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Igalia's contribution to the Mozilla project and Open Prioritization

As many web platform developer and Firefox users, I believe Mozilla’s mission is instrumental for a better Internet. In a recent Igalia’s chat about the Web Ecosystem Health, participants made the usual observation regarding this important role played by Mozilla on the one hand and the limited development resources and small Firefox’s usage share on the other hand. In this blog post, I’d like to explain an experimental idea we are launching at Igalia to try and make browser development better match the interest of the web developer and user community.

Open Prioritization by Igalia. An experiment in crowd-funding prioritization.

Igalia’s contribution to browser repositories

As mentioned in the past in this blog, Igalia has contributed to different part of Firefox such as multimedia (e.g. <video> support), layout (e.g. Stylo, WebRender, CSS, MathML), scripts (e.g. BigInt, WebAssembly) or accessibility (e.g. ARIA). But is it enough?

Although commit count is an imperfect metric it is also one of the easiest to obtain. Let’s take a look at how Igalia’s commits repositories of the Chromium (chromium, v8), Mozilla (mozilla-central, servo, servo-web-render) and WebKit projects were distributed last year:

pie chart
Diagram showing, the distribution of Igalia's contributions to browser repositories in 2019 (~5200 commits). Chromium (~73%), Mozilla (~4%) and WebKit (~23%).

As you can see, in absolute value Igalia contributed roughly 3/4 to Chromium, 1/4 to WebKit, with a small remaining amount to Mozilla. This is not surprising since Igalia is a consulting company and our work depends on the importance of browsers in the market where Chromium dominates and WebKit is also quite good for iOS devices and embedded systems.

This suggests a different way to measure our contribution by considering, for each project, the percentage relative to the total amount of commits:

Bar graph
Diagram showing, for each project, the percentage of Igalia's commits in 2019 relative to the total amount of the project. From left to right: Chromium (~3.96%), Mozilla (~0.43%) and WebKit (~10.92%).

In the WebKit project, where ~80% of the contributions were made by Apple, Igalia was second with ~10% of the total. In the Chromium project, the huge Google team made more than 90% of the contributions and many more companies are involved, but Igalia was second with about 4% of the total. In the Mozilla project, Mozilla is also doing ~90% of the contributions but Igalia only had ~0.5% of the total. Interestingly, the second contributing organization was… the community of unindentified gmail.com addresses! Of course, this shows the importance of volunteers in the Mozilla project where a great effort is done to encourage participation.

Open Prioritization

From the commit count, it’s clear Igalia is not contributing as much to the Mozilla project as to Chromium or WebKit projects. But this is expected and is just reflecting the priority set by large companies. The solid base of Firefox users as well as the large amount of volunteer contributors show that the Mozilla project is nevertheless still attractive for many people. Could we turn this into browser development that is not funded by advertising or selling devices?

Another related question is whether the internet can really be shaped by the global community as defended by the Mozilla’s mission? Is the web doomed to be controlled by big corporations doing technology’s “evangelism” or lobbying at standardization committees? Are there prioritization issues that can be addressed by moving to a more collective decision process?

At Igalia, we internally try and follow a more democratic organization and, at our level, intend to make the world a better place. Today, we are launching a new Open Prioritization experiment to verify whether crowdfunding could be a way to influence how browser development is prioritized. Below is a short (5 min) introductory video:

I strongly recommend you to take a look at the proposed projects and read the FAQ to understand how this is going to work. But remember this is an experiment so we are starting with a few ideas that we selected and tasks that are relatively small. We know there are tons of user reports in bug trackers and suggestions of standards, but we are not going to solve everything in one day !

If the process is successful, we can consider generalizing this approach, but we need to test it first, check what works and what doesn’t, consider whether it is worth pursuing, analyze how it can be improved, etc

Two Crowdfunding Tasks for Firefox

CIELAB color space*
Representation of the CIELAB color space (top view) by Holger Everding, under CC-SA 4.0.

As explained in the previous paragraph, we are starting with small tasks. For Firefox, we selected the following ones:

  • CSS lab() colors. This is about giving web developers a way to express colors using the CIELAB color space which approximates better the human perception. My colleague Brian Kardell wrote a blog with more details. Some investigations have been made by Apple and Google. Let’s see what we can do for Firefox !

  • SVG path d attribute. This is about expressing SVG path using the corresponding CSS syntax for example <path style="d: path('M0,0 L10,10,...')">. This will likely involve a refactoring to use the same parser for both SVG and CSS paths. It’s a small feature but part of a more general convergence effort between SVG and CSS that Igalia has been involved in.

Conclusion

Is this crowd-funded experiment going to work? Can this approach solve the prioritization problems or at least help a bit? How can we improve that idea in the future?…

There are many open questions but we will only be able to answer them if we have enough people participating. I’ll personally pledge for the two Firefox projects and I invite you to at least take a look and decide whether there is something there that is interesting for you. Let’s try and see!

Contributions to Web Platform Interoperability (First Half of 2020)

Note: This blog post was co-authored by AMP and Igalia teams.

Web developers continue to face challenges with web interoperability issues and a lack of implementation of important features. As an open-source project, the AMP Project can help represent developers and aid in addressing these challenges. In the last few years, we have partnered with Igalia to collaborate on helping advance predictability and interoperability among browsers. Standards and the degree of interoperability that we want can be a long process. New features frequently require experimentation to get things rolling, course corrections along the way and then, ultimately as more implementations and users begin exploring the space, doing really interesting things and finding issues at the edges we continue to advance interoperability.

