I'm using numba's #guvectorize to change two different arrays. The code is:
#guvectorize([(int64[:], int64[:], int64[:], int64[:])], '(n),(n)->(n),(n)', target= 'parallel')
def g(x, y, res, res_two):
res = x
for i in range(x.shape[0]-1):
var = np.random.poisson((2),1)[0]
res_two[i] = var
res[i+1] = res[i] + res_two[i]
print("res[i+1] is", res[i+1], "for x[i] is", x[i])
q = (np.arange(5)) * 0
q[0] = 5
r = (np.arange(5)) * 0
g(q,r)
print("q is", q)
print("r is", r)
And the results printed out are:
As one can see, q is changing, but r isn't.
What must I do to use guvectorize to input two arrays and change those two arrays?
Related
I am trying to run a model using the GPU, no problem with the CPU. I think somehow using measured boundary conditions is causing the issue but I am not sure. I am following this example: https://docs.sciml.ai/dev/modules/NeuralPDE/tutorials/gpu/. I am following this example for using measured boundary conditions: https://docs.sciml.ai/dev/modules/MethodOfLines/tutorials/icbc_sampled/
using Random
using NeuralPDE, Lux, CUDA, Random
using Optimization
using OptimizationOptimisers
using NNlib
import ModelingToolkit: Interval
using Interpolations
# Measured Boundary Conditions (Arbitrary For Example)
bc1 = 1.0:1:1001.0 .|> Float32
bc2 = 1.0:1:1001.0 .|> Float32
ic1 = zeros(101) .|> Float32
ic2 = zeros(101) .|> Float32;
# Interpolation Functions Registered as Symbolic
itp1 = interpolate(bc1, BSpline(Cubic(Line(OnGrid()))))
up_cond_1_f(t::Float32) = itp1(t)
#register_symbolic up_cond_1_f(t)
itp2 = interpolate(bc2, BSpline(Cubic(Line(OnGrid()))))
up_cond_2_f(t::Float32) = itp2(t)
#register_symbolic up_cond_2_f(t)
itp3 = interpolate(ic1, BSpline(Cubic(Line(OnGrid()))))
init_cond_1_f(x::Float32) = itp3(x)
#register_symbolic init_cond_1_f(x)
itp4 = interpolate(ic2, BSpline(Cubic(Line(OnGrid()))))
init_cond_2_f(x::Float32) = itp4(x)
#register_symbolic init_cond_2_f(x);
# Parameters and differentials
#parameters t, x
#variables u1(..), u2(..)
Dt = Differential(t)
Dx = Differential(x);
# Arbitrary Equations
eqs = [Dt(u1(t, x)) + Dx(u2(t, x)) ~ 0.,
Dt(u1(t, x)) * u1(t,x) + Dx(u2(t, x)) + 9.81 ~ 0.]
# Boundary Conditions with Measured Data
bcs = [
u1(t,1) ~ up_cond_1_f(t),
u2(t,1) ~ up_cond_2_f(t),
u1(1,x) ~ init_cond_1_f(x),
u2(1,x) ~ init_cond_2_f(x)
]
# Space and time domains
domains = [t ∈ Interval(1.0,1001.0),
x ∈ Interval(1.0,101.0)];
# Neural network
input_ = length(domains)
n = 10
chain = Chain(Dense(input_,n,NNlib.tanh_fast),Dense(n,n,NNlib.tanh_fast),Dense(n,4))
strategy = GridTraining(.25)
ps = Lux.setup(Random.default_rng(), chain)[1]
ps = ps |> Lux.ComponentArray |> gpu .|> Float32
discretization = PhysicsInformedNN(chain,
strategy,
init_params=ps)
# Model Setup
#named pdesystem = PDESystem(eqs,bcs,domains,[t,x],[u1(t, x),u2(t, x)])
prob = discretize(pdesystem,discretization);
sym_prob = symbolic_discretize(pdesystem,discretization);
# Losses and Callbacks
pde_inner_loss_functions = sym_prob.loss_functions.pde_loss_functions
bcs_inner_loss_functions = sym_prob.loss_functions.bc_loss_functions
callback = function (p, l)
println("loss: ", l)
println("pde_losses: ", map(l_ -> l_(p), pde_inner_loss_functions))
println("bcs_losses: ", map(l_ -> l_(p), bcs_inner_loss_functions))
return false
end;
# Train Model (Throws Error)
res = Optimization.solve(prob,Adam(0.01); callback = callback, maxiters=5000)
phi = discretization.phi;
I get the following error:
GPU broadcast resulted in non-concrete element type Union{}.
