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Q.Explain the procedure of the Cyclic method of drawing a league fixture for an even number of teams, and state how the method is modified when the number of teams is odd.

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The cyclic method draws a league (round-robin) fixture: fix one team in place and rotate the rest by one position each round.

Even N: rounds = N-1, no team is idle. Odd N: add a dummy "bye" in the fixed position, so rounds = N and one team rests each round.

Total matches always = N(N-1)/2.

Concept understanding: what a league fixture must achieve

In a knock-out tournament a team is eliminated on losing, so only N-1 matches are needed. In a league (round-robin) tournament every team meets every other team once — nobody is eliminated, and the winner is decided on points. The fixture-maker's job is therefore harder: he must schedule all the pairings, in rounds, with no team playing twice in the same round and no pair meeting twice in the tournament.

League tournament, N teams

  • Total number of matches: Matches = (N(N-1))/2 (each team plays the other N-1 teams; dividing by 2 removes the double-count of each pair)
  • Number of rounds: Rounds = N - 1 (N even), Rounds = N (N odd)

Fixtures for a league may be drawn by the staircase, cyclic or tabular method. The cyclic method is the one asked for here.

Part 1 — The cyclic method for an EVEN number of teams

Step 1. Take N teams and number them 1, 2, …, N.

Step 2. Write them in two rows: half the teams left-to-right along the top row, the other half right-to-left along the bottom row. Each vertical column is one match.

Step 3. Fix the first team in its corner. It never moves.

Step 4. For the next round, rotate every other team by one position clockwise around the fixed team. Read off the new columns as the new matches.

Step 5. Repeat until N-1 rounds are complete. Every pair will have met exactly once.

Worked fixture: N = 6 teams

Matches = (6 × 5)/2 = 15, Rounds = 6 - 1 = 5

Round 1 is laid out like this (team 1 fixed at the top-left):

Top row123
Bottom row654

Each column is one Round-1 match: 1 v 6, 2 v 5 and 3 v 4.

Now hold 1 still and rotate 2, 3, 4, 5, 6 one place clockwise each round:

RoundMatch 1Match 2Match 3
11 v 62 v 53 v 4
21 v 56 v 42 v 3
31 v 45 v 36 v 2
41 v 34 v 25 v 6
51 v 23 v 64 v 5

Check: 5 rounds × 3 matches = 15 matches ✓, and all 15 possible pairs (1-2, 1-3, 1-4, 1-5, 1-6, 2-3, 2-4, 2-5, 2-6, 3-4, 3-5, 3-6, 4-5, 4-6, 5-6) appear exactly once ✓.

Note

Because 6 is even, the teams pair off perfectly: all six teams play in every round and no team sits out.

Part 2 — The modification for an ODD number of teams

With an odd N the teams cannot be paired off — one team is always left without an opponent. The cyclic method handles this by a simple device:

  1. Add a dummy team called the "bye" so that the total becomes even (N + 1).
  2. Place the bye in the fixed position — the position that never rotates.
  3. Draw the fixture exactly as for an even number.
  4. In each round, whichever real team is drawn against the bye does not play — it rests for that round.
  5. Because each of the N teams must take one turn opposite the bye, the fixture now needs N rounds instead of N - 1.
  6. The number of matches is unchanged at N(N-1)/2 — pairings with the dummy are not matches.

Worked fixture: N = 5 teams (odd)

Matches = (5 × 4)/2 = 10, Rounds = 5 (= N)

Add the dummy "Bye" to make 6 places, and put the Bye in the fixed corner:

Top rowBye12
Bottom row543

Each column is one pairing: the team opposite the Bye (Team 5) rests, while 1 v 4 and 2 v 3 are played.

RoundTeam resting (drawn v Bye)Match 1Match 2
1Team 51 v 42 v 3
2Team 45 v 31 v 2
3Team 34 v 25 v 1
4Team 23 v 14 v 5
5Team 12 v 53 v 4

Check: 5 rounds × 2 matches = 10 matches ✓, all 10 pairs appear once ✓, and each of the five teams rests exactly once ✓ — which is what makes the method fair.

Comparison at a glance

| | N even | N odd |

|---|---|---| …

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