15 karma · joined February 28, 2025
In any case, the key point in my view is that there are 19,321 neighbors at distance ≤4. If we assume as an input condition that their values can be arbitrary—that is, the value of one neighbor has no relation to the others—then regardless of the implementation or mathematical identity used, we’ll end up performing 19,320 summations.
It’s a different story if we want to repeat this process for multiple points. In that case, we can optimize, since some neighbors might be shared and summed only once. This is exactly what my algorithm does: by handling everything in a matrix-based way, it reduces the number of summations per point to just 101 instead of 19,321. I’m not sure if there’s a specific mathematical identity behind this. In fact, I asked on StackExchange but haven’t had much success: https://math.stackexchange.com/questions/5040947/efficient-a...
The issue is that our problem isn’t exactly like that. I haven’t gone into too much detail, but you should be able to create a function that takes a pre-built space as input and sums over it, even if you don’t know how the space was built or whether it was constructed with random values.
For details on why this is the case:
Step 1.1: First, calculate, based on a table 3x14 like ‘probas’, which represents how many people have bet on event j for match i, the number of winners in each category if a certain prediction occurs (I assume homogeneity here). This is also a Hamming neighborhood sum, but you can use the boxes algorithm (~1 sec).
Step 1.2: The prizes depend (inversely) on the number of winners. So, apply a formula like this to the previous space: bet_price * coefficients / (winners + bet_price / revenue)
Where 'winners' are the values calculated in 1.1, and the rest are inputs: REVENUE = 1000000.0 PRICE = 0.75 COEFFICIENTS = [0.16, 0.075, 0.075, 0.075, 0.09] //percentaje of revenue correspondy to each category follow game rules
Step 1.3: Now we have the prizes for each prediction. To calculate the value of betting on a specific prediction, we need to do a sum product of the prizes corresponding to that bet (distance <= 4), each multiplied by its probability. So, we multiply the results from the previous formula by the probability of occurrence (using another table similar to ‘probas’). So it only least make the summation.
Step 2: Now that the space is constructed, we need to sum Hamming neighbors, and this is where there’s no shortcut that I know of. You have to assume the space contains randomly generated values. This is the computational bottleneck, and this is where the algorithm I mentioned applies. In fact, as you can see, it doesn’t only go to one space but five, one for each category. So, the sum shouldn’t go to r <= x but should sum the neighbors exactly at x in the corresponding space.
Once all values are computed, one approach is to search for the highest expected value, but that’s not the only criterion to manage: (1) there's a huge variance issue (you’ll win a lot, but with very low probability), (2) if you're placing multiple bets overlap reduces their combined value.
Great work on your approach! I’ll try to understand the code you linked, but I suspect it’s not doing exactly the same thing (or only for a very specific case). In my code, there's a function called SHN_boxes (which takes ~1s in Colab for this problem), and it’s a shortcut applicable only in some cases (not in this one). Did you use a similar approach?
4 seconds in google apps script by handling indices in loops instead of rearranging elements