02/9/18

When Julia is faster than C, digging deeper

In my earlier post I showed an example where Julia is significantly faster than c. I got this insightful response

So I decided to dig deeper. Basically the standard c rand() 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.

/* eurler2.c */
#include <stdio.h>      /* printf, NULL */
#include <stdlib.h>     /* srand, rand */
#include "SFMT.h"       /* fast Mersenne Twister */
 
sfmt_t sfmt;
 
double r2()
{
    return sfmt_genrand_res53(&sfmt);
 
}
 
double euler(long int n)
{
    long int m=0;
    long int i;
    for(i=0; i<n; i++){
        double the_sum=0;
        while(1) {
            m++;
            the_sum+=r2();
            if(the_sum>1.0) break;
        }
    }
    return (double)m/(double)n;
}
 
 
int main ()
{
  sfmt_init_gen_rand(&sfmt,123456);
  printf ("Euler : %2.5f\n", euler(1000000000));
 
  return 0;
}

I had to compile with a whole bunch of flags which I induced from SFMT‘s Makefile to get faster performance.

gcc -O3 -finline-functions -fomit-frame-pointer -DNDEBUG -fno-strict-aliasing --param max-inline-insns-single=1800  -Wall -std=c99 -msse2 -DHAVE_SSE2 -DSFMT_MEXP=1279 -ISFMT-src-1.5.1 -o eulerfast SFMT.c euler2.c

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.

02/8/18

When Julia is faster than C

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

So I spend a few minutes coding this into Julia

function euler(n)
    m=0
    for i=1:n
        the_sum=0.0
        while true
            m+=1
            the_sum+=rand()
            (the_sum>1.0) && break;
        end
    end
    m/n
end

Timing this on my machine, I got

julia> @time euler(1000000000)
 15.959913 seconds (5 allocations: 176 bytes)
2.718219862

Gave a little under 16 seconds.

Tried a c implementation

#include <stdio.h>      /* printf, NULL */
#include <stdlib.h>     /* srand, rand */
#include <time.h>       /* time */
 
double r2()
{
    return (double)rand() / (double)((unsigned)RAND_MAX + 1);
}
 
double euler(long int n)
{
    long int m=0;
    long int i;
    for(i=0; i<n; i++){
        double the_sum=0;
        while(1) {
            m++;
            the_sum+=r2();
            if(the_sum>1.0) break;
        }
    }
    return (double)m/(double)n;
}
 
 
int main ()
{
  printf ("Euler : %2.5f\n", euler(1000000000));
 
  return 0;
}

and compiling with either gcc

gcc  -Ofast euler.c

or clang

clang  -Ofast euler.c

gave a timing twice as long

$ 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.

01/27/18

On the 7th anniversary of…a noble dream

Give me a reason to hang on
Give the strength to continue
In this land full of pain

How can I continue
to walk among the lifeless…the cowards and the ill-bred
while knowing that the noble ones are dying…or dead

In this reign of fear, you bend…or disappear
Ozymandias now has pathetic pretender of silly sneer and cold command
His colossal wrecks stretch everywhere, further spreading despair

The servile and hungry rise in thunderous applause to Works of their new god
The smell of decay stifling, but the fouls reckon it ambrosia
Cowardice begetting dystopia

A reckoning is overdue….

01/11/18

Factorizations to get a very special number

Mathematical recreation is fun. I came across a nice Mind your Decisions video and thought it would be cool to use Julia to find that special number. Have a look at the video here before we continue

  1. using Primes
  2.  
  3. function getBigestFactors{T}(a::T,b::T)
  4.     c=T()
  5.     for k in keys(a)
  6.         c[k]=max(a[k],b[k]) 
  7.     end
  8.     for k in keys(b)
  9.         c[k]=max(a[k],b[k]) 
  10.     end
  11.     c
  12. end
  13.  
  14. superfactors=reduce(getBigestFactors,map(factor,BigInt.(1:100))) 
  15.  
  16. n=prodfactors(superfactors)

Julia can be amazingly expressive. In line 14 are getting all the prime factors of all the numbers from 1 to 100 using the map function. We are using the type BigInt since will be generating pretty large numbers. The standard Int will simply overflow. We then use reduce to get all the common factors for the numbers from 1 to 100.

If your are slight confused lets take a bit slower. First we generate a one dimensional array of BigInts from 1 to 100.

BigInt.(1:100)

This makes use of Julia‘s very powerful dot factorization syntax. We then map that array into another array of Primes.Factorization{BigInt} through the factor function in the Primes package. The Primes.Factorization type is subtype of Associative. I learned that from reading the implementation.

The cool thing about Associative types that the can be very conveniently accessed. Lets have a closer look.

a=factor(24)
println(a)

yields

Primes.Factorization(2=>3,3=>1)

That is 2^3 3

You can then do the following

print(a[2])

yielding

3

This means that the number 24 has a 2 raised to the third power as a factor.

Or, be more adventurous and try

print(a[5])

which will print a

0

This makes perfect sense as 5^0=1 and 1 is always a factor of any number. We will always get 0 exponent for any number that is not a prime factor of a given number. The Primes.jl package authors have done great use of Julia‘s very powerful type system.

To solve this problem we need to look at all the prime factors for all the numbers from 1 to 100 find the highest exponent of the of each of those prime factors. To do that we implement the getBigestFactor function when does that for any two prime factorizations. We plug that into reduce and et voilà!

The superfactors in line 14 are

2^63^45^27^2111317192329313741434753596167717379838997

Finally we multiply them to get our special number at line 16

n=69720375229712477164533808935312303556800

Are we done yet?

