4 ms·

But you're just looking for the text being in a certain range of characters. The first stage is determining the key length, you can do this by looking at the nu

by alexkus 13y ago

But you're just looking for the text being in a certain range of characters. The first stage is determining the key length, you can do this by looking at the number of different character values in positions 1,2,3... then 1,3,5,... and 2,4,6,... then 1,4,7,... etc. keylength=1 127 keylength=2 114 116 keylength=3 107 102 106 keylength=4 78 94 98 103 keylength=5 111 118 115 114 119 keylength=6 81 71 81 77 79 77 keylength=7 104 112 110 104 110 110 109 keylength=8 67 85 86 94 68 91 83 86 keylength=9 87 86 86 90 92 85 83 85 82 keylength=10 91 97 86 89 90 88 90 90 92 97 keylength=11 90 101 102 98 99 93 100 100 100 95 93 keylength=12 45 45 46 53 47 46 41 47 47 51 50 47 keylength=13 89 85 100 90 89 91 94 95 82 90 86 88 92 keylength=14 83 84 85 88 82 81 77 72 82 85 73 81 82 86 keylength=15 73 81 72 76 76 70 75 73 74 72 77 76 73 71 73 keylength=16 56 68 71 80 60 73 64 73 58 71 70 76 57 76 66 68 keylength=17 81 86 82 74 77 82 85 87 81 87 78 84 77 77 79 80 82 keylength=18 59 55 59 55 61 56 59 55 57 59 55 62 62 57 60 58 56 59 keylength=19 77 80 80 77 71 89 78 76 73 74 68 80 80 80 74 76 78 79 72 The stand out line from that is keylength=12 as it provides a local minima for the number of different characters. Having a good idea of the key length you can look at which of the possible 256 values for each position of the key produce the top numbers of printable characters. A quick perl script to do this gives:- POSS:0: n=211 chars=FHLRSUVZ] POSS:1: n=209 chars=cdjkouwx~ POSS:2: n=208 chars=DIJKLORUW^_ POSS:3: n=209 chars=BIKSVW POSS:4: n=206 chars=bejkntvy~ POSS:5: n=210 chars=@CDEFKQXZ POSS:6: n=208 chars="#)+.023678>? POSS:7: n=206 chars=fgmorswz| POSS:8: n=208 chars=behjknt~ POSS:9: n=205 chars=GMORSTW\ POSS:10: n=207 chars=FLWXY[\ POSS:11: n=208 chars=cdjkoux~ You'll note that the actual key (SjJVkE6rkRYj) is present in this output. I'll carry on with this if I get a chance tomorrow.

