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Ruff in pre-commit

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This commit is contained in:
David Doblas Jiménez
2026-04-05 21:45:24 +02:00
parent 83515c4682
commit b485914e01
82 changed files with 4317 additions and 34 deletions
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45
.pre-commit-config.yaml
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@@ -1,20 +1,31 @@
repos:
- repo: https://github.com/pre-commit/pre-commit-hooks
rev: v4.4.0
- repo: https://github.com/pre-commit/pre-commit-hooks
rev: v5.0.0
hooks:
- id: trailing-whitespace
- id: end-of-file-fixer
- id: check-yaml
- id: check-added-large-files
- repo: https://github.com/PyCQA/flake8
rev: 6.0.0
- id: check-added-large-files
- id: check-yaml
- id: end-of-file-fixer
- id: trailing-whitespace
- repo: local
hooks:
- id: flake8
- repo: https://github.com/psf/black
rev: 23.1.0
hooks:
- id: black
- repo: https://github.com/PyCQA/isort
rev: 5.12.0
hooks:
- id: isort
- id: ruff-check
name: ruff check
entry: uv run ruff check --fix
language: system
types_or: [python, pyi]
require_serial: true
- id: ruff-format
name: ruff format
entry: uv run ruff format
language: system
types_or: [python, pyi]
require_serial: true
- id: mypy
name: mypy
entry: uv run mypy src
language: system
pass_filenames: false
require_serial: true

2
.python-version
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@@ -1 +1 @@
system
3.14

2
how-to-python Normal file
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@@ -0,0 +1,2 @@
from the src folder:
$ python -m project_euler_python.problems_001_005.Problem001

36
pyproject.toml
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@@ -7,17 +7,16 @@ authors = [
]
requires-python = ">=3.14"
dependencies = [
"pytablereader>=0.30.0,<0.35.0",
"pytablewriter>=0.64.1,<0.65.0",
"sympy>=1.12,<2.0",
]
[dependency-groups]
dev = [
"pre-commit>=2.16.0,<3.0.0",
"flake8>=4.0.1,<5.0.0",
"mypy>=0.920,<1.0.0",
"isort>=5.10.1,<6.0.0",
"black>=21.12b0,<22.0",
"mypy>=1.11.0,<2.0.0",
"pre-commit>=4.0.0,<5.0.0",
"ruff>=0.6.0,<1.0.0",
]
[build-system]
@@ -30,3 +29,30 @@ package-dir = {"" = "src"}
[tool.setuptools.packages.find]
where = ["src"]
include = ["project_euler_python*"]
[tool.ruff]
line-length = 88
target-version = "py314"
[tool.ruff.lint]
select = [
"E",
"F",
"I",
"UP",
]
[tool.ruff.format]
quote-style = "double"
indent-style = "space"
[tool.mypy]
python_version = "3.14"
[[tool.mypy.overrides]]
module = ["sympy", "sympy.*"]
ignore_missing_imports = true
[[tool.mypy.overrides]]
module = ["pytablereader", "pytablereader.*"]
ignore_missing_imports = true

147
src/Julia/Problem054.jl Normal file
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@@ -0,0 +1,147 @@
#=
Created on 02 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 54 of Project Euler
https://projecteuler.net/problem=54 =#
using BenchmarkTools
# function replace_values_in_string(text, args_dict)
# for (k, v) in args_dict
# text = text.replace(k, string(v))
# end
# return text
# end
# function n_of_a_kind(hand, n)
# return max([x for x in hand if hand.count(x) == n] || [0])
# end
# function to_numerical(hand)
# return sorted([int(x[:end]) for x in hand], reverse=True)
# end
# # 10 Ranks functions.
# function high_card(str_hand)
# return to_numerical(str_hand)
# end
# function one_pair(hand)
# return n_of_a_kind(hand, 2)
# end
# function two_pair(hand)
# pairs = set([x for x in hand if hand.count(x) == 2])
# return 0 if len(pairs) < 2 else max(pairs)
# end
# function three_of_a_kind(hand)
# return n_of_a_kind(hand, 3)
# end
# function straight(hand)
# return 0 if not list(range(hand[0], hand[-1]-1, -1)) == hand else max(hand)
# end
# function flush(str_hand)
# return len(set([x[-1] for x in str_hand])) == 1
# end
# function full_house(hand)
# return three_of_a_kind(hand) if one_pair(hand) and three_of_a_kind(hand) else 0
# end
# function four_of_a_kind(hand)
# return n_of_a_kind(hand, 4)
# end
# function straight_flush(str_hand)
# straight_result = straight(to_numerical(str_hand))
# return straight_result if straight_result and flush(str_hand) else 0
# end
# function royal_flush(str_hand)
# return flush(str_hand) && list(range(14, 9, -1)) == to_numerical(str_hand)
# end
function Problem54()
#=
In the card game poker, a hand consists of five cards and are ranked,
from lowest to highest, in the following way:
High Card: Highest value card.
One Pair: Two cards of the same value.
Two Pairs: Two different pairs.
Three of a Kind: Three cards of the same value.
Straight: All cards are consecutive values.
Flush: All cards of the same suit.
Full House: Three of a kind and a pair.
Four of a Kind: Four cards of the same value.
Straight Flush: All cards are consecutive values of same suit.
Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
If two players have the same ranked hands then the rank made up of the
highest value wins; for example, a pair of eights beats a pair of fives
(see example 1 below). But if two ranks tie, for example, both players
have a pair of queens, then highest cards in each hand are compared
(see example 4 below); if the highest cards tie then the next highest
cards are compared, and so on.
Consider the following five hands dealt to two players:
Hand Player 1 Player 2 Winner
1 5H 5C 6S 7S KD 2C 3S 8S 8D TD Player 2
Pair of Fives Pair of Eights
2 5D 8C 9S JS AC 2C 5C 7D 8S QH Player 1
Highest card Ace Highest card Queen
3 2D 9C AS AH AC 3D 6D 7D TD QD Player 2
Three Aces Flush with Diamonds
4 4D 6S 9H QH QC 3D 6D 7H QD QS Player 1
Pair of Queens Pair of Queens
Highest card Nine Highest card Seven
5 2H 2D 4C 4D 4S 3C 3D 3S 9S 9D Player 1
Full House Full House
With Three Fours with Three Threes
The file, poker.txt, contains one-thousand random hands dealt to two
players. Each line of the file contains ten cards (separated by a single
space): the first five are Player 1's cards and the last five are Player
2's cards. You can assume that all hands are valid (no invalid characters
or repeated cards), each player's hand is in no specific order, and in
each hand there is a clear winner.
How many hands does Player 1 win? =#
# replace_map = {"T"=> 10, "J"=> 11, "Q"=> 12, "K"=> 13, "A"=> 14}
# score = [0, 0]
# for line in open("../files/Problem54.txt", "r").read().splitlines():
# line = replace_values_in_string(line, replace_map).split()
# hands = line[:5], line[5:]
# for rank in (royal_flush, straight_flush, four_of_a_kind, full_house, flush,
# straight, three_of_a_kind, two_pair, one_pair, high_card):
# should_convert_hand = "str" not in rank.__code__.co_varnames[0] # Checks parameter name.
# result = [rank(to_numerical(hand) if should_convert_hand else hand) for hand in hands]
# if result[0] != result[1]:
# score[0 if result[0] > result[1] else 1] += 1
# break
# end
# end
# end
# return score[0]
return 0
end
println("Time to evaluate Problem $(lpad(54, 3, "0")):")
@btime Problem54()
println("")
println("Result for Problem $(lpad(54, 3, "0")): ", Problem54())

23
src/Julia/Problem060.jl Normal file
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@@ -0,0 +1,23 @@
#=
Created on 30 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 60 of Project Euler
https://projecteuler.net/problem=60 =#
using BenchmarkTools
function Problem60()
#=
Statement =#
return
end
println("Time to evaluate Problem $(lpad(60, 3, "0")):")
@btime Problem60()
println("")
println("Result for Problem $(lpad(60, 3, "0")): ", Problem60())

88
src/Julia/Problems001-050/Problem011.jl Normal file
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@@ -0,0 +1,88 @@
#=
Created on 07 Jul 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 11 of Project Euler
https://projecteuler.net/problem=11 =#
using BenchmarkTools
GRID = [
[8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8],
[49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0],
[81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65],
[52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91],
[22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80],
[24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50],
[32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70],
[67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21],
[24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72],
[21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95],
[78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92],
[16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57],
[86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58],
[19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40],
[4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66],
[88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69],
[4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36],
[20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16],
[20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54],
[1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48],
]
function grid_product(x, y, dx, dy, n)
result = 1
for i = 0:n
println("Accessing GRID: ", x, y, dx, dy)
println(y + i * dy, x + i * dx)
result *= GRID[y+i*dy][x+i*dx]
println("result from grid product: ", result)
end
return result
end
function Problem11()
#=
In the 20x20 grid above, four numbers along a diagonal line have been
marked in red.
The product of these numbers is 26 x 63 x 78 x 14 = 1788696.
What is the greatest product of four adjacent numbers in any direction
(up, down, left, right, or diagonally) in the 20x20 grid? =#
ans = 0
width = height = length(GRID)
adjacent_nums = 4
for y in 1:height
for x in 1:width
println("ans before: ", ans)
println(y, x)
if x + adjacent_nums <= width
ans = max(grid_product(x, y, 1, 0, adjacent_nums), ans)
end
if y + adjacent_nums <= height
ans = max(grid_product(x, y, 0, 1, adjacent_nums), ans)
end
if (x + adjacent_nums <= width) & (y + adjacent_nums <= height)
ans = max(grid_product(x, y, 1, 1, adjacent_nums), ans)
end
if (x - adjacent_nums >= -1) & (y + adjacent_nums <= height)
ans = max(grid_product(x, y, -1, 1, adjacent_nums), ans)
end
println("ans after: ", ans)
end
end
return ans
end
# println("Time to evaluate Problem 11:")
# @btime Problem11()
# println("")
# println("Result for Problem 11: ", Problem11())
Problem11()

10
src/Julia/create_template.py
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@@ -18,13 +18,13 @@ def create_problem():
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem {args['problem']:0>3} of Project Euler
https://projecteuler.net/problem={args['problem']}
Solution for Problem {args["problem"]:0>3} of Project Euler
https://projecteuler.net/problem={args["problem"]}
=#
using BenchmarkTools
function Problem{args['problem']:0>3}()
function Problem{args["problem"]:0>3}()
#=
Statement
=#
@@ -35,9 +35,9 @@ def create_problem():
println("Took:")
@btime Problem{args['problem']:0>3}()
@btime Problem{args["problem"]:0>3}()
println("")
println("Result for Problem $(lpad({args['problem']}, 3, "0")): ", Problem{args['problem']:0>3}())
println("Result for Problem $(lpad({args["problem"]}, 3, "0")): ", Problem{args["problem"]:0>3}())
""" # noqa: E501
)

2
src/files/Problem13.txt
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@@ -97,4 +97,4 @@
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690
53503534226472524250874054075591789781264330331690

