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Proof.md

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Proofs

1: OBOC solution is identical to original connectome.

Let C be our original connectome; consisting of different cells and a directed graph G of their connectivity. Let A be the set of all barcodes, and B the set of all barcode pairs. Each cell in C has a barcode unique to the cell, so use barcodes to denote cells. Denote the barcodes with lowercase letters, k, l, m. Each edge in G has a unique barcode-pairs, and can be identified with one. Denote barcode pairs with capital letters X, Y, Z. (Barcode-pairs are directed)

Let there be some distribution of the barcode pairs on a separate network that is OBOC. Call the new connectome C', with synapses exist iff there is a barcode-pair X in between two cells.

Cells in C' have the same barcodes as C since the barcode-pairs contain only the barcodes of C. The C' is OBOC, and each cell with a connection can be identified with the unique barcode in it. Identify these cells k', l', m', prime denoting that they represent the cells in C' containing barcodes of their letter.

For all pairs of barcodes k and l, there is a barcode-pair X with the label k->l if and only if there is a synapse from cell k to cell l. For all pairs of cells m' and n', there is a connection between them if and only if there is a barcode-pair Y with label m->n. Thus there is a connection from cell k' to l' if and only if there is a synapse from cell k to cell l. Which means, if there is an OBOC solution, that solution has the same connectivity as the original.