Online Python classes in Herne

1. Minutes 0 to 5: the warm-up. The mentor opens the video call, greets each learner by name and puts a two-line puzzle on the shared screen. Nobody has to speak first; anyone can type a guess in the chat.
2. Minutes 5 to 20: one new idea. The mentor writes a short program live, explains one new thing (a loop, a dictionary, reading a file) and runs it. Then the program is broken on purpose, and the group says what will go wrong before it does.
3. Minutes 20 to 45: the learner types. Each learner writes their own version in their own editor and shares the screen when stuck. The mentor reads the code line by line with them, asks what they expected and lets them find the mistake.
4. Minutes 45 to 55: show and explain. Two or three learners show a working program and say in their own words why it works. Explaining is where most of the learning settles.
5. Minutes 55 to 60: what comes next. A small task for the days before the next class, and for younger learners a short note to parents on what was built.

The mentors work from India over a live video link, so the slots further down land in Herne afternoons and at weekends. In a group the same mentor stays with the batch; one-to-one learners keep the same mentor too, so the person reading your code next month already knows how you think.

This table is ours: it describes how our Python courses are ordered, not any school curriculum. The pit-shaft project below is the point where all three routes meet: a child can count shafts on paper, a teenager can write Version 1, and an adult can write Version 2 and explain why it is faster.

import csv

shafts = []
with open("herne_shafts.csv", encoding="utf-8") as f:
    for row in csv.DictReader(f):
        if row["opened"] and row["closed"]:
            shafts.append((row["name"], int(row["opened"]), int(row["closed"])))

print("Shafts with both years:", len(shafts))

working = {}
for year in range(1850, 2001):
    working[year] = sum(1 for _, a, b in shafts if a <= year <= b)

peak = max(working.values())
peak_years = [y for y in working if working[y] == peak]
print("Most shafts at work at once:", peak)
print("In the years:", peak_years[0], "to", peak_years[-1])

The peak is 1930 and 1931, with 75 of the 79 shafts working, not the 1950s. By 1950 the count had already fallen to 68. Version 1 is correct but wasteful: it makes 151 passes over the list, one per year, and most years nothing changes. A professional would ask a different question: when does the count change at all?

The count changes only when a shaft opens or closes. So Version 2 turns each shaft into two events: plus one in the year it opened, and minus one in the year after it closed, because a shaft that closed in 1967 was still working during 1967. Sort the events, walk through them once with a running total, and keep the highest value seen.

import csv

events = []
with open("herne_shafts.csv", encoding="utf-8") as f:
    for row in csv.DictReader(f):
        if row["opened"] and row["closed"]:
            events.append((int(row["opened"]), +1))
            events.append((int(row["closed"]) + 1, -1))

events.sort()
running = 0
peak = 0
for year, change in events:
    running += change
    if running > peak:
        peak, peak_year = running, year

print("Events to sort:", len(events))
print("Peak:", peak, "first reached in", peak_year)
print("Still working after the last event:", running)

The answers agree, which is the first thing to check whenever you replace a slow program with a clever one. The hidden detail is the sort. Python sorts the pairs by year first and then by the second number, so in any year a closing event, minus one, comes before an opening event, plus one. Had the openings come first, the running total would briefly count a shaft that had already stopped, and in a different dataset the peak could be one too high. Deciding how ties sort is part of the program, not a detail.

Data: Stadt Herne, Historische Zechenstandorte, Datenlizenz Deutschland Zero 2.0, downloaded 3 November 2026. Counts are our own, from the 79 sites with both years, counting a shaft as working in its first and last year. The single busiest closing year in the list is 1967, with 16 shafts; the last one, Ewald 2, is listed as working until 2000.

1. Ages 9 to 11. Take ten shafts from the table on paper, draw each as a bar on a timeline, and count the bars over 1930. Then over 1970.
2. Ages 12 to 15. Run Version 1, then change it to print the year with the most closures. Predict the answer from the source note before you run it.
3. Ages 16 and up. Swap the minus one and plus one so openings sort first, and find out whether the peak changes on this data. Then build a tiny test file where it does.
4. Adults. Time both versions on a list of 80,000 made-up shafts and explain the difference in terms of how many steps each one takes.

Source: bwinf.de pages for the Biber, the Jugendwettbewerb (2026 edition) and the Bundeswettbewerb, read on 2 November 2026. In round 1 of the Bundeswettbewerb, group work is allowed and welcome; round 2, from early 2027, must be done alone.

We never use or copy the organisers' tasks. In class we write our own problems in the same spirit, like the shaft sweep above, and a learner who wants to enter is pointed to the organisers' free archive of past rounds. For the Biber, ask the class teacher whether the school is registered.

German law (the Einheitenzeitgesetz) sets MEZ at UTC+1 for winter and MESZ at UTC+2 for summer. The clocks went back on 25 October 2026 and go forward again on 28 March 2027. Indian clocks stay at UTC+5:30 all year: a gap of four and a half hours to Herne in winter, three and a half in summer. The rows are examples only; no batch time is fixed here.

When you book, the times you see are already in your own clock. Adults in Herne often prefer an early evening or a weekend morning; school pupils usually take a slot after lessons end.