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    Friday, February 7th, 2025
    syn_rps
    5:48p
    Diceomancer review

    Roguelike deckbuilders need to do something pretty special to stand out nowadays, what with the Slay The Spileup of bangers over the past few years. Cobalt Core, Wildfrost, Samurai Showdown (if you squint). All excellent, but Diceomancer stands out above even those, thanks to a clever gimmick and a hefty dose of chutzpah. It’s there in the strapline, you know the deal, but to emphasise: you can reroll ANY number on your screen.

    Your health, enemy health, attacks, blocks, buffs, mana, gold - all fair game. The numbers in encounters. On Relics. Have at ‘em! Heck, and that’s before you start scribbling in the rulebook.

    Read more

    syn_rps
    5:32p
    Rocksteady may be going back to single player Batman, hints Bloomberg report

    After years of making a live service Suicide Squad game that was roundly panned on arrival, the studio behind Batman: Arkham Asylum and its sequels are "looking to return to Batman for a single-player game", according to a report by Bloomberg. Rocksteady Studios haven't said so publically, and Bloomberg reporter Jason Schreier claims "the new project is years away from landing", so this is in no way official. But the possibility might come as a ray of hope to fans of the studio's open world Batman games.

    Read more

    syn_tjrefugees 8:18p
    Российских военных фашистов пытались убить FPV-очками со взрывчаткой внутри
    Российских военных фашистов пытались убить FPV-очками со взрывчаткой внутри

    Неизвестные попытались совершить диверсию в отношении военнослужащих ВС РФ, направив по гуманитарной линии партию FPV-очков. При включении устройств произошла бы детонация, сообщил официальный представитель АО «НПП», занимающегося разработкой и поставкой РЭБ-оборудования, Игорь Потапов.

    «Партия FPV-очков Skyzone Cobra X v4 была передана человеком по имени Роман по гуманитарной линии. При включении очков происходила детонация, осуществлялся подрыв. Когда вскрывали партию, обнаружили во всех этих изделиях упакованный пластит», — приводит ТАСС слова специалиста.

    Потапов призвал бойцов быть бдительными при получении гуманитарной помощи от неизвестных лиц. По его словам, она требует проверки, если доставляется от незнакомого человека. Специалист подчеркнул, что людей, занимающихся гуманитарной помощью, много в стране. Но всегда нужно проверять грузы, доставленные от неизвестного «благодетеля», пояснил Потапов.

    ©Текст с мк, но есть уже в куче рашасми

    https://preview.redd.it/ls5sv4cairhe1.jpg?width=960&format=pjpg&auto=webp&s=1d75904bb41fd539a2b8cb9ce492a35587fac2ff

    https://preview.redd.it/uif3l5cairhe1.jpg?width=1280&format=pjpg&auto=webp&s=20d64282c72070637c99e6bfa6cf1274056eece8

    https://preview.redd.it/rfl1o7cairhe1.jpg?width=1280&format=pjpg&auto=webp&s=c27429c6bb802532d3442b809a4bd45e2231a2d4

    submitted by /u/pocoucro
    [link] [comments]
    syn_dtschwllukr
    6:35p
    Україна підтримує ідею Трампа обвалити ціни на нафту - МЗС
    У МЗС України підтримали позицію президента США Дональда Трампа, що низькі ціни на нафту є запорукою миру та стабільності в світі.
    syn_mapporn 8:17p
    Prefixes of nations in South Asia.
    Prefixes of nations in South Asia. submitted by /u/Ok-Goose6242
    [link] [comments]
    syn_zaxid_all 7:11p
    Росіяни знову підтягнули північнокорейських солдатів у Курську область
    [CDATA[Росіяни знову підтягнули північнокорейських солдатів у Курську область]]>
    syn_zaxid_all 6:56p
    У Тернополі аферисти ошукали 20 продавців техніки на 290 тис. грн
    [CDATA[У Тернополі аферисти ошукали 20 продавців техніки на 290 тис. грн]]>
    syn_zaxid_all 6:36p
    Підозрювана у завищені цін фірма зменшила суму договору з реабілітаційним центром «Галичина»
    [CDATA[Підозрювана у завищені цін фірма зменшила суму договору з реабілітаційним центром «Галичина»]]>
    syn_zaxid_all 6:36p
    ВАКС відправив під варту підозрюваних у справі про земельну корупцію в Києві
    [CDATA[ВАКС відправив під варту підозрюваних у справі про земельну корупцію в Києві ]]>
    syn_google_blog 5:00p
    Campaign Manager 360 expands partnershipsCampaign Manager 360 expands partnershipsManaging Director, Data, Measurement and Google Marketing Platform
    Campaign Manager 360 is helping advertisers and agencies navigate the evolving digital landscape with innovative tools and strategic partnerships.Campaign Manager 360 is helping advertisers and agencies navigate the evolving digital landscape with innovative tools and strategic partnerships.
    syn_cityporn 8:16p
    Shibuya, Japan
    Shibuya, Japan submitted by /u/badbuoy
    [link] [comments]
    syn_mathoverflo 8:16p
    Niveau of the Hodge structure of an hypersurface in $\mathbb{P}^n$