Both AMP and Igalia are very pleased to have been able to play important roles at all stages of this process and help drive things forward. During the first half of this year, here’s what we’ve been up to…

Default Aspect Ratio of Images

In our previous blog post we mentioned our experiment to implement the intrinsic size attribute in WebKit. Although this was a useful prototype for standardization discussions, at the end there was a consensus to switch to an alternative approach. This new approach addresses the same use case without the need of a new attribute. The idea is pretty simple: use specified width and height attributes of an image to determine the default aspect ratio. If additional CSS is used e.g. “width: 100%; height: auto;”, browsers can then compute the final size of the image, without waiting for it to be downloaded. This avoids any relayout that could cause bad user experience. This was implemented in Firefox and Chromium and we did the same in WebKit. We implemented this under a flag which is currently on by default in Safari Tech Preview and the latest iOS 14 beta.

Scrolling

We continued our efforts to enhance scroll features. In WebKit, we began with scroll-behavior, which provides the ability to do smooth scrolling. Based on our previous patch, it has landed and is guarded by an experimental flag “CSSOM View Smooth Scrolling” which is disabled by default. Smooth scrolling currently has a generic platform-independent implementation controlled by a timer in the web process, and we continue working on a more efficient alternative relying on the native iOS UI interfaces to perform scrolling.

We have also started to work on overscroll and overscroll customization, especially for the scrollend event. The scrollend event, as you might expect, is fired when the scroll is finished, but it lacked interoperability and required some additional tests. We added web platform tests for programmatic scroll and user scroll including scrollbar, dragging selection and keyboard scrolling. With these in place, we are now working on a patch in WebKit which supports scrollend for programmatic scroll and Mac user scroll.

On the Chrome side, we continue working on the standard scroll values in non-default writing modes. This is an interesting set of challenges surrounding the scroll API and how it works with writing modes which was previously not entirely interoperable or well defined. Gaining interoperability requires changes, and we have to be sure that those changes are safe. Our current changes are implemented and guarded by a runtime flag “CSSOM View Scroll Coordinates”. With the help of Google engineers, we are trying to collect user data to decide whether it is safe to enable it by default.

Another minor interoperability fix that we were involved in was to ensure that the scrolling attribute of frames recognizes values “noscroll” or “off”. That was already the case in Firefox and this is now the case in Chromium and WebKit too.

Intersection and Resize Observers

As mentioned in our previous blog post, we drove the implementation of IntersectionObserver (enabled in iOS 12.2) and ResizeObserver (enabled in iOS 14 beta) in WebKit. We have made a few enhancements to these useful developer APIs this year.

Users reported difficulties with observe root of inner iframe and the specification was modified to accept an explicit document as a root parameter. This was implemented in Chromium and we implemented the same change in WebKit and Firefox. It is currently available Safari Tech Preview, iOS 14 beta and Firefox 75.

A bug was also reported with ResizeObserver incorrectly computing size for non-default zoom levels, which was in particular causing a bug on twitter feeds. We landed a patch last April and the fix is available in the latest Safari Tech Preview and iOS 14 beta.

Resource Loading

Another thing that we have been concerned with is how we can give more control and power to authors to more effectively tell the browser how to manage the loading of resources and improve performance.

The work that we started in 2019 on lazy loading has matured a lot along with the specification.

The lazy image loading implementation in WebKit therefore passes the related WPT tests and is functional and comparable to the Firefox and Chrome implementations. However, as you might expect, as we compare uses and implementation notes it becomes apparent that determining the moment when the lazy image load should start is not defined well enough. Before this can be enabled in releases some more work has to be done on improving that. The related frame lazy loading work has not started yet since the specification is not in place.

We also added an implementation for stale-while-revalidate. The stale-while-revalidate Cache-Control directive allows a grace period in which the browser is permitted to serve a stale asset while the browser is checking for a newer version. This is useful for non-critical resources where some degree of staleness is acceptable, like fonts. The feature has been enabled recently in WebKit trunk, but it is still disabled in the latest iOS 14 beta.

Contributions were made to improve prefetching in WebKit taking into account its cache partitioning mechanism. Before this work can be enabled some more patches have to be landed and possibly specified (for example, prenavigate) in more detail. Finally, various general Fetch improvements have been done, improving the fetch WPT score. Examples are:

What’s next

There is still a lot to do in scrolling and resource loading improvements and we will continue to focus on the features mentioned such as scrollend event, overscroll behavior and scroll behavior, lazy loading, stale-while-revalidate and prefetching.

As a continuation of the work done for aspect ratio calculation of images, we will consider the more general CSS aspect-ratio property. Performance metrics such as the ones provided by the Web Vitals project is also critical for web developers to ensure that their websites provide a good user experience and we are willing to investigate support for these in Safari.

We love doing this work to improve the platform and we’re happy to be able to collaborate in ways that contribute to bettering the web commons for all of us.

Transcendental number from the oscillating zeno's paradox

Oscillating Zeno’s paradox

In a previous blog post, I described an oscillating Zeno’s paradox, which can be formalized as follows. Atalanta moves on the real line. She leaves from 0 and at step nn she decides to move forward or backward by 12n+1\frac{1}{2^{n+1}}. If ϵn{1,1}\epsilon_n \in {\{-1, 1\}} corresponds to the chosen direction then writing S0=0S_0 = 0 and

Sn+1=Sn+ϵn2n+1 S_{n+1} = S_n + \frac{\epsilon_n}{2^{n+1}}

we obtain sequence of Atalatan’s position on the real line.

In the non-oscillating case (ϵn=1\epsilon_n = 1 for all nn) then this just corresponds to a geometric series of common ration 12\frac{1}{2}. SnS_n converges to 1 by lower values, with remaining distance being 12n\frac{1}{2^n}. More generally, SnS_n is just the partial sum of an absolutely convergent series and so is convergent to a destination S=n=0+ϵn2n+1S_\infty = {\sum_{n=0}^{+\infty} \frac{\epsilon_n}{2^{n+1}}}. It is easy to see that |Sn|<1\left|S_n\right| < 1, |S|1\left|S_\infty\right| \leq 1 and |SSn|12n{\left| S_\infty - S_n \right| \leq \frac{1}{2^n}}. If additionally ϵn\epsilon_n is chosen so that it is not ultimately constant (i.e. Atalatan’s position keeps oscillating) then this becomes a strict inequality |SSn|<12n{\left| S_\infty - S_n \right| < \frac{1}{2^n}}.

