## Project Euler: Problem 21

By | 3. April 2015

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

```public class Problem21 {

public static void main(String[] args) {
int sum = 0;
for (int i = 1; i < 10000; i++) {
int d = d(i);
if((d(d) == i && i != d)){
sum += i;
System.out.println(i);
}
}
System.out.println(sum);
}

public static int d (int number){
int sum = 0;
for (int i = 1; i < number; i++) {
if (number%i==0) {
sum += i;
}
}
return sum;
}

}```

Lösung: 31626

## Project Euler: Problem 19

By | 3. April 2015

You are given the following information, but you may prefer to do some research for yourself.

• 1 Jan 1900 was a Monday.
• Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

```public class Problem19 {

public static void main(String[] args) {

int[] months = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int days = 0;
int sundays = 0;

for (int i = 1901; i <= 2000; i++) {
for (int j = 0; j < months.length; j++) {
if (days % 7 == 0) {
sundays++;
}
if (j == 1 && isLeapYear(i)) {
days += 29;
} else {
days += months[j];
}

}
}
System.out.println(sundays);

}
public static boolean isLeapYear(int year) {
return ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0);
}
}```

Lösung: 171

## Project Euler: Problem 15

By | 2. April 2015

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20×20 grid?

Das Ganze wird diesmal ohne ein Java-Programm gelöst, da es den ganzen Aufwand nicht wert wäre.

Entscheidend ist hier folgende Formel

wobei n der Kantenlänge des Quadrats entspricht.

Also ergibt ein Quadrat der Kantenlänge n=20

Lösung: 137846528820

## Project Euler: Problem 11

By | 2. April 2015

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

```08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48```

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

```int[][] numbers = {
{ 8,  2, 22, 97, 38, 15,  0, 40,  0, 75,  4,  5,  7, 78, 52, 12, 50, 77, 91,  8},
{49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48,  4, 56, 62,  0},
{81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30,  3, 49, 13, 36, 65},
{52, 70, 95, 23,  4, 60, 11, 42, 69, 24, 68, 56,  1, 32, 56, 71, 37,  2, 36, 91},
{22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80},
{24, 47, 32, 60, 99,  3, 45,  2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50},
{32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70},
{67, 26, 20, 68, 02, 62, 12, 20, 95, 63, 94, 39, 63,  8, 40, 91, 66, 49, 94, 21},
{24, 55, 58,  5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72},
{21, 36, 23,  9, 75,  0, 76, 44, 20, 45, 35, 14,  0, 61, 33, 97, 34, 31, 33, 95},
{78, 17, 53, 28, 22, 75, 31, 67, 15, 94,  3, 80,  4, 62, 16, 14,  9, 53, 56, 92},
{16, 39,  5, 42, 96, 35, 31, 47, 55, 58, 88, 24,  0, 17, 54, 24, 36, 29, 85, 57},
{86, 56,  0, 48, 35, 71, 89,  7,  5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58},
{19, 80, 81, 68,  5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77,  4, 89, 55, 40},
{ 4, 52,  8, 83, 97, 35, 99, 16,  7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66},
{88, 36, 68, 87, 57, 62, 20, 72,  3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69},
{ 4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18,  8, 46, 29, 32, 40, 62, 76, 36},
{20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74,  4, 36, 16},
{20, 73, 35, 29, 78, 31, 90,  1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57,  5, 54},
{ 1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52,  1, 89, 19, 67, 48}
};

int max = 0;

//horizontal
for (int i = 0; i < numbers.length; i++) {
for (int j = 0; j < numbers[0].length-3; j++) {
int temp = numbers[i][j] * numbers[i][j+1] * numbers[i][j+2] * numbers[i][j+3];
if (temp>max) {
max = temp;
System.out.println("i: " + i + " j: " + j);
}
}
}

//vertikal
for (int i = 0; i < numbers[0].length; i++) {
for (int j = 0; j < numbers.length-3; j++) {
int temp = numbers[j][i] * numbers[j+1][i] * numbers[j+2][i] * numbers[j+3][i];
if (temp>max) {
max = temp;
System.out.println("i: " + i + " j: " + j);
}
}
}

//diagonal left up to down right
for (int i = 0; i < numbers[0].length-3; i++) {
for (int j = 0; j < numbers.length-3; j++) {
int temp = numbers[j][i] * numbers[j+1][i+1] * numbers[j+2][i+2] * numbers[j+3][i+3];
if (temp>max) {
max = temp;
System.out.println("i: " + i + " j: " + j);
}
}
}

//diagonal right up to left down
for (int i = numbers[0].length-1; i > 2; i--) {
for (int j = 0; j < numbers.length-3; j++) {
int temp = numbers[j][i] * numbers[j+1][i-1] * numbers[j+2][i-2] * numbers[j+3][i-3];
if (temp>max) {
max = temp;
System.out.println("i: " + i + " j: " + j);
}
}
}

System.out.println(max);```

Lösung: 70600674

## Neues CMS online

By | 9. Januar 2015

Vor einiger Zeit hatte ich mich dazu entschlossen, von WordPress nach Drupal zu wechseln. Leider brachte das nicht die erwünschten Verbesserungen. Also wurde kurzerhand WordPress wieder eingespielt.