This probably means that the function you are broadcasting contains an error or type instability.
Please Advise.
Given the concrete syntax of a language, I would like to define a function "instance" with signature str (type[&T]) that could be called with the reified type of the syntax and return a valid instance of the language.
For example, with this syntax:
lexical IntegerLiteral = [0-9]+;
start syntax Exp
= IntegerLiteral
| bracket "(" Exp ")"
> left Exp "*" Exp
> left Exp "+" Exp
;
A valid return of instance(#Exp) could be "1+(2*3)".
The reified type of a concrete syntax definition does contain information about the productions, but I am not sure if this approach is better than a dedicated data structure. Any pointers of how could I implement it?
The most natural thing is to use the Tree data-type from the ParseTree module in the standard library. It is the format that the parser produces, but you can also use it yourself. To get a string from the tree, simply print it in a string like so:
str s = "<myTree>";
A relatively complete random tree generator can be found here: https://github.com/cwi-swat/drambiguity/blob/master/src/GenerateTrees.rsc
The core of the implementation is this:
Tree randomChar(range(int min, int max)) = char(arbInt(max + 1 - min) + min);
Tree randomTree(type[Tree] gr)
= randomTree(gr.symbol, 0, toMap({ <s, p> | s <- gr.definitions, /Production p:prod(_,_,_) <- gr.definitions[s]}));
Tree randomTree(\char-class(list[CharRange] ranges), int rec, map[Symbol, set[Production]] _)
= randomChar(ranges[arbInt(size(ranges))]);
default Tree randomTree(Symbol sort, int rec, map[Symbol, set[Production]] gr) {
p = randomAlt(sort, gr[sort], rec);
return appl(p, [randomTree(delabel(s), rec + 1, gr) | s <- p.symbols]);
}
default Production randomAlt(Symbol sort, set[Production] alts, int rec) {
int w(Production p) = rec > 100 ? p.weight * p.weight : p.weight;
int total(set[Production] ps) = (1 | it + w(p) | Production p <- ps);
r = arbInt(total(alts));
count = 0;
for (Production p <- alts) {
count += w(p);
if (count >= r) {
return p;
}
}
throw "could not select a production for <sort> from <alts>";
}
Tree randomChar(range(int min, int max)) = char(arbInt(max + 1 - min) + min);
It is a simple recursive function which randomly selects productions from a reified grammar.
The trick towards termination lies in the weight of each rule. This is computed a priori, such that every rule has its own weight in the random selection. We take care to give the set of rules that lead to termination at least 50% chance of being selected (as opposed to the recursive rules) (code here: https://github.com/cwi-swat/drambiguity/blob/master/src/Termination.rsc)
Grammar terminationWeights(Grammar g) {
deps = dependencies(g.rules);
weights = ();
recProds = {p | /p:prod(s,[*_,t,*_],_) := g, <delabel(t), delabel(s)> in deps};
for (nt <- g.rules) {
prods = {p | /p:prod(_,_,_) := g.rules[nt]};
count = size(prods);
recCount = size(prods & recProds);
notRecCount = size(prods - recProds);
// at least 50% of the weight should go to non-recursive rules if they exist
notRecWeight = notRecCount != 0 ? (count * 10) / (2 * notRecCount) : 0;
recWeight = recCount != 0 ? (count * 10) / (2 * recCount) : 0;
weights += (p : p in recProds ? recWeight : notRecWeight | p <- prods);
}
return visit (g) {
case p:prod(_, _, _) => p[weight=weights[p]]
}
}
#memo
rel[Symbol,Symbol] dependencies(map[Symbol, Production] gr)
= {<delabel(from),delabel(to)> | /prod(Symbol from,[_*,Symbol to,_*],_) := gr}+;
Note that this randomTree algorithm will not terminate on grammars that are not "productive" (i.e. they have only a rule like syntax E = E;
Also it can generate trees that are filtered by disambiguation rules. So you can check this by running the parser on a generated string and check for parse errors. Also it can generated ambiguous strings.
