A paper collage can teach dependency order without a stopwatch

Define the arrows before arranging the cards

NIST’s Dictionary of Algorithms and Data Structures defines topological sorting as arranging items according to a partial order, where some pairs have no comparison. This original workshop explores that idea using invented paper tasks.

It describes no actual school assignment, local event, measured work time or classroom result. Families can use the fictional problem as a shared reading exercise; no accounts, equipment purchases or personal information are needed.

Imagine making one paper collage. Give six task cards these labels: A chooses a theme; B writes a caption; C draws an emblem; D arranges the page; E writes the emblem credit; F checks the completed sheet.

These names make the exercise memorable, but the stated rules decide the answer. B and C each require A.

D requires both B and C. E requires C.

F requires both D and E. Assume those are the complete rules for this invented problem.

Write a rule that another reader can check

Draw an arrow from a prerequisite to the task that needs it: A to B, A to C, B to D, C to D, C to E, D to F and E to F. An arrow means the card at its tail must appear earlier in the final order.

It does not mean the tasks take equal time, require the same person or happen immediately one after another. Keep that meaning beside the drawing so the picture remains understandable when someone else receives it.

Begin with no cards placed. A is ready because it has no prerequisites.

B and C are not ready yet, even if their names sound easy. Place A first and mark it completed on the paper record.

Now B and C are both ready. Choosing either one is permitted.

The absence of an arrow between them matters: the exercise does not ask us to decide whether writing is more valuable than drawing, or whether one imaginary child should always go first. It asks only whether the prerequisites have been placed.

Follow two different successful routes

Try A, B, C, D, E, F. After A, choose B.

D still cannot follow because C is missing. Place C next.

D and E are now ready, so choose D and then E. Finally place F.

Check the seven arrows individually. Each tail occurs before its head.

That establishes a valid order under these rules without requiring the order to be the only possible answer. The check uses the original constraints, rather than trusting the appearance of a tidy row of cards.

Now try A, C, E, B, D, F. Placing C after A makes E ready even though B is unfinished.

E does not require B, so it may appear third. D must wait until B is placed because D needs both B and C.

F remains last because it needs D and E. This second order illustrates a useful distinction: completing one branch does not require completing every other branch first.

Adding an unstated rule would change the problem instead of solving the one on the page.

Count the alternatives without guessing

A must be first: every other card depends on it directly or through another card. F must be last because its prerequisites cover all the remaining work through their own prerequisites.

We therefore need to arrange only B, C, D and E in the middle. If B is second, C must be third; neither D nor E is ready before C.

After B and C, either D then E or E then D works. That branch contributes two complete orders.

If C is second, either B or E can be third. Choosing B makes both D and E ready, producing two more orders.

Choosing E leaves B as the next available card, followed by D, producing one more. The five middle sequences are B C D E, B C E D, C B D E, C B E D and C E B D.

Put A before each and F after each. Two plus two plus one gives five valid complete orders, rather than the twenty-four unrestricted arrangements of four distinct middle cards.

Find the exact reason an order fails

Consider A, B, D, C, E, F. Its problem is specific: D appears before prerequisite C.

Moving C before D repairs that violation while preserving the other stated rules. In contrast, A, C, B, E, D, F succeeds even though it differs from our first example.

A useful explanation names the broken arrow when rejecting an answer, and checks all arrows before accepting one. A different-looking answer is not automatically an incorrect answer.

Change the model by adding one new arrow, F to C. C must now come before D, D before F, and F before C.

Starting at C and following those arrows returns to C. No linear arrangement can put every card in this loop before the next one: it would require C to precede itself.

Sorting more carefully will not repair contradictory prerequisites. Remove the added rule only if the exercise author agrees that it was mistaken; do not quietly erase a troublesome requirement and present the result as a solution to the unchanged problem.

Separate an order from a timetable

Our five orders establish permitted sequences, not completion times. They contain no durations, staffing limits, drying time or shared-tool rules.

One valid order cannot be called fastest from this information. If two imaginary people work simultaneously, the single row also does not by itself specify their assignments.

Those would be new questions requiring additional assumptions. The small model is useful precisely because its conclusion stays attached to the rules it actually contains.

For a final paper check, ask a partner to mark each prerequisite beside a proposed order before seeing the five answers. Then ask which pair can exchange places in A, B, C, D, E, F without breaking a rule.

Swapping D and E works; swapping C and D does not. Explain the difference using the arrows, then restore the original drawing without the added loop.

The result is a reproducible reasoning exercise, not a promise that real creative work can be reduced to six cards or that a valid sequence guarantees a good collage.