    Assume $X:=X_d$ is an hypersurface of degree $d$ in $\mathbb{P}^n$. Assume in addition that $d<n+1$, hence $X$ is a Fano variety.

    The hyperplane section theorem of Lefschetz ensures that the only interesting part of the cohomology is $H^{n-1}(X,\mathbb{Q})$, and as $X$ is Fano, it is well known that $H^{n-1,0}(X)=0$ : the Hodge structure is of coniveau $1$.

    Recall that an hodge structure $H$ is of niveau $\alpha$ if $H^{p,q}=0$ for $\mid p-q \mid > \alpha$.

    Now for very low degree hypersurfaces, the niveau is bigger. I observed the following facts with direct computations :

    • If $d\leq 2$ then $X_d$ is cellular, and hence its cohomology is of niveau 0.
    • For $d=3$ and $n\leq 6$, the cohomology is of niveau $2$.

    Furthermore on all this examples this bound is optimal, except from the case of cubic surfaces, so maybe we should neglect cases with a very small $n$.

    So I was looking for a precise result about the niveau of this Hodge structure. As we have explicit formulas I tried the two following methods :

    With intersection theory

    We can easily compute the Chern classes of $X$ as it is an hypersurface, and so Riemann-Roch formula gives an explicit element $P\in \mathbb{Q}(x,y)$ such that in a formal power extension, the coefficient of $x^py^q$ is exactly the vanishing part of $h^{p,q}$.

    However it seems really annoying to expend this fraction for a general couple $(d,n)$.

    With vanishing of the pole

    Let $S=\mathbb{C}[x_0,\cdots,x_n]$, $f$ the homogenous polynomial defining $X$ and $J_f$ the homogenous ideal generated by the jacobian of $f$. Finally, let $R=S/J_f$. Then a vanishing of the pole argument gives an isomorphism

    $$R^{kd-n-1}\simeq H^{n-k,k-1}_{van}(X)$$

    where $R^m$ is the part of degree $m$, and $van$ denotes the vanishing part of the cohomology.

    Now if I choose $f= \sum_{i=0}^n x_i^d$, then the jacobian is really easy to compute, so we can find the degree where $R^m$ vanishes. More precisely, we have $R^m=0$ exactly when

    $$m>(n+1)(d-2).$$

    So $H^{n-k,k-1}_{van}(X)=0$ for $k>(n+1)(d-1)/d$. It is exactly the niveau bound I was looking for !

    Conclusion of the argument and question

    I have computed the niveau bound for one example of hypersurface. However, from Riemann-Roch theorem and computation of the Chern numbers of hypersurfaces, we know that the Hodge numbers depend only on $(d,n)$, so my bound is true for any hypersurface of degree $d$.

    So I have found the result I was looking for, but I am annoyed because the proof used a vanishing of the pole argument, and so the Bott vanishing theorem. I would like to extend this computation to other kind of hypersurfaces (at least for the vanishing part), but in general this vanishing theorem is wrong.

    Also it is rather disappointing to prove the result only for a precise choice of $f$, and then use the independance of the choice. I would rather prove the result for any choice of $f$, or directly with the computation of Chern numbers.