As the saying goes “there is more than one way to skin a cat”. Some are more elegant than others.
Here is a more elegant and computationally more efficient way.

  1. Using Primes
  2. function getAllFactors{T}(n::T)
  3.     p_factors=filter(isprime,T.(1:n))
  4.     exponents=[floor(T,log(p,n)) for p in p_factors]
  5.     zip(p_factors,exponents) |> Dict |> Primes.Factorization
  6. end

Here we just get the prime numbers in the range from 1 to n (line 3). Using those primes we then get the highest exponents that will not yield a number outside the range (line 4). Finally, we package everything as a Primes.Factorization (line 5).

In Julia REPL we get

julia> getAllFactors(big"100")
2^63^45^27^2111317192329313741434753596167717379838997
 
julia> prodfactors(ans)
69720375229712477164533808935312303556800

At that point, you might ask, why stop at a 100. Julia is lightening fast and powerful. Just for fun we can do the same for numbers from 1 to 1000.

julia> getAllFactors(big"1000")
2^93^65^47^311^213^217^219^223^229^231^23741434753596167717379838997101103107109113127131137139149151157163167173179181191193197199211223227229233239241251257263269271277281283293307311313317331337347349353359367373379383389397401409419421431433439443449457461463467479487491499503509521523541547557563569571577587593599601607613617619631641643647653659661673677683691701709719727733739743751757761769773787797809811821823827829839853857859863877881883887907911919929937941947953967971977983991997
 
julia> prodfactors(ans)
7128865274665093053166384155714272920668358861885893040452001991154324087581111499476444151913871586911717817019575256512980264067621009251465871004305131072686268143200196609974862745937188343705015434452523739745298963145674982128236956232823794011068809262317708861979540791247754558049326475737829923352751796735248042463638051137034331214781746850878453485678021888075373249921995672056932029099390891687487672697950931603520000

Now we get super special number that is 443 digits long. Lets finish by a Julia one liner to get the number of digits

julia> getAllFactors(big"1000") |> prodfactors|> digits |>length
433
01/2/18

Visualizing the Inscribed Circle and Square Puzzle

Recently, I watched a cool mind your decsions video on an inscribed circle and rectangle puzzle. In the video they showed a diagram that was not scale. I wanted to get a sense of how these differently shaped areas will match.

There was a cool ratio between the outer and inner circle radii that is expressed as

\frac{R}{r}=\sqrt{\frac{\pi-2}{4-\pi}} .

I used Compose.jl to rapidly do that.

using Compose
set_default_graphic_size(20cm, 20cm)
ϕ=sqrt((pi -2)/(4-pi))
R=10
r=R/ϕ
ctx=context(units=UnitBox(-10, -10, 20, 20))
composition = compose(ctx,
    (ctx, rectangle(-r/√2,-r/√2,r*2,r*2),fill("white")),
    (ctx,circle(0,0,r),fill("blue")),
    (ctx,circle(0,0,R),fill("white")),
    (ctx,rectangle(-10,-10,20,20),fill("red")))
composition |> SVG("inscribed.svg")

08/6/17

Solving the code lock riddle with Julia

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.

  1. function unique_combs(n=4)
  2.     pat_lookup=Dict{String,Bool}()
  3.     for i=0:10^n-1
  4.         d=digits(i,10,n) # The digits on an integer in an array with padding
  5.         ds=d |> sort |> join # putting the digits in a string after sorting
  6.         get(pat_lookup,ds,false) || (pat_lookup[ds]=true)
  7.     end
  8.     println("The number of unique digits is $(length(pat_lookup))")
  9. 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.

07/14/17

Julia calling C: A more minimal example

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…

06/29/17

Solving the Fish Riddle with JuMP

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.

06/12/17

Reading DataFrames with non-UTF8 encoding in Julia

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.

06/12/17

Tupper’s self-referential formula in Julia

I was surprised when I came across on Tupper’s formula on twitter. I felt the compulsion to implement it in Julia.

The formula is expressed as

{1\over 2} < \left\lfloor \mathrm{mod}\left(\left\lfloor {y \over 17} \right\rfloor 2^{-17 \lfloor x \rfloor - \mathrm{mod}(\lfloor y\rfloor, 17)},2\right)\right\rfloor

and yields bitmap facsimile of itself.

In [1]:
k=big"960 939 379 918 958 884 971 672 962 127 852 754 715 004 339 660 129 306 651 505 519 271 702 802 395 266 424 689 642 842 174 350 718 121 267 153 782 770 623 355 993 237 280 874 144 307 891 325 963 941 337 723 487 857 735 749 823 926 629 715 517 173 716 995 165 232 890 538 221 612 403 238 855 866 184 013 235 585 136 048 828 693 337 902 491 454 229 288 667 081 096 184 496 091 705 183 454 067 827 731 551 705 405 381 627 380 967 602 565 625 016 981 482 083 418 783 163 849 115 590 225 610 003 652 351 370 343 874 461 848 378 737 238 198 224 849 863 465 033 159 410 054 974 700 593 138 339 226 497 249 461 751 545 728 366 702 369 745 461 014 655 997 933 798 537 483 143 786 841 806 593 422 227 898 388 722 980 000 748 404 719"
setprecision(BigFloat,10000);

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) in enumerate(0.0:1:105.0), (iy,y) in enumerate(k:k+16)
        field[iy,107-ix]=1/2<floor(mod(floor(y/17)*2^(-17*floor(x)-mod(floor(y),17)),2))
    end
   field 
end
In [3]:
f=tupper_field(k);
using Images
img = colorview(Gray,.!f)
Out[3]:

I just inverted the boolean array here to get the desired bitmap output.