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src/files/Problem22.txt
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src/files/Problem42.txt
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2
src/files/Problem59.txt
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@@ -1 +1 @@
36,22,80,0,0,4,23,25,19,17,88,4,4,19,21,11,88,22,23,23,29,69,12,24,0,88,25,11,12,2,10,28,5,6,12,25,10,22,80,10,30,80,10,22,21,69,23,22,69,61,5,9,29,2,66,11,80,8,23,3,17,88,19,0,20,21,7,10,17,17,29,20,69,8,17,21,29,2,22,84,80,71,60,21,69,11,5,8,21,25,22,88,3,0,10,25,0,10,5,8,88,2,0,27,25,21,10,31,6,25,2,16,21,82,69,35,63,11,88,4,13,29,80,22,13,29,22,88,31,3,88,3,0,10,25,0,11,80,10,30,80,23,29,19,12,8,2,10,27,17,9,11,45,95,88,57,69,16,17,19,29,80,23,29,19,0,22,4,9,1,80,3,23,5,11,28,92,69,9,5,12,12,21,69,13,30,0,0,0,0,27,4,0,28,28,28,84,80,4,22,80,0,20,21,2,25,30,17,88,21,29,8,2,0,11,3,12,23,30,69,30,31,23,88,4,13,29,80,0,22,4,12,10,21,69,11,5,8,88,31,3,88,4,13,17,3,69,11,21,23,17,21,22,88,65,69,83,80,84,87,68,69,83,80,84,87,73,69,83,80,84,87,65,83,88,91,69,29,4,6,86,92,69,15,24,12,27,24,69,28,21,21,29,30,1,11,80,10,22,80,17,16,21,69,9,5,4,28,2,4,12,5,23,29,80,10,30,80,17,16,21,69,27,25,23,27,28,0,84,80,22,23,80,17,16,17,17,88,25,3,88,4,13,29,80,17,10,5,0,88,3,16,21,80,10,30,80,17,16,25,22,88,3,0,10,25,0,11,80,12,11,80,10,26,4,4,17,30,0,28,92,69,30,2,10,21,80,12,12,80,4,12,80,10,22,19,0,88,4,13,29,80,20,13,17,1,10,17,17,13,2,0,88,31,3,88,4,13,29,80,6,17,2,6,20,21,69,30,31,9,20,31,18,11,94,69,54,17,8,29,28,28,84,80,44,88,24,4,14,21,69,30,31,16,22,20,69,12,24,4,12,80,17,16,21,69,11,5,8,88,31,3,88,4,13,17,3,69,11,21,23,17,21,22,88,25,22,88,17,69,11,25,29,12,24,69,8,17,23,12,80,10,30,80,17,16,21,69,11,1,16,25,2,0,88,31,3,88,4,13,29,80,21,29,2,12,21,21,17,29,2,69,23,22,69,12,24,0,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,67,80,10,10,80,7,1,80,21,13,4,17,17,30,2,88,4,13,29,80,22,13,29,69,23,22,69,12,24,12,11,80,22,29,2,12,29,3,69,29,1,16,25,28,69,12,31,69,11,92,69,17,4,69,16,17,22,88,4,13,29,80,23,25,4,12,23,80,22,9,2,17,80,70,76,88,29,16,20,4,12,8,28,12,29,20,69,26,9,69,11,80,17,23,80,84,88,31,3,88,4,13,29,80,21,29,2,12,21,21,17,29,2,69,12,31,69,12,24,0,88,20,12,25,29,0,12,21,23,86,80,44,88,7,12,20,28,69,11,31,10,22,80,22,16,31,18,88,4,13,25,4,69,12,24,0,88,3,16,21,80,10,30,80,17,16,25,22,88,3,0,10,25,0,11,80,17,23,80,7,29,80,4,8,0,23,23,8,12,21,17,17,29,28,28,88,65,75,78,68,81,65,67,81,72,70,83,64,68,87,74,70,81,75,70,81,67,80,4,22,20,69,30,2,10,21,80,8,13,28,17,17,0,9,1,25,11,31,80,17,16,25,22,88,30,16,21,18,0,10,80,7,1,80,22,17,8,73,88,17,11,28,80,17,16,21,11,88,4,4,19,25,11,31,80,17,16,21,69,11,1,16,25,2,0,88,2,10,23,4,73,88,4,13,29,80,11,13,29,7,29,2,69,75,94,84,76,65,80,65,66,83,77,67,80,64,73,82,65,67,87,75,72,69,17,3,69,17,30,1,29,21,1,88,0,23,23,20,16,27,21,1,84,80,18,16,25,6,16,80,0,0,0,23,29,3,22,29,3,69,12,24,0,88,0,0,10,25,8,29,4,0,10,80,10,30,80,4,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,86,80,35,23,28,9,23,7,12,22,23,69,25,23,4,17,30,69,12,24,0,88,3,4,21,21,69,11,4,0,8,3,69,26,9,69,15,24,12,27,24,69,49,80,13,25,20,69,25,2,23,17,6,0,28,80,4,12,80,17,16,25,22,88,3,16,21,92,69,49,80,13,25,6,0,88,20,12,11,19,10,14,21,23,29,20,69,12,24,4,12,80,17,16,21,69,11,5,8,88,31,3,88,4,13,29,80,22,29,2,12,29,3,69,73,80,78,88,65,74,73,70,69,83,80,84,87,72,84,88,91,69,73,95,87,77,70,69,83,80,84,87,70,87,77,80,78,88,21,17,27,94,69,25,28,22,23,80,1,29,0,0,22,20,22,88,31,11,88,4,13,29,80,20,13,17,1,10,17,17,13,2,0,88,31,3,88,4,13,29,80,6,17,2,6,20,21,75,88,62,4,21,21,9,1,92,69,12,24,0,88,3,16,21,80,10,30,80,17,16,25,22,88,29,16,20,4,12,8,28,12,29,20,69,26,9,69,65,64,69,31,25,19,29,3,69,12,24,0,88,18,12,9,5,4,28,2,4,12,21,69,80,22,10,13,2,17,16,80,21,23,7,0,10,89,69,23,22,69,12,24,0,88,19,12,10,19,16,21,22,0,10,21,11,27,21,69,23,22,69,12,24,0,88,0,0,10,25,8,29,4,0,10,80,10,30,80,4,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,86,80,36,22,20,69,26,9,69,11,25,8,17,28,4,10,80,23,29,17,22,23,30,12,22,23,69,49,80,13,25,6,0,88,28,12,19,21,18,17,3,0,88,18,0,29,30,69,25,18,9,29,80,17,23,80,1,29,4,0,10,29,12,22,21,69,12,24,0,88,3,16,21,3,69,23,22,69,12,24,0,88,3,16,26,3,0,9,5,0,22,4,69,11,21,23,17,21,22,88,25,11,88,7,13,17,19,13,88,4,13,29,80,0,0,0,10,22,21,11,12,3,69,25,2,0,88,21,19,29,30,69,22,5,8,26,21,23,11,94
36,22,80,0,0,4,23,25,19,17,88,4,4,19,21,11,88,22,23,23,29,69,12,24,0,88,25,11,12,2,10,28,5,6,12,25,10,22,80,10,30,80,10,22,21,69,23,22,69,61,5,9,29,2,66,11,80,8,23,3,17,88,19,0,20,21,7,10,17,17,29,20,69,8,17,21,29,2,22,84,80,71,60,21,69,11,5,8,21,25,22,88,3,0,10,25,0,10,5,8,88,2,0,27,25,21,10,31,6,25,2,16,21,82,69,35,63,11,88,4,13,29,80,22,13,29,22,88,31,3,88,3,0,10,25,0,11,80,10,30,80,23,29,19,12,8,2,10,27,17,9,11,45,95,88,57,69,16,17,19,29,80,23,29,19,0,22,4,9,1,80,3,23,5,11,28,92,69,9,5,12,12,21,69,13,30,0,0,0,0,27,4,0,28,28,28,84,80,4,22,80,0,20,21,2,25,30,17,88,21,29,8,2,0,11,3,12,23,30,69,30,31,23,88,4,13,29,80,0,22,4,12,10,21,69,11,5,8,88,31,3,88,4,13,17,3,69,11,21,23,17,21,22,88,65,69,83,80,84,87,68,69,83,80,84,87,73,69,83,80,84,87,65,83,88,91,69,29,4,6,86,92,69,15,24,12,27,24,69,28,21,21,29,30,1,11,80,10,22,80,17,16,21,69,9,5,4,28,2,4,12,5,23,29,80,10,30,80,17,16,21,69,27,25,23,27,28,0,84,80,22,23,80,17,16,17,17,88,25,3,88,4,13,29,80,17,10,5,0,88,3,16,21,80,10,30,80,17,16,25,22,88,3,0,10,25,0,11,80,12,11,80,10,26,4,4,17,30,0,28,92,69,30,2,10,21,80,12,12,80,4,12,80,10,22,19,0,88,4,13,29,80,20,13,17,1,10,17,17,13,2,0,88,31,3,88,4,13,29,80,6,17,2,6,20,21,69,30,31,9,20,31,18,11,94,69,54,17,8,29,28,28,84,80,44,88,24,4,14,21,69,30,31,16,22,20,69,12,24,4,12,80,17,16,21,69,11,5,8,88,31,3,88,4,13,17,3,69,11,21,23,17,21,22,88,25,22,88,17,69,11,25,29,12,24,69,8,17,23,12,80,10,30,80,17,16,21,69,11,1,16,25,2,0,88,31,3,88,4,13,29,80,21,29,2,12,21,21,17,29,2,69,23,22,69,12,24,0,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,67,80,10,10,80,7,1,80,21,13,4,17,17,30,2,88,4,13,29,80,22,13,29,69,23,22,69,12,24,12,11,80,22,29,2,12,29,3,69,29,1,16,25,28,69,12,31,69,11,92,69,17,4,69,16,17,22,88,4,13,29,80,23,25,4,12,23,80,22,9,2,17,80,70,76,88,29,16,20,4,12,8,28,12,29,20,69,26,9,69,11,80,17,23,80,84,88,31,3,88,4,13,29,80,21,29,2,12,21,21,17,29,2,69,12,31,69,12,24,0,88,20,12,25,29,0,12,21,23,86,80,44,88,7,12,20,28,69,11,31,10,22,80,22,16,31,18,88,4,13,25,4,69,12,24,0,88,3,16,21,80,10,30,80,17,16,25,22,88,3,0,10,25,0,11,80,17,23,80,7,29,80,4,8,0,23,23,8,12,21,17,17,29,28,28,88,65,75,78,68,81,65,67,81,72,70,83,64,68,87,74,70,81,75,70,81,67,80,4,22,20,69,30,2,10,21,80,8,13,28,17,17,0,9,1,25,11,31,80,17,16,25,22,88,30,16,21,18,0,10,80,7,1,80,22,17,8,73,88,17,11,28,80,17,16,21,11,88,4,4,19,25,11,31,80,17,16,21,69,11,1,16,25,2,0,88,2,10,23,4,73,88,4,13,29,80,11,13,29,7,29,2,69,75,94,84,76,65,80,65,66,83,77,67,80,64,73,82,65,67,87,75,72,69,17,3,69,17,30,1,29,21,1,88,0,23,23,20,16,27,21,1,84,80,18,16,25,6,16,80,0,0,0,23,29,3,22,29,3,69,12,24,0,88,0,0,10,25,8,29,4,0,10,80,10,30,80,4,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,86,80,35,23,28,9,23,7,12,22,23,69,25,23,4,17,30,69,12,24,0,88,3,4,21,21,69,11,4,0,8,3,69,26,9,69,15,24,12,27,24,69,49,80,13,25,20,69,25,2,23,17,6,0,28,80,4,12,80,17,16,25,22,88,3,16,21,92,69,49,80,13,25,6,0,88,20,12,11,19,10,14,21,23,29,20,69,12,24,4,12,80,17,16,21,69,11,5,8,88,31,3,88,4,13,29,80,22,29,2,12,29,3,69,73,80,78,88,65,74,73,70,69,83,80,84,87,72,84,88,91,69,73,95,87,77,70,69,83,80,84,87,70,87,77,80,78,88,21,17,27,94,69,25,28,22,23,80,1,29,0,0,22,20,22,88,31,11,88,4,13,29,80,20,13,17,1,10,17,17,13,2,0,88,31,3,88,4,13,29,80,6,17,2,6,20,21,75,88,62,4,21,21,9,1,92,69,12,24,0,88,3,16,21,80,10,30,80,17,16,25,22,88,29,16,20,4,12,8,28,12,29,20,69,26,9,69,65,64,69,31,25,19,29,3,69,12,24,0,88,18,12,9,5,4,28,2,4,12,21,69,80,22,10,13,2,17,16,80,21,23,7,0,10,89,69,23,22,69,12,24,0,88,19,12,10,19,16,21,22,0,10,21,11,27,21,69,23,22,69,12,24,0,88,0,0,10,25,8,29,4,0,10,80,10,30,80,4,88,19,12,10,19,9,29,80,18,16,31,22,29,80,1,17,17,8,29,4,0,10,80,12,11,80,84,86,80,36,22,20,69,26,9,69,11,25,8,17,28,4,10,80,23,29,17,22,23,30,12,22,23,69,49,80,13,25,6,0,88,28,12,19,21,18,17,3,0,88,18,0,29,30,69,25,18,9,29,80,17,23,80,1,29,4,0,10,29,12,22,21,69,12,24,0,88,3,16,21,3,69,23,22,69,12,24,0,88,3,16,26,3,0,9,5,0,22,4,69,11,21,23,17,21,22,88,25,11,88,7,13,17,19,13,88,4,13,29,80,0,0,0,10,22,21,11,12,3,69,25,2,0,88,21,19,29,30,69,22,5,8,26,21,23,11,94

