# Math and Computing Olympiad Practice

Happy New Year, everyone! Let me tell you about the epic New Year's Eve that I had. I got into a fight with the last problem from the December 2016 USA Computing Olympiad contest. I struggled mightily, felt beaten down at times, lost all hope, but I finally overcame. It was a marathon. We started sparring around noon, and I did not vanquish my foe until the final hour of 2017.

Having a long weekend in a strange new city, I've had to get creative with things to do. I decided to tackle some olympiad problems. For those who are not familiar with the competitive math or programming scene, the USAMO and USACO are math and programming contests targeted towards high school students.

So, as an old man, what am I doing whittling away precious hours tackling these problems? I wish that I could say that I was reliving my glory days from high school. But truth be told, I've always been a lackluster student, who did the minimal effort necessary. I can't recall ever having written a single line of code in high school, and I maybe solved 2 or 3 AIME problems (10 years later, I can usually do the first 10 with some consistency, the rest are a toss-up). Of course, I never trained for the competitions, so who knows if I could have potentially have done well.

We all have regrets from our youth. For me, I have all the familiar ones: mistreating people, lost friends, not having the best relationship with my parents, losing tennis matches, quitting the violin, and of course, the one girl that got away. However, what I really regret the most was not having pursued math and computer science earlier. I'm not sure why. Even now, 10 years older, it's quite clear that I am not talented enough to have competed in the IMO or IOI: I couldn't really hack it as a mathematician, and you can scroll to the very bottom to see my score of 33.

Despite the lack of talent, I just really love problem solving. In many ways it's become an escape for me when I feel lonely and can't make sense of the world. I can get lost in my own abstract world and forget about my physical needs like sleep or food. On solving a problem, I wake up from my stupor, realize that the world has left me behind, and find everyone suddenly married with kids.

There is such triumph in solving a hard problem. Of course, there are times of struggle and hopelessness. Such is my fledging sense of self-worth that it depends on my ability to solve abstract problems that have no basis in reality. Thus, I want to write up my solution to Robotic Cow Herd and 2013 USAMO Problem 2

## Robotic Cow Herd

In the Platinum Division December 2016 contest, there were 3 problems. In contest, I was completely stuck on Lots of Triangles and never had a chance to look at the other 2 problems. This past Friday, I did Team Building in my own time. It took me maybe 3 hours, so I suspect if I started with that problem instead, I could have gotten a decent amount of points on the board.

Yesterday, I attempted Robotic Cow Herd. I was actually able to solve this problem on my own, but I worked on it on and off over a period of 12 hours, so I definitely wouldn't have scored anything in this case.

My solution is quite different than the given solution, which uses binary search. I did actually consider such a solution, but only gave it 5 minutes of though before abandoning it, far too little time to work out the details. Instead, my solution is quite similar to the one that they describe using priority queue before saying such a solution wouldn't be feasible. However, if we are careful about how we fill our queue it can work.

We are charged with assembling $K$ different cows that consist of $N$ components, where each component will have $M$ different types. Each type of component has an associated cost, and cow $A$ is different than cow $B$ if at least one of the components is of a different type.

Of course, we aren't going to try all $M^N$ different cows. It's clear right away that we can take greedy approach, start with the cheapest cow, and get the next cheapest cow by varying a single component. Since each new cow that we make is based on a previous cow, it's only necessary to store the deltas rather than all $N$ components. Naturally, this gives way to a tree representation shown in the title picture.

Each node is a cow prototype. We start with the cheapest cow as the root, and each child consists of a single delta. The cost of a cow can be had by summing the deltas from the root to the node. Now every prototype gives way to $N$ new possible prototypes. $NK$ is just too much to fit in a priority queue. Hence, the official solution says this approach isn't feasible.

However, if we sort our components in the proper order, we know the next two cheapest cows based off this prototype. Moreover, we have to handle a special case, where instead of a cow just generating children, it also generates a sibling. We sort by increasing deltas. In the given sample data, our base cost is $4$, and our delta matrix (not a true matrix) looks like $$\begin{pmatrix} 1 & 0 \\ 2 & 1 & 2 & 2\\ 2 & 2 & 5 \end{pmatrix}.$$

Also, we add our microcontrollers in increasing order to avoid double counting. Now, if we have just added microcontroller $(i,j)$, the cheapest thing to do is to change it to $(i + 1, 0)$ or $(i, j + 1)$. But what about the case, where we want to skip $(i+1,0)$ and add $(i + 2, 0), (i+3,0),\ldots$? Since we're lazy about pushing into our priority queue and only add one child at a time, when a child is removed, we add its sibling in this special case where $j = 0$.

Parent-child relationships are marked with solid lines. Creation of a node is marked with a red arrow. Nodes still in the queue are blue. The number before the colon denotes the rank of the cow. In this case, the cost for 10 cows is $$4 + 5 + 5 + 6 + 6 + 7 + 7 + 7 + 7 + 7 = 61.$$

Dashed lines represent the special case of creating a sibling. The tuple $(1,-,0)$ means we used microcontrollers $(0,1)$ and $(2,0)$. For component $1$, we decided to just use cheapest one. Here's the code.

import java.io.*;
import java.util.*;

public class roboherd {
/**
* Microcontrollers are stored in a matrix-like structure with rows and columns.
* Use row-first ordering.
*/
private static class Position implements Comparable<Position> {
private int row;
private int column;

public Position(int row, int column) {
this.row = row; this.column = column;
}

public int getRow() { return this.row; }

public int getColumn() { return this.column; }

public int compareTo(Position other) {
if (this.getRow() != other.getRow()) return this.getRow() - other.getRow();
return this.getColumn() - other.getColumn();
}

@Override
public String toString() {
return "{" + this.getRow() + ", " + this.getColumn() + "}";
}
}

/**
* Stores the current cost of a cow along with the last microcontroller added. To save space,
* states only store the last delta and obscures the rest of the state in the cost variable.
*/
private static class MicrocontrollerState implements Comparable<MicrocontrollerState> {
private long cost;
private Position position; // the position of the last microcontroller added

public MicrocontrollerState(long cost, Position position) {
this.cost = cost;
this.position = position;
}

public long getCost() { return this.cost; }

public Position getPosition() { return this.position; }

public int compareTo(MicrocontrollerState other) {
if (this.getCost() != other.getCost()) return (int) Math.signum(this.getCost() - other.getCost());
return this.position.compareTo(other.position);
}
}

public static void main(String[] args) throws IOException {
PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("roboherd.out")));
int N = Integer.parseInt(st.nextToken()); // number of microcontrollers per cow
int K = Integer.parseInt(st.nextToken()); // number of cows to make
assert 1 <= N && N <= 100000 : N;
assert 1 <= K && K <= 100000 : K;
ArrayList<int[]> P = new ArrayList<int[]>(N); // microcontroller cost deltas
long minCost = 0; // min cost to make all the cows wanted
for (int i = 0; i < N; ++i) {
int M = Integer.parseInt(st.nextToken());
assert 1 <= M && M <= 10 : M;
int[] costs = new int[M];
for (int j = 0; j < M; ++j) {
costs[j] = Integer.parseInt(st.nextToken());
assert 1 <= costs[j] && costs[j] <= 100000000 : costs[j];
}
Arrays.sort(costs);
minCost += costs[0];
// Store deltas, which will only exist if there is more than one type of microcontroller.
if (M > 1) {
int[] costDeltas = new int[M - 1];
for (int j = M - 2; j >= 0; --j) costDeltas[j] = costs[j + 1] - costs[j];
}
}
in.close();
N = P.size(); // redefine N to exclude microcontrollers of only 1 type
--K; // we already have our first cow
// Identify the next best configuration in log(K) time.
PriorityQueue<MicrocontrollerState> pq = new PriorityQueue<MicrocontrollerState>(3*K);
// Order the microcontrollers in such a way that if we were to vary the prototype by only 1,
// the best way to do would be to pick microcontrollers in the order
// (0,0), (0,1),...,(0,M_0-2),(1,0),...,(1,M_1-2),...,(N-1,0),...,(N-1,M_{N-1}-2)
Collections.sort(P, new Comparator<int[]>() {
@Override
public int compare(int[] a, int[] b) {
for (int j = 0; j < Math.min(a.length, b.length); ++j)
if (a[j] != b[j]) return a[j] - b[j];
return a.length - b.length;
}
});
pq.add(new MicrocontrollerState(minCost + P.get(0)[0], new Position(0, 0)));
// Imagine constructing a tree with K nodes, where the root is the cheapest cow. Each node contains
// the delta from its parent. The next cheapest cow can always be had by taking an existing node on
// the tree and varying a single microcontroller.
for (; K > 0; --K) {
MicrocontrollerState currentState = pq.remove(); // get the next best cow prototype.
long currentCost = currentState.getCost();
minCost += currentCost;
int i = currentState.getPosition().getRow();
int j = currentState.getPosition().getColumn();
// Our invariant to avoid double counting is to only add microcontrollers with "greater" position.
// Given a prototype, from our ordering, the best way to vary a single microcontroller is replace
// it with (i,j + 1) or add (i + 1, 0).
if (j + 1 < P.get(i).length) {
pq.add(new MicrocontrollerState(currentCost + P.get(i)[j + 1], new Position(i, j + 1)));
}
if (i + 1 < N) {
// Account for the special case, where we just use the cheapest version of type i microcontrollers.
// Thus, we remove i and add i + 1. This is better than preemptively filling the priority queue.
if (j == 0) pq.add(new MicrocontrollerState(
currentCost - P.get(i)[j] + P.get(i + 1)[0], new Position(i + 1, 0)));
pq.add(new MicrocontrollerState(currentCost + P.get(i + 1)[0], new Position(i + 1, 0)));
}
}
out.println(minCost);
out.close();
}
}


