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Help me find more/pattern to values


NoFear
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Please, need help with determining how values are converted. Here is a list of values in game and the Dword value of it:

10        1076101120

7458   1086136832

7468   1086139392

7478   1086141952

 

2978   1084703680

3056   1084743680

3234   1084834816

3254   1084845056

3289   1084862976

3299   1084868096

3309   1084873216

3334  1084886016

3354   1084896256

 

15492   1087259136

15570   1087269120

 

 

Hopefully @Enyby you can help me :)

Edited by NoFear1183
Fixed typo in value
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I'll try. Sorta difficult to obtain a value I want. 

 

Edit: 

Game is I Love Coffee.  It is by Line. 

Edited by NoFear1183
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Counting up from 1

1072693248

1073741824

1074266112

1074790400

1075052544

1075314688

1075576832

1075838976

1075970048

1076101120

1076232192

1076363264

1076494336

1076625408

1076887552

1076953088

1077018624

1077084160

1077149696

1077215232

1077280768

1077346304

 

Values 1 through 23. If something doesn't seem right, I could've mistyped... 

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Answer: same as previous is double.

double - double hex - first 32 bit as dword.

10 = 4024000000000000 1076101120
2978 = 40a7440000000000 1084703744
3056 = 40a7e00000000000 1084743680
3234 = 40a9440000000000 1084834816
3254 = 40a96c0000000000 1084845056
3289 = 40a9b20000000000 1084862976
3299 = 40a9c60000000000 1084868096
3309 = 40a9da0000000000 1084873216
3334 = 40aa0c0000000000 1084886016
3354 = 40aa340000000000 1084896256
7458 = 40bd220000000000 1086136832
7468 = 40bd2c0000000000 1086139392
7478 = 40bd360000000000 1086141952
15492 = 40ce420000000000 1087259136 
15570 = 40ce690000000000 1087269120

Code for http://www.tutorialspoint.com/compile_java_online.php

public class HelloWorld{

     public static void main(String []args){
         int[] arr = new int[] {10,2978,3056,3234,3254,3289,3299,3309,3334,3354,7458,7468,7478,15492,15570};
         for (int j: arr) {
             long l = Double.doubleToLongBits((double)j);
             System.out.println(j + " = " + Long.toHexString(l) + " " + ((int)(l >> 32)));
         }
        
     }
}

For your new data:

1 = 3ff0000000000000 1072693248 
2 = 4000000000000000 1073741824 
3 = 4008000000000000 1074266112 
4 = 4010000000000000 1074790400 
5 = 4014000000000000 1075052544 
6 = 4018000000000000 1075314688 
7 = 401c000000000000 1075576832 
8 = 4020000000000000 1075838976 
9 = 4022000000000000 1075970048 
10 = 4024000000000000 1076101120
11 = 4026000000000000 1076232192
12 = 4028000000000000 1076363264
13 = 402a000000000000 1076494336
14 = 402c000000000000 1076625408
15 = 402e000000000000 1076756480
16 = 4030000000000000 1076887552
17 = 4031000000000000 1076953088
18 = 4032000000000000 1077018624
19 = 4033000000000000 1077084160
20 = 4034000000000000 1077149696
21 = 4035000000000000 1077215232
22 = 4036000000000000 1077280768
23 = 4037000000000000 1077346304 

 

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Usually cifer worked with xor - this is very comfortable for this operation. Because xor A twice got same number.

And some shift: to left, to right or cyclic shift.

Of course can be another biwise operators - and, or, not but it rarely.

 

First step in any investigation - write source and destination in bit form and compare visually.

Another step - compare changes from similar numbers and how it changes reflect in destination.

Next steps dependent on getted results.

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