Thoughts on CTWC2025

Hello and welcome to my CTWC thoughts 2025 edition.

I finally watched it live again. That's probably due to my huge interest on TGM4 listening to its OST and watching every pushing attempt, but that's topic for another day. Youtube on the other side, sensed my taste and started pushing everything about CTWC to me, from player interviews to day 1-2 qualis and of course, the finals on Sunday.

There were about 2k livestream watchers early in the finals. That number is surely much lower than the numbers we saw during 20-22 twitch live streaming, but that isn't a bad number either. It is hard to say whether it is better or worse comparing to 2024, this is something we can say much much later.

This year however, I would not cover anything further about change in popularity. I felt bad about missing technical contents in 2024. Sure I missed the tournament in live and surely wouldn't spend more than a few hours to re-watch the whole thing, but it is bad in the sense that the missing technical content created a hole, something hard to recover when I come back trying to figure out what was classic Tetris in 2024. Think about it -- I mentioned "lv29" ZERO times in my 2024 thoughts. How was that even possible?!?

I believe my argument on the tournament and my vision over the "endgame" is pretty well laid in the past years, so there is no need to emphasize that every year. Is the lv39 super kill screen killing all the fun? I might say yes in 2024 but this year I started to appreciate that -- we will talk more about that later.

As a compensation I want to simply focus on the players' performance and the shift in the meta. Let's start, shall we?

Gameplay before lv29

The community is getting so good that lv19 speed does not pose too much additional threat comparing to lv18 speed, so we may as well talk about both of them in one go.

Although I didn't say anything about gameplay last year, the progression from 2023 or since rolling is clear: optimize up to lv28 as much as possible. Given their ability to roll, it is possible to take near maximum risk and survive consistently. 

This was a goal, but this is precisely what top players are doing in 2025.

The best example is Tristop with how he maintains crazy (80%+) tetris rate in the majority of games. He even entered second transition once at 1.3M+ score and another time at near 1.3M, both would be tournament record breaking. It is unfortunate that we don't see him progressing further, because his ability of taking huge lead would be such a big threat aginst those who survives up to high 30s to hope opponents to top out.

Second transition and so on

I think this is what truly separates very good players (say, the top 32) and champion contenders. Assume that everyone survives up to lv28 what would you do?

They would have taken the risk if they can afford to, but the speed is simply too much for most players. What you see in the tournament is basically what they think is the best balance between risk and reward for players' self. If their strategy is not aggressive enough they can only hope that opponent would unfortunately top out -- if your opponent out-maneuvered you from lv29, chances are there he holds the lead before lv29 as well.

And for those title contenders, what's fascinating is that each of them take enough risk so that they probably top out before lv39 aka the second kill screen. If you think carefully, reaching lv39 is actually a waste because that means there is space for extra aggressiveness for potential extra tetrises that are worth more than potential score you can get up to lv39. Very few games actually ends with double second kill screen. I shall appreciate that as medal of players seriously taking this game and format to a very competitive level.

In short, a player's potent under such format is decided by their effectiveness in the last 10 levels. If you read the games in that way, you won't be surprised that Alex T won the tournament unbeaten at all. He is the only guy who can not only comfortably survive, but also consistently turning mess into tetris ready shapes. Again I have to mention Tristop here, he is the only one who can outperform Alex before lv29 and then try to survive Alex T's counterattack. (What about Meme? I think he is very well-rounded, unfortunately weaker than Alex on every edge, putting him in a very bad position in case of 1v1 matchup...)

Qualification and seeding

Remember how seeding works? It is all about getting maximum number of maxouts within a time limit. For some time I do not treat an accurate way to assign seeding, possibly due to impression from distant past where all you need is to get a few maxouts and chill.

Not anymore now.

Getting a few maxouts merely gets you into the gold bracket...before getting kicked out by top rankers even if you survived the top 48 stage. You need to press really hard to score 12 maxouts which all top 16 did. The aggressiveness is precisely what is needed in the knockout stage. 

I wasn't sure about the correlation between aggressiveness before and after lv29. Like...is it possible for a player to be very skillful that mastered lv19 speed but simply can't react after lv29? The tournament proves that these two are properly correlated i.e. it also properly seeds the player in the knockout stage.

