Year: 2015

Scatter Plot Bug Fix in Dato’s GraphLab Create ML Package in Python

Anirudh Technical , , , , , ,

I have been using Dato’s GraphLab Create for Coursera’s new Machine Learning Specialization that uses Python. Like me, if you’ve been facing trouble obtaining scatter plots on your canvas in GraphLab Create despite the following code:

graphlab.canvas.set_target('ipynb')

…then no worries, there is a quick fix. I’ve been deliberately lousy with the presentation, so sorry about that. Chances are that no one’s going to end up reading this anyway. I saw this problem being discussed on a Dato forum, so I decided to blog about the fix.

EDIT: Note that this problem is in GraphLab Create v1.6 only. They came up with v1.6.1 a few days after the problem was escalated on their forum, so a good option would be to upgrade GraphLab Create.

The problem you face should looks something like this (click images below to enlarge):Screenshot from 2015-09-25 14:10:20

To solve the problem:

Locate sframe.py from your home directory by searching for it from your desktop environment (applies to Windows users too). I found it in the following path on my computer:

~/anaconda/lib/python2.7/site-packages/graphlab/canvas/views

The file sframe.py should look like this:

Screenshot from 2015-09-25 15:49:44

Then replace the code in lines 255-227 of the opened .py file with the code highlighted below:

Screenshot from 2015-09-25 15:53:19

This should take care of the problem for good.

Now you have your desired result:

Screenshot from 2015-09-25 16:18:41

Machine Learning — New Coursera Specialization from the University of Washington

Anirudh Non Technical , , , , , , , , , ,

I have finally embarked on my first machine learning MOOC / Specialization. I love Python, and this course uses Python as the language of choice. Also, the instructors assert that Python is widely used in industry, and is becoming the de facto language for data science in industry. They use IPython Notebook in their assignments and videos.

The specialization offered by the University of Washington consists of 5 courses and a capstone project spread across about 8 months (September through April). The specialization’s first iteration kicked off yesterday.

washingtonMachineLearningThe first course, Machine Learning Foundations: A Case Study Approach is 6 weeks long, running from September 22 through November 9.

The Instructors:

Emily Fox and Carlos Guestrin
EmilyFoxguestrin-dato

Key Learning Outcomes
– Identify potential applications of machine learning in practice.
– Describe the core differences in analyses enabled by regression, classification, and clustering.
– Select the appropriate machine learning task for a potential application.
– Apply regression, classification, clustering, retrieval, recommender systems, and deep learning.
– Represent your data as features to serve as input to machine learning models.
– Assess the model quality in terms of relevant error metrics for each task.
– Utilize a dataset to fit a model to analyze new data.
– Build an end-to-end application that uses machine learning at its core.
– Implement these techniques in Python.

Week-by-Week
Week 1: Introductory welcome videos and the instructors’ views on the future of intelligent applications
Week 2: Predicting House Prices (Regression)
Week 3: Classification (Sentiment Analysis)
Week 4: Clustering and Similarity: Retrieving Documents
Week 5: Recommending Products
Week 6: Deep Learning: Searching for Images

EDIT

It’s been 3 days since the course began, and here’s how the classmate demographic looks like:

Classmates09252015

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

Anirudh Non Technical , , , , , , ,

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 “magic” 5-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

magic5gon

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.

magic5gonInstance

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!

