# py.test-3 --log-cli-level=DEBUG -vv share.py

import copy
import math
import logging
import itertools

#
# Implementation of the Hostea revenue sharing model
# See https://forum.hostea.org/t/decision-revenue-sharing-model/92
#
# The members argument is a list of Hostea members, represented as a list:
# [
#  [
#   ID,      # username in the https://gitea.hostea.org/Hostea/organization repository
#   EXPENSE, # integer, expense elligible for payment
#   INCOME,  # integer, total available income
#  ],
#  ...
# ]
#
# The function returns which share of the pending expenses each member
# is allowed to redeem and, if necessary, how much to invoice other
# peer members to get the rest.
#
# share_income(members) => (shares, payments)
#
# The 'shares' are as follows:
#
#  [
#    [ID, EXPENSE, INCOME],
#    ...
#  }
#
# For instance in the if the 'members' argument is [['isabelle', 200, 10]], the
# shares will be [['isabelle', 10, 10]] where the 200 expense was reduced to a
# share of 10 because there only is 10 income available.
#
# The 'payments' maps member IDs to the amount other peer members must be invoiced:
#
#  {
#    ID: [(ID, integer), (ID, integer), ...],
#    ...
#  }
#
# For instance in the following marie will invoice isable for 200
# and john for 50 while lucie will invoice john for 10:
#
#  {
#    'marie': [('isabelle', 200), ('john', 50)],
#    'lucie': [('john', 10)],
#  }
#
# If a member has both income and expenses, they are not required to invoice
# themselves, it is implicit. For instance, if the 'payable' return value is
# [['isabelle', 10, 10]], the 'payments' return value will be empty and isabelle
# will be expected to reduce both her income and her expenses by 10.
#
def share_income(members):
    payable = payable_expenses(members)
    payments = {
        member[ID]: paying_members
        for member, paying_members in distribute(copy.deepcopy(payable))
    }
    return (payable, payments)


def test_share_income():
    members = [["a", 10, 5], ["b", 450, 25], ["c", 0, 0]]
    payable, payments = share_income(members)
    assert payable == [["a", 10, 5], ["b", 20, 25]]
    assert payments == {"a": [("b", 5)]}


######################################################################
#
# Structure layout and utilities
#
######################################################################

ID = 0
EXPENSE = 1
INCOME = 2
OPTIMUM_TRANSACTIONS = 2


def total(members, position):
    return sum(member[position] for member in members)


def total_expense(members):
    return total(members, EXPENSE)


def total_income(members):
    return total(members, INCOME)


######################################################################
#
# Figure out the share each member is entitled to
#
######################################################################
#
# For a given income, return a list of the members who will not be paid in full
# and how much will remain in their expense balance.
#
# The members list given in argument is a list of pairs of
#
#   ['member', EXPENSE]
#
# Where 'member' is the unique id of the member and EXPENSE is a
# positive integer.
#
# The income argument must be an integer lower or equal to the sum of
# all members EXPENSE.
#
# The returned member list is a subset of the members list given in
# argument and only contains the members that cannot be paid in full
# with the income given in argument. For instance:
#
# unpaid_expenses(3, [['a', 2], ['b', 2]]) == [['b', 1]]
#
# Means that with an income of 3, only member 'a' can be paid in full and
# the remaining expense for member 'b' is 1.
#
def unpaid_expenses(income, members):
    if income <= 0:
        #
        # Recursion termination: no more income to share, none of the
        # remaining members get anything
        #
        return members
    if income <= len(members):
        #
        # The income is an integer number: when it cannot be divided
        # among members (2/3 == 0), some of them get 1 and the others
        # get nothing. If the income is consistently very low some
        # members to never get paid but the amount of money involved
        # would then be so low that no member would care.
        #
        share = 1
        count = income
    else:
        #
        # The share that each member will get at this stage of the
        # recursion is the lowest expense. If there is not enough
        # income to give each member the lowest expense, it is divided
        # evenly.
        #
        share = min(
            int(income / len(members)), min(members, key=lambda m: m[EXPENSE])[EXPENSE]
        )
        count = len(members)
    #
    # remaining is the list of members that cannot be paid in full
    # with the share.
    #
    remaining = []
    for i in range(count):
        m = members[i]
        m[EXPENSE] -= share
        if m[EXPENSE] > 0:
            #
            # If the member needs more to be paid in full, they are
            # elligible to participate in the next recursion round.
            #
            remaining.append(m)
    #
    # This is the border case when income <= len(members): some
    # members cannot get their share because there is not enough
    # income to provide an equal share of 1 to everyone and were
    # excluded from the loop above.  They are added because they could
    # not be paid in full.
    #
    remaining.extend(members[count:])
    #
    # Now that each member got an equal share (except for the border
    # case above) recurse, but only with the members that expect to be
    # paid more.
    #
    return unpaid_expenses(income - share * count, remaining)


def test_unpaid_expenses():
    assert unpaid_expenses(1, [["a", 1], ["b", 1]]) == [["b", 1]]
    assert unpaid_expenses(5, [["a", 10], ["b", 2]]) == [["a", 7]]
    assert unpaid_expenses(5, [["a", 2], ["b", 10], ["c", 1]]) == [["b", 8]]
    assert unpaid_expenses(5, [["a", 2], ["b", 10], ["c", 1], ["d", 40]]) == [
        ["b", 9],
        ["d", 39],
    ]
    assert unpaid_expenses(5, [["a", 2], ["b", 10], ["c", 1], ["d", 40], ["e", 3]]) == [
        ["a", 1],
        ["b", 9],
        ["d", 39],
        ["e", 2],
    ]
    assert unpaid_expenses(
        5, [["a", 2], ["b", 10], ["c", 1], ["d", 40], ["e", 3], ["f", 1]]
    ) == [["a", 1], ["b", 9], ["d", 39], ["e", 2], ["f", 1]]


def payable_expenses(members):
    members = copy.deepcopy(members)
    id2members = {m[ID]: m for m in members}
    for unpaid in unpaid_expenses(
        min(total_expense(members), total_income(members)), copy.deepcopy(members)
    ):
        id2members[unpaid[ID]][EXPENSE] -= unpaid[EXPENSE]
    return [m for m in id2members.values() if m[EXPENSE] > 0 or m[INCOME] > 0]


def test_payable_expenses():
    members = [["a", 1, 0], ["b", 1, 1], ["c", 0, 0]]
    expected = [["a", 1, 0], ["b", 0, 1]]
    assert payable_expenses(members) == expected


######################################################################
#
# Distribute the income from senders to receivers, nothing tricky here
#
######################################################################


def distribute_to_receivers(receivers, senders):
    result = []
    for receiver in receivers:
        paying_members, senders = get_paid(receiver, senders)
        result.append((receiver, paying_members))
    return result


def test_distribute_to_receivers():
    receivers = [["a", 5, 0], ["b", 10, 0]]
    senders = [["c", 0, 12], ["d", 0, 3]]
    assert distribute_to_receivers(receivers, senders) == [
        (["a", 0, 0], [("c", 5)]),
        (["b", 0, 0], [("c", 7), ("d", 3)]),
    ]

    receivers = [["b", 12, 0], ["a", 5, 0]]
    senders = [["c", 0, 10], ["d", 0, 5], ["e", 0, 1], ["f", 0, 1]]
    expected = [
        (["b", 0, 0], [("c", 10), ("d", 2)]),
        (["a", 0, 0], [("d", 3), ("e", 1), ("f", 1)]),
    ]
    assert distribute_to_receivers(receivers, senders) == expected

