Coding and maths classes in Hamburg

Source: Gesellschaft für Informatik, Informatik-Monitor 2026/27, Hamburg page, read 2 November 2026. The written Informatik Abitur at both levels is developed centrally and is the same across Hamburg; at present no practical computer work is allowed in it, and the permitted aids are a non-programmable, non-graphing calculator and a spelling dictionary.

That last rule matters for how a pupil prepares. A Hamburg Abitur candidate writes and reads programs on paper, without a compiler to point out the mistakes. We train exactly that: predicting what code does before running it, tracing variables by hand, and writing a class or an algorithm cleanly in one go.

The monitor's figures for Hamburg are higher than in most Länder: in 2024/25 about 22 percent of Sekundarstufe II pupils took Informatik and about 2 percent of Abitur candidates had it as an exam subject. In 2025/26 Hamburg had 610 qualified Informatik teachers, about 5 percent of all teachers.

Our mentors are in India and teach the same way for every learner, wherever they are: a mentor who gets to know each pupil, the pupil doing the work on their own keyboard or paper, and every problem argued out before it is solved. We teach pupils to think, not to copy a worked answer or an AI tool's output.

import json

for name in ["strassenbruecken", "fussgaengerbruecken"]:
    with open(name + ".json", encoding="utf-8") as src:
        parts = json.load(src)["features"]
    print(name, "rows:", len(parts))

Version 1 counts parts. A bridge with separate decks, ramps or footways has one record per part. Version 2 collects the internal structure numbers in a set, so each bridge is counted once however many parts it has.

import json

bridges = set()
for name in ["strassenbruecken", "fussgaengerbruecken"]:
    with open(name + ".json", encoding="utf-8") as src:
        parts = json.load(src)["features"]
    numbers = {p["properties"]["internebauwerksnummer"] for p in parts}
    print(name, "bridges:", len(numbers))
    bridges |= {(name, n) for n in numbers}

print("Bridges in both files:", len(bridges))

So the two files hold 1,497 records but 1,344 bridges: 128 road-bridge records and 25 foot-bridge records are extra parts of a bridge already counted. Neither number is "wrong". They answer different questions, and a report that does not say which one it gives has not finished its job.

import json
from statistics import median

with open("strassenbruecken.json", encoding="utf-8") as src:
    parts = json.load(src)["features"]

first_year = {}
for p in parts:
    props = p["properties"]
    year = props.get("baujahr")
    if not year:
        continue
    key = props["internebauwerksnummer"]
    first_year[key] = min(int(year), first_year.get(key, 9999))

years = sorted(first_year.values())
print("Road bridges with a year:", len(years))
print("Oldest:", years[0], "Median:", median(years))

Data: Landesbetrieb Straßen, Brücken und Gewässer, via Transparenzportal Hamburg, Brücken und sonstige Ingenieurbauwerke Hamburg, Datenlizenz Deutschland Namensnennung 2.0; we fetched both files on 3 November 2026 and every count in the table is ours. The register covers the LSBG's area of responsibility, so it is not a count of every bridge in the city. The 1802 record is named Franzosenbrücke.

Version 3 is where the maths joins in. For each bridge it keeps the earliest year among its parts, then takes the median. The median, 1976, is the year that splits the dated road bridges into two halves; the mean, 1969.5 in our own check, sits earlier because 29 of the dated road bridges were built before 1900 and drag it back. Ask a pupil to compute both and explain the gap, and you have a statistics lesson that no textbook had to invent.

1. Without a computer, grades 6 and 7. A club list has a row per pupil per club. Ten pupils, fourteen rows. How do you find how many pupils there are? What would you need on every row?
2. Grade 8 or 9, Python. Run Version 1. Find one bridge that appears more than once and write down which column tells its parts apart. Then write Version 2.
3. Grade 10, maths and Python. Compute the mean year as well as the median in Version 3. Which is earlier, and why? Draw a rough sketch of the distribution.
4. Oberstufe. Thirty-one road-bridge records have no year. Could leaving them out move the median? Argue for the largest and the smallest median they could produce.

Germany keeps MEZ, UTC+1, in winter and MESZ, UTC+2, in summer, as the Einheitenzeitgesetz provides; this winter began on 25 October 2026 and summer time returns on 28 March 2027. India has no clock change and stays at UTC+5:30, four and a half hours ahead of Hamburg now and three and a half from late March.