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Merging Responsibly

from Quentin Santos [alt+shift+b] in programming

tl;dr: I do not like merging the main branch into feature branches, and I do not like squashing MR/PRs git commit git has emerged as the obvious choice for SCCS (Source Code Control System). The consensus is that it is unambiguously better than the previous standard, SVN. There is some debate as to what is … Continue reading Merging Responsibly → The post Merging Responsibly appeared first on Quentin Santos.
28th Apr 2024

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More from Quentin Santos

My first PCB: AIC3204 eval board

I had some fun with the Adafruit TLV320DAC3100. This is a small board built around the (TLV320)DAC3100 audio chip. “DAC” stands for “Digital-to-Analog Conversion”; that is, it takes some audio signal (here, I²S) and converts it to the analog equivalent (differential for speakers, single-ended for headphones). Microcontrollers typically don’t have analog output, and getting analog […] The post My first PCB: AIC3204 eval board appeared first on Quentin Santos.

3 weeks ago • 1 votes
A few spikes in Pico cos()

A picture is worth a thousand LLM prompts: Horizontal value is the input value x for cosf(x), ranging from 0 to 2π. The orange line shows the result of evaluating cosf(x), which looks exactly as expected. The blue line shows the time it took to evaluate cosf(x) for each x, in nanoseconds. I discovered this […] The post A few spikes in Pico cos() appeared first on Quentin Santos.

15th Aug 2026 • 1 votes
The RP2350 is too fast

I had a few hours of “fun” figuring out why my RP2350 could use my ILI9341 display, but not the ST7796 one. To be more specific, displaying stuff with the first display worked flawlessly, while the ST7796 would display a single frame correctly, and then glitch. Depending on the exact code running, the glitches would […] The post The RP2350 is too fast appeared first on Quentin Santos.

15th May 2026 • 1 votes
The RP2350-USB-A cannot see devices disconnect

What’s better for New Year’s Eve than debugging some USB hardware issues? In my previous article, I explained how the design of Waveshare’s RP2350-USB-A can make the USB-A host connector incapable of detecting certain USB devices, and of any hot-plugging. The solution was to desolder R13, a pull-up resistor on D+: With this fix, my […] The post The RP2350-USB-A cannot see devices disconnect appeared first on Quentin Santos.

1st Jan 2026 • 1 votes
Fixing the RP2350-USB-A not working as USB host

tl;dr: you need to desolder R13, the resistor closest the pin 6 of the board, as indicated by the red arrow in the cover picture of this article Context I am currently toying around with emulating a USB device. However, I also wanted to be able to plug a keyboard in, so I needed a … Continue reading Fixing the RP2350-USB-A not working as USB host → The post Fixing the RP2350-USB-A not working as USB host appeared first on Quentin Santos.

20th Nov 2025 • 3 votes

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Two-Stack Sliding-Window Aggregation