Sorting is $O(NM\log N)$. Polling from the priority queue is $O(K\log K)$ since each node will at most generate 3 additional nodes to put in the priority queue. So, total running time is $O(NM\log N + K\log K)$.

## 2013 USAMO Problem 2

Math has become a bit painful for me. While it was my first love, I have to admit that a bachelor's and master's degree later, I'm a failed mathematician. I've recently overcome my disappointment and decided to persist in learning and practicing math despite my lack of talent. This is the first USAMO problem that I've been able to solve, which I did on Friday. Here's the problem.

For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$.

The solution on the AOPS wiki uses tiling. I use a different strategy that leads to the same result.

Let the points on the cricle be $P_1,P_2, \ldots,P_n$. First, we prove that each point on the circle is visited either $1$ or $2$ times, except for $A = P_1$, which can be visited $3$ times since it's our starting and ending point. It's clear that $2$ times is upper bound for the other points. Suppose a point is never visited, though. We can only move in increments of $1$ and $2$, so if $P_k$ was never visited, we have made a move of $2$ steps from $P_{k-1}$ twice, which is not allowed.

In this way, we can index our different paths by tuples $(m_1,m_2,\ldots,m_n)$, where $m_i$ is which move we make the first time that we visit $P_i$, so $m_i \in \{1,2\}$. Since moves have to be distinct, the second move is determined by the first move. Thus, we have $2^n$ possible paths.

Here are examples of such paths.

Both paths are valid in the sense that no move is repeated. However, we only count the one on the left since after two cycles we must return to $P_1$.

The path on the left is $(1,2,2,2,1)$, which is valid since we end up at $A = P_1$. The path on the right is $(1,1,1,1,1)$, which is invalid, since miss $A = P_1$ the second time. The first step from a point is black, and the second step is blue. The edge labels are the order in which the edges are traversed.

Now, given all the possible paths with distinct moves for a circle with $n - 1$ points, we can generate all the possible paths for a circle with $n$ points by appending a $1$ or a $2$ to the $n - 1$ paths if we consider their representation as a vector of length $n - 1$ of $1$s and $2$s. In this way, the previous $2^{n-1}$ paths become $2^n$ paths.

Now, we can attack the problem in a case-wise manner.

1. Consider an invalid path, $(m_1,m_2,\ldots,m_{n-1})$. All these paths must land at $P_1$ after the first cycle around the circle. Why? Since the path is invalid, that means we touch $P_{n-1}$ in the second cycle and make a jump over $P_1$ by moving $2$ steps. Thus, if we previously touched $P_{n-1},$ we moved to $P_1$ since moves must be distinct. If the first time that we touch $P_{n-1}$ is the second cycle, then, we jumped over it in first cycle by going moving $P_{n-2} \rightarrow P_1$.
1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is now valid. $P_n$ is now where $P_1$ would have been in the first cycle, so we hit $P_n$ and move to $P_1$. Then, we continue as we normally did. Instead of ending like $P_{n-1} \rightarrow P_2$ by jumping over $P_1$, we jump over $P_n$ instead, so we end up making the move $P_{n-1} \rightarrow P_1$ at the end.
2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is now valid. This case is easier. We again touch $P_n$ in the first cycle. Thus, next time we hit $P_n$, we'll make the move $P_n \rightarrow P_1$ since we must make distinct moves. If we don't hit $P_n$ again, that means we jumped $2$ from $P_{n-1}$, which means that we made the move $P_{n - 1} \rightarrow P_1$.
2. Consider an existing valid path, now, $(m_1,m_2,\ldots,m_{n-1})$. There are $a_{n-1}$ of these.

1. Let it be a path where we touch $P_1$ $3$ times.
1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is invalid. $P_n$ will be where $P_1$ was in the first cycle. So, we'll make the move $P_n \rightarrow P_1$ and continue with the same sequence of moves as before. But instead of landing at $P_1$ when the second cycle ends, we'll land at $P_n$, and jump over $P_1$ by making the move $P_n \rightarrow P_2$.
2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is valid. Again, we'll touch $P_n$ in the first cycle, so the next time that we hit $P_n$, we'll move to $P_1$. If we don't touch $P_n$ again, we jump over it onto $P_1$, anyway, by moving $P_{n-1} \rightarrow P_1$.
2. Let it be a path where we touch $P_1$ $2$ times.

1. Make it $(m_1,m_2,\ldots,m_{n-1}, 1)$. This path is valid. Instead of jumping over $P_1$ at the end of the first cycle, we'll be jumping over $P_n$. We must touch $P_n$, eventually, so from there, we'll make the move $P_n \rightarrow P_1$.
2. Make it $(m_1,m_2,\ldots,m_{n-1}, 2)$. This path is invalid. We have the same situation where we skip $P_n$ the first time. Then, we'll have to end up at $P_n$ the second time and make the move $P_{n} \rightarrow P_2$.

In either case, old valid paths lead to $1$ new valid path and $1$ new invalid path.

Thus, we have that $a_n = 2^n - a_{n-1} \Rightarrow \boxed{a_{n - 1} + a_n = 2^n}$ for $n \geq 4$ since old invalid paths lead to $2$ new valid paths and old valid paths lead to $1$ new valid path. And actually, this proof works when $n \geq 3$ even though the problem only asks for $n \geq 4$. Since we have $P_{n-2} \rightarrow P_1$ at one point in the proof, anything with smaller $n$ is nonsense.

Yay, 2017!

# Welcoming 2017

It's been a long time since I posted anything. Mostly because I haven't been up to anything interesting, lately. I don't really cook anymore, but I did manage to bake these Double Chocolate Cookies. I love chocolate, but I think these were too much for me.

For Thanksgiving, I got to see my older cousin Daniel Pham. When I was younger, I thought that he was the coolest guy ever. He played high school football and went to Stanford. Still a pretty cool guy, but I guess through no fault of his own, he could never live up to the "colossal vitality of [my] illusion." Or I guess life just gets more boring once you have kids. By the way, recognize the quote?

I remade my Apple Pie for the holiday. It was definitely easier and came out better this time around. There's still some work that I could do to make the crust flakier. Maybe I really need the vodka? Given enough opportunities I hope to figure it out.

Recently, I was back in Philly for the first time since I moved to Seattle. It was great to see family again. Amish and I don't really agree on foods very much, for he hates fish sauce, a staple of Vietnamese cuisine. However, we both like Mozzarella Sticks. I guess it might be a tradition to make these with him when I'm back in Philly

Anyway, it's New Year's Eve. Given that I'm in a strange new city, instead of ringing in the new year with loved ones, I have a quiet night and extended weekend to reflect on this new chapter of my life in Seattle.

