Linear Algebra behind the lm() function in R

This post comes out of the blue, nearly 2 years since my last one. I realize I’ve been lazy, so here’s hoping I move from an inertia of rest to that of motion, implying, regular and (hopefully) relevant posts. I also chanced upon some wisdom while scrolling through my Twitter feed:

This blog post in particular was meant to be a reminder to myself and other R users that the much used lm() function in R (for fitting linear models) can be replaced with some handy matrix operations to obtain regression coefficients, their standard errors and other goodness-of-fit stats printed out when summary() is called on an lm object.

Linear regression can be formulated mathematically as follows:
$\mathbf{y} = \mathbf{X} \mathbf{\beta} + \mathbf{\epsilon}$,
$\mathbf{\epsilon} \sim N(0, \sigma^2 \mathbf{I})$

$\mathbf{y}$ is the $\mathbf{n}\times \mathbf{1}$ outcome variable and $\mathbf{X}$ is the $\mathbf{n}\times \mathbf{(\mathbf{k}+1)}$ data matrix of independent predictor variables (including a vector of ones corresponding to the intercept). The ordinary least squares (OLS) estimate for the vector of coefficients $\mathbf{\beta}$ is:

$\hat{\mathbf{\beta}} = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \mathbf{y}$

The covariance matrix can be obtained with some handy matrix operations:
$\textrm{Var}(\hat{\mathbf{\beta}}) = (\mathbf{X}^{\prime} \mathbf{X})^{-1} \mathbf{X}^{\prime} \;\sigma^2 \mathbf{I} \; \mathbf{X} (\mathbf{X}^{\prime} \mathbf{X})^{-1} = \sigma^2 (\mathbf{X}^{\prime} \mathbf{X})^{-1}$
given that $\textrm{Var}(AX) = A \times \textrm{Var}X \times A^{\prime}; \textrm{Var}(\mathbf{y}) = \mathbf{\sigma^2}$

The standard errors of the coefficients are basically $\textrm{Diag}(\sqrt{\textrm{Var}(\hat{\mathbf{\beta}})}) = \textrm{Diag}(\sqrt{\sigma^2 (\mathbf{X}^{\prime} \mathbf{X})^{-1}})$ and with these, one can compute the t-statistics and their corresponding p-values.

Lastly, the F-statistic and its corresponding p-value can be calculated after computing the two residual sum of squares (RSS) statistics:

• $\mathbf{RSS}$ – for the full model with all predictors
• $\mathbf{RSS_0}$ – for the partial model ($\mathbf{y} = \mathbf{\mu} + \mathbf{\nu}; \mathbf{\mu} = \mathop{\mathbb{E}}[\mathbf{y}]; \mathbf{\nu} \sim N(0, \sigma_0^2 \mathbf{I})$) with the outcome observed mean as estimated outcome

$\mathbf{F} = \frac{(\mathbf{RSS_0}-\mathbf{RSS})/\mathbf{k}}{\mathbf{RSS}/(\mathbf{n}-\mathbf{k}-1)}$

I wrote some R code to construct the output from summarizing lm objects, using all the math spewed thus far. The data used for this exercise is available in R, and comprises of standardized fertility measures and socio-economic indicators for each of 47 French-speaking provinces of Switzerland from 1888. Try it out and see for yourself the linear algebra behind linear regression.

view raw lm_linear_algebra.R hosted with ❤ by GitHub

Hope this was useful and worth your time!

MITx: 6.008.1x Computational Probability and Inference

I got really interested in Computational Probability and Inference (6.008.1x) for the following reasons:

1. I love probability and have solved countless problems on probability ever since I learned math
2. …and yet I’ve never coded up probabilistic models!
3. The assignments and project work for this course are to be implemented in Python!

You don’t need to have prior experience in either probability or inference, but you should be comfortable with basic Python programming and calculus.

