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Pages: 1
00 - 195225786
01 - 195225787
02 - 195225784
03 - 195225785
04 - 195225790
05 - 195225791
06 - 195225788
07 - 195225789
08 - 195225778
09 - 195225779
10 - 195225776
11 - 195225777
12 - 195225782
13 - 195225783
14 - 195225780
15 - 195225781
16 - 195225770
17 - 195225771
I recently came accross a new algo (not exactly rfid related) and I think I find it out; on the left is the shown value, on the right the real written value (find the relation was not easy but I am almost sure it is correct); my "progression" seems to be correct, can someone explain it mathematically ? It should be "base4-something" but I am not able to implement it in excel...
Last edited by asper (2014-10-23 18:23:48)
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11 is listed twice on the left, is that a typo or correct?
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assuming the first 11 is 10, i believe it is a basic xor. convert to hex both numbers and xor them and you will find the constant.
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if it helps, add this function to excel to calculate xor
Public Function BitXorH(x As String, y As String, Optional z As String = "0", Optional w As String = "0", Optional v As String = "0", Optional u As String = "0", Optional t As String = "0", Optional s As String = "0") As String
BitXorH = Hex(Val("&H" & x) Xor Val("&H" & y) Xor Val("&H" & z) Xor Val("&H" & w) Xor Val("&H" & v) Xor Val("&H" & u))
End FunctionOffline
Yes double 11 is a typo, sorry!
I converted to bin and saw a 2-bits-base4 correlation-progression obviously starting frim the last 2 bits:
10
11
00
01
try to decode in bin; i think it better solve the problem.
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it is xor. (xor last 2 hex values of large number and the hex value of the small value)
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Sorry man an example:
06 - 195225788
06 => H06 ==> 0xor6 = 6
195225788 => HBA2E8BC ==> BxorC = 7
Maybe is "xor last 2 hex values MINUS ONE of large number and the hex value of the small value"... or not ?
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xor the last2 with the hex of the num2
Num Hex Last2 Num2 Hex Xor
195225786 BA2E8BA BA 0 0 BA
195225787 BA2E8BB BB 1 1 BA
195225784 BA2E8B8 B8 2 2 BA
195225785 BA2E8B9 B9 3 3 BA
195225790 BA2E8BE BE 4 4 BA
195225791 BA2E8BF BF 5 5 BA
195225788 BA2E8BC BC 6 6 BA
195225789 BA2E8BD BD 7 7 BA
195225778 BA2E8B2 B2 8 8 BA
195225779 BA2E8B3 B3 9 9 BA
195225776 BA2E8B0 B0 10 A BA
195225777 BA2E8B1 B1 11 B BA
195225782 BA2E8B6 B6 12 C BA
195225783 BA2E8B7 B7 13 D BA
195225780 BA2E8B4 B4 14 E BA
195225781 BA2E8B5 B5 15 F BA
195225770 BA2E8AA AA 16 10 BA
195225771 BA2E8AB AB 17 11 BALast edited by marshmellow (2014-10-23 21:18:44)
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Ok i understand now, thank you for the clarification.
My findings also work:
00 - 195225786 = 1011101000101110100010111010
01 - 195225787 = 1011101000101110100010111011
02 - 195225784 = 1011101000101110100010111000
03 - 195225785 = 1011101000101110100010111001
04 - 195225790 = 1011101000101110100010111110
05 - 195225791 = 1011101000101110100010111111
06 - 195225788 = 1011101000101110100010111100
07 - 195225789 = 1011101000101110100010111101
08 - 195225778 = 1011101000101110100010110010
09 - 195225779 = 1011101000101110100010110011
10 - 195225776 = 1011101000101110100010110000
11 - 195225777 = 1011101000101110100010110001
12 - 195225782 = 1011101000101110100010110110
13 - 195225783 = 1011101000101110100010110111
14 - 195225780 = 1011101000101110100010110100
15 - 195225781 = 1011101000101110100010110101
16 - 195225770 = 1011101000101110100010101010
17 - 195225771 = 1011101000101110100010101011
You can see that there is a sort of "base4" coding; consider last 6 bits only:
111010
111011
111000
111001
111110
111111
111100
111101
110010
110011
110000
110001
100110
100111
100100
100101
and so on with higher bits so in my opinion there is a base4 coding of the decimal short value.
Last edited by asper (2014-10-23 22:36:14)
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I think you may be looking too hard.
It is doubtful that it is a coincidence that you can xor the last 2 words of the written number with BA and always get the value your looking for. Unless there are additional samples that tell a more complex story.
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Man can I contact you via mail or some social media ?
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I think you still have my ICQ
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