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Add solution for Project Euler problem 101 (#4033)
* Added solution for Project Euler problem 101 * Got rid of map functions * updating DIRECTORY.md * Better function/variable names * Better variable names * Type hints * Doctest for nested function Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_097/sol1.py)
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_097/sol1.py)
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* Problem 099
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* Problem 099
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_099/sol1.py)
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_099/sol1.py)
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* Problem 101
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_101/sol1.py)
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* Problem 112
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* Problem 112
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_112/sol1.py)
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_112/sol1.py)
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* Problem 113
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* Problem 113
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0
project_euler/problem_101/__init__.py
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project_euler/problem_101/__init__.py
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project_euler/problem_101/sol1.py
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project_euler/problem_101/sol1.py
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"""
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If we are presented with the first k terms of a sequence it is impossible to say with
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certainty the value of the next term, as there are infinitely many polynomial functions
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that can model the sequence.
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As an example, let us consider the sequence of cube
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numbers. This is defined by the generating function,
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u(n) = n3: 1, 8, 27, 64, 125, 216, ...
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Suppose we were only given the first two terms of this sequence. Working on the
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principle that "simple is best" we should assume a linear relationship and predict the
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next term to be 15 (common difference 7). Even if we were presented with the first three
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terms, by the same principle of simplicity, a quadratic relationship should be
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assumed.
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We shall define OP(k, n) to be the nth term of the optimum polynomial
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generating function for the first k terms of a sequence. It should be clear that
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OP(k, n) will accurately generate the terms of the sequence for n ≤ k, and potentially
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the first incorrect term (FIT) will be OP(k, k+1); in which case we shall call it a
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bad OP (BOP).
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As a basis, if we were only given the first term of sequence, it would be most
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sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u(1).
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Hence we obtain the
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following OPs for the cubic sequence:
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OP(1, n) = 1 1, 1, 1, 1, ...
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OP(2, n) = 7n-6 1, 8, 15, ...
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OP(3, n) = 6n^2-11n+6 1, 8, 27, 58, ...
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OP(4, n) = n^3 1, 8, 27, 64, 125, ...
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Clearly no BOPs exist for k ≥ 4.
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By considering the sum of FITs generated by the BOPs (indicated in red above), we
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obtain 1 + 15 + 58 = 74.
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Consider the following tenth degree polynomial generating function:
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1 - n + n^2 - n^3 + n^4 - n^5 + n^6 - n^7 + n^8 - n^9 + n^10
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Find the sum of FITs for the BOPs.
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"""
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from typing import Callable, List, Union
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Matrix = List[List[Union[float, int]]]
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def solve(matrix: Matrix, vector: Matrix) -> Matrix:
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"""
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Solve the linear system of equations Ax = b (A = "matrix", b = "vector")
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for x using Gaussian elimination and back substitution. We assume that A
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is an invertible square matrix and that b is a column vector of the
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same height.
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>>> solve([[1, 0], [0, 1]], [[1],[2]])
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[[1.0], [2.0]]
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>>> solve([[2, 1, -1],[-3, -1, 2],[-2, 1, 2]],[[8], [-11],[-3]])
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[[2.0], [3.0], [-1.0]]
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"""
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size: int = len(matrix)
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augmented: Matrix = [[0 for _ in range(size + 1)] for _ in range(size)]
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row: int
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row2: int
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col: int
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col2: int
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pivot_row: int
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ratio: float
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for row in range(size):
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for col in range(size):
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augmented[row][col] = matrix[row][col]
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augmented[row][size] = vector[row][0]
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row = 0
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col = 0
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while row < size and col < size:
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# pivoting
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pivot_row = max(
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[(abs(augmented[row2][col]), row2) for row2 in range(col, size)]
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)[1]
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if augmented[pivot_row][col] == 0:
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col += 1
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continue
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else:
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augmented[row], augmented[pivot_row] = augmented[pivot_row], augmented[row]
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for row2 in range(row + 1, size):
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ratio = augmented[row2][col] / augmented[row][col]
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augmented[row2][col] = 0
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for col2 in range(col + 1, size + 1):
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augmented[row2][col2] -= augmented[row][col2] * ratio
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row += 1
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col += 1
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# back substitution
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for col in range(1, size):
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for row in range(col):
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ratio = augmented[row][col] / augmented[col][col]
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for col2 in range(col, size + 1):
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augmented[row][col2] -= augmented[col][col2] * ratio
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# round to get rid of numbers like 2.000000000000004
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return [
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[round(augmented[row][size] / augmented[row][row], 10)] for row in range(size)
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]
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def interpolate(y_list: List[int]) -> Callable[[int], int]:
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"""
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Given a list of data points (1,y0),(2,y1), ..., return a function that
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interpolates the data points. We find the coefficients of the interpolating
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polynomial by solving a system of linear equations corresponding to
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x = 1, 2, 3...
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>>> interpolate([1])(3)
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1
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>>> interpolate([1, 8])(3)
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15
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>>> interpolate([1, 8, 27])(4)
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58
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>>> interpolate([1, 8, 27, 64])(6)
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216
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"""
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size: int = len(y_list)
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matrix: Matrix = [[0 for _ in range(size)] for _ in range(size)]
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vector: Matrix = [[0] for _ in range(size)]
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coeffs: Matrix
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x_val: int
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y_val: int
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col: int
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for x_val, y_val in enumerate(y_list):
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for col in range(size):
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matrix[x_val][col] = (x_val + 1) ** (size - col - 1)
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vector[x_val][0] = y_val
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coeffs = solve(matrix, vector)
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def interpolated_func(var: int) -> int:
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"""
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>>> interpolate([1])(3)
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1
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>>> interpolate([1, 8])(3)
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15
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>>> interpolate([1, 8, 27])(4)
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58
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>>> interpolate([1, 8, 27, 64])(6)
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216
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"""
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return sum(
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round(coeffs[x_val][0]) * (var ** (size - x_val - 1))
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for x_val in range(size)
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)
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return interpolated_func
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def question_function(variable: int) -> int:
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"""
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The generating function u as specified in the question.
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>>> question_function(0)
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1
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>>> question_function(1)
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1
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>>> question_function(5)
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8138021
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>>> question_function(10)
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9090909091
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"""
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return (
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1
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- variable
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+ variable ** 2
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- variable ** 3
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+ variable ** 4
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- variable ** 5
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+ variable ** 6
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- variable ** 7
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+ variable ** 8
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- variable ** 9
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+ variable ** 10
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)
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def solution(func: Callable[[int], int] = question_function, order: int = 10) -> int:
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"""
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Find the sum of the FITs of the BOPS. For each interpolating polynomial of order
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1, 2, ... , 10, find the first x such that the value of the polynomial at x does
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not equal u(x).
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>>> solution(lambda n: n ** 3, 3)
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74
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"""
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data_points: List[int] = [func(x_val) for x_val in range(1, order + 1)]
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polynomials: List[Callable[[int], int]] = [
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interpolate(data_points[:max_coeff]) for max_coeff in range(1, order + 1)
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]
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ret: int = 0
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poly: int
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x_val: int
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for poly in polynomials:
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x_val = 1
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while func(x_val) == poly(x_val):
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x_val += 1
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ret += poly(x_val)
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return ret
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if __name__ == "__main__":
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print(f"{solution() = }")
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