WHAT YOU’LL LEARN
– Basic discrete probability theory
– Graphical models as a data structure for representing probability distributions
– Algorithms for prediction and inference
– How to model real-world problems in terms of probabilistic inference

The course started on September 12, is 12-weeks long and is structured in the following manner:

Week 1 (9/12 – 9/16): Introduction to probability and computation
A first look at basic discrete probability, how to interpret it, what probability spaces and random variables are, and how to code these up and do basic simulations and visualizations.

Week 2 (9/19 – 9/23): Incorporating observations
Incorporating observations using jointly distributed random variables and using events. Three classic probability puzzles are presented to help elucidate how to interpret probability: Simpson’s paradox, Monty Hall, boy or girl paradox.

Week 3 (9/26 – 9/30): Introduction to inference, structure in distributions, and information measures
The product rule and inference with Bayes’ theorem. Independence: A structure in distributions. Measures of randomness: entropy and information divergence. Mutual information.

Week 4 (10/3 – 10/7): Expectations, and driving to infinity in modeling uncertainty
Expected values of random variables. Classic puzzle: the two envelope problem. Probability spaces and random variables that take on a countably infinite number of values and inference with these random variables.

Week 5 (10/10 – 10/14): Efficient representations of probability distributions on a computer
Introduction to undirected graphical models as a data structure for representing probability distributions and the benefits/drawbacks of these graphical models. Incorporating observations with graphical models.

Week 6 (10/17 – 10/21): Inference with graphical models, part I
Computing marginal distributions with graphical models in undirected graphical models including hidden Markov models..

Week 7 (10/24 – 10/28): Inference with graphical models, part II
Computing most probable configurations with graphical models including hidden Markov models.

Week 8 (10/31 – 11/4): Introduction to learning probability distributions
Learning an underlying unknown probability distribution from observations using maximum likelihood. Three examples: estimating the bias of a coin, the German tank problem, and email spam detection.

Week 9 (11/7 – 11/11): Parameter estimation in graphical models
Given the graph structure of an undirected graphical model, we examine how to estimate all the tables associated with the graphical model.

Week 10 (11/14 – 11/18): Model selection with information theory
Learning both the graph structure and the tables of an undirected graphical model with the help of information theory. Mutual information of random variables.

Week 11 (11/21 – 11/25): Final project
Final project assigned

Week 12 (11/28 – 12/2): Final project

I’m SO taking this course. Hope this interests you as well!

Implementing Undirected Graphs in Python

There are 2 popular ways of representing an undirected graph.

Each list describes the set of neighbors of a vertex in the graph.

The elements of the matrix indicate whether pairs of vertices are adjacent or not in the graph.

Here’s an implementation of the above in Python:

view raw graphUndirected.py hosted with ❤ by GitHub

Output:

 {} {} ["A:['B', 'C', 'E']", "C:['A', 'B', 'D', 'E']", "B:['A', 'C', 'D']", "E:['A', 'C']", "D:['B', 'C']"] [[ 0. 1. 1. 0. 1.] [ 1. 0. 1. 1. 0.] [ 1. 1. 0. 1. 1.] [ 0. 1. 1. 0. 0.] [ 1. 0. 1. 0. 0.]]

Computing Work Done (Total Pivot Comparisons) by Quick Sort

A key aspect of the Quick Sort algorithm is how the pivot element is chosen. In my earlier post on the Python code for Quick Sort, my implementation takes the first element of the unsorted array as the pivot element.

However with some mathematical analysis it can be seen that such an implementation is O(n2) in complexity while if a pivot is randomly chosen, the Quick Sort algorithm is O(nlog2n).

To witness this in action, one can measure the work done by the algorithm comparing two cases, one with a randomized pivot choice – and one with a fixed pivot choice, say the first element of the array (or the last element of the array).

Implementation

A decent proxy for the amount of work done by the algorithm would be the number of pivot comparisons. These comparisons needn’t be computed one-by-one, rather when there is a recursive call on a subarray of length m, you should simply add m−1 to your running total of comparisons.

3 Cases

To put things in perspective, let’s look at 3 cases. (This is basically straight out of a homework assignment from Tim Roughgarden’s course on the Design and Analysis of Algorithms).
Case I with the pivot being the first element.
Case II with the pivot being the last element.
Case III using the “median-of-three” pivot rule. The primary motivation behind this rule is to do a little bit of extra work to get much better performance on input arrays that are nearly sorted or reverse sorted.

