3 x 2 4, x 2 3 x 4x 5, x 2 3 x 2 5, x 3x 3 4, x 3x 2 4 x 5, x 3x 4x 2 5, x 3x 3 5, x 4 4, x 3 4 x 5, x 2 4 x 2 5, x 4x 3 5, x 4 5, x 5 1, x 4 1 x 2, x 4 1 x 3, x 4 1 x 4, x 4 1 x 5, x 3 1 x 2 2, x 3 1 x 2x 3, x 3 1 x 2x 4, x 3 1 x 2x 5, x 3 1 x 2 3, x 3 1 x 3x 4, x 3 1 x 3x 5, x 3 1 x 2 4, x 3 1 x 4x 5, x 3 1 x 2 5, x 2 1 x 3 2, x 2 1 x 2 2 x 3, x 2 1 x 2 2 x 4, x 2 1 x 2 2 x 5, x 2 1 x 2x 2 3, x 2 1 x 2x 3x 4, x 2 1 x 2x 3x 5, x 2 1 x 2x 2 4, x 2 1 x 2x 4x 5, x 2 1 x 2x 2 5, x 2 1 x 3 3, x 2 1 x 2 3 x 4, x 2 1 x 2 3 x 5, x 2 1 x 3x 2 4, x 2 1 x 3x 4x 5, x 2 1 x 3x 2 5, x 2 1 x 3 4, x 2 1 x 2 4 x 5, x 2 1 x 4x 2 5, x 2 1 x 3 5, x 1x 4 2, x 1x 3 2 x 3, x 1x 3 2 x 4, x 1x 3 2 x 5, x 1x 2 2 x 2 3, x 1x 2 2 x 3x 4, x 1x 2 2 x 3x 5, x 1x 2 2 x 2 4, x 1x 2 2 x 4x 5, x 1x 2 2 x 2 5, x 1x 2x 3 3, x 1x 2x 2 3 x 4, x 1x 2x 2 3 x 5, x 1x 2x 3x 2 4, x 1x 2x 3x 4x 5, x 1x 2x 3x 2 5, x 1x 2x 3 4, x 1x 2x 2 4 x 5, x 1x 2x 4x 2 5, x 1x 2x 3 5, x 1x 4 3, x 1x 3 3 x 4, x 1x 3 3 x 5, x 1x 2 3 x 2 4, x 1x 2 3 x 4x 5, x 1x 2 3 x 2 5, x 1x 3x 3 4, x 1x 3x 2 4 x 5, x 1x 3x 4x 2 5, x 1x 3x 3 5, x 1x 4 4, x 1x 3 4 x 5, x 1x 2 4 x 2 5, x 1x 4x 3 5, x 1x 4 5, x 5 2, x 4 2 x 3, x 4 2 x 4, x 4 2 x 5, x 3 2 x 2 3, x 3 2 x 3x 4, x 3 2 x 3x 5, x 3 2 x 2 4, x 3 2 x 4x 5, x 3 2 x 2 5, x 2 2 x 3 3, x 2 2 x 2 3 x 4, x 2 2 x 2 3 x 5, x 2 2 x 3x 2 4, x 2 2 x 3x 4x 5, x 2 2 x 3x 2 5, x 2 2 x 3 4, x 2 2 x 2 4 x 5, x 2 2 x 4x 2 5, x 2 2 x 3 5, x 2x 4 3, x 2x 3 3 x 4, x 2x 3 3 x 5, x 2x 2 3 x 2 4, x 2x 2 3 x 4x 5, x 2x 2 3 x 2 5, x 2x 3x 3 4, x 2x 3x 2 4 x 5, x 2x 3x 4x 2 5, x 2x 3x 3 5, x 2x 4 4, x 2x 3 4 x 5, x 2x 2 4 x 2 5, x 2x 4x 3 5, x 2x 4 5, x 5 3, x 4 3 x 4, x 4 3 x 5, x 3 3 x 2 4, x 3 3 x 4x 5, x 3 3 x 2 5, x 2 3 x 3 4, x 2 3 x 2 4 x 5, x 2 3 x 4x 2 5, x 2 3 x 3 5, x 3x 4 4, x 3x 3 4 x 5, x 3x 2 4 x 2 5, x 3x 4x 3 5, x 3x 4 5, x 5 4, x 4 4 x 5, x 3 4 x 2 5, x 2 4 x 3 5, x 4x 4 5, x / 17

59 random monomials selected with probability 0.05 each x 1, x 2, x 3, x 4, x 5, x 2 1, x 1x 2, x 1x 3, x 1x 4, x 1x 5, x 2 2, x 2x 3, x 2x 4, x 2x 5, x 2 3, x 3x 4, x 3x 5, x 2 4, x 4x 5, x 2 5, x 3 1, x 2 1 x 2, x 2 1 x 3, x 2 1 x 4, x 2 1 x 5, x 1x 2 2, x 1x 2x 3, x 1x 2x 4, x 1x 2x 5, x 1x 2 3, x 1x 3x 4, x 1x 3x 5, x 1x 2 4, x 1x 4x 5, x 1x 2 5, x 3 2, x 2 2 x 3, x 2 2 x 4, x 2 2 x 5, x 2x 2 3, x 2x 3x 4, x 2x 3x 5, x 2x 2 4, x 2x 4x 5, x 2x 2 5, x 3 3, x 2 3 x 4, x 2 3 x 5, x 3x 2 4, x 3x 4x 5, x 3x 2 5, x 3 4, x 2 4 x 5, x 4x 2 5, x 3 5, x 4 1, x 3 1 x 2, x 3 1 x 3, x 3 1 x 4, x 3 1 x 5, x 2 1 x 2 2, x 2 1 x 2x 3, x 2 1 x 2x 4, x 2 1 x 2x 5, x 2 1 x 2 3, x 2 1 x 3x 4, x 2 1 x 3x 5, x 2 1 x 2 4, x 2 1 x 4x 5, x 2 1 x 2 5, x 1x 3 2, x 1x 2 2 x 3, x 1x 2 2 x 4, x 1x 2 2 x 5, x 1x 2x 2 3, x 1x 2x 3x 4, x 1x 2x 3x 5, x 1x 2x 2 4, x 1x 2x 4x 5, x 1x 2x 2 5, x 1x 3 3, x 1x 2 3 x 4, x 1x 2 3 x 5, x 1x 3x 2 4, x 1x 3x 4x 5, x 1x 3x 2 5, x 1x 3 4, x 1x 2 4 x 5, x 1x 4x 2 5, x 1x 3 5, x 4 2, x 3 2 x 3, x 3 2 x 4, x 3 2 x 5, x 2 2 x 2 3, x 2 2 x 3x 4, x 2 2 x 3x 5, x 2 2 x 2 4, x 2 2 x 4x 5, x 2 2 x 2 5, x 2x 3 3, x 2x 2 3 x 4, x 2x 2 3 x 5, x 2x 3x 2 4, x 2x 3x 4x 5, x 2x 3x 2 5, x 2x 3 4, x 2x 2 4 x 5, x 2x 4x 2 5, x 2x 3 5, x 4 3, x 3 3 x 4, x 3 3 x 5, x 2 3 x 2 4, x 2 3 x 4x 5, x 2 3 x 2 5, x 3x 3 4, x 3x 2 4 x 5, x 3x 4x 2 5, x 3x 3 5, x 4 4, x 3 4 x 5, x 2 4 x 2 5, x 4x 3 5, x 4 5, x 5 1, x 4 1 x 2, x 4 1 x 3, x 4 1 x 4, x 4 1 x 5, x 3 1 x 2 2, x 3 1 x 2x 3, x 3 1 x 2x 4, x 3 1 x 2x 5, x 3 1 x 2 3, x 3 1 x 3x 4, x 3 1 x 3x 5, x 3 1 x 2 4, x 3 1 x 4x 5, x 3 1 x 2 5, x 2 1 x 3 2, x 2 1 x 2 2 x 3, x 2 1 x 2 2 x 4, x 2 1 x 2 2 x 5, x 2 1 x 2x 2 3, x 2 1 x 2x 3x 4, x 2 1 x 2x 3x 5, x 2 1 x 2x 2 4, x 2 1 x 2x 4x 5, x 2 1 x 2x 2 5, x 2 1 x 3 3, x 2 1 x 2 3 x 4, x 2 1 x 2 3 x 5, x 2 1 x 3x 2 4, x 2 1 x 3x 4x 5, x 2 1 x 3x 2 5, x 2 1 x 3 4, x 2 1 x 2 4 x 5, x 2 1 x 4x 2 5, x 2 1 x 3 5, x 1x 4 2, x 1x 3 2 x 3, x 1x 3 2 x 4, x 1x 3 2 x 5, x 1x 2 2 x 2 3, x 1x 2 2 x 3x 4, x 1x 2 2 x 3x 5, x 1x 2 2 x 2 4, x 1x 2 2 x 4x 5, x 1x 2 2 x 2 5, x 1x 2x 3 3, x 1x 2x 2 3 x 4, x 1x 2x 2 3 x 5, x 1x 2x 3x 2 4, x 1x 2x 3x 4x 5, x 1x 2x 3x 2 5, x 1x 2x 3 4, x 1x 2x 2 4 x 5, x 1x 2x 4x 2 5, x 1x 2x 3 5, x 1x 4 3, x 1x 3 3 x 4, x 1x 3 3 x 5, x 1x 2 3 x 2 4, x 1x 2 3 x 4x 5, x 1x 2 3 x 2 5, x 1x 3x 3 4, x 1x 3x 2 4 x 5, x 1x 3x 4x 2 5, x 1x 3x 3 5, x 1x 4 4, x 1x 3 4 x 5, x 1x 2 4 x 2 5, x 1x 4x 3 5, x 1x 4 5, x 5 2, x 4 2 x 3, x 4 2 x 4, x 4 2 x 5, x 3 2 x 2 3, x 3 2 x 3x 4, x 3 2 x 3x 5, x 3 2 x 2 4, x 3 2 x 4x 5, x 3 2 x 2 5, x 2 2 x 3 3, x 2 2 x 2 3 x 4, x 2 2 x 2 3 x 5, x 2 2 x 3x 2 4, x 2 2 x 3x 4x 5, x 2 2 x 3x 2 5, x 2 2 x 3 4, x 2 2 x 2 4 x 5, x 2 2 x 4x 2 5, x 2 2 x 3 5, x 2x 4 3, x 2x 3 3 x 4, x 2x 3 3 x 5, x 2x 2 3 x 2 4, x 2x 2 3 x 4x 5, x 2x 2 3 x 2 5, x 2x 3x 3 4, x 2x 3x 2 4 x 5, x 2x 3x 4x 2 5, x 2x 3x 3 5, x 2x 4 4, x 2x 3 4 x 5, x 2x 2 4 x 2 5, x 2x 4x 3 5, x 2x 4 5, x 5 3, x 4 3 x 4, x 4 3 x 5, x 3 3 x 2 4, x 3 3 x 4x 5, x 3 3 x 2 5, x 2 3 x 3 4, x 2 3 x 2 4 x 5, x 2 3 x 4x 2 5, x 2 3 x 3 5, x 3x 4 4, x 3x 3 4 x 5, x 3x 2 4 x 2 5, x 3x 4x 3 5, x 3x 4 5, x 5 4, x 4 4 x 5, x 3 4 x 2 5, x 2 4 x 3 5, x 4x 4 5, x / 17

60 A method for generating random instances of this problem: Choose our number of variables, n. Choose a maximum total degree for the monomials, D. (Total degree = sum of exponents, e.g., tdeg(x 5 1 x 2x 2 3 ) = 8.) Choose some probability 0 < p < 1. Finally, for each of the ( ) n+d n monomials in n variables of total degree up to D, we select it to be in our set with probability p. For each monomial the choice is independent. For example, if n = 5, D = 5, and p = 0.05, then... In this instance, we generated the random set {x 2 x 4, x 2 1 x 4, x 2 x 3 x 5, x 3 4, x 1x 2 2 x 3, x 1 x 2 3 x 4, x 1 x 3 5, x 3 2 x 4, x 2 x 3 4, x 3 4 x 5, x 2 1 x 2 3 x 5, x 1 x 2 x 2 3 x 5, x 1 x 2 x 4 x 2 5, x 4 2 x 5, x 2 x 3 x 3 4, x 2 3 x 4x 2 5 }. 15 / 17

61 Dimension of solution space for random instances For the random model, we can accurately predict the dimension from the values of n, D, and p. 16 / 17

62 Dimension of solution space for random instances For the random model, we can accurately predict the dimension from the values of n, D, and p. Theorem (De Loera, Petrović, Stasi, S., Wilburne 2017) If 1 D t+1 << p << 1 D t, then with high probability dim(v (M)) = t. The statement holds for any t, 0 t n. Here, with high probability means the following: if you fix n and let D, then the probability that dim(v (M)) = t converges to 1, for p in this range. 16 / 17

63 Thanks for your attention! Alon and Spencer, The Probabilistic Method, 4th ed, Wiley 2016 De Loera, Petrović, S, Stasi, and Wilburne, Random Monomial Ideals, arxiv , 2017 For a probabilistic proof of existence in algebra, check out: Daniel Erman and Jay Yang, Random Flag Complexes and Asymptotic Syzygies, arxiv , / 17