Logic:
eploy(list, constant)
if list is empty then
return:
0;
else
return:
(first_element + constant*eploy(rest_of_the_elements, constant)
I have written following code:
fun eploy(xs, x1:int) =
if null xs
then (0)
else (x::xs') => x + x1*eploy(xs',x1)
eploy([1,2],4);
If you want to do pattern matching then you need to use case:
fun eploy(xs, x1) =
case xs of
nil => 0
| x::xs' => x + x1*eploy(xs', x1)
You can also merge that into the function definition by using clauses:
fun eploy(nil, x1) = 0
| eploy(x::xs', x1) = x + x1*eploy(xs', x1)
Related
How to get coefficients for ALL combinations of the variables of a multivariable polynomial using sympy.jl or another Julia package for symbolic computation?
Here is an example from MATLAB,
syms a b y
[cxy, txy] = coeffs(ax^2 + by, [y x], ‘All’)
cxy =
[ 0, 0, b]
[ a, 0, 0]
txy =
[ x^2y, xy, y]
[ x^2, x, 1]
My goal is to get
[ x^2y, xy, y]
[ x^2, x, 1]
instead of [x^2, y]
I asked the same question at
https://github.com/JuliaPy/SymPy.jl/issues/482
and
https://discourse.julialang.org/t/symply-jl-for-getting-coefficients-for-all-combination-of-the-variables-of-a-multivariable-polynomial/89091
but I think I should ask if this can be done using Sympy.py.
Using Julia, I tried the following,
julia> #syms x, y, a, b
julia> ff = sympy.Poly(ax^2 + by, (x,y))
Poly(ax**2 + by, x, y, domain='ZZ[a,b]')
julia> [prod(ff.gens.^i) for i in ff.monoms()]
2-element Vector{Sym}:
x^2
y
This is a longer form rewrite of the one-liner in the comment.
It uses Pipe.jl to write expressions 'functionally', so familiarity with pipe operator (|>) and Pipe.jl will help.
using SymPy
using Pipe
#syms x, y, a, b
ff = sympy.Poly(a*x^2 + b*y, (x,y))
max_degrees =
#pipe ff.monoms() .|> collect |> hcat(_...) |>
reduce(max, _, dims=2) |> vec
degree_iter =
#pipe max_degrees .|> UnitRange(0, _) |>
tuple(_...) |> CartesianIndices
result = [prod(ff.gens.^Tuple(I)) for I in degree_iter] |>
reverse |> eachcol |> collect
or using more of the python methods:
[prod(ff.gens.^I) for
I in Iterators.product((0:d for d in ff.degree.(ff.gens))...)] |>
reverse |> eachcol |> collect
Both give the desired result:
2-element Vector{...}:
[x^2*y, x*y, y]
[x^2, x, 1]
UPDATE:
In case there are more than 2 generators, the result needs to be a Array with higher dimension. The last bits of matrix transposes is immaterial and the expressions become:
Method 1:
max_degrees =
#pipe ff.monoms() .|> collect |> hcat(_...) |>
reduce(max, _, dims=2) |> vec
degree_iter =
#pipe max_degrees .|> UnitRange(0, _) |>
tuple(_...) |> CartesianIndices
result = [prod(ff.gens.^Tuple(I)) for I in degree_iter]
Method 2:
result = [prod(ff.gens.^Tuple(I)) for I in degree_iter]
Thanks a lot #Dan Getz. Your solution works for the TOY example from MATLAB. My real case is more complicated, which has more variables and polynominals. I tried your method for 3 variables,
using SymPy
#syms x, y, z, a, b
ff = sympy.Poly(a*x^2 + b*y + z^2 + x*y + y*z, (x, y, z))
[prod(ff.gens.^Tuple(I)) for I in CartesianIndices(tuple(UnitRange.(0,vec(reduce(max, hcat(collect.(ff.monoms())...), dims=1)))...))]
I got the following error,
ERROR: LoadError: DimensionMismatch: arrays could not be broadcast to a common size; got a dimension with lengths 3 and 5
Stacktrace:
How to generate your method to any number of variables with different degrees, e.g., x^3 + y^3 + z^3 + xyz + xy^2z?
You can find the degree of each of the two variables of interest and then use them to create the matrix of generators; you can use them to get the coefficients of interest. I am not sure what you expect if the equation were like a*x**2 + b*y + c...
>>> from sympy import *
>>> from sympy.abc import a, b, x, y
>>> eq = a*x**2 + b*y
>>> deg = lambda x: Poly(eq, x).degree() # helper to give degree in "x"
>>> v = (Matrix([x**i for i in range(deg(x),-1,-1)]
... )*Matrix([y**i for i in range(deg(y),-1,-1)]).T).T; v
Matrix([[x**2*y, x*y, y], [x**2, x, 1]])
>>> Matrix(*v.shape, [eq.coeff(i) if i.free_symbols else eq.as_coeff_Add()[0]
... for i in v])
Matrix([[0, 0, b], [a, 0, 0]])
From #jverzani (thanks)
using SymPy;
#syms a b x y;
eq = a*x^2 + b*y;
deg = x -> sympy.Poly(eq, x).degree();
xs, ys = [x^i for i ∈ range(deg(x):-1:0], [y^i for i ∈ deg(y):-1:0];
v = permutedims(xs .* permutedims(ys));
M = [x^2*y x*y y; x^2 x 1];
[length(free_symbols(i)) > 0 ? eq.coeff(i) : eq.as_coeff_add()[1] for i ∈ v];
[0 0 b; a 0 0]
Is there any way to run name resolution on an arbitrary expression without running it? e.g. I would like to take an expression such as
quote
x = 1
y = 2*x + 1
z = x^2 - 1
f(x) = 2*x + 1
end
and be told that the names defined in the scope of this block are x, y, z, f and the names *, +, ^, - are pulled in from outside the scope of this block. Bonus points if it can tell me that there's a sub-scope defined in the body of f which creates it's own name x and pulls in + from an enclosing scope.
This question appeared in the Julia Zulip community
Thanks to Takafumi for showing me how to solve this on Zulip
We can get a list of locally defined names in the outermost scope of a julia expression like so:
ex = quote
x = 1
y = 2*x + 1
z = x^2 - 1
f(x) = 2*x + 1
end
using JuliaVariables, MLStyle
function get_locals(ex::Expr)
vars = (solve_from_local ∘ simplify_ex)(ex).args[1].bounds
map(x -> x.name, vars)
end
julia> get_locals(ex)
4-element Array{Symbol,1}:
:f
:y
:z
:x
and we can get the symbols pulled in from outside the scope like this:
_get_outers(_) = Symbol[]
_get_outers(x::Var) = x.is_global ? [x.name] : Symbol[]
function _get_outers(ex::Expr)
#match ex begin
Expr(:(=), _, rhs) => _get_outers(rhs)
Expr(:tuple, _..., Expr(:(=), _, rhs)) => _get_outers(rhs)
Expr(_, args...) => mapreduce(_get_outers, vcat, args)
end
end
get_outers(ex) = (unique! ∘ _get_outers ∘ solve_from_local ∘ simplify_ex)(ex)
julia> get_outers(ex)
6-element Array{Symbol,1}:
:+
:*
:-
:^
This question already has answers here:
Value of bindings in SML?
(2 answers)
Closed 6 years ago.
Could someone please help. I don't get the sequence of evaluation here and how we got values of "ans". e.g. in the first example there's no value of y and I'm not sure whether this returns a pair or calls x ! (fn y => y x). It would be very helpful if you can Trace each expression.
val x = 1
val f = (fn y => y x)
val x = 7
val g = (fn y => x - y)
val ans = f g
val ans = 6 : int
=====================================
fun f p =
let
val x = 3
val y = 4
val (z,w) = p
in
(z (w y)) + x
end
val x = 1
val y = 2
val ans = f((fn z => x + z), (fn x => x + x + 0))
val ans = 12 : int
There are a few things which help make problems like this much clearer
when trying understand an alien function Lexical scoping works.
add in types to the parameters and return values without modifying the program, the compiler will tell you if you get it wrong...
replace anonymous functions with named ones.
rename variable bindings that have the same names but refer to different lexical scope.
remove variable bindings that only get used once.
binding a value to a name does not actually perform any computation,
so is merely for the benefit of the reader, if it is not doing that job
it merely serves to obfuscate, then by all means remove it.
fun f (y1 : int -> 'a) = y1 1 : 'a;
fun g (y2 : int) = 7 - y2 : int;
val ans : int = f g;
so g is given as a parameter to f, f calls g giving it the parameter x having the value 1 making y2 = 1, which g subtracts 7 - 1 returning 6.
the return value of g is an int, thus f's 'a type when g is applied to it is an int.
for the 2nd one clean it up a bit, I pulled the anonymous fn's out into their own and named values and call f (foo, bar) to make it more readable...
fun f p =
let val x = 3
val y = 4
val (z, w) = p
in (z (w y)) + x end
fun foo z = z + 1;
fun bar x = x * 2;
val ans = f(foo, bar);
Finally, we can get rid of the let values which are only used once
and replace the (z,w) = p with just (z, w) as a parameter to the function which should be much easier to follow
fun f (z, w) = (z (w 4)) + 3
fun foo z = z + 1;
fun bar x = x * 2;
val ans = f(foo, bar);
val ans = ((4 * 2) + 1) + 3
I'm trying to write a function in SML to compute the partial sum of an alternating harmonic series, and for the life of me I can't figure out why the compiler says one of the cases is redundant. I haven't used case statements before(or local, for that matter), but the order of these cases seems right to me.