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#!/usr/bin/env python
"""
Created on 24 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 051 of Project Euler
https://projecteuler.net/problem=51
"""
from project_euler_python.utils import is_prime, list_primes, timeit
@timeit("Problem 051")
def compute():
"""
By replacing the 1st digit of the 2-digit number *3, it turns out that
six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
By replacing the 3rd and 4th digits of 56**3 with the same digit, this
5-digit number is the first example having seven primes among the ten
generated numbers, yielding the family:
56003, 56113, 56333, 56443, 56663, 56773, and 56993.
Consequently 56003, being the first member of this family, is the smallest
prime with this property.
Find the smallest prime which, by replacing part of the number (not
necessarily adjacent digits) with the same digit, is part of an eight prime
value family.
"""
primes = sorted(set(list_primes(1_000_000)) - set(list_primes(57_000)))
digits = {
"0": [],
"1": [],
"2": [],
"3": [],
"4": [],
"5": [],
"6": [],
"7": [],
"8": [],
"9": [],
}
for d in digits.keys():
for p in primes:
p = str(p)
if p.count(d) == 3 and p[-1] != d:
digits[d].append(p)
for d in {"0", "1", "2"}:
for p in digits[d]:
res = 0
i = 10
for D in {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9"} - {d}:
i -= 1
q = int(p.replace(d, D))
if is_prime(q) and q > 57_000:
res += 1
if i + res < 7:
break
if res == 7:
return p
if __name__ == "__main__":
print(f"Result for Problem 051: {compute()}")

38
src/project_euler_python/Problem052.py Normal file
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#!/usr/bin/env python
"""
Created on 26 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 052 of Project Euler
https://projecteuler.net/problem=52
"""
from project_euler_python.utils import timeit
@timeit("Problem 052")
def compute():
"""
It can be seen that the number, 125874, and its double, 251748,
contain exactly the same digits, but in a different order.
Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x,
and 6x, contain the same digits.
"""
for number in range(123_456, 1_000_000):
if (
sorted(str(number))
== sorted(str(2 * number))
== sorted(str(3 * number))
== sorted(str(4 * number))
== sorted(str(5 * number))
== sorted(str(6 * number))
):
return number
if __name__ == "__main__":
print(f"Result for Problem 052: {compute()}")

45
src/project_euler_python/Problem053.py Normal file
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#!/usr/bin/env python
"""
Created on 26 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 053 of Project Euler
https://projecteuler.net/problem=53
"""
from math import comb
from project_euler_python.utils import timeit
@timeit("Problem 053")
def compute():
"""
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, (5 over 3) = 10.
In general, (n over r) = n!/r!*(n-r)!, where r<=n, n!=n*(n-1)*...*2*1,
and 0!=1.
It is not until, that a value exceeds one-million: (23 over 10) = 1144066.
How many, not necessarily distinct, values of (n over r) for 1<=n<=100,
are greater than one-million?
"""
ans = 0
for x in range(101):
for y in range(101):
if comb(x, y) > 1_000_000:
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 053: {compute()}")

163
src/project_euler_python/Problem054.py Normal file
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#!/usr/bin/env python
"""
Created on 27 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 054 of Project Euler
https://projecteuler.net/problem=54
"""
from pathlib import Path
from project_euler_python.utils import timeit
# 3 help functions.
def replace_values_in_string(text: str, args_dict: dict) -> str:
for k, v in args_dict.items():
text = text.replace(k, str(v))
return text
def n_of_a_kind(hand: list, n: int) -> int:
return max([x for x in hand if hand.count(x) == n] or [0])
def to_numerical(hand: list) -> list:
return sorted([int(x[:-1]) for x in hand], reverse=True)
# 10 Ranks functions.
def high_card(str_hand: list) -> list:
return to_numerical(str_hand)
def one_pair(hand: list) -> int:
return n_of_a_kind(hand, 2)
def two_pair(hand: list) -> int:
pairs = set([x for x in hand if hand.count(x) == 2])
return 0 if len(pairs) < 2 else max(pairs)
def three_of_a_kind(hand: list) -> int:
return n_of_a_kind(hand, 3)
def straight(hand: list) -> int:
return 0 if not list(range(hand[0], hand[-1] - 1, -1)) == hand else max(hand)
def flush(str_hand: list) -> bool:
return len(set([x[-1] for x in str_hand])) == 1
def full_house(hand: list) -> int:
return three_of_a_kind(hand) if one_pair(hand) and three_of_a_kind(hand) else 0
def four_of_a_kind(hand: list) -> int:
return n_of_a_kind(hand, 4)
def straight_flush(str_hand: list) -> int:
straight_result = straight(to_numerical(str_hand))
return straight_result if straight_result and flush(str_hand) else 0
def royal_flush(str_hand: list) -> bool:
return flush(str_hand) and list(range(14, 9, -1)) == to_numerical(str_hand)
@timeit("Problem 054")
def compute():
"""
In the card game poker, a hand consists of five cards and are ranked,
from lowest to highest, in the following way:
High Card: Highest value card.
One Pair: Two cards of the same value.
Two Pairs: Two different pairs.
Three of a Kind: Three cards of the same value.
Straight: All cards are consecutive values.
Flush: All cards of the same suit.
Full House: Three of a kind and a pair.
Four of a Kind: Four cards of the same value.
Straight Flush: All cards are consecutive values of same suit.
Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.
The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.
If two players have the same ranked hands then the rank made up of the
highest value wins; for example, a pair of eights beats a pair of fives
(see example 1 below). But if two ranks tie, for example, both players
have a pair of queens, then highest cards in each hand are compared
(see example 4 below); if the highest cards tie then the next highest
cards are compared, and so on.
Consider the following five hands dealt to two players:
Hand Player 1 Player 2 Winner
1 5H 5C 6S 7S KD 2C 3S 8S 8D TD Player 2
Pair of Fives Pair of Eights
2 5D 8C 9S JS AC 2C 5C 7D 8S QH Player 1
Highest card Ace Highest card Queen
3 2D 9C AS AH AC 3D 6D 7D TD QD Player 2
Three Aces Flush with Diamonds
4 4D 6S 9H QH QC 3D 6D 7H QD QS Player 1
Pair of Queens Pair of Queens
Highest card Nine Highest card Seven
5 2H 2D 4C 4D 4S 3C 3D 3S 9S 9D Player 1
Full House Full House
With Three Fours with Three Threes
The file, poker.txt, contains one-thousand random hands dealt to two
players. Each line of the file contains ten cards (separated by a single
space): the first five are Player 1's cards and the last five are Player
2's cards. You can assume that all hands are valid (no invalid characters
or repeated cards), each player's hand is in no specific order, and in
each hand there is a clear winner.
How many hands does Player 1 win?
"""
replace_map = {"T": 10, "J": 11, "Q": 12, "K": 13, "A": 14}
score = [0, 0]
file = Path("files/Problem54.txt")
for line in open(file).read().splitlines():
line = replace_values_in_string(line, replace_map).split()
hands = line[:5], line[5:]
for rank in (
royal_flush,
straight_flush,
four_of_a_kind,
full_house,
flush,
straight,
three_of_a_kind,
two_pair,
one_pair,
high_card,
):
should_convert_hand = (
"str" not in rank.__code__.co_varnames[0]
) # Checks parameter name.
result = [
rank(to_numerical(hand) if should_convert_hand else hand)
for hand in hands
]
if result[0] != result[1]:
score[0 if result[0] > result[1] else 1] += 1
break
return score[0]
if __name__ == "__main__":
print(f"Result for Problem 054: {compute()}")

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#!/usr/bin/env python
"""
Created on 02 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 055 of Project Euler
https://projecteuler.net/problem=55
"""
from project_euler_python.utils import is_palindrome, timeit
@timeit("Problem 055")
def compute():
"""
If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.
Not all numbers produce palindromes so quickly. For example,
349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337
That is, 349 took three iterations to arrive at a palindrome.
Although no one has proved it yet, it is thought that some numbers, like
196, never produce a palindrome. A number that never forms a palindrome
through the reverse and add process is called a Lychrel number. Due to the
theoretical nature of these numbers, and for the purpose of this problem,
we shall assume that a number is Lychrel until proven otherwise. In
addition you are given that for every number below ten-thousand, it will
either:
(i) become a palindrome in less than fifty iterations, or,
(ii) no one, with all the computing power that exists, has managed so far
to map it to a palindrome.
In fact, 10677 is the first number to be shown to require over fifty
iterations before producing a palindrome:
4668731596684224866951378664 (53 iterations, 28-digits).
Surprisingly, there are palindromic numbers that are themselves Lychrel
numbers; the first example is 4994.
How many Lychrel numbers are there below ten-thousand?
"""
ans = 0
for n in range(11, 10_000):
num = n
is_lychrel = True
for _ in range(50):
num += int(str(num)[::-1])
if is_palindrome(num):
is_lychrel = False
break
if is_lychrel:
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 055: {compute()}")

