Thursday 28 December 2023

Thoughts on CTWC2023 finals

CTWC2023 has been over for some time. If you follow the tourney or even my blog posts covering it each year, you'd know it's held somewhere at the bottom of November. Could be a bit later but definitely not after Christmas.

And I completely missed it, only noticing something else which I will also cover here that YouTube started pushing CTWC clips to me. I received zero pushes on CTWC23 clips at all. Yes, zero.

YouTube is of course partially for the blame cause YouTube must have known my interest on CTWC. But it also because the hit rate on classic tetris that is dropping that YouTube decided not to push? But before getting into that let us talk about the tournament itself.

Sadly I don't have time to watch all match ups one by one -- I kind of treasure the time in 2020 where lock down gave me enough time to really go through every single ones. I ended up watching the finals only, so this is my brief thoughts on the finals instead of the whole tourney. Sidnev and Fractals are good representatives of the top circle so my conclusion may apply to those (single digit number of) top players, but not to the larger base like all CTWC finalists.

Very close match between the two. It is very hard for me to tell the difference in style -- the grip may differ but they way they controlled the pieces are too similar. Some luck kicked in and Fractal ended up with a reverse sweep. That requires unparalleled precision but also heart of steel. Good sportsmanship between the two but it had been tradition among the top classic tetris players circles anyway.

...and that's all. Giving further comments is even harder than doing that on the 22 finals where there is at least one unforgettable game that you can talk about. All 5 games in the finals are basically smooth all the way up to Lv.29 then kill screen (or second transition if you like) hits where a roll of dice decides the winner. One may wonder why would it be a roll of dice when the sequence is identical for the two. That is because placement difference is chaotic so eventually the boards would be totally different. Some configurations are weak to a particular sequence and that's the moment that kills.

But wait! Even when you find yourself in trouble due to bad sequence there is still room for rebounce right? Unfortunately we have the Lv.39 hard cap this time. Sounds irrelevant but a hard cap means pure survival is no longer the primal goal in the play. Or rather, a strategy that is relative safe for one to stay at Lv.29 speed over a prolonged period is no longer desirable. Instead you take a high risk high reward route where you would probably die somewhere between Lv.29 and 39 but gives a higher average score (assuming the hard cap at Lv.39) -- if you can't score enough you lose either way hitting the cap or not, so it's better to be aggressive from the beginning.

Last year I said the meta was optimization up to Lv.29 then survive, partly because the cap wasn't there but also because rollers are not super comfortable with Lv.29 speed...yet. This year they got themselves enough space to optimize under such hostile condition, and this is probably what most top players did in recent months.

As the style homogenizes, what is left is basically consistency (ideal input vs actual input), mentality (concentration etc.) and tiniest bit of efficiency. Sure if you count the stats over a bigger sample like the accumulated masters heads to heads you will be able to see the difference among the top players, but in a first to 3 this is just a roll of dice where the better side has a slightly better chance of winning (not to mention lower rounds are just first to 2 sets). (Kind of reminds me about a math question that I just analyzed not long ago -- Simon Marais 2023 B2. I am not going to cover the math here to check my entry if you want to have a look at it.)

This is not by means undermining the credibility of the tournament. A winner is still a winner and I truly believe that every player deserves their final ranking in this tournament. The point is the fun is gone. The commentaries on how close the scores are and how close to top out their players had been are enthralling, but it becomes repetitive soon after some games. The commentators are doing their jobs well, but the game is losing its value to be spectated. 

It might be unfair to compare views count on a fresh video but it is surely dropping, and they might be the reason that they weren't pushed in the first place. It could be a death spiral -- less attention means less new blood, which means less attention in turns. Granted the game has been low key for 20+ years before the current community bought it back, but the spiral is still something one wants to avoid. 

What is the future of the competition, if you ask me? 

There is no reverting in skill and meta progression. Players have proven themselves that Lv.29 speed is no longer a hurdle in general. From this point the remaining progression would be to find the sweet spot balancing aggression and average performance. On the non-capped track the game is even 'cleared' -- Blue Scuti, a very young new comer to the game reaching semifinals this year, survived up to Lv.157 in a game where the game crashes due to memory overflow. Such ending could be viewed as a 'game clear' in the same way as donkey kong or pac-man. I considered the game 'solved' in the sense that all remaining milestones are reachable given enough tries/luck. 

(I am quite surprised that the color shifts didn't cause too much trouble to Blue Scuti though. Other than the really dark blocks at Lv.157 which Blue Scuti almost topped out there are a few more notorious color combinations like dark green and blue at 146 or black and white at 148 but he breezed through the levels fine. Maybe players have practiced enough for that? They recall me of TGM invisible blocks too...)

Classic Tetris will remain as a competitive sport fine despite the lost in visible progression/meta shifts that makes it less appealing. New variations is essential if we want to revitalize the game, but each change will bend it further away from the original game. If it's too far away from NES Tetris people will just as well play modern interpretations like Tetris Effect, so the issue is pretty delicate here.

With top players really comfortable up to Lv.29, I think it will be fine to start from Lv.19 speed just to spare time for later stages. The new Lv.19 phase could last more than 100 lines, somewhere between 100 and 230, but this is up to debate. The time spared can be spent at Lv.29 speed -- actually I prefer a longer Lv.29 speed section like up to Lv.49 considering how quick players topped out with their current aggression. 

Also perhaps a change in game format? Just like how chess is categorized into classic, rapid and blitz we can create multiple rule sets for classic Tetris matches as well. For example the 'uncapped' with no extra cap after Lv.29, a 'rapid' as described above that starts from Lv.19 to Lv.49, or 'first to million' where players optimize to reach million as quick as possible. Some requires slight tweaks to the game but not beyond where modifications have already taken place. The rest are merely change of rules but not changing the game itself. The common theme is to force players into diversified strategy or optimization bias. I expect a much more interesting game in that way. I called a change last year and I would do the same here, and hopefully the drop at anticipation if any, would alert the organizers to evaluate what's going on.

I still look forward to CTWC next year, and if you are interested in the game it's definitely worth a peek into the fantastic world of classic Tetris. It's never too late!

Friday 15 December 2023

15/12/2023: 完結撒花








DDR沒啥好說的,弄個低能活動只想白嫖玩家那幾個代幣。三道的錢打四首歌真當大家是白痴?啊抱歉我還真看到有人上釣而且不少……Flare Gauge跟encore extra stage有點像不過把判定難度分成三六九等的話至少普羅大眾可以喝點湯。不過對我這種低acc玩家來說沒甚麼分別就是,反正用上flare gauge要解鎖18怎樣也打不過,解鎖低等級的譜面可以但沒甚麼意義。想想我連Snow Garland Fairy ESP都沒興趣去解了,你還叫我去打活動?Konmai你還是去騙那些會掏錢買GP的人吧,有本事你就把歌曲一直鎖住,這樣我也省去刷Lv18全通的功夫。


另一個jubeat的問題是我的jubility已經上不去,這個問題我一年前已經提過。事實上我的jubility從clan到現在beyond Ave都沒動過。其實除了間距問題以外更麻煩的是那個30+30的制度,也就是jubility來是最高分的30首自選和30首pick up譜面的總和,而pick up池就是當下版本的新增曲目。pick up池裡面的曲目他媽的少,而且多半都要通過漫長的遊玩來解鎖。Ave我也累積200道以上,pick up池裡卻只有20首不到根本沒法填滿pick up分數,還是你覺得我應該把Advanced打一打把分數算進去呢?每次改版後pick up池裡面的10.6~10.9稀缺到玩家再強也暫時拿不到金名,那jubility的鑑別度又在哪裡呢?我就遇過幾次紫(Violet)名一上來就隨手在10.6 10.7打個99+的。當然你可以說jubility根本從頭到尾都只是騙錢的機制,這樣說我也可以接受。