Since I moved here for work, perhaps it's no surprise that I spend most of my time here working, and admittedly, I quite enjoy my job. It seems that software engineering rather suits me. Given a problem, I often find myself obsessed with solving it and unable to put it aside. Then, there's that rush of dopamine when your code passes the tests and everything changes from red to green. Admittedly, most of these problems are rather meaningless, but living "with the bearing of one who was going to give his days and nights to Ecclesiastes for ever", these bugs and math problems have become a kind of escape for me.

We all imagine ourselves as heroes of our personal story. Lost in the abstract world of code and math, I seem to have forgotten that everyone else has a story, too, not just me. I read something of that sort in a book once, where we always expect everyone around us to stay as static characters while we, personally, plow on with our story and overcome many obstacles. Perhaps, this is why mothers cry when their children go to college, or every family gathering, we're shocked how old are nieces and nephews are. When I was in college, I remember we would joke about certain friends, saying "I could never see him/her married", or "could you imagine so-and-so as a doctor?" Not in a sexist or racist way, but they were such jokesters or so irresponsible, it was hard to imagine that scenario.

Seeing my cousin Daniel and visiting Philly, I've started to see this lack of imagination in myself. Everyone has rich and varied lives that proceed with or without me, and to be honest, I'm mostly a nonfactor in their stories. I guess there's a selfish part of me that makes it hard to accept changes that I wasn't directly part of. Buried under work, I've finally emerged into a new world where everyone is getting married, having kids, and moving to new cities.

We sort of imagine a life for ourselves evolving as a person by forming new relationships, getting new hobbies, or advancing our career, yet for some reason, we're often taken aback when others' lives evolve in the same way. I guess that's because we often describe our loved ones as our rocks that we rely on when times are tough. Unfotunately, these rocks are quite amorphous, and we can't exactly expect that person to fit into the neat little box that we made for them in our mind.

In some ways, this is a good thing. I'm continually amazed by what others accomplish. I might remember them as a struggling college student or socially awkward. It's great to see people exceed expectations. In other ways, it's disappointing. People that you thought were close friends will go ghost and disappear.

Anyway, I'm not sure if there's any point to this rambling. I don't really have any resolutions for 2017. I guess that I'm just becoming more cognizant of how little that I understand about the world around me.

# Policy-Based Data Structures in C++

Certain problems in competitive programming call for more advanced data structure than our built into Java's or C++'s standard libraries. Two examples are an order statistic tree and a priority queue that lets you modify priorities. It's questionable whether these implementations are useful outside of competitive programming since you could just use Boost.

## Order Statistic Tree

Consider the problem ORDERSET. An order statistic tree trivially solves this problem. And actually, implementing an order statistic tree is not so difficult. You can find the implementation here. Basically, you have a node invariant

operator()(node_iterator node_it, node_const_iterator end_nd_it) const {
node_iterator l_it = node_it.get_l_child();
const size_type l_rank = (l_it == end_nd_it) ? 0 : l_it.get_metadata();

node_iterator r_it = node_it.get_r_child();
const size_type r_rank = (r_it == end_nd_it) ? 0 : r_it.get_metadata();

}


where each node contains a count of nodes in its subtree. Every time you insert a new node or delete a node, you can maintain the invariant in $O(\log N)$ time by bubbling up to the root.

With this extra data in each node, we can implement two new methods, (1) find_by_order and (2) order_of_key. find_by_order takes a nonnegative integer as an argument and returns the node corresponding to that index, where are data is sorted and we use $0$-based indexing.

find_by_order(size_type order) {
node_iterator it = node_begin();
node_iterator end_it = node_end();

while (it != end_it) {
node_iterator l_it = it.get_l_child();
const size_type o = (l_it == end_it)? 0 : l_it.get_metadata();

if (order == o) {
return *it;
} else if (order < o) {
it = l_it;
} else {
order -= o + 1;
it = it.get_r_child();
}
}

return base_type::end_iterator();
}


It works recursively like this. Call the index we're trying to find $k$. Let $l$ be the number of nodes in the left subtree.

• $k = l$: If you're trying to find the $k$th-indexed element, then there will be $k$ nodes to your left, so if the left child has $k$ elements in its subtree, you're done.
• $k < l$: The $k$-indexed element is in the left subtree, so replace the root with the left child.
• $k > l$: The $k$ indexed element is in the right subtree. It's equivalent to looking for the $k - l - 1$ element in the right subtree. We subtract away all the nodes in the left subtree and the root and replace the root with the right child.

order_of_key takes whatever type is stored in the nodes as an argument. These types are comparable, so it will return the index of the smallest element that is greater or equal to the argument, that is, the least upper bound.

order_of_key(key_const_reference r_key) const {
node_const_iterator it = node_begin();
node_const_iterator end_it = node_end();

const cmp_fn& r_cmp_fn = const_cast<PB_DS_CLASS_C_DEC*>(this)->get_cmp_fn();
size_type ord = 0;
while (it != end_it) {
node_const_iterator l_it = it.get_l_child();

if (r_cmp_fn(r_key, this->extract_key(*(*it)))) {
it = l_it;
} else if (r_cmp_fn(this->extract_key(*(*it)), r_key)) {
ord += (l_it == end_it)? 1 : 1 + l_it.get_metadata();
it = it.get_r_child();
} else {
ord += (l_it == end_it)? 0 : l_it.get_metadata();
it = end_it;
}
}
return ord;
}


This is a simple tree traversal, where we keep track of order as we traverse the tree. Every time we go down the right branch, we add $1$ for every node in the left subtree and the current node. If we find a node that it's equal to our key, we add $1$ for every node in the left subtree.

While not entirely trivial, one could write this code during a contest. But what happens when we need a balanced tree. Both Java implementations of TreeSet and C++ implementations of set use a red-black tree, but their APIs are such that the trees are not easily extensible. Here's where Policy-Based Data Structures come into play. They have a mechanism to create a node update policy, so we can keep track of metadata like the number of nodes in a subtree. Conveniently, tree_order_statistics_node_update has been written for us. Now, our problem can be solved quite easily. I have to make some adjustments for the $0$-indexing. Here's the code.

#include <functional>
#include <iostream>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

namespace phillypham {
template<typename T,
typename cmp_fn = less<T>>
using order_statistic_tree =
__gnu_pbds::tree<T,
__gnu_pbds::null_type,
cmp_fn,
__gnu_pbds::rb_tree_tag,
__gnu_pbds::tree_order_statistics_node_update>;
}

int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);

int Q; cin >> Q; // number of queries

phillypham::order_statistic_tree<int> orderStatisticTree;
for (int q = 0; q < Q; ++q) {
char operation;
int parameter;
cin >> operation >> parameter;
switch (operation) {
case 'I':
orderStatisticTree.insert(parameter);
break;
case 'D':
orderStatisticTree.erase(parameter);
break;
case 'K':
if (1 <= parameter && parameter <= orderStatisticTree.size()) {
cout << *orderStatisticTree.find_by_order(parameter - 1) << '\n';
} else {
cout << "invalid\n";
}
break;
case 'C':
cout << orderStatisticTree.order_of_key(parameter) << '\n';
break;
}
}
cout << flush;
return 0;
}


## Dijkstra's algorithm and Priority Queues

Consider the problem SHPATH. Shortest path means Dijkstra's algorithm of course. Optimal versions of Dijkstra's algorithm call for exotic data structures like Fibonacci heaps, which lets us achieve a running time of $O(E + V\log V)$, where $E$ is the number of edges, and $V$ is the number of vertices. In even a fairly basic implementation in the classic CLRS, we need more than what the standard priority queues in Java and C++ offer. Either, we implement our own priority queues or use a slow $O(V^2)$ version of Dijkstra's algorithm.

Thanks to policy-based data structures, it's easy to use use a fancy heap for our priority queue.

#include <algorithm>
#include <climits>
#include <exception>
#include <functional>
#include <iostream>
#include <string>
#include <unordered_map>
#include <utility>
#include <vector>

#include <ext/pb_ds/priority_queue.hpp>

using namespace std;

namespace phillypham {
template<typename T,
typename cmp_fn = less<T>> // max queue by default
class priority_queue {
private:
struct pq_cmp_fn {
bool operator()(const pair<size_t, T> &a, const pair<size_t, T> &b) const {
return cmp_fn()(a.second, b.second);
}
};
typedef typename __gnu_pbds::priority_queue<pair<size_t, T>,
pq_cmp_fn,
__gnu_pbds::pairing_heap_tag> pq_t;
typedef typename pq_t::point_iterator pq_iterator;
pq_t pq;
vector<pq_iterator> map;

public:
class entry {
private:
size_t _key;
T _value;

public:
entry(size_t key, T value) : _key(key), _value(value) {}

size_t key() const { return _key; }

T value() const { return _value; }
};

priority_queue() {}

priority_queue(int N) : map(N, nullptr) {}

size_t size() const {
return pq.size();
}

size_t capacity() const {
return map.size();
}

bool empty() const {
return pq.empty();
}

/**
* Usually, in C++ this returns an rvalue that you can modify.
* I choose not to allow this because it's dangerous, however.
*/
T operator[](size_t key) const {
return map[key] -> second;
}

T at(size_t key) const {
if (map.at(key) == nullptr) throw out_of_range("Key does not exist!");
return map.at(key) -> second;
}

entry top() const {
return entry(pq.top().first, pq.top().second);
}

int count(size_t key) const {
if (key < 0 || key >= map.size() || map[key] == nullptr) return 0;
return 1;
}

pq_iterator push(size_t key, T value) {
// could be really inefficient if there's a lot of resizing going on
if (key >= map.size()) map.resize(key + 1, nullptr);
if (key < 0) throw out_of_range("The key must be nonnegative!");
if (map[key] != nullptr) throw logic_error("There can only be 1 value per key!");
map[key] = pq.push(make_pair(key, value));
return map[key];
}

void modify(size_t key, T value) {
pq.modify(map[key], make_pair(key, value));
}

void pop() {
if (empty()) throw logic_error("The priority queue is empty!");
map[pq.top().first] = nullptr;
pq.pop();

}

void erase(size_t key) {
if (map[key] == nullptr) throw out_of_range("Key does not exist!");
pq.erase(map[key]);
map[key] = nullptr;
}

void clear() {
pq.clear();
fill(map.begin(), map.end(), nullptr);
}
};
}


By replacing __gnu_pbds::pairing_heap_tag with __gnu_pbds::binomial_heap_tag, __gnu_pbds::rc_binomial_heap_tag, or __gnu_pbds::thin_heap_tag, we can try different types of heaps easily. See the priority_queue interface. Unfortunately, we cannot try the binary heap because modifying elements invalidates iterators. Conveniently enough, the library allows us to check this condition dynamically .

#include <iostream>
#include <functional>
#include <ext/pb_ds/priority_queue.hpp>

using namespace std;

int main(int argc, char *argv[]) {
__gnu_pbds::priority_queue<int, less<int>, __gnu_pbds::binary_heap_tag> pq;
cout << (typeid(__gnu_pbds::container_traits<decltype(pq)>::invalidation_guarantee) == typeid(__gnu_pbds::basic_invalidation_guarantee)) << endl;
// prints 1
cout << (typeid(__gnu_pbds::container_traits<__gnu_pbds::priority_queue<int, less<int>, __gnu_pbds::binary_heap_tag>>::invalidation_guarantee) == typeid(__gnu_pbds::basic_invalidation_guarantee)) << endl;
// prints 1
return 0;
}


See the documentation for basic_invalidation_guarantee. We need at least point_invalidation_guarantee for the below code to work since we keep a vector of iterators in our phillypham::priority_queue.

vector<int> findShortestDistance(const vector<vector<pair<int, int>>> &adjacencyList,
int sourceIdx) {
phillypham::priority_queue<int, greater<int>> minDistancePriorityQueue(N);
for (int i = 0; i < N; ++i) {
minDistancePriorityQueue.push(i, i == sourceIdx ? 0 : INT_MAX);
}
vector<int> distances(N, INT_MAX);
while (!minDistancePriorityQueue.empty()) {
phillypham::priority_queue<int, greater<int>>::entry minDistanceVertex =
minDistancePriorityQueue.top();
minDistancePriorityQueue.pop();
distances[minDistanceVertex.key()] = minDistanceVertex.value();
for (pair<int, int> nextVertex : adjacencyList[minDistanceVertex.key()]) {
int newDistance = minDistanceVertex.value() + nextVertex.second;
if (minDistancePriorityQueue.count(nextVertex.first) &&
minDistancePriorityQueue[nextVertex.first] > newDistance) {
minDistancePriorityQueue.modify(nextVertex.first, newDistance);
}
}
}
return distances;
}


Fear not, I ended up using my own binary heap that wrote from Dijkstra, Paths, Hashing, and the Chinese Remainder Theorem. Now, we can benchmark all these different implementations against each other.

int main(int argc, char *argv[]) {
ios::sync_with_stdio(false); cin.tie(NULL);
int T; cin >> T;              // number of tests
for (int t = 0; t < T; ++t) {
int N; cin >> N;            // number of nodes
unordered_map<string, int> cityIdx;
for (int i = 0; i < N; ++i) {
string city;
cin >> city;
cityIdx[city] = i;
int M; cin >> M;
for (int j = 0; j < M; ++j) {
int neighborIdx, cost;
cin >> neighborIdx >> cost;
--neighborIdx; // convert to 0-based indexing
}
}
// compute output
int R; cin >> R;            // number of subtests
for (int r = 0; r < R; ++r) {
string sourceCity, targetCity;
cin >> sourceCity >> targetCity;
int sourceIdx = cityIdx[sourceCity];
int targetIdx = cityIdx[targetCity];
cout << distances[targetIdx] << '\n';
}
}
cout << flush;
return 0;
}


I find that the policy-based data structures are much faster than my own hand-written priority queue.

Algorithm Time (seconds)
PBDS Pairing Heap, Lazy Push 0.41
PBDS Pairing Heap 0.44
PBDS Binomial Heap 0.48
PBDS Thin Heap 0.54
PBDS RC Binomial Heap 0.60
Personal Binary Heap 0.72

Lazy push is small optimization, where we add vertices to the heap as we encounter them. We save a few hundreths of a second at the expense of increased code complexity.

vector<int> findShortestDistance(const vector<vector<pair<int, int>>> &adjacencyList,
int sourceIdx) {
vector<int> distances(N, INT_MAX);
phillypham::priority_queue<int, greater<int>> minDistancePriorityQueue(N);
minDistancePriorityQueue.push(sourceIdx, 0);
while (!minDistancePriorityQueue.empty()) {
phillypham::priority_queue<int, greater<int>>::entry minDistanceVertex =
minDistancePriorityQueue.top();
minDistancePriorityQueue.pop();
distances[minDistanceVertex.key()] = minDistanceVertex.value();
for (pair<int, int> nextVertex : adjacencyList[minDistanceVertex.key()]) {
int newDistance = minDistanceVertex.value() + nextVertex.second;
if (distances[nextVertex.first] == INT_MAX) {
minDistancePriorityQueue.push(nextVertex.first, newDistance);
distances[nextVertex.first] = newDistance;
} else if (minDistancePriorityQueue.count(nextVertex.first) &&
minDistancePriorityQueue[nextVertex.first] > newDistance) {
minDistancePriorityQueue.modify(nextVertex.first, newDistance);
distances[nextVertex.first] = newDistance;
}
}
}
return distances;
}


All in all, I found learning to use these data structures quite fun. It's nice to have such easy access to powerful data structures. I also learned a lot about C++ templating on the way.

# Emma and Other Thoughts

I just finished reading Jane Austen's Emma. Despite it being full of romance, I didn't particularly like this book. Several of the characters practiced various forms of duplicity and were rewarded with happy endings. There seems to be some attempt at lesson warning against youthful arrogance and the obsession with social class, but such admonishments come across as weak since everyone ends up happily ever after anyway.

For some reason the quote that stuck out most to me was made by Frank Churchill:

It is always the lady's right to decide on the degree of acquaintance.

I suppose that this is true. All the gentleman can do is ask, but the lady has the final say. I just found it strange since it seemed to imply that women have the upper hand in dating even in the 1800s.