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#!/usr/bin/env python
"""
Created on 07 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 056 of Project Euler
https://projecteuler.net/problem=56
"""
from project_euler_python.utils import timeit
@timeit("Problem 056")
def compute():
"""
A googol (10^100) is a massive number: one followed by one-hundred zeros;
100100 is almost unimaginably large: one followed by two-hundred zeros.
Despite their size, the sum of the digits in each number is only 1.
Considering natural numbers of the form, a^b, where a, b < 100, what is the
maximum digital sum?
"""
ans = 0
for a in range(100):
for b in range(100):
num = sum([int(digit) for digit in str(a**b)])
if num > ans:
ans = num
return ans
if __name__ == "__main__":
print(f"Result for Problem 056: {compute()}")

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#!/usr/bin/env python
"""
Created on 09 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 057 of Project Euler
https://projecteuler.net/problem=57
"""
from fractions import Fraction
from project_euler_python.utils import timeit
@timeit("Problem 057")
def compute():
"""
It is possible to show that the square root of two can be expressed
as an infinite continued fraction.
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?
"""
ans = 0
f = Fraction(1, 2)
for _ in range(1000):
f = 1 / (2 + f)
result = 1 + f
if len(str(result.numerator)) > len(str(result.denominator)):
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 057: {compute()}")

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#!/usr/bin/env python
"""
Created on 10 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 058 of Project Euler
https://projecteuler.net/problem=58
"""
from project_euler_python.utils import is_prime, timeit
@timeit("Problem 058")
def compute():
"""
Starting with 1 and spiralling anticlockwise in the following way,
a square spiral with side length 7 is formed.
37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 5 4 3 12 29
40 19 6 1 2 11 28
41 20 7 8 9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49
It is interesting to note that the odd squares lie along the bottom right
diagonal, but what is more interesting is that 8 out of the 13 numbers
lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.
If one complete new layer is wrapped around the spiral above, a square
spiral with side length 9 will be formed. If this process is continued,
what is the side length of the square spiral for which the ratio of primes
along both diagonals first falls below 10%?
"""
ratio = 1
corners = []
side = 0
num = 1
while ratio > 0.1:
side += 2
for _ in range(4):
num += side
corners.append(is_prime(num))
ratio = sum(corners) / len(corners)
return side + 1
if __name__ == "__main__":
print(f"Result for Problem 058: {compute()}")

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#!/usr/bin/env python
"""
Created on 13 Oct 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 059 of Project Euler
https://projecteuler.net/problem=59
"""
from itertools import permutations
from string import ascii_lowercase
from project_euler_python.utils import timeit
@timeit("Problem 059")
def compute():
"""
Each character on a computer is assigned a unique code and the preferred
standard is ASCII (American Standard Code for Information Interchange).
For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
A modern encryption method is to take a text file, convert the bytes to
ASCII, then XOR each byte with a given value, taken from a secret key.
The advantage with the XOR function is that using the same encryption key
on the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
then 107 XOR 42 = 65.
For unbreakable encryption, the key is the same length as the plain text
message, and the key is made up of random bytes. The user would keep the
encrypted message and the encryption key in different locations, and
without both "halves", it is impossible to decrypt the message.
Unfortunately, this method is impractical for most users, so the modified
method is to use a password as a key. If the password is shorter than the
message, which is likely, the key is repeated cyclically throughout the
message. The balance for this method is using a sufficiently long password
key for security, but short enough to be memorable.
Your task has been made easy, as the encryption key consists of three
lower case characters. Using p059_cipher.txt, a file containing the
encrypted ASCII codes, and the knowledge that the plain text must contain
common English words, decrypt the message and find the sum of the ASCII
values in the original text.
"""
with open("files/Problem59.txt") as f:
encrypted = [int(char) for char in f.read().split(",")]
plain_text = len(encrypted) // 3
for key in permutations(ascii_lowercase, 3):
decrypted = ""
for k, i in zip(list(key) * plain_text, encrypted):
decrypted += chr(ord(k) ^ i)
# assuming Euler will be in the text
if "Euler" in decrypted:
return sum([ord(c) for c in decrypted])
if __name__ == "__main__":
print(f"Result for Problem 059: {compute()}")

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#!/usr/bin/env python
"""
Created on 04 Dic 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 060 of Project Euler
https://projecteuler.net/problem=60
"""
from project_euler_python.utils import is_prime, list_primes, timeit
def solve_backtrack(max_chain_length, chain):
primes_list = list_primes(10_000)
for p in primes_list[1:]:
if p not in chain and is_concatenable(chain, p):
chain.append(p)
if len(chain) == max_chain_length:
return chain
solve_backtrack(max_chain_length, chain)
chain.pop()
return chain
def is_concatenable(chain, candidate):
for n in chain:
if not is_prime(int(str(n) + str(candidate))):
return False
if not is_prime(int(str(candidate) + str(n))):
return False
return True
@timeit("Problem 060")
def compute():
"""
The primes 3, 7, 109, and 673, are quite remarkable. By taking any two
primes and concatenating them in any order the result will always be prime.
For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of
these four primes, 792, represents the lowest sum for a set of four primes
with this property.
Find the lowest sum for a set of five primes for which any two primes
concatenate to produce another prime.
"""
ans = solve_backtrack(5, [])
return sum(ans)
if __name__ == "__main__":
print(f"Result for Problem 060: {compute()}")

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#!/usr/bin/env python
"""
Created on 16 Jul 2022
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 061 of Project Euler
https://projecteuler.net/problem=61
"""
from itertools import permutations
from project_euler_python.utils import timeit
def get_numbers(order: int, number: int) -> int:
return (number * (number - 1) * (order - 2) // 2) + number
poligonals: dict[int, set[str]] = {
o: set(str(get_numbers(o, n)) for n in range(150)) for o in range(3, 9)
}
four_digits_numbers: set[str] = {str(n) for n in range(10**3, 10**4)}
eligibles: dict[int, list[str]] = {
k: list(v & four_digits_numbers) for k, v in poligonals.items()
}
@timeit("Problem 061")
def compute():
"""
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers
are all figurate (polygonal) numbers and are generated by the following
formulae:
Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, ...
Square P4,n=n2 1, 4, 9, 16, 25, ...
Pentagonal P5,n=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal P6,n=n(2n-1) 1, 6, 15, 28, 45, ...
Heptagonal P7,n=n(5n-3)/2 1, 7, 18, 34, 55, ...
Octagonal P8,n=n(3n-2) 1, 8, 21, 40, 65, ...
The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three
interesting properties.
The set is cyclic, in that the last two digits of each number is the
first two digits of the next number (including the last number with the
first).
Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and
pentagonal (P5,44=2882), is represented by a different number in the
set.
This is the only set of 4-digit numbers with this property.
Find the sum of the only ordered set of six cyclic 4-digit numbers for
which each polygonal type: triangle, square, pentagonal, hexagonal,
heptagonal, and octagonal, is represented by a different number in the set.
"""
perms = permutations(eligibles.keys(), 6)
for perm in perms:
for a in eligibles.get(perm[0]):
ans = []
a1, a2 = a[:2], a[2:]
for b in filter(lambda n: n[:2] == a2, eligibles.get(perm[1])):
b2 = b[2:]
for c in filter(lambda n: n[:2] == b2, eligibles.get(perm[2])):
c2 = c[2:]
for d in filter(lambda n: n[:2] == c2, eligibles.get(perm[3])):
d2 = d[2:]
for e in filter(lambda n: n[:2] == d2, eligibles.get(perm[4])):
e2 = e[2:]
for f in filter(
lambda n: n[:2] == e2 and n[2:] == a1,
eligibles.get(perm[5]),
):
ans.append([a, b, c, d, e, f])
return sum(map(int, ans[0]))
if __name__ == "__main__":
print(f"Result for Problem 061: {compute()}")

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#!/usr/bin/env python
"""
Created on 25 Jul 2022
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 062 of Project Euler
https://projecteuler.net/problem=62
"""
from collections import defaultdict
from project_euler_python.utils import timeit
@timeit("Problem 062")
def compute():
"""
The cube, 41063625 (345^3), can be permuted to produce two other cubes:
56623104 (384^3) and 66430125 (405^3). In fact, 41063625 is the smallest
cube which has exactly three permutations of its digits which are also
cube.
Find the smallest cube for which exactly five permutations of its digits
are cube.
"""
cubes = defaultdict(list)
for number in range(345, 10_000):
cube_str = "".join(sorted(list(str(number**3))))
cubes[cube_str].append(number**3)
if len(cubes[cube_str]) == 5:
return min(cubes[cube_str])
if __name__ == "__main__":
print(f"Result for Problem 062: {compute()}")

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#!/usr/bin/env python
"""
Created on 05 Aug 2022
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 063 of Project Euler
https://projecteuler.net/problem=63
"""
from project_euler_python.utils import timeit
@timeit("Problem 063")
def compute():
"""
The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the
9-digit number, 134217728=8^9, is a ninth power.
How many n-digit positive integers exist which are also an nth power?
"""
ans = 0
# no need to go higher than 10, because 10**2 = 100
for number in range(1, 10):
for exp in range(1, 30):
if len(str(number**exp)) == exp:
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 063: {compute()}")

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#!/usr/bin/env python
"""
Created on 26 Oct 2022
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 64 of Project Euler
https://projecteuler.net/problem=64
"""
from math import floor, sqrt
from project_euler_python.utils import timeit
def continuedFraction(n):
m = 0
d = 1
a = floor(sqrt(n))
i = 0
while True:
m = d * a - m
d = (n - m**2) / d
a = floor((floor(sqrt(n)) + m) / d)
if i == 0:
m0 = m
d0 = d
a0 = a
if i > 0:
if m == m0 and d == d0 and a == a0:
return True if i % 2 != 0 else False
i += 1
def checkIfSquare(n):
return True if floor(sqrt(n)) == sqrt(n) else False
@timeit("Problem 64")
def compute():
"""
# Statement
"""
# https://en.wikipedia.org/wiki/Periodic_continued_fraction#Canonical_form_and_repetend
result = 0
for i in range(2, 10001):
if not checkIfSquare(i):
if continuedFraction(i):
result += 1
return result
if __name__ == "__main__":
print(f"Result for Problem {64:003d} is {compute()}")