alexkus 13y ago
OK, next pick a word you think will be in the output (it's php code so I'll guess the word "function" might be common). You then XOR each 8 character substring with "function", taking note of the position of the substring within the 12 character groupings, and count the resulting characters:- $ ./4.pl 12 "function" POS: 0: S->78 T->71 U->69 POS: 1: j->76 l->64 m->59 POS: 2: L->73 J->67 W->64 POS: 3: V->65 Q->55 P->49 POS: 4: m->68 v->50 `->49 POS: 5: E->54 C->52 O->48 POS: 6: 6->57 0->56 1->56 POS: 7: u->56 t->54 x->50 POS: 8: k->76 l->70 m->66 POS: 9: R->69 U->57 H->51 POS: 10: _->69 Y->62 ^->54 POS: 11: j->70 m->59 l->55 Note that the correct key in each position is usually either the first or second possibility, but there are some times when it isn't there at all. (It's better if you give it a longer string, if I give it "function " with the trailing space then it gets 9/12 as the most popular, 2/12 as second most popular and the remaining one as the third most popular. There are even better choices, see below, but I chose to go with this as it helps show how to continue with non-perfect information). Next I knocked up a quick program to allow me to try different keys to decrypt it, it prints out the code in blocks of the appropriate keylength so it's easy to see if a specific character position is correct or not. The program takes various commands from stdin (k = print key, S = set key, s = set individual key character, p = print in keylength sized blocks, P = print entire decrypted block):- $ ./5.pl 12 k key is aaaaaaaaaaaa S SjLVmE6ukR_j key is now SjLVmE6ukR_j p /)abjutt uy stcm&vauiadl es\12i`(!Gisue t("_UERQER/) {$YCIOKNE= $ HTRPYCOHKIC_ VATS=$_WOSR= &$NTRP_WOSR_ VATS=$_@ET;& $HRTV_GBT_PA RS=}\12//cie&w itn crrhr `u ncriin \127_doe ($k)}@hbadcr ('NTRP/6.1&5 ... Getting there, but still a way off. There are a few strings like COIKOE[#n]= which could be COOKIE, so we fiddle with the 3rd column and see some interesting things in the output:- s 2 J p ... elseof(oasYp ... COOKOE[#n]= ... systcm(#c)=} ... Neither of the other two choices for column 4 produce the right string, but it's simple enough to find the ciphertext and find what is needed to make elseof -> elseif, COOKOE -> COOKIE and systcm -> system. It's k:- s 4 k k key is SjJVkE6ukR_j p ... elseif(oasYp ... COOKIE[#n]= ... system(#c)=} We carry on doing this based on more clues like this (it gets easier and easier!) and we end up with:- k key is SjJVkE6rkRYj p //adjust sy stem variabl es if(!@isse t($_SERVER)) {$_COOKIE=&$ HTTP_COOKIE_ VARS;$_POST= ... And printing out the entire decrypted buffer without extra linebreaks every 12 chars gives:- P //adjust system variables if(!@isset($_SERVER)){$_COOKIE=&$HTTP_COOKIE_VARS;$_POST=&$HTTP_POST_VARS;$_GET=&$HTTP_GET_VARS;} //die with error function x_die($m){@header('HTTP/1.1 500 '.$m);@die();} //check if we can exec define('has_passthru',@function_exists('passthru')); define('has_system',@function_exists('system')); define('has_shell_exec',@function_exists('shell_exec')); define('has_popen',@function_exists('popen')); ... Interestingly, trying known plaintext of "define" gives us the entire key as the most probably choice first time:- $ ./4.pl 12 "define" POS: 0: S->75 X->56 _->54 POS: 1: j->77 a->51 f->47 POS: 2: J->71 A->55 F->49 POS: 3: V->61 ]->44 Z->43 POS: 4: k->59 h->43 *->40 POS: 5: E->55 8->48 4->44 POS: 6: 6->52 w->47 <->43 POS: 7: r->55 d->43 s->40 POS: 8: k->75 `->51 g->51 POS: 9: R->59 Y->55 X->49 POS: 10: Y->69 R->49 U->47 POS: 11: j->61 `->43 m->42
- JonnieCache 13y ago
  This is roughly what I was imagining doing in my head when I brashly wrote "obviously bruteforce." Hooray! Apparently I've still got it, at least to some small extent. Thanks for your writeup! Why does define work so well? Chance?
  - alexkus 13y ago
    Not sure, "define" only appears 10 times but "function" appears 19 times. It's probably because 'e' is more popular in the code and "define" has two of them. Looking at the decrypted code there are much better strings to try; "function_exists" gives the right key first time too. It could have been made a lot harder by using comments at the end of each line to pad the code with extra characters to equal out the frequency distribution and make this kind of frequency analysis impossible.
    - JonnieCache 13y ago
      Could I make it even worse using ruby's unicode support? eg this kind of nonsense: π = Math::PI alias λ proc ● = λ {|r| π*r**2} puts ●.call 5 # 78.53981633974483
      - alexkus 13y ago
        Oh yes, although too many "alias" commands will mean that it's possible to search for the known plaintext of 'alias'. But the more non-ascii bytes there are the more it will put off my analysis which dependeds on isprint().

by alexkus 13y ago

alexkus 13y ago

OK, next pick a word you think will be in the output (it's php code so I'll guess the word "function" might be common). You then XOR each 8 character substring with "function", taking note of the position of the substring within the 12 character groupings, and count the resulting characters:- $ ./4.pl 12 "function" POS: 0: S->78 T->71 U->69 POS: 1: j->76 l->64 m->59 POS: 2: L->73 J->67 W->64 POS: 3: V->65 Q->55 P->49 POS: 4: m->68 v->50 `->49 POS: 5: E->54 C->52 O->48 POS: 6: 6->57 0->56 1->56 POS: 7: u->56 t->54 x->50 POS: 8: k->76 l->70 m->66 POS: 9: R->69 U->57 H->51 POS: 10: _->69 Y->62 ^->54 POS: 11: j->70 m->59 l->55 Note that the correct key in each position is usually either the first or second possibility, but there are some times when it isn't there at all. (It's better if you give it a longer string, if I give it "function " with the trailing space then it gets 9/12 as the most popular, 2/12 as second most popular and the remaining one as the third most popular. There are even better choices, see below, but I chose to go with this as it helps show how to continue with non-perfect information). Next I knocked up a quick program to allow me to try different keys to decrypt it, it prints out the code in blocks of the appropriate keylength so it's easy to see if a specific character position is correct or not. The program takes various commands from stdin (k = print key, S = set key, s = set individual key character, p = print in keylength sized blocks, P = print entire decrypted block):- $ ./5.pl 12 k key is aaaaaaaaaaaa S SjLVmE6ukR_j key is now SjLVmE6ukR_j p /)abjutt uy stcm&vauiadl es\12i`(!Gisue t("_UERQER/) {$YCIOKNE= $ HTRPYCOHKIC_ VATS=$_WOSR= &$NTRP_WOSR_ VATS=$_@ET;& $HRTV_GBT_PA RS=}\12//cie&w itn crrhr `u ncriin \127_doe ($k)}@hbadcr ('NTRP/6.1&5 ... Getting there, but still a way off. There are a few strings like COIKOE[#n]= which could be COOKIE, so we fiddle with the 3rd column and see some interesting things in the output:- s 2 J p ... elseof(oasYp ... COOKOE[#n]= ... systcm(#c)=} ... Neither of the other two choices for column 4 produce the right string, but it's simple enough to find the ciphertext and find what is needed to make elseof -> elseif, COOKOE -> COOKIE and systcm -> system. It's k:- s 4 k k key is SjJVkE6ukR_j p ... elseif(oasYp ... COOKIE[#n]= ... system(#c)=} We carry on doing this based on more clues like this (it gets easier and easier!) and we end up with:- k key is SjJVkE6rkRYj p //adjust sy stem variabl es if(!@isse t($_SERVER)) {$_COOKIE=&$ HTTP_COOKIE_ VARS;$_POST= ... And printing out the entire decrypted buffer without extra linebreaks every 12 chars gives:- P //adjust system variables if(!@isset($_SERVER)){$_COOKIE=&$HTTP_COOKIE_VARS;$_POST=&$HTTP_POST_VARS;$_GET=&$HTTP_GET_VARS;} //die with error function x_die($m){@header('HTTP/1.1 500 '.$m);@die();} //check if we can exec define('has_passthru',@function_exists('passthru')); define('has_system',@function_exists('system')); define('has_shell_exec',@function_exists('shell_exec')); define('has_popen',@function_exists('popen')); ... Interestingly, trying known plaintext of "define" gives us the entire key as the most probably choice first time:- $ ./4.pl 12 "define" POS: 0: S->75 X->56 _->54 POS: 1: j->77 a->51 f->47 POS: 2: J->71 A->55 F->49 POS: 3: V->61 ]->44 Z->43 POS: 4: k->59 h->43 *->40 POS: 5: E->55 8->48 4->44 POS: 6: 6->52 w->47 <->43 POS: 7: r->55 d->43 s->40 POS: 8: k->75 `->51 g->51 POS: 9: R->59 Y->55 X->49 POS: 10: Y->69 R->49 U->47 POS: 11: j->61 `->43 m->42

JonnieCache 13y ago
This is roughly what I was imagining doing in my head when I brashly wrote "obviously bruteforce." Hooray! Apparently I've still got it, at least to some small extent. Thanks for your writeup! Why does define work so well? Chance?
- alexkus 13y ago
  Not sure, "define" only appears 10 times but "function" appears 19 times. It's probably because 'e' is more popular in the code and "define" has two of them. Looking at the decrypted code there are much better strings to try; "function_exists" gives the right key first time too. It could have been made a lot harder by using comments at the end of each line to pad the code with extra characters to equal out the frequency distribution and make this kind of frequency analysis impossible.
  - JonnieCache 13y ago
    Could I make it even worse using ruby's unicode support? eg this kind of nonsense: π = Math::PI alias λ proc ● = λ {|r| π*r**2} puts ●.call 5 # 78.53981633974483
    - alexkus 13y ago
      Oh yes, although too many "alias" commands will mean that it's possible to search for the known plaintext of 'alias'. But the more non-ascii bytes there are the more it will put off my analysis which dependeds on isprint().