# 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!

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
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:

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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 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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
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!
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):

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 # 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

# 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:

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 # 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

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 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
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