In theory, it is possible to reach any destination x[1,1]x \in {[-1,1]} using this approach. Just define ϵn=1\epsilon_n = 1 if and only if SnxS_n \leq x (i.e. “move toward xx”). It is easy to prove by recurrence that |xSn|12n{\left| x - S_n \right| \leq \frac{1}{2^n}} for all nn and so S=xS_\infty = x. However, this assumes we already know xx in order to decide the sequence of directions ϵn\epsilon_n. Let’s see how to build sequences without knowing the destination.

Avoiding an at most countable subset

Suppose that XX \subseteq \mathbb R is at most countable and let (xn)n{(x_n)}_{n \in \mathbb N} be a sequence of real numbers such that X={xn:n}X = \left\{ x_n : n \in \mathbb N \right\}. We want to find a sequence such that SXS_\infty \notin X.

Assuming additionally that it has infinitely many terms above 11 and infinitely many terms below 1-1, then we obtain a non-ultimately-constant sequence by defining ϵn=1\epsilon_n = 1 if and only if SnxnS_n \geq x_n (i.e. “move away from xx”). This is because 1<Sn<1-1 < S_n < 1 so there are always infinitely many terms in front of and behind Atalanta.

Incidentally, X=X = \mathbb N with trivial enumeration xn=nx_n = n shows a simple counter-example with no term below 1-1 and ultimately constant ϵn\epsilon_n. In that case, S0=0S_0 = 0 and S1=12S_{1} = \frac{1}{2} and Sn+1=Sn12n+1S_{n+1} = S_{n} - \frac{1}{2^{n+1}} for n1n \geq 1. But S=0S_\infty = 0 \in \mathbb N.

To workaround that issue, we can alternatively define the sequence ϵ2n=1\epsilon_{2n} = 1 if and only if SnxS_n \geq x and ϵ2n+1=ϵ2n\epsilon_{2n+1} = -\epsilon_{2n}. This still moves away from all xnx_n once, keeps oscillating and does not require additional assumption on XX.

Whatever the decision chosen, we can show that SXS_{\infty} \notin X. Otherwise, consider one NN such that ϵN\epsilon_{N} is defined by moving away from SS_{\infty}. Then either SNS0{S_N - S_\infty} \geq 0 and SN+1=SN+12N+1S+12N+1S_{N+1} = {S_N + \frac{1}{2^{N+1}}} \geq {S_\infty + \frac{1}{2^{N+1}}} or SN+1=SN12N+1<S12N+1S_{N+1} = {S_N - \frac{1}{2^{N+1}}} < {S_\infty - \frac{1}{2^{N+1}}}. But in any case, we would have |SSN+1|12N+1{\left| S_\infty - S_{N+1} \right| \geq \frac{1}{2^{N+1}}} which contradicts our strict inequality for not ultimately constant ϵn\epsilon_n.

As explained in the previous blog, this shows that \mathbb R (or any complete subspace containing \mathbb Q) is uncountable. Otherwise, we could use a sequence enumerating it to arrive to a contradiction.

Similarly, because \mathbb Q is countable then we can find a sequence such that {xn:n}=\left\{ x_n : n \in \mathbb N \right\} = \mathbb Q, which gives us an irrational number SS_\infty. If instead XX is the set of algebraic numbers (which is countable and contains \mathbb Q) then we obtain a transcendental SS_\infty.

An explicit computation

The big difference between the case of \mathbb Q and the set of algebraic numbers is that it’s much easier to actually compute a list of (xn)n{(x_n)}_{n \in \mathbb N} that lists all the \mathbb Q. For example, a simple way to enumerate the fractions pq\frac{p}{q} without trying to avoid duplicate values is, for each M=0,1,2,...M=0, 1, 2, ... to enumerate the integers pp and q0q \neq 0 bounded by MM in the lexicographical order. One can similarly enumerate for each MM the non-constant polynomial of degree at most MM and integer coefficient bounded by MM but it’s not obvious how to actually compute their roots.

One can rely on root finding algorithm to estimate the roots α\alpha in order to be able to calculate the sign SnαS_n - \alpha and so determine ϵn\epsilon_n. But in theory, α\alpha can be arbitrary close to SnS_n so it’s not clear how much precision to request.

Instead, let’s consider (Pn)n{(P_n)}_{n \in \mathbb N} an enumeration of the non-constant polynomial with integer cofficients and define ϵn=1\epsilon_n = 1 if and only if (PnPn)(Sn)0{(P_n\prime P_n)}{(S_n)} \geq 0. With that new formula, ϵn\epsilon_n can be calculated very easily in O(degPn)O(\deg P_n) from the cofficients of PnP_n using Horner’s method.

For any integer M1M \geq 1, let’s consider QM±=(X±2)MQ_M^\pm = {(X \pm 2)}^M. Then its product with its derivative is M(X±2)2M1M {(X \pm 2)}^{2M -1}. Its value at any xx has the same sign as x±2x \pm 2 and if |x|<1\left|x\right| < 1 it just ±\pm. For any nn, we can find MM large enough such that QM+Q_M^+ and QMQ_M^- have not been enumerated yet. So ϵn\epsilon_n is not ultimately constant. Alternatively, we could have used the trick from the previous section to force that property.

Now let’s prove that SS_\infty is still algebraic. If that’s not the case, let nn such that Pn(S)=0{P_n(S_\infty)} = 0. We can choose nn such that PnP_n is of minimal degree and so Pn(S)0{P_n\prime(S_\infty)} \neq 0. Since PnP_n has only finitely many roots, there is knk \geq n such that at maximum distance 12k\frac{1}{2^k} around SS_\infty, SS_\infty is the only root of PnP_n and PnP_n\prime does not vanish. Let’s MM large enough such that the polynomial RM=Pn2M+1R_M = P_n^{2M+1} has not been enumerated before PkP_k. Let’s ll such that Pl=RMP_l = R_M.

In the considered neighbourhood of SS_\infty, PlP_l has only one root SS_\infty, has the same sign as PnP_n and its derivative (2M+1)Pn2MPn{(2M+1)}P_n^{2M} P_n\prime has the same nonzero constant sign as PnP_n\prime. Analyzing each combination PlP_l\prime positive/negative and SlS_l greater/smaller than SS_\infty, we verify that ϵl\epsilon_l is chosen such that |SSl+1|12l+1{\left| S_\infty - S_{l+1} \right| \geq \frac{1}{2^{l+1}}} which is obviously also true if Sl=SS_l = S_\infty. This agains contradict our strict inequality. Q.E.D.

A Zeno paradox to prove the reals are uncountable

A simple Zeno’s paradox

A simple variant of Zeno’s paradoxes can be described as follows. Atalanta wishes to walk to the end of a path. When she gets halfway, she still have to walk the remaining half of the path ; When she gets halfway of that remaining half, she still needs to walk the remaining quarter of the path ; When she gets halfway of that remaining quarter, she still needs to walk the remaining eight of the path ; and so on. Hence it seems she will never reach the end of the path, which is paradoxical.

Schema for classical Zeno's paradox

Obviously, this is not true: Atalanta is going to reach the end of the path after a certain amount of time. The issue in the previous reasoning is that (assuming she walks at constant velocity) walking these subpaths also takes her respectively half, a quarter, a eight, etc of the total time she actually needs to arrive to her destination… and the observations are only done for a total time that is less than the one needed.

A modern variant

Now suppose the ground is an infinite real line, itself covered by an infinite treadmill on which Atalanta is walking. This treadmill brings Atalanta in the opposite direction by twice her velocity. She can decide to turn the treadmill on or off, but its state must remain the same during the whole walk of a subpath. In the following schema, the treadmill was enabled when she was on the second subpath:

Schema for modern variant of Zeno's paradox

After the total duration considered in the previous problem, Atalanta has walked exactly the same total distance on the treadmill. The conveyor belt also moves by summing up distances that are twice the distances on subpaths. Because the treadmill can be on or off, the conveyor’s belt position is not following a simple linear function of the time. Let’s admit for a while it still converges to a certain distance when getting closer to the total duration, I will come back to this at the end of the blog post.

With respect to the ground, Atalanta’s position is basically given by her departure position, moved backward by the distance of the conveyor belt with respect to the ground and moved forward by the distance she walked on the treadmill. As a consequence of the previous paragraph, Atalanta is also converging to a destination with respect to the ground.

Modern Zeno’s paradox (with a flaw)

Now consider an enumeration of all the points of the real line. At the beginning the treadmill is off. Before walking the first half of the path, Atalanta picks the first point in that enumeration and enables the treadmill if she can see that point in front of her. Before walking the next quarter of the path, she picks the second point and enables or disables the treadmill according to whether she can see that second point in front of her. She continues that way for each subpath and each point of the real line.

When a given point is picked there are two options. Either it is not in front of her and so she will just walk forward to a certain distance ; or otherwise the the treamill will additionnaly take her backward by twice that distance. In any case, with respect to the ground, she is moving away from the picked point by the distance she walks during that subpath.

Schema for modern variant of Zeno's paradox (picking a point behind or in front of Atalanta)

At some step, the point of the real line corresponding to Atalanta’s destination will be picked and she will move away from that destination by a certain distance. But the remaining subpaths are half, a quarter, a eight, etc that distance. To have a chance to converge to the destination again she must now always keep the same direction. But so far, only finitely many points have been picked and so there is at least one point beyond the destination that has not been chosen yet (actually infinitely many). This means she will have to change her direction at least once in the future and so will never be able to reach her destination!

Uncountability of the real line

How to solve the paradox in this modern variant?

First, let’s go back to why the distance of the conveyor belt with respect to the ground is convergent. Consider the binary number whose digits describe the sequences of states of the treadmill: 0 if turned off and 1 if turned on. Place the binary point after the first digit to obtain a real number. Then one can check that the distance of the conveyor belt is given by that real number multiplied by Atalanta’s total distance on the treadmill. Of course, one must admit that such a binary number with infinitely many digits after the binary point is well defined!

For a rigorous proof, note that the conveyor belt is always going in the same direction and at most twice Atalanta’s total walking distance on the treadmill. So the convergence of the conveyor belt just comes from the monotone convergence theorem. This assumption was correct.

However, reading more carefully the argumentation, we actually assumed that there is a countable enumeration of all the points of the real line. But Cantor proved at the end of the 19th century that this is wrong, the real line is uncountable!

It is interesting to actually turn this modern Zeno’s paradox into an apparently “elementary” alternative of Cantor’s diagonal argument, in a way that involves only concepts understandable by the Ancient Greeks and no advanced mathematical formalisms. In particular, this is not mentioning explicitly the positional notation invented by Indian mathematicians. As a comparison, Cantor’s proof assumes that a decimal notation with infinitely many digits after the decimal point is well defined (similar to what we used for our proof with binary numbers) ; or it must deal with the edge case of a real number with two different decimal notations (e.g. 0.4999999… = 0.5).

For completeness

Nevertheless, we still had to use some kind of completeness property in order to prove that the moving distance of the conveyor belt is convergent. Actually, Cauchy completeness is enough.

It is also interesting to note that all the moves as well as Atalanta’s total distance on the treadmill are rational numbers. Since the rational numbers are countable, this modern Zeno’s paradox can also be performed by considering only rational points on the ground. In that case, the flaw is then in the assumption that the conveyor belt converges to a rational point: the set of rational numbers is not complete.

More generally, we can also applie this modern Zeno’s paradox to any ordered field since they include a subset isomorphic to the rational numbers. The conclusion becomes that if the field is Cauchy-complete then it is uncountable.

