Modular arithmetic and chess problems

week01/01. Problems/README.md

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+Python Problems==============## Divisible by a given numberFind all numbers in the list `n` which are divisible by a given integer `d`.### Signature```pythondef divisible_by_d(n: list, d: int) -> list:    pass```### Test examples```python>>> divisible_by_d([1,2,3,4,5,6], 3)[3, 6]>>> divisible_by_d([0,1,2,3,4,5,6], 4)[0, 4]>>> divisible_by_d([0], 4)[0]>>> divisible_by_d([1,3,5], 2)[]```## Knight valid moves countGiven a tuple with two indexes that denote the knight position on a matrix board, return the count of all valid moves which the knight can made.### Signature```pythondef knight_moves(pos: tuple) -> int:    # pos[0] denotes the knights row    # pos[1] denotes the knights column    pass```### Test examples```python>>> knight_moves((2, 6))6>>> knight_moves((0, 7))2>>> knight_moves((5, 0))4>>> knight_moves((4, 3))8>>> knight_moves((6, 7))3```Consider these graphical examples`Eight legal moves where pos = (4, 3)`|M|0|1|2|3|4|5|6|7||-|-|-|-|-|-|-|-|-||**0**| | | | | | | | ||**1**| | | | | | | | ||**2**| | | \*| | \*| | | ||**3**| | \*| | | | \*| | ||**4**| | | | N| | | | ||**5**| | \*| | | | \*| | ||**6**| | | \*| | \*| | | ||**7**| | | | | | | | |`Two legal moves where pos = (7, 7)`|M|0|1|2|3|4|5|6|7||-|-|-|-|-|-|-|-|-||**0**| | | | | | | | ||**1**| | | | | | | | ||**2**| | | | | | | | ||**3**| | | | | | | | ||**4**| | | | | | | | ||**5**| | | | | | | \*| ||**6**| | | | | | | | ||**7**| | | | | | \*| | N|

week01/01. Problems/solutions.py

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+"""
+Copyright (c) 2019 Hack Bulgaria
+
+The project is released under MIT License
+https://github.com/HackBulgaria/Programming-101-Python-2020-Spring
+"""
+
+
+def divisible_by_d(n: list, d: int) -> list:
+	'''Returns a list of all numbers from n divislbe by d.'''
+	return [i for i in n if i % d == 0]
+
+
+def knight_moves(pos: tuple) -> int:
+	'''Returns the count of all legal moves which the knight can make.'''
+	offsets = ((-2, -1), (-2, 1), (-1, -2), (-1, 2),
+			   (1, -2), (1, 2), (2, -1), (2, 1))
+
+	return [ 0 <= (pos[0] + o[0]) < 8 and
+			 0 <= (pos[1] + o[1]) < 8 
+			for o in offsets].count(True)