log☇︎
18900+ entries in 0.02s
mircea_popescu: BingoBoingo that article's call-list also reminded me of like, http://trilema.com/2013/of-mice-and-men-revisited/ and such great lulz
mircea_popescu: BingoBoingo also weirdly i don't get your pingbacks
mircea_popescu: so i'd say needs a rewrite.
mircea_popescu: but the main issue is that this reads like an approximately sorted succession of conclusions. painfully absent a) the logical justification for the various sentences and b) a very clear indication of structure, how they all work together ; from this b there would then flow a much better a.
mircea_popescu: also, dubious expressions, a collection : "to understand that why they want" ; "if your words actual words are avaible" ; "manufacter a pretext" ; "lack anyway to know" ; "the manufactered insecurity" ; "maladapted in ways invite bullying."
mircea_popescu: and don't be the sixth.
mircea_popescu: also i wouldn't give http://trilema.com/2016/and-they-wont-fucking-yield/#selection-67.90-67.101 & co obscure grupuscle any voice. freeze peach, really ? five morons talking about "things" "as if they were things" dun make "things" anything.
mircea_popescu: BingoBoingo fledling / fledgling ?
mircea_popescu: on the other hand, entirely unclear to me why the fuck we're discussing some [evidently buggy/misimplemented] program or my own inept handcalculus. either we agree a system of n equations with n unknowns is determinate or we don't and that's the end of the matter.
mircea_popescu: if we manage to fixate on a specific datapoint we'd prolly benefit here.
mircea_popescu: i don't actually follow your proggy any ; regardless, the values it spits dun seem to agree with what i expect.
mircea_popescu: by now, covered all the possible combinations of 4 bits (ain't that many) -- but never fell on the mp.py 1010.
mircea_popescu: K = 0101, P = 1110. RB = (1110), (1101), (1011), (0111). E1 = 1 x 1 x 0 = 0 ; E2 = 1 x 1 x 1 = 1 ; E3 = 1 x 0 x 1 = 0 ; E4 = 0 x 1 x 1 = 0 ; E = 0100. ☟︎
mircea_popescu: mno, actually still bs.
mircea_popescu: asciilifeform ^
mircea_popescu: K = 0101, P = 1110. RB = (1110), (1101), (1011), (0111). E1 = 1 x 1 x 0 = 0 ; E2 = 1 x 1 x 1 = 1 ; E3 = 1 x 0 x 1 = 0 ; E4 = 0 x 1 x 1 = 1 ; E = 0101 ?
mircea_popescu: lmao ima have to redo this. apparently im terrible at handmaffs
mircea_popescu: ok, so make 7 10 yield 1000 not 1010. as above!
mircea_popescu: unless you're willing do debug the .py, it's spitting out wrong values.
mircea_popescu: i guess this'll have to rest then.
mircea_popescu: notrly.
mircea_popescu: i honestly have nfi what you're talking aboot.
mircea_popescu: why is your E 1010 ?
mircea_popescu: asciilifeform K = 0101, P = 1110. RB = (1110), (1101), (1011), (0111). E1 = 1 x 1 x 0 = 1 ; E2 = 1 x 1 x 1 = 0 ; E3 = 1 x 0 x 1 = 0 ; E4 = 0 x 1 x 1 = 0 ; E = 1000.
mircea_popescu: asciilifeform why "new buf: [1, 1, 0, 1]" ?
mircea_popescu: i confess i dunno what teh breakdown is, here.
mircea_popescu: there's a reason i say " Consider K = 01010"
mircea_popescu: :) *
mircea_popescu: only if 1st bit of key is set :(
mircea_popescu: asciilifeform you agree that if i give you 5 equations with 5 unknowns, this is in fact resolvable ?
mircea_popescu: to be clear : it's exactly the same scheme ; showing you how P ~must~ be obtainable from known K and E.
mircea_popescu: so if K[q] = 0, then that line's skipped ; otherwise it's applied.
mircea_popescu: no no, x is xor * is multiplication (in the sense that if the key is 0 at that offset, the rb dun get applied)
mircea_popescu: yes.
mircea_popescu: E[1] = P[1] x P[2] x P[4] ; P[2] = P[2] x P[3] x P[5] is what i meant.
mircea_popescu: neh.
mircea_popescu: s-boxes, if you prefer.
mircea_popescu: ring buffers
mircea_popescu: for a 5 bit key you only have to try 2^3 permutations or so, it's true. but anyways.
mircea_popescu: so : with K it's a system of 5 equations with 5 unknowns ; without K it's a system of 5 equations with 10 unknowns.
mircea_popescu: To obtain P back from E without K : P[1] = P[1] x K[1] * P[1] x K[2] * P[2] x K[3] * P[3] x K[4] * P[4] x K[5] * P[5] ; P[2] = P[2] x K[1] * P[2] x K[2] * P[3] x K[3] * P[4] x K[4] * P[5] x K[5] * P[6] ; and so following all the way down.
mircea_popescu: To now obtain P back from E and K : P[1] = P[1] x P[2] x P[4] ; P[2] = P[2] x P[3] x P[5] ; P[3] = P[3] x P[4] x P[1] ; P[4] = P[4] x P[5] x P[2] ; P[5] = P[5] x P[0] x P[3].
