Quote from: nomachine on Today at 12:37:06 PM

Quote from: Andzhig on Today at 11:19:57 AM

from the standpoint of probability, this doesn’t seem to mean anything

Here is my prediction script. I have +- 0-40% deviations
With correction factor for puzzle 66
alpha = 2.7931615407112251695923857484645003617613050861

more decimals in correction factor more precise result

Code:

import numpy as np
from mpmath import mp
import argparse

# Set higher decimal precision for mpmath
mp.dps = 60 # Adjust precision as needed

# Target numbers (ordinal, value)
target_numbers = [
   (1, 1), (2, 3), (3, 7), (4, 8), (5, 21), (6, 49), (7, 76), (8, 224), (9, 467), (10, 514),
   (11, 1155), (12, 2683), (13, 5216), (14, 10544), (15, 26867), (16, 51510),
   (17, 95823), (18, 198669), (19, 357535), (20, 863317), (21, 1811764),
   (22, 3007503), (23, 5598802), (24, 14428676), (25, 33185509),
   (26, 54538862), (27, 111949941), (28, 227634408), (29, 400708894),
   (30, 1033162084), (31, 2102388551), (32, 3093472814), (33, 7137437912),
   (34, 14133072157), (35, 20112871792), (36, 42387769980), (37, 100251560595),
   (38, 146971536592), (39, 323724968937), (40, 1003651412950),
   (41, 1458252205147), (42, 2895374552463), (43, 7409811047825),
   (44, 15404761757071), (45, 19996463086597), (46, 51408670348612),
   (47, 119666659114170), (48, 191206974700443), (49, 409118905032525),
   (50, 611140496167764), (51, 2058769515153876), (52, 4216495639600700),
   (53, 6763683971478124), (54, 9974455244496707), (55, 30045390491869460),
   (56, 44218742292676575), (57, 138245758910846492), (58, 199976667976342049),
   (59, 525070384258266191), (60, 1135041350219496382), (61, 1425787542618654982),
   (62, 3908372542507822062), (63, 8993229949524469768),
   (64, 17799667357578236628), (65, 30568377312064202855),
   (66, 46346217550346335726)
]

def main(puzzle_number):
   # Extracting the ordinal (x) and values (y) up to the puzzle before the target puzzle
   if puzzle_number in [x[0] for x in target_numbers]:
   ordinals = np.array([x[0] for x in target_numbers if x[0] < puzzle_number], dtype=float)
   values = np.array([x[1] for x in target_numbers if x[0] < puzzle_number], dtype=float)
   else:
   # Use all available data for prediction if puzzle_number is not in the list
   ordinals = np.array([x[0] for x in target_numbers], dtype=float)
   values = np.array([x[1] for x in target_numbers], dtype=float)

   # Compute log of the values using mpmath for high precision
   log_values = np.array([float(mp.log(val)) for val in values], dtype=float)

   # Introduce weights to give more importance to recent values
   weights = np.linspace(0.5, 1, len(ordinals)) # Linear weighting

   # Perform linear regression on log(values) to fit the exponential model, using weights
   coefficients = np.polyfit(ordinals, log_values, 1, w=weights)

   # Extracting the coefficients
   log_a = coefficients[1]
   log_b = coefficients[0]

   # Convert coefficients back to the exponential form
   a = mp.exp(log_a)
   b = mp.exp(log_b)

   # Correction factor
   alpha = 2.7931615407112251695923857484645003617613050861

   # Predict the value for the given puzzle using the exponential model
   predicted_value = (a * mp.power(b, puzzle_number)) - ((((2 ** puzzle_number) - 1) - (2 ** (puzzle_number - 1))) / alpha)

   # Check if the real value is available
   real_value = None
   if puzzle_number in [x[0] for x in target_numbers]:
   real_value = [x[1] for x in target_numbers if x[0] == puzzle_number][0]

   # Function to format mpmath values to avoid scientific notation
   def format_mp(value, decimals=None):
   if decimals is not None:
   # Use string formatting to keep decimal places without thousands separators
   return f"{float(value):,.{decimals}f}".replace(',', '')
   return mp.nstr(value, mp.dps)

   # Compare the corrected value with the real value for the given puzzle
   if real_value is not None:
   print(f"Puzzle {puzzle_number} Real Value: {format_mp(real_value)}")
   print(f"Puzzle {puzzle_number} Predicted Value: {format_mp(predicted_value, 2)}")

   # Calculate the percentage error in the corrected prediction if real value is available
   if real_value is not None:
   error_percentage = abs((predicted_value - real_value) / real_value) * 100
   print(f"Percentage error in the corrected prediction for Puzzle {puzzle_number}: {format_mp(error_percentage, 2)}%")

if __name__ == "__main__":
   parser = argparse.ArgumentParser(description="Predict puzzle values based on an exponential model.")
   parser.add_argument("puzzle_number", type=int, help="Puzzle number to predict")
   args = parser.parse_args()
   main(args.puzzle_number)

python3 predict.py 20
Puzzle 20 Real Value: 863317
Puzzle 20 Predicted Value: 580555.48
Percentage error in the corrected prediction for Puzzle 20: 32.75%

python3 predict.py 27
Puzzle 27 Real Value: 111949941
Puzzle 27 Predicted Value: 88704196.49
Percentage error in the corrected prediction for Puzzle 27: 20.76%

python3 predict.py 45
Puzzle 45 Real Value: 19996463086597
Puzzle 45 Predicted Value: 20580213444471.79
Percentage error in the corrected prediction for Puzzle 45: 2.92%

python3 predict.py 50
Puzzle 50 Real Value: 611140496167764
Puzzle 50 Predicted Value: 635143323281525.25
Percentage error in the corrected prediction for Puzzle 50: 3.93%

python3 predict.py 63
Puzzle 63 Real Value: 8993229949524469768
Puzzle 63 Predicted Value: 5468647699742324736.00
Percentage error in the corrected prediction for Puzzle 63: 39.19%

python3 predict.py 66
Puzzle 66 Real Value: 46346217550346335726
Puzzle 66 Predicted Value: 46346217550346338304.00
Percentage error in the corrected prediction for Puzzle 66: 0.00%

python3 predict.py 67
Puzzle 67 Predicted Value: 90626458014188470272.00

This is like looking into a crystal ball. It doesn't mean anything. Grin

try predict all upper known ranges,what error will be ?