I am adding edges to a simple weighted directed graph (from SimpleWeightedDiGraph() that is part of LightGraphs package) in Julia. Some of the arcs are "free" (null weight). However, when specifying the weight of 0 it is not added as a new edge and a shortest path problem does not include it in the possible solution. Is there an easy way to add "free" edges/arcs to a graph in Julia?
The key issue is how zero values are represented in a sparse matrix (which is the underlying data store for SimpleWeightedGraphs. While it is true that the underlying zero value is preserved once it's explicitly set:
julia> g = SimpleWeightedGraph(6)
{6, 0} undirected simple Int64 graph with Float64 weights
julia> add_edge!(g, 1, 2, 1.0)
true
julia> add_edge!(g, 1, 3, 1.0)
true
julia> add_edge!(g, 1, 3, 0.0)
true
julia> weights(g)
6×6 SparseMatrixCSC{Float64,Int64} with 4 stored entries:
[2, 1] = 1.0
[3, 1] = 0.0
[1, 2] = 1.0
[1, 3] = 0.0
this will fail if you have to do anything with the edges:
julia> collect(edges(g))
1-element Array{SimpleWeightedGraphs.SimpleWeightedEdge{Int64,Float64},1}:
Edge 1 => 2 with weight 1.0
There's no really good solution to this. My advice is to use a sufficiently small weight as proposed above to approximate a zero value.
(PS: the reason the initial add_edge!(g, 1, 3, 0.0) doesn't work is because in Julia, setting the value of a new sparsematrix element to zero is a no-op.)
This modification of the SimpleWeightedGraphs README example works for me:
using LightGraphs, SimpleWeightedGraphs
# set the size of the graph
g = SimpleWeightedDiGraph(3)
add_edge!(g, 1, 2, 0.5)
add_edge!(g, 2, 3, 0.8)
add_edge!(g, 1, 3, 2.0)
# find the shortest path from vertex 1 to vertex 3 taking weights into account.
enumerate_paths(dijkstra_shortest_paths(g, 1), 3) # gives [1,2,3]
# reweight the edge from 1 to 3 to be "free"
add_edge!(g, 1, 3, 0.0)
enumerate_paths(dijkstra_shortest_paths(g, 1), 3) # gives [1,3]
Notice that the vertices must be in the graph (according to its size) to be able to set their weights, as stated in the docs: ?add_edge!.
Related
I want to calculate the Euclidean distance between a tuple and each tuple within a Vector in Julia using the map function, like below (but I get two values instead of three):
julia> tups = [
(1, 3),
(11, 2),
(0, 1)
];
julia> map((x, y) -> √(sum((x.-y).^2)), tups, (3, 3))
2-element Vector{Float64}:
2.0
8.06225774829855
How can I make it work correctly?
Julia has the Distances package especially for these types of calculations. The 'Julian way' encourages interoperability between packages to allow benefitting from future development of the ecosystem. For example, new metric definitions, or specialized hardware code to compute distances.
For the problem in the post, the code would look:
julia> using Distances
julia> tups = [
(1, 3),
(11, 2),
(0, 1)
];
julia> euclidean.(tups,Ref((3,3)))
3-element Vector{Float64}:
2.0
8.06225774829855
3.605551275463989
Notice the use of broadcasting instead of map with dot syntax euclidean.. The Ref((3,3)) causes broadcasting to consider (3,3) as a single element to broadcast and not break it to a pair of Ints.
The code you've written is pretty equal to this:
[
func((1, 3), 3),
func((11, 2), 3)
]
The map function iterates over the given collections iter times equal to the lowest length:
julia> length((3, 3)), length(tups)
(2, 3)
So it iterates two times, not three. To make that work, you can repeat the (3, 3), three times or even omit the (3, 3) argument:
julia> map((x, y) -> √(sum((x.-y).^2)), tups, ((3, 3), (3, 3), (3, 3)))
3-element Vector{Float64}:
2.0
8.06225774829855
3.605551275463989
# OR
julia> map((x, y) -> √(sum((x.-y).^2)), tups, ((3, 3) for _∈1:3))
3-element Vector{Float64}:
2.0
8.06225774829855
3.605551275463989
# Or omit the last argument
julia> map(arg -> √((3 - arg[1])^2 + (3 - arg[2])^2), tups)
3-element Vector{Float64}:
2.0
8.06225774829855
3.605551275463989
Assuming I have a vector of say four dimensions in which every variable lays in a special interval. Thus we got:
Vector k = (x1,x2,x3,x4) with x1 = (-2,2), x2 = (0,2), x3 = (-4,1), x4 = (-1,1)
I am only interested in the points constraint by the intervals.