from itertools import permutations
from itertools import combinations
# array of candidate solutions empty at the beginning
record = []
# choose 5 numbers for inner cells between 1 and 9; there are 9C5 combinations
# the problem ask for a 16-digit number, so 10 is not to be included in inner cells
cells = range(1,10)
inner_cells = [map(int,comb) for comb in combinations(cells,5)]
# code to calculate min and max couple in an array
def minCouple(array):
answer = array[0]+array[-1]
for i in xrange(len(array)-1):
coupleSum = array[i] + array[i+1]
if coupleSum < answer:
answer = coupleSum
return answer
def maxCouple(array):
answer = 0
for i in xrange(len(array)-1):
if i==0:
coupleSum = array[0]+ array[-1]
if coupleSum > answer:
answer = coupleSum
else:
coupleSum = array[i]+ array[i+1]
if coupleSum > answer:
answer = coupleSum
return answer
# Algorithm
for array in inner_cells:
pivot = array[0]
perm_array = array[1:]
perms = [map(int,perm) for perm in permutations(perm_array,4)]
for perm in perms:
checkArray = perm
checkArray.insert(0,pivot)
outerRing = [el for el in range(1,11) if el not in checkArray]
xMax = minCouple(checkArray) + max(outerRing)
xMin = maxCouple(checkArray) + min(outerRing)
if xMax >= xMin:
for x in xrange(xMin, xMax+1):
i = checkArray[0]
j = checkArray[1]
k = checkArray[2]
l = checkArray[3]
m = checkArray[4]
alpha = x-i-j
beta = x-j-k
gamma = x-k-l
delta = x-l-m
theta = x-m-i
outerCalculated = [alpha, beta, gamma, delta, theta]
if sorted(outerCalculated) == sorted(outerRing):
a = [alpha, i, j]
b = [beta, j, k]
c = [gamma, k, l]
d = [delta, l, m]
e = [theta, m, i]
min_val = min(alpha, beta, gamma, delta, theta)
if alpha == min_val:
append = a+b+c+d+e
elif beta == min_val:
append = b+c+d+e+a
elif gamma == min_val:
append = c+d+e+a+b
elif delta == min_val:
append = d+e+a+b+c
elif theta == min_val:
append = e+a+b+c+d
l = [str(i) for i in append]
s = ''.join(l)
integer_list = int(s)
record.append(integer_list)
print max(record)
view raw euler68.py hosted with ❤ by GitHub