By the way, this code was inspired by the PhD thesis of Naveneetha Vasudevan of King's College, London.
Suppose S is a set with t elements modulo n. There are indeed, 2^t subsets of any length. Illustrate a PARI/GP program which finds the smallest subset U (in terms of length) of distinct elements such that the sum of all elements in U is 0 modulo n. It is easy to write a program which searches via brute force, but brute force is infeasible as t and n get larger, so would appreciate help writing a program which doesn't use brute force to solve this instance of the subset sum problem.
Dynamic Approach:
def isSubsetSum(st, n, sm) :
# The value of subset[i][j] will be
# true if there is a subset of
# set[0..j-1] with sum equal to i
subset=[[True] * (sm+1)] * (n+1)
# If sum is 0, then answer is true
for i in range(0, n+1) :
subset[i][0] = True
# If sum is not 0 and set is empty,
# then answer is false
for i in range(1, sm + 1) :
subset[0][i] = False
# Fill the subset table in botton
# up manner
for i in range(1, n+1) :
for j in range(1, sm+1) :
if(j < st[i-1]) :
subset[i][j] = subset[i-1][j]
if (j >= st[i-1]) :
subset[i][j] = subset[i-1][j] or subset[i - 1][j-st[i-1]]
"""uncomment this code to print table
for i in range(0,n+1) :
for j in range(0,sm+1) :
print(subset[i][j],end="")
print(" ")"""
return subset[n][sm];
I got this code from here I don't know weather it seems to work.
function getSummingItems(a,t){
return a.reduce((h,n) => Object.keys(h)
.reduceRight((m,k) => +k+n <= t ? (m[+k+n] = m[+k+n] ? m[+k+n].concat(m[k].map(sa => sa.concat(n)))
: m[k].map(sa => sa.concat(n)),m)
: m, h), {0:[[]]})[t];
}
var arr = Array(20).fill().map((_,i) => i+1), // [1,2,..,20]
tgt = 42,
res = [];
console.time("test");
res = getSummingItems(arr,tgt);
console.timeEnd("test");
console.log("found",res.length,"subsequences summing to",tgt);
console.log(JSON.stringify(res));
I'm looking to create a function that returns a solve for x math equation that can be preformed in ones head (Clearly thats a bit subjective but I'm not sure how else to phrase it).
Example problem: (x - 15)/10 = 6
Note: Only 1 x in the equation
I want to use the operations +, -, *, /, sqrt (Only applied to X -> sqrt(x))
I know that let mathExpression = NSExpression(format: question) converts strings into math equations but when solving for x I'm not sure how to go about doing this.
I previously asked Generating random doable math problems swift for non solving for x problems but I'm not sure how to convert that answer into solving for x
Edit: Goal is to generate an equation and have the user solve for the variable.
Since all you want is a string representing an equation and a value for x, you don't need to do any solving. Just start with x and transform it until you have a nice equation. Here's a sample: (copy and paste it into a Playground to try it out)
import UIKit
enum Operation: String {
case addition = "+"
case subtraction = "-"
case multiplication = "*"
case division = "/"
static func all() -> [Operation] {
return [.addition, .subtraction, .multiplication, .division]
}
static func random() -> Operation {
let all = Operation.all()
let selection = Int(arc4random_uniform(UInt32(all.count)))
return all[selection]
}
}
func addNewTerm(formula: String, result: Int) -> (formula: String, result: Int) {
// choose a random number and operation
let operation = Operation.random()
let number = chooseRandomNumberFor(operation: operation, on: result)
// apply to the left side
let newFormula = applyTermTo(formula: formula, number: number, operation: operation)
// apply to the right side
let newResult = applyTermTo(result: result, number: number, operation: operation)
return (newFormula, newResult)
}
func applyTermTo(formula: String, number:Int, operation:Operation) -> String {
return "\(formula) \(operation.rawValue) \(number)"
}
func applyTermTo(result: Int, number:Int, operation:Operation) -> Int {
switch(operation) {
case .addition: return result + number
case .subtraction: return result - number
case .multiplication: return result * number
case .division: return result / number
}
}
func chooseRandomNumberFor(operation: Operation, on number: Int) -> Int {
switch(operation) {
case .addition, .subtraction, .multiplication:
return Int(arc4random_uniform(10) + 1)
case .division:
// add code here to find integer factors
return 1
}
}
func generateFormula(_ numTerms:Int = 1) -> (String, Int) {
let x = Int(arc4random_uniform(10))
var leftSide = "x"
var result = x
for i in 1...numTerms {
(leftSide, result) = addNewTerm(formula: leftSide, result: result)
if i < numTerms {
leftSide = "(" + leftSide + ")"
}
}
let formula = "\(leftSide) = \(result)"
return (formula, x)
}
func printFormula(_ numTerms:Int = 1) {
let (formula, x) = generateFormula(numTerms)
print(formula, " x = ", x)
}
for i in 1...30 {
printFormula(Int(arc4random_uniform(3)) + 1)
}
There are some things missing. The sqrt() function will have to be implemented separately. And for division to be useful, you'll have to add in a system to find factors (since you presumably want the results to be integers). Depending on what sort of output you want, there's a lot more work to do, but this should get you started.
