Enable ruff RUF002 rule (#11377)

* Enable ruff RUF002 rule

* Fix

---------

Co-authored-by: Christian Clauss <cclauss@me.com>

backtracking/sudoku.py

@@ -1,7 +1,7 @@
 """
-Given a partially filled 9×9 2D array, the objective is to fill a 9×9
+Given a partially filled 9x9 2D array, the objective is to fill a 9x9
 square grid with digits numbered 1 to 9, so that every row, column, and
-and each of the nine 3×3 sub-grids contains all of the digits.
+and each of the nine 3x3 sub-grids contains all of the digits.
 
 This can be solved using Backtracking and is similar to n-queens.
 We check to see if a cell is safe or not and recursively call the

bit_manipulation/single_bit_manipulation_operations.py

@@ -8,8 +8,8 @@ def set_bit(number: int, position: int) -> int:
     Set the bit at position to 1.
 
     Details: perform bitwise or for given number and X.
-    Where X is a number with all the bits – zeroes and bit on given
-    position – one.
+    Where X is a number with all the bits - zeroes and bit on given
+    position - one.
 
     >>> set_bit(0b1101, 1) # 0b1111
     15
@@ -26,8 +26,8 @@ def clear_bit(number: int, position: int) -> int:
     Set the bit at position to 0.
 
     Details: perform bitwise and for given number and X.
-    Where X is a number with all the bits – ones and bit on given
-    position – zero.
+    Where X is a number with all the bits - ones and bit on given
+    position - zero.
 
     >>> clear_bit(0b10010, 1) # 0b10000
     16
@@ -42,8 +42,8 @@ def flip_bit(number: int, position: int) -> int:
     Flip the bit at position.
 
     Details: perform bitwise xor for given number and X.
-    Where X is a number with all the bits – zeroes and bit on given
-    position – one.
+    Where X is a number with all the bits - zeroes and bit on given
+    position - one.
 
     >>> flip_bit(0b101, 1) # 0b111
     7
@@ -79,7 +79,7 @@ def get_bit(number: int, position: int) -> int:
     Get the bit at the given position
 
     Details: perform bitwise and for the given number and X,
-    Where X is a number with all the bits – zeroes and bit on given position – one.
+    Where X is a number with all the bits - zeroes and bit on given position - one.
     If the result is not equal to 0, then the bit on the given position is 1, else 0.
 
     >>> get_bit(0b1010, 0)

compression/burrows_wheeler.py

@@ -1,7 +1,7 @@
 """
 https://en.wikipedia.org/wiki/Burrows%E2%80%93Wheeler_transform
 
-The Burrows–Wheeler transform (BWT, also called block-sorting compression)
+The Burrows-Wheeler transform (BWT, also called block-sorting compression)
 rearranges a character string into runs of similar characters. This is useful
 for compression, since it tends to be easy to compress a string that has runs
 of repeated characters by techniques such as move-to-front transform and

compression/lempel_ziv.py

@@ -1,5 +1,5 @@
 """
-One of the several implementations of Lempel–Ziv–Welch compression algorithm
+One of the several implementations of Lempel-Ziv-Welch compression algorithm
 https://en.wikipedia.org/wiki/Lempel%E2%80%93Ziv%E2%80%93Welch
 """
 
@@ -43,7 +43,7 @@ def add_key_to_lexicon(
 
 def compress_data(data_bits: str) -> str:
     """
-    Compresses given data_bits using Lempel–Ziv–Welch compression algorithm
+    Compresses given data_bits using Lempel-Ziv-Welch compression algorithm
     and returns the result as a string
     """
     lexicon = {"0": "0", "1": "1"}

compression/lempel_ziv_decompress.py

@@ -1,5 +1,5 @@
 """
-One of the several implementations of Lempel–Ziv–Welch decompression algorithm
+One of the several implementations of Lempel-Ziv-Welch decompression algorithm
 https://en.wikipedia.org/wiki/Lempel%E2%80%93Ziv%E2%80%93Welch
 """
 
@@ -26,7 +26,7 @@ def read_file_binary(file_path: str) -> str:
 
 def decompress_data(data_bits: str) -> str:
     """
-    Decompresses given data_bits using Lempel–Ziv–Welch compression algorithm
+    Decompresses given data_bits using Lempel-Ziv-Welch compression algorithm
     and returns the result as a string
     """
     lexicon = {"0": "0", "1": "1"}

data_structures/binary_tree/red_black_tree.py

@@ -17,7 +17,7 @@ class RedBlackTree:
     and slower for reading in the average case, though, because they're
     both balanced binary search trees, both will get the same asymptotic
     performance.
-    To read more about them, https://en.wikipedia.org/wiki/Red–black_tree
+    To read more about them, https://en.wikipedia.org/wiki/Red-black_tree
     Unless otherwise specified, all asymptotic runtimes are specified in
     terms of the size of the tree.
     """

digital_image_processing/edge_detection/canny.py

@@ -74,9 +74,9 @@ def detect_high_low_threshold(
     image_shape, destination, threshold_low, threshold_high, weak, strong
 ):
     """
-    High-Low threshold detection. If an edge pixel’s gradient value is higher
+    High-Low threshold detection. If an edge pixel's gradient value is higher
     than the high threshold value, it is marked as a strong edge pixel. If an
-    edge pixel’s gradient value is smaller than the high threshold value and
+    edge pixel's gradient value is smaller than the high threshold value and
     larger than the low threshold value, it is marked as a weak edge pixel. If
     an edge pixel's value is smaller than the low threshold value, it will be
     suppressed.

digital_image_processing/index_calculation.py

@@ -182,7 +182,7 @@ class IndexCalculation:
         Atmospherically Resistant Vegetation Index 2
         https://www.indexdatabase.de/db/i-single.php?id=396
         :return: index
-            −0.18+1.17*(self.nir−self.red)/(self.nir+self.red)
+            -0.18+1.17*(self.nir-self.red)/(self.nir+self.red)
         """
         return -0.18 + (1.17 * ((self.nir - self.red) / (self.nir + self.red)))
 

dynamic_programming/combination_sum_iv.py

@@ -18,7 +18,7 @@ Approach:
 The basic idea is to go over recursively to find the way such that the sum
 of chosen elements is “tar”. For every element, we have two choices
     1. Include the element in our set of chosen elements.
-    2. Don’t include the element in our set of chosen elements.
+    2. Don't include the element in our set of chosen elements.
 """
 
 

electronics/coulombs_law.py

@@ -20,8 +20,8 @@ def couloumbs_law(
 
     Reference
     ----------
-    Coulomb (1785) "Premier mémoire sur l’électricité et le magnétisme,"
-    Histoire de l’Académie Royale des Sciences, pp. 569–577.
+    Coulomb (1785) "Premier mémoire sur l'électricité et le magnétisme,"
+    Histoire de l'Académie Royale des Sciences, pp. 569-577.
 
     Parameters
     ----------

hashes/fletcher16.py

@@ -1,6 +1,6 @@
 """
 The Fletcher checksum is an algorithm for computing a position-dependent
-checksum devised by John G. Fletcher (1934–2012) at Lawrence Livermore Labs
+checksum devised by John G. Fletcher (1934-2012) at Lawrence Livermore Labs
 in the late 1970s.[1] The objective of the Fletcher checksum was to
 provide error-detection properties approaching those of a cyclic
 redundancy check but with the lower computational effort associated

linear_algebra/lu_decomposition.py

@@ -1,5 +1,5 @@
 """
-Lower–upper (LU) decomposition factors a matrix as a product of a lower
+Lower-upper (LU) decomposition factors a matrix as a product of a lower
 triangular matrix and an upper triangular matrix. A square matrix has an LU
 decomposition under the following conditions:
     - If the matrix is invertible, then it has an LU decomposition if and only

linear_algebra/src/schur_complement.py

@@ -18,7 +18,7 @@ def schur_complement(
     the pseudo_inv argument.
 
     Link to Wiki: https://en.wikipedia.org/wiki/Schur_complement
-    See also Convex Optimization – Boyd and Vandenberghe, A.5.5
+    See also Convex Optimization - Boyd and Vandenberghe, A.5.5
     >>> import numpy as np
     >>> a = np.array([[1, 2], [2, 1]])
     >>> b = np.array([[0, 3], [3, 0]])

machine_learning/polynomial_regression.py

@@ -11,7 +11,7 @@ for polynomial regression:
 
 β = (XᵀX)⁻¹Xᵀy = X⁺y
 
-where X is the design matrix, y is the response vector, and X⁺ denotes the Moore–Penrose
+where X is the design matrix, y is the response vector, and X⁺ denotes the Moore-Penrose
 pseudoinverse of X. In the case of polynomial regression, the design matrix is
 
     |1  x₁  x₁² ⋯ x₁ᵐ|
@@ -106,7 +106,7 @@ class PolynomialRegression:
 
         β = (XᵀX)⁻¹Xᵀy = X⁺y
 
-        where X⁺ denotes the Moore–Penrose pseudoinverse of the design matrix X. This
+        where X⁺ denotes the Moore-Penrose pseudoinverse of the design matrix X. This
         function computes X⁺ using singular value decomposition (SVD).
 
         References:

maths/chudnovsky_algorithm.py

@@ -5,7 +5,7 @@ from math import ceil, factorial
 def pi(precision: int) -> str:
     """
     The Chudnovsky algorithm is a fast method for calculating the digits of PI,
-    based on Ramanujan’s PI formulae.
+    based on Ramanujan's PI formulae.
 
     https://en.wikipedia.org/wiki/Chudnovsky_algorithm
 

maths/entropy.py

@@ -21,10 +21,10 @@ def calculate_prob(text: str) -> None:
     :return: Prints
     1) Entropy of information based on 1 alphabet
     2) Entropy of information based on couples of 2 alphabet
-    3) print Entropy of H(X n∣Xn−1)
+    3) print Entropy of H(X n|Xn-1)
 
     Text from random books. Also, random quotes.
-    >>> text = ("Behind Winston’s back the voice "
+    >>> text = ("Behind Winston's back the voice "
     ...         "from the telescreen was still "
     ...         "babbling and the overfulfilment")
     >>> calculate_prob(text)

maths/lucas_lehmer_primality_test.py

@@ -1,12 +1,12 @@
 """
-In mathematics, the Lucas–Lehmer test (LLT) is a primality test for Mersenne
+In mathematics, the Lucas-Lehmer test (LLT) is a primality test for Mersenne
 numbers.  https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test
 
 A Mersenne number is a number that is one less than a power of two.
 That is M_p = 2^p - 1
 https://en.wikipedia.org/wiki/Mersenne_prime
 
-The Lucas–Lehmer test is the primality test used by the
+The Lucas-Lehmer test is the primality test used by the
 Great Internet Mersenne Prime Search (GIMPS) to locate large primes.
 """
 

maths/modular_division.py

@@ -9,7 +9,7 @@ def modular_division(a: int, b: int, n: int) -> int:
     GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
 
     Given three integers a, b, and n, such that gcd(a,n)=1 and n>1, the algorithm should
-    return an integer x such that 0≤x≤n−1, and  b/a=x(modn) (that is, b=ax(modn)).
+    return an integer x such that 0≤x≤n-1, and  b/a=x(modn) (that is, b=ax(modn)).
 
     Theorem:
     a has a multiplicative inverse modulo n iff gcd(a,n) = 1

maths/numerical_analysis/bisection_2.py

@@ -1,5 +1,5 @@
 """
-Given a function on floating number f(x) and two floating numbers ‘a’ and ‘b’ such that
+Given a function on floating number f(x) and two floating numbers `a` and `b` such that
 f(a) * f(b) < 0 and f(x) is continuous in [a, b].
 Here f(x) represents algebraic or transcendental equation.
 Find root of function in interval [a, b] (Or find a value of x such that f(x) is 0)

maths/numerical_analysis/nevilles_method.py

@@ -1,7 +1,7 @@
 """
 Python program to show how to interpolate and evaluate a polynomial
 using Neville's method.
-Neville’s method evaluates a polynomial that passes through a
+Neville's method evaluates a polynomial that passes through a
 given set of x and y points for a particular x value (x0) using the
 Newton polynomial form.
 Reference:

maths/simultaneous_linear_equation_solver.py

@@ -2,10 +2,10 @@
 https://en.wikipedia.org/wiki/Augmented_matrix
 
 This algorithm solves simultaneous linear equations of the form
-λa + λb + λc + λd + ... = γ as [λ, λ, λ, λ, ..., γ]
-Where λ & γ are individual coefficients, the no. of equations = no. of coefficients - 1
+λa + λb + λc + λd + ... = y as [λ, λ, λ, λ, ..., y]
+Where λ & y are individual coefficients, the no. of equations = no. of coefficients - 1
 
-Note in order to work there must exist 1 equation where all instances of λ and γ != 0
+Note in order to work there must exist 1 equation where all instances of λ and y != 0
 """
 
 

matrix/largest_square_area_in_matrix.py

@@ -31,7 +31,7 @@ Explanation: There is no 1 in the matrix.
 
 Approach:
 We initialize another matrix (dp) with the same dimensions
-as the original one initialized with all 0’s.
+as the original one initialized with all 0's.
 
 dp_array(i,j) represents the side length of the maximum square whose
 bottom right corner is the cell with index (i,j) in the original matrix.
@@ -39,7 +39,7 @@ bottom right corner is the cell with index (i,j) in the original matrix.
 Starting from index (0,0), for every 1 found in the original matrix,
 we update the value of the current element as
 
-dp_array(i,j)=dp_array(dp(i−1,j),dp_array(i−1,j−1),dp_array(i,j−1)) + 1.
+dp_array(i,j)=dp_array(dp(i-1,j),dp_array(i-1,j-1),dp_array(i,j-1)) + 1.
 """
 
 

matrix/spiral_print.py

@@ -89,7 +89,7 @@ def spiral_traversal(matrix: list[list]) -> list[int]:
     Algorithm:
         Step 1. first pop the 0 index list. (which is [1,2,3,4] and concatenate the
                 output of [step 2])
-        Step 2. Now perform matrix’s Transpose operation (Change rows to column
+        Step 2. Now perform matrix's Transpose operation (Change rows to column
                 and vice versa) and reverse the resultant matrix.
         Step 3. Pass the output of [2nd step], to same recursive function till
                 base case hits.

neural_network/back_propagation_neural_network.py

@@ -2,10 +2,10 @@
 
 """
 
-A Framework of Back Propagation Neural Network（BP） model
+A Framework of Back Propagation Neural Network (BP) model
 
 Easy to use:
-    * add many layers as you want ！！！
+    * add many layers as you want ! ! !
     * clearly see how the loss decreasing
 Easy to expand:
     * more activation functions

other/davis_putnam_logemann_loveland.py

@@ -1,7 +1,7 @@
 #!/usr/bin/env python3
 
 """
-Davis–Putnam–Logemann–Loveland (DPLL) algorithm is a complete, backtracking-based
+Davis-Putnam-Logemann-Loveland (DPLL) algorithm is a complete, backtracking-based
 search algorithm for deciding the satisfiability of propositional logic formulae in
 conjunctive normal form, i.e, for solving the Conjunctive Normal Form SATisfiability
 (CNF-SAT) problem.

other/fischer_yates_shuffle.py

@@ -1,6 +1,6 @@
 #!/usr/bin/python
 """
-The Fisher–Yates shuffle is an algorithm for generating a random permutation of a
+The Fisher-Yates shuffle is an algorithm for generating a random permutation of a
 finite sequence.
 For more details visit
 wikipedia/Fischer-Yates-Shuffle.

physics/archimedes_principle_of_buoyant_force.py

@@ -3,7 +3,7 @@ Calculate the buoyant force of any body completely or partially submerged in a s
 fluid.  This principle was discovered by the Greek mathematician Archimedes.
 
 Equation for calculating buoyant force:
-Fb = ρ * V * g
+Fb = p * V * g
 
 https://en.wikipedia.org/wiki/Archimedes%27_principle
 """

physics/center_of_mass.py

@@ -16,8 +16,8 @@ assumed to be concentrated to visualize its motion. In other words, the center o
 is the particle equivalent of a given object for the application of Newton's laws of
 motion.
 
-In the case of a system of particles P_i, i = 1, ..., n , each with mass m_i that are
-located in space with coordinates r_i, i = 1, ..., n , the coordinates R of the center
+In the case of a system of particles P_i, i = 1, ..., n , each with mass m_i that are
+located in space with coordinates r_i, i = 1, ..., n , the coordinates R of the center
 of mass corresponds to:
 
 R = (Σ(mi * ri) / Σ(mi))
@@ -36,8 +36,8 @@ def center_of_mass(particles: list[Particle]) -> Coord3D:
     Input Parameters
     ----------------
     particles: list(Particle):
-    A list of particles where each particle is a tuple with it´s (x, y, z) position and
-    it´s mass.
+    A list of particles where each particle is a tuple with it's (x, y, z) position and
+    it's mass.
 
     Returns
     -------

physics/centripetal_force.py

@@ -6,7 +6,7 @@ or centre of curvature.
 The unit of centripetal force is newton.
 
 The centripetal force is always directed perpendicular to the
-direction of the object’s displacement. Using Newton’s second
+direction of the object's displacement. Using Newton's second
 law of motion, it is found that the centripetal force of an object
 moving in a circular path always acts towards the centre of the circle.
 The Centripetal Force Formula is given as the product of mass (in kg)

physics/lorentz_transformation_four_vector.py

@@ -12,13 +12,13 @@ two inertial reference frames and X' moves in the x direction with velocity v
 with respect to X, then the Lorentz transformation from X to X' is X' = BX,
 where
 
-    | γ  -γβ  0  0|
-B = |-γβ  γ   0  0|
+    | y  -γβ  0  0|
+B = |-γβ  y   0  0|
     | 0   0   1  0|
     | 0   0   0  1|
 
 is the matrix describing the Lorentz boost between X and X',
-γ = 1 / √(1 - v²/c²) is the Lorentz factor, and β = v/c is the velocity as
+y = 1 / √(1 - v²/c²) is the Lorentz factor, and β = v/c is the velocity as
 a fraction of c.
 
 Reference: https://en.wikipedia.org/wiki/Lorentz_transformation
@@ -63,7 +63,7 @@ def beta(velocity: float) -> float:
 
 def gamma(velocity: float) -> float:
     """
-    Calculate the Lorentz factor γ = 1 / √(1 - v²/c²) for a given velocity
+    Calculate the Lorentz factor y = 1 / √(1 - v²/c²) for a given velocity
     >>> gamma(4)
     1.0000000000000002
     >>> gamma(1e5)
@@ -90,12 +90,12 @@ def transformation_matrix(velocity: float) -> np.ndarray:
     """
     Calculate the Lorentz transformation matrix for movement in the x direction:
 
-    | γ  -γβ  0  0|
-    |-γβ  γ   0  0|
+    | y  -γβ  0  0|
+    |-γβ  y   0  0|
     | 0   0   1  0|
     | 0   0   0  1|
 
-    where γ is the Lorentz factor and β is the velocity as a fraction of c
+    where y is the Lorentz factor and β is the velocity as a fraction of c
     >>> transformation_matrix(29979245)
     array([[ 1.00503781, -0.10050378,  0.        ,  0.        ],
            [-0.10050378,  1.00503781,  0.        ,  0.        ],

physics/reynolds_number.py

@@ -8,10 +8,10 @@ pipe. Reynolds number is defined by the ratio of inertial forces to that of
 viscous forces.
 
 R = Inertial Forces / Viscous Forces
-R = (ρ * V * D)/μ
+R = (p * V * D)/μ
 
 where :
-ρ = Density of fluid (in Kg/m^3)
+p = Density of fluid (in Kg/m^3)
 D = Diameter of pipe through which fluid flows (in m)
 V = Velocity of flow of the fluid (in m/s)
 μ = Viscosity of the fluid (in Ns/m^2)

physics/terminal_velocity.py

@@ -8,13 +8,13 @@ and buoyancy is equal to the downward gravity force acting on the
 object. The acceleration of the object is zero as the net force acting on
 the object is zero.
 
-Vt = ((2 * m * g)/(ρ * A * Cd))^0.5
+Vt = ((2 * m * g)/(p * A * Cd))^0.5
 
 where :
 Vt = Terminal velocity (in m/s)
 m = Mass of the falling object (in Kg)
 g = Acceleration due to gravity (value taken : imported from scipy)
-ρ = Density of the fluid through which the object is falling (in Kg/m^3)
+p = Density of the fluid through which the object is falling (in Kg/m^3)
 A = Projected area of the object (in m^2)
 Cd = Drag coefficient (dimensionless)
 

project_euler/problem_004/sol1.py

@@ -4,7 +4,7 @@ Project Euler Problem 4: https://projecteuler.net/problem=4
 Largest palindrome product
 
 A palindromic number reads the same both ways. The largest palindrome made
-from the product of two 2-digit numbers is 9009 = 91 × 99.
+from the product of two 2-digit numbers is 9009 = 91 x 99.
 
 Find the largest palindrome made from the product of two 3-digit numbers.
 

project_euler/problem_004/sol2.py

@@ -4,7 +4,7 @@ Project Euler Problem 4: https://projecteuler.net/problem=4
 Largest palindrome product
 
 A palindromic number reads the same both ways. The largest palindrome made
-from the product of two 2-digit numbers is 9009 = 91 × 99.
+from the product of two 2-digit numbers is 9009 = 91 x 99.
 
 Find the largest palindrome made from the product of two 3-digit numbers.
 

project_euler/problem_008/sol1.py

@@ -4,7 +4,7 @@ Project Euler Problem 8: https://projecteuler.net/problem=8
 Largest product in a series
 
 The four adjacent digits in the 1000-digit number that have the greatest
-product are 9 × 9 × 8 × 9 = 5832.
+product are 9 x 9 x 8 x 9 = 5832.
 
     73167176531330624919225119674426574742355349194934
     96983520312774506326239578318016984801869478851843

project_euler/problem_008/sol2.py

@@ -4,7 +4,7 @@ Project Euler Problem 8: https://projecteuler.net/problem=8
 Largest product in a series
 
 The four adjacent digits in the 1000-digit number that have the greatest
-product are 9 × 9 × 8 × 9 = 5832.
+product are 9 x 9 x 8 x 9 = 5832.
 
     73167176531330624919225119674426574742355349194934
     96983520312774506326239578318016984801869478851843

project_euler/problem_008/sol3.py

@@ -4,7 +4,7 @@ Project Euler Problem 8: https://projecteuler.net/problem=8
 Largest product in a series
 
 The four adjacent digits in the 1000-digit number that have the greatest
-product are 9 × 9 × 8 × 9 = 5832.
+product are 9 x 9 x 8 x 9 = 5832.
 
     73167176531330624919225119674426574742355349194934
     96983520312774506326239578318016984801869478851843

project_euler/problem_015/sol1.py

@@ -1,9 +1,9 @@
 """
 Problem 15: https://projecteuler.net/problem=15
 
-Starting in the top left corner of a 2×2 grid, and only being able to move to
+Starting in the top left corner of a 2x2 grid, and only being able to move to
 the right and down, there are exactly 6 routes to the bottom right corner.
-How many such routes are there through a 20×20 grid?
+How many such routes are there through a 20x20 grid?
 """
 
 from math import factorial

project_euler/problem_020/sol1.py

@@ -1,9 +1,9 @@
 """
 Problem 20: https://projecteuler.net/problem=20
 
-n! means n × (n − 1) × ... × 3 × 2 × 1
+n! means n x (n - 1) x ... x 3 x 2 x 1
 
-For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
 and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
 
 Find the sum of the digits in the number 100!

project_euler/problem_020/sol2.py

@@ -1,9 +1,9 @@
 """
 Problem 20: https://projecteuler.net/problem=20
 
-n! means n × (n − 1) × ... × 3 × 2 × 1
+n! means n x (n - 1) x ... x 3 x 2 x 1
 
-For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
 and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
 
 Find the sum of the digits in the number 100!

project_euler/problem_020/sol3.py

@@ -1,9 +1,9 @@
 """
 Problem 20: https://projecteuler.net/problem=20
 
-n! means n × (n − 1) × ... × 3 × 2 × 1
+n! means n x (n - 1) x ... x 3 x 2 x 1
 
-For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
 and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
 
 Find the sum of the digits in the number 100!

project_euler/problem_020/sol4.py

@@ -1,9 +1,9 @@
 """
 Problem 20: https://projecteuler.net/problem=20
 
-n! means n × (n − 1) × ... × 3 × 2 × 1
+n! means n x (n - 1) x ... x 3 x 2 x 1
 
-For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
+For example, 10! = 10 x 9 x ... x 3 x 2 x 1 = 3628800,
 and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
 
 Find the sum of the digits in the number 100!

project_euler/problem_022/sol1.py

@@ -10,7 +10,7 @@ score.
 
 For example, when the list is sorted into alphabetical order, COLIN, which is
 worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would
-obtain a score of 938 × 53 = 49714.
+obtain a score of 938 x 53 = 49714.
 
 What is the total of all the name scores in the file?
 """

project_euler/problem_022/sol2.py

@@ -10,7 +10,7 @@ score.
 
 For example, when the list is sorted into alphabetical order, COLIN, which is
 worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would
-obtain a score of 938 × 53 = 49714.
+obtain a score of 938 x 53 = 49714.
 
 What is the total of all the name scores in the file?
 """

project_euler/problem_025/sol1.py

@@ -1,7 +1,7 @@
 """
 The Fibonacci sequence is defined by the recurrence relation:
 
-    Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+    Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
 
 Hence the first 12 terms will be:
 

project_euler/problem_025/sol2.py

@@ -1,7 +1,7 @@
 """
 The Fibonacci sequence is defined by the recurrence relation:
 
-    Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+    Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
 
 Hence the first 12 terms will be:
 

project_euler/problem_025/sol3.py

@@ -1,7 +1,7 @@
 """
 The Fibonacci sequence is defined by the recurrence relation:
 
-    Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+    Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
 
 Hence the first 12 terms will be:
 

project_euler/problem_027/sol1.py

@@ -9,12 +9,12 @@ n2 + n + 41
 It turns out that the formula will produce 40 primes for the consecutive values
 n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible
 by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41.
-The incredible formula  n2 − 79n + 1601 was discovered, which produces 80 primes
-for the consecutive values n = 0 to 79. The product of the coefficients, −79 and
-1601, is −126479.
+The incredible formula  n2 - 79n + 1601 was discovered, which produces 80 primes
+for the consecutive values n = 0 to 79. The product of the coefficients, -79 and
+1601, is -126479.
 Considering quadratics of the form:
 n² + an + b, where |a| < 1000 and |b| < 1000
-where |n| is the modulus/absolute value of ne.g. |11| = 11 and |−4| = 4
+where |n| is the modulus/absolute value of ne.g. |11| = 11 and |-4| = 4
 Find the product of the coefficients, a and b, for the quadratic expression that
 produces the maximum number of primes for consecutive values of n, starting with
 n = 0.

project_euler/problem_031/sol1.py

@@ -2,14 +2,14 @@
 Coin sums
 Problem 31: https://projecteuler.net/problem=31
 
-In England the currency is made up of pound, £, and pence, p, and there are
+In England the currency is made up of pound, f, and pence, p, and there are
 eight coins in general circulation:
 
-1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
-It is possible to make £2 in the following way:
+1p, 2p, 5p, 10p, 20p, 50p, f1 (100p) and f2 (200p).
+It is possible to make f2 in the following way:
 
-1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
-How many different ways can £2 be made using any number of coins?
+1xf1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
+How many different ways can f2 be made using any number of coins?
 """
 
 

project_euler/problem_031/sol2.py

@@ -3,17 +3,17 @@ Problem 31: https://projecteuler.net/problem=31
 
 Coin sums
 
-In England the currency is made up of pound, £, and pence, p, and there are
+In England the currency is made up of pound, f, and pence, p, and there are
 eight coins in general circulation:
 
-1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
-It is possible to make £2 in the following way:
+1p, 2p, 5p, 10p, 20p, 50p, f1 (100p) and f2 (200p).
+It is possible to make f2 in the following way:
 
-1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
-How many different ways can £2 be made using any number of coins?
+1xf1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
+How many different ways can f2 be made using any number of coins?
 
 Hint:
-    > There are 100 pence in a pound (£1 = 100p)
+    > There are 100 pence in a pound (f1 = 100p)
     > There are coins(in pence) are available: 1, 2, 5, 10, 20, 50, 100 and 200.
     > how many different ways you can combine these values to create 200 pence.
 

project_euler/problem_032/sol32.py

@@ -3,7 +3,7 @@ We shall say that an n-digit number is pandigital if it makes use of all the
 digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through
 5 pandigital.
 
-The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
+The product 7254 is unusual, as the identity, 39 x 186 = 7254, containing
 multiplicand, multiplier, and product is 1 through 9 pandigital.
 
 Find the sum of all products whose multiplicand/multiplier/product identity can

project_euler/problem_038/sol1.py

@@ -3,9 +3,9 @@ Project Euler Problem 38: https://projecteuler.net/problem=38
 
 Take the number 192 and multiply it by each of 1, 2, and 3:
 
-192 × 1 = 192
-192 × 2 = 384
-192 × 3 = 576
+192 x 1 = 192
+192 x 2 = 384
+192 x 3 = 576
 
 By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call
 192384576 the concatenated product of 192 and (1,2,3)

project_euler/problem_040/sol1.py

@@ -11,7 +11,7 @@ It can be seen that the 12th digit of the fractional part is 1.
 If dn represents the nth digit of the fractional part, find the value of the
 following expression.
 
-d1 × d10 × d100 × d1000 × d10000 × d100000 × d1000000
+d1 x d10 x d100 x d1000 x d10000 x d100000 x d1000000
 """
 
 

project_euler/problem_044/sol1.py

@@ -1,14 +1,14 @@
 """
 Problem 44: https://projecteuler.net/problem=44
 
-Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten
+Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2. The first ten
 pentagonal numbers are:
 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
 It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
-70 − 22 = 48, is not pentagonal.
+70 - 22 = 48, is not pentagonal.
 
 Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference
-are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
+are pentagonal and D = |Pk - Pj| is minimised; what is the value of D?
 """
 
 

project_euler/problem_045/sol1.py

@@ -3,8 +3,8 @@ Problem 45: https://projecteuler.net/problem=45
 
 Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
 Triangle	 	T(n) = (n * (n + 1)) / 2	 	1, 3, 6, 10, 15, ...
-Pentagonal	 	P(n) = (n * (3 * n − 1)) / 2	 	1, 5, 12, 22, 35, ...
-Hexagonal	 	H(n) = n * (2 * n − 1)	 	1, 6, 15, 28, 45, ...
+Pentagonal	 	P(n) = (n * (3 * n - 1)) / 2	 	1, 5, 12, 22, 35, ...
+Hexagonal	 	H(n) = n * (2 * n - 1)	 	1, 6, 15, 28, 45, ...
 It can be verified that T(285) = P(165) = H(143) = 40755.
 
 Find the next triangle number that is also pentagonal and hexagonal.

project_euler/problem_046/sol1.py

@@ -4,12 +4,12 @@ Problem 46: https://projecteuler.net/problem=46
 It was proposed by Christian Goldbach that every odd composite number can be
 written as the sum of a prime and twice a square.
 
-9 = 7 + 2 × 12
-15 = 7 + 2 × 22
-21 = 3 + 2 × 32
-25 = 7 + 2 × 32
-27 = 19 + 2 × 22
-33 = 31 + 2 × 12
+9 = 7 + 2 x 12
+15 = 7 + 2 x 22
+21 = 3 + 2 x 32
+25 = 7 + 2 x 32
+27 = 19 + 2 x 22
+33 = 31 + 2 x 12
 
 It turns out that the conjecture was false.
 

project_euler/problem_047/sol1.py

@@ -5,14 +5,14 @@ Problem 47
 
 The first two consecutive numbers to have two distinct prime factors are:
 
-14 = 2 × 7
-15 = 3 × 5
+14 = 2 x 7
+15 = 3 x 5
 
 The first three consecutive numbers to have three distinct prime factors are:
 
-644 = 2² × 7 × 23
-645 = 3 × 5 × 43
-646 = 2 × 17 × 19.
+644 = 2² x 7 x 23
+645 = 3 x 5 x 43
+646 = 2 x 17 x 19.
 
 Find the first four consecutive integers to have four distinct prime factors each.
 What is the first of these numbers?

project_euler/problem_053/sol1.py

@@ -10,7 +10,7 @@ In combinatorics, we use the notation, 5C3 = 10.
 
 In general,
 
-nCr = n!/(r!(n−r)!),where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
+nCr = n!/(r!(n-r)!),where r ≤ n, n! = nx(n-1)x...x3x2x1, and 0! = 1.
 It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
 
 How many, not necessarily distinct, values of nCr, for 1 ≤ n ≤ 100, are greater

project_euler/problem_097/sol1.py

@@ -1,7 +1,7 @@
 """
 The first known prime found to exceed one million digits was discovered in 1999,
-and is a Mersenne prime of the form 2**6972593 − 1; it contains exactly 2,098,960
-digits. Subsequently other Mersenne primes, of the form 2**p − 1, have been found
+and is a Mersenne prime of the form 2**6972593 - 1; it contains exactly 2,098,960
+digits. Subsequently other Mersenne primes, of the form 2**p - 1, have been found
 which contain more digits.
 However, in 2004 there was found a massive non-Mersenne prime which contains
 2,357,207 digits: (28433 * (2 ** 7830457 + 1)).

project_euler/problem_104/sol1.py

@@ -3,7 +3,7 @@ Project Euler Problem 104 : https://projecteuler.net/problem=104
 
 The Fibonacci sequence is defined by the recurrence relation:
 
-Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
+Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
 It turns out that F541, which contains 113 digits, is the first Fibonacci number
 for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9,
 but not necessarily in order). And F2749, which contains 575 digits, is the first

project_euler/problem_120/sol1.py

@@ -3,7 +3,7 @@ Problem 120 Square remainders: https://projecteuler.net/problem=120
 
 Description:
 
-Let r be the remainder when (a−1)^n + (a+1)^n is divided by a^2.
+Let r be the remainder when (a-1)^n + (a+1)^n is divided by a^2.
 For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49.
 And as n varies, so too will r, but for a = 7 it turns out that r_max = 42.
 For 3 ≤ a ≤ 1000, find ∑ r_max.

project_euler/problem_123/sol1.py

@@ -4,7 +4,7 @@ Problem 123: https://projecteuler.net/problem=123
 Name: Prime square remainders
 
 Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and
-let r be the remainder when (pn−1)^n + (pn+1)^n is divided by pn^2.
+let r be the remainder when (pn-1)^n + (pn+1)^n is divided by pn^2.
 
 For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.
 The least value of n for which the remainder first exceeds 10^9 is 7037.

project_euler/problem_135/sol1.py

@@ -3,9 +3,9 @@ Project Euler Problem 135: https://projecteuler.net/problem=135
 
 Given the positive integers, x, y, and z, are consecutive terms of an arithmetic
 progression, the least value of the positive integer, n, for which the equation,
-x2 − y2 − z2 = n, has exactly two solutions is n = 27:
+x2 - y2 - z2 = n, has exactly two solutions is n = 27:
 
-342 − 272 − 202 = 122 − 92 − 62 = 27
+342 - 272 - 202 = 122 - 92 - 62 = 27
 
 It turns out that n = 1155 is the least value which has exactly ten solutions.
 

project_euler/problem_144/sol1.py

@@ -6,7 +6,7 @@ works its way back out.
 The specific white cell we will be considering is an ellipse with the equation
 4x^2 + y^2 = 100
 
-The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, allowing the
+The section corresponding to -0.01 ≤ x ≤ +0.01 at the top is missing, allowing the
 light to enter and exit through the hole.
 
 The light beam in this problem starts at the point (0.0,10.1) just outside the white
@@ -20,7 +20,7 @@ In the figure on the left, the red line shows the first two points of contact be
 the laser beam and the wall of the white cell; the blue line shows the line tangent to
 the ellipse at the point of incidence of the first bounce.
 
-The slope m of the tangent line at any point (x,y) of the given ellipse is: m = −4x/y
+The slope m of the tangent line at any point (x,y) of the given ellipse is: m = -4x/y
 
 The normal line is perpendicular to this tangent line at the point of incidence.
 

project_euler/problem_174/sol1.py

@@ -14,7 +14,7 @@ t = 32 is type L(2).
 Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for example,
 N(15) = 832.
 
-What is ∑ N(n) for 1 ≤ n ≤ 10?
+What is sum N(n) for 1 ≤ n ≤ 10?
 """
 
 from collections import defaultdict

pyproject.toml

@@ -10,6 +10,7 @@ lint.ignore = [    # `ruff rule S101` for a description of that rule
   "PLW2901",  # PLW2901: Redefined loop variable -- FIX ME
   "PT011",    # `pytest.raises(Exception)` is too broad, set the `match` parameter or use a more specific exception
   "PT018",    # Assertion should be broken down into multiple parts
+  "RUF001",   # String contains ambiguous {}. Did you mean {}?
   "RUF002",   # Docstring contains ambiguous {}. Did you mean {}?
   "RUF003",   # Comment contains ambiguous {}. Did you mean {}?
   "S101",     # Use of `assert` detected -- DO NOT FIX

strings/jaro_winkler.py

@@ -3,7 +3,7 @@
 
 def jaro_winkler(str1: str, str2: str) -> float:
     """
-    Jaro–Winkler distance is a string metric measuring an edit distance between two
+    Jaro-Winkler distance is a string metric measuring an edit distance between two
     sequences.
     Output value is between 0.0 and 1.0.
 

strings/manacher.py

@@ -5,7 +5,7 @@ def palindromic_string(input_string: str) -> str:
     >>> palindromic_string('ababa')
     'ababa'
 
-    Manacher’s algorithm which finds Longest palindromic Substring in linear time.
+    Manacher's algorithm which finds Longest palindromic Substring in linear time.
 
     1. first this convert input_string("xyx") into new_string("x|y|x") where odd
         positions are actual input characters.

strings/prefix_function.py

@@ -1,7 +1,7 @@
 """
 https://cp-algorithms.com/string/prefix-function.html
 
-Prefix function Knuth–Morris–Pratt algorithm
+Prefix function Knuth-Morris-Pratt algorithm
 
 Different algorithm than Knuth-Morris-Pratt pattern finding