Games, Tournaments and Networks

1Round-robin match count

In a single round-robin with n teams, every unordered pair meets exactly once, so the number of matches is \dfrac{n(n-1)}{2}. Equivalently, each team plays n-1 opponents and dividing the team–match incidences by 2 removes the double count. Divide by 2 because each pair of teams meets only once, not twice — skipping that half doubles the schedule and breaks every points check that follows.

Imagine n dots with one undirected edge for every pair — that edge set is the match list. The teaching move is to count those edges as n(n-1)/2, not to redraw the clique.

How it works

1. Name the pairsA single round-robin is one match per unordered pair of distinct teams.
2. Count incidencesEach of the n teams plays n-1 others, giving n(n-1) team–match slots.
3. Halve onceEach match fills two slots, so matches = n(n-1)/2.

2Knockout needs n-1 matches

A single-elimination tournament with n players needs exactly n-1 matches to produce one champion. Each match eliminates exactly one player, and crowning a winner means eliminating everyone else. In any single-elimination event, matches equal players minus one, regardless of the bracket shape — byes and bye rounds change when matches happen, not how many eliminations are required.

Picture a bracket that funnels n names down to one champion: every match drops one name from the board. Counting those drops gives n-1 matches before any round labels are drawn.

How it works

1. Count survivors wantedOne champion must remain; n-1 players must leave.
2. One exit per matchEach match removes exactly one player from contention.
3. Match totalTherefore matches = n-1, independent of how the bracket is drawn.

3Round-robin is not knockout

The standing trap is to feed a knockout stem the round-robin formula, or the reverse. All-play-all uses \dfrac{n(n-1)}{2}; single elimination uses n-1. Confusing round-robin match count with knockout count breaks the whole schedule — every later points or wins claim is then checked against the wrong total.

No spatial figure: write "every pair meets" or "a loss eliminates" beside the stem, then take the matching row of the Format → match count table.

Order of attack

1. Read the formatDoes every pair meet, or does a loss remove a player?
2. Pick the identityAll-play-all → n(n-1)/2. Single elimination → n-1.
3. Lock totals firstFix match count before assigning points, wins, or remaining fixtures.

4Total points must match the schedule

Once the match count is fixed, translate the scoring rule into a global check. In a pure win/loss system where every match distributes a fixed number of points p (for example win = 2, loss = 0), total points across the table equal that fixed amount times the number of matches. That identity is a consistency filter: if the standings sum to something else, a result or a fixture count is wrong.

Figure. Four teams, single round-robin: 4×3/2 = 6 matches. Each match puts 2 points into the table, so the required total is 2×6 = 12. Listed points A 6, B 4, C 2, D 0 sum to 12 — consistent with the schedule.

How it works

1. Fix matchesFrom the format, compute how many matches the league actually plays.
2. Points per matchUnder a fixed win/loss rule, each match puts the same p points into the table.
3. Cross-check the sumSum every team's points and compare with p \times matches before solving finer questions.

5Results into standings equations

Assign variables for wins, draws and losses (or for games still to play) and turn each standing line into an equation. With a 3–1–0 rule, a decisive match puts 3 points into the table and a draw puts 2, so the global sum is not a single constant times matches unless you already know how many draws occurred. Convert the points table into equations and use the total-points check for consistency after the draw count is known.

Figure. A beat B, B drew with C, A beat C. Scoring is win = 3, draw = 1, loss = 0. A has two wins (6). B has one loss and one draw (1). C has one draw and one loss (1). Decisive matches put 3+3 into the table and the draw puts 2, total 8, matching 6+1+1.

Order of attack

1. Score ruleWrite win / draw / loss points explicitly before inventing symbols.
2. One team, one equationFor each team, points = 3W + D (under 3–1–0) with W + D + L = matches played.
3. Close the booksSum points and compare with 3 \times (decisive matches) + 2 \times (draws).

6Shortest route on a weighted network

Model cities as nodes and routes as edges with travel costs as edge weights. A path's length is the sum of its edge weights; the shortest route between two nodes is the path whose sum is smallest. List candidate paths and sum weights — the smallest total is the shortest route — rather than trusting the path with the fewest hops when weights differ.

Figure. Schematic network for the worked example. Edge labels are costs. The cheapest A→D route is A–C–B–D with total 6, not the two-hop routes that each cost 7.

How it works

1. Draw the graphOne node per place; one edge per allowed road with its weight labelled.
2. Enumerate simple pathsOn a small DILR network, list the distinct routes from start to end that do not revisit a node.
3. Compare sumsAdd weights along each path; the minimum sum is the shortest route.

7Count paths by systematic branching

When the question asks how many routes exist rather than which is cheapest, treat the network as a branching tree of choices from the start node. At each node, list the unused legal next edges, and reject a branch that revisits a node if the stem forbids it. Count paths using systematic branching — depth-first or level-by-level — so every simple path is listed once and none are double-counted.

Order of attack

1. Fix start and endName the origin and destination; decide whether revisits are allowed (usually not).
2. Branch outwardFrom the current node, open one branch per unused outgoing edge.
3. Tally completionsA branch that reaches the destination adds one path; a dead end or revisit dies.