# Solutions to Machine Learning Programming Assignments

This post contains links to a bunch of code that I have written to complete Andrew Ng’s famous machine learning course which includes several interesting machine learning problems that needed to be solved using the Octave / Matlab programming language. I’m not sure I’d ever be programming in Octave after this course, but learning Octave just so that I could complete this course seemed worth the time and effort. I would usually work on the programming assignments on Sundays and spend several hours coding in Octave, telling myself that I would later replicate the exercises in Python.

If you’ve taken this course and found some of the assignments hard to complete, I think it might not hurt to go check online on how a particular function was implemented. If you end up copying the entire code, it’s probably your loss in the long run. But then John Maynard Keynes once said, ‘In the long run we are all dead‘. Yeah, and we wonder why people call Economics the dismal science!

Most people disregard Coursera’s feeble attempt at reigning in plagiarism by creating an Honor Code, precisely because this so-called code-of-conduct can be easily circumvented. I don’t mind posting solutions to a course’s programming assignments because GitHub is full to the brim with such content. Plus, it’s always good to read others’ code even if you implemented a function correctly. It helps understand the different ways of tackling a given programming problem.

Enjoy!

# Spot the Difference — It’s NumPy!

My first brush with NumPy happened over writing a block of code to make a plot using pylab. ⇣

pylab is part of matplotlib (in matplotlib.pylab) and tries to give you a MatLab like environment. matplotlib has a number of dependencies, among them numpy which it imports under the common alias np. scipy is not a dependency of matplotlib.

I had a tuple (of lows and highs of temperature) of lengh 2 with 31 entries in each (the number of days in the month of July), parsed from this text file:

 Boston July Temperatures ------------------------- Day High Low ------------ 1 91 70 2 84 69 3 86 68 4 84 68 5 83 70 6 80 68 7 86 73 8 89 71 9 84 67 10 83 65 11 80 66 12 86 63 13 90 69 14 91 72 15 91 72 16 88 72 17 97 76 18 89 70 19 74 66 20 71 64 21 74 61 22 84 61 23 86 66 24 91 68 25 83 65 26 84 66 27 79 64 28 72 63 29 73 64 30 81 63 31 73 63
view raw julyTemps.txt hosted with ❤ by GitHub

Given below, are 2 sets of code that do the same thing; one without NumPy and the other with NumPy. They output the following graph using PyLab:

Code without NumPy

 import pylab def loadfile(): inFile = open('julyTemps.txt', 'r') high =[]; low = [] for line in inFile: fields = line.split() if len(fields) < 3 or not fields[0].isdigit(): pass else: high.append(int(fields[1])) low.append(int(fields[2])) return low, high def producePlot(lowTemps, highTemps): diffTemps = [highTemps[i] - lowTemps[i] for i in range(len(lowTemps))] pylab.title('Day by Day Ranges in Temperature in Boston in July 2012') pylab.xlabel('Days') pylab.ylabel('Temperature Ranges') return pylab.plot(range(1,32),diffTemps) producePlot(loadfile()[1], loadfile()[0])
view raw withoutNumPy.py hosted with ❤ by GitHub

Code with NumPy
 import pylab import numpy as np def loadFile(): inFile = open('julyTemps.txt') high = [];vlow = [] for line in inFile: fields = line.split() if len(fields) != 3 or 'Boston' == fields[0] or 'Day' == fields[0]: continue else: high.append(int(fields[1])) low.append(int(fields[2])) return (low, high) def producePlot(lowTemps, highTemps): diffTemps = list(np.array(highTemps) - np.array(lowTemps)) pylab.plot(range(1,32), diffTemps) pylab.title('Day by Day Ranges in Temperature in Boston in July 2012') pylab.xlabel('Days') pylab.ylabel('Temperature Ranges') pylab.show() (low, high) = loadFile() producePlot(low, high)
view raw withNumPy.py hosted with ❤ by GitHub

The difference in code lies in how the variable diffTemps is calculated.

diffTemps = list(np.array(highTemps) - np.array(lowTemps))


diffTemps = [highTemps[i] - lowTemps[i] for i in range(len(lowTemps))]


Notice how straight forward it is with NumPy. At the core of the NumPy package, is the ndarray object. This encapsulates n-dimensional arrays of homogeneous data types, with many operations being performed in compiled code for performance. element-by-element operations are the “default mode” when an ndarray is involved, but the element-by-element operation is speedily executed by pre-compiled C code.