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#!/usr/bin/env python
"""
Created on 08 Sep 2023
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 066 from Project Euler
https://projecteuler.net/problem=66
"""
from sympy import simplify
from sympy.abc import t, x, y
from sympy.solvers.diophantine.diophantine import diop_quadratic
from project_euler_python.utils import timeit
@timeit("Problem 066")
def compute():
"""
Consider quadratic Diophantine equations of the form:
x^2 - Dy^2 = 1
For example, when D = 13, the minimal solution in x is
649^2 - 13 * 180^2 = 1
It can be assumed that there are no solutions in positive integers when D is
square.
By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the
following:
3^2 - 2 * 2^2 = 1
2^2 - 3 * 1^2 = 1
9^2 - 5 * 4^2 = 1
5^2 - 6 * 2^2 = 1
8^2 - 7 * 3^2 = 1
Hence, by considering minimal solutions in x for D <= 7, the largest x is
obtained when D = 5.
Find the value of D <= 1000 in minimal solutions of x for which the largest
value of x is obtained.
"""
max_d, max_x = 0, 0
for d in range(2, 1000):
solve = diop_quadratic(x**2 - d * y**2 - 1, t)
for i in solve:
sol = simplify(i.subs({t: 0}))
xx = sol[0]
if xx > max_x:
max_x = xx
max_d = d
return max_d
if __name__ == "__main__":
print(f"Result for Problem 066 is {compute()}")

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#!/usr/bin/env python
"""
Creation of templates for the problems of Project Euler
"""
import datetime
import inspect
from argparse import ArgumentParser
def create_problem():
with open(Problem, "w+") as f:
template = inspect.cleandoc(
f'''#!/usr/bin/env python
"""
Created on {today}
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem {(args["problem"]):0>3} from Project Euler
https://projecteuler.net/problem={args["problem"]}
"""
from utils import timeit
@timeit("Problem {(args["problem"]):0>3}")
def compute():
"""
# Statement
"""
# Your code goes here
if __name__ == "__main__":
print(f"Result for Problem {(args["problem"]):0>3} is {{compute()}}")
''' # noqa: E501
)
f.write(template)
if __name__ == "__main__":
today = datetime.datetime.now().strftime("%d %b %Y")
parser = ArgumentParser(description=__doc__)
# Add your arguments here
parser.add_argument("-p", "--problem", help="number of the problem to solve")
args = vars(parser.parse_args())
Problem = f"Problem{(args['problem']):0>3}.py"
create_problem()

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#!/usr/bin/env python
"""
Created on 14 Mar 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 001 of Project Euler
https://projecteuler.net/problem=1
"""
from project_euler_python.utils import timeit
@timeit("Problem 001")
def compute():
"""
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
"""
limit = 1000
ans = sum(n for n in range(limit) if (n % 3 == 0 or n % 5 == 0))
return ans
if __name__ == "__main__":
print(f"Result for problem 001: {compute()}")

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#!/usr/bin/env python
"""
Created on 14 Mar 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 002 of Project Euler
https://projecteuler.net/problem=2
"""
from project_euler_python.utils import timeit
@timeit("Problem 002")
def compute():
"""
Each new term in the Fibonacci sequence is generated by adding the
previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
Find the sum of all the even-valued terms in the sequence which do not
exceed four million.
"""
ans = 0
limit = 4_000_000
x, y = 1, 1
z = x + y # Because every third Fibonacci number is even
while z <= limit:
ans += z
x = y + z
y = z + x
z = x + y
return ans
if __name__ == "__main__":
print(f"Result for problem 002: {compute()}")

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#!/usr/bin/env python
"""
Created on 18 Mar 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 003 of Project Euler
https://projecteuler.net/problem=3
"""
from project_euler_python.utils import timeit
@timeit("Problem 003")
def compute():
"""
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?
"""
ans = 600851475143
factor = 2
while factor * factor < ans:
while ans % factor == 0:
ans = ans // factor
factor += 1
return ans
if __name__ == "__main__":
print(f"Result for problem 003: {compute()}")

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#!/usr/bin/env python
"""
Created on 18 Mar 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 004 of Project Euler
https://projecteuler.net/problem=4
"""
from project_euler_python.utils import timeit
@timeit("Problem 004")
def compute():
"""
A palindromic number reads the same both ways. The largest palindrome made
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.
"""
ans = 0
for i in range(100, 1_000):
for j in range(100, 1_000):
palindrome = i * j
s = str(palindrome)
if s == s[::-1] and palindrome > ans:
ans = palindrome
return ans
if __name__ == "__main__":
print(f"Result for problem 004: {compute()}")

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#!/usr/bin/env python
"""
Created on 23 Apr 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 005 of Project Euler
https://projecteuler.net/problem=5
"""
from math import gcd
from project_euler_python.utils import timeit
# The LCM of two natural numbers x and y is given by:
# def lcm(x, y):
# return x * y // math.gcd(x, y)
# It is possible to compute the LCM of more than two numbers by iteratively
# computing the LCM of two numbers, i.e. LCM(a, b, c) = LCM(a, LCM(b, c))
@timeit("Problem 005")
def compute():
"""
2520 is the smallest number that can be divided by each of the numbers
from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of
the numbers from 1 to 20?
"""
ans = 1
for i in range(1, 21):
ans *= i // gcd(i, ans)
return ans
if __name__ == "__main__":
print(f"Result for problem 005: {compute()}")

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#!/usr/bin/env python
"""
Created on 17 Jun 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 006 of Project Euler
https://projecteuler.net/problem=6
"""
from project_euler_python.utils import timeit
@timeit("Problem 006")
def compute():
"""
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 55^2 = 3025
Hence the difference between the sum of the squares of the first ten
natural numbers and the square of the sum is 3025 − 385 = 2640.
Find the difference between the sum of the squares of the first one
hundred natural numbers and the square of the sum.
"""
limit = 100 + 1
square_of_sum = sum(n for n in range(limit)) ** 2
sum_squares = sum(n**2 for n in range(limit))
diff = square_of_sum - sum_squares
return diff
if __name__ == "__main__":
print(f"Result for Problem 006: {compute()}")

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#!/usr/bin/env python
"""
Created on 28 Jun 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 007 of Project Euler
https://projecteuler.net/problem=7
"""
from project_euler_python.utils import is_prime, timeit
@timeit("Problem 007")
def compute():
"""
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see
that the 6th prime is 13.
What is the 10001st prime number?
"""
target = 10_001
number = 2
primes_list = []
while len(primes_list) < target:
if is_prime(number):
primes_list.append(number)
number += 1
return primes_list[-1]
if __name__ == "__main__":
print(f"Result for Problem 007: {compute()}")

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#!/usr/bin/env python
"""
Created on 28 Abr 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 008 of Project Euler
https://projecteuler.net/problem=8
"""
import collections
from itertools import islice
from project_euler_python.utils import timeit
# recipe from more-itertools
def sliding_window(iterable, n):
# sliding_window('ABCDEFG', 4) -> ABCD BCDE CDEF DEFG
it = iter(iterable)
window = collections.deque(islice(it, n), maxlen=n)
if len(window) == n:
yield tuple(window)
for x in it:
window.append(x)
yield tuple(window)
@timeit("Problem 008")
def compute():
"""
The four adjacent digits in the 1000-digit number that have the
greatest product are 9 x 9 x 8 x 9 = 5832.
731671...963450
Find the thirteen adjacent digits in the 1000-digit number that have
the greatest product. What is the value of this product?
"""
NUM = """
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
"""
num = NUM.replace("\n", "").replace(" ", "")
adjacent_digits = 13
ans = 0
for nums in sliding_window(num, adjacent_digits):
prod = 1
for num in nums:
prod *= int(num)
if prod > ans:
ans = prod
return ans
if __name__ == "__main__":
print(f"Result for Problem 008: {compute()}")

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#!/usr/bin/env python
"""
Created on 26 Aug 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 009 of Project Euler
https://projecteuler.net/problem=9
"""
from project_euler_python.utils import timeit
@timeit("Problem 009")
def compute():
"""
A Pythagorean triplet is a set of three natural numbers, a < b < c,
for which a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
upper_limit = 1000
for a in range(1, upper_limit + 1):
for b in range(a + 1, upper_limit + 1):
c = upper_limit - a - b
if a * a + b * b == c * c:
# It is now implied that b < c, because we have a > 0
return a * b * c
if __name__ == "__main__":
print(f"Result for Problem 009: {compute()}")

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#!/usr/bin/env python
"""
Created on 26 Aug 2017
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 010 of Project Euler
https://projecteuler.net/problem=10
"""
from project_euler_python.utils import list_primes, timeit
@timeit("Problem 010")
def compute():
"""
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
ans = sum(list_primes(1_999_999))
return ans
if __name__ == "__main__":
print(f"Result for Problem 010: {compute()}")

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#!/usr/bin/env python
"""
Created on 07 Jul 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 011 of Project Euler
https://projecteuler.net/problem=11
"""
from project_euler_python.utils import timeit
GRID = [
[8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8],
[49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0],
[81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65],
[52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91],
[22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80],
[24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50],
[32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70],
[67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21],
[24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72],
[21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95],
[78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92],
[16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57],
[86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58],
[19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40],
[4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66],
[88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69],
[4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36],
[20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16],
[20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54],
[1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48],
]
def grid_product(x, y, dx, dy, n):
result = 1
for i in range(n):
result *= GRID[y + i * dy][x + i * dx]
return result
@timeit("Problem 011")
def compute():
"""
In the 20x20 grid above, four numbers along a diagonal line have been
marked in red.
The product of these numbers is 26 x 63 x 78 x 14 = 1788696.
What is the greatest product of four adjacent numbers in any direction
(up, down, left, right, or diagonally) in the 20x20 grid?
"""
ans = 0
width = height = len(GRID)
adjacent_nums = 4
for y in range(height):
for x in range(width):
if x + adjacent_nums <= width:
ans = max(grid_product(x, y, 1, 0, adjacent_nums), ans)
if y + adjacent_nums <= height:
ans = max(grid_product(x, y, 0, 1, adjacent_nums), ans)
if x + adjacent_nums <= width and y + adjacent_nums <= height:
ans = max(grid_product(x, y, 1, 1, adjacent_nums), ans)
if x - adjacent_nums >= -1 and y + adjacent_nums <= height:
ans = max(grid_product(x, y, -1, 1, adjacent_nums), ans)
return ans
if __name__ == "__main__":
print(f"Result for Problem 011: {compute()}")

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#!/usr/bin/env python
"""
Created on 01 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 012 of Project Euler
https://projecteuler.net/problem=12
"""
from itertools import count
from math import floor, sqrt
from project_euler_python.utils import timeit
# Returns the number of integers in the range [1, n] that divide n.
def num_divisors(n):
end = floor(sqrt(n))
divs = []
for i in range(1, end + 1):
if n % i == 0:
divs.append(i)
if end**2 == n:
divs.pop()
return 2 * len(divs)
@timeit("Problem 012")
def compute():
"""
The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number would be:
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
"""
triangle = 0
for i in count(1):
# This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
# = i*(i+1)/2
triangle += i
if num_divisors(triangle) > 500:
return str(triangle)
if __name__ == "__main__":
print(f"Result for Problem 012: {compute()}")

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#!/usr/bin/env python
"""
Created on 1 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 013 of Project Euler
https://projecteuler.net/problem=13
"""
from pathlib import Path
from project_euler_python.utils import timeit
@timeit("Problem 013")
def compute():
"""
Work out the first ten digits of the sum of the following one-hundred
50-digit numbers.
"""
file = Path("files/Problem13.txt")
with open(file) as f:
num = f.readlines()
ans = 0
for line in num:
ans += int(line)
return str(ans)[:10]
if __name__ == "__main__":
print(f"Result for Problem 013: {compute()}")

