hacker.org

someone hinted there is no key, but if 1=a, 2=b, 3=c ... that is still a key of a sort even if a simple one. the binary data represented by the 2 color image, i double checked there is no 01 or fe intensity colors hiding a faint image :( though the binary might still make an image of some sort with ...

i used a big hash table 16111 prime probably overkill but all the work i did, i may as well go for least collisions its the biggest prime that i felt safely fit in the memory

79+g0^<2^1^-4?d1+$p!953**2557***8+95**1+882**1-0^<0^0^4^/4^*-4^+0^<8?4c09-4-g>0^6?259*-g!

yeah i could make it more efficient but it works

8g0^<g1+2c5g1+<p!0<1<00>01>61^2*>0^2*1+2^1v>62^2*>1v2*1+>02c

i solved it two ways

0<1<00>01>81v>81v>884**0^<g1-09-3-g

and something like this that i tossed later

0<1<-4*8882***+g {spam 512~ bytes} 0>p {spam 512~ bytes} 1>p

or something close to that anyways

i cheated with divide
if smallnumber > 0 and < bignumber then ( smallnumber + bignumber - 1) / bignumber = 1
else = 0

0<1</88888****1-+8/8/8/8/8/4*g1<p!0<p

yeah this wont work if the test has a big number in it but who cares

lol seems i got the same idea as another, hurrah division

0<1</4?0<p!1<p

i abuse c to avoid negative jumps

00^1+1^<1^<1^:2?2gp!d0^29+:2?3cd1+1c

i guessed how the cipher worked then dictionary brute forced ctihuolhlacat against all known English 13 letter words thankfully the key repeated mostly over the length this word

the answer has two words with a space between

my one complaint is the file size of the image is a bit large, takes a while to download. this could easily be reduced by say, making it a gif with 50% width and height. it should still be readable.

the first 8 bytes is
5499fa99
am i correct in assessing that this translates to the 8 byte number
0x99fa9954 with 99 the most significant
brute force failed when assuming
it was 0x5499fa99
i wanted to know if the fail was my algorithm or just endianness.

i did not brute force adding 0s myself but i think it can be done. you could take say the first 10 hex digits and just brute force all the x b and add 0 combos on that rather then the whole thing. you should get 5 - 10 letters that would be more then enough to tell if a x b pair is good. if you get ...

seems several different layer filters will get the answer. xor and negate being much more readable then additive. for those curious i used Paint.net

guassian would be N^2 time with packed bools if each row could fit into an int. naturally its still an N^3 problem just done parallel. Am I correct in guessing that some form of algebra, letting the highly regular nature of the N unknowns trim down to a simplified matrix with (N ^ (2/3)) unknowns th...

mine machines tend to look similar to each other, i tried to keep it small 0s/,x?\1+s/x^?\s/1-xvP\ \===/-*8|4^x/!\?=-1x/ \?\s+s/1-xvP\ /s\1//\?-1x=/ \0P?=/ version 2 for the tiny challenge 88 but need 44 ? grr how do i cut this in half even if i take out the second print loop its not half the size :...