local
fun altHarmAux (x:int, y:real) =
case x of
1 => 1.0
| evenP => altHarmAux(x-1, y - y/(real x))
| oddP => altHarmAux(x-1, y + y/(real x))
in
fun altHarmonic (a:int) = altHarmAux(a, real a)
end
Even if you have defined the two predicate functions somewhere, they can't be used in a case like that.
whatever you write on the left hand of => will be bound to the value you are matching on, thus the two last matches in your case will match the same input, rendering the last one useless, as the first one will always be used
You will have to apply your predicate function to the value directly, and then match on the result
local
fun altHarmAux (x, y) =
case (x, evenP x) of
(1, _) => 1.0
| (_ true) => altHarmAux(x-1, y - y/(real x))
| (_, false) => altHarmAux(x-1, y + y/(real x))
in
fun altHarmonic a = altHarmAux(a, real a)
end
or perhaps simpler
local
fun altHarmAux (1, _) = 1.0
| altHarmAux (x, y) =
altHarmAux (x-1, y + (if evenP x then ~y else y) / (real x))
in
fun altHarmonic a = altHarmAux (a, real a)
end
or
local
fun altHarmAux (1, _) = 1.0
| altHarmAux (x, y) =
if evenP x then
altHarmAux (x-1, y - y/(real x))
else
altHarmAux (x-1, y + y/(real x))
in
fun altHarmonic a = altHarmAux (a, real a)
end
I'm doing some homework but I've been stuck for hours on something.
I'm sure it's really trivial but I still can't wrap my head around it after digging through the all documentation available.
Can anybody give me a hand?
Basically, the exercise in OCaml programming asks to define the function x^n with the exponentiation by squaring algorithm.
I've looked at the solution:
let rec exp x = function
0 -> 1
| n when n mod 2 = 0 -> let y = exp x (n/2) in y*y
| n when n mod 2 <> 0 -> let y = exp x ((n-1)/2) in y*y*x
;;
What I don't understand in particular is how the parameter n can be omitted from the fun statement and why should it be used as a variable for a match with x, which has no apparent link with the definition of exponentiation by squaring.
Here's how I would do it:
let rec exp x n = match n with
0 -> 1
| n when (n mod 2) = 1 -> (exp x ((n-1)/2)) * (exp x ((n-1)/2)) * x
| n when (n mod 2) = 0 -> (exp x (n/2)) * (exp x (n/2))
;;
Your version is syntaxically correct, yields a good answer, but is long to execute.
In your code, exp is called recursively twice, thus yielding twice as much computation, each call yielding itself twice as much computation, etc. down to n=0. In the solution, exp is called only once, the result is storred in the variable y, then y is squared.
Now, about the syntax,
let f n = match n with
| 0 -> 0
| foo -> foo-1
is equivalent to:
let f = function
| 0 -> 0
| foo -> foo-1
The line let rec exp x = function is the begging of a function that takes two arguments: x, and an unnammed argument used in the pattern matching. In the pattern matching, the line
| n when n mod 2 = 0 ->
names this argument n. Not that a different name could be used in each case of the pattern matching (even if that would be less clear):
| n when n mod 2 = 0 -> let y = exp x (n/2) in y*y
| p when p mod 2 <> 0 -> let y = exp x ((p-1)/2) in y*y*x
The keyword "function" is not a syntaxic sugar for
match x with
but for
fun x -> match x with
thus
let rec exp x = function
could be replaced by
let rec exp x = fun y -> match y with
which is of course equivalent with your solution
let rec exp x y = match y with
Note that i wrote "y" and not "n" to avoid confusion. The n variable introduced after the match is a new variable, which is only related to the function parameter because it match it. For instance, instead of
let y = x in ...
you could write :
match x with y -> ...
In this match expression, the "y" expression is the "pattern" matched. And like any pattern, it binds its variables (here y) with the value matched. (here the value of x) And like any pattern, the variables in the pattern are new variables, which may shadow previously defined variables. In your code :
let rec exp x n = match n with
0 -> 1
| n when (n mod 2) = 1 -> (exp x ((n-1)/2)) * (exp x ((n-1)/2)) * x
| n when (n mod 2) = 0 -> (exp x (n/2)) * (exp x (n/2))
;;
the variable n in the two cases shadow the parameter n. This isn't a problem, though, since the two variable with the same name have the same value.