Recently I came across a fascinating Numberphile video on truncatable primes

I immediately thought it would be cool to whip a quick Julia code to get the full enumeration of all left truncatable primes, count the number of branches and also get the largest left truncatable prime.

using Primes
function get_left_primes(s::String)
p_arr=Array{String,1}()for i=1:9
number_s="$i$s"if isprime(parse(BigInt, number_s))push!(p_arr,number_s)endend
p_arr
endfunction get_all_left_primes(l)
r_l= Array{String,1}()
n_end_points=0for i in l
new_l=get_left_primes(i)
isempty(new_l)&&(n_end_points+=1)append!(r_l,new_l)
next_new_l,new_n=get_all_left_primes(new_l)
n_end_points+=new_n # counting the chainsappend!(r_l,next_new_l)end
r_l, n
end

The first function just prepends a number (expressed in String for convenience) and checks for it possible primes that can emerge from a single digit prepending. For example:

Julia has good documentation on dealing with Dates and Time, however that is often in the context constructing and Date and Time objects. In this post, I am focus on the ability to iterate over Dates and Times. This is very useful in countless application.

We start of by capturing this moment and moving ahead into the future

Notice that Julia was clever enough properly interpret that we will be on the in another day after exactly one hour. Thanks to it multiple dispatch of the DateTime type to be able to do TimeType period arithmatic.

You can then write a nice for loop that does something every four hours for the next two days.

julia>for t=this_moment:Dates.Hour(4):this_moment+Dates.Day(2)println(t)#or somethings special with that timeend2018-04-01T23:13:33.4372018-04-02T03:13:33.4372018-04-02T07:13:33.4372018-04-02T11:13:33.4372018-04-02T15:13:33.4372018-04-02T19:13:33.4372018-04-02T23:13:33.4372018-04-03T03:13:33.4372018-04-03T07:13:33.4372018-04-03T11:13:33.4372018-04-03T15:13:33.4372018-04-03T19:13:33.4372018-04-03T23:13:33.437

Often we are not so interested in the full dates. For example if we are reading a video file and we want to get a frame every 5 seconds while using VideoIO.jl. We can deal here with the simpler Time type.

julia> video_start=Dates.Time(0,5,20)
00:05:20

Here we are interested in starting 5 minutes and 20 seconds into the video.
Now we can make a nice loop from the start to finish

for t=video_start:Dates.Second(5):video_start+Dates.Hour(2)
h=Dates.Hour(t).value
m=Dates.Minute(t).value
s=Dates.Second(t).value
ms=Dates.Millisecond(t).value# Do something interesting with ffmpeg seek on the videoend

So I decided to dig deeper. Basically the standard crand() is not that good. So instead I searched for the fastest Mersenne Twister there is. I downloaded the latest code and compiled it in the fastest way for my architecture.

And after all that trouble we got the performance down to 18 seconds. Still slower that Julia‘s 16 seconds.

$ time ./eulerfast
Euler : 2.71824
real 0m18.075s
user 0m18.085s
sys 0m0.001s

Probably, we could do a bit better with more tweaks, and probably exceed Julia‘s performance with some effort. But at that point, I got tired of pushing this further. The thing I love about Julia is how well it is engineered and hassle free. It is quite phenomenal the performance you get out of it, with so little effort. And for basic technical computing things, like random number generation, you don’t have to dig hard for a better library. The “batteries included” choices in the Julia‘s standard library are pretty good.

it is quite nice though that you can rely on the quality of Base for numerics.

On e-day, I came across this cool tweet from Fermat’s library

Happy e-day of the century! 2 7 18

Curious fact about Euler's Number: Pick a uniformly random number in [0,1] and repeat until the sum of the numbers picked is >1. You'll on average pick e≈2.718… numbers! pic.twitter.com/P0wx9hQu79

$ time ./a.out
Euler : 2.71829
real 0m36.213s
user 0m36.238s
sys 0m0.004s

For the curios, I am using this version of Julia

julia> versioninfo()
Julia Version 0.6.3-pre.0
Commit 93168a6 (2017-12-18 07:11 UTC)
Platform Info:
OS: Linux (x86_64-linux-gnu)
CPU: Intel(R) Core(TM) i7-4770HQ CPU @ 2.20GHz
WORD_SIZE: 64
BLAS: libopenblas (USE64BITINT DYNAMIC_ARCH NO_AFFINITY Haswell)
LAPACK: libopenblas64_
LIBM: libopenlibm
LLVM: libLLVM-3.9.1 (ORCJIT, haswell)

Now one should not put too much emphasis on such micro benchmarks. However, I found this a very curious examples when a high level language like Julia could be twice as fast a c. The Julia language authors must be doing some amazing mojo.

