Mainly Tech projects on Python and Electronic Design Automation.

Sunday, July 27, 2014

Does the Californian dressing down, jeans-n-T-shirt attire of most software professionals hurt us?

Earlier generations have said  that "You are what you wear" but that doesn't seem to apply in the software industry wear billionaires are seen in the media with hardly a jacket, let alone a tailored suit and tie.

Everybody thinks that they too can be just as successful and, of course, you want to fit in with peers, so most dress down in comparison with other professions.

But what is the effect on the individual?  We can't all be Steve Jobs. Most of us will end up in the median of the profession. Could we  do better if we valued ourselves higher and signalled that to our employers by a mass change in our attire?

We have already seen that large tech companies have been found guilty of colluding to hold back our wages; and that Software outsourcing is not the panacea it was wished to be. Managers know that it can be difficult to recruit good developers too.

So, in this capitalist environment we find ourselves in, maybe we should follow the legal profession and recognise that, in some ways, we are what we wear, and to our employers - currently, they are happy for us to wear jeans.

Thursday, July 10, 2014

Unique numbers in multiplication tables

 I answered a Stack overflow question that asked for the number of unique numbers in an n-by-n multiplication table and got it horribly wrong.
I decided to look into the maths of it a little further last night and now know a bit more about the question but still have no formula for generating the number of distinct integers in an n-by-n multiplication table.

Table generator

In [1]:
def ptable(n):
    width = len(str(n*n))
    nrange = range(1, n+1)
    for i in nrange:
        print(' '.join('%*i' % (width, i * j) for j in nrange))

for n in range(1, 4):
    n2 = 2**n
    print('\n%i-by-%i times table' % (n2, n2))

2-by-2 times table
1 2
2 4

4-by-4 times table
 1  2  3  4
 2  4  6  8
 3  6  9 12
 4  8 12 16

8-by-8 times table
 1  2  3  4  5  6  7  8
 2  4  6  8 10 12 14 16
 3  6  9 12 15 18 21 24
 4  8 12 16 20 24 28 32
 5 10 15 20 25 30 35 40
 6 12 18 24 30 36 42 48
 7 14 21 28 35 42 49 56
 8 16 24 32 40 48 56 64

Unique number counter

The only way i could think of counting the number of unique integers is to generate the table then get the size of the set of the tables numbers:
In [2]:
def mult_table_numbs(n):
    nrange = range(1, n+1)
    return len(set(n0 * n1 for n0 in nrange for n1 in nrange))

for n in range(1,11):
    print('%2i: %i' % (n, mult_table_numbs(n)))
 1: 1
 2: 3
 3: 6
 4: 9
 5: 14
 6: 18
 7: 25
 8: 30
 9: 36
10: 42

So for n of 4 there are 9 distinct numbers counted.
Stick the numbers 1,3,6,9,14,18,25,30,36,42 into OEIS and the integer sequence search engine indeed recognises it as A027424
I simplified. I actually first used Pythons itertools.product and reduce. for product I had to use product(range(1,n+1), repeat=2). The repeat=2 is because it is for a two-dimensional multiplication table.
That got me thinking and I quickly realised that I could create a k-dimensional multiplication table unique number counter by setting the repeat to a new argument k with default value 2.

The k-dimensional multiplication table unique number counter

In [3]:
from itertools import product
from functools import reduce

mul = int.__mul__

def mult_table_numbers(n, k=2):
    return len(set(reduce(mul, p, 1) 
                   for p in product(range(1,n+1), repeat=k)))

Dimension hopping

OEIS mentioned no formulae for generating the unique number counts except by direct counting so I thought "what if I explore the dimensions k as well as varying n"?
print('n,k=2.. matrix')

for n in range(2,13):

    print('%2i: %s' % (n, repr([mult_table_numbers(n, k)

                                for k in range(2,10)]).replace(' ','')))
The above takes several tens of minutes to run but finally gives the following results:
n,k=2.. matrix

 2: [3,4,5,6,7,8,9,10]

 3: [6,10,15,21,28,36,45,55]

 4: [9,16,25,36,49,64,81,100]

 5: [14,30,55,91,140,204,285,385]

 6: [18,40,75,126,196,288,405,550]

 7: [25,65,140,266,462,750,1155,1705]

 8: [30,80,175,336,588,960,1485,2200]

 9: [36,100,225,441,784,1296,2025,3025]

10: [42,120,275,546,980,1632,2565,3850]
The 2: [3,4,5,6,7,8,9,10] line reads that for n=2: a 2-dimensional multiplication table has 3 distinct numbers; a 3D table 4; a 5D table 6 distinct numbers, and so on...
Now each line with its series of distinct numbers for different k- dimensions is itself a sequence. The n=2 line is the sequence k+1 for example. I could look up the sequences in OEIS and extract the formulas for each row. If I could then link the formulas for different rows I might be able to finally generate a formula for the unique count, lets call it m(n, k) for dimension k=2.

OEIS extraction

Time on OEIS revealed the following information:
m(n,k)PolynomialOEIS Description
m(2,k)(k+1)Positive integers A000027
m(3,k)(k+1)*(k+2)/2Triangular numbers A000217
m(4,k)(k+1)*(k+1)Squares A000290
m(5,k)(k+1)(k+2)(2*(k+1)+1)/6Square pyramidal numbers A000330
m(6,k)(k+1)*2(k+2)/2Pentagonal pyramidal numbers A002411
m(7,k)(k+1)(2+k)(3+k)(1+3(k+1))/244-dimensional pyramidal numbers A001296
m(8,k)(k+1)^2(k+2)(k+3)/64-dimensional figurate numbers: A002417
m(9,k)((k+1)*(k+2)/2)^2Sum of first n cubes; or n-th triangular number squared A000537
Whilst playing around with the polynomials using SymPy Gamma and the Wolfram sites I rearranged them to help find patterns

Constant multipliers for the polynomials above (when expanded)

I could find no patterns in this:

nrearranged polynomial for m(n,k)
1(1/(1)) * (1+0k)
2(1/(1)) * (1+k)
3(1/(1*2)) * (1+k)(2+k)
4(1/(1*1)) * (1+k)(1+k)
5(1/(1*2*3)) * (1+k)(2+k)(3+2k)
6(1/(1*2*1)) * (1+k)(2+k)(1+k)
7(1/(1*2*3*4)) * (1+k)(2+k)(3+k)(4+3k)
8(1/(1*2*1*3)) * (1+k)(2+k)(1+k)(3+k)
9(1/(1*2*1*2)) * (1+k)(2+k)(1+k)(2+k)
10(1/(1*2*1*3)) * (1+k)(2+k)(1+k)(3+2k)
There are some patterns in the above. the divisor is always all the constant terms (x+yk) of the polynomial multiplied together.
I then saw that some later polynomials were multiples of earlier ones so decided on using the shorthand @n for m(n,k) to produce the following:

shortformfactored polynomialAlternate factorisation
@3@2*(2+k) / 2
@5@3*(3+2k) / 3
@7@3*(3+k)(4+3k) / 6
@8(@6 = @2*@3)*(3+k) / 3@4*(2+k)(3+k) / 6
@9(@6 = @2*@3)*(2+k) / 2@3*@3
@10(@6 = @2*@3)*(3+2k) / 3@2*@5

Unfortunately I can see no pervasive patterns here either.


I've looked. I've enjoyed the journey, but I've still to find what I'm looking for!


Subscribe Now: google

Add to Google Reader or Homepage

Go deh too!

Blog Archive