I recall the quote from William Thackeray's Vanity Fair:

And oh, what a mercy it is that these women do not exercise their powers oftener! We can’t resist them, if they do. Let them show ever so little inclination, and men go down on their knees at once: old or ugly, it is all the same. And this I set down as a positive truth. A woman with fair opportunities, and without an absolute hump, may marry WHOM SHE LIKES. Only let us be thankful that the darlings are like the beasts of the field, and don’t know their own power. They would overcome us entirely if they did.

Another author asserts that women have power over men.

In light of Donald Trump's election, I've been thinking a lot about misogny and feminism. It's clear that women put up with a lot of misogny and objectification from Donald Trump's comments, the Harvard Mens Soccer Team's "Scouting Report", and the harsh judgement lopped on Hilary Clinton for everything from her looks to the way she speaks. I could offer even more evidence like personal anecdotes and the gender pay gap, but this post would go on forever.

Despite all the overwhelming evidence of the challenges that women face, many men and even some women don't find much to like in feminism, for Jill Filipovic writes in The Men Feminists Left Behind that

...young women are soaring, in large part because we are coming of age in a kind of feminist sweet spot: still exhibiting many traditional feminine behaviors — being polite, cultivating meaningful connections, listening and communicating effectively — and finding that those same qualities work to our benefit in the classroom and workplace, opening up more opportunities for us to excel.

The fact of the matter is that while men like Donald Trump and the Harvard Mens Soccer team exhibit despicable behavior, most men are not billionaires or star athletes and are not in the position of power to get away with such actions. Even if we do harbor such hateful attitudes, we're not in a position to act on them, which we may feel absolves us of our guilt. Thus, it can feel that feminist are falsely accusing us of wrongdoing.

On the contrary, a typical young man's interaction with women often puts him in a losing position:

• if the woman is a classmate, she probably has a higher grade as women do much better in school on average.
• in dating, it's rejection after rejection for men. You might go on a few dates, pay for a couple of dinners, and never receive an answer to a text, and
• while female body image problems receive the most attention, male attractiveness is actually judged more harshly according to Dataclysm, where Christian Rudder writes,

When you consider the supermodels, the porn, the cover girls, the Lara Croft–style fembots, the Bud Light ads, and, most devious of all, the Photoshop jobs that surely these men see every day, the fact that male opinion of female attractiveness is still where it’s supposed to be is, by my lights, a small miracle. It’s practically common sense that men should have unrealistic expectations of women’s looks, and yet here we see it’s just not true.

Now, I know many women will tell me that a lot of those men are jerks and deserved to be rejected. All they really wanted was sex. They'll certainly have personal anecdotes of being used for sex or being ghosted themselves. There seems to be some type of selection bias, however, where women focus on men in positions of power. According to Robin W. Simon in Teaching Men to Be Emotionally Honest, boys are "more invested in ongoing romantic relationships," so the typical male actually takes heart break more severely but is left with no outlet for his emotions because men don't feel safe to be emotionally vulnerable.

So, maybe Jane Austen and William Thackeray have a point. Many men can feel rather oppressed and might even see misogyny as somewhat justified on that account. Objectification of women becomes a sort of defense mechanism to make rejection more bearable, for it's easier to take rejection from someone you don't respect. I know that at times, I've had these thoughts.

Now, to be perfectly clear, I no longer feel this way. The issues that women face are very real, and while I will probably never fully understand them, I know there are very real forces of oppression that women fight against, and that feminism is necessary. I wrote this so that women might understand where some men are coming from. I just thought that there could be better understanding between the two sides because it's important for us men to fight for gender equality, too.

Men actually have a lot to gain from gender equality. While women might feel forced to be spend a lot of time on childcare, many men feel forced into careers they might not have chosen otherwise, but for the pressure to provide for a family. In the United States, men commit suicide much more often, and use of mental health resources may be a contributor, for men don't feel comfortable seeking help.

I guess it just occured to me that a lot of the animosity between the two sexes can be attributed to outmoded dating rituals, so maybe we can start there?

# Factor Analysis and Matrix Calculus

Enough about my abs. Back to more important stuff, you know, like math. I've been slowly working my way through Machine Learning: a Probabilistic Perspective.

Since my last post on this topic Nearest Neighbors and Discriminant Analysis, I've gotten to do some cool problems on spam classification, Spam classification using logistic regression and Spam classification with naive Bayes. It's somewhat surprising how logistic regression performs much better than Naive Bayes with less parameters (5.8% versus 11% misclassification rate). Of course, logistic regression is a discriminative model, while Naive Bayes is generative. Estimating the conditional distribution is in some sense a smaller, and hence, easier problem. Generative models do have some advantages, though, especially when there is missing data.

Other problems have been a bit of drag. Now that I'm reading about Latent linear models and Sparse linear models, I've been getting killed by matrix calculus. I've decided to write down some of the more useful identities as a reference to myself. As evidence of how tedious these exercises are, the derivations in the textbooks and solutions manual are riddle with errors.

Some resources:

## Factor Analysis

The basic idea factor analysis model actually seemed quite intuitive when I first saw it. The idea is that underlying data is just a vector independent standard normals, that is, $$\mathbf{z}_i \sim \mathcal{N}(\mathbf{0}, \mathbf{I}_L).$$ However, we actually observe $$\mathbf{x}_i \mid \mathbf{z}_i \sim \mathcal{N}(\boldsymbol\mu + \mathbf{W}\mathbf{z}_i, \boldsymbol\Psi),$$ where $\boldsymbol\Psi$ is diagonal.

In general, I'll denote observed values with $\mathbf{x}_i$ and hidden variables with $\mathbf{z}_i$. Intuitively, I think of $\mathbf{z}_i$ as the "genetics." A single gene may affect many different traits eye color, height, hair color, bicep size, and intelligence. So if we observed 2 genes that affect those 5 traits, the entry $\mathbf{W}_{ij}$ is the effect that gene $j$ has on trait $i$. We'll denote the number of latent factors, the length of $\mathbf{z}_i$, as $L$, the number of observed factors, the length of $\mathbf{x}_i$ as $D$, and finally, the number of observations as $N$. Thus, $\mathbf{W}$ is a $D \times L$ matrix.

But why stop there? To generalize this model, we can consider a mixture of factor analysis models. Now, the underlying data is $(\mathbf{z}_i, q_i),$ where $$q_i \sim \operatorname{Cat}(\pi_1,\pi_2,\ldots,\pi_K),$$ and for each category $k \in \{1,2,\ldots,K\}$, we a separate $\mathbf{W}_k$ and $\boldsymbol\mu_k$, so that $$\mathbf{x}_i \mid (\mathbf{z}_i, q_i = k) \sim \mathcal{N}(\boldsymbol\mu_k + \mathbf{W}_k\mathbf{z}_i, \boldsymbol\Psi).$$

One helpful way to view this is a graphical model, which I've included in the title picture. One can easily sort out the dependencies. The observed variables are shaded. The deterministic parameters are in diamonds. The latent factors are the unshaded circles with thin borders. The parameters that we are trying to estimate are given thick borders, so we want to estimate $\boldsymbol\theta = \left(\mathbf{W}_k, \boldsymbol\mu_k,\boldsymbol\Psi,\pi_k\right)$ in this case.

### Fitting the model with the EM Algorithm

From the graphical model, one can write down the probability or likelihood of the data, $$L(\boldsymbol\theta) = p\left(\mathcal{D} \mid \mathbf{\theta}\right) = \prod_{i=1}^N \mathcal{N}\left(\mathbf{z}_i \mid \mathbf{0}, \mathbf{I}_L\right) \prod_{k^\prime=1}^K\left(\pi_{k^\prime} \mathcal{N}\left(\mathbf{x}_i \mid \mathbf{W}_k\mathbf{z}_i + \boldsymbol\mu_k, \boldsymbol\Psi\right) \right)^{I(q_i = k)}.$$

Typically, one fits models by maximizing this likelihood, or equivalently, the log-likelihood. The problem is that we can't evaluate the function if we don't know $\mathbf{z}_i$ and $q_i$. This is where the expectation–maximization (EM) algorithm comes in. We replace the unknown values with their expectation and then maximize. We do this iteratively until we achieve convergence.

