Reverse engineering using computational algebra

Transcription

1 Reverse engineering using computational algebra Matthew Macauley Department of Mathematical Sciences Clemson University Math 4500, Fall 2016 M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

2 What is reverse engineering? Sometimes, complex biological systems can seem a bit like this: (click here!). Systems biology is the study of systems of biological components. A central problem in systems biology is to use experimental data to infer the structure of a system such as a gene regulatory network. Modeling approaches Bottom-up: Build a network from the known local information about every single object. Top-down ( Reverse-engineering ): View the system as a black box, then use the available data to make a model. Previously, we ve mostly studied the first approach to modeling. In this lecture, we ll focus on the second approach. Many problems in statistics (e.g., linear regression) deal with the second approach. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

3 The blind men and the elephant An old parable from India tells of several blind men who try to determine what an elephant looks like just by touch. The blind men are trying to reverse engineer an elephant from just a few data points. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

4 Inferring a Boolean network model (elephant) from data (observations) Consider a Boolean network model on n nodes, with update function F : F n 2 Fn 2. There are 2 n input states. Suppose we don t know the actual function F, but through experimental data, we are able to observe several transitions: s 1 = (s 11, s 12,..., s 1n ) s 2 = (s 21,..., s 2n ) s m = (s m1,..., s mn) t 1 = (t 11, t 12,..., t 1n ) t 2 = (t 21,..., t 2n ) t m = (t m1,..., t mn) Reverse engineering Start with experimental data (observations) and reconstruct the model (elephant). The two main features are: (i) the network topology, or wiring diagram, (ii) the Boolean functions at each node: F = (f 1,..., f n). M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

5 Inferring a Boolean network model (elephant) from data (observations) Consider the following polynomial dynamical system: f 1 (x 1, x 2, x 3 ) = x 1 x 2 = x 1 x 2 f 2 (x 1, x 2, x 3 ) = x 1 x 2 x 3 = x 1 x 2 x 3 f 3 (x 1, x 2, x 3 ) = x 1 x 2 = x 1 x 2. The state space of the FDS map F = (f 1, f 2, f 3 ) is the following graph: Question What if we only knew part of this state space, e.g., (1, 1, 0) (1, 0, 1) (0, 0, 0) (0, 0, 0). Could we recover the individual functions? How many possible models could yield this fragment? M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

6 Reverse engineering Broad goal Find the best model F = (f 1,..., f n) that fits the data: Input states: s 1,..., s m F n Output states: t 1,..., t m F n with F (s i ) = t i Note that: F (s i ) = (f 1 (s i ), f 2 (s i ),..., f n(s i )) = (t i1, t i2,..., t in ) = t i. Question What if no models fit the data? What if many models fit the data? (This is more likely.) First, we ll find all models that fit the data. This is called the model space: { } F 1 F 2 F n = (f 1,..., f n) f j (s i ) = t ij for all i and j. Once we do this, the new problem becomes choosing the best one. This is called model selection. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

7 Similar problems in other areas of mathematics 1. Parametrize a line in R n. 2. Parametrize a plane in R n. 3. Solve the underdetermined system Ax = b. 4. Solve the differential equation x + x = 2. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

8 Parametrize a line in R n Suppose we want to write the equation for a line that contains a vector v R n : w z v+w v t v+w t v x y This line, which contains the zero vector, is tv = {tv : t R}. Now, what if we want to write the equation for a line parallel to v? This line, which does not contain the zero vector, is tv + w = {tv + w : t R}. Note that ANY particular w on the line will work!!! M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

9 Solve an underdetermined system Ax = b Suppose we have a system of equations that has too many variables, so there are infinitely many solutions. For example: How to solve: 2x + y 3z = 4 3x 5y + 2z = 6 Ax = b form : [ ] x y = z [ 4 6]. 1. Solve the related homogeneous equation Ax = 0 (this is null space, NS(A)); 2. Find any particular solution x p to Ax = b; 3. Add these together to get the general solution: x = NS(A) + x p. This works because geometrically, the solution space is just a line, plane, etc. Here are two possible ways to write the solution: C , C M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

