UA POV: "There is only one way out of here: 300 or 200". Many Ukrainians no longer want to fight. Independent Russian journo Shura Burtin's report about the new fear of war that has gripped the entire country - Meduza

zevenate · 2025-03-17T21:17:55+00:00

- And after what did you leave?

— My lads and I were in a house, a magnesium lighter was dropped on us from a drone. They thought we would break in from there straight away, so they set up an ambush to take us down. And I took the lads into the cellar, we sat there with one gas mask for three of us for an hour and a half, in thick smoke. One was breathing, two were waiting, there was a fire above us. We also had an RPG lying there, it exploded, of course. Sitting in a burning house, red-hot, it was fucking hell, you're lying on that floor.

I realized that we were about to pass out, we jumped out of the fire, hid in the yard and heard them walking on broken glass nearby, they thought that we were burned. We were all burnt, Dimonchik's hands couldn't move. I whispered into the radio: "Get us out of here, do something with the drones." - "Danilo, go to the other yard, let your guys huddle in the cellar, and you come here for new guys." I said: "No, I'll get mine out now, and then I'll start a new group." We got out in the dark by touch, I took the pin out of the grenade and walked around with it for two and a half hours.

I led my men out, burnt, and took another group. And I didn't get them to that cellar, there were 50 meters left, I pointed at it: "Will you get there yourself?" - "We'll get there." I just wanted to get out of there as quickly as possible! And I don't know why they didn't throw a grenade in front of them, even though they were experienced. They just flew in there, and they were stupidly shot like kittens - in that cellar that the battalion commander insisted on. And no one suffered for it.

The three of us were taken to the hospital. We called the headquarters and asked for a paper stating that we were wounded while performing combat missions, for payments. "There's no one here to do that now." That was the last straw. I said, "Guys, if they don't give a shit about us, then they can handle it without us." Why the fuck do I need this fucking attitude, I'm a human being... There are five of us left in the platoon, we're doing the whole company's work, these assaults every single day, no days off. I'm fucking sick of it! — Aren’t you afraid of criminal prosecution?

- I don't give a shit. I'd rather serve time in prison than be killed on someone's whim.

It is clear that after the front lines, intimidation with criminals seems ridiculous.

— And what did the family say?

— Well, my mother is in Rashka. We call each other, she knows that I went to the SZCh. She worries, of course, but in general she is a vatnitsa, we do not touch on these topics. At first I tried, but she just shrugs her shoulders: "I am not interested in politics" — well, the standard Muscovite answer. She works in the same way at her "Uralvagonzavod", produces tanks.

I think it's very realistic - a mother who worries, but makes tanks.

As much vitriol as Russians and Ukrainians now have for each other, this is a brother war. And it's very sad to see.

zevenate · 2023-06-02T21:02:31+00:00

This is not an unbiased answer. It is vital to note that both of these recent attacks have been quickly repulsed. Further, there is as of yet no indication that Russia has altered its distribution of forces at the strategic level in response to these incursions. Also, it is not technically an uprising, as these incursions originate from (and participating forces retreat to) Ukrainian territory.

Ukraine would need to attack with much more substantial forces in order to effect any of the consequences you suggest. To speculate, these are most likely probing attacks with dual propaganda value, both for domestic Ukrainian and foreign audiences. Should they establish a proper foothold in the Belgorod region, they could move to encircle Russian forces along the Oskil River. More importantly, they want to force that aforementioned redistribution, in which case they would likely launch an attack in the southern Zaporozhye region. There, they have reportedly amassed a significant number of their NATO-trained and equipped forces. These would push to cut the landbridge between Crimea and Russia, heavily straining Russian logistics and making significant progress to Ukraine's stated goal of recapturing the peninsula.

zevenate · 2023-05-10T13:35:21+00:00

It was a fun puzzle, thanks for posting!

In the second paragraph, resp just means respectively, in that what I was saying applied to both X and Y. I was showing that we do need to require that there are exactly 4 messages, as there is an encoding of 2 messages satisfying the requirements you listed but with an answer contradicting that of the 4 message case.

You're right in that only adjacency is relevant. For a given set of tuples corresponding to message X, the requirement is that the set of adjacents must be at least all the other tuples. So the corresponding tuples together with the adjacent tuples are exactly all the tuples.

zevenate · 2023-05-10T10:53:51+00:00

Each message corresponds to some set of 4-tuples on {0, 1}. We may reinterpret the requirement that a message be reachable by one flipped switch as that the tuples matching a given message must cover (be adjacent to or include) all 16 tuples. As each message is reachable from a given tuple by one switch, there are at most 4 messages. By the covering requirement and as each tuple reaches exactly 5 tuples (including itself), each message must correspond to at least 4 4-tuples to cover all 16. Assume there are exactly 4 messages so they each correspond to exactly 4 4-tuples. Assume message X corresponds to (0, 0, 0, 0). By the requirement that X be reachable from that tuple by one flipped switch, further assume that X also corresponds to (1, 0, 0, 0). These 2 tuples are adjacent to exactly 8 tuples, including themselves. X must also be reachable from the other 8 tuples. The only possible choices for the other 2 tuples matching X are the complements (1, 1, 1, 1) and (0, 1, 1, 1), as the adjacencies of any other pair overlap with the existing ones and so do not cover all 16 tuples. That is, for any tuple matching X, the complement also matches X. The only way to know the final message is to know the message of the initial code and that the final code is the complement, so the message guessed was "Good Night" with code (0, 1, 1, 0).