mircea_popescu: E therefore is : E1 = 0 x 0 x 1 = 0 ; E2 = 0 x 0 x 1 = 0 ; E3 = 1 x 1 x 0 = 0 ; E4 = 1 x 1 x 0 = 1 ; E5 = 1 x 1 x 1 = 0 thus E = 00010.
mircea_popescu: Consider K = 01010 ; P = 00111 ; RB = (00111), (01110), (11100), (11001), (10011).
mircea_popescu: as long as P doesn't have more bits than K, this is a determinate system.
mircea_popescu: so this is, in fact, a system of (here) 512 xor-equations, with ak nown result (e[x]) and a known parameter matrix (K)
mircea_popescu: E[i] = K[0]*P[a] x K[1] * P[b] x ...
mircea_popescu: damn.
mircea_popescu: E is just what these equal to, see.
mircea_popescu: P[n] = K[0]*P[a] x K[1] * P[b] x ...
mircea_popescu: well, we're trying to get P back right.
mircea_popescu: (provided of course k>=n, which yes, it's a block cipher)
mircea_popescu: asciilifeform check this out : as per the "chipher", let there be a plaintext P of n bits ; and a key K of k bits. given a ciphertext E of n bits, it is a fact that any one bit of P is the result of xoring of up to k bits of P. if you know K you know ~which ones~, and as you have E you know ~what they must xor to~. this results in a message-wide system of k equations which is determinate.
mircea_popescu: now tell me, "why can't tv be more entertaining".
mircea_popescu: this has been a very fun morning.
mircea_popescu: asciilifeform this is so much better than having to correct the title.
mircea_popescu: "our block cipher has to have backdoor because we built a paper shredder and well..."
mircea_popescu: HA.
mircea_popescu: and no, "we have pre-padded ring buffers with THIS particular message that's worth hardwiring because it's ever so magical" ain't a reasoning.
mircea_popescu: and yes, you've inspired me.
mircea_popescu: asciilifeform like it or not, this is the ur-blockchipher. ALL OTHERS, not only casually but NECESSARILY, are mini-clipped versions of this. "competition or no competition", if it got shitboxes it's this and naught else.
mircea_popescu: rotates, it's a ring buffer.
mircea_popescu: same procedure, xor the Pmessage as per key.
mircea_popescu: works.
mircea_popescu: do you mind ?
mircea_popescu: show me!
mircea_popescu: weaker than serpent.
mircea_popescu: are you saying this is weak ?
mircea_popescu: ow shit, i mystypoed in the title, wtf is a "chipher".
mircea_popescu: didja read that whole thing
mircea_popescu: isn't it ?!
mircea_popescu: i'd have said more, but the editor showed wordcount 777, and i deem this a very good omen re cipher quality.
mircea_popescu: asciilifeform http://trilema.com/2018/so-i-designed-a-block-chipher/
mircea_popescu: seems like it.
mircea_popescu: the court will only consider positive cases.
mircea_popescu: bright kid walks in with a convincing story as to how he's gonna walk in with item != bright kid walks into here one day with that item.
mircea_popescu: the issue is that ~however many~.
mircea_popescu: he GOTTA WALK WITH IT THO
mircea_popescu: aha!
mircea_popescu: so far, we're still ~in the same position as http://btcbase.org/log/2018-10-26#1866400 / http://btcbase.org/log/2018-10-26#1866449 ☝︎☝︎
mircea_popescu: suppose... you actually do that ?
mircea_popescu: bitter days of http://trilema.com/2016/btmsr-block-cipher-competition/#comment-123804
mircea_popescu: asciilifeform likely explanation being "whisperer", ie, technical quimby sinking it behind the doors because "it's fucking stupid".
mircea_popescu: asciilifeform consider also that if a, b, c, d, e, f are rng words, then (P xor a) xor b) etc does not in fact substract anything.
mircea_popescu: right.
mircea_popescu: which finally got some attention from ppls.
mircea_popescu: the reason being that it wasn't gonna move till i baked it in some pie.
mircea_popescu: http://btcbase.org/log/2018-10-29#1867088 << not controversial. can substract, though doesn't necessarily have to ; can never add. ☝︎
mircea_popescu: right.
mircea_popescu: yes 1:1 by count. but by occurence ?
mircea_popescu: because if not, then VERY likely what we're seeing is a subclass of http://btcbase.org/log/2018-10-29#1867048 ☝︎
mircea_popescu: right. anyway, back to the practicals -- are sbox outputs ~equiprobable~ ?
mircea_popescu: asciilifeform the objection you had, you know, "this is trivially true but is not what we want" goes very much to the core of the issue -- cryptographic notions of strength are very hard to meaningfully describe mathematically because htey're always "obviously this has ONE trivial solution, but does it have more than the one ?!" math deals in 0 and 1s, it's ill equipped to deal with this batshit nonsense.
mircea_popescu: heh.
mircea_popescu: it's what i do for a living.
mircea_popescu: i have problems with statements.
mircea_popescu: what do you want me to do ?
mircea_popescu: i agreed it is not civilised!
mircea_popescu: the function which takes all numbers <4 to themselves and all numbers >4 to 4.
mircea_popescu: cuz E4, E5 are the indistinct sysnonyms of P4.