在中二分數容許一堆Justice的情況下能否在高等級譜面拿到同樣分數取決於你能不能看懂譜面然後採用正確的打法而非每一個音符都打得很準。在這方面我的潛力明顯還有提升的空間:本來看不懂的Jade Star現在已經可以FC,蒼穹舞楽在yt看過譜面後回去就打了個接近1000000的分數。現在15.6的我要再上大概就只能開始讀14.6 14.7的譜面了吧?能不能有朝一日摸到16.0的大門,我十分期待。

Monday 27 November 2023

古早遊戲BGM巡遊(6): Csikos Post

在遊戲裡使用古典樂remix作為配樂可謂屢見不鮮,一個遠古的例子是任天堂俄羅斯方塊C模式的音樂實際上來自巴哈的BWV 814 Minuett。那麼,如果要你選一首古典樂配樂的代表作的話你又會選哪首呢?

我會選Csikos Post。說英文可能還不夠明確,但如果我叫它郵遞馬車呢?還是不懂?那如果我叫它熱血行進曲呢?


非常具張力的一首樂曲而且長度適中,很適合放在遊戲裡作配樂使用。最特別的是其原曲作者並不怎樣出名也沒有別的代表作:比如給愛麗絲你會叫給愛麗絲或者貝多芬的給愛麗絲,叫不出來像上面的BMV 814你至少也會叫巴哈那首;可這首郵遞馬車作曲家叫不出名字,曲名也不容易叫出來,於是命名權就落到將其發揚光大的遊戲裡。

但這首音樂被使用的廣泛程度絕對不限於熱血高校。遠的有耀西曲奇(Yoshi's cookie/類似瑪莉醫生的spin off)或GBC哈姆太郎,近的則有NDS右腦爽解2012瑪莉奧x索尼克奧運會等。其在音遊上的版本就更多了:pop'nReflec BeatO2JamDDR Mario MixNostalgia都有自己的版本,但我最愛的還是PIU的Banya版本






Saturday 18 November 2023

Simon Marais 23B4 revisited + comments on official solution

(SM2023, B4) Let $n$ be a non-square number. 

(a) Find all pairs of natural numbers $(a,b)$ so that $r^a+\sqrt{n}, r^b+\sqrt{n}$ are both rational for some positive real $r$.
(b) Find all pairs of natural numbers $(a,b)$ so that $r^a+\sqrt{n}, r^b+\sqrt{n}$ are both rational for some real $r$. This is an open question currently.

When I first drafted my quick comment I simply omitted this question because a Q4 is a Q4 and has to be respected, so I only left one or two sentences there. Later on I found this one of interest so I dived in and wrote a simple answer. But then some gap in the solution had been lingering in my mind, and unsurprisingly I missed out something important so the answer is not fully correct. But I feel like I am close to the answer since the math behind is nothing overly complicated.

Below is my later attempt. There is a gap still, not recommended for readers.


Obviously, every $a=b$ pair is a solution because you can easily make up $r = (n-\sqrt{n})^{1/a}$. Next, if $(a,b)$ is a solution then so as its multiples $(ka,kb)$ because you can take $r' = r^{1/k}$. 

As a result we can now assume that $a,b$ are coprime. As $r^a, r^b \in \mathbb{Q}[\sqrt{n}]$, we know that $r\in \mathbb{Q}[\sqrt{n}]$ as well. I think this is well known -- but in case that is not, think about Euclidean algorithm (recall that a field is Euclidean...or even simpler the algorithm where you find gcd) that you can divide each other until you reach $r^1 = r$. 

Now write $r= p-q\sqrt{n}$ for some positive rational $p,q$ (the proofs are similar for different sign combination. Like you can write $r = p+q\sqrt{n}$ can the argument below proceeds the same). If you still remember your linear algebra lesson you can solve the recurrence by writing $p_k-q_k \sqrt{n} = (p-q\sqrt{n})^k$ for rational $(p_k), (q_k)$. For $v_k = (p_k,q_k)^t$, we have $v_k = A^{k}v_0$ where $A = \begin{bmatrix}p & nq\\ q & p\end{bmatrix}$.

The general formula for $p_k,q_k$ can be easily obtained by either method of eigenvectors or if you know how expand binomial powers. Anyway we have: 
$p_k = \frac{1}{2}((p+q\sqrt{n})^k+(p-q\sqrt{n})^k)$
$q_k = \frac{1}{2\sqrt{n}}((p+q\sqrt{n})^k-(p-q\sqrt{n})^k)$.
Notice that the formula is the same for $(p+q\sqrt{n})^k$. For negative $p,q$ the sign just alternates. 

Clearly $q_k$ is an alternating almost exponential series. More precisely we expect $q_{2k}-1 \approx q(p+q\sqrt{n})^{2k-2}$ and $q_{2k}\approx 2pq(p+q\sqrt{n})^{2k-2}$. The two subsequences are surely strictly monotonic and would not give us any non-trivial answer. The difficult task is however to find $k,k'$ so that $q_{2k} = q_{2k'-1}$.

A few approaches are there: first you may want to show that $q_n$ may not even be an integer if $p,q$ are (non-integral) fractions past a certain $k$. Secondly you may consider something even stronger that $q_k$ can't be 1 for largr $k$ because exponential sequences are either diverging or goes to zero, but the problem is you can always make very marginal case where $(p+q\sqrt{n})$ is very close to 1 (e.g. using continued fractions) so that the 'rate of divergence' is not properly bounded.

The problem for us is (1) we can't calculate terms manually for anything past $k=4$ -- even that is nasty enough as you can see below, and (2) what is the meaning of (b) being unsolved? Most arguments, if we assume that $r\in \mathbb{Q}[\sqrt{n}]$, does not really care about its positivity. In fact, we care if $(p-q\sqrt{n})$ being negative or not more so $(1-2\sqrt{2})^k$ is easier to handle than $(1+2\sqrt{2})^k$!

My approach is basically the observation that the last $k$ where $q_k$ can easily be made as an integer when $p,q\notin \mathbb{Z}$ is $q_4$ with where $(\frac{1}{2} + \frac{3}{2}\sqrt{3}) = \frac{223}{4} + 21\sqrt{3}$. One suspects that for larger $k$ the expansion looks like $q_k = kp^{k-1}q + cnp^{k-3}q^3 + c'n^2q^5(...)$, where the $n^2$ is causing problems when we want to solve integrality or doing mod checks. But at the end the approach does not distinguish between (a) and (b). Now I am really confused...and I see nothing wrong in the Euclidean argument too. 

Still, let us try to check the lower $q_k$'s:

$q_1 = q$
$q_2 = 2pq$
$q_3 = 3p^2q + nq^3$
$q_4 = 4pq(p^2+nq^2)$

As claimed, $q_1\neq q_3$ and $q_2\neq q_4$ so we only need to solve the 12,14,23,34 pairs.

$q_1=q_2=1$: this is easy with $p = \frac{1}{2}$ and $q=1$. Possibly a solution that students would have noticed without going through all these hassle.

