06/14/16

Martinkysel.com is now HTTPS

I would like to give a quick shoutout to the folks at Let’s Encrypt that have done a great job at making the web a better place. Do you ask why I care about SSL protected web pages?

Because HTTPS:

  • prevents possible intruders from tampering with the communications between this website and you, my fellow readers. Intruders include intentionally malicious attackers, and legitimate but intrusive companies, such as ISPs or hotels that inject ads into pages.
  • prohibits intrusions. It is possible to exploit unprotected communications and to trick users into giving up sensitive information or installing malware, or to insert their own advertisements into website resources. For example, some third-parties inject advertisements into websites that potentially break user experiences and create security vulnerabilities. It is probably not my case, but the greater Encrypt the web initiative is a cause worth pursuing.
  • protects resources that travel between this website and your browser. Images, cookies, scripts, HTML… they’re all exploitable. Intrusions can occur at any point in the network, including a user’s machine, a Wi-Fi hotspot, or a compromised ISP, just to name a few.

Would you give your PC to a complete stranger in Starbucks and leave? I would not and therefore I will not do the same to my page. I agree that reading some page on the web is different than giving someone access to your Facebook pictures. But is it really? How often do you open web pages over a insecure Wi-fi? Do you trust each and every wire you are connected over?

Letsencrypt is a free SSL certificate authority by the folks from Mozilla, Cisco, Akamai, Facebook and many others. If you own, manage or run a page, give them a try. Make the web a better place.

Without further ado I announce that martinkysel.com is now HTTPS.


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06/10/16

Demystifying Virtual Tables in C++ – Part 2 Non-virtual inheritance

Introduction

In this post we will continue our dive into the C++ dynamic dispatch. So far we have verified that gdb does not create a virtual table for simple classes and default constructors [Part 1 Trivial Constructors].

In part 2 we will look at non-virtual derived classes, their construction and memory layout.

When creating a class, instead of writing completely new data members and member functions, the programmer can designate that the new class should inherit the members of an existing class. This existing class is called the base class, and the new class is referred to as the derived class.

I assume that readers have a familiarity with C++. I will not be explaining why a class should be derived from another class neither will I explain assembly in detail. We will verify common knowledge about c++ using the gdb debugger on a 64bit Ubuntu linux machine.

Continue reading


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05/21/16

Demystifying Virtual Tables in C++ – Part 1 Trivial Constructors

Introduction

This won’t be an easy read. After each claim I will provide supporting code in both C++ and assembly.

Recently I was working on a particularly sneaky bug. A thing happened to me in C++, that should not be possible under normal circumstances. By normal, I mean that 1+1 is guaranteed to 2; the sun rises in the East and sets in the West; all water is wet and the earth is dry…. Yet is still happened. Namely: the vptr of a fully constructed virtual object was pointing to the virtual table of it’s base class. To debug this particular issue one needs to dive deep into the compiler world and into the dreaded waters of assembly.

I hope that I do not have to point out that when I started I had absolutely no clue what is going on. I knew what virtual pointers do and that the compilers are reasonably smart about them. As every modern software engineer I consulted the web oracle only to realise that there is very little written online. I wanted to know how vptrs and vtables are implemented. Not your typical high level theory explaining how dynamic dispatch works. I wanted to know what makes it tick. And while doing so I might improve the world a bit by sharing my journey. So I hopped on my trusted GDB and rode of to the assembly land.

Continue reading


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05/13/16

HackerRank ‘Bigger is Greater’ Solution

Short Problem Definition:

Given a word w, rearrange the letters of w to construct another word in such a way that is lexicographically greater than w. In case of multiple possible answers, find the lexicographically smallest one among them.

Link

Bigger is Greater

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

This task challenges us to find the next permutation of any given array. There are many implementations available online and it is worthwhile comparing them . I would recommend reading the article by Nayuki or re-implementing the std::next_permutation.

Solution:
def next_permutation(arr):
    # Find non-increasing suffix
    i = len(arr) - 1
    while i > 0 and arr[i - 1] >= arr[i]:
        i -= 1
    if i <= 0:
        return False
    
    # Find successor to pivot
    j = len(arr) - 1
    while arr[j] <= arr[i - 1]:
        j -= 1
    arr[i - 1], arr[j] = arr[j], arr[i - 1]
    
    # Reverse suffix
    arr[i : ] = arr[len(arr) - 1 : i - 1 : -1]
    return True 
        
def main():
    t = input()
    for _ in xrange(t):
        s = list(raw_input())
        if next_permutation(s):
            print "".join(s)
        else:
            print "no answer"
    
if __name__ == '__main__':
    main()

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03/4/16

HackerRank ‘Castle on the Grid’ Solution

Short Problem Definition:

You are given a grid with both sides equal to N/N. Rows and columns are numbered from 0/0 to N−1/N−1. There is a castle on the intersection of the aath row and the bbth column.

Your task is to calculate the minimum number of steps it would take to move the castle from its initial position to the goal position (c/d).

