Another popular question of mine. Well here's to self-help, starting with a sketch illustrating the 3 paths in the $N = 4, k =2$ situation:

The number of ways to arrange the 2 test sets to occur in 4 time periods is ${4 \choose 2} = 6$, and $\frac{k}{N} = .5$ is the fraction of the combinations that will start with a test set. Since a "path" is a continuous group of blocks from the first to the last sequential group, there are .5 * 6 = 3 paths, which aligns with $\phi(N, k)$ from the question.

Here's a sketch for a more complicated example with N = 5 and k = 2, which leads to 4 paths:

[Edit 2023-08-17: I answered this question myself after a while with no activity. Nice to see it's picked up, but I forgot a lot. I had to remind myself that this is all one time series, and that the time groups are on the vertical (1, 2, 3, 4). The horizontal (1,...,6) would be the CV index. I don't know what the intuitive explantion for the paths are, except I can see how they are combinatorially exhaustive].

Starting with a sketch illustrating the 3 paths in the $N = 4, k =2$ situation:

The number of ways to arrange the 2 test sets to occur in 4 time periods is ${4 \choose 2} = 6$, and $\frac{k}{N} = .5$ is the fraction of the combinations that will start with a test set. Since a "path" is a continuous group of blocks from the first to the last sequential group, there are .5 * 6 = 3 paths, which aligns with $\phi(N, k)$ from the question.

Here's a sketch for a more complicated example with N = 5 and k = 2, which leads to 4 paths:

Another popular question of mine. Well here's to self-help, starting with a sketch illustrating the 3 paths in the $N = 4, k =2$ situation:

The number of ways to arrange the 2 test sets to occur in 4 time periods is ${4 \choose 2} = 6$, and $\frac{k}{N} = .5$ is the fraction of the combinations that will start with a test set. Since a "path" is a continuous group of blocks from the first to the last sequential group, there are .5 * 6 = 3 paths, which aligns with $\phi(N, k)$ from the question.

Here's a sketch for a more complicated example with N = 5 and k = 2, which leads to 4 paths:

You might not like my question, but at least I have great handwriting.

Another popular question of mine. Well here's to self-help, starting with a sketch illustrating the 3 paths in the $N = 4, k =2$ situation:

The number of ways to arrange the 2 test sets to occur in 4 time periods is ${4 \choose 2} = 6$, and $\frac{k}{N} = .5$ is the fraction of the combinations that will start with a test set. Since a "path" is a continuous group of blocks from the first to the last sequential group, there are .5 * 6 = 3 paths, which aligns with $\phi(N, k)$ from the question.

Here's a sketch for a more complicated example with N = 5 and k = 2, which leads to 4 paths: