F12.4 Copyright and licences

Write allowed(licence, use) that returns True or False for a proposed use, given LICENCES where each entry is (may change, may sell, must share changes). A use of "change" needs may change; "sell" needs may sell; "use as is" is always allowed. For each (licence, use) in requests, print <licence> / <use>: yes or : no. If the licence is not in LICENCES, print <licence> / <use>: unknown licence. At the end print allowed: <n> of <total>.

# the two lines every program starts with: the commands, then the robot
from bugbot import *
connect()

# licence: (may change, may sell, must share changes)
LICENCES = {"MIT": (True, True, False), "GPL": (True, True, True), "All rights reserved": (False, False, False)}
requests = [
    ("MIT", "change"),
    ("MIT", "sell"),
    ("GPL", "change"),
    ("All rights reserved", "change"),
    ("All rights reserved", "use as is"),
    ("CC BY", "sell"),
]

The hint students can ask for: If the licence is not in the table, say so. Otherwise look up what that licence allows and check it against the use asked for, remembering that using it unchanged is always allowed. Count the ones you allowed.

A solution

from bugbot import *
connect()
LICENCES = {"MIT": (True, True, False), "GPL": (True, True, True), "All rights reserved": (False, False, False)}
requests = [
    ("MIT", "change"),
    ("MIT", "sell"),
    ("GPL", "change"),
    ("All rights reserved", "change"),
    ("All rights reserved", "use as is"),
    ("CC BY", "sell"),
]

def allowed(licence, use):
    change, sell, share = LICENCES[licence]
    if use == "change":
        return change
    if use == "sell":
        return sell
    return True

yes = 0
for licence, use in requests:
    if licence not in LICENCES:
        print(f"{licence} / {use}: unknown licence")
    elif allowed(licence, use):
        yes = yes + 1
        print(f"{licence} / {use}: yes")
    else:
        print(f"{licence} / {use}: no")
print(f"allowed: {yes} of {len(requests)}")

Any program that meets the task's checks is marked correct in the simulator; this is one way, not the only way.