S[α] for strings of ordinals

Update 2020/03/10: Added complexity analysis for S.substr(α, N)

In a previous blog post, I defined for each ordinal α\alpha a string SαS_\alpha (made of the characters for the empty set, comma, opening brace and closing brace) that enumerates the element of α\alpha. I gave a simple formulas to calculate the length LαL_\alpha of this string.

My colleague Ioanna was a bit disappointed that I didn’t provide a script for calculating the infinite SαS_\alpha strings. Obviously, the complexity would be “Ω(ω)\Omega(\omega)” but it is still possible to evaluate the string at a given position: Given β<Lα\beta < L_\alpha, what is the character of SαS_\alpha at position β\beta?

Since the initial segments of the strings are compatible, another way to express this is by introducing the class S=αOrdSαS = \bigcup_{\alpha \in {\mathrm{Ord}}} S_\alpha corresponding to a giant string enumerating the class Ord\mathrm{Ord} of all ordinals. Given an ordinal α\alpha, what is S[α]S[\alpha]?

A small generalization of this S.charAt(α) operation is S.substr(α, N) calculating the substring of length NN starting at α\alpha.

Example

  • S2S_{2} is "∅,{∅}", so S[0]S[0] is the character ∅, S[1]S[1] is a comma, S[2]S[2] is an opening brace and S[4]S[4] is a closing brace.
  • Sω+1S_{\omega+1} is made of of an ω\omega-concatenation of finite strings (the character ∅, a comma, S1S_1 surrounded by braces, a comma, S2S_2 surrounded by braces, a comma, S4S_4 surrounded by braces, a comma, etc), followed by a comma, an opening brace, the same ω\omega-concatenation of finite strings and finally a closing brace. So S[ω]S[\omega] is a comma, S[ω+1]S[\omega+1] is an opening brace, S[ω+2]S[\omega+2] is the character "∅" and S[ω2]S[\omega2] is a closing brace.

In the previous example, we have basically analyzed the string Sα+1S_{\alpha+1} at a given successor ordinal, splitting it into two copies of SαS_\alpha, comma and braces. This suggests some easy values of SS:

Lemma

For any n<ωn < \omega and β1\beta \geq 1, S[α]S[\alpha] is:

  • A comma if α\alpha can be written 2n32^n - 3 or ωβ2n+n\omega^\beta 2^n + n
  • An opening brace if α\alpha can be written 2n22^n - 2 or ωβ2n+n+1\omega^\beta 2^n + n + 1
  • A closing brace if α\alpha can be written 2n+142^{n+1} - 4 or ωβ2n+1+n\omega^\beta 2^{n+1} + n

Proof: For any ordinal α\alpha, by viewing the string Sα+1S_{\alpha+1} as a concatenation of SαS_\alpha, a comma, an opening fence, SαS_\alpha and a closing fence, we deduce that:

  • S[Lα]{S\left[L_{\alpha}\right]} is a comma.
  • S[Lα+1]{S\left[L_{\alpha} + 1 \right]} is an opening brace.
  • Lα+1L_{\alpha+1} is a successor ordinal and S[Lα+11]{S\left[L_{\alpha+1} - 1 \right]} is a closing brace.

The lemma follows immediately from the calculation of string lengths performed in the previous blog post. □

Warning: The rest of the blog post gives the solution to this puzzle, so you might want to have fun solving it yourself first and then go back checking my proposed solution later 😉...

More generally, the proof of the lemma can be extended by saying that if we find nn such that 2n1α2n+15{2^n - 1} \leq \alpha \leq {2^{n+1} - 5} then δ=α(2n1)2n4\delta = {\alpha - {(2^n - 1)}} \leq {2^n - 4} is the index of S[α]{S[\alpha]} in the second substring SαS_\alpha of Sα+1S_{\alpha+1} and so S[α]=S[δ]{S[\alpha]} = {S[\delta]}.

Details will be provided in the theorem below but one can already write a simple JavaScript recursive program to evalute SS at finite ordinals:

Script for α<ω\alpha < \omega

The character of SS at position α=\alpha =

is

The following intermediary step will be helpful to evaluate SS at infinite ordinal α\alpha:

Proposition

Let α\alpha is infinite and β=logω(α)1\beta = \log_{\omega}(\alpha) \geq 1. Let 1q<ω1 \leq q < \omega and 0ρ<ωβ0 \leq \rho < \omega^\beta be the quotient and remainder of the euclidean division of α\alpha by ωβ\omega^\beta. Let’s define:

n={log2(q) if q is not a power of 2 or this value is ρlog2(q)1 otherwise.n = \begin{cases} \left\lfloor {\log_2(q)} \right\rfloor & \text{ if } q \text{ is not a power of 2 or this value is } \leq \rho \\ \left\lfloor {\log_2(q)} \right\rfloor - 1 & \text{ otherwise.} \end{cases}

Then S[α]S[\alpha] is:

  • A comma if α=ωβ2n+n\alpha = {\omega^{\beta} 2^n + n}
  • An opening brace if α=ωβ2n+n+1\alpha = {\omega^{\beta} 2^n + n + 1}
  • An closing brace if α=ωβ2n+1+n\alpha = {\omega^{\beta} 2^{n+1} + n}
  • S[δ]S[\delta] otherwise where δ\delta is the unique ordinal such that α=ωβ2n+n+2+δ\alpha = {\omega^\beta 2^n + n + 2 + \delta} and δ<ωβ2n+n\delta < {\omega^{\beta} 2^n + n}.

Proof: First by construction we have α=ωβq+ρ\alpha = { {\omega^\beta q } + \rho}.