# Karatsuba Multiplication Algorithm – Python Code

Motivation for this blog post

I’ve enrolled in Stanford Professor Tim Roughgarden’s Coursera MOOC on the design and analysis of algorithms, and while he covers the theory and intuition behind the algorithms in a surprising amount of detail, we’re left to implement them in a programming language of our choice.

And I’m ging to post Python code for all the algorithms covered during the course!

The Karatsuba Multiplication Algorithm

Karatsuba’s algorithm reduces the multiplication of two n-digit numbers to at most $n^{\log_23}\approx n^{1.585}$ single-digit multiplications in general (and exactly $n^{\log_23}$ when n is a power of 2). Although the familiar grade school algorithm for multiplying numbers is how we work through multiplication in our day-to-day lives, it’s slower ($\Theta(n^2)\,\!$) in comparison, but only on a computer, of course!

Here’s how the grade school algorithm looks:
(The following slides have been taken from Tim Roughgarden’s notes. They serve as a good illustration. I hope he doesn’t mind my sharing them.)

…and this is how Karatsuba Multiplication works on the same problem:

A More General Treatment

Let $x$ and $y$ be represented as $n$-digit strings in some base $B$. For any positive integer $m$ less than $n$, one can write the two given numbers as

$x = x_1B^m + x_0$
$y = y_1B^m + y_0$,

where $x_0$ and $y_0$ are less than $B^m$. The product is then

$xy = (x_1B^m + x_0)(y_1B^m + y_0)$
$xy = z_2B^{2m} + z_1B^m + z_0$

where

$z_2 = x_1y_1$
$z_1 = x_1y_0 + x_0y_1$
$z_0 = x_0y_0$

These formulae require four multiplications, and were known to Charles Babbage. Karatsuba observed that $xy$ can be computed in only three multiplications, at the cost of a few extra additions. With $z_0$ and $z_2$ as before we can calculate

$z_1 = (x_1 + x_0)(y_1 + y_0) - z_2 - z_0$

which holds since

$z_1 = x_1y_0 + x_0y_1$
$z_1 = (x_1 + x_0)(y_1 + y_0) - x_1y_1 - x_0y_0$

A more efficient implementation of Karatsuba multiplication can be set as $xy = (b^2 + b)x_1y_1 - b(x_1 - x_0)(y_1 - y_0) + (b + 1)x_0y_0$, where $b = B^m$.

### Example

To compute the product of 12345 and 6789, choose B = 10 and m = 3. Then we decompose the input operands using the resulting base (Bm = 1000), as:

12345 = 12 · 1000 + 345
6789 = 6 · 1000 + 789

Only three multiplications, which operate on smaller integers, are used to compute three partial results:

z2 = 12 × 6 = 72
z0 = 345 × 789 = 272205
z1 = (12 + 345) × (6 + 789) − z2z0 = 357 × 795 − 72 − 272205 = 283815 − 72 − 272205 = 11538

We get the result by just adding these three partial results, shifted accordingly (and then taking carries into account by decomposing these three inputs in base 1000 like for the input operands):

result = z2 · B2m + z1 · Bm + z0, i.e.
result = 72 · 10002 + 11538 · 1000 + 272205 = 83810205.

Pseudocode and Python code

 procedure karatsuba(num1, num2) if (num1 < 10) or (num2 < 10) return num1*num2 /* calculates the size of the numbers */ m = max(size_base10(num1), size_base10(num2)) m2 = m/2 /* split the digit sequences about the middle */ high1, low1 = split_at(num1, m2) high2, low2 = split_at(num2, m2) /* 3 calls made to numbers approximately half the size */ z0 = karatsuba(low1,low2) z1 = karatsuba((low1+high1),(low2+high2)) z2 = karatsuba(high1,high2) return (z2*10^(2*m2))+((z1-z2-z0)*10^(m2))+(z0)

 def karatsuba(x,y): """Function to multiply 2 numbers in a more efficient manner than the grade school algorithm""" if len(str(x)) == 1 or len(str(y)) == 1: return x*y else: n = max(len(str(x)),len(str(y))) nby2 = n / 2 a = x / 10**(nby2) b = x % 10**(nby2) c = y / 10**(nby2) d = y % 10**(nby2) ac = karatsuba(a,c) bd = karatsuba(b,d) ad_plus_bc = karatsuba(a+b,c+d) - ac - bd # this little trick, writing n as 2*nby2 takes care of both even and odd n prod = ac * 10**(2*nby2) + (ad_plus_bc * 10**nby2) + bd return prod
view raw karatsuba.py hosted with ❤ by GitHub

# Why Parselmouth Harry Potter is also Parsermouth Harry Potter

If you’re a Pythonista or just a coder, you may have come across this web cartoon:

Its creator Ryan Sawyer has been working as a full-time graphic designer and freelance illustrator for the past 10 years. His projects have been featured on websites such as /Film, io9, BoingBoing, Uproxx, MusicRadar, SuperPunch, IGN, and PackagingDigest.

