How to Solve Scheduling Puzzles: Orders, Rounds and Clashes

To solve a scheduling or ordering puzzle, turn every clue into a simple comparison between two items, such as before, after, straight after, same round or different round. Then place the items that the comparisons pin down first, usually at the ends of the order or in the rounds with the fewest options, and let each placement squeeze the rest into place.

This guide explains the two main kinds of these puzzles, a seven-step method, a fully worked ordering puzzle, the two-row trick for round schedules and a practice puzzle with home and away rounds.

What is a scheduling or ordering puzzle?

An ordering puzzle gives you a handful of items and asks for the one sequence that fits every clue. The items might be jobs done one after another, books on a shelf or runners crossing a line. Typical clues say that one item comes before another, straight after it, or not first.

A scheduling puzzle adds rounds or time slots. Several things can happen in the same round, but some pairs must never share one. The classic version asks who meets whom in which round when every player must meet every other player exactly once.

Both kinds rest on the same idea. Each item gets a position or a round, and the clues are rules about how those positions compare.

How do round robin schedules work?

A round robin is a format in which every pair of players or teams meets exactly once. With 2n teams, all the pairs can be fitted into 2n − 1 rounds, with n games happening at the same time in every round ^[1].

For a small group the numbers are easy to check by hand. Four players A, B, C and D form six pairs, and two games fit in each round, so three rounds are needed. One answer is A against B with C against D, then A against C with B against D, then A against D with B against C.

Puzzles built on this format hide the pairings and give you clues instead. You might learn that A meets D before meeting C, or that B rests in the round when C is busy. Your job is to rebuild the full table.

What does a home and away table add?

Real leagues add a second layer. Each team is marked home or away for every round, and two teams can only meet in a round where one of them is at home and the other is away ^[1]. A table of these marks is called a home and away pattern set ^[1].

A paper published on 30 April 2009 studied when such a table allows a complete schedule at all ^[1]. It gives a four-team table and a three-day schedule that fits it, then turns to much larger questions ^[1]. Using a computer, the paper confirmed a proposed rule about these tables for six teams ^[1]. It also showed that the natural extension of that rule fails for some partial tournaments, where only certain pairs have to meet ^[1].

For puzzle solvers the lesson is simple and does not need the heavy mathematics. A home and away table is a list of forbidden meetings. Two players who are both home, or both away, in a round cannot meet in that round.

Here is a small count made for this guide, by trying every table for the smallest case. With four players over three rounds, and two home and two away in every round, there are two hundred and sixteen possible tables. Twenty-four of them allow no schedule at all, and every other one allows exactly two.

Why is a schedule really a colouring problem?

Think of each item as a dot, and join two dots with a line whenever those items must not share a round. Now give each round its own colour. A good schedule is then a colouring in which no line joins two dots of the same colour, which mathematicians call a proper colouring ^[2].

A paper published on 22 April 2020 made this link precise ^[2]. In it, a scheduling problem is a logical rule built from comparisons, each saying that one item's slot number is at most another's ^[2]. Strictly before, equal and different can all be written using those comparisons ^[2].

The paper calls a condition edge-like when it is a list of pairs joined by or, saying that the two items of at least one listed pair get different slot numbers ^[2]. It calls a scheduling problem graph-like when it joins edge-like conditions with and ^[2]. It then proved that a scheduling problem is graph-like exactly when it is the problem of properly colouring some generalised graph ^[2].

Here is a tiny clash example in that spirit. Four workshops W, X, Y and Z share some helpers. W clashes with X, Y and Z, and X also clashes with Y. How many time slots are needed?

W, X and Y all clash with each other, so they need three different slots, and two slots can never be enough. Z only clashes with W, so it can join the slot of X or of Y. Three slots are needed, and no fewer will do.

The same paper also shows how hard colouring gets ^[2]. For objects called simplicial complexes, it proved that deciding whether k colours are enough is NP-complete, except when k is 1, or when k is 2 and s is 1, where s + 1 is the size of the groups that must never be all one colour ^[2]. In those exceptional cases the question lies in the class P ^[2].

That is one reason hand puzzles are kept small. A few items can be reasoned through with a pencil, while large timetables are better left to computers.

Which clue types appear, and how do you write them down?

Before you place anything, rewrite each clue in a short form. This table lists the common types.

Writing the short forms in one column next to the puzzle saves you rereading long sentences. It also makes clear at once which items are mentioned most often, and those are usually the best place to start.

What is the step-by-step method?

1. List the items and the positions. Write the positions as a row of empty boxes, or the rounds as columns.
2. Rewrite every clue in short form. Use the table above.
3. Glue the blocks. Any straight-after clue joins two items into one tile. Fewer tiles means fewer orders to think about.
4. Draw the before chains. Write X then an arrow then Y for each before clue. Items with nothing in front of them are candidates for first place, and items with nothing after them are candidates for last place.
5. Work on the ends first. Ask which item can possibly be first and which can be last. The ends usually have the fewest candidates.
6. Fill the middle by elimination. Once the ends are fixed, list the few orders left for the middle and test each against the clues.
7. Check every clue. Read the clues one more time against the finished order or table.

For round puzzles, steps three to five change slightly. Look for the pair with the fewest possible rounds and place it first, because every placement removes both players from that round.

How does the method solve a five-job ordering puzzle?

Here is an original puzzle. Five garden jobs are done one after another on a single morning: weeding, raking, pruning, digging and planting.

1. Raking happens at some point before pruning.
2. Planting happens straight after digging.
3. Weeding happens at some point before digging.
4. Pruning is not the last job.
5. Raking is not the first job.

In what order are the jobs done? Have a go first. A computer check of all one hundred and twenty possible orders confirms that these five clues allow exactly one answer, and that removing any single clue lets in more than one.

Step 1: short forms. Raking before pruning. Digging then planting as a block. Weeding before digging. Pruning is not fifth. Raking is not first.

Step 2: glue the block. Digging and planting now travel together as one tile. That leaves four tiles to arrange: weeding, raking, pruning and the digging-planting tile. The block alone cuts the possible orders from one hundred and twenty down to twenty-four.

Step 3: work on the last place. Which job can be fifth? Weeding cannot, because it must come before digging. Raking cannot, because it must come before pruning. Pruning cannot, by clue four. Digging cannot be fifth, because planting must follow it. Only planting is left, so planting is fifth and digging is fourth.

Step 4: fill the first three places. Weeding, raking and pruning fill places one to three, with raking before pruning. Raking is not first, so raking cannot be ahead of both others. That leaves only weeding first, raking second and pruning third.

Step 5: check every clue. Raking is second and pruning third. Planting follows digging. Weeding is before digging. Pruning is not last. Raking is not first. Every clue holds.

How do you build a round robin by hand?

When a puzzle asks you to build a full round robin, the two-row trick does the work for you. For six players, write A, B and C in a top row and F, E and D in a bottom row, so that F sits under A, E under B and D under C. Each letter plays the letter directly above or below it.

After each round, A stays in the top left corner. The other five letters each move one place clockwise round the rectangle: along the top row to the right, down the right side, along the bottom row to the left, and up into the top row next to A. Then pair the columns again.

A quick search confirms that this table has fifteen different games and that every player appears once in every round. Puzzle setters often start from a table like this and then hide most of it behind clues.

How do scheduling and ordering puzzles compare?

What are the most common mistakes?

• Reading before as straight before. X before Y allows other items in between unless the clue says straight after or next to.
• Forgetting the reverse clue. If X is before Y, then Y is not first and X is not last. Mark both.
• Placing a pair in a round that is already full. In a round puzzle, each player appears only once per round.
• Guessing too early. Exhaust the ends and the most limited pairs before you try a guess.
• Skipping the final check. One misread arrow can give an order that looks right but breaks a clue.

You can practise the ends-first habit on a live puzzle about which tank fills up first.

Can you solve this practice puzzle?

Four players A, B, C and D each meet every other player exactly once over three rounds, with two games in each round. Every round, each player is marked home or away, and a game can only be played between a home player and an away player.

One extra clue: A and B meet in the last round. Who plays whom in each round?

Answer: Round one is A against D and B against C. Round two is A against C and B against D. Round three is A against B and C against D.

Why: B and C are both home in round two and both away in round three, so they can only meet in round one. That leaves A and D for the other game of round one.

A and B can meet in round two or round three, and the clue puts them in round three, with C and D as the other game. Round two then takes the last two pairs, A with C and B with D.

A full search confirms this is the only schedule that fits the table and the clue. Without the clue, the table allows exactly two schedules.

For one more ordering challenge, try this live riddle about a pile of sheets, where you must work out which one was placed second.