Ans: 6531031914842725

Maximum Path Sum — Dynamic Programming Algorithm

Anirudh Technical , , , , , , , ,

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

Problem18
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
52 09 61 30 61 97 66 21 96 92 98 90 06 34 96 60 32 69 68 33 75 84 18 31 71 50 84 63 03 03 19 11 28 42 75 45 45
61 31 61 68 96 34 49 39 05 71 76 59 62 67 06 47 96 99 34 21 32 47 52 07 71 60 42 72 94 56 82 83 84 40 94 87 82 46
01 20 60 14 17 38 26 78 66 81 45 95 18 51 98 81 48 16 53 88 37 52 69 95 72 93 22 34 98 20 54 27 73 61 56 63 60 34 63
93 42 94 83 47 61 27 51 79 79 45 01 44 73 31 70 83 42 88 25 53 51 30 15 65 94 80 44 61 84 12 77 02 62 02 65 94 42 14 94
32 73 09 67 68 29 74 98 10 19 85 48 38 31 85 67 53 93 93 77 47 67 39 72 94 53 18 43 77 40 78 32 29 59 24 06 02 83 50 60 66
32 01 44 30 16 51 15 81 98 15 10 62 86 79 50 62 45 60 70 38 31 85 65 61 64 06 69 84 14 22 56 43 09 48 66 69 83 91 60 40 36 61
92 48 22 99 15 95 64 43 01 16 94 02 99 19 17 69 11 58 97 56 89 31 77 45 67 96 12 73 08 20 36 47 81 44 50 64 68 85 40 81 85 52 09
91 35 92 45 32 84 62 15 19 64 21 66 06 01 52 80 62 59 12 25 88 28 91 50 40 16 22 99 92 79 87 51 21 77 74 77 07 42 38 42 74 83 02 05
46 19 77 66 24 18 05 32 02 84 31 99 92 58 96 72 91 36 62 99 55 29 53 42 12 37 26 58 89 50 66 19 82 75 12 48 24 87 91 85 02 07 03 76 86
99 98 84 93 07 17 33 61 92 20 66 60 24 66 40 30 67 05 37 29 24 96 03 27 70 62 13 04 45 47 59 88 43 20 66 15 46 92 30 04 71 66 78 70 53 99
67 60 38 06 88 04 17 72 10 99 71 07 42 25 54 05 26 64 91 50 45 71 06 30 67 48 69 82 08 56 80 67 18 46 66 63 01 20 08 80 47 07 91 16 03 79 87
18 54 78 49 80 48 77 40 68 23 60 88 58 80 33 57 11 69 55 53 64 02 94 49 60 92 16 35 81 21 82 96 25 24 96 18 02 05 49 03 50 77 06 32 84 27 18 38
68 01 50 04 03 21 42 94 53 24 89 05 92 26 52 36 68 11 85 01 04 42 02 45 15 06 50 04 53 73 25 74 81 88 98 21 67 84 79 97 99 20 95 04 40 46 02 58 87
94 10 02 78 88 52 21 03 88 60 06 53 49 71 20 91 12 65 07 49 21 22 11 41 58 99 36 16 09 48 17 24 52 36 23 15 72 16 84 56 02 99 43 76 81 71 29 39 49 17
64 39 59 84 86 16 17 66 03 09 43 06 64 18 63 29 68 06 23 07 87 14 26 35 17 12 98 41 53 64 78 18 98 27 28 84 80 67 75 62 10 11 76 90 54 10 05 54 41 39 66
43 83 18 37 32 31 52 29 95 47 08 76 35 11 04 53 35 43 34 10 52 57 12 36 20 39 40 55 78 44 07 31 38 26 08 15 56 88 86 01 52 62 10 24 32 05 60 65 53 28 57 99
03 50 03 52 07 73 49 92 66 80 01 46 08 67 25 36 73 93 07 42 25 53 13 96 76 83 87 90 54 89 78 22 78 91 73 51 69 09 79 94 83 53 09 40 69 62 10 79 49 47 03 81 30
71 54 73 33 51 76 59 54 79 37 56 45 84 17 62 21 98 69 41 95 65 24 39 37 62 03 24 48 54 64 46 82 71 78 33 67 09 16 96 68 52 74 79 68 32 21 13 78 96 60 09 69 20 36
73 26 21 44 46 38 17 83 65 98 07 23 52 46 61 97 33 13 60 31 70 15 36 77 31 58 56 93 75 68 21 36 69 53 90 75 25 82 39 50 65 94 29 30 11 33 11 13 96 02 56 47 07 49 02
76 46 73 30 10 20 60 70 14 56 34 26 37 39 48 24 55 76 84 91 39 86 95 61 50 14 53 93 64 67 37 31 10 84 42 70 48 20 10 72 60 61 84 79 69 65 99 73 89 25 85 48 92 56 97 16
03 14 80 27 22 30 44 27 67 75 79 32 51 54 81 29 65 14 19 04 13 82 04 91 43 40 12 52 29 99 07 76 60 25 01 07 61 71 37 92 40 47 99 66 57 01 43 44 22 40 53 53 09 69 26 81 07
49 80 56 90 93 87 47 13 75 28 87 23 72 79 32 18 27 20 28 10 37 59 21 18 70 04 79 96 03 31 45 71 81 06 14 18 17 05 31 50 92 79 23 47 09 39 47 91 43 54 69 47 42 95 62 46 32 85
37 18 62 85 87 28 64 05 77 51 47 26 30 65 05 70 65 75 59 80 42 52 25 20 44 10 92 17 71 95 52 14 77 13 24 55 11 65 26 91 01 30 63 15 49 48 41 17 67 47 03 68 20 90 98 32 04 40 68
90 51 58 60 06 55 23 68 05 19 76 94 82 36 96 43 38 90 87 28 33 83 05 17 70 83 96 93 06 04 78 47 80 06 23 84 75 23 87 72 99 14 50 98 92 38 90 64 61 58 76 94 36 66 87 80 51 35 61 38
57 95 64 06 53 36 82 51 40 33 47 14 07 98 78 65 39 58 53 06 50 53 04 69 40 68 36 69 75 78 75 60 03 32 39 24 74 47 26 90 13 40 44 71 90 76 51 24 36 50 25 45 70 80 61 80 61 43 90 64 11
18 29 86 56 68 42 79 10 42 44 30 12 96 18 23 18 52 59 02 99 67 46 60 86 43 38 55 17 44 93 42 21 55 14 47 34 55 16 49 24 23 29 96 51 55 10 46 53 27 92 27 46 63 57 30 65 43 27 21 20 24 83