If the first case of the definition of nn, 2nq<2n+12^n \leq q < 2^{n+1} so we always have α<ωβ(q+1)ωβ2n+1Lωβ+n+1\alpha < {\omega^\beta {(q+1)}} \leq \omega^\beta 2^{n+1} \leq L_{\omega{\beta} + n + 1}. If additionnaly qq is not a power then 2n<q2^n < q and so αωβqωβ(2n+1)Lωβ+n\alpha \geq {\omega^{\beta}q} \geq {\omega^{\beta}{(2^n+1)}} \geq L_{\omega{\beta} + n}. Otherwise q=2nq = 2^n and nρn \leq \rho and so again αLωβ+n\alpha \geq L_{\omega{\beta} + n}.

In the second case of the definition of nn, log2(q)ρ+1\left\lfloor {\log_2(q)} \right\rfloor \geq \rho + 1 so nρ0n \geq \rho \geq 0. qq is a power of 2 and more precisely q=2n+1>2nq = 2^{n+1} > 2^n so we deduce the same way as in the previous case that αLωβ+n\alpha \geq L_{\omega{\beta} + n}. Moreover, ρ<n+1\rho < n + 1 so again α=ωβ2n+1+ρ<Lωβ+n+1\alpha = {\omega^\beta 2^{n+1} + \rho} < L_{\omega{\beta} + n + 1}.

We can thus view Sωβ+n+1S_{\omega \beta + n + 1} as a concatenation of Sωβ+nS_{\omega \beta+n}, a comma, an opening brace, Sωβ+nS_{\omega \beta+n} and a closing brace. We assume that ωβ2n+n+2α<ωβ2n+1+n { {\omega^{\beta} {2^n}} + n + 2} \leq \alpha < {\omega^{\beta} 2^{n+1} + n} as the three other cases are handled by the lemma. Then δ\delta is well-defined and is actually the index of S[α]S[\alpha] in the second copy of Sωβ+nS_{\omega \beta + n} so S[α]=S[δ]{S[\alpha]} = S{[\delta]}. □

We are now ready to give a nice way to evaluate SS at any ordinal α\alpha:

Theorem

S[α]S[\alpha] can be calculated inductively as follows:

  • If α=0\alpha = 0, S[α]S[\alpha] is the character "∅".
  • If 0<αω0 < \alpha \leq \omega then 2log2(α+3)3α2log2(α+3)+14 \right\rfloor}} - 3} \leq \alpha \leq \right\rfloor + 1}} - 4} and S[α]S[\alpha] is equal to:
    • A comma if α=2log2(α+3)3\alpha = {2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor} - 3}
    • An opening brace if α=2log2(α+3)2\alpha = {2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor} - 2}
    • A closing brace if α=2log2(α+3)+14\alpha = {2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor + 1}} - 4
    • S[δ]S[\delta] otherwise where δ=α(2log2(α+3)1)2log2(α+3)4α3\delta = {\alpha - \left( 2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor} - 1 \right)} \leq {2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor} - 4} \leq \alpha - 3.
      Moreover δ\delta compares against α\alpha as follows: δα12+12log2(α+3)+134\frac{\delta}{\alpha} \leq \frac{1}{2} + \frac{1}{2^{\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor + 1}} \leq \frac{3}{4} and log2(δ+3)<log2(α+3) {\left\lfloor {\log_2{(\delta+3)}} \right\rfloor} < {\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor}
  • If α\alpha is infinite, let 0<c1<ω0 < c_1 < \omega be the coefficient in its Cantor normal form corresponding to the smallest infinite term and c0<ωc_0 < \omega be finite term if there is one, or zero otherwise. Let kk be the exponent corresponding to the smallest nonzero term in the binary decomposition of c1c_1. Then S[α]S[\alpha] is equal to:
    • A closing brace if c0k1c_0 \leq k - 1.
    • A comma if c0=kc_0 = k.
    • An opening brace if c0=k+1c_0 = k + 1.
    • S[c0(k+2)]S[{c_0 - {(k+2)}}] if c0k+2c_0 \geq k + 2.

Proof:

The case α=0\alpha = 0 is clear. Suppose 0<α<ω0 < \alpha < \omega and let l=log2(α+3)l = {\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor}. By definition 2lα+3<2l+12^l \leq \alpha + 3 < 2^{l+1} so 2l3α2l+142^l - 3 \leq \alpha \leq 2^{l+1} - 4. Let’s consider the case 2l1α2l+152^l - 1 \leq \alpha \leq 2^{l+1} - 5 as the three other cases are already known from the Lemma. Sl+1S_{l+1} is made of the concatenation of SlS_l, a comma and SlS_l surrounded by braces. δ\delta is actually the index of S[α]S[\alpha] in the second copy of SlS_l so S[α]=S[δ]{S[\alpha]} = S{[\delta]}. The first equality is straighforward:

δ=α(2l1)2l+15(2l1)=2l4=2l13α3\delta = {\alpha - {(2^l -1)}} \leq {2^{l+1} - 5 - {(2^l -1)}} = {2^l - 4} = {2^l - 1 - 3}\leq \alpha - 3

Morever we have:

δα12l12l+1+512l12l+112+12l+1 \frac{\delta}{\alpha} \leq {1 - \frac{2^l - 1}{2^{l+1}+5}} \leq {1 - \frac{2^l - 1}{2^{l+1}}} \leq {\frac{1}{2} + \frac{1}{2^{l+1}}}

and so the second inequality follows from the fact that l1l \geq 1. Finally, the third inequality comes from:

2log2(δ+3)δ+32l4+3<2l 2^{\left\lfloor {\log_2{(\delta+3)}} \right\rfloor} \leq {\delta + 3} \leq {2^l - 4 + 3} < 2^l

Let’s consider the case of an infinite α\alpha. For some N1N \geq 1, βN>βN1>>β11\beta_N > \beta_{N-1} > \dots > \beta_1 \geq 1 and c0<ωc_0 < \omega and 0<c1,c2,cN<ω0 < c_1, c_2, c_N < \omega, we can write Cantor’s Normal form as:

α=ωβNcN+ωβN1cN1++ωβ1c1+c0\alpha = { \omega^{\beta_N} c_N} + { \omega^{\beta_{N-1}} c_{N-1} } + \dots + { \omega^{\beta_{1}} c_{1} } + c_0

With the notation of the proposition, we have β=βN\beta = \beta_N, q=cNq = c_N and

ρ=ωβN1cN1++ωβ1c1+c0 \rho = { \omega^{\beta_{N-1}} c_{N-1} } + \dots + { \omega^{\beta_{1}} c_{1} } + c_0

If N2N \geq 2, then ρ\rho is infinite and so n=log2(q)n = \left\lfloor {\log_2(q)} \right\rfloor and we are in the fouth bullet of the proposition. Moreover q2nq \geq 2^n and we can write:

α=ωβ2n+ωβ(q2n)+ρ=ωβ2n+n+2+δ\alpha = { \omega^{\beta} 2^n} + {\omega^{\beta} \left(q-2^n\right)} + \rho = { \omega^{\beta} 2^n} + n + 2 + \delta

where δ=ωβ(q2n)+ρ\delta = {\omega^{\beta} \left(q-2^n\right)} + \rho is infinite and so cancels out the n+2n + 2 term. It follows that S[α]=S[ωβ(q2n)+ρ]S{[\alpha]} = S\left[ {\omega^{\beta} \left(q-2^n\right)} + \rho \right]. Essentially, we have just removed from cNc_{N} its term of highest exponent in its binary decomposition!

By repeated application of the theorem, we can remove each binary digit of the cic_i for ii going from NN to 22. When then arrive at i=1i = 1:

S[α]=S[ωβ1c1+c0]{S[\alpha]} = S\left[ \omega^{\beta_1} c_1 + c_0\right]

With the notation of the proposition, we now have β=β1\beta = \beta_1, q=c1q = c_1 and ρ=c0\rho = c_0. If the binary decomposition of c1c_1 has more than one nonzero digit then so qq is not a power of 2. So although ρ\rho is now finite, we are still in the first case of the proposition and δ\delta remains infinite. So we can remove all but the last digit of c1c_1 by repeated application of the proposition:

S[α]=S[ωβ12k+c0]{S[\alpha]} = S\left[ \omega^{\beta_1} 2^k + c_0\right]

where kk is the exponent corresponding to the smallest nonzero term in the binary decomposition of c1c_1.

Using the lemma, S[α]S[\alpha] is a comma if k=c0k = c_0, an opening brace if k=c01k = c_0 - 1 and a closing brace if k=c0+1k = c_0 + 1.

If kc02k \leq c_0 - 2 then k=log2(2k)c0k = \left\lfloor {\log_2(2^k)} \right\rfloor \leq c_0 and writing

ωβ12k+c0=ωβ12k+k+2+(c0(k+2)){\omega^{\beta_1} 2^k + c_0} = \omega^{\beta_1} 2^k + k + 2 + \left(c_0 - {(k + 2)}\right)

we deduce from the proposition that S[α]=S[c0(k+2)]{S[\alpha]} = {S[{c_0 - {(k+2)}}]}.

Finally, if kc0+2k \geq c_0 + 2 then k=log2(2k)>c0k = \left\lfloor {\log_2(2^k)} \right\rfloor > c_0 and writing

ωβ12k+c0=ωβ12k1+k1+2+ωβ12k1+c0{\omega^{\beta_1} 2^k + c_0} = {\omega^{\beta_1} 2^{k-1}} + k - 1 + 2 + {\omega^{\beta_1} 2^{k-1}} + c_0

we deduce from the proposition that

S[α]=S[ωβ12k1+c0]{S[\alpha]} = S\left[ \omega^{\beta_1} 2^{k-1} + c_0\right]

We have essentially decremented kk and we can repeat this until we reach the case k=c0+1k = c_0 + 1 for which we already said that the character is a closing brace. □

As an application of this theorem, here is a few simple exercises:

Exercise 1

  1. S[272+2]S\left[2^72 + 2\right] is an empty set.
  2. S[ω72]S\left[\omega^72\right] is a comma.
  3. S[ω7272+72]S\left[\omega^72 72 + 72\right] is a closing brace.
  4. S[ω7272+ω4242+12]S\left[\omega^72 72 + \omega^42 42 + 12\right] is an opening brace.

Exercise 2

The evaluation of SS at

Corollary 1: Time complexity

Let α\alpha be an ordinal. Let c1<ωc_1 < \omega be the coefficient in its Cantor normal form corresponding to the smallest infinite term if there is one, or zero otherwise. Let c0<ωc_0 < \omega be its finite term if there is one, or zero otherwise. Then S[α]S[\alpha] can be evaluated in:

  • O(log2(c0+2)2+log2(c1+2))O\left( \log_2{(c_0+2)}^2 + \log_2{(c_1+2)} \right) elementary arithmetic operations and comparisons on integers.
  • O(log2(c0+2))O\left( \log_2{(c_0+2)}\right) elementary operations if integers are represented in binary and leading/trailing zero counting and bit shifts are elementary operations.

Proof: First note that the “+ 2” is just to workaround for the edge cases c0=0c_0 = 0 or c1=0c_1 = 0.

For the infinite case c1>0c_1 > 0 we need to calculate kk that is performing the find first set. A naive implementation can be done in O(log2(c1+2))O(\log_2{(c_1+2)}) steps by browsing the digits of c1c_1 to find the first nonzero for example by calculating the remainder modulo increasing power of 2. Each iteration requires only O(1)O(1) elementary integer operations %, * and comparisons. Then returning the result of moving to the finite case requires O(1)O(1) integer operations +, − and comparisons.