Use the graphical model as a reference, we can plug in the values that are in diamonds or shaded, we are taking the expectation of the terms that involve variables in circles with a thin border, and we are choosing the values for the variables in thick borders such that the likelihood is maximized.

I won't go into the mathematical and convergence properties of this algorithm, but I'll show how it's performed. To take the expectation, we need some value for $\boldsymbol\theta$, so we choose some initial $\boldsymbol\theta_0$. At each iteration, we use $\boldsymbol\theta_l = \left(\pi_k^{(l)},\boldsymbol\mu_k^{(l)},\mathbf{W}_k^{(l)}, \boldsymbol\Psi^{(l)}\right)$ to create a better estimate $\boldsymbol\theta_{l + 1} = \left(\pi_k^{(l+1)},\boldsymbol\mu_k^{(l+1)},\mathbf{W}_k^{(l+1)}, \boldsymbol\Psi^{(l+1)}\right).$

Let's write the log-likelihood \begin{align} l(\boldsymbol\theta) &= \sum_{i=1}^N\log\mathcal{N}\left(\mathbf{z}_i \mid \mathbf{0}, \mathbf{I}_L\right) + \sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[\log\pi_{k} + \log\mathcal{N}\left(\mathbf{x}_i \mid \mathbf{W}_k\mathbf{z}_i + \boldsymbol\mu_k, \boldsymbol\Psi\right)\right] \label{eqn:loglikelihood}\\ &= \sum_{i=1}^N \left[-\frac{L}{2}\log 2\pi -\frac{1}{2}\mathbf{z}_i^\intercal\mathbf{z}_i \right] + \nonumber\\ &\sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[ \log\pi_{k} -\frac{N}{2}\log 2\pi -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left(\mathbf{x}_i - \mathbf{W}_k\mathbf{z}_i - \boldsymbol\mu_k\right)^\intercal{\boldsymbol\Psi}^{-1}\left(\mathbf{x}_i - \mathbf{W}_k\mathbf{z}_i- \boldsymbol\mu_k\right) \right]. \nonumber \end{align}

We need to eliminate anything with $q_i$ and $\mathbf{z}_i$ by taking expectation. Let's first start with $I(q_i = k).$

\begin{align} \mathbb{E}\left[I(q_i = k) \mid \mathbf{x}_i, \boldsymbol\theta_l\right] &= p(q_i = k \mid \mathbf{x}_i, \boldsymbol\theta_l) \nonumber\\ &= \frac{p(\mathbf{x}_i \mid q_i = k, \boldsymbol\theta_l)p(q_i = k \mid \boldsymbol\theta_l)}{\sum_{k^\prime = 1}^K p(\mathbf{x}_i \mid q_i = k^\prime, \boldsymbol\theta_l)p(q_i = k^\prime \mid \boldsymbol\theta_l)n} \label{eqn:Iq_initial} \end{align} by Bayes' rule.

Recall that $p(q_i = k \mid \boldsymbol\theta_l) = \pi_k^{(l)}$, and \begin{align} p(\mathbf{x}_i \mid q_i = k, \boldsymbol\theta_l) &= \int p(\mathbf{x}_i,\mathbf{z}_i \mid q_i = k, \boldsymbol\theta_l)d\mathbf{z}_i \nonumber\\ &= \int \mathcal{N}\left(\mathbf{x}_i \mid \mathbf{W}_k^{(l)}\mathbf{z}_i + \boldsymbol\mu_k^{(l)}, \boldsymbol\Psi^{(l)}\right)\mathcal{N}\left(\mathbf{z}_i \mid \mathbf{0}, \mathbf{I}_L\right)d\mathbf{z}_i \nonumber\\ &= \mathcal{N}\left( \mathbf{x}_i \mid \boldsymbol\mu_k^{(l)}, \boldsymbol\Psi^{(l)} + \mathbf{W}_k^{(l)}\left(\mathbf{W}_k^{(l)}\right)^\intercal\right) \label{eqn:px} \end{align} by Equation 4.126 of Murphy's textbook. Plugging in Equation \ref{eqn:px} into Equation \ref{eqn:Iq_initial}, we find that $$r_{ik}^{(l)} = \mathbb{E}\left[I(q_i = k) \mid \mathbf{x}_i, \boldsymbol\theta_l\right] = \frac{\pi_k^{(l)}\mathcal{N}\left( \mathbf{x}_i \mid \boldsymbol\mu_k^{(l)}, \boldsymbol\Psi^{(l)} + \mathbf{W}_k^{(l)}\left(\mathbf{W}_k^{(l)}\right)^\intercal\right)} {\sum_{k^\prime=1}^K\pi_{k^\prime}^{(l)}\mathcal{N}\left( \mathbf{x}_i \mid \boldsymbol\mu_{k^\prime}^{(l)}, \boldsymbol\Psi^{(l)} + \mathbf{W}_{k^\prime}^{(l)}\left(\mathbf{W}_{k^\prime}^{(l)}\right)^\intercal\right)}. \label{eqn:Iq}$$

Next, we need to take care of the terms with $\mathbf{z}_i$. To do this, we find the conditional distribution for $\mathbf{z}_i$. Note that we can condition on both $\mathbf{x}_i$ and $q_i$ since we only care about the $\mathbf{z}_i$ terms multiplied by $I(q_i = k)$, for the other $\mathbf{z}_i$ terms disappear when we take the derivative with respect $\pi_k$, $\mathbf{W}_k$, $\boldsymbol\mu_k$, or $\boldsymbol\Psi$.

If we note that $\mathbf{x}_i \mid (\mathbf{z}_i, q_i = k, \boldsymbol\theta_l) \sim \mathcal{N}\left(\mathbf{W}_k^{(l)}\mathbf{z}_i + \boldsymbol\mu_k^{(l)}, \boldsymbol\Psi^{(l)}\right)$, and $\mathbf{z}_i \mid (q_i = k, \boldsymbol\theta_l) \sim \mathcal{N}\left(\mathbf{0}, \mathbf{I}_L\right),$ \begin{align} p(\mathbf{z}_i \mid \mathbf{x}_i, q_i = k,\boldsymbol\theta_l) &= \mathcal{N}\left(\mathbf{z}_i \mid \mathbf{m}_{ik}^{(l)}, \boldsymbol\Sigma_{ik}^{(l)}\right), \\ \text{where}~\boldsymbol\Sigma_{ik}^{(l)} &= \left(\mathbf{I}_L + \left(\mathbf{W}_k^{(l)}\right)^\intercal\left(\boldsymbol\Psi^{(l)}\right)^{-1}\mathbf{W}_k^{(l)}\right)^{-1} \nonumber\\ \text{and}~\mathbf{m}_{ik}^{(l)} &= \boldsymbol\Sigma_{ik}^{(l)} \mathbf{W}_k^{(l)}\left(\boldsymbol\Psi^{(l)}\right)^{-1}\left(\mathbf{x}_i - \boldsymbol\mu_k^{(l)}\right)\nonumber \end{align} by Equation 4.125 in Murphy's textbook.

Now, let's do a couple of things to clean up notation. First, to simply Equation \ref{eqn:loglikelihood}, we'll drop all terms that aren't functions of $\boldsymbol\theta$, so we have $$\tilde{l}(\boldsymbol\theta) = \sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[ \log\pi_{k} -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left(\mathbf{x}_i - \mathbf{W}_k\mathbf{z}_i - \boldsymbol\mu_k\right)^\intercal{\boldsymbol\Psi}^{-1}\left(\mathbf{x}_i - \mathbf{W}_k\mathbf{z}_i- \boldsymbol\mu_k\right) \right]. \label{eqn:loglikelihood1}$$

In the next step, we define $$\tilde{\mathbf{z}}_i = \begin{pmatrix} \mathbf{z}_i \\ 1 \end{pmatrix},~\text{and}~ \tilde{\mathbf{W}}_k = \begin{pmatrix} \mathbf{W}_k & \boldsymbol\mu_k \end{pmatrix}.$$ Now $\boldsymbol\theta = (\tilde{\mathbf{W}}_k, \boldsymbol\Psi, \pi_k)$, and we can rewrite Equation \ref{eqn:loglikelihood1} as \begin{align} \tilde{l}(\boldsymbol\theta) &= \sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[ \log\pi_{k} -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left(\mathbf{x}_i - \tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i\right)^\intercal{\boldsymbol\Psi}^{-1}\left(\mathbf{x}_i - \tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i\right) \right] \nonumber\\ &= \sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[ \log\pi_{k} -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left( \mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\mathbf{x}_i - 2\mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i + \tilde{\mathbf{z}}_i^\intercal\tilde{\mathbf{W}}_k^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i \right) \right] \nonumber\\ &= \sum_{i=1}^N\sum_{k = 1}^K I(q_i = k)\left[ \log\pi_{k} -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left( \mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\mathbf{x}_i - 2\mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i + \operatorname{tr}\left(\tilde{\mathbf{W}}_k^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\tilde{\mathbf{z}}_i\tilde{\mathbf{z}}_i^\intercal \right)\right) \right] \end{align} by cyclic property the trace.

