How To Scramble Numbers On These Tables?

[fera]

Hi to all of you guys here…
May this thread fits on this section. A friend of mine gave me this enigma. It is written in Excel format. Since here I can’t attach .xls file, I don't know how to put the file, name Enigma.xls.
There are infinite amount of tables, with ten rows (row 0,1,2,….9) each. Inside of each tables, there are numbers from 1 to 92, 93 to 184, 185 to 276, and 277 to 284, which lie on their certain rows. Here I gave the example tables that have been filled in for 40 tables. By finding the patterns/ formulas, my friend asked me to extend the tables to fill in the blank tables 41,42,43,etc as given beneath of Table 40. Just like SUDOKU, in each tables there will be no same numbers vertically, horizontally and diagonally. If these tables are using permutations from an ideal table that you can see beneath Table 40 (supposed the ideal table was right), then how to find the formulas of its permutations?
Can somebody help me about this?
Thx.
Hope my English is good enough for explaining this.

[nobunaga209]

Hmn, you may wanna load the file on rapidshare or something so that people can take a look at it. Kinda hard to solve a puzzle if you can't see the pieces.
:p

[fera]

Hi nobunaga209,
Ok, I put the file at Mediafire.com (a file hosting service) name Enigma.xls:
http://www.mediafire.com/?sharekey=3111bea8a00eb0494c17ca8801618ef7e04e75f6e8ebb871
Thx.

[fera]

Someone has made a program in Java like this:

import java.util.*;
public class Table {
static Scanner console = new Scanner(System.in);
public static void main (String[] args)
{

String list = "010509131741454953572125293337616569737702030406070810111214" +
"151618192022232426272830313234353638394042434446474850515254" +
"5556585960626364666768707172747576787980818283848586878889909192";
String number;
int counter = 0;
int randomNumber = 0;
int rowPlacement = 0;


Vector row_0 = new Vector();
Vector row_1 = new Vector();
Vector row_2 = new Vector();
Vector row_3 = new Vector();
Vector row_4 = new Vector();
Vector row_5 = new Vector();
Vector row_6 = new Vector();
Vector row_7 = new Vector();
Vector row_8 = new Vector();
Vector row_9 = new Vector();

for (counter=0; counter<184; counter = counter + 2)
{
number = list.substring(counter, counter + 2);

//-------------------------------------------------------------------
if (counter == 10 || counter == 20 || counter == 30 || counter >= 40)
{
rowPlacement = 0;
}
//-------------------------------------------------------------------
do
{
randomNumber = (int) ( 10 * Math.random() );
}
while (rowPlacement > randomNumber);
//-------------------------------------------------------------------

if (randomNumber == 0)
{
row_0.addElement(number);
rowPlacement = 0;
}

else if (randomNumber == 1)
{
row_1.addElement(number);
rowPlacement = 1;
}

else if (randomNumber == 2)
{
row_2.addElement(number);
rowPlacement = 2;
}

else if (randomNumber == 3)
{
row_3.addElement(number);
rowPlacement = 3;
}

else if (randomNumber == 4)
{
row_4.addElement(number);
rowPlacement = 4;
}

else if (randomNumber == 5)
{
row_5.addElement(number);
rowPlacement = 5;
}

else if (randomNumber == 6)
{
row_6.addElement(number);
rowPlacement = 6;
}

else if (randomNumber == 7)
{
row_7.addElement(number);
rowPlacement = 7;
}

else if (randomNumber == 8)
{
row_8.addElement(number);
rowPlacement = 8;
}

else if (randomNumber == 9)
{
row_9.addElement(number);
rowPlacement = 9;
}

}
System.out.println(row_0);
System.out.println(row_1);
System.out.println(row_2);
System.out.println(row_3);
System.out.println(row_4);
System.out.println(row_5);
System.out.println(row_6);
System.out.println(row_7);
System.out.println(row_8);
System.out.println(row_9);
}
}

But it didn't work correctly yet :)

[r1ch4rd]

You can make that a lot nicer by using a Vector (or even array) with the other Vectors in, makes a nicer neater loop.

Don't know the answer to the question though! :D

[fera]

Hi too richardjames,
Someone has made it here:
http://hypography.com/forums/physics-and-mathematics/19098-how-to-find-the-formula-permutations.html#post260764
Thx for your attension.