CPSC 427 Problem Set 4 solution

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1 Consensus Problem
In this and following assignments, we will be developing a simulator for a distributed consensus algorithm. Consensus is at the heart of maintaining consistency in distributed databases
as well as in cryptocurrencies and blockchain algorithms.
We consider the consensus problem in a simple setting. The players are trying to reach
agreement on a course of action. Each player has a current preference called her choice,
which is the current value stored in her choice register. We assume a simple binary choice,
so the choice value is either 0 or 1. The players communicate with each other, and from
time to time a player may change her choice. The goal is for the players to arrive at a
stage where all players are making the same choice. In this case, we say the players have
reached consensus, and we call the common choice the consensus value. We also require
that the consensus value be stable, meaning that once consensus has been reached, nobody
can subsequently ever change her choice.
We assume the agents communicate using the random-pair communication model. In
this model, a communication round consists of a randomly chosen player (called the sender )
sending a message to another randomly chosen player (called the receiver ). Sender and
receiver must be distinct. For the algorithms considered here, the sender’s message is
always her current choice value. The receiver, depending on the message received, may
change her current choice and her internal state.
A population of players solves the consensus problem if following is true:
1. For all possible initial choices of the players, if the players start in their designated
initial states, the computation eventually reaches consensus with probability 1.
2. Once consensus has been reached, no player can subsequently ever change her choice.
Note that if all players start with the same choice, then that choice is the consensus value,
and no player ever changes her choice.
There are many possible algorithms for reaching consensus. Here are a couple very
simple ones that we will be exploring.
1.1 Fickle
Whenever a fickle player receives a message, she changes her choice, if necessary, to agree
with the sender’s choice. That is, she sets her choice register to the value contained in the
message.
It is easy to see that there is some sequence of message transmissions that causes the
system to reach consensus. Once consensus has been reached, no player will change her
choice since every subsequent message contains the consensus value.
It is also easy to believe that it might take a very large number of random communication
rounds to reach consensus.
2 Problem Set 4
1.2 Follow the Crowd
A follow-the-crowd player has a one-bit state register in which she saves the last message
received. She changes her choice only when she gets two messages in a row that both
disagree with her current choice. Thus, she waits until she gets a sense of the crowd before
deciding to follow. We assume that each player starts with her state register set equal to
her choice.
In greater detail, when a follow-the-crowd player receives a message m, she compares it
with her current state. If it differs, she replaces the current state with m. If it is the same,
she replaces the current choice with m.
It is believable that this might converge to a consensus value faster than fickle since it
is less likely for a player holding the majority choice to change to the minority value.
2 Assignment Goals
1. To learn how to organize a simulation of a large system.
2. To learn about a simple model of asynchronous distributed computing.
3. To learn how to generate uniformly distributed random numbers from a finite interval.
4. To experience a computationally-intensive application where efficiency matters.
3 Problem
In this assignment, you will implement a simulation of a large number of agents attempting
to reach consensus using the fickle algorithm under the random-pair communication model.
The follow-the-crowd algorithm will be used in a later assignment.
You are required to implement two classes and a main program.
• class Agent models an agent running the fickle algorithm. The public interface must
support these functions:
– Agent(int ch) constructs an agent with choice ch.
– void update(int m) performs the update to the agent as specified by algorithm
fickle upon receipt of the message m.
– int choice() const returns the agent’s current choice.
• class Simulator simulates a collection of n agents trying to reach consensus using
the random communication model described in section 1. Its public interface consists
of the following:
– Simulator( int numAgents, int numOne, unsigned int seed ) constructs
a simulator for numAgents agents. The first numOne of these have initial choice 1;
the remainder have initial choice 0. seed is used to initialize the random number
generator random().
– int run( int& rounds ) runs the simulation for as many rounds as it takes
to reach consensus. The number of communication rounds used is stored in the
output parameter rounds. The consensus value is returned.
Handout #6—October 22, 2018 3
To carry out the simulation requires the ability to select a random pair of distinct
agents j and k to serve as sender and receiver in a communication round. Further
details are given in section 4.
To know when consensus is reached requires the simulator to keep track of the number
of agents having each of the two possible choice values. Since the only way an agent
might change its choice is because of update(), you should just update the counts
of agents having a given choice after each communication round. Do not poll every
agent after every round.
• main.cpp implements a command
> consensus numAgents numOne [seed]
that takes two required argumets, numAgents and numOne, and one optional argument,
seed. These three arguments should be converted to numbers and passed to the
Simulator constructor. If seed is omitted, the result of time(0) should be used
instead.
The run() function in main.cpp should instatiate a Simulator with the given parameters and then run it. When Simulator::run() returns, you should print a single
line to cout consisting of five whitespace-separated numbers: The number of agents,
the number of agents initially choosing one, the actual seed used, the number of
communication rounds required to reach consensus, and the final consensus value.
You should test your code using various combinations of parameters. Increasing the
population size numPlayers will cause a big increase in run time as will having numOne be
close to numPlayers/2. You may terminate your experiments once the run time grows to
more than a few seconds. This may come rather quickly with fickle, but that is for you
to find out. As usual, you should submit test input files and the corresponding outputs
produce by your program.
4 Program Notes
I will furnish some test cases on the Zoo in /c/cs427/code/ps4/, and you should test
it with some parameter combinations of your own. However, you might only be able to
duplicate my output if you run your code on the Zoo and your program uses the random
number generator in the same way, namely, at each round, first select the sender and then
select the receiver.
To select a random sender from among n agents, you can use the function
RandomUniform(n), which returns a uniformly-distributed random integer in the range
[0 . . . n − 1].
int RandomUniform( int n ) {
long int usefulMax = RAND_MAX – (RAND_MAX+1)%n;
long int r;
do { r = random(); }
while ( r > usefulMax );
return r % n;
}
4 Problem Set 4
The purpose of this code is to make all numbers in the given range equally likely.1
To make sure you use the same number of calls on the random number generator as I
do when choosing the sender and receiver, you should first chose the sender j from among
the n agents. Now there are only n − 1 eligible receivers, so you should choose a a number
k in the range [0 . . . n − 2] and adjust k to avoid j by incrementing k if k ≥ j.
The submission guidelines are the same as in previous assignments. Submit all files
needed to compile your project along with a Makefile. Include a notes.txt file, a file of
sample inputs and a file of the corresponding outputs.
5 Grading Rubric
Your assignment will be graded according to the scale given in Figure 1 (see below).
# Pts. Item
1. 4 All relevant standards from previous problem sets are followed regarding submission, identification of authorship on all files, and so forth. A
well-formed Makefile or makefile is submitted that specifies compiler
options -O1 -g -Wall -std=c++17. Running make successfully compiles
and links the project and results in an executable file consensus. Each
function definition is preceded by a comment that describes clearly what
it does.
2. 2 Required sample input and output files are submitted.
3. 4 The program shows good style. All functions are clean and concise. Inline
initializations, inline functions, and const are used where appropriate.
Variable names are appropriate to the context. Programs are consistently
indented according to the course indenting style. Each class has a separate
.hpp file and, if needed, a separate .cpp file.
4. 2 Everything is private in all classes except for the specified public interface
and any needed special functions (constructors, destructor, move and
copy constructors and assignments).
5. 8 All of the functionality in section 3 is correctly implemented.
20 Total points.
Figure 1: Grading rubric.
1Note that it is not sufficient to just take random()%n since if n does not divide RAND_MAX + 1, some
numbers will have a greater probability of being chosen than others.
For example, ifrandom() were to produce numbers in the range [0 . . . 9] and we wanted numbers in
the range [0 . . . 3], then reducing each of the numbers in the range [0 . . . , 9] mod 4 gives the sequence
0, 1, 2, 3, 0, 1, 2, 3, 0, 1. We see that 0 and 1 each occur 3 times, whereas 2 and 3 each occur only twice. Thus,
0 and 1 are each generated with probability 0.3 and 2 and 3 are each generated with probability only 0.2. To
be uniformly distributed, all probabilities should be 0.25. In this example, where we pretend RAND_MAX==9
and n = 4, my code computes usefulMax to be 9 – 10%4 = 7. Then whenever random() returns 8 or 9, the
program loops and tries again.