Now, we note that $$\tilde{\mathbf{z}}_i \mid \left(\mathbf{x}_i, q_i = k,\boldsymbol\theta_l\right) \sim \mathcal{N}\left( \begin{pmatrix} \mathbf{m}_{ik}^{(l)} \\ 1 \end{pmatrix}, \begin{pmatrix} \boldsymbol\Sigma_{ik}^{(l)} & \mathbf{0} \\ \mathbf{0}^\intercal & 0 \end{pmatrix} \right).$$ Using this, we'll have that \begin{align} \mathbf{b}_{ik}^{(l)} &= \mathbb{E}\left[\tilde{\mathbf{z}}_i \mid \mathbf{x}_i ,q_i = k, \boldsymbol\theta_l\right] =\begin{pmatrix} \mathbf{m}_{ik}^{(l)} \\ 1 \end{pmatrix} \\ \mathbf{C}_{ik}^{(l)} &= \mathbb{E}\left[\tilde{\mathbf{z}}_i\tilde{\mathbf{z}}_i^\intercal \mid \mathbf{x}_i ,q_i = k, \boldsymbol\theta_l\right] = \begin{pmatrix} \boldsymbol\Sigma_{ik}^{(l)} & \mathbf{b}_{ik}^{(l)} \\ \left(\mathbf{b}_{ik}^{(l)}\right)^\intercal & 1 \end{pmatrix}. \end{align}

Thus, E-step becomes writing our objective function as $$Q_{\boldsymbol\theta_l}(\boldsymbol\theta) = \sum_{i=1}^N\sum_{k = 1}^K r_{ik}^{(l)}\left[ \log\pi_{k} -\frac{1}{2}\log |\boldsymbol\Psi| -\frac{1}{2}\left( \mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\mathbf{x}_i - 2\mathbf{x}_i^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\mathbf{b}_{ik}^{(l)} + \operatorname{tr}\left(\tilde{\mathbf{W}}_k^\intercal\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)} \right)\right) \right]. \label{eqn:estep}$$

We want to maximize $Q$ to obtain $$\boldsymbol\theta_{l+1} = \operatorname*{arg\,max}_{\boldsymbol\theta} Q_{\boldsymbol\theta_l}(\boldsymbol\theta)$$ for the M-step. This can be done by taking derivatives.

For $\pi_k$ this, is not that hard. Let $\pi_K = 1 - \sum_{k=1}^{K-1}\pi_k$, which gives us that $$\frac{\partial}{\partial\pi_k}Q_{\boldsymbol\theta_l}(\boldsymbol\theta) = \sum_{i=1}^N \left(\frac{r_{ik}^{(l)}}{\pi_k} - \frac{r_{iK}^{(l)}}{\pi_K}\right)$$ for $k < K$. Setting this equal to $0$, we can solve for $\hat{\boldsymbol\pi}$. \begin{align*} \hat\pi_K\sum_{i=1}^N r_{ik}^{(l)} &= \hat\pi_k\sum_{i=1}^N r_{iK}^{(l)} \\ \hat\pi_K\sum_{i=1}^N\left(r_{i1}^{(l)} + r_{i2}^{(l)} + \cdots + r_{i,K-1}^{(l)}\right) &= \left(\hat\pi_1 + \hat\pi_2 + \cdots + \hat\pi_{K-1}\right)\sum_{i=1}^N r_{iK}^{(l)} \\ \hat\pi_K\sum_{i=1}^N\left(1 - r_{iK}^{(l)}\right) &= \left(1 - \hat\pi_K\right)\sum_{i=1}^N r_{iK}^{(l)} \\ N\hat\pi_K - \hat\pi_K\sum_{i=1}^N r_{iK}^{(l)} &= \sum_{i=1}^N r_{iK}^{(l)} - \hat\pi_K\sum_{i=1}^N r_{iK}^{(l)} \\ \hat\pi_K &= \frac{1}{N}\sum_{i=1}^N r_{iK}^{(l)}. \end{align*}

By symmetry, we have that $$\pi_k^{(l + 1)} = \hat\pi_k = \frac{1}{N}\sum_{i=1}^N r_{ik}^{(l)} \label{eqn:mpi}$$ for any $k$.

For $\boldsymbol\Psi$, we need several matrix identities. The first one is $$\frac{\partial}{\partial \mathbf{A}} \log |\mathbf{A}| = \left(\mathbf{A}^{-1}\right)^\intercal. \label{eqn:mat_det}$$

To see this, note that we can write $|\mathbf{A}| = \sum_{i=1}^N \left(-1\right)^{i+j}\mathbf{A}_{ij}\left|\mathbf{A}_{-i,-j}\right|$ for any $j$, so we have that \begin{align*} \frac{\partial}{\partial \mathbf{A}_{ij}} \log |\mathbf{A}| &= \frac{1}{|\mathbf{A}|}(-1)^{i+j}\left|\mathbf{A}_{-i,-j}\right| \\ &= \frac{1}{|\mathbf{A}|}\mathbf{C}_{ij} \\ &= \left(\mathbf{A}^{-1}\right)^\intercal_{ij}, \end{align*} where we have used the definition of the matrix inverse in terms of the adjugate matrix, and $\mathbf{C}$ is the cofactor matrix.

Next, we prove that $$\frac{\partial}{\partial \mathbf{A}}\mathbf{x}^\intercal \mathbf{A}\mathbf{y} = \mathbf{x}\mathbf{y}^\intercal. \label{eqn:mat_quad}$$ To see this we rewrite \begin{equation*} \mathbf{x}^\intercal \mathbf{A}\mathbf{y} = \operatorname{tr}\left(\mathbf{x}^\intercal \mathbf{A}\mathbf{y}\right) = \operatorname{tr}\left(\mathbf{A}\mathbf{y}\mathbf{x}^\intercal\right) = \sum_{i=1}^N\sum_{k=1}^N \mathbf{A}_{ik}y_kx_i, \end{equation*} which implies that \begin{equation*} \frac{\partial}{\partial \mathbf{A}_{ij}}\mathbf{x}^\intercal \mathbf{A}\mathbf{y} = x_iy_j. \end{equation*}

Now, the last trick is to rewrite $\log|\boldsymbol\Psi| = -\log\left|\boldsymbol\Psi^{-1}\right|$, and note that the MLE is preserved by parametrization. So, using Equations \ref{eqn:mat_det} and \ref{eqn:mat_quad}, we can take the derivative of Equation \ref{eqn:estep} with respect to $\boldsymbol\Psi^{-1}$ to get \begin{align*} \frac{\partial}{\partial\boldsymbol\Psi^{-1}}Q_{\boldsymbol\theta_l}(\boldsymbol\theta) &= \sum_{i=1}^N\sum_{k = 1}^K r_{ik}^{(l)}\left[ \frac{1}{2}\boldsymbol\Psi -\frac{1}{2}\left(\mathbf{x}_i\mathbf{x}_i^\intercal - 2\tilde{\mathbf{W}}_k\mathbf{b}_{ik}^{(l)}\mathbf{x}_i^\intercal + \tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)}\tilde{\mathbf{W}}_k^\intercal\right) \right]. \end{align*} If we set this equal to $0$, we find that \begin{align*} \hat{\boldsymbol\Psi} = \frac{1}{N}\sum_{i=1}^N\sum_{k = 1}^K r_{ik}^{(l)}\left( \mathbf{x}_i\mathbf{x}_i^\intercal - 2\tilde{\mathbf{W}}_k\mathbf{b}_{ik}^{(l)}\mathbf{x}_i^\intercal + \tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)}\tilde{\mathbf{W}}_k^\intercal \right), \end{align*} where we have used the symmetry of $\boldsymbol\Psi$.