81 72 93 19 69 52 48 01 13 83 92 69 20 48 69 59 20 62 05 42 28 89 90 99 32 72 84 17 08 87 36 03 60 31 36 36 81 26 97 36 48 54 56 56 27 16 91 08 23 11 87 99 33 47 02 14 44 73 70 99 43 35 33
90 56 61 86 56 12 70 59 63 32 01 15 81 47 71 76 95 32 65 80 54 70 34 51 40 45 33 04 64 55 78 68 88 47 31 47 68 87 03 84 23 44 89 72 35 08 31 76 63 26 90 85 96 67 65 91 19 14 17 86 04 71 32 95
37 13 04 22 64 37 37 28 56 62 86 33 07 37 10 44 52 82 52 06 19 52 57 75 90 26 91 24 06 21 14 67 76 30 46 14 35 89 89 41 03 64 56 97 87 63 22 34 03 79 17 45 11 53 25 56 96 61 23 18 63 31 37 37 47
77 23 26 70 72 76 77 04 28 64 71 69 14 85 96 54 95 48 06 62 99 83 86 77 97 75 71 66 30 19 57 90 33 01 60 61 14 12 90 99 32 77 56 41 18 14 87 49 10 14 90 64 18 50 21 74 14 16 88 05 45 73 82 47 74 44
22 97 41 13 34 31 54 61 56 94 03 24 59 27 98 77 04 09 37 40 12 26 87 09 71 70 07 18 64 57 80 21 12 71 83 94 60 39 73 79 73 19 97 32 64 29 41 07 48 84 85 67 12 74 95 20 24 52 41 67 56 61 29 93 35 72 69
72 23 63 66 01 11 07 30 52 56 95 16 65 26 83 90 50 74 60 18 16 48 43 77 37 11 99 98 30 94 91 26 62 73 45 12 87 73 47 27 01 88 66 99 21 41 95 80 02 53 23 32 61 48 32 43 43 83 14 66 95 91 19 81 80 67 25 88
08 62 32 18 92 14 83 71 37 96 11 83 39 99 05 16 23 27 10 67 02 25 44 11 55 31 46 64 41 56 44 74 26 81 51 31 45 85 87 09 81 95 22 28 76 69 46 48 64 87 67 76 27 89 31 11 74 16 62 03 60 94 42 47 09 34 94 93 72
56 18 90 18 42 17 42 32 14 86 06 53 33 95 99 35 29 15 44 20 49 59 25 54 34 59 84 21 23 54 35 90 78 16 93 13 37 88 54 19 86 67 68 55 66 84 65 42 98 37 87 56 33 28 58 38 28 38 66 27 52 21 81 15 08 22 97 32 85 27
91 53 40 28 13 34 91 25 01 63 50 37 22 49 71 58 32 28 30 18 68 94 23 83 63 62 94 76 80 41 90 22 82 52 29 12 18 56 10 08 35 14 37 57 23 65 67 40 72 39 93 39 70 89 40 34 07 46 94 22 20 05 53 64 56 30 05 56 61 88 27
23 95 11 12 37 69 68 24 66 10 87 70 43 50 75 07 62 41 83 58 95 93 89 79 45 39 02 22 05 22 95 43 62 11 68 29 17 40 26 44 25 71 87 16 70 85 19 25 59 94 90 41 41 80 61 70 55 60 84 33 95 76 42 63 15 09 03 40 38 12 03 32
09 84 56 80 61 55 85 97 16 94 82 94 98 57 84 30 84 48 93 90 71 05 95 90 73 17 30 98 40 64 65 89 07 79 09 19 56 36 42 30 23 69 73 72 07 05 27 61 24 31 43 48 71 84 21 28 26 65 65 59 65 74 77 20 10 81 61 84 95 08 52 23 70
47 81 28 09 98 51 67 64 35 51 59 36 92 82 77 65 80 24 72 53 22 07 27 10 21 28 30 22 48 82 80 48 56 20 14 43 18 25 50 95 90 31 77 08 09 48 44 80 90 22 93 45 82 17 13 96 25 26 08 73 34 99 06 49 24 06 83 51 40 14 15 10 25 01
54 25 10 81 30 64 24 74 75 80 36 75 82 60 22 69 72 91 45 67 03 62 79 54 89 74 44 83 64 96 66 73 44 30 74 50 37 05 09 97 70 01 60 46 37 91 39 75 75 18 58 52 72 78 51 81 86 52 08 97 01 46 43 66 98 62 81 18 70 93 73 08 32 46 34
96 80 82 07 59 71 92 53 19 20 88 66 03 26 26 10 24 27 50 82 94 73 63 08 51 33 22 45 19 13 58 33 90 15 22 50 36 13 55 06 35 47 82 52 33 61 36 27 28 46 98 14 73 20 73 32 16 26 80 53 47 66 76 38 94 45 02 01 22 52 47 96 64 58 52 39
88 46 23 39 74 63 81 64 20 90 33 33 76 55 58 26 10 46 42 26 74 74 12 83 32 43 09 02 73 55 86 54 85 34 28 23 29 79 91 62 47 41 82 87 99 22 48 90 20 05 96 75 95 04 43 28 81 39 81 01 28 42 78 25 39 77 90 57 58 98 17 36 73 22 63 74 51
29 39 74 94 95 78 64 24 38 86 63 87 93 06 70 92 22 16 80 64 29 52 20 27 23 50 14 13 87 15 72 96 81 22 08 49 72 30 70 24 79 31 16 64 59 21 89 34 96 91 48 76 43 53 88 01 57 80 23 81 90 79 58 01 80 87 17 99 86 90 72 63 32 69 14 28 88 69
37 17 71 95 56 93 71 35 43 45 04 98 92 94 84 96 11 30 31 27 31 60 92 03 48 05 98 91 86 94 35 90 90 08 48 19 33 28 68 37 59 26 65 96 50 68 22 07 09 49 34 31 77 49 43 06 75 17 81 87 61 79 52 26 27 72 29 50 07 98 86 01 17 10 46 64 24 18 56
51 30 25 94 88 85 79 91 40 33 63 84 49 67 98 92 15 26 75 19 82 05 18 78 65 93 61 48 91 43 59 41 70 51 22 15 92 81 67 91 46 98 11 11 65 31 66 10 98 65 83 21 05 56 05 98 73 67 46 74 69 34 08 30 05 52 07 98 32 95 30 94 65 50 24 63 28 81 99 57
19 23 61 36 09 89 71 98 65 17 30 29 89 26 79 74 94 11 44 48 97 54 81 55 39 66 69 45 28 47 13 86 15 76 74 70 84 32 36 33 79 20 78 14 41 47 89 28 81 05 99 66 81 86 38 26 06 25 13 60 54 55 23 53 27 05 89 25 23 11 13 54 59 54 56 34 16 24 53 44 06
13 40 57 72 21 15 60 08 04 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