I recently came across an interesting thread on Reddit on the origins of this cartoon. Basically, the cartoonist, ergo Python-speaking-Harry, got their code from this Stack Overflow forum for short, useful Python code snippets! Convenient, right?!

What’s funny is that the forum later got closed as it was deemed not constructive!

The code is supposed to print a recursive count of lines of python source code from the current working directory, including an ignore list – so as to print total sloc. Don’t blame me though, if the code doesn’t work!

 # prints recursive count of lines of python source code from current directory # includes an ignore_list. also prints total sloc import os cur_path = os.getcwd() ignore_set = set(["__init__.py", "count_sourcelines.py"]) loclist = [] for pydir, _, pyfiles in os.walk(cur_path): for pyfile in pyfiles: if pyfile.endswith(".py") and pyfile not in ignore_set: totalpath = os.path.join(pydir, pyfile) loclist.append( ( len(open(totalpath, "r").read().splitlines()), totalpath.split(cur_path)[1]) ) for linenumbercount, filename in loclist: print "%05d lines in %s" % (linenumbercount, filename) print "\nTotal: %s lines (%s)" %(sum([x[0] for x in loclist]), cur_path)
view raw sloc.py hosted with ❤ by GitHub

# Magic 5-gon Ring — Project Euler (Problem 68)

Yet another exciting math problem that requires an algorithmic approach to arrive at a quick solution! There is a pen-paper approach to it too, but this post assumes we’re more interested in discussing the programming angle.

First, the problem:

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.

Total Solution Set:
9 4,2,3; 5,3,1; 6,1,2
9 4,3,2; 6,2,1; 5,1,3
10 2,3,5; 4,5,1; 6,1,3
10 2,5,3; 6,3,1; 4,1,5
11 1,4,6; 3,6,2; 5,2,4
11 1,6,4; 5,4,2; 3,2,6
12 1,5,6; 2,6,4; 3,4,5
12 1,6,5; 3,5,4; 2,4,6
By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.

Problem

Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a “magic5-gon ring?

Algorithm

In attempting this problem, I choose to label the 5 inner nodes as i, j, k, l, and m.
α, β, γ, δ, and θ being the corresponding outer nodes.

Let x be the sum total of each triplet line, i.e.,

x = α + i + j = β + j + k = γ + k + l = δ + l + m = θ + m + i

First Observation:
For the string to be 16-digits, 10 has to be in the outer ring, as each number in the inner ring is included in the string twice. Next, we fill the inner ring in an iterative manner.

Second Observation:
There 9 numbers to choose from for the inner ring — 1, 2, 3, 4, 5, 6, 7, 8 and 9.
5 have to be chosen. This can be done in 9C5 = 126 ways.
According to circular permutation, if there are n distinct numbers to be arranged in a circle, this can be done in (n-1)! ways, where (n-1)! = (n-1).(n-2).(n-3)…3.2.1. So 5 distinct numbers can be arranged in 4! permutations, i.e., in 24 ways around a circle, or pentagonal ring, to be more precise.
So in all, this problem can be solved in 126×24 = 3024 iterations.

Third Observation:
For every possible permutation of an inner-ring arrangement, there can be one or more values of x (triplet line-sum) that serve as a possible contenders for a “magic” string whose triplets add up to the same number, x. To ensure this, we only need that the values of α through θ of the outer ring are distinct, different from the inner ring, with the greatest of these equal to 10.
Depending on the relative positioning of the numbers in the inner ring, one can narrow the range of x-values one might have to check for each permutation. To zero-down on such a range, let’s look at an example. Shown in the figure below is a randomly chosen permutation of number in the inner ring – 7, 2, 3, 4 and 5, in that order.

So 10, 9, 8, 6 and 1 must fill the outer circle. It’s easy to notice that the 5, 7 pair is the greatest adjacent pair. So whatever x is, it has to be at least 5 + 7 + 1 = 13 (1 being the smallest number of the outer ring). Likewise,  2, 3 is the smallest adjacent pair, so whatever x is, it can’t be any more than 2 + 3+ 10 = 15 (10 being the largest number of the outer ring). This leaves us with a narrow range of x-values to check – 13, 14 and 15.

Next, we arrange the 5 triplets in clock-wise direction starting with the triplet with the smallest number in the outer ring to form a candidate string. This exercise when done for each of the 3024 permutations will shortlist a range of candidates, of which, the maximum is chosen.

That’s all there is to the problem!

Here’s the Python Code. It executes in about a tenth of a second!

view raw euler68.py hosted with ❤ by GitHub

Ans: 6531031914842725

# Maximum Path Sum — Dynamic Programming Algorithm

I came across this problem recently that required solving for the maximum-sum path in a triangle array.

To copy the above triangle array:

 75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
view raw euler18.txt hosted with ❤ by GitHub

As can be seen, there are 15 levels to this tree (including the top most node). Therefore, there are 214 possible routes to scan in order to check for the maximum sum using brute force. As there are only 214 (16384) routes, it is possible to solve this problem by trying every route. However, doing the same using brute force on a triangle array of 100 levels would take several billion years to solve using a computer that checks through say, 1012 routes per second. A greedy algorithm might per-chance work for the particular 4-level example problem stated above, but will not always work, and in most cases won’t. For instance, for the 100-level problem:

 59 73 41 52 40 09 26 53 06 34 10 51 87 86 81 61 95 66 57 25 68 90 81 80 38 92 67 73 30 28 51 76 81 18 75 44 84 14 95 87 62 81 17 78 58 21 46 71 58 02 79 62 39 31 09 56 34 35 53 78 31 81 18 90 93 15 78 53 04 21 84 93 32 13 97 11 37 51 45 03 81 79 05 18 78 86 13 30 63 99 95 39 87 96 28 03 38 42 17 82 87 58 07 22 57 06 17 51 17 07 93 09 07 75 97 95 78 87 08 53 67 66 59 60 88 99 94 65 55 77 55 34 27 53 78 28 76 40 41 04 87 16 09 42 75 69 23 97 30 60 10 79 87 12 10 44 26 21 36 32 84 98 60 13 12 36 16 63 31 91 35 70 39 06 05 55 27 38 48 28 22 34 35 62 62 15 14 94 89 86 66 56 68 84 96 21 34 34 34 81 62 40 65 54 62 05 98 03 02 60 38 89 46 37 99 54 34 53 36 14 70 26 02 90 45 13 31 61 83 73 47 36 10 63 96 60 49 41 05 37 42 14 58 84 93 96 17 09 43 05 43 06 59 66 57 87 57 61 28 37 51 84 73 79 15 39 95 88 87 43 39 11 86 77 74 18 54 42 05 79 30 49 99 73 46 37 50 02 45 09 54 52 27 95 27 65 19 45 26 45 71 39 17 78 76 29 52 90 18 99 78 19 35 62 71 19 23 65 93 85 49 33 75 09 02 33 24 47 61 60 55 32 88 57 55 91 54 46 57 07 77 98 52 80 99 24 25 46 78 79 05 92 09 13 55 10 67 26 78 76 82 63 49 51 31 24 68 05 57 07 54 69 21 67 43 17 63 12 24 59 06 08 98 74 66 26 61 60 13 03 09 09 24 30 71 08 88 70 72 70 29 90 11 82 41 34 66 82 67 04 36 60 92 77 91 85 62 49 59 61 30 90 29 94 26 41 89 04 53 22 83 41 09 74 90 48 28 26 37 28 52 77 26 51 32 18 98 79 36 62 13 17 08 19 54 89 29 73 68 42 14 08 16 70 37 37 60 69 70 72 71 09 59 13 60 38 13 57 36 09 30 43 89 30 39 15 02 44 73 05 73 26 63 56 86 12 55 55 85 50 62 99 84 77 28 85 03 21 27 22 19 26 82 69 54 04 13 07 85 14 01 15 70 59 89 95 10 19 04 09 31 92 91 38 92 86 98 75 21 05 64 42 62 84 36 20 73 42 21 23 22 51 51 79 25 45 85 53 03 43 22 75 63 02 49 14 12 89 14 60 78 92 16 44 82 38 30 72 11 46 52 90 27 08 65 78 03 85 41 57 79 39 52 33 48 78 27 56 56 39 13 19 43 86 72 58 95 39 07 04 34 21 98 39 15 39 84 89 69 84 46 37 57 59 35 59 50 26 15 93 42 89 36 27 78 91 24 11 17 41 05 94 07 69 51 96 03 96 47 90 90 45 91 20 50 56 10 32 36 49 04 53 85 92 25 65 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19 11 98 34 45 09 97 86 71 03 15 56 19 15 44 97 31 90 04 87 87 76 08 12 30 24 62 84 28 12 85 82 53 99 52 13 94 06 65 97 86 09 50 94 68 69 74 30 67 87 94 63 07 78 27 80 36 69 41 06 92 32 78 37 82 30 05 18 87 99 72 19 99 44 20 55 77 69 91 27 31 28 81 80 27 02 07 97 23 95 98 12 25 75 29 47 71 07 47 78 39 41 59 27 76 13 15 66 61 68 35 69 86 16 53 67 63 99 85 41 56 08 28 33 40 94 76 90 85 31 70 24 65 84 65 99 82 19 25 54 37 21 46 33 02 52 99 51 33 26 04 87 02 08 18 96 54 42 61 45 91 06 64 79 80 82 32 16 83 63 42 49 19 78 65 97 40 42 14 61 49 34 04 18 25 98 59 30 82 72 26 88 54 36 21 75 03 88 99 53 46 51 55 78 22 94 34 40 68 87 84 25 30 76 25 08 92 84 42 61 40 38 09 99 40 23 29 39 46 55 10 90 35 84 56 70 63 23 91 39 52 92 03 71 89 07 09 37 68 66 58 20 44 92 51 56 13 71 79 99 26 37 02 06 16 67 36 52 58 16 79 73 56 60 59 27 44 77 94 82 20 50 98 33 09 87 94 37 40 83 64 83 58 85 17 76 53 02 83 52 22 27 39 20 48 92 45 21 09 42 24 23 12 37 52 28 50 78 79 20 86 62 73 20 59 54 96 80 15 91 90 99 70 10 09 58 90 93 50 81 99 54 38 36 10 30 11 35 84 16 45 82 18 11 97 36 43 96 79 97 65 40 48 23 19 17 31 64 52 65 65 37 32 65 76 99 79 34 65 79 27 55 33 03 01 33 27 61 28 66 08 04 70 49 46 48 83 01 45 19 96 13 81 14 21 31 79 93 85 50 05 92 92 48 84 59 98 31 53 23 27 15 22 79 95 24 76 05 79 16 93 97 89 38 89 42 83 02 88 94 95 82 21 01 97 48 39 31 78 09 65 50 56 97 61 01 07 65 27 21 23 14 15 80 97 44 78 49 35 33 45 81 74 34 05 31 57 09 38 94 07 69 54 69 32 65 68 46 68 78 90 24 28 49 51 45 86 35 41 63 89 76 87 31 86 09 46 14 87 82 22 29 47 16 13 10 70 72 82 95 48 64 58 43 13 75 