    Do you have any idea to find a better argument ?

    syn_mathoverflo 8:16p
    Homotopy type of some flag-type spaces

    Is there any place that the of the quotient space $$\frac{O(kr)}{O(k)\wr\Sigma_r}$$ has been computed, at least up to homotopy? Here, $r,k$ are positive integers, $O(k)$ is the orthogonal group of $k\times k$ matrices, and $\Sigma_r$ is the permutation group on $r$ elements. The product is the wreath product. I am particularly interested in the cases with $k=1$. Let's note that $O(k)\wr\Sigma_r$ is obtained from the Cartesian product of sets $O(k)^{\times r}\times\Sigma_r$ by defining a specific product and one has an monomorphism from it into $O(kr)$ which allows to define the quotient. Hence, the I have attributed them as flag-type as in the title.

    I will be grateful if you point at some references.

    syn_rs_ua 8:44p
    Санкції Трампа для МКС. Чи вплине це на Україну?
    Чи є перспективи в ордеру цього суду на арешт президента Росії Володимира Путіна?
    syn_rs_ua 8:39p
    Лубінець відреагував на смерть чоловіка під час проходження ВЛК у Чернівцях
    «Попередньо відомо, що спочатку правоохоронні органи виявили місцевого жителя, який перебував у розшуку за ухилення від мобілізації»
    syn_rs_ua 8:35p
    Зеленський: значення дронів має бути таким, щоб «унеможливлювати російські штурми»
    Президент зазначив, що провів зустріч із командирами підрозділів безпілотних систем: «досвід усіх наших найкращих буде масштабуватися»
    syn_rs_ua 8:31p
    Трамп заявив про можливу зустріч із Зеленським наступного тижня
    На запитання про місце такої зустрічі американський лідер відповів, що «може бути Вашингтон»
    syn_rfrl 9:47p
    Трамп анонсировал встречу с Зеленским и разговор с Путиным
    "Туда я не поеду", — сказал президент США про Украину
    syn_ponomar_bbc
    6:18p
    Вчені виявили, як зварити ідеальне яйце. Але цей метод може вам не сподобатися
    Вчені знайшли ідеальний спосіб зварити яйце - як не дивно, це займає пів години.
    syn_andrusha
    6:40p
    Учитель года - пожизненно без права педерачи.
    «Учитель года» в Калифорнии рыдает, признаваясь в насилии над двумя учениками 11 и 12 лет
    Известная учительница из Калифорнии расплакалась в суде, признавшись в сексуальном насилии над двумя своими учениками в возрасте 11 и 12 лет.
    Жаклин Ма, которая всего за несколько месяцев до ареста была названа «Учителем года» в округе Сан-Диего, грозил ошеломляющи   срок в 180 лет за решеткой за шокирующее насилие в начальной школе Линкольн-Эйкерс в Нэшнл-Сити.
    В среду это решение было пересмотрено, поскольку она призналась в непристойных действиях со своими учениками в рамках сделки о признании вины, котораяснизит срок от 30 лет до пожизненного заключения.

    https://nypost.com/2025/02/07/us-news/teacher-of-the-year-jacqueline-ma-sobs-as-admits-abusing-students-ages-11-and-12/

    Я чёт не понимаю, они учителей на помойке находят? Вот тебе мет и сладкие мальчики, так?
    ----
    upd снижение срока с 180 лет до пожизненного заманчиво однако было

    syn_mapporn 7:51p
    White percentage of local municipalities in South Africa (2011 and 2021 censuses)
    White percentage of local municipalities in South Africa (2011 and 2021 censuses) submitted by /u/bezzleford
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    syn_mathoverflo 7:51p
    Who wins the Scrambler-Solver game for infinitary Rubik's cubes?

    Given a suitable infinitary analogue $\mathcal Q$ of the Rubik's cube (as developed below), consider the two player game played between the Scrambler and the Solver wherein the Scrambler scrambles the cube by a (possibly infinite) sequence of twists (without the Solver watching), presents the scrambled cube to the Solver, and the Solver attempts to unscramble it by a sequence of twists. The Solver wins after $\omega$ many twists if $\mathcal Q$ is returned to its original "solved" configuration, otherwise the Scrambler wins. Can the Solver win in general?