So to say v1 = (0,1,2,0) is important where v2 = (-5,-5,5,5) is not.
In additon to that the point i+1 should be relatively close to point i among my journey. Therefore I dont want to jump around in space.
Is there a proper way of walking through those interesting points?
For example in 2D space with x1,x2 = (-2,2) like so:
Note: The frequenz of the red line could be higher
There are many ways to create a space-filling curve while preserving closeness. See the Wikipedia article for a few examples (some have associated algorithms for generating them): https://en.wikipedia.org/wiki/Space-filling_curve
Regardless, let's work with your zig-zag pattern for 2D and work on extending it to 3D and 4D. To extend it into 3D, we just add another zig to the zig-zag. Take a look at the (rough) diagram below:
Essentially, we repeat the pattern that we had in 2D but we now have multiple layers that represent the third dimension. The extra zig that we need to add is the switch between bottom-to-top and top-to-bottom every layer. This is pretty simple to abstract:
In 2D, we have x and y axes.
We move across the x domain switching between positive and negative
directions most frequently.
We move across the y domain once.
In 3D, we have x, y, and z axes.
We move across the x domain switching between positive and negative directions most frequently.
We move across the y domain switching between positive and negative directions second most frequently.
We move across the z domain once.
It should be clear how this generalizes to higher dimensions. Now, I'll present some (Python 3) code that implements the zig-zag pattern for 4D. Let's represent the position in 4D space as (x, y, z, w) and the ranges in each dimension as (x0, x1), (y0, y1), (z0, z1), (w0, w1). These are our inputs. Then, we also define xdir, ydir, and zdir to keep track of the direction of the zig-zag.
x, y, z, w = x0, y0, z0, w0
xdir, ydir, zdir = +1, +1, +1
for iw in range(w1 - w0):
for iz in range(z1 - z0):
for iy in range(y1 - y0):
for ix in range(x1 - x0):
print(x, y, z, w)
x = x + xdir
xdir = -xdir
print(x, y, z, w)
y = y + ydir
ydir = -ydir
print(x, y, z, w)
z = z + zdir
zdir = -zdir
print(x, y, z, w)
w = w + 1
This algorithm has the guarantee that no two points printed out after each other have a distance greater than 1.
Using recursion, you can clean this up to make a very nice generalizable method. I hope this helps; let me know if you have any questions.
With the work of #Matthew Miller I implemented this generalization for any given multidimenisonal space:
'''assuming that we take three points out of our intervals [0,2] for a,b,c
which every one of them is corresponding to one dimension i.e. a 3D-space'''
a = [0,1,2]
b = [0,1,2]
c = [0,1,2]
vec_in = []
vec_in.append(a)
vec_in.append(b)
vec_in.append(c)
result = []
hold = []
dir = [False] * len(vec_in)
def create_points(vec , index, temp, desc):
if (desc):
loop_x = len(vec[index])-1
loop_y = -1
loop_z = -1
else:
loop_x = 0
loop_y = len(vec[index])
loop_z = 1
for i in range(loop_x,loop_y,loop_z):
temp.append(vec[index][i])
if (index < (len(vec) - 1)):
create_points(vec, index + 1, temp, dir[index])
else:
u = []
for k in temp:
u.append(k)
result.append(u)
temp.pop()
if (dir[index] == False):
dir[index] = True
else:
dir[index] = False
if len(temp) != 0:
temp.pop()
#render
create_points(vec_in, 0, hold, dir[0])
for x in (result):
print(x)
The result is a journey which covers every possible postion in a continous way:
[0, 0, 0]
[0, 0, 1]
[0, 0, 2]
[0, 1, 2]
[0, 1, 1]
[0, 1, 0]
[0, 2, 0]
[0, 2, 1]
[0, 2, 2]
[1, 2, 2]
[1, 2, 1]
[1, 2, 0]
[1, 1, 0]
[1, 1, 1]
[1, 1, 2]
[1, 0, 2]
[1, 0, 1]
[1, 0, 0]
[2, 0, 0]
[2, 0, 1]
[2, 0, 2]
[2, 1, 2]
[2, 1, 1]
[2, 1, 0]
[2, 2, 0]
[2, 2, 1]
[2, 2, 2]
How can I create a Static Directed Graph from an array of tuples in Julia without having to create a Simple Directed Graph first. An example edge list I have is [(1,2),(2,3),(3,4)]. The documentation of StaticGraphs.jl is limited.
