Tootfinder

heise | "Donkey Kong Bananza" angespielt: Affengeil oder total Banane?
Nintendo bringt endlich einen neuen Plattformer heraus – aber ohne Mario. Wir haben ausprobiert, ob "Donkey Kong Bananza" ein Kaufgrund für die Switch 2 ist.

Trump slammed as the
‘slowest learner in presidential history'
amid 'humiliating failures'
https://www.rawstory.com/2671938487/?cx_te

Trump slammed as ‘slowest learner in presidential history' amid 'humiliating failures'
MSNBC analyst Steve Benen pulled no punches Monday in a stinging op-ed that dissected Donald Trump’s latest presidential frustrations, calling the MAGA leader “gobsmacked” by the realities of foreign policy.“I’m surprised by the degree to which the president is surprised,” Benen, a producer for "The...

JD Vance on Republican megabill: ICE funding matters, 'everything else' is 'immaterial' (Steve Benen/MSNBC)
https://www.msnbc.com/rachel-maddow-show/maddowblog/jd-vance-republican-megabill-ice-funding-matters-everything-else-immat-rcna216181
http://www.memeorandum.com/250701/p38#a250701p38

Relative discrepancy of hypergraphs
Diep Luong-Le, Tuan Tran, Dilong Yang
https://arxiv.org/abs/2506.23264 https://arxiv.org/pdf/2506…

Relative discrepancy of hypergraphs
Given $k$-uniform hypergraphs $G$ and $H$ on $n$ vertices with densities $p$ and $q$, their relative discrepancy is defined as $\hbox{disc}(G,H)=\max\big||E(G')\cap E(H')|-pq\binom{n}{k}\big|$, where the maximum ranges over all pairs $G',H'$ with $G'\cong G$, $H'\cong H$, and $V(G')=V(H')$. Let $\hbox{bs}(k)$ denote the smallest integer $m \ge 2$ such that any collection of $m$ $k$-uniform hypergraphs on $n$ vertices with moderate densities contains a pair $G,H$ for which $\…

Penger styrer makta, uansett.
Å sette en grense på 5000,- per person eller organisasjon og kreve full åpenhet er bare plaster på såret i et system som er korrupt fra bunnen av.
Partipolitikk handler om å holde folket nede, ikke gi dem reell makt. En felles plattform for politikk? Bare enda en måte å kontrollere og temme opprŸr på.
La oss rive hele greia ned i stedet!

#MastodonPoll: Who gave Elon Musk his black eye?
1. "little X" (as Elon claimed)
2. Stephen Miller (as is popularly assumed)
3. Scott Bessent (as Bannon claimed)
4. Elon himself
"little X"
Stephen Miller
Scott Bessent
Elon himself

NIMBY war mir bekannt, aber
BANANA (an acronym for "build absolutely nothing anywhere near anything -or anyone") und CAVE people (citizens against virtually everything) kannte ich noch nicht.
https://en.wikipedia.org/wiki/NIMBY#BANANA_and_CAVE

NIMBY - Wikipedia

"A new aviation solidarity coalition on premium flyers (first- and business- class tickets, and private jets) has been launched today at FFD4 by France, Kenya, Barbados, Spain, Somalia, Benin, Sierra Leone and Antigua & Barbuda. It will be supported by the European Commission"
Eight countries launch solidarity coalition for levies on premium flyers | Global Solidarity Levies Task Force

Eight countries launch solidarity coalition for levies on premium flyers | Global Solidarity Levies Task Force
The Global Solidarity Levies Task Force brings together countries from around the world to advance options for international levelies that will raise much needed revenue for the fight against climate change, and to support development and nature.

Elon Musk vs. Steve Bannon War Explodes With Prison and Deportation Threats (Tom Sanders/The Daily Beast)
https://www.thedailybeast.com/elon-musk-vs-steve-bannon-war-explodes-with-prison-and-deportation-threats/
http://www.memeorandum.com/250701/p162#a250701p162

Halfway to induced saturation for even cycles
Xinyue Fan, Sahab Hajebi, Sepehr Hajebi, Sophie Spiral
https://arxiv.org/abs/2505.24100 https://

Halfway to induced saturation for even cycles
For graphs $G$ and $H$, we say that $G$ is $H$-free if no induced subgraph of $G$ is isomorphic to $H$, and that $G$ is $H$-induced saturated if $G$ is $H$-free but removing or adding any edge in $G$ creates an induced copy of $H$. A full characterization of graphs $H$ for which $H$-induced saturated graphs exist remains elusive. Even the case where $H$ is a path -- now settled by the collective results of Martin and Smith, Bonamy et al., and Dvoŕǎk -- was already quite challenging. What if…