Comment #2

hypersoniq — Mon, 11 Mar 2024 12:20:16 GMT

Before the function returns to process the next iteration, it will call another function that will handle the top 5 zero counts. This will be a list of lists that is ranked by that 11th element, the hit count. That means 10 billion iterations per column, where it will check the last draw, make a replacement "pick" and test it on the next result. It is a strange hobby, but I sharpen coding skills with each new idea.... [ More ]

Comment #1

hypersoniq — Mon, 11 Mar 2024 12:10:19 GMT

Once we have a run time on the first pick 2 game, we can extrapolate estimated run times for the rest. Why run all 10 billion combinations? That way I can say I left no stone unturned in the search. Each column in each game will run from the second draw, since the first will be needed to base the first pick. The initial test will run the first 1,000 iterations of the list, just to be sure the logic holds out and the top 5 ranking and retention algorithms work as... [ More ]

Original Blog Entry: Planning the ultimate search for the top replacement values.

hypersoniq — Sun, 10 Mar 2024 14:05:32 GMT

I had started with raw follower data, moved to a sequentially indexed mirror system, then moved to followers within that indexed system.

The basic premise, given the digit in the last draw, which one has the highest history of making a match?

With follower data plugged in, so far the zero count (where there is a match) is the highest, but there is yet to be a hit.

The next thing to do would be to check ALL possibilities of replacement values for each digit to find the best one for each column of each game.

For the basic mirror system, the replacement scheme is

0=5

1=6

2=7

3=8

4=9

5=0

6=1

7=2

8=3

9=4

On one side is the index for the last drawn number, on the other, the number to replace it with.

The total possible combinations of replacement values comes out to ten Billion!

So, why not test them all?

For each column, that means recording the number of hits , and that will result in running through 28 columns for the mid and eve PA pick 2 through pick 5 data, 280 billion iterations of a recursive algorithm in total...

How do we do that without running out of memory? By only recording the top 5 sets for each column.

How to do it? Lists!

Starting with [0, 0, 0, 0, 0, 0, 0, 0, 0] and ending with [9, 9, 9, 9, 9, 9, 9, 9, 9]

Each iteration will append a zero count, so that one of the top 5 output lists might look like [5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 750] and each of these lists will be stored in a list of lists which will be sorted by the appended zero count and compared to the most recent one, if it has a higher zero count it goes into the list and the lowest count gets popped off of the list.

The output will be the sorted list of lists, containing the 5 highest zero counts. 140 total lists. The zero counts will allow sorting into the top 5 without needing to store them all.

The main design challenges...

Writing a list generation loop to generate the lists.

Looping through the history files and counting the hits , and appending that number to the current list.

Sorting and ranking logic to maintain a running top 5 list of lists.

Perhaps writing output to a csv file in case the program crashes.

Setting up a system to run continuously. (Raspberry Pi version 5 should do the trick)

Fitting the whole system into my current recursive framework.

Will take some time and small tests to get the logic right. But that is the ultimate rabbit hole dive for any replacement system.

Happy Coding!... [ More ]