10 Linear differential equations Solve the differential equation x + x = 2. How to solve: 1. Solve the related homogeneous equation x + x = 0. The solutions are x h (t) = a cos t + b sin t. 2. Find any particular solution x p(t) to x + x = 2. By inspection, we see that x p(t) = 2 works. 3. Add these together to get the general solution: x(t) = x h (t) + x p(t) = a cos t + b sin t + 2. Note that while the general solution above is unique, its presentation need not be. For example, we could write it this way: x(t) = x h (t) + x p(t) = a(2 cos t 3 sin t) + b sin t + (2 cos t + 8 sin t). Here, the particular solution has (unnecessary) extra terms that vanish on the homogeneous part, x + x = 0. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

11 Reverse engineering: Problem statement Definition A finite dynamical system (FDS) is a function F = (f 1,..., f n): X n X n where each f i : X n X is a local function and X < (usually X = F 2 = {0, 1}). Key fact If X = F is a finite field (e.g., Z 2, Z 3, Z p, etc.), then every function f i : F n F is a polynomial in x 1,..., x n. Goal Given a set of data: Input states: s 1,..., s m F n Output states: t 1,..., t m F n with F (s i ) = t i Construct the model space F 1 F n of all models F = (f 1,..., f n) that fit the data: We ll find each F 1,..., F n separately. F (s i ) = (f 1 (s i ),..., f n(s i )) = (t i1,..., t in ) = t i. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

12 Reverse engineering: How to find F j We wish to find the set F j of all local functions (polynomials!) f j that fit the data: F j = {f j : f j (s 1 ) = t 1j,..., f j (s m) = t mj }. Define the set I (it is actually an ideal of the polynomial ring F[x 1,..., x n]) I = {h : h(s i ) = 0 for all i = 1,..., m} = {all polynomials that vanish on the data}. Theorem The set of polynomials that fit the data at node j is F j = f j + I = {f j + h : h I }, where f j is any one particular polynomial that fits the data. Thus, to find F j, we need to do two things: 1. Find the ideal I ; (all solutions to {f j (s i ) = 0, i}) 2. Find any polynomial f j that fits the data. (one solution to {f j (s i ) = t ij, i}) M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

13 Reverse engineering: How to find I and f j 1. Finding I : Define I (s i ) to be the set of polynomials that vanish on s i : I (s i ) = {all polynomials h i such that h i (s i ) = 0} = {(x 1 s i1 )g 1 (x) + (x 2 s i2 )g 2 (x) + + (x n s in )g n(x)} = x 1 s i1, x 2 s i2,..., x n s in Clearly, the set I of polynomials that vanish on all s i (for i = 1,..., m) is m I = I (s i ). i=1 2. Finding f j : There are many algorithms. Lagrange interpolation is one of them. In this lecture, we will learn another method which has the Chinese remainder theorem lurking behind the scenes. We ll get started with this now. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

14 Finding f j (one method) For each data point s i (i = 1,..., m), we ll construct an r-polynomial that has the following property: { 1 x = s i r i (x) = 0 x s i Once we have these, the polynomial f j (x) we seek will be One way to construct the r-polynomials: where f j (x) = t 1j r 1 (x) + t 2j r 2 (x) + + t mj r m(x). r i (x) = m b ik (x), k=1 k i b ik (x) = (s il s kl ) p 2 (x l s kl ) and l is the first coordinate in which s i and s k differ. (Actually, any coordinate where they differ will do.) Remark When our systems are Boolean (over F 2 ), then this reduces to b ik (x) = x l s kl. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