Median-of-Three Pivot Rule

Consider the first, middle, and final elements of the given array. (If the array has odd length it should be clear what the “middle” element is; for an array with even length 2k, use the kth element as the “middle” element. So for the array 4 5 6 7, the “middle” element is the second one —- 5 and not 6! Identify which of these three elements is the median (i.e., the one whose value is in between the other two), and use this as your pivot.

Python Code

This file contains all of the integers between 1 and 10,000 (inclusive, with no repeats) in unsorted order. The integer in the ith row of the file gives you the ith entry of an input array. I downloaded this file and named it QuickSort_List.txt

You can run the code below and see for yourself that the number of comparisons for Case III are 138,382 compared to 162,085 and 164,123 for Case I and Case II respectively. You can play around with the code in an IPython / Jupyter notebook here.

 #!/usr/bin/env # Case I # First element of the unsorted array is chosen as pivot element for sorting using Quick Sort def countComparisonsWithFirst(x): """ Counts number of comparisons while using Quick Sort with first element of unsorted array as pivot """ global count_pivot_first if len(x) == 1 or len(x) == 0: return x else: count_pivot_first += len(x)-1 i = 0 for j in range(len(x)-1): if x[j+1] < x[0]: x[j+1],x[i+1] = x[i+1], x[j+1] i += 1 x[0],x[i] = x[i],x[0] first_part = countComparisonsWithFirst(x[:i]) second_part = countComparisonsWithFirst(x[i+1:]) first_part.append(x[i]) return first_part + second_part # Case II # Last element of the unsorted array is chosen as pivot element for sorting using Quick Sort def countComparisonsWithLast(x): """ Counts number of comparisons while using Quick Sort with last element of unsorted array as pivot """ global count_pivot_last if len(x) == 1 or len(x) == 0: return x else: count_pivot_last += len(x)-1 x[0],x[-1] = x[-1],x[0] i = 0 for j in range(len(x)-1): if x[j+1] < x[0]: x[j+1],x[i+1] = x[i+1], x[j+1] i += 1 x[0],x[i] = x[i],x[0] first_part = countComparisonsWithLast(x[:i]) second_part = countComparisonsWithLast(x[i+1:]) first_part.append(x[i]) return first_part + second_part # Case III # Median-of-three method used to choose pivot element for sorting using Quick Sort def middle_index(x): """ Returns the index of the middle element of an array """ if len(x) % 2 == 0: middle_index = len(x)/2 - 1 else: middle_index = len(x)/2 return middle_index def median_index(x,i,j,k): """ Returns the median index of three when passed an array and indices of any 3 elements of that array """ if (x[i]-x[j])*(x[i]-x[k]) < 0: return i elif (x[j]-x[i])*(x[j]-x[k]) < 0: return j else: return k def countComparisonsMedianOfThree(x): """ Counts number of comparisons while using Quick Sort with median-of-three element is chosen as pivot """ global count_pivot_median if len(x) == 1 or len(x) == 0: return x else: count_pivot_median += len(x)-1 k = median_index(x, 0, middle_index(x), -1) if k != 0: x[0], x[k] = x[k], x[0] i = 0 for j in range(len(x)-1): if x[j+1] < x[0]: x[j+1],x[i+1] = x[i+1], x[j+1] i += 1 x[0],x[i] = x[i],x[0] first_part = countComparisonsMedianOfThree(x[:i]) second_part = countComparisonsMedianOfThree(x[i+1:]) first_part.append(x[i]) return first_part + second_part ##################################################################### # initializing counts count_pivot_first = 0; count_pivot_last = 0; count_pivot_median = 0 ##################################################################### # Cast I # Read the contents of the file into a Python list NUMLIST_FILENAME = "QuickSort_List.txt" inFile = open(NUMLIST_FILENAME, 'r') with inFile as f: numList = [int(integers.strip()) for integers in f.readlines()] # call functions to count comparisons countComparisonsWithFirst(numList) ##################################################################### # Read the contents of the file into a Python list NUMLIST_FILENAME = "QuickSort_List.txt" inFile = open(NUMLIST_FILENAME, 'r') with inFile as f: numList = [int(integers.strip()) for integers in f.readlines()] # call functions to count comparisons countComparisonsWithLast(numList) ##################################################################### # Read the contents of the file into a Python list NUMLIST_FILENAME = "QuickSort_List.txt" inFile = open(NUMLIST_FILENAME, 'r') with inFile as f: numList = [int(integers.strip()) for integers in f.readlines()] # call functions to count comparisons countComparisonsMedianOfThree(numList) ##################################################################### print count_pivot_first, count_pivot_last, count_pivot_median
view raw countComparisons.py hosted with ❤ by GitHub

