05/27/18

This page is GDPR compliant

Important Stuff

I haven’t posted in some period of time and I think that the time has finally come to,… *drum roll*… talk about something important.

I am proud to announce that Martinkysel.com is GDPR compliant. I know you are intrigued so let us examine how I managed this:

What is GDPR?

The goal of GDPR is to protect user’s personally identifying information (PII) and hold businesses to a higher standard when it comes to how they collect, store, and use this data.

Article 4 states that personal data means any information relating to an identified or identifiable natural person (‘data subject’)”.

“… an identifiable natural person is one who can be identified, directly or indirectly, in particular by reference to an identifier such as a name, an identification number, location number, an online identifier or to one or more factors specific to the physical, physiological, genetic, mental, economic, cultural or social identity of that natural person.”

What does this page store?

Nothing. Wicked cool, right?

Moral of the story

Don’t store data you don’t need.

In Related News

I am working on a large series of articles about SQL, that should appear on the NuoDB Techblog, stay tuned!

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08/2/18

Why A Chairman Ain’t the Master Replica

INTRO

As a distributed, ACID-compliant database, we get a lot of questions about how NuoDB works, and how it compares to more well-known architectures. In particular, people want to understand how NuoDB compares to a master / master architecture, and (related) how we handle lock management. This post will discuss both. We will also discuss when shared-nothing architecture is viable and when conflicting decisions need to coordinated.

In this blog post, we will discuss the concept of Chairmanship, NuoDB’s answer to distributed coordination. We will prove that Chairmanship is both consistent and resilient to failure. But before we go into the details, let us explore the basics of NuoDB architecture.

Continue reading →

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07/5/18

Quick dive into NuoDB Architecture

Introduction

Traditionally, relational databases were designed for scale-up architectures. Supporting more clients or higher throughput required an upgrade to a larger server. Until recently, this meant that implementing a read & write scale-out architecture either required a NoSQL database and abandoning traditional relational database benefits, or relying on sharding and explicit replication. There were no solutions that could scale-out and still provide complete ACID (Atomicity, Consistency, Isolation, and Durability) -compliant semantics. This tension is what inspired the NewSQL movement and ultimately led to today’s modern “elastic SQL” databases.

NuoDB is an elastic SQL database designed with distributed application deployment challenges in mind. It’s a full SQL solution that provides all the properties of ACID-compliant transactions and standard relational SQL language support. It’s also designed from the start as a distributed system that scales the way a cloud service has to scale, providing high availability and resiliency with no single points of failure. Different from traditional shared-disk or shared nothing architectures, NuoDB’s patented database presents a new kind of peer-to-peer, on-demand independence that yields high availability, low-latency, and a deployment model that is easy to manage.

This article highlights the key concepts and architectural differences that set NuoDB apart from traditional relational databases and even other elastic SQL databases.

Continue reading →

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08/5/16

HackerRank ‘Non-Divisible Subset’ Solution

Short Problem Definition:

Given a set S of n distinct integers, print the size of a maximal subset S’ of S where the sum of any 2 numbers in S’ are not evenly divisible by k.

Link

Non-Divisible Subset

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

This is by all means not an easy task and is also reflected by the high failure ratio of the participants. For a sum of two numbers to be evenly divisible by k the following condition has to hold. If the remainder of N1%k == r then N2%k = k-r for N1+N2 % k == 0. Let us calculate the set of all numbers with a remainder of r and k-r and pick the larger set. If the remainder is half of k such as 2 % 4 = 2 or exactly k such as 4 % 4 = 0, just one number from each of these sets can be contained in S’.

Solution:

def solveSubset(S, k, n):
    r = [0] * k
    
    for value in S:
        r[value%k] += 1
    
    result = 0
    for a in xrange(1, (k+1)//2):
        result += max(r[a], r[k-a])
    if k % 2 == 0 and r[k//2]:
        result += 1
    if r[0]:
        result += 1
    return result
    
n, k = map(int, raw_input().split())
S = map(int, raw_input().split())
print solveSubset(S, k, n)

use std::io;

fn get_numbers() -> Vec<u32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn calculate_nondivisible(a: Vec<u32>, n: usize, k: usize) -> u32 {
    let mut result = 0;
    
    let mut r = vec![0; k];
    for val in a {
        r[(val as usize)%k] += 1;
    }
    
    for idx in 1..(k+1)/2 {
        result += std::cmp::max(r[idx as usize], r[(k-idx) as usize]);
    }
    
    if k % 2 == 0 && r[k/2] != 0 {
        result += 1;
    }
    if r[0] != 0 {
        result += 1;
    }
    
    result
}

fn main() {
    let line = get_numbers();
    let a = get_numbers();
 
    println!("{}", calculate_nondivisible(a, line[0] as usize, line[1] as usize) );
}

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08/4/16

HackerRank ‘Flatland Space Station’ Solution

Short Problem Definition:

For each city, determine its distance to the nearest space station and print the maximum of these distances.

Link

Flatland Space Station

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

This is a two pass algorithm. First, measure the distance to the last station on the left. And on the second pass measure the distance to the nearest station on the right. Pick the minimum of both values. Remember that the first and last position are not necessarily stations.

Solution:

use std::io;
use std::cmp;

fn get_numbers() -> Vec<u32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn find_distance(c: Vec<u32>, n: usize) -> u32 {
    let mut solution = 0;
    let mut distances = vec![n as u32; n];
    
    // first pass
    let mut last_seen = 0;
    let mut seen_one = false;
    for i in 0..n {
        if c[i] == 1 {
            seen_one = true;
            last_seen = 0;
        } else {
            last_seen += 1;
        }
        if seen_one {
            distances[i] = last_seen;
        }
    }
    
    // second pass
    let mut last_seen = 0;
    let mut seen_one = false;
    for i in (0..n).rev() {
        if c[i] == 1 {
            seen_one = true;
            last_seen = 0;
        } else {
            last_seen += 1;
        }
        solution = cmp::max(solution, 
                            match seen_one {
                                true => cmp::min(last_seen, distances[i]),
                                false => distances[i],
                            }
                    );
    }
    
    solution
}

fn main() {
    let line = get_numbers();
    let n = line[0] as usize;
    let c = get_numbers();
    let mut v = vec![0; n];
    for station in c {
        v[station as usize] = 1;
    }
    println!("{}", find_distance(v, n));
}

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08/3/16

HackerRank ‘Kangaroo’ Solution

Short Problem Definition:

There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location x1 and moves at a rate of v1 meters per jump. The second kangaroo starts at location x2 and moves at a rate of v2 meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they’ll ever land at the same location at the same time?

Link

Kangaroo

Complexity:

time complexity is O(1)

space complexity is O(1)

Execution:

There is no need to simulate the movement. We can reason that the two kangaroos either meat at the smallest common multiply or never.

Solution:

use std::io;
fn get_numbers() -> Vec<u32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn main() {
    let numbers = get_numbers();
    let x1 = numbers[0];
    let v1 = numbers[1];
    let x2 = numbers[2];
    let v2 = numbers[3];
    
     if x1 == x2 && v1 == v2 {
        println!("YES");
    }
    else if x1 == x2 && v1 > v2 {
        println!("NO");
    }
    else if x1 <= x2 && v1 <= v2 {
        println!("NO");
    }
    else {
        if (x2 - x1) % (v1 - v2) == 0  {
            println!("YES");
        }
        else {
            println!("NO");
        }
    }
}

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08/2/16

HackerRank ‘Save the Prisoner!’ Solution

Short Problem Definition:

A jail has N prisoners, and each prisoner has a unique id number,S , ranging from 1 to N. There are M sweets that must be distributed to the prisoners. But wait—there’s a catch—the very last sweet S is poisoned! Can you find and print the ID number of the last prisoner to receive a sweet so he can be warned?

Link

Save the Prisoner!

Complexity:

time complexity is O(1)

space complexity is O(1)

Execution:

This challenge is painlessly trivial.

Solution:

use std::io;

fn get_number() -> u32 {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn solve_prisoner(n: u32, m: u32, s: u32) -> u32 {
    ((s - 1 + m - 1 ) % n) +1
}

fn main() {
    let t = get_number();
    for _ in 0..t {
        let line = get_numbers();
        println!("{}", solve_prisoner(line[0], line[1], line[2]));
    }
}

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08/2/16

HackerRank ‘Jumping on the Clouds’ Solution

Short Problem Definition:

Emma is playing a new mobile game involving clouds numbered from 1 to n. There are two types of clouds, ordinary clouds and thunderclouds. The game ends if Emma jumps onto a thundercloud, but if she reaches the last cloud, she wins the game!

Link

Jumping on the Clouds

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

Theoretically your solution can depend on the fact that the win condition is guaranteed, but I don’t like such solutions. Here I present a semi-DP approach that keeps track of the optimal number of jumps it takes to reach each cloud.

Solution:

use std::io;
use std::cmp;

fn get_number() -> u32 {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
    let mut line = String::new();
    io::stdin().read_line(&mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn calculate_jumping(a: Vec<u32>, n: usize) -> u32{
    let mut v = vec![100; n];
    v[0] = 0;
    
    for i in 1..n {
        //println!("{} {} {:?}", i, a[i], v);
        if a[i] == 1 {
            continue;
        }
        
        if i == 1 {
            v[i] = v[i-1] + 1;
        } else {
            v[i] = cmp::min(v[i-1], v[i-2]) + 1;
        }
    }
    
    v[n-1]
}

fn main() {
    let n = get_number() as usize;
    let a = get_numbers();
 
    println!("{}", calculate_jumping(a, n) );
}

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08/1/16

HackerRank ‘Jumping on the Clouds: Revisited’ Solution

Short Problem Definition:

Aerith is playing a cloud game! In this game, there are clouds numbered sequentially from 1 to n. Each cloud is either an ordinary cloud or a thundercloud. Given the values of n and k the configuration of the clouds, can you determine the final value of e after the game ends?

Link

Jumping on the Clouds: Revisited

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

Simulate the game in a loop.

Solution:

#!/bin/python

def solveCloudRevisited(c, n, k):
    pos = 0
    cnt = 0

    while cnt == 0 or pos != 0:
        pos += k
        pos %= n
        if c[pos] == 0:
            cnt += 1
        else:
            cnt += 3
    
    return 100 - cnt

if __name__ == '__main__':
    n,k = map(int,raw_input().strip().split(' '))
    c = map(int,raw_input().strip().split(' '))
    print solveCloudRevisited(c, n, k)

use std::io;

fn get_numbers() -&gt; Vec&lt;u32&gt; {
    let mut line = String::new();
    io::stdin().read_line(&amp;mut line).ok().expect("Failed to read line");
    line.split_whitespace().map(|s| s.parse::&lt;u32&gt;().unwrap()).collect()
}

fn calculate_jumping(a: Vec&lt;u32&gt;, n: usize, k: usize) -&gt; u32 {
    let mut e = 0;
    let mut pos = 0;
    
    while e == 0 || pos != 0 {
        e += match a[pos] {
            0 =&gt; 1,
            1 =&gt; 3,
            _ =&gt; panic!("invalid input"),
        };
        
        pos += k;
        pos %= n;
    }
    
    100 - e
}

fn main() {
    let line = get_numbers();
    let (n, k) = (line[0], line[1]);
    let a = get_numbers();
    println!("{}", calculate_jumping(a, n as usize, k as usize) );
}

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07/3/16

HackerRank ‘Divisible Sum Pairs’ Solution

Short Problem Definition:

You are given an array of n integers and a positive integer, k. Find and print the number of (i,j) pairs where i < j and ai + aj is evenly divisible by k.

Link

Divisible Sum Pairs

Complexity:

time complexity is O(N^2)

space complexity is O(1)

Execution:

Brute force search.

Solution:

n,k = raw_input().strip().split(' ')
n,k = [int(n),int(k)]
a = map(int,raw_input().strip().split(' '))
count=0
for i in xrange(len(a)):
    for j in xrange(i+1,len(a)):
        if (a[i]+a[j]) % k == 0:
            count+=1
        
print count

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