We have two issues. $\boldsymbol\Psi$ is suppose to be diagonal, and we need to choose a value for $\tilde{\mathbf{W}}_k$. To enforce the diagonal constraint, we just take the diagonal of $\hat{\boldsymbol\Psi}$ since that we could have just taken derivatives component-wise. For $\tilde{\mathbf{W}}_k$, we use $\tilde{\mathbf{W}}_k^{(l+1)}$, which turns out to not depend on $\boldsymbol\Psi,$ so we have $$\boldsymbol\Psi^{(l+1)} = \frac{1}{N}\sum_{i=1}^N\sum_{k = 1}^K r_{ik}^{(l)}\operatorname{diag}\left( \mathbf{x}_i\mathbf{x}_i^\intercal - 2\tilde{\mathbf{W}}_k^{(l+1)}\mathbf{b}_{ik}^{(l)}\mathbf{x}_i^\intercal + \tilde{\mathbf{W}}^{(l+1)}_k\mathbf{C}_{ik}^{(l)}\left(\tilde{\mathbf{W}}_k^{(l+1)}\right)^\intercal \right). \label{eqn:mpsi}$$

Now, we need to deal with the various $\tilde{\mathbf{W}}_k$. I'm not going to prove this identity, but we have that $$\frac{\partial}{\partial \mathbf{X}}\operatorname{tr}\left(\mathbf{X}^\intercal \mathbf{B} \mathbf{X}\mathbf{C}\right) = \mathbf{B}\mathbf{X}\mathbf{C} + \mathbf{B}^\intercal \mathbf{X} \mathbf{C}^\intercal, \label{eqn:trw}$$ which is Equation 117 in the The Matrix Cookbook. Taking the derivative of Equation \ref{eqn:estep} with respect to $\tilde{\mathbf{W}}_k$, we get \begin{align} \frac{\partial}{\partial\tilde{\mathbf{W}}_k}Q_{\boldsymbol\theta_l}(\boldsymbol\theta) &= -\frac{1}{2}\sum_{i=1}^N r_{ik}^{(l)} \left( 2\boldsymbol\Psi^{-1}\mathbf{x}_i\left(\mathbf{b}_{ik}^{(l)}\right)^\intercal - \boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)} - \boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)} \right) \nonumber\\ &= \sum_{i=1}^N r_{ik}^{(l)} \left(\boldsymbol\Psi^{-1}\tilde{\mathbf{W}}_k\mathbf{C}_{ik}^{(l)} - \boldsymbol\Psi^{-1}\mathbf{x}_i\left(\mathbf{b}_{ik}^{(l)}\right)^\intercal \right), \end{align} where we have used the symmetry of $\boldsymbol\Psi$ and $\mathbf{C}_{ik}^{(l)}$. Setting this equal to $0$, we find that $$\tilde{\mathbf{W}}_k^{(l+1)} = \left(\sum_{i=1}^N r_{ik}^{(l)}\mathbf{x}_i\left(\mathbf{b}_{ik}^{(l)}\right)^\intercal\right) \left(\sum_{i=1}^N r_{ik}^{(l)}\mathbf{C}_{ik}^{(l)} \right)^{-1}. \label{eqn:mw}$$

All in all, we combine Equations \ref{eqn:mpi}, \ref{eqn:mpsi}, and \ref{eqn:mw} to get $\boldsymbol\theta_{l+1}$, so our M-step is \begin{align} \pi_k^{(l + 1)} &= \hat\pi_k = \frac{1}{N}\sum_{i=1}^N r_{ik}^{(l)} \\ \tilde{\mathbf{W}}_k^{(l+1)} &= \left(\sum_{i=1}^N r_{ik}^{(l)}\mathbf{x}_i\left(\mathbf{b}_{ik}^{(l)}\right)^\intercal\right) \left(\sum_{i=1}^N r_{ik}^{(l)}\mathbf{C}_{ik}^{(l)} \right)^{-1} \\ \boldsymbol\Psi^{(l+1)} &= \frac{1}{N}\sum_{i=1}^N\sum_{k = 1}^K r_{ik}^{(l)}\operatorname{diag}\left( \mathbf{x}_i\mathbf{x}_i^\intercal - 2\tilde{\mathbf{W}}_k^{(l+1)}\mathbf{b}_{ik}^{(l)}\mathbf{x}_i^\intercal + \tilde{\mathbf{W}}^{(l+1)}_k\mathbf{C}_{ik}^{(l)}\left(\tilde{\mathbf{W}}_k^{(l+1)}\right)^\intercal \right) . \end{align}

## Principal Component Analysis

Now, $\boldsymbol\Psi$ begin diagonal is already a pretty strong assumption. Principal Component Analysis (PCA) makes the even stronger assumption that $\boldsymbol\Psi = \sigma^2 \mathbf{I}_D$. For this reason it's much easier to compute. All it takes is a singular value decomposition (SVD).

A nice application of these methods is Latent Semantic Indexing. Here, we've represented 9 documents as a word count vector. There are nearly 500 words. Using only 2 latent factors, we can cluster the documents.

Documents 1, 2, and 3 are about alien abductions.

Unfortunately, the assumption that the variance does not differ between the dimensions can be a bad one. In Probabilistic Principal Component Analysis versus Factor Analysis, we see how PCA fails to capture the relationship between $x_1$ and $x_2$ since it reduces the variance by focusing on $x_3$.

Back in 2011, I started eating Paleo, and briefly became famous with my story Quitting Rice. While I was pretty hardcore at first, I've been more 80/20, now. Perhaps, the first thing you'll notice is that I've hardly any made any gains in 5 years ☹. Such is the sad life of a natty lifter. I'm actually significantly stronger, but it doesn't show. I'm actually about 15 lbs heavier in that July 2016 picture.

Now, on the plus side, I have achieved long-term weight loss. According to this article from the The American Journal of Clinical Nutrition, only about 20% of people that lose weight keep it off. It's not clear whether that's because the diets stop working, or people can't adhere to them long-term. Despite significant life changes, I've found this diet rather easy to maintain, however.

May 2013: doing some yard work

Since I've started Paleo, I've had 3 different jobs, earned a graduate degree, and lived in 3 different cities: Boston, Philadelphia, and Seattle. I've taken plenty of vacations, which include Vietnam, Thailand, and Australia. Even when I barely lifted one year, I managed to stay lean.

2014: at a beach in Australia

Thus, in my $N = 1$ experiment, I've decided that Paleo works long-term. Cooking is not so hard, and I find that I can pretty much each as much as I want and not even think about it. Despite availing myself of Google's limitless food, I haven't really gained weight, here, in Seattle.

Of course, there's the other problem of eating all that cholesterol and saturated fat. It would seem that I've headed towards coronary heart disease. And depending on how you interpret my lipid panel, you may be right.

From August 2016

While my triglycerides and HDL are great, that red LDL certainly doesn't look good. I'm not a doctor, so I don't really know what to make of it. There are some studies saying that LDL number may not even be correct because my triglycerides are so low (see: The impact of low serum triglyceride on LDL-cholesterol estimation). It's strange that eating lots of trigylcerides leads to very low trigylcerides. I probably get 60% of my calories from fat, with most of that fat being saturated.

Moreover, in Comparison of serum lipid values in patients with coronary artery disease at 70 years of age, they say that:

Triglycerides and ratio of triglycerides to HDL cholesterol were the most powerful, independent variables related to precocity of CAD.

According to their model, my risk of coronary artery disease should be very low, so who knows? Is it not common knowledge that lots of LDL is bad?

I guess my insulin levels are fine, too? Blood pressure was 110/72 for what it's worth.

Anyway, if you've never given Paleo a try. I highly recommend it. Besides the weight loss, there are plenty of other benefits like not being hungry as often and more stable energy levels. I'm not a medical professional or dietician, but I'd be happy to answer any questions about the lifestyle. Looking at all these pictures of myself, I've realized that the light plays cruel tricks on the eye.