Jim Nastos

Educator, Algorithms and Data Scientist, Networks and Graph Theory

[Learning on Graphs Conference 2026 @LogConference]

Here are the top 10 reviewers for LoG 2023, as chosen by the community! Congratulations to everyone, and thank you for your contribution to LoG! 🤗

@DobrikG @MoreheadAlex @Francesco_dgv @luoyunan @RishiSonthalia @gbouritsas @arjunsubgraph J. Nastos C. Bravo-Hermdorff, R. Abboud

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@petergyang

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@TheFlatEartherr do you really think that is the whole height of the CNtower?

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@dpernoux @KH2020H0

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@ilarrosac @KH2020H0 Do you also think that 2/2/2 = 1/1/1 ?

[Algorithms MDPI @Algorithms_MDPI]

New #specialissue #ArtificialIntelligence in #NetworkScience

Editors: Dr. Yong Gao from University of British Columbia - Okanagan; Dr. Jim Nastos from Okanagan College

https://t.co/vquY9Ifuzn #mdpialgorithms via MDPI

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@fos_ubco @ubcokanagan @A_Wiebe A new combinatorialist in Kelowna - amazing! Well done Amy, and well done UBCO

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@aimsterlemieux @mrhooker I'm pretty sure my late policy was harsher than that :-/

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@cretiredroy People don't trust math they don't understand. Better to argue by pictures:

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@Margare46581420 @MAGRAHO @cretiredroy Do you honestly think two 5-inch boxes are enough to compensate for a 10-inch box? Here is a picture to help you:

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@neurocyc @eipim1i0 @_Abhishek99 @pickover Or I guess I should rather say that squaring both sides and getting a correct statement does not mean the original statement was correct

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@neurocyc @eipim1i0 @_Abhishek99 @pickover the point is that squaring both sides (left side and right side) and getting equality does not mean the original left side and right side are equal. If that were true, it would mean -1 = 1. Squaring both sides introduces erroneous solutions

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@OSTBear @mezzancholyhill A clown and her dolly

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@DAlperovitch Thank you for your realtime updates and analysis/commentary... can you recommend other reliable twitter accounts/people to follow for more of the same?

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@RickysMaths @Cshearer41 Great picture on a grid: Pick's Theorem now applies! Area of a polygon = p + q/2 - 1. p = number of interior integer points = 7. q = number of integer points on the boundary = 8. So A = p + q/2 - 1 = 7+4-1 = 10.

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@gkalinkat @Cataranea @Maratus_spider @BonteDries @TrineBilde @ForTheSpiders @erincpow this paper
https://t.co/ATcWCo1Dux
counts families and genera found in various habitat areas

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@keithburgoyne @CBCPEI @nicolammacleod It's not a barn funnel weaver. Those would have banded legs. This is a giant house spider (eratigena atrica or duelica)

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@chadkoh ResearchGate - you can request papers directly from authors (if they don't have a version already available on the site)

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Community-finding with quasi-threshold graphs applied to the coopetition scene in the Canadian winery industry: now an open-access paper at EJM: https://t.co/jnUNKAz7a5

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@CBC @CBCKids What's happening is that 2( ) is understood to be a unary operator, and people eventually learn that unary operators bind more tightly than multiplication. BEDMAS and PEMDAS insufficiently handle unary operators. Unary operators have precedence on par with exponentiation