$q_2=q_3=1$: Substitution gives the quintic equation $8p^4 - 4p^3 + n = 0$ with discriminant $256n^2(512n-27)$. By checking $(512n-27)$ mod 8 we know this is a non-square so that gives no rational solution.

$q_1=q_4=1$: Substitution gives the equation $4p(p^2+n)-1 = 4p^3 + 4np-1 = 0$. Surely one real root but the discriminant of a cubic equation doesn't tell much about rationality. Instead we convert $p$ into a fraction $p = \frac{s}{t}$ so $4p(p^2+n)$ becomes $\frac{4s(s^2+nt^2)}{t^3}$ for some $(s,t)=1$ but then we require $t^2\mid s^2$ which is absurd.

$q_3=q_4=1$: this is of course the hardest...but the technique never changes. From $q_3=1$ we obtain $q = (4p-1)/8p^3$. Substitution into $q_4=1$ gives a horrible degree 9 equation 
$3p^2((4p-1)/(8p^3)) + n((4p-1)/(8p^3))^3$
$ = (n(4p-1)^3 - 3\times 64p^2 + 3\times 256p^3)/512p^9 = 1$. 
Looks horrible but all we need is to take mod $p^2$ which eliminates the last two terms. Since $(p,4p-1)=1$ that forces $n$ to be non-square-free, hence the contradiction. So $(a,b) = (k,k)$ or $(k,2k)$ are all the possible pairs.

Other than discriminant checking, the main trick is to force a term to be multiple of $p^2$ (or anything else) if you want the sum to be a multiple of $n^2$ and so does the rest of the terms, then we can apply the square-free assumption. I believe this can be applied for higher $q_k$'s because we can check higher powers of $p$ against the $n^2$ factor. But clearly (b) being open does not support that.

So, basically the sure answers are in forms of $(a,b) = (k,k)$ or $(k,2k)$. But are there more? I don't know.

The more I think about this question the more I like it with so many different scattered technique that were used. This is the type of question we would like to see more. I would not hide the fact that my first attempt is a miserable fail as I overlooked a large part of it. I wrote that we look at $A^2$ again we know from there that the denominator would sure to stack up every 2 steps, but what about consecutive terms $q_k$ and $q_{k+1}$? I stared at what I wrote before I finally realized what should be corrected...but nope I am probably not getting 7 this time.

Just food for thought...what if we are now in $\mathbb{Q}[\sqrt{-n}]$, or that $(a,b)$ are simply integers instead? 

--------------------(NEW 08/12/2023)---------------------------

The solution is out and below is my answer-checking...

A1. No surprise, but their solution 3 is neat. If you skipped that while reading the solution you should go and have a check.

A2. The idea is right but I got the lower bound wrong for not considering $S$ to be set of all possible convex functions, but that is still easy.

A3. Completely wrong...hmmm I totally missed the 'unequal size' condition. Instead I took the sets $A_i$ in the way that $A_i$'s are not pairwise subsets and supsets. The correct version of A3 is much easier. Still I appreciate they put all the details unlike me (especially A2 and A3).

A4. As expected but their approach of finding the 'crtical point' is more elegant than calculus bashing.

B1-B3 are all as expected. But they spent more words on B2 that I would have thought. At this point I think all Simon Marais contestants should learn generating functions before they come -- such question appears almost every year!

B4. Here is the big thing -- I say that I can't see how should I tweak the solution to accustom the condition that distinguishes (a) and (b). How about scrapping the whole approach? It is now clear that my approach is more for (b) the open problem. (a) is much easier. I also missed that fact that for $(a,b) = (1,2)$ we actually got a negative solution so that does not count in (a) but in (b)...

No surprise in C1-C3 again, though C3 should deserve only a one-line solution if any.

Monday 13 November 2023

Rains and returning period

Auckland's 1-in-200 years flood was given the name Anniversary Weekend flood 2023 on wiki...or simply just the Anniversary flood. The scars are still visible around, and recently a review has just been concluded from the Metservice side. The conclusion is that their rain prediction is poor.

Um...thank you.

I think most Aucklanders aren't surprised as the forecast tends to underestimate rains heavily all the time, where longer forecasts on precipitation are merely better than flip of a coin. We used to read radar directly because we can pinpoint the location of interest and is more accurate in general. We can at least tell whether a region is going to rain in an hour while the forecast can't!

When I wrote about the flood earlier this year, I said it's not a good time to talk about the 1-in-N-years thing because it's not the right time to do so. I think we are in the position to do it now especially when a comparable event happened later this year: the torrential rainfall over Hong Kong on early September 2023: record breaking 158mm in one hour, 800mm+ widely over the Island over 12-24 hours. A rare city wide flood that immediately recalls what happened over Auckland some times ago. It ended up being described as a 1-in-500-years event but was taken as excuse to inability of the government.

There are too many things we can talk about the two storms here like crisis management or building standards but I just want to focus on the numerical and scientific bits here as below.

Auckland's major flood on 27 January was caused by atmospheric river dragged near NZ by remnants of Tropical Depression 06F and blocked or forced stationary by a nearby anticyclone/high. At the same time, Hong Kong's flood was caused by the remnants of Typhoon Hanna (11W), stuck in a saddle field right above Hong Kong so that rain bands kept sweeping in.

The similarity is clear: both floods were caused by remnants of tropical systems that are stuck so that the affecting period is prolonged. But then it seems like this is the only recipe for such rainfall: normal lows can't be that powerful so it must be something related to tropical systems; mature systems (e.g. typhoons) are primarily driven by higher atmospheric features and seldom stuck like that so it has to be as weak as depressions or lows as remnants of the previous systems. Even so we need that to get stuck over an extended period (e.g. 12h) and provide enough water vapor to the system for that to work. Without considering other more extreme events like volcanic eruption/nuclear winters etc, is there any other possible way to produce such rainfall?

I suspect that the long tail on rainfall distribution over considerable period of time -- is not normal, because it is only produced under a specific combination of meteorological events. The distribution is composed of some 'common raining events', then events like what caused the above floods that is responsible for part of the tail, then more extreme event for the rest of the tail. Just think of adding three unequal Poisson (not saying these distribution is Poisson of course) distributions together and what would you get? Of course these are more speculation than anything, but that should allow us to get into what causes these rainfall outliers.

The next thing we should think about is the returning period which is the most talked about. If the distribution is not even normal how can we calculate the returning period? Another problem is, even with the returning period on our hands how should we set standards using that? 

The 1-in-500-years claim of the Hong Kong flood is based on the peak 1 hour rainfall 158.1mm which exceeds the 500 years mark (~155mm) on the 1hr rainfall returning period table. Auckland's 1-in-200-years flood is based on NIWA's claim but I can't find any numerical evidence supporting that. They were referring to the specific flood on 27 Jan so it might be 24hrs cumulative rainfall (or least 12, considering this is how long it lasted). Of course the flood is record breaking in all timeframes like the monthly precipitation of 539mm easily exceeding any of the months in the last 170 years. 

It is clear that the officials picked 200 and 500 respectively because this is the greatest that can be interpreted out of the numbers so that it sounds as serious as possible. But to those who want to use the returning period for risk management what's the proper timeframe to look at?

Unlike water level for dams where the returning period is duration independent, drainage capacity isn't and could have very different implication over different timeframe. A 1-in-200-years 2mins rainfall may sounds scary but is nothing to the drainage. The returning period of monthly rainfall is also pretty useless for drainage system because all you care is whether or not you can handle the spike. If the 538mm rainfall is distributed evenly then we will receive ~0.7mm of precipitation per hour nonstop for a month. Bad to have in the summer but no big harm at all.

