Games and tournaments in CAT DILR: matches, points and extreme cases

Tournament sets look like sport and behave like algebra. A league of teams, a table of final points, and questions such as "what was the result of the match between A and B?" or "what is the maximum number of teams that could have qualified?" Almost every one of them opens with the same two steps: count the matches, then count the points those matches can produce.

This guide gives you the counting rules, the total-points trick, a points table solved to a unique result, and the extreme-case reasoning behind CAT's favourite tournament questions. Every answer below was checked by listing all possible results by computer.

Counting matches

The knockout rule has a one-line proof: every match eliminates exactly one player, and all but the champion must be eliminated.

Knockouts with byes

When the number of players is not a power of 2, some players get a bye (a free pass) in round 1. With n players, find the next power of 2 at or above n, call it N. Then:

• Byes in round 1 = N − n.
• Rounds = the power k where N = 2 to the power k.
• Matches are still n − 1.

Example. 20 players. The next power of 2 is 32, so 32 − 20 = 12 byes and 5 rounds. Round 1: the other 8 players play 4 matches, and 4 winners join the 12 who had byes, making 16. Then 8, 4, 2 and 1 matches follow. Total 4 + 8 + 4 + 2 + 1 = 19 = 20 − 1.

The total-points trick

In a league, every match hands out a fixed number of points, and that number depends only on whether it was drawn.

Under win 2 / draw 1, total points are always 2 × matches, so the total tells you nothing new. Under win 3 / draw 1, the total tells you exactly how many matches were drawn. That is why CAT sets usually use 3-1-0.

Worked set 1: a points table with one solution

Four teams, A, B, C and D, play a single round-robin. A win earns 3 points, a draw 1, a loss 0. Final points: A 5, B 4, C 3, D 2.

Step 1: count. 4 teams give 4 × 3 / 2 = 6 matches. Total points = 5 + 4 + 3 + 2 = 14. With 6 matches, 3 × 6 − d = 14, so d = 4 draws and 2 decisive matches.

Step 2: break each score into wins, draws and losses. Each team played 3 matches.

Step 3: use the total of 2 wins. A and B have one win each, so C has none: C drew all three. D's record is two draws and one loss.

Step 4: place the results. C drew with A, B and D. A's other two matches (against B and D) are one win and one draw. B's other two (against A and D) are one win and one loss.

• If B lost to D, D would have a win, but D has none. So B lost to A, and B beat D.
• A beat B, so A's remaining match, against D, was a draw.

A check of all 3 to the power 6 = 729 possible result patterns confirms this is the only one that gives 5, 4, 3, 2.

Q1. How many matches were drawn? 4.

Q2. Which team did B beat? D.

Q3. Which teams finished unbeaten? A and C. A won one and drew two; C drew all three. A solver who looks only at wins misses C.

Worked set 2: extreme cases in a six-team league

Six teams play a single round-robin with no draws. Every match has a winner.

There are 6 × 5 / 2 = 15 matches, so 15 wins in total.

Q1. What is the maximum number of teams that can win 4 or more matches?

• Four such teams would need at least 16 wins, more than the 15 available. So the answer is at most 3.
• Three is achievable. Call the strong teams T1, T2, T3 and the others T4, T5, T6. Let T1 beat T2, T2 beat T3 and T3 beat T1, so each strong team has one win among themselves. Let each strong team beat all three weak teams. Each strong team has 1 + 3 = 4 wins.
• Answer: 3.

Q2. What is the fewest wins the top team can have? 15 wins over 6 teams average 2.5, so someone has at least 3. Three is achievable (three teams on 3 wins, three on 2). Answer: 3.

Q3. The top three teams by wins qualify. How many wins guarantee qualification, whatever the other results and however ties are broken?

• With 3 wins, a team can finish tied with three others on 3 wins (four teams on 3, the other two sharing 3 wins), and a tie-break could knock it out.
• With 4 wins, at most two other teams can also reach 4 or more. If three others did, those four teams would need 16 wins.
• Answer: 4 wins. An exhaustive check of all 2 to the power 15 = 32,768 result patterns agrees: no pattern leaves a 4-win team with three teams at or above it.

Common mistakes

• Using n(n − 1) for a single round-robin. That counts each match twice.
• Forgetting that byes do not change the match count. A knockout always has n − 1 matches.
• Treating 3-1-0 like 2-1-0. Only under 3-1-0 does the points total reveal the number of draws.
• Answering a maximum without building the case. Showing "4 teams is impossible" is half the answer; showing "3 teams is possible" is the other half.
• Reading "unbeaten" as "won every match". Unbeaten includes draws.

Practice set

1. A tournament has 24 teams in four groups of six. Each group plays a single round-robin, and the top two from each group enter a knockout. How many matches are played in all?
2. Five teams play a single round-robin under 3-1-0 scoring. The points total is 26. How many matches were drawn?
3. In worked set 2, what is the largest number of wins the team with the fewest wins can have?

Answers:

1. 67. Each group has 15 matches, so 4 × 15 = 60. Eight teams in a knockout play 8 − 1 = 7. Total 60 + 7 = 67.
2. 4. Five teams play 10 matches. 3 × 10 − d = 26, so d = 4.
3. 2. The average is 2.5, so someone has at most 2. The pattern with three teams on 3 and three on 2 achieves it.

What to do next

• Memorise the match-count table and the byes rule; you should never spend time deriving them in the exam.
• On every points-table set, write "total points → number of draws" as your first line.
• For every maximum or minimum question, write both halves: why one more is impossible, and a case that achieves the answer.
• Continue with maxima and minima in LR, which uses the same bound-then-build logic.