There's a way to do this but it requires you to have the edges and their reverses already sorted into two vectors. Assume you have a directed path graph 1 -> 2 -> 3 -> 4:
fwd = [(1, 2), (2, 3), (3, 4)] # these are your forward edges, sorted
rev = [(2, 1), (3, 2), (4, 3)] # these are the reverse of the forward edges, sorted
# also, sort(reverse.(fwd)) will do this easily.
g = StaticDiGraph(4, fwd, rev) # number of vertices is the first argument
testing:
julia> h = StaticDiGraph(path_digraph(4))
{4, 3} directed simple static {UInt8, UInt8} graph
julia> g == h
true
Let's say I have a vector of unique integers, for example [1, 2, 6, 4] (sorting doesn't really matter).
Given some n, I want to get all possible values of summing n elements of the set, including summing an element with itself. It is important that the list I get is exhaustive.
For example, for n = 1 I get the original set.
For n = 2 I should get all values of summing 1 with all other elements, 2 with all others etc. Some kind of memory is also required, in the sense that I have to know from which entries of the original set did the sum I am facing come from.
For a given, specific n, I know how to solve the problem. I want a concise way of being able to solve it for any n.
EDIT: This question is for Julia 0.7 and above...
This is a typical task where you can use a dictionary in a recursive function (I am annotating types for clarity):
function nsum!(x::Vector{Int}, n::Int, d=Dict{Int,Set{Vector{Int}}},
prefix::Vector{Int}=Int[])
if n == 1
for v in x
seq = [prefix; v]
s = sum(seq)
if haskey(d, s)
push!(d[s], sort!(seq))
else
d[s] = Set([sort!(seq)])
end
end
else
for v in x
nsum!(x, n-1, d, [prefix; v])
end
end
end
function genres(x::Vector{Int}, n::Int)
n < 1 && error("n must be positive")
d = Dict{Int, Set{Vector{Int}}}()
nsum!(x, n, d)
d
end
Now you can use it e.g.
julia> genres([1, 2, 4, 6], 3)
Dict{Int64,Set{Array{Int64,1}}} with 14 entries:
16 => Set(Array{Int64,1}[[4, 6, 6]])
11 => Set(Array{Int64,1}[[1, 4, 6]])
7 => Set(Array{Int64,1}[[1, 2, 4]])
9 => Set(Array{Int64,1}[[1, 4, 4], [1, 2, 6]])
10 => Set(Array{Int64,1}[[2, 4, 4], [2, 2, 6]])
8 => Set(Array{Int64,1}[[2, 2, 4], [1, 1, 6]])
6 => Set(Array{Int64,1}[[2, 2, 2], [1, 1, 4]])
4 => Set(Array{Int64,1}[[1, 1, 2]])
3 => Set(Array{Int64,1}[[1, 1, 1]])
5 => Set(Array{Int64,1}[[1, 2, 2]])
13 => Set(Array{Int64,1}[[1, 6, 6]])
14 => Set(Array{Int64,1}[[4, 4, 6], [2, 6, 6]])
12 => Set(Array{Int64,1}[[4, 4, 4], [2, 4, 6]])
18 => Set(Array{Int64,1}[[6, 6, 6]])
EDIT: In the code I use sort! and Set to avoid duplicate entries (remove them if you want duplicates). Also you could keep track how far in the index on vector x in the loop you reached in outer recursive calls to avoid generating duplicates at all, which would speed up the procedure.