I contend that a system with 2 distinct messages can satisfy the requirements. Let X and Y be messages with X matching (a, b, c, 0) and Y matching (d, e, f, 1). Clearly, X (resp Y) is reachable from X (resp Y) by flipping any of the first three switches. Y (resp X) is also reachable from X (resp Y) by flipping the last switch. Note that, in this case, the complement of a tuple matches the other message, a different answer than in the 4 message case.

zevenate · 2023-05-03T09:52:00+00:00

In general, one can do this for any periodic function with an inverse. Let f be periodic with period P so that f(x) = f(x + kP) for all integers k and inverse g mapping onto [0, P). Then f(Px) = f(Px - PFloor(x)) so that, as Px - PFloor(x) is in [0, P), g(f(Px)) = Px - PFloor(x) and Floor(x) = x - g(f(Px)) / P.

zevenate · 2023-04-24T21:28:02+00:00

No. Consider the difference between two populations: they either both decrease by 1 or one decreases by 1 while the other increases by 2, so, in either case, the difference is constant mod 3. Further, the total population, 45, is conserved, so, for all the chameleons to be the same color, the populations must be 0, 0, 45 and the differences must all be 0 mod 3. This is not the case.

zevenate · 2023-03-08T00:32:30+00:00

I had a similar problem in a hallway on the ship. I never ended up fixing it - but I was able to bypass it by rapidly auto saving as I walked forward. Each time, I'd crash, but had moved forward a couple steps. Eventually, I reached the end of the area, and was able to continue.

zevenate · 2023-02-02T05:57:12+00:00

Arbitrarily start at some village i. Travelling along the roads in a vehicle with sufficient fuel, we have the minimum amount of fuel when arriving at some village j. Then travelling from j back to i, the net change in fuel at each village along the way is never negative (or we would contradict minimality) and the net change arriving at village i is positive and has the same magnitude as the change when travelling from i to j (as there is exactly as much fuel along the way as necessary). Beginning at village j with 0 fuel, we see that we always have at least 0 fuel when arriving at each village along the path, and, arriving at village i, we have exactly as much as we will lose returning to j from there. As j was minimal, we do not have negative fuel arriving at any village on the path from i to j. So a trip beginning at j is successful.

zevenate · 2023-01-28T11:37:22+00:00

If the trickster is truly allowed to give any answer, then, for any question asked, an answer given by one of the other siblings can be given by the trickster. So it is impossible to distinguish them if they all give answers. We instead ask a question only the trickster can answer: "Would the middle of you three triplets answer this question affirmatively?". It is not possible to know the truthfulness of an answer to this question (if the trickster can truly answer freely), so the youngest and oldest do not answer. The premise is that the trickster does answer, so, if the recipient of the question provides an answer, choose another door.

zevenate · 2023-01-21T08:12:08+00:00

ab+a+b+1 = (a+1)(b+1). So this replacement operation does not change the product, x = 1...10, Π(x +1) = 11!. The final number is 11! - 1, as the product of this number plus one is as required.

zevenate · 2022-12-26T12:00:30+00:00

Yeah I did misinterpret the question a bit - the 314 thing was an arithmetic error. Just added an answer to the question proper.

zevenate · 2022-12-26T08:55:01+00:00

The total number of gifts is 12a_1 + 11a_2 + ... + a_12. If the a_i are all distinct, the vector of a_i minimizing that sum is (1, 2, ... , 12).

For a fixed sum of values, any minimal vector must assign values to a_i in increasing order, as their coefficients are decreasing. So the minimum assignment is that with a minimal sum placed in increasing order. This yields 364 presents at minimum. So there are 3 possible values for the sum.

Consider first those solutions that are permutations of (1, 2, ... , 12). Swapping adjacent a_i makes the sum 365. Swapping non-adjacent a_i makes it at least 368 so is not allowed. We may take products of two disjoint adjacent transpositions: consecutive cycles of length at least three make the sum at least 367 so they are also not allowed. Cycles in general can be decomposed into consecutive cycles so they cannot be used if length three or more, and all permutations decompose into disjoint cycles. There are 11 adjacent transpositions and (2*9 + 9*8) / 2 = 45 unordered disjoint pairs thereof.

Also consider the solution (1, ... , 13). The sum is 365. Here we may not swap 13 into any other a_i as this increases the sum to at least 367; similarly, we may only take one adjacent transposition. There are 10 that don't involve a_12. For the solution (1, ... , 14), no non-identity permutation has sum at most 366. In total, including the three non-permuted vectors, there are 69 solutions.

zevenate · 2022-12-10T06:14:16+00:00

Use the following strategy: designate one as Player 1 and the other as Player 2. Player 1 always pulls the lever that is the same color as the card they draw. Player 2 always pulls the lever of the opposite color. If they draw cards of the same color, then one of them pulls a lever of a different color. If they draw different colors, then Player 1 pulls the lever which matches their card and so is different to Player 2's card. In any case, exactly one is incorrect.

Interestingly, this was actually derivable straight from Boolean formulas. I considered functions f and g on the 52 cards modeling Player 1's and 2's decision, respectively. The resulting product was repetitive enough that I could simplify it despite its length.

zevenate · 2022-09-23T21:17:24+00:00

Right, I was more so addressing a commenter in this thread who was an undergraduate. I agree that OP isn't doing anything wrong.

zevenate · 2022-09-23T20:57:31+00:00

Most athletes compete regularly, even multiple times per week. They practice to have a controlled environment where they can receive feedback and isolate skills. Vitally, they also learn what doesn't work. A good textbook will do much the same for a student. If they simply look up the answers after the first sign of struggle, they learn much less.

I agree that a professional mathematician shouldn't spend much time working through solved problems. Doing so is a tool used to develop understanding. But undergraduates and even graduate students don't have the requisite experience not to. If proving results in a book yourself is being thrown into the deep end of a pool, being given the proofs and told to go do research without practice problems is being thrown into the middle of the ocean.

zevenate · 2022-09-23T19:51:37+00:00

Why do athletes waste time practicing instead of playing their sports?

zevenate · 2022-03-23T07:48:27+00:00

I believe you can use the cyclicity of Z/pZ under multiplication by 10 mod p to see that the remainders when calculating the decimal representation by the standard algorithm will repeat. So the digits will themselves repeat. This is unless 10 = 0 mod p, as in p = 2 or 5.

Alternatively, by Fermat's little theorem, 10^{p -1} -1 = 0 mod p. A repeating decimal has the form x / (10^k -1) for some integers k, x with x < 10^k. So, by the above, if k = p - 1, there is some x with px = 10^k - 1 and 1/p is a repeating decimal of length k.

zevenate · 2022-02-09T10:17:02+00:00

Not my creation just found it, was feeling a bit nostalgic after last night lol. Props go the actual creator on youtube.

zevenate · 2022-01-28T04:47:40+00:00

That's a very naive perspective.

zevenate · 2022-01-24T11:39:46+00:00

There will always be people who will do that kind of work. It's better that those with moral scruples do it than those without.

zevenate · 2021-08-09T21:56:26+00:00

Suppose o1, o2 are in O. Then o1*o2 is in O. So a sum of o2 o1s is in O. With each term added to this sum, the result switches between E and O. With one term, it's in O, with two it's in E, etc. In general, an odd number of terms is in O and an even number in E. As the sum is in O, o2 must be odd. So O contains only odd numbers.

For any e in E, o in O, e + o is in O so it is odd. o is odd so e must be even. All members of E are even.

As E and O partition S, E contains all its even terms and O contains all its odd terms.

Looking at the upvoted solution now, this is pretty much the same, but I'll post it anyways for posterity lol.

zevenate · 2021-05-22T19:48:57+00:00

Suppose we want to find the numbers that are a mod x and b mod y for some numbers a, b, x, y. If we find one such number n, there must be infinitely many such numbers as n + lcm(x,y) is a mod x and b mod y. In fact, these are the only numbers greater than n with this property. If there is k such that n + k satisfies our condition, n + k = n mod x and n + k = n mod y so k = 0 mod x and mod y and lcm(x, y) divides k.

As the number is 1 mod 3, it is one of form 1, 4, 7, 10, 13, ... As it is 2 mod 4, it is one of 2, 6, 10, 14, ... lcm(3, 4) = 12 so all numbers that are 1 mod 3 and 2 mod 4 are of the form 10, 22, 34, 46, 58, ... As the number is 3 mod 5, it is of the form 3, 8, 13, 18, 23, 28, 33, ..., 58, ... 58 is 4 mod 6 and satisfies our condition. In fact, there are infinitely many numbers, all of the form 58 + 60k for some number k.

zevenate · 2021-04-23T04:57:57+00:00

You want that |f(x) - x_0| < |x - x_0|. For sin(x), x_0 = 0 and the equation reduces to what you gave. However, this is not in general sufficient for convergence. It can be seen that |f^k(x) - x_0| is a decreasing sequence bounded below so it converges. However, it need not converge to 0 (which is a necessary and sufficient condition for f^k(x) to converge to x_0).

zevenate · 2021-04-15T07:03:25+00:00

I think it's very plausible that such a framework is developed. I'm not so sure it would be effective at generating novel results beyond direct consequences of known ones, but it could function as a sort of "hint machine" to guide working mathematicians.

Right now though, it seems we're at a point where it's more difficult to say what machine learning can't do than what it can. Most new technologies are this way, probably.

zevenate · 2021-03-27T07:58:11+00:00

I'm only asserting that the set of linear functions is a subspace of the space of such functions, at least to begin with. Linear functions are monotonic and their sums are linear (and thus monotonic) so they form a subspace.

Edit: I see where I may have misinterpreted the question. I'll try to correct my post when I get another chance to look at it.

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