Python Challenge - Level 15

Link: http://www.pythonchallenge.com/pc/return/uzi.html
Username: huge
Password: file

Problem

The image shows a calendar page for January with several critical clues:

The year has a burnt hole showing 1 and ending with 6: year 1xx6.
January 1 is a Thursday (hence January 26 is a Monday).
February has 29 days, so it is a leap year.
A red circle marks January 26.
In HTML comment: 
In HTML comment: 
Title: whom?

Clues & Deduction

We need a leap year 1xx6 where January 26 was a Monday (or January 1 was a Thursday). Among qualifying years:

1176, 1356, 1576, 1756, 1976
The youngest is 1976.
The second youngest is 1756.
The note says: buy flowers for tomorrow (January 27, 1756).
Wolfgang Amadeus Mozart was born on January 27, 1756.

The answer is mozart.

Solution

Python

Using Python's standard calendar and datetime modules:

import calendar
import datetime

candidates = []
for year in range(1006, 2000, 10):
    if calendar.isleap(year):
        d = datetime.date(year, 1, 26)
        if d.isoweekday() == 1:  # Monday
            candidates.append(year)

print("Candidates:", candidates)
# Candidates: [1176, 1356, 1576, 1756, 1976]

second_youngest = candidates[-2]
print("Second youngest:", second_youngest)  # 1756

Go

Go's time package handles dates and weekdays:

package main

import (
	"fmt"
	"time"
)

func isLeapYear(year int) bool {
	return (year%4 == 0 && year%100 != 0) || (year%400 == 0)
}

func main() {
	var candidates []int
	for year := 1006; year < 2000; year += 10 {
		if isLeapYear(year) {
			t := time.Date(year, time.January, 26, 0, 0, 0, 0, time.UTC)
			if t.Weekday() == time.Monday {
				candidates = append(candidates, year)
			}
		}
	}

	fmt.Println("Candidates:", candidates)
	if len(candidates) >= 2 {
		targetYear := candidates[len(candidates)-2]
		fmt.Printf("Second youngest: %d\n", targetYear) // 1756
	}
}

Rust

In Rust, we can compute the weekday and leap year directly without external crates using Zeller's congruence or Tomohiko Sakamoto's day-of-week algorithm, or with chrono:

fn is_leap_year(year: i32) -> bool {
    (year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)
}

// Sakamoto's algorithm for day of week (0 = Sunday, 1 = Monday, ...)
fn day_of_week(y: i32, m: i32, d: i32) -> i32 {
    let t = [0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4];
    let y = if m < 3 { y - 1 } else { y };
    (y + y / 4 - y / 100 + y / 400 + t[(m - 1) as usize] + d) % 7
}

fn main() {
    let mut candidates = Vec::new();
    for year in (1006..2000).step_by(10) {
        if is_leap_year(year) && day_of_week(year, 1, 26) == 1 {
            candidates.push(year);
        }
    }

    println!("Candidates: {:?}", candidates);
    if candidates.len() >= 2 {
        let second_youngest = candidates[candidates.len() - 2];
        println!("Second youngest: {}", second_youngest); // 1756
    }
}

Language Comparison

Feature	Python	Go	Rust
Calendar / Date stdlib	`calendar` and `datetime` built-in with `isleap()` and `.isoweekday()`	`time.Date()` and `time.Weekday()` in stdlib; leap year logic written manually	No calendar/date in stdlib; use standard formula or `chrono` crate
Historical Dates	Proleptic Gregorian calendar extends back to year 1	Proleptic Gregorian calendar; handles historical years without limitation	`chrono::NaiveDate` handles dates from year -262145 to +262144
Step Iteration	`range(1006, 2000, 10)`	Standard `for year := 1006; year < 2000; year += 10`	`(1006..2000).step_by(10)`

Key insight: Python's calendar.isleap makes calendar riddles pleasant to solve quickly in a single script. Go's time package is surprisingly robust for historical dates, while Rust's zero-dependency pure arithmetic approach demonstrates that Gregorian weekday math is just a short, fast formula.