Now suppose we take 1hr and 12hr rainfall into consideration more than anything else. Does it make the returning period a reliable indicator? On one side you say yes because it covers flash floods that are  immediately visible as well as the whole raining period that the system is designed to withstand. But consider this: a typhoon just swept through Hong Kong exactly a week before the flood that introduces significant amount of trash into the drainage which of course weakens its capability to drain. Should we take the two as independent event? If not, is it necessary to take returning period of longer timeframe back into consideration? Same thing happened to Auckland as a subtropical low hit Auckland 4 days after bringing more precipitation, then there came Gabrielle early February that again broke the still sagged SH1.

The lesson here is that the returning period is a highly on-paper number, easily manipulated by the data interpreter, highly confusing to the public and hard to use as a reliable reference. The bottom line is that it's something that you can always calculate and compare...but that's it. 

Oh and before I conclude it's unfair not to say something for the forecast institutes. Precipitation forecasting is a very complicated task even in 2023: it is very random in locality and intensity. The difference between rain and no rain could just be hundreds of meters apart, and there is no point to forecast upon such precision. Even if we can tell if it's going to rain or not, the cloud may develop or dissipate at any moment. This is particularly true in case of heavy rains. Just look at the radar -- the chances of actual heavy rain is much higher than finding signals of "heavy rains" on the chart! These are true limitations of technology up to now and are things that can't improve overnight. Still I am grateful that they are thinking to improve. Let us hope that floods like that don't ever happen again.

Friday 20 October 2023

Simon Marais 2023

Another year, another set of Simon Marais problems. Let's go! *Last edited 24/10/23 -- I added my comment on B4 which turned out nicely.

A1. Simple. The coordinates of the final point can be calculated as limit of geometric series.

A2. Quite confusing that you might waste time thinking about plane partitioning problem (aka cake cutting problem), but it's much simpler. The function $g(n)$ must be of the form of piecewisely glued $f_i(x)$. Since $f_i$'s are linear functions (or affine they call not to be confused with linear in the operator sense) it is easier to interpret convexity with positive second derivative almost everywhere. 

That is, whenever we switch from $f_i$ to $f_j$, the slope value must have gone higher for the sake of convexity. In the ideal case we can have linear functions $f_i$ with increasing slopes so that the $n-1$ intersection points between $f_i$ and $f_{i+1}$ has a natural increasing order. And to construct such example is easy: just take tangent lines from a parabola (because we are taking about positive second derivative). 

The last thing we need to show is that in such case we can skip one of the linear function and the rest could retain such structure. More precisely, suppose $f_i,f_{i+1}, f_{i+2}$ are in increasing slope. Let $(x_i,f_i(x_i))$ and $(x_{i+1}, f_{i+1}(x_{i+1}))$ be the intersection points of $f_i,f_{i+1}$ and $f_{i+1}, f_{i+2}$ respectively. Then the intersection between $f_i$ and $f_{i+2}$ is $(x,f_i(x))$ where $x\in (x_i,x_{i+1})$. This is clear because everything is continuous and almost everywhere differentiable.

The weird thing is that the question asks for smallest possible value as well...isn't it 1 unless I didn't read properly?

A3. Instinctive guess: the smallest $2k$ (or $2k+1$) such that $C(2k,k) \geq n$ (or $C(2k+1,k)$). Why? Try to look at the poset $(n,\geq)$. If you choose an element then everything above AND below it would be banned, so it's a really powerful restriction. The best choice is to choose everything from the same level. That says, we can already fit $C(2k,k)$ sets with a total of $2k$ elements. The only thing that is left to show is that we can't make $C(2k,k)+1$ such sets.

What I am thinking is much stronger: given $A_1,...,A_n$ using $k$ elements and suppose that $C(k,r)\geq n$ for some $r$. Then there exists valid sets $A'_1,...,A'_n$ so that each $A'_i$ is either a subset or superset of $A_i$, each of size $r$. Is it really possible? Probably takes pages to prove.

A4. The focus is the function $f_n(x) = \frac{(n^2+1)x^2}{x^3+n^2}$. We want to split the sequence into two parts: $n=0,1$ and the rest because the function behaves more steadily for $n\geq 2$. 

The first two iterations are simple: $a_1 = a_0^{-1}$ and $a_2 = 2a_1^2(a_1^3+1)^{-1}$, so we get $a_2 = 2a_0(1+a_0^3)^{-1}$. What's so special about this formula? Well, we claim that if $a_2>1$ then the sequence diverges. If the above claim is true then the rest is easy: $x^3-2x+1=0$ has two positive roots: $\phi -1$ and $1$, so that is the range of divergence!

The proof of the claim is also about bashing the function $f$ like $f_n(x)$ peaks at $x_n = (2n^2)^{1/3}$ and $f'_n(x)>1$ between 1 and halfway to $x_n$...complicated, but not hard.

But then I figured out the significance of $a_2$ and the perturbation $\phi -1$ only using computers but participants have to do it by hand! As for difficulty of A4 it somehow makes sense but it's a bit sad to see questions like this being the 'boss' of a paper.

B1. We have seen questions of similar style for many, many times in this competition: simple question with simple solution, worded in an abstract way to confuse people. In my words it would be like "find largest possible $|| \sum v_i||/2$". Do you understand now?

B2. Notice that the whole schedule is decided before tournament. For example the first 2-bracket is always player 1 (ranked 1st) vs 2 (let's call that 1v2). The first 4-bracket (first 2 rounds) is always 1v2, 3v4 and the winner of 1v2 plays the winner of 3v4. The same goes for 8-bracket, 16-bracket and so on. The thing is, if the expected rank of winner of the first 4-bracket is $r$, then the expected rank of the winner of the second 4-bracket is $r+4$. Since the winner of the first 4-bracket is always ranked higher than that from the second 4-bracket, we can calculate the expected rank for the winner of the first 8-bracket as well. Similar calculation repeats till the 8th round aka the final.

I don't mind repeating the calculation 8 times for 1/8 of total credits in a college math competition. But it's not fun.

B3. An intuitively true statement in linear algebra. 

WLOG assume that $A,B$ have basis $e_1,...,e_n$ and $e_{n+1},...,e_{2n}$ respectively. Since $A,B$ share no non-trivial intersection, $(e_j)$ is a basis for the whole space. For a basis $v_1,...,v_n$ for $C$, we can write $v_i = \sum a_{ij}e_j$ for each $i$. Now we look at the rank of the matrix $M = (a_{ij})_{i,j=1,...n}$ and similarly $N = (a_{ij})_{i=1,...,n, j=n+1,...,2n}$. If $M$ does not have full rank, that means a linear combination of $c_i$ is in $B$ causing contradiction. Similarly $N$ must have full rank as well. That is, we can take $c_i$'s component in $A$ as the basis for $A$ and the same for $B$. Call that $a_i$ and $b_i$ respectively then we have $a_i+b_i=c_i$ for each $i$, hence the linear dependence.

Linear algebra is just that elegant.

B4. (EDIT 24/10/23) Ok I am back to solve B4 when my friend pings me back for discussion.

Obviously, every $a=b$ pair is a solution because you can easily make up $r = (n-\sqrt{n})^{1/a}$. Next, if $(a,b)$ is a solution then so as its multiples $(ka,kb)$ because you can take $r' = r^{1/k}$. 

