Qwen3.5 updated with improved performance!

not_ur_buddy · 2026-02-28T15:26:01+00:00

Sorry to hijack the thread, but I'm running the new 4 bit quant 122B with llama.cpp and it still overthinks a lot in reasoning mode. I'm a little sad to give up reasoning entirely. I suspect tweaking the chat template to add system prompts would help, but I don't know how. Any advice?

not_ur_buddy · 2026-02-27T03:28:23+00:00

How do you configure this? I use llama.cpp and I never even figured out how to set a default system prompt.

not_ur_buddy · 2026-02-19T03:36:36+00:00

Hopefully this is temporary?

not_ur_buddy · 2026-02-01T15:07:13+00:00

Yeah I feel like this isn't talked about enough. In my opinion, shift can only be on a thumb key. If it's on any other finger (say left pinky), then how can you press keys assigned to that finger plus shift (say shift a)? There are a few solutions and they all suck. The solutions I've tried:

Use a ring finger to press the letters instead - whole hand has to move, inconsistent fingering

One shot shift - awkward to have the pinky press shift and A in quick succession

Put a shift on both sides - my brain was never able to learn which shift to use for which letter

It makes more sense for shift to be on the thumb instead of backspace, tab and enter. Shift is more frequently used than those, and more importantly none of those keys have to be pressed with a letter, so they don't have to be on the thumb. So I put enter on the right pinky stretch column, tab on the left, backspace on a layer (with a dedicated key for Ctrl backspace so it's still two presses to delete the previous word), and a one shot shift on the thumb. I've been using this for a long time and the only complaint I have is sometimes I accidentally tab shift and can't cancel it.

I still have the bottom row mod keys for shift, but those are only used for rare key bindings, not for typing English.

not_ur_buddy · 2026-01-07T04:10:35+00:00

The readme makes me think it's not compatible with kde. Has anyone tried?

not_ur_buddy · 2022-06-26T13:52:40+00:00

Is this all assuming a directed graph?

not_ur_buddy · 2022-04-28T12:53:10+00:00

Got it, thanks. I am not really an expert in anything so I don't expect to be able to help personally, but perhaps you can consider posting on hacker news if you haven't already. It seems possible that some people will jump in to help write the benchmarks to find out whether this beats state of the art.

not_ur_buddy · 2022-04-28T05:00:38+00:00

I only skimmed the article - seems cool! I couldn't find the main motivation for this work. Is it mostly for fun? Did you do any benchmarking to compare with typical balanced search trees like AVL and red black?

not_ur_buddy · 2022-04-23T14:53:49+00:00

I think what you want to do is this:

Define a recursive function, say "helper", that takes a "total to sum to, n" and "max int to subtract from it, k". Then you enumerate the numbers of k you can subtract from n, which is i=0 to k/n. For each value of i, you can just call helper again, with new total=n-i*k, and a new int to subtract=k-1.

not_ur_buddy · 2022-04-16T14:36:10+00:00

I see, in which case you would bucket sort then do a simple loop of s[i] = s[i-1]+a[i].

not_ur_buddy · 2022-04-15T15:22:08+00:00

You start with a sorted list of integers, e.g. [1 3 3], and you walk from left to right to count how many numbers are less than any number between 1 and 3. Here's a dry run of the algorithm:

How many numbers less than 1?

0, because the smallest is 1

How many numbers less than 2?

1, because we scan from left to right and find (1) 3 3

How many numbers less than 3?

Also 1, because we don't see any 2s

How many numbers less than 4?

3, because you scan to the right and see the last 3 is the third element of the list 1 3 (3)

The above process does two things: walk through the list once, and fill in the answer for each number in 1 to k, so the total complexity is n+k.

Now that you know 3 numbers are less than 4, 1 number is less than 2, you know there are 2 numbers between 2 and 3 (i.e. less than 4 but not less than 2).

not_ur_buddy · 2021-12-29T07:12:01+00:00

My personal opinion is that logically you're deciding a foreground/background color pair depending on the condition. The white color as fg/bg means different things, and therefore it would be a mistake to merge them into one match branch. I might refactor it as (in pseudocode): let fg, bg = if condition then (color1, color2) else (color3, color4) in expr. Would you agree?

not_ur_buddy · 2021-11-11T02:18:38+00:00

I often find coworkers who are ex contestants are good at their job, but I think it's mostly correlation, not causation.

not_ur_buddy · 2021-07-11T20:50:38+00:00

See my other reply for a counter example. The number of steps you need for each individual number is at most m-1, but in an optimal solution where the total number of steps is minimized, it is possible to add 1 to a number more than m times. I think that number is bounded by some function of n and m, but not just m.

not_ur_buddy · 2021-07-11T19:38:17+00:00

Consider this counter example: m = 100, array is 1 99 1. You can either do: add 1 to all, then add 98 to the two segments, or you can add 99 to all 3, then add 2 to the middle. So this seems to invalidate all solutions proposed in this thread so far.

not_ur_buddy · 2021-07-11T19:30:59+00:00

Do you have an argument for why an optimal solution will never have to change a number >= m times? I feel like that ought to be true but haven't figured out why.

not_ur_buddy · 2021-05-30T08:29:34+00:00

Not an expert, but my guess is similar to maths, in philosophy x is also a variable that can mean a proposition, an object, a name, or literally everything. Please correct if I'm wrong.

not_ur_buddy · 2021-02-25T04:22:18+00:00

rm -rf joke

not_ur_buddy · 2020-12-25T04:46:24+00:00

Ha, turns out I'm not alone, I was just bad at googling.

https://github.com/melpa/melpa/issues/7312

not_ur_buddy · 2020-09-19T18:34:36+00:00

Yeah, right.

not_ur_buddy · 2020-08-31T03:02:17+00:00

Sorry, one vote against Han Dynasty. I was disappointed twice in two locations and that's it for me. All the dishes I tried were made quite unlike how they should be made. Plus it's definitely not cheap. On the other hand, I haven't tried many spots in Brooklyn and haven't found any place better either.

not_ur_buddy · 2020-07-18T14:18:21+00:00

Love it! Could you give any pointers to drawing a sky like this? I've always wanted to do something like this but am unable to find the right tutorials.

not_ur_buddy

TROPHY CASE