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#!/usr/bin/env python
"""
Created on 7 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 014 of Project Euler
https://projecteuler.net/problem=14
"""
from project_euler_python.utils import timeit
def chain_length(n, terms):
length = 0
while n != 1:
if n in terms:
length += terms[n]
break
if n % 2 == 0:
n = n / 2
else:
n = 3 * n + 1
length += 1
return length
@timeit("Problem 014")
def compute():
"""
The following iterative sequence is defined for the set of positive
integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following
sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen that this sequence (starting at 13 and finishing at 1)
contains 10 terms. Although it has not been proved yet (Collatz Problem),
it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.
"""
ans = 0
limit = 1_000_000
score = 0
terms = dict()
for i in range(1, limit):
terms[i] = chain_length(i, terms)
if terms[i] > score:
score = terms[i]
ans = i
return ans
if __name__ == "__main__":
print(f"Result for Problem 014: {compute()}")

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#!/usr/bin/env python
"""
Created on 7 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 015 of Project Euler
https://projecteuler.net/problem=15
"""
from math import factorial
from project_euler_python.utils import timeit
@timeit("Problem 015")
def compute():
"""
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 20x20 grid?
"""
n = 20
ans = int(factorial(2 * n) / (factorial(n) * factorial(2 * n - n)))
return ans
if __name__ == "__main__":
print(f"Result for Problem 015: {compute()}")

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#!/usr/bin/env python
"""
Created on 13 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 016 of Project Euler
https://projecteuler.net/problem=16
"""
from project_euler_python.utils import timeit
@timeit("Problem 016")
def compute():
"""
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
"""
n = 1000
ans = sum(int(digit) for digit in str(2**n))
return ans
if __name__ == "__main__":
print(f"Result for Problem 016: {compute()}")

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#!/usr/bin/env python
"""
Created on 13 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 017 of Project Euler
https://projecteuler.net/problem=17
"""
from project_euler_python.utils import timeit
def num_to_letters(num):
nums = {
0: "",
1: "one",
2: "two",
3: "three",
4: "four",
5: "five",
6: "six",
7: "seven",
8: "eight",
9: "nine",
10: "ten",
11: "eleven",
12: "twelve",
13: "thirteen",
14: "fourteen",
15: "fifteen",
16: "sixteen",
17: "seventeen",
18: "eighteen",
19: "nineteen",
20: "twenty",
30: "thirty",
40: "forty",
50: "fifty",
60: "sixty",
70: "seventy",
80: "eighty",
90: "ninety",
100: "hundred",
1000: "thousand",
}
if num <= 20:
return len(nums[num])
elif num < 100:
tens, units = divmod(num, 10)
return len(nums[tens * 10]) + num_to_letters(units)
elif num < 1000:
hundreds, rest = divmod(num, 100)
if rest:
return (
num_to_letters(hundreds)
+ len(nums[100])
+ len("and")
+ num_to_letters(rest)
)
else:
return num_to_letters(hundreds) + len(nums[100])
else:
thousands, _ = divmod(num, 1000)
return num_to_letters(thousands) + len(nums[1000])
@timeit("Problem 017")
def compute():
"""
If the numbers 1 to 5 are written out in words: one, two, three, four,
five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written
out in words, how many letters would be used?
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
20 letters. The use of "and" when writing out numbers is in compliance
with British usage.
"""
n = 1000
ans = 0
for num in range(1, n + 1):
ans += num_to_letters(num)
return ans
if __name__ == "__main__":
print(f"Result for Problem 017: {compute()}")

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#!/usr/bin/env python
"""
Created on 15 Sep 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 018 of Project Euler
https://projecteuler.net/problem=18
"""
from project_euler_python.utils import timeit
triangle = [
[75],
[95, 64],
[17, 47, 82],
[18, 35, 87, 10],
[20, 4, 82, 47, 65],
[19, 1, 23, 75, 3, 34],
[88, 2, 77, 73, 7, 63, 67],
[99, 65, 4, 28, 6, 16, 70, 92],
[41, 41, 26, 56, 83, 40, 80, 70, 33],
[41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
[53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
[70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
[91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
[63, 66, 4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
[4, 62, 98, 27, 23, 9, 70, 98, 73, 93, 38, 53, 60, 4, 23],
]
@timeit("Problem 018")
def compute():
"""
By starting at the top of the triangle below and moving to adjacent
numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle above
"""
for i in reversed(range(len(triangle) - 1)):
for j in range(len(triangle[i])):
triangle[i][j] += max(triangle[i + 1][j], triangle[i + 1][j + 1])
return triangle[0][0]
if __name__ == "__main__":
print(f"Result for Problem 018: {compute()}")

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#!/usr/bin/env python
"""
Created on 15 Sep 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 019 of Project Euler
https://projecteuler.net/problem=19
"""
from datetime import date
from project_euler_python.utils import timeit
@timeit("Problem 019")
def compute():
"""
You are given the following information, but you may prefer to do some
research for yourself.
1 Jan 1900 was a Monday.
Thirty days has September, April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
A leap year occurs on any year evenly divisible by 4, but not on a century
unless it is divisible by 400.
How many Sundays fell on the first of the month during the twentieth
century (1 Jan 1901 to 31 Dec 2000)?
"""
ans = 0
for year in range(1901, 2001):
for month in range(1, 13):
if date(year, month, 1).weekday() == 6:
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 019: {compute()}")

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#!/usr/bin/env python
"""
Created on 15 Sep 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 020 of Project Euler
https://projecteuler.net/problem=20
"""
from math import factorial
from project_euler_python.utils import timeit
@timeit("Problem 020")
def compute():
"""
n! means n x (n - 1) x ... x 3 x 2 x 1
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!
"""
fact = factorial(100)
ans = sum(int(digit) for digit in str(fact))
return ans
if __name__ == "__main__":
print(f"Result for Problem 020: {compute()}")

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#!/usr/bin/env python
"""
Created on 15 Sep 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 021 of Project Euler
https://projecteuler.net/problem=21
"""
from project_euler_python.utils import timeit
def sum_divisors(n):
return sum(i for i in range(1, n // 2 + 1) if n % i == 0)
@timeit("Problem 021")
def compute():
"""
Let d(n) be defined as the sum of proper divisors of n (numbers
less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are
1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
"""
n = 10_000
ans = 0
for i in range(1, n):
value = sum_divisors(i)
if i != value and sum_divisors(value) == i:
ans += i
return ans
if __name__ == "__main__":
print(f"Result for Problem 021: {compute()}")

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#!/usr/bin/env python
"""
Created on 31 Dec 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 022 of Project Euler
https://projecteuler.net/problem=22
"""
from pathlib import Path
from project_euler_python.utils import timeit
@timeit("Problem 022")
def compute():
"""
Using names.txt, a 46K text file containing over five-thousand first names,
begin by sorting it into alphabetical order. Then working out the
alphabetical value for each name, multiply this value by its alphabetical
position in the list to obtain a name 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 x 53 = 49714.
What is the total of all the name scores in the file?
"""
file = Path("files/Problem22.txt")
with open(file) as f:
names = sorted(f.read().replace('"', "").split(","))
ans = 0
for idx, name in enumerate(names, 1):
ans += sum(ord(char) - 64 for char in name) * idx
return ans
if __name__ == "__main__":
print(f"Result for Problem 022: {compute()}")

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#!/usr/bin/env python
"""
Created on 05 Jan 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 023 of Project Euler
https://projecteuler.net/problem=23
"""
from project_euler_python.utils import timeit
@timeit("Problem 023")
def compute():
"""
A perfect number is a number for which the sum of its proper divisors is
exactly equal to the number. For example, the sum of the proper divisors
of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
number.
A number n is called deficient if the sum of its proper divisors is less
than n and it is called abundant if this sum exceeds n.
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
number that can be written as the sum of two abundant numbers is 24. By
mathematical analysis, it can be shown that all integers greater than 28123
can be written as the sum of two abundant numbers. However, this upper
limit cannot be reduced any further by analysis even though it is known
that the greatest number that cannot be expressed as the sum of two
abundant numbers is less than this limit.
Find the sum of all the positive integers which cannot be written as the
sum of two abundant numbers.
"""
limit = 28124
divisor_sum = [0] * limit
for i in range(1, limit):
for j in range(i * 2, limit, i):
divisor_sum[j] += i
abundant_nums = [i for (i, x) in enumerate(divisor_sum) if x > i]
expressible = [False] * limit
for i in abundant_nums:
for j in abundant_nums:
if i + j < limit:
expressible[i + j] = True
else:
break
ans = sum(i for (i, x) in enumerate(expressible) if not x)
return ans
if __name__ == "__main__":
print(f"Result for Problem 023: {compute()}")

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#!/usr/bin/env python
"""
Created on 11 Sep 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 024 of Project Euler
https://projecteuler.net/problem=24
"""
from itertools import permutations
from project_euler_python.utils import timeit
@timeit("Problem 024")
def compute():
"""
A permutation is an ordered arrangement of objects. For example, 3124 is
one possible permutation of the digits 1, 2, 3 and 4. If all of the
permutations are listed numerically or alphabetically, we call it
lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
"""
digits = list(range(10))
_permutations = list(permutations(digits))
ans = "".join(str(digit) for digit in _permutations[999_999])
return ans
if __name__ == "__main__":
print(f"Result for Problem 024: {compute()}")

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#!/usr/bin/env python
"""
Created on 11 Sep 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 025 of Project Euler
https://projecteuler.net/problem=25
"""
from project_euler_python.utils import timeit
@timeit("Problem 025")
def compute():
"""
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn-1 + Fn-2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The 12th term, F12, is the first term to contain three digits.
What is the index of the first term in the Fibonacci sequence to
contain 1000 digits?
"""
a, b = 1, 1
ans = 2
while len(str(b)) < 1000:
a, b = b, b + a
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 025: {compute()}")

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#!/usr/bin/env python
"""
Created on 11 Sep 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 026 of Project Euler
https://projecteuler.net/problem=26
"""
from project_euler_python.utils import timeit
@timeit("Problem 026")
def compute():
"""
A unit fraction contains 1 in the numerator. The decimal representation
of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring
cycle in its decimal fraction part.
"""
cycle_length = 0
ans = 0
for number in range(3, 1000, 2):
if number % 5 == 0:
continue
p = 1
while 10**p % number != 1:
p += 1
if p > cycle_length:
cycle_length, ans = p, number
return ans
if __name__ == "__main__":
print(f"Result for Problem 026: {compute()}")

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#!/usr/bin/env python
"""
Created on 15 Sep 2019
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 027 of Project Euler
https://projecteuler.net/problem=27
"""
from project_euler_python.utils import is_prime, timeit
@timeit("Problem 027")
def compute():
"""
Euler discovered the remarkable quadratic formula:
n^2 + n + 41
It turns out that the formula will produce 40 primes for the consecutive
integer values 0≤n≤39. However, when n=40, 40^2+40+41=40(40+1)+41 is
divisible by 41, and certainly when n=41,41^2+41+41 is clearly divisible
by 41.
The incredible formula n^2-79n+1601 was discovered, which produces 80
primes for the consecutive values 0≤n≤79. The product of the coefficients,
-79 and 1601, is -126479.
Considering quadratics of the form:
n^2 + an + b
where |a|<1000, |b|≤1000 and |n| is the modulus/absolute value of n
e.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.
"""
limit = 1000
consecutive_values = 0
for a in range(-999, limit):
for b in range(limit + 1):
n = 0
while is_prime(abs((n**2) + (a * n) + b)):
n += 1
if n > consecutive_values:
consecutive_values = n
ans = a * b
return ans
if __name__ == "__main__":
print(f"Result for Problem 027: {compute()}")

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#!/usr/bin/env python
"""
Created on 3 Jan 2020
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 028 of Project Euler
https://projecteuler.net/problem=28
"""
from project_euler_python.utils import timeit
@timeit("Problem 028")
def compute():
"""
Starting with the number 1 and moving to the right in a clockwise
direction a 5 by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral
formed in the same way?
"""
size = 1001 # Must be odd
ans = 1 # Special case for size 1
step = 0
i, current = 1, 1
while step < size - 1:
step = i * 2
for _ in range(1, 5):
current += step
ans += current
i += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 028: {compute()}")

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#!/usr/bin/env python
"""
Created on 3 Jan 2020
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 029 of Project Euler
https://projecteuler.net/problem=29
"""
from itertools import product
from project_euler_python.utils import timeit
@timeit("Problem 029")
def compute():
"""
Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
2^2=4, 2^3=8, 2^4=16, 2^5=32
3^2=9, 3^3=27, 3^4=81, 3^5=243
4^2=16, 4^3=64, 4^4=256, 4^5=1024
5^2=25, 5^3=125, 5^4=625, 5^5=3125
If they are then placed in numerical order, with any repeats removed, we
get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2≤a≤100
and 2≤b≤100?
"""
ans = len(set(a**b for a, b in product(range(2, 101), repeat=2)))
return ans
if __name__ == "__main__":
print(f"Result for Problem 029: {compute()}")

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#!/usr/bin/env python
"""
Created on 3 Jan 2020
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 030 of Project Euler
https://projecteuler.net/problem=30
"""
from project_euler_python.utils import timeit
def power_digit_sum(pow, n):
return sum(int(c) ** pow for c in str(n))
@timeit("Problem 030")
def compute():
"""
Surprisingly there are only three numbers that can be written as the sum
of fourth powers of their digits:
1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4
As 1 = 14 is not a sum it is not included.
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
Find the sum of all the numbers that can be written as the sum of fifth
powers of their digits.
"""
ans = sum(i for i in range(2, 1_000_000) if i == power_digit_sum(5, i))
return ans
if __name__ == "__main__":
print(f"Result for Problem 030: {compute()}")

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#!/usr/bin/env python
"""
Created on 24 Feb 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 031 of Project Euler
https://projecteuler.net/problem=31
"""
from itertools import product
from project_euler_python.utils import timeit
@timeit("Problem 031")
def compute():
"""
In the United Kingdom the currency is made up of pound (£) and pence (p).
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:
1x£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
How many different ways can £2 be made using any number of coins?
"""
ans = 0
coins = [2, 5, 10, 20, 50, 100]
bunch_of_coins = product(*[range(0, 201, i) for i in coins])
for money in bunch_of_coins:
if sum(money) <= 200:
ans += 1
# consider also the case for 200 coins of 1p
return ans + 1
if __name__ == "__main__":
print(f"Result for Problem 031: {compute()}")

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#!/usr/bin/env python
"""
Created on 26 Feb 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 032 of Project Euler
https://projecteuler.net/problem=32
"""
from project_euler_python.utils import timeit
@timeit("Problem 032")
def compute():
"""
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 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 be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only
include it once in your sum.
"""
ans = set()
pandigital = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
for x in range(1, 100):
for y in range(100, 10_000):
# product = x * y
if sorted(str(x) + str(y) + str(x * y)) == pandigital:
ans.add(x * y)
return sum(ans)
if __name__ == "__main__":
print(f"Result for Problem 032: {compute()}")

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#!/usr/bin/env python
"""
Created on 04 Mar 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 033 of Project Euler
https://projecteuler.net/problem=33
"""
from project_euler_python.utils import timeit
@timeit("Problem 033")
def compute():
"""
The fraction 49/98 is a curious fraction, as an inexperienced mathematician
in attempting to simplify it may incorrectly believe that 49/98 = 4/8,
which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
There are exactly four non-trivial examples of this type of fraction, less
than one in value, and containing two digits in the numerator and
denominator.
If the product of these four fractions is given in its lowest common terms,
find the value of the denominator.
"""
numerator = 1
denominator = 1
for x in range(10, 100):
for y in range(10, 100):
if x < y:
if str(x)[1] == str(y)[0]:
if int(str(y)[1]) != 0:
if int(str(x)[0]) / int(str(y)[1]) == x / y:
numerator *= x
denominator *= y
ans = int(denominator / numerator)
return ans
if __name__ == "__main__":
print(f"Result for Problem 033: {compute()}")

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#!/usr/bin/env python
"""
Created on 02 Apr 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 034 of Project Euler
https://projecteuler.net/problem=34
"""
from math import factorial
from project_euler_python.utils import timeit
@timeit("Problem 034")
def compute():
"""
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial
of their digits.
Note: As 1! = 1 and 2! = 2 are not sums they are not included.
"""
ans = 0
for num in range(10, 2_540_160):
sum_of_factorial = 0
for digit in str(num):
sum_of_factorial += factorial(int(digit))
if sum_of_factorial == num:
ans += sum_of_factorial
return ans
if __name__ == "__main__":
print(f"Result for Problem 034: {compute()}")

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#!/usr/bin/env python
"""
Created on 02 Apr 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 035 of Project Euler
https://projecteuler.net/problem=35
"""
from project_euler_python.utils import is_prime, timeit
def circular_number(number):
num_str = str(number)
ans = []
for i in range(len(num_str)):
ans.append(int(num_str[i:] + num_str[:i]))
return ans
@timeit("Problem 035")
def compute():
"""
The number, 197, is called a circular prime because all rotations of the
digits: 197, 971, and 719, are themselves prime.
There are thirteen such primes below 100:
2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
"""
ans = []
for i in range(2, 1_000_000):
if is_prime(i):
all_primes = True
for j in circular_number(i):
if not is_prime(j):
all_primes = False
break
if all_primes:
ans.append(i)
return len(ans)
if __name__ == "__main__":
print(f"Result for Problem 035: {compute()}")

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#!/usr/bin/env python
"""
Created on 08 Apr 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 036 of Project Euler
https://projecteuler.net/problem=36
"""
from project_euler_python.utils import timeit
def is_palidrome(num):
return str(num) == str(num)[::-1]
@timeit("Problem 036")
def compute():
"""
The decimal number, 585 = 1001001001_2 (binary), is palindromic
in both bases.
Find the sum of all numbers, less than one million, which are palindromic
in base 10 and base 2.
"""
ans = 0
for i in range(1, 1_000_001, 2):
if is_palidrome(i) and is_palidrome(bin(i)[2:]):
ans += i
return ans
if __name__ == "__main__":
print(f"Result for Problem 036: {compute()}")

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#!/usr/bin/env python
"""
Created on 09 Apr 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 037 of Project Euler
https://projecteuler.net/problem=37
"""
from project_euler_python.utils import is_prime, list_primes, timeit
def is_truncatable_prime(number):
num_str = str(number)
for i in range(1, len(num_str)):
if not is_prime(int(num_str[i:])) or not is_prime(int(num_str[:-i])):
return False
return True
@timeit("Problem 037")
def compute():
"""
The number 3797 has an interesting property. Being prime itself, it is
possible to continuously remove digits from left to right, and remain
prime at each stage: 3797, 797, 97, and 7.
Similarly we can work from right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left
to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
"""
ans = 0
primes = list_primes(1_000_000)
# Statement of the problem says this
for number in primes[4:]:
if is_truncatable_prime(number):
ans += number
return ans
if __name__ == "__main__":
print(f"Result for Problem 037: {compute()}")

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#!/usr/bin/env python
"""
Created on 03 Jun 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 038 of Project Euler
https://projecteuler.net/problem=38
"""
from project_euler_python.utils import timeit
@timeit("Problem 038")
def compute():
"""
Take the number 192 and multiply it by each of 1, 2, and 3:
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)
The same can be achieved by starting with 9 and multiplying by
1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is
the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number that can
be formed as the concatenated product of an integer with
(1,2, ... , n) where n > 1?
"""
ans = []
# Number must 4 digits (exactly) to be pandigital
# if n > 1
for i in range(1, 10_000):
integer = 1
number = ""
while len(number) < 9:
number += str(integer * i)
if sorted(number) == list("123456789"):
ans.append(number)
integer += 1
return max(ans)
if __name__ == "__main__":
print(f"Result for Problem 038: {compute()}")

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#!/usr/bin/env python
"""
Created on 05 Jun 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 039 of Project Euler
https://projecteuler.net/problem=39
"""
from project_euler_python.utils import timeit
@timeit("Problem 039")
def compute():
"""
If p is the perimeter of a right angle triangle with integral length sides,
{a,b,c}, there are exactly three solutions for p = 120:
{20,48,52}, {24,45,51}, {30,40,50}
For which value of p ≤ 1000, is the number of solutions maximised?
"""
ans, val = 0, 0
for p in range(2, 1001, 2):
sol = 0
for a in range(1, p):
for b in range(a + 1, p - 2 * a):
c = p - (a + b)
if a**2 + b**2 == c**2:
sol += 1
elif a**2 + b**2 > c**2:
# As we continue our innermost loop, the left side
# gets bigger, right gets smaller, so we're done here
break
if sol > ans:
ans, val = sol, p
return val
if __name__ == "__main__":
print(f"Result for Problem 039: {compute()}")

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#!/usr/bin/env python
"""
Created on 23 Jun 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 040 of Project Euler
https://projecteuler.net/problem=40
"""
from project_euler_python.utils import timeit
@timeit("Problem 040")
def compute():
"""
An irrational decimal fraction is created by concatenating the positive
integers:
0.123456789101112131415161718192021...
It can be seen that the 12th digit of the fractional part is 1.
If d_n represents the n^th digit of the fractional part, find the value of
the following expression.
d_1 x d_{10} x d_{100} x d_{1_000} x d_{10_000} x d_{100_000}
x d_{1_000_000}
"""
fraction = ""
for i in range(1, 1_000_000):
fraction += str(i)
ans = 1
for i in range(7):
ans *= int(fraction[10**i - 1])
return ans
if __name__ == "__main__":
print(f"Result for Problem 040: {compute()}")