15 A Boolean example Consider the following model of the lac operon, which implicitly assumes that A degrades slower than M or B. f M = x A f B = x M f A = L (B L m) (A B). If lactose levels are low, then L = L m = 0, and this model in polynomial form becomes the following PDS (left) and state space (right): f 1 = x f 2 = x 1 f 3 = (x 2 + 1)x Exercise Let s reverse engineer the Boolean network from just knowing the 6 red nodes and 5 transitions. In other words, find all triples of polynomials f = (h 1, h 2, h 3 ) such that: f (0, 0, 1) = (h 1(0, 0, 1), h 2(0, 0, 1), h 3(0, 0, 1)) = (1, 0, 1), f (1, 0, 1) = (h 1(1, 0, 1), h 2(1, 0, 1), h 3(1, 0, 1)) = (1, 1, 1), f (1, 1, 1) = (h 1(1, 1, 1), h 2(1, 1, 1), h 3(1, 1, 1)) = (1, 1, 0), f (1, 1, 0) = (h 1(1, 1, 0), h 2(1, 1, 0), h 3(1, 1, 0)) = (0, 1, 0), f (0, 1, 0) = (h 1(0, 1, 0), h 2(0, 1, 0), h 3(0, 1, 0)) = (0, 0, 0), M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

16 A Boolean example (cont.) Since we know the original functions a priori, we secretly know the answer to this. (f 1, f 2, f 3 ) = (x 3, x 1, (x 2 + 1)x 3 ) The ideal of polynomials that vanish on each s k is: I1 = ideal(x1, x2, x3-1); I2 = ideal(x1-1, x2, x3-1); I3 = ideal(x1-1, x2-1, x3-1); I4 = ideal(x1-1, x2-1, x3); I5 = ideal(x1, x2-1, x3); The ideal of polynomials that vanish on every s k is: I = intersect{i1,i2,i3,i4,i5}; Thus, the complete model space is F 1 F 2 F 3, F j = f j + I If we hadn t known f 1, f 2, f 3 a priori, then we d have to find our own particular solution that fits the data. We ll do that now. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

17 A Boolean example (cont.) We re looking for a single solution f = (f 1, f 2, f 3 ) that fits the data. We know how to do this. For example: f 1 (x) = t 11 r 1 (x) + t 21 r 2 (x) + t 31 r 3 (x) + t 41 r 4 (x) + t 51 r 5 (x) = 1r 1 (x) + 1r 2 (x) + 1r 3 (x) + 0r 4 (x) + 0r 5 (x) = r 1 (x) + r 2 (x) + r 3 (x), where s 1 = (0, 0, 1)= t 0 r 1 (x) = 5 b 1k (x) = b 12 (x)b 13 (x)b 14 (x)b 15 (x). k=1 k 1 s 2 = (1, 0, 1) = t 1 s 3 = (1, 1, 1) = t 2 s 4 = (1, 1, 0) = t 3 Recall that b 1k (x) = x l s kl, where l is the first coordinate that s 1 differs from s k. skip k = 1 b 12 (x) = (x 1 s 21 ) = x b 13 (x) = (x 1 s 31 ) = x b 14 (x) = (x 1 s 41 ) = x b 15 (x) = (x 2 s 52 ) = x s 5 = (0, 1, 0) = t 4 s 6 =(0, 0, 0) = t 5 Now, r 1 (x) = (x 1 + 1) 3 (x 2 + 1) = (x 1 + 1)(x 2 + 1). M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

18 A Boolean example (cont.) Recall that b ik (x) = x l s kl, where l is the first coordinate that s i differs from s k. s 1 = (0, 0, 1)= t 0 s 2 = (1, 0, 1) = t 1 s 3 = (1, 1, 1) = t 2 b 21 (x) = (x 1 s 11 ) = x 1 skip k = 2 b 23 (x) = (x 2 s 32 ) = x b 24 (x) = (x 2 s 42 ) = x b 25 (x) = (x 1 s 51 ) = x 1 b 31 (x) = (x 1 s 11 ) = x 1 b 32 (x) = (x 2 s 22 ) = x 2 skip k = 3 b 34 (x) = (x 3 s 43 ) = x 3 b 35 (x) = (x 1 s 51 ) = x 1 s 4 = (1, 1, 0) = t 3 s 5 = (0, 1, 0) = t 4 s 6 = (0, 0, 0) = t 5 b 41 (x) = (x 1 s 11 ) = x 1 b 42 (x) = (x 2 s 22 ) = x 2 b 43 (x) = (x 3 s 33 ) = x skip k = 4 b 45 (x) = (x 1 s 51 ) = x 1 b 51 (x) = (x 2 s 12 ) = x 2 b 52 (x) = (x 1 s 21 ) = x b 53 (x) = (x 1 s 31 ) = x b 54 (x) = (x 1 s 42 ) = x skip k = 5 Recall that x k i = x i, and (x j + 1) k = x j + 1, so the r-polynomials are r 1 (x) = (x 1 + 1)(x 2 + 1) r 2 (x) = x 1 (x 2 + 1) r 3 (x) = x 1 x 2 x 3 r 4 (x) = x 1 x 2 (x 3 + 1) r 5 (x) = (x 1 + 1)x 2 M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

