# 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:

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 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:

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 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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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.

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 # 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

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)

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 # 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)

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 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)

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
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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!
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters
 # 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)

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.

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:

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 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

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):

This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters
 # 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

Ans: 76576500

# 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

This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters
 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
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters. Learn more about bidirectional Unicode characters
 # 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

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

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.

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).

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.

# Which Programming Languages Get Used Most At Hackathons?

For a quick peek into the list:

The Top 10 Languages At Devpost’s Hackathons:

1. HTML/CSS (see note below)
2. JavaScript
3. Python
4. Java
5. C/C++
6. PHP
7. Objective-C
8. C#
9. Swift
10. JSON (which isn’t … really a programming language, but is on their list for some reason, so I’m including #11 too)
11. Ruby

Read the full Techcrunch article to know why.

In stark contrast:

The Top 10 Languages according to IEEE Spectrum’s 2015 Rankings:

1. Java
2. C
3. C++
4. Python
5. C#
6. R
7. PHP
8. JavaScript
9. Ruby
10. Matlab

Note: HTML isn’t quite a “programming” language — it’s a markup language, meaning it’s a means of laying out the elements of a document. But it’s a “language” none the less, and one that pretty much every web developer taps endlessly, so we’ll let the semantic stuff slide

# Which Programming Language Should I Learn First? [Infographic]

Here’s a pretty interesting flow chart to determine which programming language would suit you:

or

# Review: An Introduction to Interactive Programming in Python (Part 1)

This class (Part 1 of a 2-part course on interactive programming using Python – and the first course of the Fundamentals of Computing Specialization offered by RICE Unviersity) was an excellent introduction to programming because of its focus on building interactive (and fun) applications with the lessons learned each week. Most introductory coding classes start with text based (boring?) programs, while all through this course you’re required to build a series of projects that get progressively complicated with every passing week. I’m not to be mistaken to be trashing conventional pedagogy, but then again, how many gifted coders do you know who learned to code after completing all the exercises, cover-to-cover of some programming textbook? The best way to learn to enjoy coding would be to build interactive stuff, and this course scores full points on that.

A short introduction to the class in a charmingly nerdy way

The mini-projects / assignments during the course are implemented on a cloud-based environment called CodeSkulptor (built by Scott Rixner, one of the instructors for this course). I found CodeSkulptor unique, in that it allows you to share your code (because it’s browser based) with just about anyone with an Internet connection and makes you work with a graphic user interface (GUI) module similar to Pygame, called Simplegui. It also had a debugging tool, called Viz Mode that helped visualize the process. It eases the task of debugging your code and you’ll realize how cool it is as you start using it more.

Since the course mini-projects were peer-reviewed, evaluating other people’s code also became a more straight-jacket affair, as everyone has their code on the same version of Python. This ensures that the focus is on learning to code, without wasting time on the logistics of programming environment (tuning differences in versions or IDEs). I especially enjoyed peer grading – for each mini project we completed, we had to evaluate and grade the work of 5 others. This was very rewarding – because I got the opportunity to fix bugs in others’ code (which makes you a better coder, I guess) and also got to see better implementations than the ones I had coded, further enriching the learning experience. Indeed, the benefits of peer grading and assessment have been well studied and documented.

Of all the assignments, the one I loved the most was implementing the classic arcade game Pong. You could try playing a version of the game I implemented here. It is a 2-player implementation, but you can play it as a single-player game, only if you imagine yourself to be answering this somewhat cheeky question! Which Pong character are you? Left or Right?

The principal reason behind my joining this course was the way it is structured and taught. We had to watch two sets of videos (part a and part b) and then complete one quiz for each set. The main task for each week was to complete a mini-project that was due along with the quizzes early Sunday morning, followed by assessment of peers’ mini-projects on the following Sunday-Wednesday. The instructors clearly put in A LOT OF WORK to make the lecture videos interesting, laced with humor, with just enough to get you going on your own with the week’s mini-project. That way you’d spend less time viewing the lecture videos, spending more time on actually getting the code for your mini-project to work. So in a way, one might say this course doesn’t follow standard pedagogy for an introductory programming course, but then, as Scott Rixner assures, “You’d know enough to be dangerous!

