Task assignment: check every allowed person pairing

Task assignment checks whether every simultaneous job can receive a different person through a supplied allowed pair. Complete this exercise to save all six candidate assignments, every valid result and a concrete reason an alternative card has no complete solution. Eligibility is stipulated teaching data, not a real employee qualification or safety assessment.

Content checked: 18 October 2026.

What does task assignment mean in this exercise?

Imagine three low-risk internal tasks labelled U1, U2 and U3. People J, K and L are fictional identifiers, not actual staff. Every person is available throughout the same abstract window. No one is contacted, no live task board changes and no work is performed by following this note.

Every task is mandatory and needs exactly one eligible person. Each person can take at most one task in that window; splitting a task or scheduling the tasks one after another is outside the model. No duration, cost, preference, permission or staffing recommendation is supplied. We test feasibility, not which employee is best.

Allowed pairs must come from the source owner. A job title cannot create permission for a pair marked no. Nor can an attractive complete-looking table justify using the same person twice. Preserve the full source before annotating an answer, so a later reader can distinguish a calculation from an edited requirement.

The artificial intelligence guide provides broader starting points. This small core needs only a text editor and the supplied matrix, not an assistant account or solver installation.

Read a complete three-by-three eligibility card

In source V1, U1 allows J or K. U2 also allows J or K. U3 allows only L. The matrix below is complete: yes means the pair is allowed, no means it is not allowed under this exercise. There is no unknown entry in the main card.

Read a row as one task and a column as one person. U1–J and U1–K both pass individually. U3–J does not. Equal-looking candidate lists for U1 and U2 do not merge those two mandatory jobs. They need two different people at the same time, even though their eligible-person sets are identical.

Listing three people for three tasks is only a count check. Finding at least one yes in each row is another useful check, but it is not yet a complete assignment. The same person might be the only candidate for several rows. Distinct-person constraints must be checked across the whole proposed result.

The delegation guide covers skills, workload, authority and review agreements. We do not repeat that selection conversation. This exercise begins with eligibility and availability already supplied and checks whether the entire simultaneous pairing can coexist.

Why do two complete assignments survive all six checks?

For U1 there are three initial person choices; then two unused choices for U2 and one remaining choice for U3. That produces six distinct assignments. With three tasks and three people, covering every task without reusing anyone also uses every person. Other staffing problems may leave people unused.

The first candidate U1→J, U2→K, U3→L has three allowed pairs and passes. Swapping J and K gives U1→K, U2→J, U3→L, which also passes. The jobs receive different people, so these are distinct results.

Every other candidate fails at least one pair. In particular U3 cannot receive J or K, so any row that gives L to U1 or U2 leaves U3 with an ineligible person. Preserve the pair-check sequence in task order U1, U2, U3. An assignment is valid only when all three pair flags are yes and person identities remain unique.

Keep both passing results. No cost, preference or name-based priority tells us to choose one over the other. Selecting the alphabetically first result as the uniquely best allocation would add an unsupported objective. The responsible owner may supply another decision rule later, but that would be a separate input.

How can every task have candidates and still fail?

Create a separate V2 card where U1, U2 and U3 each allow J or K, while L is not eligible for any of them. Availability is unchanged: all three people are present. Every task still has two eligible people, yet no complete distinct-person assignment exists. The matrix is known, not incomplete.

Take the whole task subset U1,U2,U3. Its combined candidate set is only J,K. Three simultaneous tasks need three distinct people, but this subset can draw on only two. That count obstruction is enough to reject a complete assignment. L being present does not make L an allowed candidate.

All six candidate assignments fail on V2 because every one-to-one arrangement uses L somewhere. That task–L pair is marked no. The table keeps those failures visible instead of quietly replacing no with yes or leaving one task unassigned. Passing two of the three tasks is not a complete solution to a three-task contract.

A smaller obstruction can occur too: if two tasks both allow only the same person, those two tasks already need more distinct candidates than they have. You need not call the entire workforce unsuitable. State the exact task subset and its combined eligible-person identifiers, then ask which input the owner can clarify or legitimately change.

The shift coverage check counts available people through time windows. Here the count can look sufficient while eligibility blocks assignment. The resource conflict audit inspects fixed equipment intervals; our result neither shifts task times nor supplies a new equipment order.

Separate a known forbidden pair from an unanswered one

An unanswered eligibility entry is not automatically no, and certainly not yes. Suppose a revised card leaves U3–L unknown. Both previously valid V1 assignments relied on that pair. Their confirmation now needs the missing source answer; copying the old yes into a new version would invent evidence.

A different revised card might retain a complete assignment using only known yes entries while leaving an unused pair unknown. That known witness remains valid under the known permissions. However, the unknown entry could introduce another solution, so do not claim the complete list of all assignments or general impossibility without resolving the relevant gaps.

Duplicate task or person identifiers create another source problem. Confirm identity rather than merging records or treating one person under two codes as two available people. An allowed-pair reference to an unlisted person also requires clarification. The six-row method assumes three distinct tasks and three distinct known people.

The minimum demo set selects overlapping feature coverage; repeated coverage is not repeated use of a person. The slot allocation rule distributes identical position counts among categories. Here every task–person pair has its own source status, with no weights, quotas or minimum-count objective.

If tasks can run at different times, people can take multiple jobs or qualifications must be assessed, this simultaneous allowed-pair contract no longer answers the whole question. Preserve those conditions as owner questions. A successful fictional matching is not hiring advice, a fairness finding or permission to assign real work.

Save both source cards and their different conclusions

Open an available plain-text editor and create a document headed “Task assignment V1/V2: fictional eligibility”. Copy the simultaneous-task contract, both complete matrices and each six-row candidate ledger. Under V1 list both passing assignments. Under V2 record the three-task subset and its two-person candidate union.

Keep the source version beside every conclusion. Add a boundary note saying no real qualifications, safety conditions, authority or employee agreement were verified. The record is an inspectable feasibility check, not a task notification or operating instruction. Keep the failed rows.

Use the editor's save command, choose a folder you recognise and name the file task-assignment-V1-V2.txt. Close the document. Open that folder in your file manager and reopen the saved file. Check that U3–L is yes in V1 but no in V2, both valid V1 mappings remain and V2 still names J,K as the only candidates.

A proposed fifteen-minute allowance gives three minutes to the source, four to the six-row checks, three to the failed-card subset witness, two to conclusions and the owner question, then three to saving and reopening. This is not a measured reader trial. Supporting calculations used two independent enumerations and files were saved and read back through filesystem operations; no graphical interface, real staff assignment or optimisation solver was tested.

• Preserve three distinct task and person IDs.
• Every task appears once and each person at most once.
• Check each assigned pair against its source entry.
• Retain every valid mapping and the failed-card witness.