As a result we can now assume that $a,b$ are coprime. As $r^a, r^b \in \mathbb{Q}[\sqrt{n}]$, we know that $r\in \mathbb{Q}[\sqrt{n}]$ as well. I think this is well known -- but in case that is not, think about Euclidean algorithm (recall that a field is Euclidean...or even simpler the algorithm where you find gcd) that you can divide each other until you reach $r^1 = r$. 

For the rest see my updated post.

C1. Cover most cup by adjusting the modulo offset. No quick way to solution but to check 30 combinations...or 15.

C2. Quite disgusting in my view...but not super hard. The following should be clear once you draw the diagram:

Here red and blue are not the underlying colors but for easier reading. For odd $2n+1$, you can along the axis draw consecutive triangles, exactly half of it colored blue and half red. Above those triangles are parallelograms with slopes 1 and $(k-1)/(k+1)$ (below the diagonal line, reflect to the other side for similar result), covering both colors half each.

And for even $n$, I think similar argument gives a no but I didn't write down concretely.


Well I know most unis do not put inequalities in standard materials but this is an insult to those with competitive exposure.

Rewrite sequence as product so that we minimize $\sum x_i$ where $\prod x_i = 2$. AM-GM gives the answer $n2^{1/n}$ and it suffices to take the limit. Write $2^{1/n} = e^{\log (2)/n} = 1 + \log (2) /n + \log ^2(2)/(2n^2) + ...$ we clearly sees that $n2^{1/n}-n \to \log 2$.

Easiest of all.

C4. Hmmm chess and game theory, gonna skip this one unless I want my puzzle rating falls further tonight.


If they are not going to change their way of drafting questions it does not make sense for me to criticize from the perspective of comparing against Putnam and other major competitions. Instead I shall compare this year's problems to that from the previous years.

Q1 is what I called the "7 points to everyone" or "free points so that they come back next year" question. That's fine. All three of them serves as an appropriate free question this year.

Q2 is the warm up question. B2 is perhaps a bit too straightforward but A2 and C2 brings abstractness to quite a level. In terms of difficulty they are consistent with previous Q2s.

Q3 is the real deal usually involving very long but reasonable (cf. Q4 being overly fancy) answers. I think A3 is up to that level (or slightly lower than if we go back to the graph problems few years ago). B3 is clearly not at Q3 level and C3 is a joke.

Q4 is the really hard question. A4 is nice (the "non open question Q4 being easier"), B4 looks very niche and C4 is crazy. All three are typical.

In overall it's quite consistent against past tests all in terms of scope, difficulty, and problems that are completely messed up, although I don't have full confident on my solutions for those Q3 and 4s that I have solved given how shockingly short it is. Surely this won't be the Pacific Putnam in another 50 years, but we can at least expect a few nice problems every year.

See you in 2024.

Friday 13 October 2023

夢.十夜 (9) Fatigue






這張七星對應的是現在的爬塔活動妖精之森(The Elder Tree),顧名思義這就是一直爬塔闖關的活動。每一層都有隨機生成的怪物或寶箱,每十層和百層都有對應的中BOSS和大BOSS。此外還有一些特殊規則比如收集一些碎片通往隱藏樓層有不同的掉落和分數加成等,但其實核心玩法就是訓練活動那樣。跟訓練活動一樣每爬一層都會消耗若干體力,不同的是這裡打輸了的話玩家會掉到下一層,怪物會保持殘血而重試的時候可以呼叫好友支援。










重點是在活動七星的加持下,整個活動變成了可以無腦刷下去就能贏的遊戲。不只這樣,這活動同時也是無課玩家絕對贏不了的遊戲。縱然有複雜的遊戲機制掩護,這仍然是赤裸的P2W(Pay to Win)活動。













咖啡廳以黑色為基調將店面分割成一個個小空間,每個空間裡都以黑色金屬椅搭配木製桌子,這種對比予人一種頗為自在的感覺。縱使店裡坐了不少兩人以上的顧客,坐在獨立空間裡面我也能毫不在意地享受自己獨處的時間。店裡播放的音樂是鋼琴版的Speak Softly, Love,原曲的那種厚重感被爵士節奏重構成一首輕快的小曲。不過能認出這首的顧客應該不多,大概是店主或者店員的口味吧?


我一邊一小口一小口地品著咖啡,一邊百無聊賴地在手機上滑著。如果不玩逆戰幻想,那我還可以玩甚麼呢?把Play Store裡面的營收排行榜拉出來看,逆戰幻想排在第五名。除它以外榜上並沒有太多社交式網遊,更多的是靠著社交平台傳播但本身沒有社交功能的小遊戲。比如某個把三消糖果的闖關遊戲,只要在facebook上求助就能收到體力的設計深得中年大叔大嬸的喜愛,輕鬆讓他們真的很需要過關時掏出錢包買那些貴死人的道具。又比如那個推銀機,單是可以不用錢在家裡玩這點已經吸引無數玩家,然而他們不知道的是他們所看的廣告已經讓開發者賺得盆滿砵滿了。


































回到主人身邊 等待第二朝的到來




祭典之後 百年之約即將完成
在血池之中 等待第二朝的到來









首先一個倒了的遊戲(連遊戲公司都改行了)實在沒甚麼好忌諱的,當然我不認為裡面有甚麼冒犯性的內容。對整個遊戲的評價我留到整篇收尾再講,但我想起坂口博信講到FF當初命名問題的報導:當初的FF本來是想叫Fighting Fantasy,後來因為商標問題而只能改成Final Fantasy卻造就一代經典。這個遊戲的中文名字逆戰幻想直譯自日文,好端端地硬要改成終戰也太奇怪了。



第一、「草藥獵人」瑪喬(Marjoram)這張卡,其實是我私心加進這個活動裡的卡片。在現實中它屬於「Mandragora March!」的活動卡片,這個是「萬聖Parade」外另一個帶我進坑和認識miso老師的訓練活動。不過章節所限我已經沒地方寫這個活動了,所以決定以這個形式把這張卡插進來。




最後說到那些刷榜玩家我想起一個類似的例子:偶像大師星光舞台(Im@s Cinderella Girl Starlight Stage)。不知是哪個喪心病狂發明了打音遊刷分的活動,而且稱號之類的獎勵只限前三甚麼的,於是激起了玩家之間慘烈的競爭……前幾名幾乎是七天二十四小時刷分,刷分的歌曲難度高但準確度不能掉。這種刷分法連Osu分數榜的觸手(望)都甘拜下風好嗎?最後他們通通筋膜炎入院就是。




Wednesday 20 September 2023

DDR diary: the BOOSTED mind

How silly was I. Here are the evidences:

June 12 -- passed Bitter Chocolate Strike CSP 2.25x. "Now with the crappy 2nd gen cab I can only do BPM 350 boost."
June 24 -- "I suddenly find BPM360 boost too slow and BPM300 intolerable."
August 30 -- Passed perditus†paradisus CSP 2.75x with an exhilarating 6:1 fast-slow ratio, a score of 550k. Lowest ever passing score.

Early September -- Every play ended up with notes overwhelmingly hit too fast. Change in offset (this is nice to have btw) helps nothing.

What happened?

Well, a closer inspection on the captures reveals the answer that took me too long to figure out: the damn boost is gone when I played perditus†paradisus. I has been using the default option since July till recently, and I played so bad in the period (not over) that I almost give up on playing any further.