42 69 21 12 67 13 64 85 58 23 98 09 37 76 05 22 31 12 66 50 29 99 86 72 45 25 10 28 19 06 90 43 29 31 67 79 46 25 74 14 97 35 76 37 65 46 23 82 06 22 30 76 93 66 94 17 96 13 20 72 63 40 78 08 52 09 90 41 70 28 36 14 46 44 85 96 24 52 58 15 87 37 05 98 99 39 13 61 76 38 44 99 83 74 90 22 53 80 56 98 30 51 63 39 44 30 91 91 04 22 27 73 17 35 53 18 35 45 54 56 27 78 48 13 69 36 44 38 71 25 30 56 15 22 73 43 32 69 59 25 93 83 45 11 34 94 44 39 92 12 36 56 88 13 96 16 12 55 54 11 47 19 78 17 17 68 81 77 51 42 55 99 85 66 27 81 79 93 42 65 61 69 74 14 01 18 56 12 01 58 37 91 22 42 66 83 25 19 04 96 41 25 45 18 69 96 88 36 93 10 12 98 32 44 83 83 04 72 91 04 27 73 07 34 37 71 60 59 31 01 54 54 44 96 93 83 36 04 45 30 18 22 20 42 96 65 79 17 41 55 69 94 81 29 80 91 31 85 25 47 26 43 49 02 99 34 67 99 76 16 14 15 93 08 32 99 44 61 77 67 50 43 55 87 55 53 72 17 46 62 25 50 99 73 05 93 48 17 31 70 80 59 09 44 59 45 13 74 66 58 94 87 73 16 14 85 38 74 99 64 23 79 28 71 42 20 37 82 31 23 51 96 39 65 46 71 56 13 29 68 53 86 45 33 51 49 12 91 21 21 76 85 02 17 98 15 46 12 60 21 88 30 92 83 44 59 42 50 27 88 46 86 94 73 45 54 23 24 14 10 94 21 20 34 23 51 04 83 99 75 90 63 60 16 22 33 83 70 11 32 10 50 29 30 83 46 11 05 31 17 86 42 49 01 44 63 28 60 07 78 95 40 44 61 89 59 04 49 51 27 69 71 46 76 44 04 09 34 56 39 15 06 94 91 75 90 65 27 56 23 74 06 23 33 36 69 14 39 05 34 35 57 33 22 76 46 56 10 61 65 98 09 16 69 04 62 65 18 99 76 49 18 72 66 73 83 82 40 76 31 89 91 27 88 17 35 41 35 32 51 32 67 52 68 74 85 80 57 07 11 62 66 47 22 67 65 37 19 97 26 17 16 24 24 17 50 37 64 82 24 36 32 11 68 34 69 31 32 89 79 93 96 68 49 90 14 23 04 04 67 99 81 74 70 74 36 96 68 09 64 39 88 35 54 89 96 58 66 27 88 97 32 14 06 35 78 20 71 06 85 66 57 02 58 91 72 05 29 56 73 48 86 52 09 93 22 57 79 42 12 01 31 68 17 59 63 76 07 77 73 81 14 13 17 20 11 09 01 83 08 85 91 70 84 63 62 77 37 07 47 01 59 95 39 69 39 21 99 09 87 02 97 16 92 36 74 71 90 66 33 73 73 75 52 91 11 12 26 53 05 26 26 48 61 50 90 65 01 87 42 47 74 35 22 73 24 26 56 70 52 05 48 41 31 18 83 27 21 39 80 85 26 08 44 02 71 07 63 22 05 52 19 08 20 17 25 21 11 72 93 33 49 64 23 53 82 03 13 91 65 85 02 40 05 42 31 77 42 05 36 06 54 04 58 07 76 87 83 25 57 66 12 74 33 85 37 74 32 20 69 03 97 91 68 82 44 19 14 89 28 85 85 80 53 34 87 58 98 88 78 48 65 98 40 11 57 10 67 70 81 60 79 74 72 97 59 79 47 30 20 54 80 89 91 14 05 33 36 79 39 60 85 59 39 60 07 57 76 77 92 06 35 15 72 23 41 45 52 95 18 64 79 86 53 56 31 69 11 91 31 84 50 44 82 22 81 41 40 30 42 30 91 48 94 74 76 64 58 74 25 96 57 14 19 03 99 28 83 15 75 99 01 89 85 79 50 03 95 32 67 44 08 07 41 62 64 29 20 14 76 26 55 48 71 69 66 19 72 44 25 14 01 48 74 12 98 07 64 66 84 24 18 16 27 48 20 14 47 69 30 86 48 40 23 16 61 21 51 50 26 47 35 33 91 28 78 64 43 68 04 79 51 08 19 60 52 95 06 68 46 86 35 97 27 58 04 65 30 58 99 12 12 75 91 39 50 31 42 64 70 04 46 07 98 73 98 93 37 89 77 91 64 71 64 65 66 21 78 62 81 74 42 20 83 70 73 95 78 45 92 27 34 53 71 15 30 11 85 31 34 71 13 48 05 14 44 03 19 67 23 73 19 57 06 90 94 72 57 69 81 62 59 68 88 57 55 69 49 13 07 87 97 80 89 05 71 05 05 26 38 40 16 62 45 99 18 38 98 24 21 26 62 74 69 04 85 57 77 35 58 67 91 79 79 57 86 28 66 34 72 51 76 78 36 95 63 90 08 78 47 63 45 31 22 70 52 48 79 94 15 77 61 67 68 23 33 44 81 80 92 93 75 94 88 23 61 39 76 22 03 28 94 32 06 49 65 41 34 18 23 08 47 62 60 03 63 33 13 80 52 31 54 73 43 70 26 16 69 57 87 83 31 03 93 70 81 47 95 77 44 29 68 39 51 56 59 63 07 25 70 07 77 43 53 64 03 94 42 95 39 18 01 66 21 16 97 20 50 90 16 70 10 95 69 29 06 25 61 41 26 15 59 63 35
view raw euler67.txt hosted with ❤ by GitHub

The Algorithm

Solving such a problem would require a powerful approach – and surely enough, there is an algorithm that solves the 100-level problem in a fraction of a second. Here’s a brief sketch of the algorithm:

You have such triangle:

   3
7 4
2 4 6
8 5 9 3


Let’s say you’re on the penultimate level 2 4 6 and you have to iterate over it.

From 2, you can go to either 8 or 5, so 8 is better (maximize you result by 3) so you calculate the first sum 8 + 2 = 10

From 4, you can go to either 5 or 9, so 9 is better (maximize you result by 4) so you calculate the second sum 9 + 4 = 13

From 6, you can go to either 9 or 3, so 9 is better again (maximize you result by 6) so you calculate the third sum 9 + 6 = 15

This is the end of first iteration and you got the line of sums 10 13 15.

Now you’ve got triangle of lower dimension:

      3
7    4
10   13    15


Keep going this way…

         3
20    19


…and you finally arrive at 23 as the answer.

The Code

Now for the Python code. I first store the 100-level triangle array in a text file, euler67.txt
I read the triangle array into Python and successively update the penultimate row and delete the last row according to the algorithm discussed above.

 # Read the problem matrix into a triangle array in python filename = 'euler67.txt' with open(filename, "r") as ins: array = [] for line in ins: array.append(line) # Convert the triangle arry entries into integers newArray = [] for i in array: j = i.split(' ') k = [int(n) for n in j] newArray.append(k) l = len(newArray) # Algorithm to calculate Maximum Path Sum for i in range(l-1): array1 = newArray[-1] array2 = newArray[-2] for j in range(len(array2)): array2[j] += max(array1[j], array1[j+1]) newArray.pop(-1) newArray[-1] = array2 print newArray[0][0]
view raw euler67.py hosted with ❤ by GitHub

This code is the key to solving problems 18 and 67 of Project Euler.
Problem 18
Ans: 1074
Problem 67
Ans: 7273

# Consecutive Prime Sum — Project Euler (Problem 50)

Many problems in Project Euler relate to working with primes. I use primesieve-python to help solve such problems. It consists of Python bindings for the primesieve C++ library. Generates primes orders of magnitude faster than any pure Python code. Features:

• Generate a list of primes
• Count primes and prime k-tuplets
• Print primes and prime k-tuplets
• Find the nth prime
• Iterate over primes using little memory

Anyway, here’s Problem 50 from Project Euler:

Here’s how I did it:

 # Question: Which prime, below one-million, can be written as the sum of the most consecutive primes from primesieve import * from math import * # Generate list of primes under a million primes_under_million = generate_primes(10**6) # Sum of consecutive primes is of order 0.5(n^2)(logn) # Calculate 'n' so that sum of consecutive primes is less than a million (and not necessarily prime) nsum = 1 n = 1 while nsum < 10**6: nsum = 0.5*(n**2)*(log(n, e)) n += 1 # Calculate index so that sum of first 'index' consecutive primes is under a million and also prime primes_subset = primes_under_million[:n] nsum = sum(primes_under_million[:n]) while nsum > 10**6: n -= 1 nsum = sum(primes_under_million[:n]) primes_sum = 0 index = 0 for i in range(len(primes_subset)): if i % 2 == 1: pass else: sumprimes = sum(primes_subset[:i]) if sumprimes > primes_sum and sumprimes < 10**6 and sumprimes in primes_under_million: primes_sum = sumprimes index = i # Print out sum of consecutive primes till 'index', index, n # print primes_sum, index, n # Check consecutive primes within a range (index to n) such that their number is greater than index and maximum j = index + 1 start = 0 while j <= n: while (j-start) >= (n-index): sumprimes = sum(primes_subset[start:j]) if sumprimes > primes_sum and sumprimes in primes_under_million: primes_sum = sumprimes start += 1 j += 1 start = 0 print primes_sum
view raw euler50.py hosted with ❤ by GitHub

# Largest Product in a Grid — Project Euler (Problem 11)

I started solving Project Euler problems this month. Check out the Project Euler tab of this blog for a list of the problems I’ve solved (with solutions) till date. Here’s a problem you might find interesting:

Here’s my solution using Python (I basically search through the entire matrix which is of O() complexity):

I first copy the maxtrix into a text file euler11.txt so that it can be later read into Python

 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48
view raw euler11.txt hosted with ❤ by GitHub