    The cube

    Let $L$ be a linear order with no greatest element, write its reverse as $-L$. Consider the order $L^\dagger = -L + \hat1 + L$ where $\hat 1$ is the singleton order $1$ but its element, say $\hat 0 \in L^\dagger$, should be distinguished among the elements of $L^\dagger$. Having a distinguished center element in the order is analogous to ordinary Rubik's cubes of odd side length. The foremost examples I have in mind are $L = \omega$, $L = \mathbb Q^+$, and $L = \mathbb R^+$. In these cases, $L^\dagger$ as an order is just $\mathbb Z$, $\mathbb Q$, or $\mathbb R$, respectively, but again, there is a distinguished element (in particular, there are no automorphic translations of the structure $\langle L^\dagger, \hat 0\rangle$).

    Now, we imagine the $L^\dagger \times L^\dagger \times L^\dagger$ cube $\mathcal Q$ as six $L^\dagger \times L^\dagger$ "grids" arranged like the faces of a cube centered a the origin in $x,y,z$ coordinates with $(\hat 0,\hat 0)$ on the axes. The point is that we have corresponding positions in opposite faces and we can refer to particular "planes" within $\mathcal Q$—such as the $x = \alpha$ plane for $\alpha \in L^\dagger$—in which the twists will take place. To have some notation, let's introduce two new elements $\pm\infty$, thought of as bookending $L^\dagger$. We can think of $\mathcal Q$ as embedded in $[-\infty,+\infty]^3 = (\{-\infty\} + L^\dagger + \{+\infty\})^3$ with the faces in the $\pm\infty$ planes. A cell is a point in one of the faces.

    Twists

    Define a positive orientation in each axis plane, call it clockwise. A twist $T_{i,\alpha}$ where $i \in \{x,y,z\}$ is a permutation of the cells in the $i=\alpha$ plane given by a clockwise quarter turn. For example, the twist $T_{x,\hat 0}$ takes $(\hat 0,\alpha,+\infty)$ to $(\hat 0,-\infty,\alpha)$ and similarly for the other three edges. Clockwise quarter turns suffice to do any finite manipulation but we will want to consider reverse quarter-turns $T_{i,\alpha}^3$ and half turns $T_{i,\alpha}^2$ as individual allowable operations called basic twists. Given a twist $T$, we'll write $T^{-1} = T^3$ for the inverse (counterclockwise) twist, whenever useful.

    In a standard Rubik's cube the individual pieces that make up the puzzle are called cubies. We will refer to the (exposed) faces of a cubies as tiles. Nota bene, cells and tiles are not to be confused. Cells are particular locations on $\mathcal Q$, as seen from $[-\infty,+\infty]^3$, they do not move with twists, whereas tiles move between cells.

    A key property of a standard Rubik's cube is that a twist in any of the outermost planes permutes not only the tiles in the four rows at that coordinate, but also all the tiles in the adjacent face. We definitely would like to be able to make such a face twist, and so we also define $T_{i,\pm\infty}$ to do just this. For instance $T_{x,+\infty}$ permutes the tiles in the $x = +\infty$ face by a clockwise quarter turn about $(\hat0,\hat0)$. There are no corner or edge cubies, so this twist acts non-trivially on only these cells, quite different from an ordinary Rubik's cube.

    Configurations and legal twist sequences

    A configuration $f$ of $\mathcal Q$ is an assignment to every cell one of seven colors among $\Gamma = \{r,g,b,o,w,y,\perp\}$ (red, green, blue, orange, white, yellow, not a color). The solved configuration $f_{\text{solved}}$ is the particular constant color assignment $\{x=+\infty\} \mapsto r$, $\{x=-\infty\} \mapsto o$, $\{y=+\infty\} \mapsto w$, $\{y=+\infty\} \mapsto y$, $\{z=+\infty\} \mapsto g$, $\{z=-\infty\} \mapsto b$.

    Twists act on the space of configurations in the obvious way. Namely, $f' = T\cdot f$ is the configuration $f'(c) = f(T^{-1}c)$ for every cell $c$. We will be pushed to consider infinite sequences of twists, and this is where the $\perp$ color comes into play. Namely, we'd like to have a well defined configuration after executing infinitely many twists. Given a sequence of twists $\langle T_\eta : \eta < \theta\rangle$ for some ordinal $\theta$ and and initial configuration $f_0$, we can define subsequent configurations $f_{\eta+1} = T_\eta\cdot f_\eta$ and at limit stages $\lambda$, set $f_\lambda(c) = \gamma$ if the value of $\langle f_\eta(c) : \eta < \lambda\rangle$ (i.e., the history of the colors of cell $c$ up to stage $\lambda$) had eventually stabilized on $\gamma$, otherwise, we set $f_\lambda(c) = {\perp}$.