19 A Boolean example (cont.) We can now compute our particular solution (f 1, f 2, f 3 ) that fits the data, using: s 1 = (0, 0, 1)= t 0 s 2 = (1, 0, 1) = t 1 s 3 = (1, 1, 1) = t 2 s 4 = (1, 1, 0) = t 3 s 5 = (0, 1, 0) = t 4 s 6 = (0, 0, 0) = t 5 f j (x) = t 1j r 1 (x) + t 2j r 2 (x) + + t mj r m(x). f 1 (x) = t 11 r 1 (x) + t 21 r 2 (x) + t 31 r 3 (x) + t 41 r 4 (x) + t 51 r 5 (x) = r 1 (x) + r 2 (x) + r 3 (x) = 1 + x 2 + x 1 x 2 x 3 f 2 (x) = t 12 r 1 (x) + t 22 r 2 (x) + t 32 r 3 (x) + t 42 r 4 (x) + t 52 r 5 (x) = r 2 (x) + r 3 (x) + r 4 (x) = x 1 f 3 (x) = t 13 r 1 (x) + t 23 r 2 (x) + t 33 r 3 (x) + t 43 r 4 (x) + t 53 r 5 (x) = r 1 (x) + r 2 (x) = 1 + x 2. Our original PDS was (f 1, f 2, f 3 ) = (x 3, x 1, x 3 + x 2 x 3 ), but our algorithm yielded (f 1, f 2, f 3 ) = (1 + x 2 + x 1 x 2 x 3, x 1, 1 + x 2 ) = (x 3, x 1, x 3 + x 2 x 3 ) + (1 + x 2 + x 3 + x 1 x 2 x 3, 0, 1 + x 2 + x 3 + x 2 x 3 ) Remark Each polynomial in the 2nd term above is in the vanishing ideal I. (Why?) M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

20 A Boolean example (cont.) Figure: Left. Phase space of the original PDS: (f 1, f 2, f 3) = (x 3, x 1, x 3(1 + x 2)). Middle. Phase space of the reverse-engineered PDS: (f 1, f 2, f 3) = (1 + x 2 + x 1x 2x 3, x 1, 1 + x 2). Right: Phase space of the reverse-engineered PDS, modulo I : (f 1, f 2, f 3) = (x 3, x 1, 1 + x 2) M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

21 A Boolean example (cont.) Now that we found a particular solution f = (f 1, f 2, f 3 ) that fits the data, we need to (re)compute the ideal I of polynomials that vanish on the data. We ll use Macaulay2 in Sage: %default_mode macaulay2 R=ZZ/2[x1,x2,x3] / ideal(x1^2-x1, x2^2-x2, x3^2-x3); s 1 = (0, 0, 1)= t 0 s 2 = (1, 0, 1) = t 1 s 3 = (1, 1, 1) = t 2 s 4 = (1, 1, 0) = t 3 s 5 = (0, 1, 0) = t 4 s 6 =(0, 0, 0) = t 5 The ideal of polynomials that vanish on each s k is: I1 = ideal(x1, x2, x3-1); I2 = ideal(x1-1, x2, x3-1); I3 = ideal(x1-1, x2-1, x3-1); I4 = ideal(x1-1, x2-1, x3); I5 = ideal(x1, x2-1, x3); The ideal of polynomials that vanish on every s k is: I = intersect{i1,i2,i3,i4,i5} To compute a Gröbner basis: G = gens gb I The output is: x2x3+x2+x3+1 x1x2+x1x3+x1+x2+x3+1 M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

22 A Boolean example (cont.) In conclusion, our model space of all PDSs (f 1, f 2, f 3 ) that fit the data: is the set where I is the vanishing ideal F 1 F 2 F 3, F j = f j + I I = g 1, g 2 = 1 + x 2 + x 3 + x 2 x 3, 1 + x 1 + x 2 + x 3 + x 1 x 2 + x 1 x 3. Our reverse-engineered PDS is slighly different than the true model : (f 1, f 2, f 3 ) = (1 + x 2 + x 1 x 2 x 3, x 1, 1 + x 2 ) = (x 3 + x 1 g 1 + g 2, x 1, (x 2 + 1)x 3 + g 1 ) Note that x 1 g 1 + g 2, 0, and g 1 must be in the vanishing ideal I. Goal ( model selection ) We would like to be able to recover functions in F j = f j + I that have no extra terms in I. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

23 A Boolean example (cont.) Goal ( model selection ) We would like to be able to be able to recover functions in F j = f j + I that have no extra terms in I Fortunately, we can do this with Macaulay2. It s called finding the remainder of f j modulo I, and we use the % symbol. f1 = 1+x2+x1*x2*x3; f2 = x1; f3 = 1+x2; f1%i; f2%i; f3%i; The output is: x3, x1, x2+1. We successfully recovered 2 of the 3 original functions. Question What would happen if we: added the (original) self-loop at 000 to the data? removed the data point ? M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

24 An example over Z 5 Consider the following time series in a 3-node system over Z 5 : s 1 = (2, 0, 0)= t 0 s 2 = (4, 3, 1) = t 1 s 3 = (3, 1, 4) = t 2 s 4 =(0, 4, 3) = t 3 For reference, here are the input vectors s i and output vectors t i : s 1 = (s 11, s 12, s 13 ) = (2, 0, 0), t 1 = (t 11, t 12, t 13 ) = (4, 3, 1), s 2 = (s 21, s 22, s 23 ) = (4, 3, 1), t 2 = (t 21, t 22, t 23 ) = (3, 1, 4), s 3 = (s 31, s 32, s 33 ) = (3, 1, 4), t 3 = (t 31, t 32, t 33 ) = (0, 4, 3). Note that s 1 differs from s 2 and s 3 in the l = 1 coodinate, so this l will work for each of r 1, r 2, and r 3. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

25 An example over Z 5 (cont.) Since we are working in Z 5, we are taking the remainder of everything modulo 5. Particularly useful identities are: 0 = 5, 1 = 4, 2 = 3, 3 = 2, and 4 = 1. Using our formulas for b ij (x), we compute: b 12 (x) = (s 11 s 21 ) 3 (x 1 s 21 ) = (2 4) 3 (x 1 4) = 8(x 1 + 1) = 2x b 13 (x) = (s 11 s 31 ) 3 (x 1 s 31 ) = (2 3) 3 (x 1 3) = x = 4x Therefore, the first r-polynomial is r 1 (x) = b 12 (x)b 13 (x) = (2x 1 + 2)(4x 1 + 3) = 8x x = 3x x In-class Exercise Compute the other two r-polynomials in this example: r 2 (x) and r 3 (x). Solution: r 2 (x) = 3x , r 3(x) = 4x x M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

26 An example over Z 5 (cont.) In summary, we computed the r-polynomials to be: r 1 (x) = b 12 (x)b 13 (x) = (2x 1 + 2)(4x 1 + 3) = 8x x = 3x x r 2 (x) = b 21 (x)b 23 (x) = (3x 1 + 4)(x 1 + 2) = 3x x = 3x r 2 (x) = b 31 (x)b 32 (x) = (x 1 + 3)(4x 1 + 4) = 4x x = 4x1 2 + x Thus, the following functions fit the data: f 1 (x) = t 11 r 1 (x) + t 21 r 2 (x) + t 31 r 3 (x) = 4(3x x 1 + 1) + 3(3x ) + 0(4x2 1 + x 1 + 2) = x1 2 + x f 2 (x) = t 12 r 1 (x) + t 22 r 2 (x) + t 32 r 3 (x) = 3(3x x 1 + 1) + 1(3x ) + 4(4x2 1 + x 1 + 2) = 3x1 2 + x f 2 (x) = t 13 r 1 (x) + t 23 r 2 (x) + t 33 r 3 (x) = 1(3x x 1 + 1) + 4(3x ) + 3(4x2 1 + x 1 + 2) = 2x x Comments on this? [Note that only the variable x 1 appears.] M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