At a point I wasn't sure about whether kicker makes sense. Is it better just to instakill a game to get a super small kicker (so that it isn't zero) than to lose intentionally at high pre-maxout score? Well I was certainly overthinking there. Most top players ended with a high kicker, not intentionally but the aggressiveness level naturally leave them a kicker like that. What about players with zero/low kicker? Well, that means they get an extra maxout instead of every single possible high kicker, which is, at the end of the day, more important.

The same qualification format has been used for a very long time, but I am glad to verify that it is actually a good one.

***

Before closing down my thoughts I just want to raise one last observation: lv29 gameplay is extremely stamina and concentration consuming. 

This is quite easily overlooked when you thought players could play casually up to a few million points (in the unlimited format), even up to the glitched levels or even close to reborn. Not only that such high level gameplay only applies to a few players, it is also about tension and pressure from the tournament. Scuti is one example where he slipped and misplaced bad piece on column 2 forcing him into rescue mode. If that happens on a player who reaches reborn, the same could have happened to anyone else.

When it comes to extreme e-sport, I always feel stamina being a very important factor. RTS games, MUGs, and of course competitive tetris. Perhaps a secret of Alex T's domination is his training from other sports...?

That's all! Another year, another fun weekend to enjoy. I will see you again in 2026.

CTWC2020

CTWC2021

CTWC2022 - general thoughts

CTWC2022

CTWC2023

CTWC2024

Log approximation 2: analytical approach

This is a sequel to my previous entry on approximating log values without calculators.

Why would I revive such a primitive topic? These questions certainly won't appear at high level Olpmpiads anyway. Well, when I came across to questions like this I always like to flex a bit. On top of that, the last entry was in 2019: six years later let us revisit the question from a higher perspective.

So yeah, let us forget about calculators and have fun with numbers. Also, log always refer to log base 10.

Problem 1. Find the first 2 digits of $A = 41^{25}$.

Recall the few well known log constants: $\log 2 \approx 0.30103$, $\log 3 \approx 0.47712$, and $\log 7 \approx 0.84510$. The last one might be less familiar but shouldn't be too hard to approximate up to good precision (which I demonstrated in my previous entry).

From here you know $\log 40 = 1+2\log 2 \approx 1.60206$ and $\log 42 = \log 2 + \log 3 + \log 7 \approx 1.62325$, so $\log 41 \approx 1.612655$. Now $\log A = 25 \log 41 \approx 40.316375$, which we only care about the decimals. 

How do we compute $10^{0.316375}$? Well we don't even need to. Note that $\log 20 < 1.316375 < \log 21$ which shows that the first two digits are 20.

Now, what if the question asks the first 3 digits of $41^{2025}$? You may try but the above trick fails simply due to lack of precision. 5 decimal places aren't enough when you are going to multiply with 2025 and ask for 3 digits precision (recall the asymptotics of log).

In order to have a glimpse on the precision we need, let us consider the original problem again but with a twist -- we want to do it rigorously.

Problem 2. Find the first 2 digits of $A = 41^{25}$, provided that $\log 2 \approx 0.30103$, $\log 3 \approx 0.47712$, $\log 7 \approx 0.84510$, $\ln 10 \approx 2.30259$ are accurate up to 5 decimal places.

We can express the upper and lower limit of our approximations as follows:

$\log 2 \in (0.301025, 301035)$

$\log 3 \in (0.477115, 0.477125)$

$\log 7 \in (0.845095, 0.845105)$

$\log 40 = 1+2\log 2 \in (1.60205, 1.60207)$

$\log 42 = \log 2 + \log 3 + \log 7 \in (1.623235, 1.623265)$

What about the possible range of $\log 41$?

Surely we have $\log 41 > \frac{1}{2}(\log 40 + \log 42) \geq 1.612642$ by convexity but the upper bound is less obvious and we need the power of calculus again. Note that 

$\ln x^2 = \ln (x^2-1) + \ln (1 + \frac{1}{x^2-1}) \leq \ln (x^2 - 1) + \frac{1}{x^2-1}$.

Here the constant $\ln 10$ kicks in:

$\log 41 \leq \frac{1}{2}(\log 40 + \log 42 + \frac{1}{1680\ln 10}) \leq 1.612798$.

Gathering the bounds we have 

$\log 41 \in (1.612642, 1.612798)$. 

Notice how the gap suddenly widens with a single log interpolation.

From here we multiply and get

$\log A \in (40.31605, 40.31995)$.

Since $\log 21 = \log 3 + \log 7 > 1.32223$, we know that $A$ must start with 20.

Our approximation almost failed! I made a mistake and forgot to halve the difference $\frac{1}{1680 \ln 10}$. With that mistake the upper bound of $\log A$ would exceed 40.323 failing the argument. The fact that the actual value of $A$ starts with 2087... only makes it worse. 

*

Not enough?

Then how about doing the calculation from scratch, without the assumption of the log constants?

Problem 3. Estimate $\ln 2$ up to 5 decimal places. (0.693147...)

There are so little we can do without advanced calculations:

- One may think about Newton's method but forgetting we can't calculate exponents. 

- One may also think about the series of $\log (1+x)$, but the expansion at $x=1$ converges too damn slow (we need like a million terms...). 

- We can also try $x = \sqrt{2}-1$ for $\ln \sqrt{2}$ hoping for quicker convergence. It indeed converges quickly but $\sqrt{2}$ is simply another nasty number to approximate. We could group the positive and negative terms and find the sum separately but then we run into hyperbolic inverse trigonometric functions (!!).

- Back to basic we have the integration $\ln 2 = \int ^2_1 x^{-1}dx$, but the problem what kind of partition do we need for such precision? Say, the upper bound is by tripizodal estimation and the lower bound by rectangular estimation. Assume a uniform partition of $N$ parts. Then the error is given by 

$E = \sum \frac{1}{2}(\frac{1}{x_i}-\frac{1}{x_i+N^{-1}}) \cdot \frac{1}{N} \geq \frac{1}{2(N+1)}$.

And to have a $N$ large enough for $E< 10^{-5}$ is just...ridiculous for manual computation.

Is all hope lost? Not yet. We need a series that projects $(0,1)\mapsto (1,\infty)$ so that we can enjoy geometric convergence. What's better than the series $\ln \frac{1+x}{1-x} = 2 \sum _{k=0}^{\infty}\frac{x^{2k+1}}{2k+1}$? The value of $\ln 2$ corresponds to $x = \frac{1}{3}$ and a quick convergence is guaranteed. 

We compute the first five terms:

$\ln 2 \approx 2\cdot (\frac{1}{3} + \frac{1}{81} + \frac{1}{1215} + \frac{1}{15309} + \frac{1}{177147}) = \frac{4297606}{6200145} \geq 0.693146$.

The above is taken as a lower bound. As for the upper bound, we loosely say that sum of the tail is less than, say 1.5 times the first remainder term, i.e. $3 \cdot (3^{11} \cdot 11)^{-1} \leq 2 \times 10^{-6}$, meaning that $\ln 2 \leq 0.693148$. Therefore the approximation $0.69315$ is accurate up to 5 decimal places.

What about $\ln 7$ then? We can easily estimate that using $\ln 7 = \ln 8 + \ln \frac{7}{8}$ with the help of the series of $\ln (1-x)$, but the brilliance of AI kicks in: we can also flip that into $-\ln \frac{8}{7}$ and apply the series of $\ln \frac{1+x}{1-x}$ again. We would end up with a smaller ratio, hence quicker convergence. 

I am not going to do all the calculations, and I am not going to ask you to try as well -- knowing how is far more important than being able to, especially when you know you actually have more tools around than simple calculators.

Lessons for today is ummm...I guess...there was a reason behind why people needed slide rule or log tables? 

No but seriously, the same question can be asked for many common functions and have very practical applications (except that we loosen the scope from hand computable to quick algorithm for computers). 

According to Wikipedia, the last series approximation for log is indeed accepted as an accurate and quickly converging approach in approximating log values. Of course there are other perhaps better approach like the use of AG-mean, but that is out of the scope of manual calculation. The only other possible mean by hand that never came to my mind is the use of Padé approximant, as laid out in this MSE question, which looks pretty accurate too.