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Comments on Is there a way to encode a unique arrangement of vertices of a graph with a unique short word?

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Is there a way to encode a unique arrangement of vertices of a graph with a unique short word?

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I call graphs $G_1$ and $G_2$ distinct iff (i) $G_1$ has a different arrangement1 of vertices than $G_2$ and (ii) $G_1$ and $G_2$ have the same number of vertices. All other properties of $G_1$ and $G_2$ have no effect on distinguishing them. For example, $G_1$ and $G_2$ may be the same even if the number of edges of the two graphs is different.


Example:

$G_1$ $G_2$ $G_3$
graph1 graph2 graph3

Here, $G_1 = G_2$ but $G_1 \neq G_3 \iff G_2 \neq G_3$.


I'm wondering if there is a way to create a unique correspondence between a graph and a single word (any string of letters, possibly gibberish) of the English alphabet. That is to say, distinct graphs $G_1$ and $G_2$ must necessarily map to a different word. It is preferable that the word has 4-7 characters. Additionally, the word may contain a single integer from 1-9 at the end of it. Hence, for example, $G_1$ could map to the word GLBXR7 and $G_2$ could map to the word MMTR.

I'm looking for an algorithm to map every distinct graph with a unique word. You may assume that every graph has the same number of vertices.

A very crude attempt is to label each vertex of a graph with letters from left to right, top to bottom and concatenate them in that order. However, I don’t think this is sufficient for uniqueness.

Clarifications

I attempt to clarify some of the potentially ambiguous elements of my post.

  1. I agree that the notion of a "graph" is not entirely relevant here. I actually chose it because I am working with this sort of information in another project where connectivity matters. Nonetheless, viewing this as a point cloud is perfectly fine; you may edit the tag(s) if you'd like.

  2. There doesn’t need to be a bijection between words and graphs. Every unique graph should map to a unique word, but every word need not map to a graph.

  3. Since I placed a finite restriction on the length of a word (4 - 7 characters), I attempt to set an upper bound on the number of vertices of a graph.

$$\text{Number of words} = \sum_{k = 4}^7 \left(26^{k-1}\cdot (26 + 9)\right) = 11244509640$$ $$\text{Number of graphs with} \le n \; \text{vertices} = \sum_{k = 1}^n \sum_{i = 1}^k {{k - 1} \choose {i - 1}} = 2^n - 1$$ $$11244509640 = 2^n - 1 \implies \lfloor n\rfloor = 33$$

  1. I apologise for not making this point before, but the algorithm should be human computable, that is, possibly something a human can do in their head in about 15 seconds.

1I'm not sure how to perfectly formalise the notion of an arrangement, but a possible necessary and sufficient condition is to consider the number of vertices to the left, right and along the same vertical axis as each vertex of graphs $G_1$ and $G_2$—if any of the values are different, the graphs are distinct.

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Underspecified (2 comments)
Underspecified
Derek Elkins‭ wrote over 2 years ago

This doesn't really make a lot of sense currently. It's also doesn't seem to be about graphs. You don't seem to care about any of the connectivity information, while you do seem to care about geometric information that is usually not considered part of a graph. For example, images $G_2$ and $G_3$ would typically be considered different visualizations of mathematically identical graphs. As described, it seems like you just have point clouds that you want to encode but quantized to some kind of grid or something. You also need to some constraint on the "graphs" if you want to fit it into a fixed number of bits.

Your wording makes it unclear whether you need a bijection or not. That is, whether every "word" gives rise to a "graph". I suspect not, but your wording is a bit ambiguous.

The details of the representation of the "words" is mostly irrelevant. Once you can encode what you want into a computationally compact form, e.g. a short bitstring, encoding that into letters is trivial.

Carefree Explorer‭ wrote over 2 years ago

Thank you for the feedback. I have edited my post with clarifications.