27 An example over Z 5 (cont.) Recall that the ideal I is the set I of polynomials that vanish on all s i : I = I (s 1 ) I (s 2 ) I (s 3 ), where s 1 = (2, 0, 0), s 2 = (4, 3, 1), s 3 = (3, 1, 4). These are precisely the sets I (s 1 ) = x 1 2, x 2, x 3 = {(x 1 2)g 1 (x) + x 2 g 2 (x) + x 3 g 3 (x)} I (s 2 ) = x 1 4, x 2 3, x 3 1 = {(x 1 4)g 1 (x) + (x 2 3)g 2 (x) + (x 3 1)g 3 (x)} I (s 3 ) = x 1 3, x 2 1, x 3 4 = {(x 1 3)g 1 (x) + (x 2 1)g 2 (x) + (x 3 4)g 3 (x)} We ll use Macaulay2 in Sage again to compute this. Remember to work over Z 5 : %default_mode macaulay2 R=ZZ/5[x1,x2,x3] / ideal(x1^5-x1, x2^5-x2, x3^5-x3); M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

28 An example over Z 5 (cont.) As before, we can compute the vanishing ideal I : I1 = ideal(x1-2, x2, x3); I2 = ideal(x1-4, x2-3, x3-1); I3 = ideal(x1-3, x2-1, x3-4); I = intersect{i1,i2,i3}; f1 = x1^2+x1+3; f2 = 3*x1^2+x1+4; f3 = 2*x1^2+2*x1+4; We can view the Gröbner basis (w.r.t. the default monomial ordering, grevlex): flatten entries gens gb I; The output is: 2 2 {x1-2x2 - x3-2, x3 + 2x2-2x3, x2*x3 + 2x2 + x3, x2 + x3} which means G = {x 1 2x 2 x 3 2, x x 2 2x 3, x 2 x 3 + 2x 2 + x 3, x x 3}. If we reduce each f i modulo I : p1=f1%i; and p2=f2%i; and p1=f3%i; we get (p 1, p 2, p 3 ) = ( x 3 1, x 2 2, 2x 3 + 1). This is called the normal form of f j with respect to the Gröbner basis G. M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29

29 An aside: why monomial order matters! Let s do the same thing but using lex instead of grevlex: S=ZZ/5[x1,x2,x3] / ideal(x1^5-x1, x2^5-x2, x3^5-x3); p1 = sub(f1,s)%sub(i,s); p2 = sub(f2,s)%sub(i,s); p3 = sub(f3,s)%sub(i,s); These polynomials are now (p 1, p 2, p 3 ) = ( x 3 1, 2x x 3 2, 2x 3 + 1). They are different because the Gröbner basis of I is different with respect to lex: flatten entries gens gb sub(i,s); The output is G = {x 3 3 x 3, x 2 2x 2 3 x 3, x 1 + x x 3 2}. In summary, the model space is F 1 F n = f + (I I ) = (f 1 + I,..., f n + I ). Here are three choices for the particular solution f = (f 1, f 2, f 3 ) that fits the data: our algorithm: (f 1, f 2, f 3 ) = (x x 1 + 3, 3x x 1 + 4, 2x x 1 + 4). normal form w.r.t. grevlex: (f 1, f 2, f 3 ) = ( x 3 1, 2x x 3 2, 2x 3 + 1). normal form w.r.t. lex: (f 1, f 2, f 3 ) = ( x 3 1, x 2 2, 2x 3 + 1). M. Macauley (Clemson) Reverse engineering using computational algebra Math 4500, Fall / 29