In Bitter Chocolate Strike the speed was 2.25x BOOST, an effective speed of ~BPM560, slow but not intolerably slow even in the usual (i.e. good white cab) standard. And perditus†paradisus? Without BOOST would be BPM470 but if boosted would be close to 700, doable on white cab in the past (or rather, very comfortable for these hard songs), but very susceptible on this shitty cab.

Going back to what I said, I felt BPM350 BOOST comfortable in this cab, which is effectively BPM525, so BPM470 is rather slow in such standard. Together with the fact that my brain is so used with the BOOST flow that ended up with the historically low passing score. (Another possibility is that I changed my sneakers and perditus†paradisus was played in the first credit with the new sneakers, but the thickness of the new pair is similar to the old one so I don't think it's too much of a problem. As a contrast it took me months to get used with my last pair of sneakers.)

I finally realized and corrected such error last week and the result is clear. I started getting A+~AA- consistently across warm up levels (i.e. 13-15). So I started experimenting about the optimized setup seriously this time:

- Speed: BOOST is a must, now I say firmly say that. The problem is that I don't bother the extra cost for premium and BPM350 BOOST is too narrow of a range to get without the 0.25x multiplier. I found BPM380 (e.g. 御千手メディテーション or Possession) barely doable and BPM400 BOOST too much.

- Note colors: the sole reason I can't use higher BPM is the existence of unreadable blue arrows, so I really want the note option to minimize that. I tried NOTE again but then the deep blue arrows are simply impossible to read (to the point I almost failed a lv13 just by that and the most recent mystery is how did I passed 18s like Glitch Angel with that). RAINBOW is acceptable but the occasional purple-ish blue notes are still mildly annoying, and VIVID is similar. I will settle with FLAT for now, something I never would have imagined in the past.

To test if it worked I tried Fascination Eternal Love Mix ESP at 1.0x BOOST, which is BPM400 BOOST:

This is not too bad -- on a mild/bad/tiring day I might get similar result on a proper white cab. BPM400 boost is manageable with 1/1 notes at the beginning, but the double steps in the middle starts becoming hard and the last section is very hard to read given the speed and note option. The song is a clear indication the what works and what doesn't. The only thing yet to be verified is whether or not I played perditus†paradisus too nervously that I treated it more like BPM190 and hit notes too fast. To address that I should think about trying 18s again...

Monday 4 September 2023

古早遊戲BGM巡遊(5): 藏海村

最近我在把我的Youtube收藏清單從頭到尾聽一次,看看我以前都在聽甚麼東西。隨便舉例的話列表上在100、200、300和400位附近的分別是陳慧琳的星期五檔案、四月是你的謊言古典音樂集、beatstream的地方創生☆チクワクティクス和The Best of Brahms。第一千位呢?是One Winged Angel的交響樂版。即使這列表已經累積了十年以上的曲目,有些規律喜好還是始終如一:古典和遊戲音樂。



從個人角度來看,我好像對這種曠野民族風的音樂特別有感?不算Sun of Son也有Babylonia跟喜多郎的Dance Of Sarasvati(也就是尋找他鄉的故事主題曲)。也許是極度適應發達城市的我的情意結?也對啦,在Mt Cook躺在芒草上仰望著星空聽著這類音樂也十分應景。




但非純古風的關卡BGM也有一堆,比如雲夢湖底會令我想到JR的發車BGM water crown;塞外沙漠的波斯風則令人想起1987倚天屠龍記的熊熊聖火;八門八窟則以現代鼓聲將以笛子為主的旋律帶來強烈的動感。當然還有最輕快的新手流星村啦!


Saturday 19 August 2023

Formal framework of ordinal mate in infinite chess

It's quite rare for me to talk about a particular youtube video on twitter, let alone on this blog. The previous exception is Alan Becker's animation series which makes sense considering the long time nostalgia. But from a creator of 2 videos?

I guess I just can't refuse the combination of chess and maths.

Chess is my recent hobby, entering around the world championship 23'. I don't really play 10 min games on toilet tub like others, but at least I am doing the puzzles every day now rated close to 2000. And needless to say, math is my lifelong hobby (or occupation).

Ordinal is always an abstract idea and often comes with even more with its abstract applications except for the Riemannian hotel I guess. But the above video is quite a concrete realisation of what ω is. 

Let us examine the phrase "mate in ω" carefully. Consider the setup as shown in 3:02 of the video.

First, is it a mate-in-X? Yes, one side has a winning strategy in finite moves no matter what, and it's not a draw. Yes, finite moves -- this is the most fascinating concept in the whole thing, where you can create an ω with apparently finite stuffs. 

Next we would like to find X. Clearly X is larger than any natural numbers given those infinite board setups, so ω is a potential candidate. Is it ω+1, ω+2 or any aω+b? This is the first tricky bit in the whole argument. The steps we need to checkmate is a cardinal, a number. The cardinality of those infiniteness is are equal (cue Hilbert here), so we can't really say this is ω+1 and that is ω because that setup takes 1 more steps to checkmate.

Below is the framework I would suggest:

Definition 1. A setup is Mω (mate in ω) if each possible opponent move results in a mate in k where k is a positive integer, but the supremum of such k's does not exist in natural numbers.

The extension of the clock to ordinals is simply a min-max thing. For example let us think about the finite case so that we can extend it to the transfinite one. What does it mean by a setup of M6? That means for each opponent move the best you can move is to keep it at most M5 (with is at least one opponent move that makes it M5). Now in the transfinite case it would be the following:

Definition 2. A setup is Mγ if there is a corresponding self move for every opponent move such that the best self move results in a mate number γ' that is less than γ, and that supremum of all such (γ'+1) is γ.

Together with the fact that a setup has a mating clock only if the game can be won finitely, the above covers all setups with a clock by transfinite induction (unless you doesn't believe in ZFC...).

Theorem. The mating clock as in Definition 2 is well-defined for all setup that is winning. In other words, any setup is either a draw or a mate for either side.

Proof: Suppose not. Then there is a setup that is not a draw but without a mating clock. By Definition 2 that setup has an opponent move that no self move would render a new setup with well-defined mating clock. That creates an infinite descent, contradicting the fact that a winning setup is finite.

Ok we are now done with the abstract bullshit. What's more important is whether the definition fits the intuition? Like how we judge the clock in the video? The answer is a clear yes. When the clock is of non-zero constant term, the opponent's move is just a regular one: the best your opponent can do is to retain the clock. But when the clock value is a limit ordinal (with no predecessor), it is then the time for the opponent to make the 'announcement' -- which drops the limit ordinal to something lower. 

A Mω setup is clearly what we would expect intuitively: the opponent move is the announcement of Mk for k being finite.

A M(ω+1) is the setup that is one move away from Mω. The best the opponent can do is a move that keeps the board at M(ω+1) so that the self move would turn it to Mω.

A M(2ω) setup is where the opponent makes the announcement to turn it to M(ω+k) before it renders down to Mω, then to Mk.

And M(ω^2)? It is where the opponent makes the announcement to turn it into M(aω+b) (note that ω^2 is the supremum of all aω as well as of all aω+b). It's straightforward to verify that the setup at 6:58 exactly does that.

There is a separate problem though. Is there really a setup with mating clock Mγ for every γ? There are setups that are Mω and M(ω^2) respectively. Adding a constant is straightforward. How about M(2ω)? That is easy: add another double column of pawns from the example of Mω on 5:24. 

