# How Voting Systems Work 1, Exercise 3: how fragile is the plurality winner?
# Start from the nine-voter electorate. Move k of the four A-first voters
# so that they rank B first instead (A, B, C becomes B, A, C), and
# report the plurality winner and the Condorcet winner each time.
from collections import Counter
from itertools import combinations

def plurality(profile):
    counts = Counter(b[0] for b in profile)
    top = max(counts.values())
    return "/".join(sorted(c for c, v in counts.items() if v == top)), dict(sorted(counts.items()))

def condorcet(profile):
    names = sorted(profile[0])
    wins = {n: 0 for n in names}
    for x, y in combinations(names, 2):
        vx = sum(1 for b in profile if b.index(x) < b.index(y))
        vy = len(profile) - vx
        if vx > vy:
            wins[x] += 1
        elif vy > vx:
            wins[y] += 1
    found = [n for n in names if wins[n] == len(names) - 1]
    return found[0] if found else "none"

for k in range(0, 5):
    profile = (
        [("A", "B", "C")] * (4 - k)
        + [("B", "A", "C")] * k
        + [("B", "C", "A")] * 3
        + [("C", "B", "A")] * 2
    )
    win, counts = plurality(profile)
    print(f"k={k}: first choices {counts}  plurality: {win:3s}  Condorcet: {condorcet(profile)}")

# Output when run:
# k=0: first choices {'A': 4, 'B': 3, 'C': 2}  plurality: A    Condorcet: B
# k=1: first choices {'A': 3, 'B': 4, 'C': 2}  plurality: B    Condorcet: B
# k=2: first choices {'A': 2, 'B': 5, 'C': 2}  plurality: B    Condorcet: B
# k=3: first choices {'A': 1, 'B': 6, 'C': 2}  plurality: B    Condorcet: B
# k=4: first choices {'B': 7, 'C': 2}  plurality: B    Condorcet: B