Here's sample output:
(x + 10) - 5 = 11 x = 6
((x + 6) + 6) - 1 = 20 x = 9
x - 2 = 5 x = 7
((x + 3) * 5) - 6 = 39 x = 6
(x / 1) + 6 = 11 x = 5
(x * 6) * 3 = 54 x = 3
x * 9 = 54 x = 6
((x / 1) - 6) + 8 = 11 x = 9
Okay, let’s assume from you saying “Note: Only 1 x in the equation” that what you want is a linear equation of the form y = 0 = β1*x + β0, where β0 and β1 are the slope and intercept coefficients, respectively.
The inverse of (or solution to) any linear equation is given by x = -β0/β1. So what you really need to do is generate random integers β0 and β1 to create your equation. But since it should be “solvable” in someone’s head, you probably want β0 to be divisible by β1, and furthermore, for β1 and β0/β1 to be less than or equal to 12, since this is the upper limit of the commonly known multiplication tables. In this case, just generate a random integer β1 ≤ 12, and β0 equal to β1 times some integer n, 0 ≤ n ≤ 12.
If you want to allow simple fractional solutions like 2/3, just multiply the denominator and the numerator into β0 and β1, respectively, taking care to prevent the numerator or denominator from getting too large (12 is again a good limit).
Since you probably want to make y non-zero, just generate a third random integer y between -12 and 12, and change your output equation to y = β1*x + β0 + y.
Since you mentioned √ could occur over the x variable only, that is pretty easy to add; the solution (to 0 = β1*sqrt(x) + β0) is just x = (β0/β1)**2.
Here is some very simple (and very problematic) code for generating random integers to get you started:
import func Glibc.srand
import func Glibc.rand
import func Glibc.time
srand(UInt32(time(nil)))
print(rand() % 12)
There are a great many answers on this website that deal with better ways to generate random integers.
Is it somehow possible to concatenate two matlab structures recursively without iterating over all leaves of one of the structures.
For instance
x.a=1;
x.b.c=2;
y.b.d=3;
y.a = 4 ;
would result in the following
res = mergeStructs(x,y)
res.a=4
res.b.c=2
res.b.d=3
The following function works for your particular example. There will be things it doesn't consider, so let me know if there are other cases you want it to work for and I can update.
function res = mergeStructs(x,y)
if isstruct(x) && isstruct(y)
res = x;
names = fieldnames(y);
for fnum = 1:numel(names)
if isfield(x,names{fnum})
res.(names{fnum}) = mergeStructs(x.(names{fnum}),y.(names{fnum}));
else
res.(names{fnum}) = y.(names{fnum});
end
end
else
res = y;
end
Then res = mergeStructs(x,y); gives:
>> res.a
ans =
4
>> res.b
ans =
c: 2
d: 3
as you require.
EDIT: I added isstruct(x) && to the first line. The old version worked fine because isfield(x,n) returns 0 if ~isstruct(x), but the new version is slightly faster if y is a big struct and ~isstruct(x).