Before even thinking about larger ordinals common seen in logic problems like $\varepsilon _0$, the video already addressed that M(ω^3) is a problem. Is that so? My guess is that it's probably the case due to planarity (or that the chess board is 2-dimensional) and the fact that chess pieces can only move in a finite set of inclination (45 degrees for bishop, 90 for rook, and 63.4 for knights). Even if M(ω^3) is possible, there should be a concrete value k so that setup of M(ω^k) does not exist, for some small k like 4 or 5.

When I first tried to create the framework for the clock I made multiple approaches to the problem. How do we define the clock value? Mω is always obvious, but the intuition is a definition that identifies M(2ω) and M(ω^2) clearly.

I have thought of measuring the asymptotic of possible mating sequences, before realizing that one may put infinitely many garbage to disrupt the mating sequence. In that way we cannot get the constant coefficient right too, but the constant should be clear defined just like above -- regular move required before the ω-announcement. 

At the end the framework is so simple and elegant. What's more surprising is how close is it to how ordinals are built canonically. In other words, the way the mating clock is built in the exactly same way as how ordinals are constructed. How can you not love this analogy?


Before we conclude for today I just want to say the comment section is also treasury. Allow me to pick a few:

- Never realized how important the 50 moves rule is
- I want to flip the board but I can't on an infinite board
- Mathematicians finding how an infinite hotel ran out of rooms but here we are finding how a rook on infinite board ran out of checks

But the most interesting one is the one asking if we have an algorithm finding these ordinal mating value? Thinking again that we can put infinitely many garbage around I don't really have much should barely be on the horizon of being decidable if not undecidable (just my guess). Of course, another comment ask if infinite chess simulates logic gates as well. In that case, well, the halting problem will be what decides the complexity of this question.

Update (2024/1): they actually made a new video showing examples of M(ω^3) and M(ω^4)! They even claimed that M(ω^ω) or anything countably higher is possible! The examples are convincing, but the other claim requires some extra thoughts...

Tuesday 8 August 2023

夢.十夜 (X6) Fair



♪~一點一點增加濕度 滴汗是特別訊號
 很心急想找出路 但是並未做到
 似聽到看到海嘯 不需三秒便由旱土 改變做瀑布



♪~あゝ 君よ

鈴聲不合時宜地從少年的手機響起。手機被他調到震動模式,唯獨起床鬧鐘和少女來電都會響起這一首La prière的君よ。本來只是因為開頭的漸進音階作成鬧鐘鈴聲應該不錯,後來又覺得藍月なくる那個穿透的聲線十分醒神,聽著聽著就成了他的指定鈴聲。至於因為少女來電而響起這首歌,在這部手機上是第一次。










「很可惜沒有呢。我當時在對面的攤檔拿著一本George Orwell的Why I write中譯本一邊看一邊偷瞄作者到底有空了沒,一直到纏住作者的粉絲都離開再等作者回去休息室喝了口水出來我才撲過去找他,沒想到不到三五分鐘又有讀者找上門來了。只能說他的粉絲累積得越來越多了啊。」

「Why I Write你也看?不愧是Orwell的粉絲。不過這種東西回家看就好了嘛。」

「Why I Write其實很短,中文版還收錄了他幾篇其他散文。我看到裡面有政治與英語[Politics and the English Language]才打開來看的。機會難得,我想看看中文版怎樣翻譯那些例句。」



















♪~夜渡欄河再倚 北風我迎頭再遇
動盪如這海 城在兩岸凝神對視






♪~幸せ幸せよ 嬉しい気持ちが止まらない
建てよ 進め 高い塔を
建てよ 進め 高い塔を--


Thursday 27 July 2023

Enshittification of Reddit round 3: the place

We know from insiders and outsiders that Reddit management has always been a mess. But it's not necessary that every single move they make is stupid.

r/place is definitely a great idea on their perspective. Not well executed but also not bad, even pretty good in fact considering their standard in the beginning of the blackout. 

Reddit has been quite successful with their April fools. Some of the better recent ones are the button (2015), the imposter (2020), the second (2021 which I found it interesting mathematically), and of course the place (2017, 2022).

Both April fools r/place were a successful snapshot of various community hanging around Reddit. The 2022 one were welcomed and even more successful with the extended canvas got properly filled with daily active user pushed to the peak. Some may say this is such a successful event and as internet trend changes so fast these days it makes sense to make that an annual event right?

I think u/spez thought so, at least after all the attempts he tried to fought against those uncooperative mods. 

r/place was relaunched a few days ago without any warning with the tagline "right place, wrong time". Nice slogan, but people won't forget what was taken away by Spez. Many protest words were put on the canvas in multiple languages, and the most impressive one is probably the guillotine on reddit head marked spez, on a fitting French flag. This is just part of the story, though.

I lost interest after day 1, but a few conclusions hold throughout in my opinion. Let us go through one by one.

1) Dropping activity

The management may have wanted to take this as an advertisement on how attractive Reddit has been that within a single call everyone would be back. Sorry that didn't happen.

In the 2022 event, 160 millions pixels were placed in 3.5 days, but looking at the official count this year  numbers aren't even close, and this is before taking bots into account.

There is now a site that tracks Reddit activity including new post and comment counts, subs status and so on(which I forgot the website). Number wise the post and comment count remained largely unchanged since the blackout, but the owner of the site also warned that such statistics is deceptive as most posts and comments are in fact, meaningless trash in funky awkward subs. Services monitoring network flow shows that Reddit's receiving less attention since the blackout, and the ad-spaces are now filled with Reddit's own promotion. But we will never know these numbers accurately, and Reddit will always neglect such claims.

And now r/place has put things under sunlight, and it becomes the latest stone cold proof that this site is in fact not in a proper state.

2) Dwindling communities

Together with the dropping pixel count (and probably true participating user count) is the dropping of smaller communities. After the blackouts many subs were forced to reopen/unprivate/un-nsfw and so on. Some ended up archived, some had their mods replaced with the sub losing steam, some decided to let weeds grow on the sub, and some decided to move the whole community away (to lemmy or discord).

This is clearly seen on the earlier days of r/place 2023 when the number of participating niche communities is low. It only improved in later days with bigger canvas, where large communities locked their piece of land and started to help smaller communities as well as opening colour block for others to use.

Looking at Reddark, almost 2000 out of ~9000 subs are still private, and some are still John Oliver'd. If you were an investor of the Reddit IPO would you be convinced about Reddit's latest attempt of redemption?

3) Flags

National flags are always a big part of the canvas in 2017 and 2022, but this time it's even more obvious. It's not surprising that Germany, the US and France are top 3 on the global leader board considering how fast Germany and France are taking over new territories (insert WWII memes here). Turkey took fourth place, but they only managed to make a single proper flag (or a few) perhaps because a moon and a star are damn hard to draw. Canada, placed tenth, will absolutely agree(but credits to them coz they made a proper flag at the end). 

Why did flags flourish even more this time? A clear reason is that with less communities around, national identity does the best to bind users together.

4) Large ambitious communities are prepared

Together with nationalists are members of a few large non-reddit based communities (i.e. with a theme of an external entity which is well run so that the fondness is not affected by the blackout), they are prepared to make the best out of r/place whenever it's coming. 2023 July is an unexpected time, but they easily powered up the machine and dominated the canvas.

Germany has a discord server coordinating over 40k users, France are more or less similar. Osu! and Touhou fans are...simply everywhere like Finland snipers. The latter even collaborated to make a bad apple animation over the course. To accomplish that undisrupted requires tremendous effort and manpower needless to say.

