From ebc2d5d79f837931e80f7d5e7e1dece9ef48f760 Mon Sep 17 00:00:00 2001 From: Ishab Date: Sun, 2 Apr 2023 13:04:11 +0100 Subject: [PATCH] Add Project Euler problem 79 solution 1 (#8607) Co-authored-by: Dhruv Manilawala --- project_euler/problem_079/__init__.py | 0 project_euler/problem_079/keylog.txt | 50 ++++++++++++++++ project_euler/problem_079/keylog_test.txt | 16 ++++++ project_euler/problem_079/sol1.py | 69 +++++++++++++++++++++++ 4 files changed, 135 insertions(+) create mode 100644 project_euler/problem_079/__init__.py create mode 100644 project_euler/problem_079/keylog.txt create mode 100644 project_euler/problem_079/keylog_test.txt create mode 100644 project_euler/problem_079/sol1.py diff --git a/project_euler/problem_079/__init__.py b/project_euler/problem_079/__init__.py new file mode 100644 index 000000000..e69de29bb diff --git a/project_euler/problem_079/keylog.txt b/project_euler/problem_079/keylog.txt new file mode 100644 index 000000000..41f156732 --- /dev/null +++ b/project_euler/problem_079/keylog.txt @@ -0,0 +1,50 @@ +319 +680 +180 +690 +129 +620 +762 +689 +762 +318 +368 +710 +720 +710 +629 +168 +160 +689 +716 +731 +736 +729 +316 +729 +729 +710 +769 +290 +719 +680 +318 +389 +162 +289 +162 +718 +729 +319 +790 +680 +890 +362 +319 +760 +316 +729 +380 +319 +728 +716 diff --git a/project_euler/problem_079/keylog_test.txt b/project_euler/problem_079/keylog_test.txt new file mode 100644 index 000000000..2c7024bde --- /dev/null +++ b/project_euler/problem_079/keylog_test.txt @@ -0,0 +1,16 @@ +319 +680 +180 +690 +129 +620 +698 +318 +328 +310 +320 +610 +629 +198 +190 +631 diff --git a/project_euler/problem_079/sol1.py b/project_euler/problem_079/sol1.py new file mode 100644 index 000000000..d34adcd24 --- /dev/null +++ b/project_euler/problem_079/sol1.py @@ -0,0 +1,69 @@ +""" +Project Euler Problem 79: https://projecteuler.net/problem=79 + +Passcode derivation + +A common security method used for online banking is to ask the user for three +random characters from a passcode. For example, if the passcode was 531278, +they may ask for the 2nd, 3rd, and 5th characters; the expected reply would +be: 317. + +The text file, keylog.txt, contains fifty successful login attempts. + +Given that the three characters are always asked for in order, analyse the file +so as to determine the shortest possible secret passcode of unknown length. +""" +import itertools +from pathlib import Path + + +def find_secret_passcode(logins: list[str]) -> int: + """ + Returns the shortest possible secret passcode of unknown length. + + >>> find_secret_passcode(["135", "259", "235", "189", "690", "168", "120", + ... "136", "289", "589", "160", "165", "580", "369", "250", "280"]) + 12365890 + + >>> find_secret_passcode(["426", "281", "061", "819" "268", "406", "420", + ... "428", "209", "689", "019", "421", "469", "261", "681", "201"]) + 4206819 + """ + + # Split each login by character e.g. '319' -> ('3', '1', '9') + split_logins = [tuple(login) for login in logins] + + unique_chars = {char for login in split_logins for char in login} + + for permutation in itertools.permutations(unique_chars): + satisfied = True + for login in logins: + if not ( + permutation.index(login[0]) + < permutation.index(login[1]) + < permutation.index(login[2]) + ): + satisfied = False + break + + if satisfied: + return int("".join(permutation)) + + raise Exception("Unable to find the secret passcode") + + +def solution(input_file: str = "keylog.txt") -> int: + """ + Returns the shortest possible secret passcode of unknown length + for successful login attempts given by `input_file` text file. + + >>> solution("keylog_test.txt") + 6312980 + """ + logins = Path(__file__).parent.joinpath(input_file).read_text().splitlines() + + return find_secret_passcode(logins) + + +if __name__ == "__main__": + print(f"{solution() = }")