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Recently I came across a java multiplayer gaming engine called Jogre. It stands for “Java Online Gaming Real-time Engine”. I was looking for something like this for quite a time now. Basically, it lets you create a webapp just like pogo.com. They are in Alpha 0.3 release and the engine comes with 23 games. If you want to code some multiplayer java games and do not want to get involved in the details of network programming I advice you to try it. It is open source.
JOGRE
I have deployed the engine at blindboard.net. Check out if you want to see the engine in work.
This snippet solves a sudoku board inputted through the command line. The input is a list of lists where inner lists are rows of the sudoku. Each zero indicates an empty cell. The program fills up the zeros and prints out the board in the given form
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| import sys
mat = []
#Testcase, uncomment next line to test the program
#mat = [[0,5,0,0,0,0,0,0,0],[0,0,0,0,4,0,0,0,3],[0,1,0,0,9,2,6,0,0],[0,0,0,0,2,4,0,8,0],[0,0,7,0,0,0,0,6,0],[4,0,0,8,1,0,7,0,0],[0,0,1,4,0,7,0,5,6],[0,0,5,0,8,0,0,9,4],[0,0,0,0,0,0,8,0,0]]
def disp():
global mat
for row in mat:
print row
def solve():
global mat
x = y = 0
found = False
for i in xrange(0,9):
for j in xrange(0,9):
if mat[i][j] == 0:
x = i
y = j
found = True
break
if( found ): break
if(not found): return 1 # YAY! solved all cells! :)
num = [1]*10
for i in xrange(0,9):
num[mat[x][i]] = 0
num[mat[i][y]] = 0
boxx = 3 * (x/3)
boxy = 3 * (y/3)
for i in xrange(0,3):
for j in xrange(0,3):
num[mat[boxx+i][boxy+j]] = 0
for i in xrange(1,10):
if( num[i] ):
mat[x][y] = i
if ( solve() ):
return 1
mat[x][y] = 0
return 0 #crap! :(
for arg in sys.argv[1::]:
mat = eval(arg)
solve()
print mat |
Here is a question that may puzzle you initially. But it has a fairly simple solution. Consider you have 3 kinds of fruits Apples, Bananas, and Coconuts. Assume that you have an unlimited source. Then how many ways are there to pick 4 fruits? There are 15 for this case.
4A,4B,4C
3A1B,3A1C,3B1A,3B1C,3C1A,3C1B
2A1B1C,2B1A1C,2C1A1B
2A2B,2B2C,2A2C
Now think about another puzzle. How many ways to arrange 4 balls and 2 sticks? We can see that 4 balls and 2 sticks consume 6 spaces. Within these 6 spaces then we can arrange 4 balls, then the sticks go to empty spaces. So there are
$\begin{pmatrix}6\\4\end{pmatrix}$ = 15 ways.
A representation looks like this then
OO|O|O
Where O is a ball and | is a stick.
Now, there is a striking similarity between these two puzzles. Observe that the sticks part the balls into 3 groups. In every representation there are three groups of balls. So, balls and sticks question asks the same as the fruit question. There are 3 kinds or groups of fruits to choose from. So the general answer for the fruit problem is
$\begin{pmatrix}n+r-1\\n\end{pmatrix}$ or $\begin{pmatrix}n+r-1\\r-1\end{pmatrix}$ where r is the number of kinds of fruits and n is the number of fruits to be chosen. The fruit question is the same as n balls r-1 sticks question, since r kinds corresponds to r-1 sticks as r-1 sticks part the n to r groups.
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
Here is the code in Scheme
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| (define (triangular-check no)
(trian-helper no 1 2))
(define (trian-helper no start count)
(cond ((> count no) start)
((> (divisors start) 500) start)
(else (trian-helper no (+ count start) (+ 1 count)))
))
(define (divisors no)
(div-helper no 1 0))
(define (div-helper no start count)
(cond ((> start (sqrt no)) (* 2 count))
((= (remainder no start) 0)
(div-helper no (+ 1 start) (+ 1 count)))
(else (div-helper no (+ 1 start) count))
))
(triangular-check 15000) |
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
Here is the code in Java
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| public class Problem12
{
private static int numberT = 0;
private static int divCount = 0;
public static void main(String args[])
{
for(int i= 1; i<30000; i++ )
{
numberT = numberT + i;
if(check(numberT)>500)
{
System.out.println(numberT);
break;
}
}
}
public static int check(int x)
{
divCount = 0;
for(int j = 1; j<= Math.sqrt((double) x); j++)
{
if(x%j == 0)
divCount++;
}
return 2*divCount;
}
} |
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| (define (even-fib limit)
(even-fib-helper 1 2 limit 0))
(define (even-fib-helper first second limit sum)
(cond ((> first limit) sum)
((= (remainder first 2) 0)
(even-fib-helper second (+ first second) limit (+ first sum)))
(else (even-fib-helper second (+ first second) limit sum))
))
(even-fib 4000000) |
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| sum = 0
second = 2
first = 1
swap = 0
for test in 1..99
if first % 2 == 0 and first < 4000000
sum = sum + first
end
swap = first
first = second
second = swap + second
end
puts sum
gets |
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|
public class Problem2
{
private static double sum = 0;
private static double number[] = new double[2];
private static double swap;
public static void main(String args[])
{
number[0]=1;
number[1]=2;
for(int i = 2; i< 99; i++)
{
swap = number[1];
number[1] = number[0] + number[1];
number[0] =swap;
check(number[0]);
}
System.out.print((int) sum);
}
public static void check(double x)
{
if(x% 2 == 0 && x<4000000)
sum = sum + x;
}
} |
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|
(define (3-or-5 limit)
(3-or-5-helper limit 0 0))
(define (3-or-5-helper limit start sum)
(cond ((= limit start) sum)
((and (= (remainder start 3) 0)
(> (remainder start 5) 0))
(3-or-5-helper limit (+ 1 start) (+ start sum)))
((and (= (remainder start 5) 0)
(> (remainder start 3) 0))
(3-or-5-helper limit (+ 1 start) (+ start sum)))
((= (remainder start 15) 0)
(3-or-5-helper limit (+ 1 start) (+ start sum)))
(else (3-or-5-helper limit (+ 1 start) sum))
))
(3-or-5 1000) |
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| sum = 0
for count in 1...1000
if count % 3 == 0
sum = sum + count
elsif count % 5 == 0
sum = sum + count
end
end
puts sum
gets |
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