But something is quite worrying about that if r/place are to be held again anytime in the future. If these communities decides to play in their own terms, are there any survival space for the rest?

When people realized team effort is overwhelming over a game they eventually team up to optimize to the point where individual effort are insignificant and the game is not fun anymore. It happens over and over again not only in r/place -- it's also in gacha games where you see those Chinese "workshops" dominates the game is simply depressing but this is a topic for another 10000 words of discussion (in fact my novel is a partial dedication to this phenomenon).

Oh and as usual there are communities unrelated to reddit trying to leave their mark on the canvas as well. Some random twitch streamers, I reckon.

5) Bots

While large communities are mostly actual active users (claimed to be, although some participants are there using bots too), bot activity is a big part of the game. The most obvious one is the 1337 building right next to Osu! which arrogantly occupied the space, "issuing warning" to anyone who wanted to fight back and highly synchronized and coordinated to the point that is impossible to be human controlled.

Reddit admin claimed that they wanted to fight bots but they refused to take action "because it's hard to distinguish between bots and highly coordinated humans". What?

Yeah you can say that they have accounts created way back in 2017 and 2022. You can say that they can farm pointless posts, comments and Karmas as well. But What about accounts created just these days then? Can you add restriction based on actual activity because not every bot account do the karma farming? Can you restrict API access like you did to the third party apps? Captcha before pixels?Bots can always squeeze through the leaks but the more obstacles you set the less effective they would be.

But Reddit did none of that and let bots do whatever they wanted to. All of that going to the fake number of traffic that would look good for their IPO. But again, such number is pure garbage if you know what's behind.

Reddit admin isn't completely wrong though. Bots are increasingly human-like behaviorally on the Internet. Captcha challenges isn't a challenge for them anymore (instead the system catches bot on other background information). The imposter event further confirms the evolution of bots. Many are in fact quite pessimistic on whether Internet would be a useful place in the future when you are not even interacting with humans. I definitely don't want to see measures like Google's plan to DRM our browsers though.

6) Censorship

Interestingly I thought this is somewhat expected even in 2017 and 2022 but did not become an issue. A probable reason is that more communities means that the canvas is more fragmented, leaving smaller room for that. To be honest finding inappropriate drawings is pretty normal for Reddit as a whole. 

I am fine taking NSFW as a no on this event considering r/place to be a place for all. Although that does not explain some of the most controversial censoring like Ronaldo's piss and Spez's guillotine. The way they censored those problematic figures are also partly why it looked bad when you have a blob of random pixels with no username suddenly splashed on the target.

If you need to censor things please do it in a more low-key way next time, but if there is a next time and of course the best is to let the communities to sort it out...


Reddit is not dead, yet. 

There are still a number of active users and communities around. The end product still shows a wide variety of cultures on it. It surely will be an important snapshot to look at when we wonder what the Internet looked like back in 2023.

But that does not change the fact that Reddit is going down the hill one step at a time, behind all those shiny (not really...just merely satisfactory?) fake numbers. Everything going just as the enshittification cycle and we will know the fate of Reddit in coming years.

Also, a big f to spez :).

Friday 21 July 2023

IMO 2023 quick thoughts

Ok the yearly event to review, IMO 2023. This time it was in Japan, hosted exactly at where comiket is held. This is kind of strange huh?

Q1. Many different approach towards the right idea. My first thought is that prime powers certainly works. Then the next simplest type of numbers to check is of the form $pq$ which...immediately fails with very simple numbers like 6 and 15. When going for 30, the product of 3 distinct primes, you quickly realizes where the problem is, then question solved.

I'm quite shocked by how easy this is due to how approachable the idea is. Here is the approach taken by most as on AoPS: observe that $d_{n-2}|d_n$ to force a chain of divisibility. It is also easy because how easy is it to check the examples and counterexamples, starting from small integers. The question gets an average score of almost 6 (5.85) shows how easy this is.

Q2. Okay I am not going to act like smart ass and solve. But frankly this is possibly one of the few geometry questions that I can actually solve at this level. There are too many parallel and perpendicular lines, trivializing angle and similar angle tracing. The extra triangle to construct is also pretty straightforward.

Q3. Very nice looking question. FToA strikes when it comes to solution of polynomials (just like the Vandermonde matrix), so you get ideas of recurring sequence (this is already non-trivial). Solution can proceed from here, details not discussed. After all this is quite a respectable Q3.

Q4. Yes finally, inequality is back! Also not in the traditional way with difficulty stacked because you guys can do AM-GM pretty easily with 3 terms but not with 2022+n terms.

The idea is obvious here. $a_n$ is strictly increasing and also increasing by at least 1 each time. But the number in question is 3034, indicating an increment of 1.5 per term -- or 3 every two terms! It is easy to calculate how much the increment needs to be to achieve an increment of 1, and this is where the pairwise different condition kicks in (which is pretty odd before this step). 

Of course, those inequality gurus already spot the obvious pattern of Cauchy-Schwatz: square roots and reciprocals. This is solvable directly using C-S, or AM-GM, or a combination of two. In overall a suitable and nice looking Q4. 

Q5. Combinatorics question and I skipped, but I saw the word ninja. I saw what you did, Japan ;)

Q6. The moment I saw the condition about 480 degrees and scalene triangles (and it took me some time to realize what's a scalene triangle) I knew immediately this a devilish question. Not only the angle sum condition is absurd, the graph itself is absurd too: $AA_1A_2$ are almost colinear, making the corresponding circumcircle (as well as the other two) incredibly huge. 

The thing I notice is that why two? What happens if we don't have a scalene triangle -- well then the almost colinear points becomes colinear and the problem collapses. From here the radical axis would come to help. Of course, proving two is another difficult thing: you prove one, then prove another, but you also need to prove that the two are distinct. Proving one seems to be the harder part though.

I am of course not qualified to talk about this problem, but quite funny that we find many possible solutions: pure geometry, inversion, coordinate geometry, bary bash and so on. The fact that the problem allows so many solution yet not many solved right (second hardest pure geometry Q6 since 2000), is the beauty of it.


The only thing I can say is that geometry is not my cup of tea, but I can feel the beauty of the two questions assigned this year. Yet, I always love to see a Q6 in number theory. The most elegant but lethal problems. 

I am less focused on competitive maths this year, but Simon Marais is again something I will eagerly wait for.

Sunday 9 July 2023

被青梅竹馬抓來(略) (8):我站的那邊才是正確的一方

Character design: @kuonyuu, Illust: @メリー commissioned by forretrio. Skeb
Editing and re-posting are prohibited // 無断転載、無断使用禁止です


























































進一步看風花雪月的話下一個問題就是,貝老師有甚麼可以教這堆學生的呢(此處不可色色)?用數據看的話我們的主角各種熟練度都和學生們沒兩樣……按現實的話就是一個實戰很強的二十歲傭兵但似乎只有純戰鬥經驗,似乎也沒有太多可以教的。看無雙裡面貝老師比較沒那麼無嘴,會多講一點自己的過去,不過距離一個全面的教師也還有一大段距離(當然另外那對夫妻檔(?)也沒好到哪裡去)。當然打通遊戲的話答案大家都知道:「是愛啊、哈利」(Love, Harry. Love)。



Character design: @kuonyuu, Illust: @miliL commissioned by forretrio. Pixiv
Editing and re-posting are prohibited // 無断転載、無断使用禁止です