Sunday Times Teaser 3199 – County Cup

by BRG on January 12, 2024

by Edmund Marshall

Published Friday January 12 2024 (link)

In our county football competition, fewer than 100 teams compete in two stages. First, the teams are allocated to more than two equally-sized groups, and in each group there is one match between each pair of teams. The top two teams in each group proceed to the first round of the knockout stage, where a single match between two teams eliminates one of them. If the number of teams entering the knockout stage is not a power of 2, sufficiently many teams are given byes (they don’t have to play in the first round), so that the number of teams in the second round is a power of 2. The knockout stage continues until only one team remains. In one year the competition was played with a single match on every day of the year.

How many teams were in the competition that year?

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BRG permalink


# the number of groups in the first round
for G in range(3, 50):
  # the number of teams (T)
  for T in range(3 * G, 100, G):
    # the teams per group
    tpg = T // G
    # the matches in stage one
    ms1 = G * tpg * (tpg - 1) // 2
    
    # byes in the first round don't change the number of 
    # matches played since each match removes one team
    ms2 = 2 * G - 1
    if (matches := ms1 + ms2) in {365, 366}:
      print(f"{T} teams [{G} groups; matches: stage 1: {ms1}, stage 2: {ms2}]")

# the number of groups in the first round

for G in range(3, 50):

# the number of teams (T)

for T in range(3 * G, 100, G):

# the teams per group

tpg = T // G

# the matches in stage one

ms1 = G * tpg * (tpg - 1) // 2

# byes in the first round don't change the number of

# matches played since each match removes one team

ms2 = 2 * G - 1

if (matches := ms1 + ms2) in {365, 366}:

print(f"{T} teams [{G} groups; matches: stage 1: {ms1}, stage 2: {ms2}]")

Frits permalink

Based on your setup and with more analysis.


print("answer:", ' or '.join(str(tpg * G) for G in range(3, 50) 
                 for tpg in range(2, 99 // G + 1) 
                 if (tpg**2 - tpg + 4) * G in {732, 734}))

print("answer:", ' or '.join(str(tpg * G) for G in range(3, 50)

for tpg in range(2, 99 // G + 1)

if (tpg**2 - tpg + 4) * G in {732, 734}))

BRG permalink


print(''.join([str(u // 2) for g in range(3, 50) for y in (366, 367)
    if (s := g * (8 * y - 15 * g)) and (r := int(s ** .5)) ** 2 == s
                                   and (u := (g + r)) < 200]), "teams")

print(''.join([str(u // 2) for g in range(3, 50) for y in (366, 367)

if (s := g * (8 * y - 15 * g)) and (r := int(s ** .5)) ** 2 == s

and (u := (g + r)) < 200]), "teams")

Frits permalink

Nice, it took me awhile to get to understand the formula.

The code also suggest an upper limit for g as r increases when g increases.

Reply
- BRG permalink
  
  I decided to join the fun!

John Z permalink


# consider group sizes
for gs in range(2, 49):
  # consider number of groups
  for ng in range(2, 99 // gs + 1):

    # number of round robin games:
    rrg = ng * ((gs - 1) * gs  // 2)

    # number of knockout games = 2 * number of rr groups - 1
    kng = 2 * ng -1

    if (rrg + kng) in (365, 366):
      
      print(f'number of teams {gs * ng}: {ng} groups of {gs};',
            f'round robin matches {rrg}, knockout matches {kng}')

# consider group sizes

for gs in range(2, 49):

# consider number of groups

for ng in range(2, 99 // gs + 1):

# number of round robin games:

rrg = ng * ((gs - 1) * gs // 2)

# number of knockout games = 2 * number of rr groups - 1

kng = 2 * ng -1

if (rrg + kng) in (365, 366):

print(f'number of teams {gs * ng}: {ng} groups of {gs};',

f'round robin matches {rrg}, knockout matches {kng}')

Sunday Times Teaser 3199 – County Cup

by Edmund Marshall

Published Friday January 12 2024 (link)

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