The projects that were completed in Part 1 of this course were indeed exciting:

Rock Paper Scissors Lizard Spock: A simple implementation played with the computer. This project covers basics on statements, expressions and variables, functions, logic and conditionals [I’m a huge fan of The Big Bang Theory, so I was obviously eager to complete this game. Instead of a series of if-elif-else clauses, this implementation used modular logic, all of which is taught in a really fun way. A great way to start off the course].
Guess the Number: Computer chooses a random number between 1 and 100 and you guess that number. It covered event-driven programming, local and global variables, buttons and input fields [This game although fun, might have been more interesting to code if the computer had to guess the number that the player chose, using bisection search].
Stopwatch: This was the first project that used a graphic user interface, using some modular arithmetic to get the digits of the ticking seconds in place. A game was also built on it where the player had to stop the watch right at the start of a second to score points. This game tested your reaction-time. It covered static drawing, timers and interactive drawing.
Pong: The last project of Part 1 and the most fun. Creating the game required only a minor step-up from learnings from previous weeks. It covered knowledge of lists, keyboard input, motion, positional/velocity control. Coding the ball physics where you put to use high-school physics knowledge of elasticity and collisions was very enjoyable. In my game, I set elasticity = 1 (for perfectly elastic collisions)

In an interview with the founders of this MOOC, who spent they say that they spent over 1000 hours building it (Part 1 and Part 2 combined, I guess). That’s an awful lot of effort and it all shows in how brilliantly the class is executed. The support system in the class is excellent. You’ll always find help available within minutes of posting your doubts and queries on the forums. I’ve seen Joe Warren (one of the main instructors of the course) replying to forum posts quite regularly. In addition, there was enough supplementary material in the form of pages on concepts and examples, practice exercises, and video content created by students from previous iterations of the class to better explain concepts and aspects of game-building, improving upon the lecture material.

Concepts and Examples

Practice Exercises

Student-created Videos Explaining Concepts

Overall, I had a great learning experience. I completed Part 1 with a 100 per cent score even though I had a minor hiccup while building the game Pong, which was the most satisfying of all the projects in Part 1. I would review Part 2 when I’m done with that in August this year. I’d easily recommend this course to anyone wishing to start off with Python. It is a great place to be introduced to Python, but it shouldn’t be your ONLY resource. I have been taking MIT’s 6.01x introductory Python course side-by-side. I shall review that course as soon as I’m through with it. That course is pedagogically more text-bookish, and indeed they do profess the use of their textbook to complement the course. I’m 4 weeks into that course and finding that enjoyable too – albeit in a different way. I still haven’t lost a point on any of the assignments or finger exercises there, and hope the trend continues:

PS: In one of the forum threads, Joe posted a list of resources that could be referred to in addition to the class.

Python Books:

Another List of Books:

• http://pythonbooks.revolunet.com/  – about 50 books –  Another good list of free python books that is kept up to date, and I believe are all free or open-source: (I won’t repeat all the books on the list here, just go check it out! Some are also on the list above, but not all)

Further Online Learning:

# How to become a programmer, or the art of Googling well

This should serve as encouragement to all and sundry. Stop wasting time getting intimidated and learn to code like a pro.

Featured Image: http://xkcd.com/979/

*Note: Please read all italicized technical words as if they were in a foreign language.

The fall semester of my senior year, I was having some serious self-confidence issues. I had slowly come to realize that I did not, in fact, want to become a researcher. Statistics pained me, and the seemingly endless and fruitless nature of research bored me. I was someone who was driven by results – tangible products with deadlines that, upon completion, had a binary state: success, or failure. Going into my senior year, this revelation was followed by another. All of my skills thus far had been cultivated for research. If I wasn’t going into research, I had… nothing.

At a liberal arts college, being a computer science major does not mean you are a “hacker”. It can mean something as simple as, you were shopping around different departments, saw a command line for the…

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