Collatz Conjecture — What You Need to Know

Anirudh Technical , , , , , , ,

Like many of my previous posts, this post too has something to do with a Project Euler problem. Here’s a sketch of the Colatz Conjecture.

The following iterative sequence is defined for the set of positive integers:

n → n/2 (n is even)
n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

So basically, it’s just this. Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has aptly been called oneness! But perhaps oneness has its pitfalls too…

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

Question
It can be seen that the sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1. Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

HUGE HINT:

Histogram of stopping times for the numbers 1 to 100 million. Stopping time is on the x axis, frequency on the y axis.

Approach 1 (A naïve, but straigh forward method)

# Longest Collatz Sequence under a million
# Function listing collatz sequence for a number
def collatz(n):
"function listing collatz sequence for a positive integer"
coll = []
coll.append(n)
while n != 1:
if n % 2 == 0:
n = n/2
coll.append(n)
else:
n = 3*n + 1
coll.append(n)
return coll
longest = 0
j = 0
for i in xrange(1, 1000000):
lencoll = len(collatz(i))
if lencoll > longest:
longest = lencoll
j = i
print j
view raw euler14.py hosted with ❤ by GitHub

Approach 2 (Smart, quick method that uses dynamic programming with the help of dictionaries)

collatz = {1:1}
def Collatz(n):
global collatz
if not collatz.has_key(n):
if n%2 == 0:
collatz[n] = Collatz(n/2) + 1
else:
collatz[n] = Collatz(3*n + 1) + 1
return collatz[n]
for j in range(1000000,0,-1):
Collatz(j)
print collatz.keys()[collatz.values().index(max(collatz.values()))]
view raw euler14.py hosted with ❤ by GitHub

I couldn’t help appreciate the elegance of the second algorithm. It’ll be well worth perusing if you don’t get it at one go. [Hint: It keeps track of the number of terms of a particular sequence as values assigned to keys of a Python dictionary]

Ans: 837799

Large sum — Project Euler (Problem 13)

Anirudh Non Technical , , , ,

This isn’t much of a problem really, but since I’m posting solutions to all the Project Euler problems I solve, I’ve been OCD’d into posting this one too. Besides, it illustrates the simplifying power of Python as a language?

Anyway… here’s the problem:

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers:

 37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690
The solution:
I first copy the problem matrix to a .txt file, in this case, euler13.txt
The solution is cake really, and I don’t care whether this was worth posting on my blog or not coz I wasted my time solving this problem anyway, and it shouldn’t have been for nothing!
# Read the problem matrix into an array in python
filename = 'euler13.txt'
with open(filename, "r") as ins:
array = []
for line in ins:
array.append(line)
# Convert the array into an array of integers
newArray = []
for i in array:
newArray.append(int(i))
# Sum up the array and print the first 10 numbers of the sum as a string
arraySum = sum(newArray)
print str(arraySum)[:10]
view raw euler13.py hosted with ❤ by GitHub
Ans: 5537376230

Highly Divisible Triangular Number — Project Euler (Problem 12)

Anirudh Technical , , , , , ,

All ∑n numbers are Triangle Numbers. They’re called so, because they can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one.

TriangularNumber_900

Problem 12 of Project Euler asks for the first triangle number with more than 500 divisors.

These are the factors of the first seven triangle numbers:

∑1 = 1: 1
∑2 = 3: 1,3
∑3 = 6: 1,2,3,6
∑4 = 10: 1,2,5,10
∑5 = 15: 1,3,5,15
∑6 = 21: 1,3,7,21
∑7 = 28: 1,2,4,7,14,28

Here’s how I proceeded:

First Step: Find the smallest number with 500 divisors. Seems like a good starting point to begin our search.
Second Step: Starting at the number found in the previous step, search for the next triangle number. Check to see whether this number has 500+ divisors. If yes, this is the number we were looking for, else…
Third Step: Check n for which ∑n = triangle number found in the previous step
Fourth Step: Add (n+1) to the last triangle number found, to find the next triangle number. Check whether this number has 500+ divisors. If yes, this number is the answer. If not, repeat Fourth Step till the process terminates.

Now for the details:

The First Step isn’t exactly a piece of cake, but necessary to reduce computation time. I solved this with a bit of mental math. The main tool for the feat is the prime number decomposition theorem:

Every integer N is the product of powers of prime numbers

N = pαqβ· … · rγ
Where p, q, …, r are prime, while α, β, …, γ are positive integers. Such representation is unique up to the order of the prime factors.
If N is a power of a prime, N = pα, then it has α + 1 factors:
1, p, …, pα-1, pα
The total number of factors of N equals (α + 1)(β + 1) … (γ + 1)

500 = 2 x 2 x 5 x 5 x 5
So, the number in question should be of the form abq4r4s4 where a, b, q, r, s are primes that minimize abq4r4s4. This is satisfied by 7x11x24x34x54 = 62370000. This marks the end of the First Step which is where we start our search for our magic number.

The next 3 steps would need helper functions defined as below:

from math import *
# Function to calculate the number of divisors of integer n
def divisors(n):
limit = int(sqrt(n))
divisors_list = []
for i in range(1, limit+1, 1):
if n % i == 0:
divisors_list.append(i)
if i != n/i:
divisors_list.append(n/i)
return len(divisors_list)
# Function to check for triangle number
def isTriangleNumber(n):
a = int(sqrt(2*n))
return 0.5*a*(a+1) == n
# Function to calculate the last term of the series adding up to the triangle number
def lastTerm(n):
if isTriangleNumber(n):
return int(sqrt(2*n))
else:
return None
view raw euler12functions.py hosted with ❤ by GitHub

As can be seen from the above code, the algorithm to calculate divisors of an integer is as follows:
1. Start by inputting a number n
2. Let an int variable limit = √n
3. Run a loop from i = 1 to  i = limit
    3.1 if n is divisible by i
3.1.1 Add i to the list of divisors
3.1.2 if i and n/i are unequal, add n/i to the list too.
4. End

Finally, executing the 4 steps mentioned earlier can be done like so (the code took less than 2s to arrive at the answer):