For the finite case c1=0c_1 = 0 and α=c0\alpha = c_0, first notice that we only require O(log2(c0+2))O\left( \log_2{(c_0+2)}\right) recursive calls given the inequality:

log2(δ+3)<log2(α+3) {\left\lfloor {\log_2{(\delta+3)}} \right\rfloor} < {\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor}

The case α=0\alpha = 0 only requires one comparison. For the case α>0\alpha > 0, we need to calculate l=log2(α+3)l = {\left\lfloor {\log_2{(\alpha+3)}} \right\rfloor} which is integer rounding of the binary logarithm or even just 2l2^l. As above, we can provide a naive implementation by calculating the quotient modulo increasing power of 2 in O(l)O(l) comparisons and elementary integer operations /, *. Then returning the result of moving to a δ<α\delta < \alpha requires only O(1)O(1) integer operations and comparisons. In total, complexity is O(log2(c0+2)2)O\left( \log_2{(c_0+2)}^2\right).

Finally, this can be simplified if one calculates kk and ll by a simple leading/trailing zero counting (or similar) and 2l2^l by a bit shift. □

Script based on Cantor normal form

If the coefficients of Cantor normal form of α\alpha corresponding the smallest infinite term and finite term are
c1c_1 =
c0c_0 =
Then the character of SS at position α\alpha is

Once we have an algorithm for S.charAt(α), it is easy to get an algorithm of NN times that complexity for S.substr(α, N) calculating the substring of length NN starting at position α\alpha by repeated calls to S.charAt(α). Let’s analyze a bit more carefully how we can make this recursive and more efficient:

Corollary 2: Algorithm for S.substr(α, N)

Let α\alpha be an ordinal and N<ωN < \omega. Then S.substr(α, N) can be calculated as follows:

  • If N=0N = 0 then it is the empty string.
  • If α=0\alpha = 0 and N=1N = 1 then it is the character "∅".
  • If 0<α<ω0 < \alpha < \omega, let l=log2(α+N+2)l = \left\lfloor {\log_2{(\alpha+N+2)}} \right\rfloor. We have 2l3α+N12l+142^l-3 \leq {\alpha + N - 1} \leq 2^{l+1} - 4 and the result is obtained by concatenating the following strings:
    1. If α2l4\alpha \leq 2^l - 4, the substring at offset α\alpha and length 2l3α2^l - 3 - \alpha.
    2. If α2l3\alpha \leq {2^l - 3}, a comma.
    3. If α2l2α+N1\alpha \leq {2^l - 2} \leq {\alpha + N - 1}, an opening brace.
    4. If α+N12l1{\alpha + N - 1} \geq {2^l - 1}, the substring at offset δ=max(0,α(2l1))\delta = \mathrm{max}{(0, \alpha - {(2^l - 1)})} and length 1+min(α+N1,2l+15)max(α,(2l1))1 + \mathrm{min}{({\alpha + N - 1}, {2^{l+1} - 5})} - \mathrm{max}{(\alpha, {(2^l - 1)})}.
    5. If α+N1=2l+14{\alpha + N - 1} = {2^{l+1} - 4}, a closing brace.
  • If α\alpha is infinite, let 0<c1<ω0 < c_1 < \omega be the coefficient in its Cantor normal form corresponding to the smallest infinite term and c0<ωc_0 < \omega be finite term if there is one, or zero otherwise. Let kk be the exponent corresponding to the smallest nonzero term in the binary decomposition of c1c_1. Then the result is obtained by a concatenating the following strings:
    1. If c0k1c_0 \leq k - 1, min(N,kc0){\mathrm{min}{(N, k - c_0)}} closing braces.
    2. If c0kc0+N1c_0 \leq k \leq c_0 + N - 1, a comma.
    3. If c0k+1c0+N1c_0 \leq k + 1 \leq c_0 + N - 1, an opening brace.
    4. If c0+N1k+2c_0 + N - 1 \geq k + 2, the substring at offset δ=max(0,c0(k+2))\delta = \mathrm{max}{({0, {c_0 - {(k + 2)}}})} and length c0+Nmax(c0,k+2)c_0 + N - {\mathrm{max}{(c_0, k + 2)}}.

Moreover, this only adds O(N)O(N) compared to the complexity of evaluating to a single offset.

Proof: The algorithm is just direct application of the Theorem. For the case where αω\alpha \geq \omega, the only change is that we add at most NN characters before moving to the finite case. The case α<ω\alpha < \omega is essentially a divide-and-conquer algorithm and we have a relation of the form:

T(L)=2T(L2)+f(L){T(L)} = 2{T\left(\frac{L}{2}\right)} + {f(L)}

where L=2l=Θ(α+N)L = 2^l = {\Theta{(\alpha + N)}} and f(L)f{(L)} is O(1)O(1) or O(log2L)O(\log_2{L}) depending on available operations, but in any case O(L)O{(\sqrt{L})}. So from the master theorem, T(L)=O(L)=O(α+N){T{(L)}} = {O(L)} = {O(\alpha+N)}. In general, this bound is not as good as repeating NN calls to S.charAt!

However, we note that if we assume that α>N4\alpha > N - 4 then α+2+N<2(α+3){\alpha + 2 + N} < {2{(\alpha+3)}} and so

log2(α+2+N)<1+log2(α+3)\log_2\left(\alpha + 2 + N\right) < {1 + \log_2{(\alpha+3)}}

We can easily discard the edge case where these start/end offsets point to a brace surrounding the right substring of the iterative step and so we get:

log2(α+N+2)=log2(α+3)\left\lfloor {\log_2{(\alpha+N+2)}} \right\rfloor = \left\lfloor {\log_2{(\alpha+3)}} \right\rfloor

which means that the left substring is just empty and so the complexity is not changed compared to S.charAt. Finally, when αN4\alpha \leq N - 4 the previous bound tells us that the steps are done in O(N)O(N). □

Script for S.substr(α, N)

If the coefficients of Cantor normal form of α\alpha corresponding the smallest infinite term and finite term are
c1c_1 =
c0c_0 =
Then the character of SS substring of length NN =
and offset α\alpha is