Commit 3707e7e2 by SOLIMAN Sylvain

 %% Cell type:markdown id: tags: # M7 Chemical SAT Solver * Deciding the satisfiability of a Boolean formula in conjunctive normal form is NP-complete * How can we program a chemical SAT solver ? F. Fages, S. Soliman, Apr. 2020 %% Cell type:markdown id: tags: # "Generate and Test" Algorithm with Stochastic CRN * A stochastic CRN can be used to enumerate random Boolean values for a variable %% Cell type:code id: tags:  parameter(k=1).  %% Cell type:code id: tags:  MA(k) for _/a => a. % reactions with inhibitors cannot fire in presence of the inhibitor, here a  %% Cell type:code id: tags:  MA(k) for a => _.  %% Cell type:code id: tags:  option(method:spn, stochastic_conversion:1).  %% Cell type:code id: tags:  numerical_simulation. plot.  %% Output %% Cell type:markdown id: tags: ## Question 1) Write a random generator of Boolean values for a vector of 3 variables a, b, c %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question 2) Write a CRN to find values satisfying the formula (x⋁¬y)⋀(y⋁¬z)⋀(z⋁¬x) * Use an event to stop the simulation when the formula is satisfied %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: # Guided Search Algorithm with Stochastic CRN ## Question 3) Improve your CRN to decrease the probability of moves of the variables that belong to satisfied clauses %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question ) Any idea to decide unsatisfiability ? statistical test question ? %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question 4) Determine the phase transition threshold in 3-SAT * the density of a SAT instance is the ratio of the number of clauses divided by the number of variables * a phase transition phenomenon is an asymptotic result showing the existence of a density threshold * under the threshold the instances are almost surely satisfiable * above the threshold the instances are almost surely unsatisfiable * the hard instances are around the density threshold %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question ) Evaluation on 2-SAT and Horn-SAT ? %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: # Guided Search Algorithm by Continuous CRN We can see the SAT solving problem as a (continous) global optimization problem and try to solve it by gradient descent. The idea is to find an energy function $E$ that is minimal when all clauses are satisfied, and then to simply enforce that $$\frac{dx}{dt} = -\frac{\partial E}{\partial x}$$ For a SAT problem involving variables $x_i\in [0, 1], 1\leq i\leq N$ and clauses $C_j, 1\leq j\leq M$ (with $C_{ji} = 1$ if $x_i$ appears positively in $C_j$, $C_{ji} = -1$ if $x_i$ appears negatively, and $0$ otherwise), we will define our energy function as a sum of squares of sub-energies for each clause. $$E = \sum_{1\leq j\leq M}K_m^2$$ $$E = \sum_{1\leq j\leq M}K_j^2$$ %% Cell type:markdown id: tags: ## Question 5) Write $K_m$ ## Question 5) Write $K_j$ Define (formally) $K_m$ as a function of the $C_ji$ and of the $x_i$, such that $K_m = 0$ iff clause $m$ is satisfied, and $K_m = 2^N$ if all $N$ variables appear in clause $m$ and are currently at the wrong value. Define (formally) $K_j$ as a function of the $C_{ji}$ and of the $x_i$, such that $K_j = 0$ iff clause $j$ is satisfied, and $K_j = 2^N$ if all $N$ variables appear in clause $j$ and are currently at the _wrong_ value. One might want to define $s_i\in[-1, 1]$ as a function of $x_i\in[0, 1]$ for ease of writing. One might want to define $s_i\in[-1, 1]$ as a function of $x_i\in[0, 1]$ and then $K_j$ as a function of the $s_i$ for ease of writing. %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question 6) Obtain $dx_i/dt$ Obtain the formal expression for $\displaystyle\frac{dx}{dt}$ (if you have used $s_i$ just note that $\frac{\partial E}{\partial x}=\frac{\partial E}{\partial x}\frac{\partial s}{\partial x}$). %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:markdown id: tags: ## Question 7) Implement on the above example (x⋁¬y)⋀(y⋁¬z)⋀(z⋁¬x) Using the commands: new_ode_system, init (to set $x_1, x_2$ and $x_3$ initial state to 0.5), ode_parameter (to set the $c_ji$ corresponding to our 3 clauses), ode_function (for the $k_j$ and $s_i$) and add_ode (to add the above $dx_i/dt$) Run a numerical simulation and plot the $x_i$ to see the result. Experiment with different initial states, and variants of the problem (changing only the $c_ji$), what do you observe? [please leave all results and remarks *in* the notebook!] %% Cell type:code id: tags:  clear_model. new_ode_system. init(x_1 = 0.5, x_2 = 0.5, x_3 = 0). ode_function( s_1 = 2*x_1 - 1, s_2 = 2*x_2 - 1, s_3 = 2*x_3 - 1, k_1 = (1-c_11*s_1)*(1-c_12*s_2)*(1-c_13*s_3), k_2 = (1-c_21*s_1)*(1-c_22*s_2)*(1-c_23*s_3), k_3 = (1-c_31*s_1)*(1-c_32*s_2)*(1-c_33*s_3) ). ode_parameter( c_11 = 1, c_12 = -1, c_13 = 0, c_21 = 0, c_22 = 1, c_23 = -1, c_31 = -1, c_32 = 0, c_33 = 1 ). add_ode( d(x_1)/dt = c_11*k_1*(1-c_12*s_2)*(1-c_13*s_3) + c_21*k_2*(1-c_22*s_2)*(1-c_23*s_3) + c_31*k_3*(1-c_32*s_2)*(1-c_33*s_3), d(x_2)/dt = c_12*k_1*(1-c_11*s_1)*(1-c_13*s_3) + c_22*k_2*(1-c_21*s_1)*(1-c_23*s_3) + c_32*k_3*(1-c_31*s_1)*(1-c_33*s_3), d(x_3)/dt = c_13*k_1*(1-c_12*s_2)*(1-c_11*s_1) + c_23*k_2*(1-c_22*s_2)*(1-c_21*s_1) + c_33*k_3*(1-c_32*s_2)*(1-c_31*s_1) ).  %% Output %% Cell type:code id: tags:  numerical_simulation. plot(show: {x_1, x_2, x_3}).  %% Output %% Cell type:code id: tags:   %% Output \begin{align*} {x_1}_0 &= 0.5\\ {x_2}_0 &= 0.5\\ {x_3}_0 &= 0.5\\ c_11 &= 1\\ c_12 &= -1\\ c_13 &= 0\\ c_21 &= 0\\ c_22 &= 1\\ c_23 &= -1\\ c_31 &= -1\\ c_32 &= 0\\ c_33 &= 1\\ s_1 &= 2*x_1-1\\ s_2 &= 2*x_2-1\\ s_3 &= 2*x_3-1\\ k_1 &= (1-c_11*s_1)*(1-c_12*s_2)*(1-c_13*s_3)\\ k_2 &= (1-c_21*s_1)*(1-c_22*s_2)*(1-c_23*s_3)\\ k_3 &= (1-c_31*s_1)*(1-c_32*s_2)*(1-c_33*s_3)\\ \frac{dx_1}{dt} &= c_11*k_1*(1-c_12*s_2)*(1-c_13*s_3)+c_21*k_2*(1-c_22*s_2)*(1-c_23*s_3)+c_31*k_3*(1-c_32*s_2)*(1-c_33*s_3)\\ \frac{dx_2}{dt} &= c_12*k_1*(1-c_11*s_1)*(1-c_13*s_3)+c_22*k_2*(1-c_21*s_1)*(1-c_23*s_3)+c_32*k_3*(1-c_31*s_1)*(1-c_33*s_3)\\ \frac{dx_3}{dt} &= c_13*k_1*(1-c_12*s_2)*(1-c_11*s_1)+c_23*k_2*(1-c_22*s_2)*(1-c_21*s_1)+c_33*k_3*(1-c_32*s_2)*(1-c_31*s_1)\\ \end{align*} %% Cell type:markdown id: tags: ## Question 8) Improving the search To avoid getting stuck in some local minima, we can add *Lagrange multipliers* $a_j, 1\leq j\leq M$. To avoid getting stuck in some local minima, we can add *Lagrange multipliers* $a_j, 1\leq j\leq M$ so that the energy becomes $$E = \sum_{1\leq j\leq M}a_j K_j^2$$ These are new variables that will have an exponential increase proportional to $K_j$. Add the 3 new variables and their ODEs with add_ode. Do you notice any difference? %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Output \begin{align*} {x_1}_0 &= 0.5\\ {x_2}_0 &= 0.5\\ {x_3}_0 &= 0.5\\ {a_1}_0 &= 1\\ {a_2}_0 &= 1\\ {a_3}_0 &= 1\\ c_1_1 &= 0\\ c_1_2 &= 0\\ c_1_3 &= 0\\ c_2_1 &= 0\\ c_2_2 &= 0\\ c_2_3 &= 0\\ c_3_1 &= 0\\ c_3_2 &= 0\\ c_3_3 &= 0\\ s_1 &= 2*x_1-1\\ s_2 &= 2*x_2-1\\ s_3 &= 2*x_3-1\\ k_1 &= (1-c_1_1*s_1)*(1-c_1_2*s_2)*(1-c_1_3*s_3)\\ k_2 &= (1-c_2_1*s_1)*(1-c_2_2*s_2)*(1-c_2_3*s_3)\\ k_3 &= (1-c_3_1*s_1)*(1-c_3_2*s_2)*(1-c_3_3*s_3)\\ \frac{dx_1}{dt} &= a_1*c_1_1*k_1*((1-c_1_2*s_2)*(1-c_1_3*s_3))+a_2*c_2_1*k_2*((1-c_2_2*s_2)*(1-c_2_3*s_3))+a_3*c_3_1*k_3*((1-c_3_2*s_2)*(1-c_3_3*s_3))\\ \frac{dx_2}{dt} &= a_1*c_1_2*k_1*((1-c_1_1*s_1)*(1-c_1_3*s_3))+a_2*c_2_2*k_2*((1-c_2_1*s_1)*(1-c_2_3*s_3))+a_3*c_3_2*k_3*((1-c_3_1*s_1)*(1-c_3_3*s_3))\\ \frac{dx_3}{dt} &= a_1*c_1_3*k_1*((1-c_1_1*s_1)*(1-c_1_2*s_2))+a_2*c_2_3*k_2*((1-c_2_1*s_1)*(1-c_2_2*s_2))+a_3*c_3_3*k_3*((1-c_3_1*s_1)*(1-c_3_2*s_2))\\ \frac{da_1}{dt} &= a_1*k_1\\ \frac{da_2}{dt} &= a_2*k_2\\ \frac{da_3}{dt} &= a_3*k_3\\ \end{align*} %% Cell type:markdown id: tags: link to M4 chemical_logic %% Cell type:markdown id: tags: ## Question ) on the link to the first Modal session below ? %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:   %% Cell type:code id: tags:  clear_model.  %% Cell type:code id: tags:  present(A,a). present(B, b).  %% Cell type:code id: tags:  parameter(a=1, b=1).  %% Cell type:markdown id: tags: # And $A\wedge B=A*B$ %% Cell type:code id: tags:  A+B=>A+B+AandB.  %% Cell type:code id: tags:  AandB=>_.  %% Cell type:code id: tags:  list_ode.  %% Output \begin{align*} {AandB}_0 &= 0\\ {A}_0 &= 1\\ {B}_0 &= 1\\ a &= 1\\ b &= 1\\ \frac{dAandB}{dt} &= A*B-AandB\\ \frac{dA}{dt} &= 0\\ \frac{dB}{dt} &= 0\\ \end{align*} %% Cell type:code id: tags:  %slider a b  %% Output %% Cell type:markdown id: tags: # Or $A\vee B = A+B-A*B$ %% Cell type:code id: tags:  A=>A+AorB.  %% Cell type:code id: tags:  B=>B+AorB.  %% Cell type:code id: tags:  A+B+AorB=>A+B.  %% Cell type:code id: tags:  AorB=>_.  %% Cell type:code id: tags:  %slider a b  %% Output %% Cell type:markdown id: tags: # Not $\neg A=1-A$ %% Cell type:code id: tags:  _ => notA.  %% Cell type:code id: tags:  notA => _.  %% Cell type:code id: tags:  A+ notA => A.  %% Cell type:code id: tags:  list_ode.  %% Output \begin{align*} {notA}_0 &= 0\\ {A}_0 &= 0.0\\ a &= 0.0\\ \frac{dnotA}{dt} &= 1-notA*A-notA\\ \frac{dA}{dt} &= 0\\ \end{align*} %% Cell type:code id: tags:  %slider a  %% Output %% Cell type:code id: tags:   %% Cell type:code id: tags:  clear_model.  %% Cell type:code id: tags:  present(A,a). parameter(a=1).  %% Cell type:code id: tags:  present(B, b). parameter(b=1).  %% Cell type:code id: tags:  A=>D.  %% Cell type:code id: tags:  B=>D.  %% Cell type:code id: tags:  MA(k) for A+B=>_. parameter(k=2.5).  %% Cell type:code id: tags:  list_ode.  %% Output \begin{align*} {B}_0 &= 1\\ {A}_0 &= 1\\ {D}_0 &= 0\\ a &= 1\\ b &= 1\\ k &= 2.5\\ \frac{dB}{dt} &= -B-k*A*B\\ \frac{dA}{dt} &= -A-k*A*B\\ \frac{dD}{dt} &= A+B\\ \end{align*} %% Cell type:code id: tags:  %slider a b k  %% Output %% Cell type:code id: tags:  validity_domain(F(G(D=v))).  %% Output v=1.00221 %% Cell type:code id: tags: