Author Archives: perfectionatic

Julia calling C: A more minimal example

By: perfectionatic

Re-posted from: http://perfectionatic.org/?p=470

Earlier I presented a minimal example of Julia calling C. It mimics how one would go about writing C code, wrapping it a library and then calling it from Julia. Today I came across and even more minimal way of doing that while reading an excellent blog on Julia’s syntactic loop fusion. Associated with the blog was notebook that explores the matter further.

Basically, you an write you C in a string and pass it directly to the compiler. It goes something like

C_code= """
       double mean(double a, double b) {
         return (a+b) / 2;
       }
       """
const Clib=tempname()
open(`gcc -fPIC -O3 -xc -shared -o $(Clib * "." * Libdl.dlext) -`, "w") do f
     print(f, C_code)
end

The tempname function generate a unique temporary file path. On my Linux system Clib will be string like "/tmp/juliaivzRkT". That path is used to generate a library name "/tmp/juliaivzRkT.so" which will then used in the ccall:

julia> x=ccall((:mean,Clib),Float64,(Float64,Float64),2.0,5.0)
3.5

This approach would be be recommended if are writing anything sophisticated in C. However, it fun to experiment with for short bits of C code that you might like to call from Julia. Saves you the hassle of creating a Makefile, compiling, etc…

Solving the Fish Riddle with JuMP

By: perfectionatic

Re-posted from: http://perfectionatic.org/?p=438

Recently I came across a nice Ted-Ed video presenting a Fish Riddle.

I thought it would be fun to try solving it using Julia’s award winning JuMP package. Before we get started, please watch the above video-you might want to pause at 2:24 if you want to solve it yourself.

To attempt this problem in Julia, you will have to install the JuMP package.

julia> Pkg.add("JuMP")

JuMP provides an algebraic modeling language for dealing with mathematical optimization problems. Basically, that allows you to focus on describing your problem in a simple syntax and it would then take care of transforming that description in a form that can be handled by any number of solvers. Those solvers can deal with several types of optimization problems, and some solvers are more generic than others. It is important to pick the right solver for the problem that you are attempting.

The problem premises are:
1. There are 50 creatures in total. That includes sharks outside the tanks and fish
2. Each SECTOR has anywhere from 1 to 7 sharks, with no two sectors having the same number of sharks.
3. Each tank has an equal number of fish
4. In total, there are 13 or fewer tanks
5. SECTOR ALPHA has 2 sharks and 4 tanks
6. SECTOR BETA has 4 sharsk and 2 tanks
We want to find the number of tanks in sector GAMMA!

Here we identify the problem as mixed integer non-linear program (MINLP). We know that because the problem involves an integer number of fish tanks, sharks, and number of fish inside each tank. It also non-linear (quadratic to be exact) because it involves multiplying two two of the problem variables to get the total number or creatures. Looking at the table of solvers in the JuMP manual. pick the Bonmin solver from AmplNLWriter package. This is an open source solver, so installation should be hassle free.

julia> Pkg.add("AmplNLWriter")

We are now ready to write some code.

using JuMP, AmplNLWriter
 
# Solve model
m = Model(solver=BonminNLSolver())
 
# Number of fish in each tank
@variable(m, n>=1, Int)
 
# Number of sharks in each sector
@variable(m, s[i=1:3], Int)
 
# Number of tanks in each sector
@variable(m, nt[i=1:3]>=0, Int)
 
@constraints m begin
    # Constraint 2
    sharks[i=1:3], 1 <= s[i] <= 7
    numfish[i=1:3], 1 <= nt[i]
      # Missing uniqueness in restriction
    # Constraint 4
    sum(nt) <= 13
    # Constraint 5
    s[1] == 2
    nt[1] == 4
    # Constraint 6
    s[2] == 4
    nt[2] == 2
end
 
# Constraints 1 & 3
@NLconstraint(m, s[1]+s[2]+s[3]+n*(nt[1]+nt[2]+nt[3]) == 50)
 
# Solve it
status = solve(m)
 
sharks_in_each_sector=getvalue(s)
fish_in_each_tank=getvalue(n)
tanks_in_each_sector=getvalue(nt)
 
@printf("We have %d fishes in each tank.\n", fish_in_each_tank)
@printf("We have %d tanks in sector Gamma.\n",tanks_in_each_sector[3])
@printf("We have %d sharks in sector Gamma.\n",sharks_in_each_sector[3])

In that representation we could not capture the restriction that “no two sectors having the same number of sharks”. We end up with the following output:

We have 4 fishes in each tank.
We have 4 tanks in sector Gamma.
We have 4 sharks in sector Gamma.

Since the problem domain is limited, we can possible fix that by adding a constrain that force the number of sharks in sector Gamma to be greater than 4.

@constraint(m,s[3]>=5)

This will result in an answer that that does not violate any of the stated constraints.

We have 3 fishes in each tank.
We have 7 tanks in sector Gamma.
We have 5 sharks in sector Gamma.

However, this seems like a bit of kludge. The proper way go about it is represent the number of sharks in the each sector as binary array, with only one value set to 1.

# Number of sharks in each sector
@variable(m, s[i=1:3,j=1:7], Bin)

We will have to modify our constraint block accordingly

@constraints m begin
    # Constraint 2
    sharks[i=1:3], sum(s[i,:]) == 1
    u_sharks[j=1:7], sum(s[:,j]) <=1 # uniquness
    # Constraint 4
    sum(nt) <= 13
    # Constraint 5
    s[1,2] == 1
    nt[1] == 4
    # Constraint 6
    s[2,4] == 1
    nt[2] == 2
end

We invent a new variable array st to capture the number of sharks in each sector. This simply obtained by multiplying the binary array by the vector [1,2,\ldots,7]^\top

@variable(m,st[i=1:3],Int)
@constraint(m, st.==s*collect(1:7))

We rewrite our last constraint as

# Constraints 1 & 3
@NLconstraint(m, st[1]+st[2]+st[3]+n*(nt[1]+nt[2]+nt[3]) == 50)

After the model has been solved, we extract our output for the number of sharks.

sharks_in_each_sector=getvalue(st)

…and we get the correct output.

This problem might have been an overkill for using a full blown mixed integer non-linear optimizer. It can be solved by a simple table as shown in the video. However, we might not alway find ourselves in such a fortunate position. We could have also use mixed integer quadratic programming solver such as Gurobi which would be more efficient for that sort of problem. Given the small problem size, efficiency hardly matters here.

Reading DataFrames with non-UTF8 encoding in Julia

By: perfectionatic

Re-posted from: http://perfectionatic.org/?p=414

Recently I ran into problem where I was trying to read a CSV files from a Scandinavian friend into a DataFrame. I was getting errors it could not properly parse the latin1 encoded names.

I tried running

using DataFrames
dataT=readtable("example.csv", encoding=:latin1)

but the got this error

ArgumentError: Argument 'encoding' only supports ':utf8' currently.

The solution make use of (StringEncodings.jl)[https://github.com/nalimilan/StringEncodings.jl] to wrap the file data stream before presenting it to the readtable function.

f=open("example.csv","r")
s=StringDecoder(f,"LATIN1", "UTF-8")
dataT=readtable(s)
close(s)
close(f)

The StringDecoder generates an IO stream that appears to be utf8 for the readtable function.