    receivers = [["a", 5, 0], ["b", 12, 0]]
    senders = [["c", 0, 10], ["d", 0, 5], ["e", 0, 1], ["f", 0, 1]]
    expected = [
        (["a", 0, 0], [("c", 5)]),
        (["b", 0, 0], [("c", 5), ("d", 5), ("e", 1), ("f", 1)]),
    ]
    assert distribute_to_receivers(receivers, senders) == expected


def get_paid(who, members):
    paying_members = []
    result = []
    for member in sorted(members, key=lambda m: m[INCOME], reverse=True):
        pay = min(member[INCOME], who[EXPENSE])
        if pay > 0:
            paying_members.append((member[ID], pay))
            member[INCOME] -= pay
        who[EXPENSE] -= pay
        result.append(member)
    assert who[EXPENSE] == 0
    return (paying_members, result)


def test_get_paid():
    orig = [["a", 10, 0], ["c", 0, 6], ["b", 0, 10]]
    paying_members, modified = get_paid(orig[0], orig)
    assert paying_members == [("b", 10)]
    assert modified == [["b", 0, 0], ["c", 0, 6], ["a", 0, 0]]

    orig = [["a", 10, 0], ["c", 0, 6], ["b", 0, 4]]
    paying_members, modified = get_paid(orig[0], orig)
    assert paying_members == [("c", 6), ("b", 4)]
    assert modified == [["c", 0, 0], ["b", 0, 0], ["a", 0, 0]]

    orig = [["a", 10, 0], ["c", 0, 8], ["b", 0, 4], ["d", 2, 0]]
    paying_members, modified = get_paid(orig[0], orig)
    assert paying_members == [("c", 8), ("b", 2)]
    assert modified == [["c", 0, 0], ["b", 0, 2], ["a", 0, 0], ["d", 2, 0]]
    paying_members, modified = get_paid(orig[3], orig)
    assert paying_members == [("b", 2)]
    assert modified == [["b", 0, 0], ["a", 0, 0], ["c", 0, 0], ["d", 0, 0]]


def self_get_paid(members):
    for member in members:
        pay = min(member[INCOME], member[EXPENSE])
        member[INCOME] -= pay
        member[EXPENSE] -= pay
    return [m for m in members if m[EXPENSE] > 0 or m[INCOME] > 0]


def test_self_get_paid():
    members = [["a", 5, 10], ["b", 20, 10], ["c", 0, 30], ["d", 40, 0]]
    expected = [["a", 0, 5], ["b", 10, 0], ["c", 0, 30], ["d", 40, 0]]
    assert self_get_paid(members) == expected


######################################################################
#
# For each receiver (i.e. a member that will have expenses paid), set
# the number of transactions they would need in the ideal situation
# where they would get paid by the senders who have the most income
# before anyone else.
#
# It will be used to stop iterating as soon as an optimal situation
# is detected instead of exploring all permutations.
#
######################################################################


def set_optimum_transactions(receivers, senders):
    result = []
    for receiver in receivers:
        expense = receiver[EXPENSE]
        optimum = 0
        for sender in sorted(senders, key=lambda m: m[INCOME], reverse=True):
            optimum += 1
            expense -= sender[INCOME]
            if expense <= 0:
                break
        assert expense <= 0
        receiver[OPTIMUM_TRANSACTIONS] = optimum
        result.append(receiver)
    return result


def test_set_optimum_transactions():
    receivers = [["a", 5, 0], ["b", 12, 0]]
    senders = [["c", 0, 10], ["d", 0, 5]]
    expected = [["a", 5, 1], ["b", 12, 2]]
    modified = set_optimum_transactions(receivers, senders)
    assert modified == expected


def is_optimal(result):
    for receiver, paying_members in result:
        if receiver[OPTIMUM_TRANSACTIONS] > len(paying_members):
            return False
    return True