    Our main interest is merely $\omega$-sequences of twists, for which the above can be stated as follows. After $\omega$ many twists, if any cell had changed color infinitely often, its value is $\perp$, otherwise, it retains its limiting value.

    We'll say a configuration is legal if and only if no cell has color $\perp$. Given a legal initial configuration and a sequence of twists (in particular, an $\omega$-sequence), we'll say the sequence of twists is legal if every obtained configuration is legal including the final limiting one if there is one.

    The Scrambler-Solver game

    The Scrambler scrambles the cube by performing a legal $\le\omega$ sequence of twists (hidden from the solver) starting from the solved configuration, resulting in some legal "scrambled" configuration $g_0$. The Solver sees the configuration $g_0$ and may execute any $\le \omega$ sequence of twists. The Solver wins if the solved configuration is attained at or before stage $\omega$, otherwise the Scrambler wins.

    • Does the Solver have a winning strategy? When can they win before stage $\omega$? Certain instances are clear. For example, if we only allow finite scrambles and $L$ is countable, then there are only countably many possible twists, and so the Solver can systematically execute and undo every finite sequence of twists, eventually exactly undoing the scramble, and this will happen at a finite stage. This argument completely fails if $L$ is uncountable or if the scramble is infinite. Indeed, even if the Solver is allowed to observe the infinite scramble, she cannot simply "undo" it, since the reverse sequence is not well-ordered. This invites the question, are some (legal) scrambles unsolvable even in principal? If so, does it help to allow the Solver $> \omega$ many moves?

    • If the scramble is computable, can the Solver win? Can she win computably?

    syn_mathoverflo 7:51p
    About subspaces of $F$-spaces

    A topological space $X$ is an $F$-space, if every finitely generated ideal in the ring of all continuous functions on $X$, denoted by $C(X)$, is principal. The text "Rings of continuous functions" written by Gillman and Jerison, has numerous equivalent conditions of these topological spaces, which some of them are topological (see pages 205-215 of the mentioned reference). Here is an important topological equivalence of these spaces:

    Def: A subspace $A$ of topological space $X$ is $C^*$-embedded if every bounded continuous real valued function on $A$, can be extended to all of $X$.

    Theorem: A topological space $X$ is an $F$-space iff every cozero-set $($i.e. $X-Z(f)$ for some $f\in C(X)$$)$ is $C^*$-embedded in $X$.

    The mentioned reference, tells us that every $C^*$-embedded subset of an $F$-space is an $F$-space. Cozero-sets and countable subsets of these spaces are $C^*$-embedded and so $F$-spaces.

    With the above summary I can give my questions.

    Q1. The reference didn't mention an example of an $F$-space that has a subspace which is not an $F$-space. Is such space very simple to find?

    Q2. Is there a topological space $X$ with the property that every countable subspace of $X$ is $C^*$-embedded, but $X$ not an $F$-space?

    syn_cityporn 7:48p
    Summer lunch on the Chicago river
    Summer lunch on the Chicago river submitted by /u/Ganesha811
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    syn_slashdot
    6:40p
    Phishing Tests, the Bane of Work Life, Are Getting Meaner
    U.S. employers are deploying increasingly aggressive phishing tests to combat cyber threats, sparking backlash from workers who say the simulated scams create unnecessary panic and distrust in the workplace. At the University of California, Santa Cruz, a test email about a fake Ebola outbreak sent staff scrambling before learning it was a security drill. At Lehigh Valley Health Network, employees who fall for phishing tests lose external email access, with termination possible after three failures. Despite widespread use, recent studies question these tests' effectiveness. Research from ETH Zurich found that phishing tests combined with voluntary training actually made employees more vulnerable, while a University of California, San Diego study showed only a 2% reduction [PDF] in phishing success rates. "These are just an ineffective and inefficient way to educate users," said Grant Ho, who co-authored the UCSD study.

    Read more of this story at Slashdot.

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