I want a concise way of being able to solve it for any n.
Here is a concise solution using IterTools.jl:
Julia 0.6
using IterTools
n = 3
summands = [1, 2, 6, 4]
myresult = map(x -> (sum(x), x), reduce((x1, x2) -> vcat(x1, collect(product(fill(summands, x2)...))), [], 1:n))
(IterTools.jl is required for product())
Julia 0.7
using Iterators
n = 3
summands = [1, 2, 6, 4]
map(x -> (sum(x), x), reduce((x1, x2) -> vcat(x1, vec(collect(product(fill(summands, x2)...)))), 1:n; init = Vector{Tuple{Int, NTuple{n, Int}}}[]))
(In Julia 0.7, the parameter position of the neutral element changed from 2nd to 3rd argument.)
How does this work?
Let's indent the one-liner (using the Julia 0.6 version, the idea is the same for the Julia 0.7 version):
map(
# Map the possible combinations of `1:n` entries of `summands` to a tuple containing their sum and the summands used.
x -> (sum(x), x),
# Generate all possible combinations of `1:n`summands of `summands`.
reduce(
# Concatenate previously generated combinations with the new ones
(x1, x2) -> vcat(
x1,
vec(
collect(
# Cartesian product of all arguments.
product(
# Use `summands` for `x2` arguments.
fill(
summands,
x2)...)))),
# Specify for what lengths we want to generate combinations.
1:n;
# Neutral element (empty array).
init = Vector{Tuple{Int, NTuple{n, Int}}}[]))
Julia 0.6
This is really just to get a free critique from the experts as to why my method is inferior to theirs!
using Combinatorics, BenchmarkTools
function nsum(a::Vector{Int}, n::Int)::Vector{Tuple{Int, Vector{Int}}}
r = Vector{Tuple{Int, Vector{Int}}}()
s = with_replacement_combinations(a, n)
for i in s
push!(r, (sum(i), i))
end
return sort!(r, by = x -> x[1])
end
#btime nsum([1, 2, 6, 4], 3)
It runs in circa 4.154 μs on my 1.8 GHz processor for n = 3. It produces a sorted array showing the sum (which may appear more than once) and how it is made up (which is unique to each instance of the sum).
In graph theory, we use the Hungarian Algorithm to compute a weighted bipartite graph's minimum edge cover (a set of edges that is incident to every vertices, the one with the minimum total weight.)
I find that in new version 8 of Mathematica, there is a whole new package of functions for Graph Theory, (begin with Graph[].) But I've not found any function that do this job. I do find a function called FindEdgeCover[] that can only find a edge cover, not the minimum one.
I did a few experiments and, although not documented, it seems that FindEdgeCover[] does what you want.
Consider for example:
h[list_] := CompleteGraph[4, EdgeWeight -> list]
FindEdgeCover[h#Range#6]
(*
-> {1->2,1->3,1->4}
*)
But
FindEdgeCover[h#Reverse#Range#6]
(*
-> {1->2,3->4}
*)
of course no warranty ...
Edit
Here you have some code to experiment with by using different weighted adjacency matrices
adj = {{\[Infinity], 1, 1, 1, 1}, {1, \[Infinity], 2, 2, 2},
{1, 2, \[Infinity], 2, 2}, {1, 2, 2, \[Infinity], 2},
{1, 2, 2, 2, \[Infinity]}}
g = WeightedAdjacencyGraph[adj];
g = WeightedAdjacencyGraph[adj, VertexShapeFunction -> "Name",
EdgeLabels ->
MapThread[
Rule, {EdgeList#g, AbsoluteOptions[g, EdgeWeight] /. {_ -> x_} -> x}],
GraphHighlight -> FindEdgeCover[g]]
NB: The code is not good at all, but I couldn't find a way to use EdgeLabels -> “EdgeWeight”. I posted this question to see if someone can do it.