I came across a neat math puzzle involving counting the number of unique combinations in a hypothetical lock where digit order does not count. Before you continue, please watch at least the first minute of following video:

The rest of the video describes two related approaches for carrying out the counting. Often when I run into complex counting problems, I like to do a sanity check using brute force computation to make sure I have not missed anything. Julia is fantastic choice for doing such computation. It has C like speed, and with an expressiveness that rivals many other high level languages.

Without further ado, here is the Julia code I used to verify my solution the problem.

function unique_combs(n=4)

pat_lookup=Dict{String,Bool}()

for i=0:10^n-1

d=digits(i,10,n)# The digits on an integer in an array with padding

ds=d |>sort|>join# putting the digits in a string after sorting

get(pat_lookup,ds,false)||(pat_lookup[ds]=true)

end

println("The number of unique digits is $(length(pat_lookup))")

end

In line 2 we create a dictionary that we will be using to check if the number fits a previously seen pattern. The loop starting in line 3, examines all possible ordered combinations. The digits function in line 4 takes any integer and generate an array of its constituent digits. We generate the unique digit string in line 5 using pipes, by first sorting the integer array of digits and then combining them in a string. In line 6 we check if the pattern of digits was seen before and make use of quick short short-circuit evaluation to avoid an if-then statement.

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:

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…

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 winningJuMP 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] == 2end# 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

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.

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.

In the above, the big integer is the magic number that lets us generate the image of the formula. I also need to setprecision of BigFloat to be very high, as rounding errors using the default precision does not get us the desired results. The implementation was inspired by the one in Python, but I see Julia a great deal more concise and clearer.

In [2]:

function tupper_field(k)field=Array{Bool}(17,106)for(ix,x)inenumerate(0.0:1:105.0),(iy,y)inenumerate(k:k+16)field[iy,107-ix]=1/2<floor(mod(floor(y/17)*2^(-17*floor(x)-mod(floor(y),17)),2))endfieldend

Recently, I came across a fascinating blog and video from Mind your Decisions. It is about how a fraction

would show the Fibonacci numbers in order when looking at its decimal output.

On a spreadsheet and most standard programming languages, such output can not be attained due to the limited precision for floating point numbers. If you try this on R or Python, you will get an output of 1e-48. Wolfram alpha,however,allows arbitrary precision.

In Julia by default we get a little better that R and Python

We observe here that we are getting the first few Fibonacci numbers . We need more precision to get more numbers. Julia has arbitrary precision arithmetic baked into the language. We can crank up the precision of the BigFloat type on demand. Of course, the higher the precision, the slower the computation and the greater the memory we use. We do that by setprecision.

That is looking much better. However it we be nice if we could extract the Fibonacci numbers that are buried in that long decimal. Using the approach in the original blog. We define a function

y(x)=one(x)-x-x^2

and calculate the decimal

a=1/y(big"1e-24")

Here we use the non-standard string literal big"..." to insure proper interpretation of our input. Using BigFloat(1e-24)) would first construct at floating point with limited precision and then do the conversion. The initial loss of precision will not be recovered in the conversion, and hence the use of big. Now we extract our Fibonacci numbers by this function

function extract_fib(a)
x=string(a)
l=2
fi=BigInt[]push!(fi,1)for i=1:div(length(x)-24,24)
j=parse(BigInt,x[l+1+(i-1)*24:l+i*24])push!(fi,j)end
fi
end

Here we first convert our very long decimal number of a string and they we exploit the fact the Fibonacci numbers occur in blocks that 24 digits in length. We get out output in an array of BigInt. We want to compare the output with exact Fibonacci numbers, we just do a quick and non-recursive implementation.

function fib(n)
f=Vector{typeof(n)}(n+1)
f[1]=f[2]=1;for i=3:n+1
f[i]=f[i-1]+f[i-2]end
f
end

Now we compare…

fib_exact=fib(200);
fib_frac=extract_fib(a);for i in eachindex(fib_frac)println(fib_exact[i], " ", fib_exact[i]-fib_frac[i])end

We get a long sequence, we just focused here on when the discrepancy happens.

The output shows that just before the extracted Fibonacci number exceeds 24 digits, a discrepancy occurs. I am not quite sure why, but this was a fun exploration. Julia allows me to do mathematical explorations that would take one or even two orders of magnitude of effort to do in any other language.