Let us number the tasks by $1,2,3,4$ and the persons by $1,2,3,4,5$.
In the first part we actually calculate the number of functions: $$\left\{ 1,2,3,4\right\} \to \left\{ \left\{ i,j\right\} \mid i,j\in\left\{ 1,2,3,4,5\right\} \text{ and }i\neq j\right\} $$
Denote this set of functions by $F$.
For $i=1,2,3,4,5$ let $F_{i}\subseteq F$
denote the subset of functions corresponding with person $i$ doing
no task.
That means that: $$F_{i}=\left\{ f\in F\mid i\notin f\left(1\right)\cup f\left(2\right)\cup f\left(3\right)\cup f\left(4\right)\right\} $$
Then to be found is: $$\left|\bigcap_{i=1}^{5}F_{i}^{\complement}\right|=\left|F\right|-\left|\bigcup_{i=1}^{5}F_{i}\right|=\binom{5}{2}^{4}-\binom{5}{1}\left|F_{1}\right|+\binom{5}{2}\left|F_{1}\cap F_{2}\right|-\binom{5}{3}\left|F_{1}\cap F_{2}\cap F_{3}\right|$$
Here $F_1\cap F_2\cap F_3\cap F_4$ and $F_1\cap F_2\cap F_3\cap F_4\cap F_5$ are both empty because for a task you need at least $2$ person.
Can you take it from here?
This is not a solution to the current problem.
Using the standard setup, I will show that
- With at most 13 tasks per calander week, over 15 weeks, then there is a series of days with exactly 33 tasks.
- With at most 12 tasks per calander week, over 13 weeks, then there is a series of days with exactly 33 tasks.
In either case, this standard setup doesn't allow us to reduce the number of weeks.
I suspect the following statement is true (and possibly something stronger), as I cannot find a counter example:
3. With at most 13 tasks per calander week, over 13 weeks, then there is a series of days with exactly 33 tasks.
Note: Obviously if we're allowed 14 tasks per week, then we could do 2 a day and never get a series of days with exactly 33 tasks.
Proof of 1:
Let $t_i$ be the number of tasks done on day $i$.
Let $T_i = \sum_{j=1}^i t_j $ be the cumulative number of tasks done by day $i$.
We have $1 \leq T_1 < T_2 < \ldots < T_{105} \leq 195$.
Let our pigeons be $T_i$. There are 105 of them.
Let our pigeonholes be the sets of the form $\{ 66k + i, 66k+i+33 \}$. Since $66 \times 3 = 198 > 195$, there are 99 of them.
So, by PP, there are 2 piegons in 1 hole, which gives us $T_j = T_i + 33$.
Proof of 2.
Set up in a similar manner.
We have $1 \leq T_1 < T_2 < \ldots < T_{91} \leq 156$.
Let our pigeons be $T_i$. There are 91 of them.
Let our pigeonholes be the sets of the form $\{ 66k + i, 66k+i+33 \}$. Since $156 = 2\times 66 + 24$, there are $66 + 24 = 90$ of them.
So, by PP, there are 2 piegons in 1 hole, which gives us $T_j = T_i + 33$.
Thoughts on 3.
There are 91 pigeons and 99 holes.
We have 8 degrees of freedom in choosing the values of $T_i$.
Best Answer
I don't see any way to do this except exhaustive search. The problem is amenable to Donald Knuth's dancing links algorithm, in that it can be viewed as a generalized set exact cover problem.
As explained in the linked paper, (or in Wikipedia) we can represent the problem as a $0$-$1$ matrix in which we must select a subset of the rows, such that each column has exactly one $1$ (the original problem) or at most one $1$ (the generalization.)
Consider a matrix in which each row represents an assignment of two workers to a task. The row $(A, B, 1, \text{Monday})$ means that workers $A$ and $B$ are assigned to task $1$ on Monday. There are $28$ possible pairings of $8$ workers, so we have a priori $28\cdot5\cdot5=1400$ rows.
For each combination of a worker and a task gives a column that must have exactly one $1$, since each worked performs each task once. Each combination of a worker and a day gives another "exactly one" column, since each worker is assigned to exactly one job each day. Each combination of a task and a day gives an "at most" column since each task is performed at most once in a particular day. Finally, each pair gives an "at most column", since no two workers can be paired twice. A priori, we have $118$ columns.
It is easy to cut down both the rows and columns substantially, since we may make the Monday assignments arbitrarily. We may assume that on Monday, workers $1$ and $2$ do job $1$, workers $3$ and $4$ do job $2$, and so on. This allows us to remove all rows involving Monday, all rows where workers $1$ and $2$ are paired, all rows where $1$ or $2$ is assigned to task $1$, and so on. We can also remove all the columns relating to Monday, and the worker-task columns that are obviated by Monday's assignments.
We can make another substantial reduction by deciding in advance which tasks are skipped on which days. Just as we assumed that tasks $1$ though $4$ are done on Monday and task $5$ is skipped, we can assumed that task $4$ is skipped on Tuesday, task $3$ is skipped on Wednesday, and so on. This allows us to eliminate rows that assign tasks on days they are not performed. We can also remove the corresponding task-day columns, and change the ones remaining to exact columns.
That said, I don't know how feasible it is to solve the problem. If there are many solutions, a computer program may find one quickly. If there are no solutions, I have no estimate of how long the program would run before exhausting all possibilities.
If the solution isn't feasible, we can break the problem down further by looking more closely at the worker pairings. Each worker is paired with $5$ others, so there are $2$ he isn't paired with. If we make a graph whose vertices are the workers, and where two vertices are adjacent if the workers never work together, then the graph is either an $8$-cycle, or the disjoint union of two $4$-cycles, or the disjoint union of a $3$-cycle and and $5$-cycle. We can treat each of these cases separately, deciding in each case which workers aren't paired, and eliminating corresponding rows and columns, in addition to the eliminations made above. In this case, all the columns become exact columns.
Still, I don't know if it's feasible. You just have to try.
EDIT
I did an exhaustive search, and it's insoluble, as you suspected. I used the first set of reductions that I mentioned, fixing a Monday schedule, and deciding which tasks would be skipped on which days. I had second thoughts about the reduction I mentioned at the end. It's okay when we have the $8$-cycle, because all the workers are interchangeable, but in the other cases, it's not clear that we can arbitrarily assign a Monday schedule. The python script ran in an eye blink anyway, so further reduction wasn't necessary.
Here is the python script I used for dancing links. I've used it many times, and I'm confident it's correct.
This is the script I wrote to generate input for the solver. You probably ought to check its correctness, at least if this problem is important to you. Note that there is no code to print the solution. I was waiting to see if one existed before writing the pretty printing.