## Java-Programme neustarten

By | 2. Januar 2015

Im Internet taucht immer wieder die Frage auf, wie man Java-Programme im .jar-Format neustarten kann. Lösungen sind allerdings rar. Also habe ich mich umgeschaut und mit den vorhandenen Informationen eine lauffähige Version geschrieben.

```String javaBin = System.getProperty("java.home") + "/bin/java";
File jarFile;
try{
jarFile = new File
(getClass().getProtectionDomain()
.getCodeSource().getLocation().toURI());
} catch(Exception e) {
return;
}
if ( !jarFile.getName().endsWith(".jar") )
return;

String toExec[] = new String[] { javaBin, "-jar", jarFile.getPath() };
try{
@SuppressWarnings("unused")
Process p = Runtime.getRuntime().exec( toExec );
} catch(Exception e) {
e.printStackTrace();
return;
}

System.exit(0);```

## LCD an Arduino anschließen

By | 12. April 2014

Hier zeige ich, wie man ein handelsübliches 16×2 Display an den Arduino Uno anschließt.

LCD RS an Arduino Digital 12
LCD RW an Arduino Digital 11
LCD E an Arduino Digital 10
LCD D4 an Arduino Digital 6
LCD D5 an Arduino Digital 5
LCD D6 an Arduino Digital 4
LCD D7 an Arduino Digital 3
An VS0 einen 10k Poti für den Kontrast und an A einen Vorwiderstand für die Hintergrundbeleuchtung. Ich habe 1k Ohm genommen.

Alles nochmal als Bild:

Jetzt fehlt nur noch der Arduino-Sketch:

```#include <LiquidCrystal.h>
LiquidCrystal lcd(12, 11, 10, 6, 5, 4, 3); //Hier die Ports angeben an denen das Display angeschlossen wurde

void setup()
{
lcd.begin(16, 2); //Art des Displays
lcd.print("This is a test!");
}

void loop()
{
lcd.setCursor(0, 1);
lcd.print("1234567890");
}```

## LED-Matrix 29×8

By | 9. April 2014

Nach langen Stunden des Lötens und Programmierens ist die Matrix hardwaremäßig endlich fertig. An der Software werde ich noch ein bisschen basteln.

Einkaufsliste:

1 x Lochrasterplatine 229 x 127 mm
232 x LED 5mm
29 x Widerstand 150 Ohm
8 x Widerstand 1k Ohm
10 x Keramikkondensator 100 nF
2 x Keramikkondensator 22 pF
1 x Quarz 16 Mhz
1 x Attiny 85
5 x Schieberegister 74HC595
4 x LED-Treiber ULN2803 50V 0,5A
8 x Transistor NPN 2N2222A 40V 0,8A
Schaltlitze

Schaltplan:

Programm:

```#include <digitalWriteFastTiny.h>

byte i = 0;
byte j = 0;
byte k = 0;

#define dataPin 2
#define shiftPin 1
#define storePin 0

boolean leds[8][37] = {
{true, true, true, true, true,       false,   false, false, false, false, false,  false,   false, false, false, false, false,  false,   false, true, false, false, false,   false,   false, false, true, false, false,     true, false, false, false, false, false, false, false},
{false, false, true, false, false,   false,   false, false, false, false, false,  false,   false, false, false, false, false,  false,   false, true, false, false, false,   false,   false, false, true, false, false,     false, true, false, false, false, false, false, false},
{false, false, true, false, false,   false,   false, true, true, true, false,     false,   false, true, true, true, false,     false,   true, true, true, false, false,      false,   false, false, true, false, false,     false, false, true, false, false, false, false, false},
{false, false, true, false, false,   false,   true, false, false, false, true,    false,   true, false, false, false, false,   false,   false, true, false, false, false,    false,   false, false, true, false, false,     false, false, false, true, false, false, false, false},
{false, false, true, false, false,   false,   true, true, true, true, true,       false,   false, true, true, true, false,     false,   false, true, false, false, false,      false,   false, false, false, false, false,    false, false, false, false, true, false, false, false},
{false, false, true, false, false,   false,   true, false, false, false, false,   false,   false, false, false, false, true,   false,   false, true, false, false, true,    false,   false, false, false, false, false,    false, false, false, false, false, true, false, false},
{false, false, true, false, false,   false,   false, true, true, true, false,     false,   true, true, true, true, false,      false,   false, false, true, true, false,       false,   false, false, true, false, false,     false, false, false, false, false, false, true, false},
{false, false, false, false, false,  false,   false, false, false, false, false,  false,   false, false, false, false, false,  false,   false, false, false, false, false,   false,   false, false, false, false, false,    false, false, false, false, false, false, false, true}
};

void setup() {
pinMode(dataPin, OUTPUT);
digitalWrite(dataPin, LOW);
pinMode(shiftPin, OUTPUT);
digitalWrite(shiftPin, LOW);
pinMode(storePin, OUTPUT);
digitalWrite(storePin, LOW);

delay(2000);
}

void loop() {
for(k=1; k<=8; k++){
writeLineToMatrix(leds[k-1], k);
//delay(1);
}

}

void writeLineToMatrix(boolean line[], byte row){
digitalWriteFast(storePin, LOW);
for(i=0; i<37; i++){
digitalWriteFast(dataPin, line[i]);
digitalWriteFast(shiftPin, HIGH);
digitalWriteFast(shiftPin, LOW);
}
digitalWriteFast(dataPin, LOW);
digitalWriteFast(storePin, HIGH);
delayMicroseconds(1000);
}```

digitalWriteFast Library

Bilder: