Merge pull request #20 from FranzDiebold/feat/add-solution-for-proble…

…m-57

Add solution for problem 57.

README.md

@@ -1,6 +1,6 @@
 # [Project Euler](https://projecteuler.net) Solutions
 
-[![solved: 56 problems](https://img.shields.io/badge/solved-56_problems-f93.svg)](./src)
+[![solved: 57 problems](https://img.shields.io/badge/solved-57_problems-f93.svg)](./src)
 ![Python: 3.8](https://img.shields.io/badge/Python-3.8-3776ab.svg)
 [![Lint and Test](https://github.com/FranzDiebold/project-euler-solutions/workflows/Lint%20and%20Test/badge.svg)](https://github.com/FranzDiebold/project-euler-solutions/actions?query=workflow%3A%22Lint+and+Test%22)
 [![license: MIT](https://img.shields.io/badge/license-MIT-brightgreen.svg)](./LICENSE.md)

src/p057_square_root_convergents.py

@@ -0,0 +1,69 @@
+"""
+Problem 57: Square root convergents
+https://projecteuler.net/problem=57
+
+It is possible to show that the square root of two can be expressed
+as an infinite continued fraction.
+
+sqrt(2) = 1 + (1 / (2 + (1 / (2 + (1 / (2 + ...))))))
+
+By expanding this for the first four iterations, we get:
+
+1 + (1 / 2)                                     = 3 / 2   = 1.5
+1 + (1 / (2 + (1 / 2)))                         = 7 / 5   = 1.4
+1 + (1 / (2 + (1 / (2 + (1 / 2)))))             = 17 / 12 = 1.41666...
+1 + (1 / (2 + (1 / (2 + (1 / (2 + (1 / 2))))))) = 41 / 29 = 1.41379...
+
+The next three expansions are 99/70, 239/169, and 577/408,
+but the eighth expansion, 1393/985, is the first example where the number of digits
+in the numerator exceeds the number of digits in the denominator.
+
+In the first one-thousand expansions,
+how many fractions contain a numerator with more digits than the denominator?
+"""
+
+from typing import Iterable, Tuple
+
+from src.common.calculations import calculate_large_product, calculate_large_sum
+
+
+def get_square_root_expansions() -> Iterable[Tuple[str, str]]:
+    """Get square root expansions for `sqrt(2)` as tuples `(numerator, denominator)`."""
+    previous_numerator = '1'
+    previous_denominator = '1'
+    numerator = '3'
+    denominator = '2'
+    while True:
+        yield numerator, denominator
+        numerator, previous_numerator = calculate_large_sum(
+            [calculate_large_product(numerator, '2'), previous_numerator]
+        ), numerator
+        denominator, previous_denominator = calculate_large_sum(
+            [calculate_large_product(denominator, '2'), previous_denominator]
+        ), denominator
+
+
+def count_larger_numerator_expansions(threshold: int) -> int:
+    """
+    Count the number of expansion fractions, where the numerator contains more digits
+    than the denominator in the first `threshold` square root expansions.
+    """
+    square_root_expansions_iter = get_square_root_expansions()
+    count = 0
+    for _ in range(threshold):
+        numerator, denominator = next(square_root_expansions_iter)
+        if len(numerator) > len(denominator):
+            count += 1
+    return count
+
+
+def main() -> None:
+    """Main function."""
+    threshold = 1000
+    count = count_larger_numerator_expansions(threshold)
+    print(f'In the first {threshold:,} expansions, {count:,} fractions contain ' \
+          f'a numerator with more digits than the denominator.')
+
+
+if __name__ == '__main__':
+    main()

test/test_p057_square_root_convergents.py

@@ -0,0 +1,57 @@
+"""
+Problem 57: Square root convergents
+https://projecteuler.net/problem=57
+
+It is possible to show that the square root of two can be expressed
+as an infinite continued fraction.
+
+sqrt(2) = 1 + (1 / (2 + (1 / (2 + (1 / (2 + ...))))))
+
+By expanding this for the first four iterations, we get:
+
+1 + (1 / 2)                                     = 3 / 2   = 1.5
+1 + (1 / (2 + (1 / 2)))                         = 7 / 5   = 1.4
+1 + (1 / (2 + (1 / (2 + (1 / 2)))))             = 17 / 12 = 1.41666...
+1 + (1 / (2 + (1 / (2 + (1 / (2 + (1 / 2))))))) = 41 / 29 = 1.41379...
+
+The next three expansions are 99/70, 239/169, and 577/408,
+but the eighth expansion, 1393/985, is the first example where the number of digits
+in the numerator exceeds the number of digits in the denominator.
+
+In the first one-thousand expansions,
+how many fractions contain a numerator with more digits than the denominator?
+"""
+
+
+def test_get_square_root_expansions():
+    # arrange
+    from src.p057_square_root_convergents import get_square_root_expansions
+
+    # act
+    actual_result_iter = get_square_root_expansions()
+
+    # assert
+    expected_result = [
+        ('3', '2'),
+        ('7', '5'),
+        ('17', '12'),
+        ('41', '29'),
+        ('99', '70'),
+        ('239', '169'),
+        ('577', '408'),
+        ('1393', '985'),
+    ]
+    for expected_expansion in expected_result:
+        assert next(actual_result_iter) == expected_expansion
+
+
+def test_count_larger_numerator_expansions():
+    # arrange
+    from src.p057_square_root_convergents import count_larger_numerator_expansions
+
+    # act
+    actual_result = count_larger_numerator_expansions(8)
+
+    # assert
+    expected_result = 1
+    assert actual_result == expected_result