February 15th, 2023 at 1:24:17 PM
permalink

So I was a lottery agent back in the 1980's and we used to sell lottery tickets with this very unusual barcode with a 4 by 12 set of blocks (representing the binary). The thing I was never able to figure out was the serial number of the ticket was 17 digits long ( 028-23199489-167004) and the barcode could only fit 48 characters of blocks. If the each numeric number has to have 4 blocks to represent each number then the barcode would not be long enough. I also found a similar barcode on a racing ticket that had the block style barcode along with a modern barcode but on that ticket they used hexadecimal and the barcode Field was 6 X 12. I would think this technology for encoding the barcode would come from the horse racing industry that used amtote tickets at the time but I had never seen or understood how this barcode was encoded? Does anyone have any idea how this was encoded? Much appreciated.

February 15th, 2023 at 5:09:19 PM
permalink

If I were encoding it, there would be 12 hexadecimal digits.

2 representing the first block (2^8)

6 representing the second block (2^24)

4 representing the third block (2^16)

This would seem to fit if the last digit (4) were a check digit, not encoded in the punchcard style area.

Attempting to encode the decimal digits as BCD would be horribly wasteful. They don't appear to be octal.

There may well be an error checking convolution.

I'm still casually considering this interesting puzzler.

This should not deter others from jumping in with guesses or solutions.

2 representing the first block (2^8)

6 representing the second block (2^24)

4 representing the third block (2^16)

This would seem to fit if the last digit (4) were a check digit, not encoded in the punchcard style area.

Attempting to encode the decimal digits as BCD would be horribly wasteful. They don't appear to be octal.

There may well be an error checking convolution.

I'm still casually considering this interesting puzzler.

This should not deter others from jumping in with guesses or solutions.

I haven't even figured out if mark or space is 1, much less where LSB/MSB, LSD/MSD are in the layout.

May the cards fall in your favor.

February 15th, 2023 at 5:40:49 PM
permalink

Here are some more shots with the beginning date code the same on two different dates. Notice that the first 5 bits are the same from left to right on the top row

February 15th, 2023 at 5:43:47 PM
permalink

February 15th, 2023 at 5:50:50 PM
permalink

February 15th, 2023 at 6:49:54 PM
permalink

here's my screwy theory;

The first block never seems to be above 031, so it requires only 5 digits in binary. The other two blocks use the 43 binary symbols, say 10^25 for the middle block and 10^18 for the last block. To represent those three numbers in those three blocks in binary notation requires a total of 27 digits that are "1"and 21 that are "0" -and there are 27 black bars and 21 spaces.

But how can you connect those 27 bars in the 4x 12 grid to make those three numbers? I can't. But what if there is a key that enumerates which position on the 4x12 xy grid corresponds to which digit in the binary representation of the three numbers. So the spot in the upper left might correspond to the 2^16 digit in the 2nd block. And maybe the grid point immediate below it corresponds to the 2^3 digit in the first block. And so on, a randomly chosen correspondance.

The purpose of using such a scheme would be to make the tickets nearly impossible to counterfeit, because the counterfeiter would not know how to arrange the blocks to represent the winning number.

The first block never seems to be above 031, so it requires only 5 digits in binary. The other two blocks use the 43 binary symbols, say 10^25 for the middle block and 10^18 for the last block. To represent those three numbers in those three blocks in binary notation requires a total of 27 digits that are "1"and 21 that are "0" -and there are 27 black bars and 21 spaces.

But how can you connect those 27 bars in the 4x 12 grid to make those three numbers? I can't. But what if there is a key that enumerates which position on the 4x12 xy grid corresponds to which digit in the binary representation of the three numbers. So the spot in the upper left might correspond to the 2^16 digit in the 2nd block. And maybe the grid point immediate below it corresponds to the 2^3 digit in the first block. And so on, a randomly chosen correspondance.

The purpose of using such a scheme would be to make the tickets nearly impossible to counterfeit, because the counterfeiter would not know how to arrange the blocks to represent the winning number.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

February 15th, 2023 at 8:07:46 PM
permalink

I don't think they're trying to be tricky. The first three numbers correspond to a date code for the day of the year. January 1 would be 001 and January 2 would be 002. That number shows a clear pattern in the first five binary digits in the example above with the three tickets. That should give a clue as to how the tickets are encoded. If you look at the racetrack ticket above. It is hexadecimal but they use more bits in the field 6 x 12 instead of 4 x 12. But that's where I'm stumped. Maybe a lookup table that corresponds to a binary number?? Seems like too much work.

February 15th, 2023 at 9:37:19 PM
permalink

Do you have any tickets from a few months later?

I agree that the first block looks like the day of the year (+1). (That suggests 9 bits, not 8.)

All these tickets seem to have a 4 as the last digit; it could specify a format, rather than being a check digit.

I agree that the first block looks like the day of the year (+1). (That suggests 9 bits, not 8.)

All these tickets seem to have a 4 as the last digit; it could specify a format, rather than being a check digit.

May the cards fall in your favor.

February 15th, 2023 at 10:29:23 PM
permalink

All the ticket i have are from January but I see the last set of numbers begin with 16 and end with 4 and I found sets of the last three numbers which I grouped and took pics of.

February 15th, 2023 at 10:30:20 PM
permalink

February 15th, 2023 at 11:13:00 PM
permalink

Here are all the tickets for viewing

https://drive.google.com/file/d/1O9mu_O3R2mfyuD_PpFccTSLRMoQMHH0G/view

https://drive.google.com/file/d/1O9mu_O3R2mfyuD_PpFccTSLRMoQMHH0G/view

March 18th, 2023 at 10:12:45 PM
permalink

Can anyone solve this?

March 19th, 2023 at 1:50:36 PM
permalink

Quote:acura1234567890Can anyone solve this?

link to original post

Well, the first column appears to be just for the scanner to identify where each row exists.

The first five data columns on top match the first number in the identifier, with LSB first; i.e. 30 in binary is 11110, so with LSB first, it is 0-1-1-1-1; 27 is 11011, so it is 1-1-0-1-1.

I have a feeling the important numbers on the ticket are hashed, and that is the number that is in the barcode; there is another record of that ticket on some master computer, which can be verified if necessary. If the numbers themselves were on the ticket somehow, it would be far too easy to generate a fake winning ticket.

March 20th, 2023 at 9:06:56 AM
permalink

I don't think gtech was trying to be tricky or encode the bets on the serial number. There is another example that I found of a similar barcode that is used in the racetrack and it uses a larger field of blocks and then uses hexadecimal to encode the serial number. But this serial number ion the NJ lottery ticket is 17 digits and the barcode is a shorter field. Another thought I had was that since the first part of the barcode is a date code it may not need to encode all 17 digits and references it to serial number on the ticket. Any thoughts?