# First Step
# First number 'check' to have 500 divisors
check = 2**4 * 3**4 * 5**4 * 7 * 11
# Second Step
# Starting from 'check', iterate sequentially checking for the next 'triangle' number
while not isTriangleNumber(check):
check += 1
# Third and Fourth Steps
# Calculate the last term of the series ('seriesLastTerm') that adds up to the newly calculated triangle number 'check'
seriesLastTerm = lastTerm(check)
# Iterate over triangle numbers checking for divisors > 500
while divisors(check) <= 500:
# add the next term to check to get the next triangle number
check += (seriesLastTerm + 1)
seriesLastTerm += 1
print check
view raw euler12nonmeatypart.py hosted with ❤ by GitHub

Ans: 76576500

Consecutive Prime Sum — Project Euler (Problem 50)

Anirudh Technical , , , , , ,

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:

ProjectEuler50

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

Answer: 997651

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

Anirudh Technical , , , , , ,

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:

ProjectEuler11

Here’s my solution using Python (I basically search through the entire matrix which is of O(n²) 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

Answer: 70600674

MOOC Review: Introduction to Computer Science and Programming Using Python (6.00.1x)

Anirudh Non Technical , , , , ,

I enrolled in Introduction to Computer Science and Programming Using Python with the primary objective of learning to code using Python. This course, as the name suggests, is more than just about Python. It uses Python as a tool to teach computational thinking and serves as an introduction to computer science. The fact that it is a course offered by MIT, makes it special.

As a matter of fact, this course is aimed at students with little or no prior programming experience who feel the need to understand computational approaches to problem solving. Eric Grimson is an excellent teacher (also Chancellor of MIT) and he delves into the subject matter to a surprising amount of detail.

The video lectures are based on select chapters from an excellent book by John Guttag. While the book isn’t mandatory for the course (the video lectures do a great job of explaining the material on their own), I benefited greatly from reading the textbook. There are a couple of instances where the code isn’t presented properly in the slides (typos or indentation gone wrong when pasting code to the slides), but the correct code / study material can be found in the textbook. Also, for explanations that are more in-depth, the book comes in handy.

Introduction to Computation and Programming Using Python

MIT offers this course in 2 parts via edX. While 6.00.1x is is an introduction to computer science as a tool to solve real-world analytical problems, 6.00.2x is an introduction to computation in data science. For a general look and feel of the course, this OCW link may be a good starting point. It contains material including video lectures and problem sets that are closely related to 6.00.1x and 6.00.2x.

Each week’s material of 6.00.1x consists of 2 topics, followed by a Problem Set. Problem Sets account for 40% of your grade. Video lectures are followed by finger exercises that can be attempted any number of times. Finger exercises account for 10% of your grade. The Quiz (kind of like a mid-term exam) and the Final Exam account for 25% each. The course is of 8 weeks duration and covers the following topics (along with corresponding readings from John Guttag’s textbook).

course_structure_till_quiz

course_structure_till_final

From the questions posted on forums, it was apparent that the section of this course that most people found challenging, was efficiency and orders of growth – and in particular, the Big-O asymptotic notation and problems on algorithmic complexity.

Lectures on Classes, Inheritance and Object Oriented Programming (OOP) were covered really well in over 100 minutes of video time. I enjoyed the problem set that followed, requiring the student to build an Internet news filter alerting the user when it noticed a news story that matched that user’s interests.

The final week had lectures on the concept of Trees, which were done hurriedly when compared to the depth of detail the instructor had earlier gone to, while explaining concepts from previous weeks. However, this material was covered quite well in Guttag’s textbook and the code for tree search algorithms was provided for perusal as part of the courseware.

At the end of the course, there were some interesting add-on videos to tickle the curiosity of the learner on the applications of computation in diverse fields such as medicine, robotics, databases and 3D graphics.

The Wiki tab for this course (in the edX platform) is laden with useful links to complement each week of lectures. I never got around to reading those, but I’m going through them now, and they’re quite interesting. It’s a section that nerds would love to skim through.

I learnt a great deal from this course (scored well too) putting in close to 6-hours-a-week of study. It is being offered again on August 26, 2015. In the mean time, I’m keeping my eyes open for MIT’s data science course (6.00.2x) which is likely to be offered in October, in continuation to 6.00.1x.