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#!/usr/bin/env python
"""
Created on 29 Jun 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 041 of Project Euler
https://projecteuler.net/problem=41
"""
from project_euler_python.utils import is_prime, timeit
def is_pandigital(number):
number = sorted(str(number))
check = [str(i) for i in range(1, len(number) + 1)]
if number == check:
return True
return False
@timeit("Problem 041")
def compute():
"""
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, 2143 is
a 4-digit pandigital and is also prime.
What is the largest n-digit pandigital prime that exists?
"""
for ans in range(7654321, 1, -1):
if is_pandigital(ans):
if is_prime(ans):
return ans
if __name__ == "__main__":
print(f"Result for Problem 041: {compute()}")

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src/project_euler_python/problems_001_050/Problem042.py Normal file
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#!/usr/bin/env python
"""
Created on 26 Jul 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 042 of Project Euler
https://projecteuler.net/problem=42
"""
from pathlib import Path
from project_euler_python.utils import timeit
def triangle_number(num):
return int(0.5 * num * (num + 1))
def word_to_value(word):
return sum(ord(letter) - 64 for letter in word)
@timeit("Problem 042")
def compute():
"""
The nth term of the sequence of triangle numbers is given by,
tn = n(n+1)/2; so the first ten triangle numbers are:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
By converting each letter in a word to a number corresponding to its
alphabetical position and adding these values we form a word value. For
example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word
value is a triangle number then we shall call the word a triangle word.
Using words.txt, a 16K text file containing nearly two-thousand common
English words, how many are triangle words?
"""
triangular_numbers = [triangle_number(n) for n in range(27)]
ans = 0
file = Path("files/Problem42.txt")
with open(file) as f:
words = f.readline().strip('"').split('","')
for word in words:
if word_to_value(word) in triangular_numbers:
ans += 1
return ans
if __name__ == "__main__":
print(f"Result for Problem 042: {compute()}")

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src/project_euler_python/problems_001_050/Problem043.py Normal file
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#!/usr/bin/env python
"""
Created on 03 Aug 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 043 of Project Euler
https://projecteuler.net/problem=43
"""
from itertools import permutations
from project_euler_python.utils import timeit
@timeit("Problem 043")
def compute():
"""
The number, 1406357289, is a 0 to 9 pandigital number because
it is made up of each of the digits 0 to 9 in some order, but
it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
way, we note the following:
d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
"""
ans = []
pandigital = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
for n in permutations(pandigital):
n_ = "".join(n)
if n_[0] != "0" and sorted("".join(n_)) == pandigital:
if int(n_[7:]) % 17 == 0:
if int(n_[6:9]) % 13 == 0:
if int(n_[5:8]) % 11 == 0:
if int(n_[4:7]) % 7 == 0:
if int(n_[3:6]) % 5 == 0:
if int(n_[2:5]) % 3 == 0:
if int(n_[1:4]) % 2 == 0:
ans.append(int(n_))
return sum(ans)
if __name__ == "__main__":
print(f"Result for Problem 043: {compute()}")

52
src/project_euler_python/problems_001_050/Problem044.py Normal file
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#!/usr/bin/env python
"""
Created on 30 Aug 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 044 of Project Euler
https://projecteuler.net/problem=44
"""
from itertools import combinations
from operator import add, sub
from project_euler_python.utils import timeit
def pentagonal(n):
return int(n * (3 * n - 1) / 2)
@timeit("Problem 044")
def compute():
"""
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.
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?
"""
ans = 0
pentagonal_list = set(pentagonal(n) for n in range(1, 2500))
pairs = combinations(pentagonal_list, 2)
for p in pairs:
if add(*p) in pentagonal_list and abs(sub(*p)) in pentagonal_list:
ans = abs(sub(*p))
# the first one found would be the smallest
break
return ans
if __name__ == "__main__":
print(f"Result for Problem 044: {compute()}")

46
src/project_euler_python/problems_001_050/Problem045.py Normal file
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#!/usr/bin/env python
"""
Created on 09 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 045 of Project Euler
https://projecteuler.net/problem=45
"""
from project_euler_python.utils import timeit
def pentagonal(n):
return int(n * (3 * n - 1) / 2)
def hexagonal(n):
return int(n * (2 * n - 1))
@timeit("Problem 045")
def compute():
"""
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
Pentagonal Pn=n(3n-1)/2 1, 5, 12, 22, 35, ...
Hexagonal Hn=n(2n-1) 1, 6, 15, 28, 45, ...
It can be verified that T285 = P165 = H143 = 40755.
Find the next triangle number that is also pentagonal and hexagonal.
"""
pentagonal_list = set(pentagonal(n) for n in range(2, 100_000))
# all hexagonal numbers are also triangle numbers!
hexagonal_list = set(hexagonal(n) for n in range(2, 100_000))
ans = sorted(hexagonal_list & pentagonal_list)
# First one is already known
return ans[1]
if __name__ == "__main__":
print(f"Result for Problem 045: {compute()}")

49
src/project_euler_python/problems_001_050/Problem046.py Normal file
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#!/usr/bin/env python
"""
Created on 12 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 046 of Project Euler
https://projecteuler.net/problem=46
"""
from project_euler_python.utils import is_prime, timeit
def is_goldbach(number):
for i in range(number - 1, 1, -1):
if is_prime(i) and ((number - i) / 2) ** 0.5 % 1 == 0:
return True
return False
@timeit("Problem 046")
def compute():
"""
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 + 2x1^2
15 = 7 + 2x2^2
21 = 3 + 2x3^2
25 = 7 + 2x3^2
27 = 19 + 2x2^2
33 = 31 + 2x1^2
It turns out that the conjecture was false.
What is the smallest odd composite that cannot be written as the sum
of a prime and twice a square?
"""
ans = 9
while True:
ans += 2
if not is_prime(ans) and not is_goldbach(ans):
return ans
if __name__ == "__main__":
print(f"Result for Problem 046: {compute()}")

63
src/project_euler_python/problems_001_050/Problem047.py Normal file
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#!/usr/bin/env python
"""
Created on 12 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 047 of Project Euler
https://projecteuler.net/problem=47
"""
from project_euler_python.utils import timeit
def factor(n):
ans = []
d = 2
while d * d <= n:
if n % d == 0:
ans.append(d)
n //= d
else:
d += 1
if n > 1:
ans.append(n)
return ans
@timeit("Problem 047")
def compute():
"""
The first two consecutive numbers to have two distinct prime factors are:
14 = 2 x 7
15 = 3 x 5
The first three consecutive numbers to have three distinct prime factors
are:
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?
"""
ans = []
for number in range(1, 1_000_000):
if len(ans) == 4:
break
elif len(set(factor(number))) == 4:
ans.append(number)
else:
ans = []
return ans[0]
if __name__ == "__main__":
print(f"Result for Problem 047: {compute()}")

30
src/project_euler_python/problems_001_050/Problem048.py Normal file
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@@ -0,0 +1,30 @@
#!/usr/bin/env python
"""
Created on 12 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 048 of Project Euler
https://projecteuler.net/problem=48
"""
from project_euler_python.utils import timeit
@timeit("Problem 048")
def compute():
"""
The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
"""
series = sum(i**i for i in range(1, 1001))
ans = str(series)[-10:]
return ans
if __name__ == "__main__":
print(f"Result for Problem 048: {compute()}")

48
src/project_euler_python/problems_001_050/Problem049.py Normal file
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@@ -0,0 +1,48 @@
#!/usr/bin/env python
"""
Created on 18 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 049 of Project Euler
https://projecteuler.net/problem=49
"""
from project_euler_python.utils import list_primes, timeit
@timeit("Problem 049")
def compute():
"""
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
increases by 3330, is unusual in two ways:
(i) each of the three terms are prime, and,
(ii) each of the 4-digit numbers are permutations of one another.
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit
primes, exhibiting this property, but there is one other 4-digit increasing
sequence.
What 12-digit number do you form by concatenating the three terms in this
sequence?
"""
ans = []
primes_list = sorted(set(list_primes(10_000)) - set(list_primes(1_000)))
for number in primes_list:
if (
set(list(str(number)))
== set(list(str(number + 3330)))
== set(list(str(number + 6660)))
):
if number + 3330 in primes_list and number + 6660 in primes_list:
ans.append(str(number) + str(number + 3300) + str(number + 6660))
# return the second one
return ans[1]
if __name__ == "__main__":
print(f"Result for Problem 049: {compute()}")

52
src/project_euler_python/problems_001_050/Problem050.py Normal file
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#!/usr/bin/env python
"""
Created on 18 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 050 of Project Euler
https://projecteuler.net/problem=50
"""
from project_euler_python.utils import is_prime, list_primes, timeit
@timeit("Problem 050")
def compute():
"""
The prime 41, can be written as the sum of six consecutive primes:
41 = 2 + 3 + 5 + 7 + 11 + 13
This is the longest sum of consecutive primes that adds to a prime below
one-hundred.
The longest sum of consecutive primes below one-thousand that adds to a
prime, contains 21 terms, and is equal to 953.
Which prime, below one-million, can be written as the sum of the most
consecutive primes?
"""
ans = 0
result = 0
prime_list = list_primes(1_000_000)
for i in range(len(prime_list)):
sum = 0
count = 0
for j in prime_list[i:]:
sum += j
count += 1
if is_prime(sum) and count > result:
result = count
ans = sum
if sum > 1_000_000:
break
return ans
if __name__ == "__main__":
print(f"Result for Problem 050: {compute()}")

54
src/project_euler_python/utils.py Normal file
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@@ -0,0 +1,54 @@
import math
import time
from functools import wraps
def timeit(name):
def profile(original):
@wraps(original)
def wrapper(*args, **kwargs):
t0 = time.perf_counter()
result = original(*args, **kwargs)
t1 = time.perf_counter()
if (t1 - t0) > 1:
print(f"Took: {(t1 - t0):.3f} s\n")
else:
print(f"Took: {(t1 - t0) * 1000:.3f} ms\n")
return result
return wrapper
return profile
def is_prime(n):
if n < 2:
return False
for i in range(2, int(math.sqrt(n)) + 1):
if n % i == 0:
return False
return True
# Returns a list of True and False indicating whether each number is prime.
# For 0 <= i <= n, result[i] is True if i is a prime number, False otherwise.
def list_primality(n):
# Sieve of Eratosthenes
result = [True] * (n + 1)
result[0] = result[1] = False
for i in range(int(math.sqrt(n) + 1)):
if result[i]:
for j in range(i * i, len(result), i):
result[j] = False
return result
# Returns all the prime numbers less than or equal to n, in ascending order
# For example: list_primes(97) = [2, 3, 5, 7, 11, ..., 83, 89, 97].
def list_primes(n):
return [i for (i, is_prime) in enumerate(list_primality(n)) if is_prime]
def is_palindrome(num):
return str(num) == str(num)[::-1]

4
src/update_results_table.py
View File

@@ -29,13 +29,13 @@ def write_md(results):
def cache_and_save_results(func):
@functools.wraps(func)
@functools.lru_cache(maxsize=None)
@functools.cache
def wrapper(script_number):
cache_file_name = f"{func.__name__.split('_')[1]}.json"
num = int(script_number[-6:-3])
if Path(cache_file_name).exists():
with open(cache_file_name, "r") as f:
with open(cache_file_name) as f:
cache = json.load(f)
else:
cache = {}

581
uv.lock generated Normal file
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