I then execute the following code from the same working directory as euler11.txt
 # import numpy module for matrix operations from numpy import * # read the file with the matrix of numbers filename = 'euler11.txt' # store each line of the file into an array with open(filename, "r") as ins: array = [] for line in ins: array.append(line) print array # create a new array that converts the number strings into number integers newArray = [] for i in array: j = i.split(' ') k = [int(n) for n in j] newArray.append(k) print newArray # convert the array of integers into a matrix of integers problemMatrix = matrix(newArray) print problemMatrix # set initial maximum product to be a dummy number, say 1 maxProd = 1 # search all combinations for maximum product for i in range(16): for j in range(16): prod1 = problemMatrix[i,j]*problemMatrix[i+1,j]*problemMatrix[i+2,j]*problemMatrix[i+3,j] if prod1 > maxProd: maxProd = prod1 prod2 = problemMatrix[i,j]*problemMatrix[i,j+1]*problemMatrix[i,j+2]*problemMatrix[i,j+3] if prod2 > maxProd: maxProd = prod2 prod3 = problemMatrix[i,j]*problemMatrix[i+1,j+1]*problemMatrix[i+2,j+2]*problemMatrix[i+3,j+3] if prod3 > maxProd: maxProd = prod3 prod4 = problemMatrix[19-i,j]*problemMatrix[18-i,j+1]*problemMatrix[17-i,j+2]*problemMatrix[16-i,j+3] if prod4 > maxProd: maxProd = prod4 print maxProd
view raw euler11.py hosted with ❤ by GitHub

# Object Oriented Programing with Python – Particle Diffusion Simulation

I’m a newbie to the programming world. I first started programming in Python in May this year, a month after I started this blog, so I still haven’t learnt enough to contribute to economics as is the stated goal of this blog. But I know I’ll get there in a year or less.

This blog was also meant to document my learning. In May, I would have called myself Newb v0.0. Today, 3 months later, I’d like to call myself Newb v0.3 and the goal is to be at least Expert v1.0 by January 2016.

With the help of Rice University’s awesome classes on Python programming I created a cool simulation of particles diffusing into space, using the concept of Classes, which I learnt just yesterday!

Click to check out the code !

# The Merge Sort — Python Code

I have just begun working on a MOOC on algorithms offered by Stanford. Since this course gives us the liberty to choose a programming language, there isn’t any code discussed in those lectures. I plan to convert any algorithm discussed in those lectures into Python code. Since Merge Sort was the first algorithm discussed, I’m starting with that.

Merge Sort is supposedly a good introduction to divide and conquer algorithms, greatly improving upon selection, insertion and bubble sort techniques, especially when input size increases.

Pseudocode:

— Recursively sort the first half of the input array.
— Recursively sort the second half of the input array.
— Merge two sorted sub-lists into one list.

C = output [length = n]
A = 1st sorted array [n/2]
B = 2nd sorted array [n/2]
i = 0 or 1 (depending on the programming language)
j = 0 or 1 (depending on the programming language)

for k = 1 to n

if A(i) < B(j)
C(k) = A(i)
i = i + 1

else if A(i) > B(j)
C(k) = B(j)
j = j + 1

Note: the pseudocode for the merge operation ignores the end cases.

Visualizing the algorithm can be done in 2 stages — first, the recursive splitting of the arrays, 2 each 2 at a time, and second, the merge operation.

Here’s the Python code to merge sort an array.

 # Code for the merge subroutine def merge(a,b): """ Function to merge two arrays """ c = [] while len(a) != 0 and len(b) != 0: if a[0] < b[0]: c.append(a[0]) a.remove(a[0]) else: c.append(b[0]) b.remove(b[0]) if len(a) == 0: c += b else: c += a return c # Code for merge sort def mergesort(x): """ Function to sort an array using merge sort algorithm """ if len(x) == 0 or len(x) == 1: return x else: middle = len(x)/2 a = mergesort(x[:middle]) b = mergesort(x[middle:]) return merge(a,b)
view raw mergesort.py hosted with ❤ by GitHub

We can divide a list in half log2 n times where n is the length of the list. The second process is the merge. Each item in the list will eventually be processed and placed on the sorted list. So the merge operation which results in a list of size n requires n operations. The result of this analysis is that log2 n splits, each of which costs n for a total of nlog2 n operations.

Other Algorithms:
Karatsuba Integer Multiplication Algorithm
Quick Sort Python Code