######################################################################
#
# Computes which members must be invoiced to pay the expenses to other
# members, with a minimum number of transactions. The algorithm is based
# on the following assumptions:
#
# * The total of all expenses is not greater than the total of all
#   incomes.
#
# * The members with are sorted with greater income first and
#   permutations where a better solution could be found with a
#   different ordering are ignored.
#
# * Iterating over the permutations of the list of members that are
#   allowed to get a share will produce a solution that is good enough.
#
# * An optimal solution (see is_optimal) will be found before
#   iterating over a significant part of the permutations when there
#   is a large (>10) number of members.
#
# The members argument is a list of Hostea members, represented as a list:
# [
#  [
#   ID,      # username in the https://gitea.hostea.org/Hostea/organization repository
#   EXPENSE, # integer, expense elligible for payment
#   INCOME,  # integer, total available income
#  ],
#  ...
# ]
#
#
######################################################################


def distribute(members):
    members = self_get_paid(members)
    senders = [m for m in members if m[INCOME] > 0]
    receivers = [m for m in members if m[EXPENSE] > 0]
    receivers = set_optimum_transactions(receivers, senders)
    logging.debug(
        f"distribute {receivers} to {senders} with at most {math.factorial(len(receivers))} iterations"
    )
    results = []
    iter = 0
    for receivers in itertools.permutations(receivers):
        iter += 1
        result = distribute_to_receivers(receivers, senders)
        #
        # There may be other solutions but since this one is optimal,
        # no need to continue
        #
        if is_optimal(result):
            logging.debug(f"found optimal result after {iter} iterations")
            return result
        results.append(result)
    return best_result(results)


def test_distribute():
    members = [["a", 10, 5], ["b", 20, 25]]
    assert distribute(members) == [(["a", 0, 1], [("b", 5)])]

    members = [["a", 10, 5], ["b", 20, 28], ["c", 3, 0]]
    assert distribute(members) == [(["a", 0, 1], [("b", 5)]), (["c", 0, 1], [("b", 3)])]

    members = [["a", 10, 5], ["b", 20, 280], ["c", 3, 0]]
    assert distribute(members) == [(["a", 0, 1], [("b", 5)]), (["c", 0, 1], [("b", 3)])]

    members = [
        ["a", 10, 5],
        ["b", 20, 280],
        ["c", 3, 0],
        ["d", 3, 0],
        ["e", 3, 0],
        ["f", 3, 0],
        ["g", 3, 0],
        ["h", 3, 0],
        ["i", 3, 0],
        ["j", 3, 0],
        ["k", 3, 0],
    ]
    expected = [
        (["a", 0, 1], [("b", 5)]),
        (["c", 0, 1], [("b", 3)]),
        (["d", 0, 1], [("b", 3)]),
        (["e", 0, 1], [("b", 3)]),
        (["f", 0, 1], [("b", 3)]),
        (["g", 0, 1], [("b", 3)]),
        (["h", 0, 1], [("b", 3)]),
        (["i", 0, 1], [("b", 3)]),
        (["j", 0, 1], [("b", 3)]),
        (["k", 0, 1], [("b", 3)]),
    ]
    assert distribute(members) == expected


#
# The best result is the one with less transactions
#
def best_result(results):
    best = None
    best_transactions_count = None
    for result in results:
        transactions_count = sum(
            [len(transactions) for receiver, transactions in result]
        )
        if (
            best_transactions_count is None
            or transactions_count < best_transactions_count
        ):
            best_transactions_count = transactions_count
            best = result
    return best


def test_best_result():
    result1 = [(["a", 0, 0], [("c", 5)]), (["b", 0, 0], [("c", 7), ("d", 3)])]
    result2 = [(["a", 0, 0], [("c", 5)]), (["b", 0, 0], [("c", 7)])]
    result3 = [(["a", 0, 0], [("c", 5)]), (["b", 0, 0], [("c", 6), ("d", 2), ("e", 2)])]
    results = [
        result1,
        result2,
        result3,
    ]
    assert best_result(results) == result2