0-fiber

📖 Block, squash, and lift formal definitions

Let $n \geq 1$ and $m \geq 1$ .

A block of dimension $n$ is a tuple

B = (P, s, b)

where:

P = {(p_{0}, \dots, p_{n - 1}) \in N^{n} | p_{i} = b_{i} + x_{i}, 0 \leq x_{i} < s_{i} \forall i}

We denote the set of all blocks of dimension $n$ by $B_{n}$ .

The squash operation is a function

squash : B_{n} \to B_{1}

Given

B = (P, s, b) \in B_{n}

with $s = (s_{0}, \dots, s_{n - 1})$ , $b = (b_{0}, \dots, b_{n - 1})$

s^{'} = (S), where S = \prod_{i = 0}^{n - 1} s_{i}

b_{0}^{'} = \sum_{i = 0}^{n - 1} b_{i} \cdot \prod_{j = i + 1}^{n - 1} s_{j}

k = \sum_{i = 0}^{n - 1} (p_{i} - b_{i}) \cdot \prod_{j = i + 1}^{n - 1} s_{j}

Then

P^{'} = {b_{0}^{'} + k | (p_{0}, \dots, p_{n - 1}) \in P}

Finally,

squash (B) = (P^{'}, (S), (b_{0}^{'}))

The lift operation is a function

lift : B_{1} \times N^{m} \to B_{m}

defined when

\prod_{i = 0}^{m - 1} s_{i}^{'} = S

Given

B = (P, (S), (b_{0})) \in B_{1}

and a new shape

s^{'} = (s_{0}^{'}, \dots, s_{m - 1}^{'}) \in N^{m}

b_{i}^{'} = ⌊ \frac{(b_{0} mod \prod_{j = i}^{m - 1} s_{j}^{'})}{\prod_{j = i + 1}^{m - 1} s_{j}^{'}} ⌋

for each $i = 0, \dots, m - 1$

y_{i} = ⌊ \frac{((k - b_{0}) mod \prod_{j = i}^{m - 1} s_{j}^{'})}{\prod_{j = i + 1}^{m - 1} s_{j}^{'}} ⌋

and then

P^{'} = {(b_{0}^{'}, \dots, b_{m - 1}^{'}) + (y_{0}, \dots, y_{m - 1}) | k \in P}

s^{'} = (s_{0}^{'}, \dots, s_{m - 1}^{'})

Finally,

lift (B, s^{'}) = (P^{'}, s^{'}, (b_{0}^{'}, \dots, b_{m - 1}^{'}))