Frik 🏳️🌈 🎗
@frikin.in
Inútil professional | CAT/EN/ES | 26
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Good morning ❤
🐘 https://mastodon.social/@frikinin
Icona per bachidebrujas | Capçalera per @thirteight.bsky.social
Good morning ❤
🐘 https://mastodon.social/@frikinin
😭vas a tener stand en algun otro evento a parte de la indie dev day? Lo quieroooo
September 28, 2025 at 10:55 AM
😭vas a tener stand en algun otro evento a parte de la indie dev day? Lo quieroooo
I haven't been able to open a single user profile all day, it keeps saying they don't exist
September 24, 2025 at 12:48 PM
I haven't been able to open a single user profile all day, it keeps saying they don't exist
i see what you mean, you're right.
Most compression libraries allow for a level compression setting. The higher, the more compressed it ends up, but the slower it runs; with the default setting usually being a middle ground. Perhaps a higher compression level might have done what you're pointing at.
Most compression libraries allow for a level compression setting. The higher, the more compressed it ends up, but the slower it runs; with the default setting usually being a middle ground. Perhaps a higher compression level might have done what you're pointing at.
September 21, 2025 at 5:14 PM
i see what you mean, you're right.
Most compression libraries allow for a level compression setting. The higher, the more compressed it ends up, but the slower it runs; with the default setting usually being a middle ground. Perhaps a higher compression level might have done what you're pointing at.
Most compression libraries allow for a level compression setting. The higher, the more compressed it ends up, but the slower it runs; with the default setting usually being a middle ground. Perhaps a higher compression level might have done what you're pointing at.
We could achieve the same result by changing the last two operations to
x6 <-31; x6 <-6
or
x1 <- 27; x11 <-1
too, btw. I'm too lazy to calc it, but since numbers are encoded in a way that uses less bits the smaller the number is; these options might be less efficient [4/4]
x6 <-31; x6 <-6
or
x1 <- 27; x11 <-1
too, btw. I'm too lazy to calc it, but since numbers are encoded in a way that uses less bits the smaller the number is; these options might be less efficient [4/4]
September 21, 2025 at 12:45 PM
We could achieve the same result by changing the last two operations to
x6 <-31; x6 <-6
or
x1 <- 27; x11 <-1
too, btw. I'm too lazy to calc it, but since numbers are encoded in a way that uses less bits the smaller the number is; these options might be less efficient [4/4]
x6 <-31; x6 <-6
or
x1 <- 27; x11 <-1
too, btw. I'm too lazy to calc it, but since numbers are encoded in a way that uses less bits the smaller the number is; these options might be less efficient [4/4]
So what we do is
x6 <-31 - go back thirty-one chars and copy six chars in a row
And now that we have 'p' at the end of the buffer, we can use "the trick"
x6 <- 1 - go back one character and copy six
[3/4]
x6 <-31 - go back thirty-one chars and copy six chars in a row
And now that we have 'p' at the end of the buffer, we can use "the trick"
x6 <- 1 - go back one character and copy six
[3/4]
September 21, 2025 at 12:45 PM
So what we do is
x6 <-31 - go back thirty-one chars and copy six chars in a row
And now that we have 'p' at the end of the buffer, we can use "the trick"
x6 <- 1 - go back one character and copy six
[3/4]
x6 <-31 - go back thirty-one chars and copy six chars in a row
And now that we have 'p' at the end of the buffer, we can use "the trick"
x6 <- 1 - go back one character and copy six
[3/4]
so, even though it has only decoded one 'p' so far, it's cloned 5 times.
o - add a literal 'o'
x24 <- - go back one char and copy 24 chars
it's the same as before
Now, we want to encode 12 'p' chars. The longest sequence is 6 'p' long, and we can't use the trick because the last char is a 'o' [2/4]
o - add a literal 'o'
x24 <- - go back one char and copy 24 chars
it's the same as before
Now, we want to encode 12 'p' chars. The longest sequence is 6 'p' long, and we can't use the trick because the last char is a 'o' [2/4]
September 21, 2025 at 12:45 PM
so, even though it has only decoded one 'p' so far, it's cloned 5 times.
o - add a literal 'o'
x24 <- - go back one char and copy 24 chars
it's the same as before
Now, we want to encode 12 'p' chars. The longest sequence is 6 'p' long, and we can't use the trick because the last char is a 'o' [2/4]
o - add a literal 'o'
x24 <- - go back one char and copy 24 chars
it's the same as before
Now, we want to encode 12 'p' chars. The longest sequence is 6 'p' long, and we can't use the trick because the last char is a 'o' [2/4]
zip works by saying "go back N characters and copy M characters in a row", but it has a little quirk where an instruction can copy from the characters that have just been created by that same instruction.
So, in this case:
p - add a literal 'p'
x5 <-1 - go back one char and copy 5 chars [1/4]
So, in this case:
p - add a literal 'p'
x5 <-1 - go back one char and copy 5 chars [1/4]
September 21, 2025 at 12:45 PM
zip works by saying "go back N characters and copy M characters in a row", but it has a little quirk where an instruction can copy from the characters that have just been created by that same instruction.
So, in this case:
p - add a literal 'p'
x5 <-1 - go back one char and copy 5 chars [1/4]
So, in this case:
p - add a literal 'p'
x5 <-1 - go back one char and copy 5 chars [1/4]