An aggregation is some kind of summary of a set of data. This can be the sum, length, minimum, etc. It is quite common to want to calculate such a summary repeatedly, e.g. “the maximum noise level in dB for the past 30 seconds” for a nuisance detector. In such a case we say there is a sliding window over our data, and we want to aggregate over our window. If our aggregation is a binary operator with an inverse, like integer sums, there is a very easy solution using a double-ended queue: from collections import deque class SlidingWindowSum: def __init__(self): self.sum = 0 self.elems = deque() def push(self, x): self.sum += x self.elems.append(x) def pop(self): self.sum -= self.elems.popleft() def eval(self): return self.sum But what if our operator has no inverse? This is actually the case for most interesting summaries such as minimum, quantile, approximate unique count (for example using HyperLogLog), etc. In fact, even something as simple as a floating-point sum suffers from the fact that floating-point addition is not invertible. For example, if you ever have a NaN in your input data with the above naive algorithm your sum will forever remain NaN, even long after the bad value has left your window. Six years ago I came up with an algorithm for maintaining just the minimum/maximum in a sliding window and posted it to cs.stackexchange. I now consider this algorithm pointless, because it turns out there is a simple and efficient algorithm that solves this problem for a very wide class of aggregations. I’m writing this blog post to spread the word, because I feel it should be more widely known. Folklore I came across this algorithm while reading a far more advanced paper, Low-Latency Sliding-Window Aggregation in Worst-Case Constant Time by Tangwongsan et al. Why is this paper titled low-latency? Because it does the same as what I’m about to describe, but in O(1) time for each step. However, in it they also described a “two-stack” algorithm, which does it in amortized O(1), and is far, far simpler. Amortized O(1) means that across many operations the total amount of work per element is constant, but an individual operation can take much longer. This is almost always fine, unless you absolutely need a low upper bound on latency. Funnily enough that paper attributes this algorithm to “adamax” from a 2011 Stack Overflow post. They in turn credit a 2001 lecture note by D. Sleator for the inspiration. However, this lecture note does not describe a sliding window aggregate, it describes the classical two-stack algorithm for implementing a FIFO queue and does amortized analysis on it. Ultimately I would not be surprised to find that this algorithm was already described in an obscure paper from the 1970s, seeing how simple and brilliant it is. Two stacks Like the authors of the paper, I will generalize the two-stack algorithm to arbitrary associative aggregation functions. By abstracting the aggregation as a set of functions, empty(), unit(x), combine(x, y) and finalize(x), you can describe many possible aggregations, for example a mean: empty = lambda: (0, 0) unit = lambda x: (x, 1) combine = lambda x, y: (x[0] + y[0], x[1] + y[1]) finalize = lambda x: x[0] / x[1] if x[1] else None I’d like to note here that these functions have the following signatures: fn empty() -> Agg; fn unit(x: Value) -> Agg; fn combine(x: Agg, y: Agg) -> Agg; fn finalize(x: Agg) -> Out; I’m making a distinction here between Value, Agg and Out because while they seem superficially similar for something like an integer sum, for an approximate unique count on strings you would have (Value, Agg, Out) = (String, HyperLogLogSketch, u64), three wildly different types. Without further ado, the algorithm: class TwoStackAgg: def __init__(self): self.values = [] self.values_agg = empty() self.cum_aggs = [] def push(self, x): self.values.append(x) self.values_agg = combine(self.values_agg, unit(x)) def pop(self): if not self.cum_aggs: cum_agg = empty() while self.values: cum_agg = combine(unit(self.values.pop()), cum_agg) self.cum_aggs.append(cum_agg) self.values_agg = empty() self.cum_aggs.pop() def eval(self): return finalize( combine(self.cum_aggs[-1], self.values_agg) if self.cum_aggs else self.values_agg ) That’s it, the entire algorithm. There’s two stacks (values and cum_aggs) and one more aggregate, values_agg. At any point in time values_agg holds the aggregate of values, and cum_aggs contains the cumulative aggregates of all values in our window that aren’t in values, in reverse order. From this we can get the aggregate over our entire window in constant time by by combining the last value of cum_aggs with values_agg. The neat part is that (assuming w is our window size) every wth operation we drain all of values and maintain a running aggregate while pushing the partial cumulative aggregates onto cum_aggs. This is what makes it amortized O(1), doing O(w) internal operations every wth pop bounds the total amount of work per element to O(1), even though a singular operation might not be constant time. I think this is best visualized. Suppose we sum [1, 2, ..., 10] with a fixed-size sliding window of four elements, then the state on each eval() call would look like this (values_agg not shown as it is simply the aggregate of the values): cum_aggs values out [] [] = 0 [] [1] = 1 [] [1, 2] = 1 + 2 [] [1, 2, 3] = 1 + 2 + 3 [] [1, 2, 3, 4] = 1 + 2 + 3 + 4 [4, 3 + 4, 2 + 3 + 4] [5] = 2 + 3 + 4 + 5 [4, 3 + 4] [5, 6] = 3 + 4 + 5 + 6 [4] [5, 6, 7] = 4 + 5 + 6 + 7 [] [5, 6, 7, 8] = 5 + 6 + 7 + 8 [8, 7 + 8, 6 + 7 + 8] [9] = 6 + 7 + 8 + 9 [8, 7 + 8] [9, 10] = 7 + 8 + 9 + 10 [8] [9, 10] = 8 + 9 + 10 [] [9, 10] = 9 + 10 [10] [] = 10 [] [] = 0 In total the memory usage is O(w), where w is your maximum window size. Note that for simplicity of analysis and the example I assumed a fixed-size window w, but there is nothing about the two-stack algorithm that requires this. You can call push(x) and pop() as many times as you’d like between each eval(), growing and shrinking the window size as needed. Floating-point non-associativity Note that we required above that our aggregate combine is associative, meaning: combine(combine(x, y), z) = combine(x, combine(y, z)) Technically speaking, floating-point addition doesn’t respect this. Nevertheless, the above algorithm is still very useful because the results closely match the expected outcome, even more so if you use a compensated summation algorithm like Kahan summation. Another neat thing about the two-stack algorithm is that it doesn’t require commutativity, if you follow the above implementation precisely. The order of operands is maintained, which can matter for things like string concatenation. However, there is a second very useful property of the above algorithm. Each aggregate is strictly a combination of the elements in the window, and none outside the window. This means if your window contains a NaN or infinity (or some other outlier), that value only poisons the windows that contain it rather than the rest of your computation. But even without NaN or infinity it is useful, due to not propagating errors endlessly. E.g. if your sliding window starts with [1e20, 1], this is what would happen with a naive rolling sum: >>> 1e20 + 1 - 1e20 - 1 -1.0 Compensated summation will reduce these effects, but not making your result depend on values outside of the window will eliminate long-term error accumulation entirely.

20 hours ago • 1 votes
Lighthouse map

Lovely global map with animated lights sweeping the waters

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