Generating Permutation Matrices in Octave / Matlab

I have been doing Gilbert Strang’s linear algebra assignments, some of which require you to write short scripts in MatLab, though I use GNU Octave (which is kind of like a free MatLab). I was trying out this problem:

To solve this quickly, it would have been nice to have a function that would give a list of permutation matrices for every n-sized square matrix, but there was none in Octave, so I wrote a function permMatrices which creates a list of permutation matrices for a square matrix of size n.

 % function to generate permutation matrices given the size of the desired permutation matrices function x = permMatrices(n) x = zeros(n,n,factorial(n)); permutations = perms(1:n); for i = 1:size(x,3) x(:,:,i) = eye(n)(permutations(i,:),:); end endfunction
view raw permMatrices.m hosted with ❤ by GitHub

For example:

The MatLab / Octave code to solve this problem is shown below:

 % Solution for part (a) p = permMatrices(3); n = size(p,3); % number of permutation matrices v = zeros(n,1); % vector of zeros with dimension equalling number of permutation matrices % check for permutation matrices other than identity matrix with 3rd power equalling identity matrix for i = 1:n if p(:,:,i)^3 == eye(3) v(i,1) = 1; end end v(1,1) = 0; % exclude identity matrix ans1 = p(:,:,v == 1) % Solution for part (b) P = permMatrices(4); m = size(P,3); % number of permutation matrices t = zeros(m,1); % vector of zeros with dimension equalling number of permutation matrices % check for permutation matrices with 4th power equalling identity matrix for i = 1:m if P(:,:,i)^4 == eye(4) t(i,1) = 1; end end % print the permutation matrices ans2 = P(:,:,t == 0)
view raw Section2_7_13.m hosted with ❤ by GitHub

Output:

Sherlock and the Beast – HackerRank

I found myself stuck on this problem recently. I must confess, I lost a couple of hours trying to get to figure the logic for this one. Here’s the problem:

I’ve written 2 functions to solve this problem. The first one I used for smaller N, say N < 30 and the second one for N > 30. The second function is elegant, and it relies on the mathematical property that if a number N is not divisible by 3, it could either leave a remainder 1 or 2.

If it leaves a remainder 2, then subtracting 5 once would make the number divisible by 3. If it leaves a remainder 1, then subtracting 5 twice would make the number divisible by 3.

We subtract 5 from N iteratively and attempt to divide N into 2 parts, one divisible by 3 and the other divisible by 5. We want the part that is divisible by 3 to be the larger part, so that the associated Decent Number is the largest possible. This explanation might seem obtuse, but if you get pen down on paper, you’ll understand what I mean.

Solution

Karatsuba Multiplication Algorithm – Python Code

Motivation for this blog post

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

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

The Karatsuba Multiplication Algorithm

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

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

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

A More General Treatment

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

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

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

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

where

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

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

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

which holds since

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

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

Example

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

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

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

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

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

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

Pseudocode and Python code

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

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

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

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

First, the problem:

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

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

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

Problem

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

Algorithm

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

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

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

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

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

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

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

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

That’s all there is to the problem!

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

view raw euler68.py hosted with ❤ by GitHub

Ans: 6531031914842725

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)

 # 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)

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