It is guaranteed that it is possible to reach the goal position from the initial position.

Link

Castle on the Grid

Complexity:

time complexity is O(N^2) or O(N^3)

space complexity is O(N^2)

Execution:

This solution works with the task as a 2D array. There are options available where you treat the task as a graph problem. In both cases each node it visited exactly once using BFS. On each node, I generate all nodes that are connected to this node in a straight line that is not broken by a ‘X’. I keep the distance data (integer) in the array itself. I store the nodes that have to be visited in a FIFO queue. Once the top element in the queue is the end, I terminate the algorithm. The data stored in the array are these:

  • X        – blocked
  • .          – not visited yet
  • (int)   – already visited. Value is the number of steps from the beginning.
Solution:
from collections import deque

class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
    
    def __str__(self):
        return "X=%d,Y=%d" % (self.x, self.y)

def getPointsFromPoint(N, arr, point):
    x = point.x
    y = point.y
    points = []
    
    while x > 0:
        x -= 1
        if arr[x][y] == 'X':
            break
        points.append(Point(x,y))
    
    x = point.x
    while x < N-1: 
        x += 1 
        if arr[x][y] == 'X': 
            break 
        points.append(Point(x,y)) 
    
    x = point.x 
    while y > 0:
        y -= 1
        if arr[x][y] == 'X':
            break
        points.append(Point(x,y))
    
    y = point.y
    while y < N-1:
        y += 1
        if arr[x][y] == 'X':
            break
        points.append(Point(x,y))
        
    return points
    
def solveCastleGrid(N, arr, start, end):
    q = deque([start])
    arr[start.x][start.y] = 0
    
    while q:
        current_point = q.pop()
        current_distance = arr[current_point.x][current_point.y]
        
        points = getPointsFromPoint(N, arr, current_point)
        for p in points:
            if arr[p.x][p.y] == '.':
                arr[p.x][p.y] = current_distance + 1
                q.appendleft(p)
                if p.x == end.x and p.y == end.y:
                    return current_distance + 1
    return -1
    
if __name__ == '__main__':
    N = input()
    arr = [0] * N
    
    for i in xrange(N):
        arr[i] = list(raw_input())
        
    start_x, start_y, end_x, end_y = map(int, raw_input().split())
    
    print solveCastleGrid(N, arr, Point(start_x, start_y), Point(end_x, end_y))

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03/2/16

RDF parse of DBPedia showing all Death Metal band members and their connections

Death Metal members and their connections

To anyone who likes Death Metal and would like to know who played with who in a graphical form. For you there is a DBpedia RDF parse that I’ve done over the last two days.
Recently I replaced the PDF by a vector graphic to allow better scaling. I suggest you view the image in a new tab or download it. Also notice that bands have the format db:BAND_NAME.

Death Metal Parse


Tutorial

You can download the data as RDF or JSON from DBpedia using SPARQL.


SELECT *
WHERE {
{?band_uri dbo:genre dbr:Death_metal }
UNION {?band_uri dbo:genre dbr:Melodic_death_metal}
UNION {?band_uri dbo:genre dbr:Folk_metal}
UNION {?band_uri dbo:genre dbr:Pagan_metal}
UNION {?band_uri dbo:genre dbr:Black_metal}
UNION {?band_uri dbo:genre dbr:Viking_metal}
UNION {?band_uri dbo:genre dbr:Gothic_metal}
UNION {?band_uri dbo:genre dbr:Power_metal}.
{?band_uri dbpedia2:currentMembers ?member_uri}
UNION {?band_uri dbpedia2:pastMembers ?member_uri}.
}

I prefer a simple script in R

library(SPARQL)
library(igraph)
library(network)
library(ergm)

endpoint <- "http://live.dbpedia.org/sparql"
options <- NULL
prefix <- c("db","http://dbpedia.org/resource/")
sparql_prefix <- "PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX foaf: <http://xmlns.com/foaf/0.1/>
PREFIX dc: <http://purl.org/dc/elements/1.1/>
PREFIX : <http://dbpedia.org/resource/>
PREFIX dbpedia2: <http://dbpedia.org/property/>
PREFIX dbpedia: <http://dbpedia.org/>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
"
q2 <- paste(sparql_prefix,
'SELECT *
WHERE {
{?band_uri dbo:genre dbr:Death_metal }
UNION {?band_uri dbo:genre dbr:Melodic_death_metal} 
UNION {?band_uri dbo:genre dbr:Folk_metal}
UNION {?band_uri dbo:genre dbr:Pagan_metal}
UNION {?band_uri dbo:genre dbr:Black_metal}
UNION {?band_uri dbo:genre dbr:Viking_metal}
UNION {?band_uri dbo:genre dbr:Gothic_metal}
UNION {?band_uri dbo:genre dbr:Power_metal}.
{?band_uri dbpedia2:currentMembers ?member_uri}
UNION {?band_uri dbpedia2:pastMembers ?member_uri}.
}')
res <- SPARQL(endpoint,q2,ns=prefix,extra=options)$results

member_band_matrix <- as.matrix(ifelse(table(res$member_uri, res$band_uri) > 0, 1, 0))

a_m <- graph.incidence(member_band_matrix)

write.graph(a_m,'death_members.graphml',format="graphml")

Afterwards you can use tools such as Gephi to visualise the data.


