The id answers which record, not which person — 10 accounts, 6 people

# Self-authored for the EML case corpus (no external origin). The index answers
# "is there a record with this id". The question asked was "is this a new
# customer".
#
# Nothing is wrong with the index. Account id is the right key for an account
# table, it is unique, the lookup is correct, and every count computed from it
# is a correct count of accounts.
#
# The two questions come apart exactly when one person holds more than one
# account, which is not an edge case - it is what happens when someone signs up
# again with a work address, or a household shares a card.
#
# Both counts are computed from the same rows. Neither is an estimate.

# [account_id, owner, signup_month]
# Month 1 deliberately contains no returning person. That month is the control:
# where each new account is a new person the two counts coincide, so the
# account count is not simply always inflated.
[["a1", "kai@x", 1], ["a2", "mei@x", 1], ["a4", "ravi@x", 1], ["a3", "kai@work", 2], ["a5", "sol@x", 2], ["a6", "mei@home", 2], ["a7", "tam@x", 2], ["a8", "kai@x2", 2], ["a9", "ravi@alt", 2], ["a10", "wen@x", 2]] => accounts

# The owner strings differ per account. Who is actually the same person is a
# separate fact, and it is stated as data rather than guessed from the string.
[["kai@x", "kai"], ["kai@work", "kai"], ["kai@x2", "kai"], ["mei@x", "mei"], ["mei@home", "mei"], ["ravi@x", "ravi"], ["ravi@alt", "ravi"], ["sol@x", "sol"], ["tam@x", "tam"], ["wen@x", "wen"]] => identity

def person_of(owner):
    for p in identity:
        if p[0] == owner:
            return p[1]
    return owner

def accounts_in(month):
    0 => c
    for a in accounts:
        if a[2] == month:
            c + 1 => c
    return c

def people_first_seen_in(month):
    [] => seen
    for a in accounts:
        if a[2] < month:
            person_of(a[1]) => p
            0 => have
            for s in seen:
                if s == p:
                    1 => have
            if have == 0:
                seen + [p] => seen
    0 => fresh
    [] => counted
    for a in accounts:
        if a[2] == month:
            person_of(a[1]) => p
            0 => have
            for s in seen:
                if s == p:
                    1 => have
            for s in counted:
                if s == p:
                    1 => have
            if have == 0:
                fresh + 1 => fresh
                counted + [p] => counted
    return fresh

"rows" ^0
"  accounts : " + str(len(accounts)) ^0
[] => people
for a in accounts:
    person_of(a[1]) => p
    0 => have
    for s in people:
        if s == p:
            1 => have
    if have == 0:
        people + [p] => people
"  distinct people : " + str(len(people)) ^0
"" ^0

"new this month, two ways" ^0
for m in [1:2]:
    "  month " + str(m) + " : " + str(accounts_in(m)) + " accounts, " + str(people_first_seen_in(m)) + " people" ^0
"" ^0

accounts_in(1) + accounts_in(2) => acc_total
people_first_seen_in(1) + people_first_seen_in(2) => ppl_total
"  totals : " + str(acc_total) + " accounts, " + str(ppl_total) + " people" ^0
"  the account number is larger by : " + str(acc_total - ppl_total) ^0
"" ^0

# ---- who is responsible for the gap ----

"people holding more than one account" ^0
0 => multi
for p in people:
    0 => c
    for a in accounts:
        if person_of(a[1]) == p:
            c + 1 => c
    if c > 1:
        multi + 1 => multi
        "  " + p + " : " + str(c) + " accounts" ^0
"  people with more than one : " + str(multi) ^0
"" ^0

# ---- the control: a month where the two questions coincide ----
#
# Month 1 has no returning person, so both counts agree there. Without this the
# reader cannot tell whether the account count is simply always inflated.

if accounts_in(1) == people_first_seen_in(1):
    "In month 1 the two counts agree - every new account was a new person." ^0
    "In month 2 they do not: " + str(accounts_in(2)) + " accounts, " + str(people_first_seen_in(2)) + " people." ^0
"" ^0

"The index is correct and the count is correct. Both answer a question about" ^0
"accounts, and growth is usually asked about people." ^0

accounts = [["a1", "kai@x", 1], ["a2", "mei@x", 1], ["a4", "ravi@x", 1], ["a3", "kai@work", 2], ["a5", "sol@x", 2], ["a6", "mei@home", 2], ["a7", "tam@x", 2], ["a8", "kai@x2", 2], ["a9", "ravi@alt", 2], ["a10", "wen@x", 2]]
identity = [["kai@x", "kai"], ["kai@work", "kai"], ["kai@x2", "kai"], ["mei@x", "mei"], ["mei@home", "mei"], ["ravi@x", "ravi"], ["ravi@alt", "ravi"], ["sol@x", "sol"], ["tam@x", "tam"], ["wen@x", "wen"]]

def person_of(owner):
    for p in identity:
        if p[0] == owner:
            return p[1]
    return owner

def accounts_in(month):
    c = 0
    for a in accounts:
        if a[2] == month:
            c = c + 1
    return c

def people_first_seen_in(month):
    seen = []
    for a in accounts:
        if a[2] < month:
            p = person_of(a[1])
            have = 0
            for s in seen:
                if s == p:
                    have = 1
            if have == 0:
                seen = seen + [p]
    fresh = 0
    counted = []
    for a in accounts:
        if a[2] == month:
            p = person_of(a[1])
            have = 0
            for s in seen:
                if s == p:
                    have = 1
            for s in counted:
                if s == p:
                    have = 1
            if have == 0:
                fresh = fresh + 1
                counted = counted + [p]
    return fresh

print("rows")
print("  accounts : " + str(len(accounts)))
people = []
for a in accounts:
    p = person_of(a[1])
    have = 0
    for s in people:
        if s == p:
            have = 1
    if have == 0:
        people = people + [p]
print("  distinct people : " + str(len(people)))
print("")
print("new this month, two ways")
for m in range(1, 3):
    print("  month " + str(m) + " : " + str(accounts_in(m)) + " accounts, " + str(people_first_seen_in(m)) + " people")
print("")
acc_total = accounts_in(1) + accounts_in(2)
ppl_total = people_first_seen_in(1) + people_first_seen_in(2)
print("  totals : " + str(acc_total) + " accounts, " + str(ppl_total) + " people")
print("  the account number is larger by : " + str(acc_total - ppl_total))
print("")
print("people holding more than one account")
multi = 0
for p in people:
    c = 0
    for a in accounts:
        if person_of(a[1]) == p:
            c = c + 1
    if c > 1:
        multi = multi + 1
        print("  " + p + " : " + str(c) + " accounts")
print("  people with more than one : " + str(multi))
print("")
if accounts_in(1) == people_first_seen_in(1):
    print("In month 1 the two counts agree - every new account was a new person.")
    print("In month 2 they do not: " + str(accounts_in(2)) + " accounts, " + str(people_first_seen_in(2)) + " people.")
print("")
print("The index is correct and the count is correct. Both answer a question about")
print("accounts, and growth is usually asked about people.")

rows
  accounts : 10
  distinct people : 6

new this month, two ways
  month 1 : 3 accounts, 3 people
  month 2 : 7 accounts, 3 people

  totals : 10 accounts, 6 people
  the account number is larger by : 4

people holding more than one account
  kai : 3 accounts
  mei : 2 accounts
  ravi : 2 accounts
  people with more than one : 3

In month 1 the two counts agree - every new account was a new person.
In month 2 they do not: 7 accounts, 3 people.

The index is correct and the count is correct. Both answer a question about
accounts, and growth is usually asked about people.

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