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02/23/16

HackerRank ‘Sherlock and Valid String’ Solution

Short Problem Definition:

A “valid” string is a string S such that for all distinct characters in S each such character occurs the same number of times in S.

Link

Sherlock and Valid String

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

I optimised this solution to the minimal case that passes all tests on HackerRank. It seems that each character occurs 1 or 2 times. I did not pay the Hackos to verify the input :). The logic of the solution is as follows: count the character counts for each character.

  • if they are all equal – it means that all characters occur exactly N times and there is no removal needed
  • if 2 or more have less or more characters – there is no way to fix the string in just 1 removal
  • if exactly 1 char has a different count than all other characters – remove this char completely and S is fixed.
Solution:
from collections import Counter


def isValid(S):
    char_map = Counter(S)
    char_occurence_map = Counter(char_map.values())

    if len(char_occurence_map) == 1:
        return True

    if len(char_occurence_map) == 2:
        for v in char_occurence_map.values():
            if v == 1:
                return True

    return False


S = raw_input()
if isValid(S):
    print "YES"
else:
    print "NO"

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04/6/15

The Aha! Moment

We have all experienced the great moments of clarity, when suddenly everything makes sense. Your heart starts beating faster, a smile creeps up on your face and you gain real wisdom – wisdom that changes your perception of the world. Some call this feeling the Eureka Effect based on a story of Archimedes jumping out of a public bathroom and shouting Eureka! as he realized how to solve a difficult problem he was facing. I prefer to call this experience the Aha! Moment based on the German Aha-Erlebnis as I am not an ancient Greek philosopher and I certainly do not like public bathrooms.

eureka-aha

You do not only encounter these moments when finally solving coding challenges, but when hearing jokes also. Have you ever heard a joke that did not seem funny at first? And right the second silence fell on the room, your brain finally made a click, you realized why the cute girl on the other side of the table was laughing so hysterically and you start laughing too. Just like… 3 seconds too late. It maybe does not happen to you, but it surely does happen to me and the wave of emotional satisfaction feels exactly the same like solving a hard coding problem.

These Aha! moments occur when your brain redistributes and reanalyzes the available information to forms a novel conclusion. The region of the brain responsible for this change is the hippocampus, widely know for its role in the formation of long- and short-term memory as well as spatial navigation.

Theoretically, “insight” means the reorientation of one’s thinking, including breaking of the unwarranted “fixation” and forming of novel, task-related associations among the old nodes of concepts or cognitive skills. Processes closely related to these aspects have been implicated in the hippocampus.

In their fMRI study Jing Luo & Kazuhisa Niki (abstract cited above) fabricated these experiences on a study group using Japanese riddles. As far as the riddles go I do understand that there are concepts that can not move a nail, but.. I do not feel any smarter after reading that the answer to the riddle is a river. I would not be a good study candidate. (I do love participating in brain imagining studies though and whenever I get the opportunity I participate gladly). What we should learn from that research (jokes aside) is that whenever you experience a Aha! moment, your brain just integrated new information into your memory and somehow connected it to whatever already was in there. That is important. That is why we all push our boundaries and challenge ourselves with previously unseen programming tasks. To force ourselves to be better, smarter and more versatile thinkers. Because after your hippocampus formed new connections, you will later discover them again and again, when you need them most.

Hippocampus


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03/30/15

Verbalize Your Thinking!

Solving new challenges can be hard. The harder it is, the more rewarding a solution feels. Sometimes, solutions written by other people work as a miraculous food for thought that opens doors into concepts you never thought of before. I would like this blog to be more than a reference that you know to visit when you are on search for solutions that you can copy-paste into the appropriate test window. I first started it as a self reference to keep track of my learning. The user base grows with every day and I think that we can do something positive with it. Not only will it support our learning process, but it will create interesting networking opportunities as well. Continue reading


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12/18/14

Eclipse for C++ Development on Windows 7 64-bit Broken Console

It is the year 2014 and for some reason Eclipse CDT (both 32 and 64bit version) since Ganimede won’t output anything the programmer wants to print to the C++ console. The other side effect is, that it is impossible to start the gdb debugger. This problem has been posted all around the web since 2010 and yet it persists with Luna.

EmptySpace

I was able to compile the source using Eclipse, but had to switch to the Cygwin console and execute the binary manually. Doing this is a tedious process and wastes a lot of time. The bad news is, that I have found no solution for Cygwin. The good news is, that the MinGW linker will accept static gcc libraries linkage. So, for the rest of this guide, I will be assuming that you can use the MinGw GNU. Continue reading


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