# Solutions to midterm exam - CSE - USF

## Solutions to midterm exam - CSE - USF

Problem  1  (10  pts)     1. (5pts)  The  following  is  the  attribute  for  a  file  in  Unix     -­‐r-­‐srw-­‐-­‐-­‐x        1  simon        fac   ...

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Problem  1  (10  pts)     1. (5pts)  The  following  is  the  attribute  for  a  file  in  Unix     -­‐r-­‐srw-­‐-­‐-­‐x        1  simon        fac                      0  Jul  18  21:55  get_score     User  “simon”  and  “jay”  belong  to  group  “fac”,  user  “daniel”  belongs  to   group  “student”.     List  all  privileges  (r,w,x)  each  of  the  three  users  have  on  the  file  (note:   the  setuid  bit  “s”  also  indicates  execution  privilege  x).     simon:  read,  execution     jay:  read,  write     daniel:  execution         2. (5pts)  When  user  jay  executes  the  get_score  program  above,  which  user(s)’s   privilege  can  the  process  have?  What  if  user  “simon”  executes  the  program?   Explain  your  answers.     Hint:  the  program  may  manipulate  the  various  uid  fields  in  the  process’s   control  block  as  we  learned  in  the  setuid  paper.       Answer:   When   user   jay   executes   the   program,   because   the   setuid   bit   is   set,   the   process   will   have   euid   set   to   simon’s   uid   so   the   process   will   have   simon’s   privilege.   In   addition,   the   real   uid   field   will   be   set   to   jay’s   uid,   and   the   program   can  swap  it  into  the  euid,  so  the  process  could  also  have  jay’s  privilege.     When  simon  executes  the  program,  both  uid  fields  will  be  set    to  simon’s  uid,  so   it  only  has  the  owner,  simon’s  privilege.

Problem  2  (10  pts)     A   program   has   a   buffer   overflow   vulnerability   on   the   stack.   The   operating   system   adopts  stack  address  randomization  as  we  explained  in  class.  It  also  uses  canary  to   detect  overridden  saved  EIP.  For  each  of  the  following  exploits,  explain  whether  it   will  work  on  this  system  or  not;  explain  your  answers.     1. (3   pt)   A   buffer   overflow   exploit   to   overwrite   the   saved   EIP   on   the   stack   to   point  to  the  shellcode  injected  on  the  stack.     Answer:   This  exploit  cannot  work.  Before  overwriting  saved  EIP  on  the  stack   the   canary   word   would   be   corrupted,   therefore   the   program   will   be   able   to   detect   the   attack.   Also   because   of   the   ASLR   protection,   the   stack   address   changes   every   time   when   the   program   launches;   the   attack   is   unlikely   to   accurately  redirect  the  program  to  a  reliable  address  of  shellcode.       2. (3   pt)   A   buffer   overflow   exploit   to   corrupt   the   local   variables   on   the   stack   frame  and  creates  a  situation  that  allows  the  attacker  to  modify  any  arbitrary   memory  location  with  arbitrary  value.  Then  use  this  to  modify  the  saved  EIP   on  the  stack  to  point  to  a  “jump  code”  in  a  fixed  memory  location  to  indirectly   jump  into  the  buffer  where  the  shellcode  is  injected.     Answer:   This  attack  will  not  work.  Even  though  the  attackers  can  modify  any   arbitrary  memory  location,  the  attacker  is  not  able  to  predict  where  the  saved   EIP  is  on  the  stack  due  to  stack  address  randomization.       3. (4   pt)   A   buffer   overflow   exploit   to   corrupt   the   local   variables   on   the   stack   frame  and  create  a  situation  that  allows  the  attacker  to  modify  any  arbitrary   memory  location  with  arbitrary  value.  Then  use  this  to  modify  the  entrance   address  of  a  system  function  that  will  be  called  by  the  program,  to  point  to  a   “jump   code”   in   a   fixed   memory   location   to   indirectly   jump   into   the   buffer   where  the  shellcode  is  injected.     Answer:   This   attack   can   work   reliably.   Because   the   attacker   can   overwrite   arbitrary   memory,   the   attacker   can   modify   the   entrance   address   of   a   system   used  function  no  matter  how  random  the  stack  address  is.  Because  the  function

will   be   called   by   the   program,   there   is   no   need   to   overflow   the   buffer   on   the   stack  to  reach  the  saved  EIP,  so  canary  will  not  protect  this.

Problem  3  (6  pts)     Which   of   the   following   OS   protection   mechanisms   can   be   circumvented   by   return   into  libc  attack.  Explain  your  answers  for  both  the  yes  and  no  cases.     1. Memory  address  randomization  for  stack.     Answer:  Yes,  this  protection  can  be  circumvented  by  return-­‐into-­‐libc  attack.  Even   the  stack  memory  address  is  randomized,  the  attacker  can  still  figure  out  how  long   a   string   is   needed   to   exactly   overwrite   the   saved   EIP,   find   a   reliable   entrance   address  of  a  libc  function,  which  is  not  randomized.  Therefore  it  is  possible  for  the   attacker  to  take  control  and  hijack  the  program.     2. Memory  address  randomization  for  shared  libraries.     Answer:   No,   this   protection   cannot   be   circumvented   by   return-­‐into-­‐libc   attack.   Return-­‐into-­‐libc   attack   requires   a   reliable   entrance   address   for   shared   libraries.   Even   if   we   assume   the   attacker   already   has   a   way   to   overwrite   saved   EIP,   memory   address   randomization   for   shared   libraries   hinders   attacker’s   ability   to   find   a   reliable  entrance  address  in  the  libc  library.       3.  Non-­‐executable  stack     Answer:   Yes,  this  protection  can  be  circumvented  by  return-­‐into-­‐libc  attack.  Even   though   the   stack   is   non-­‐executable,   the   shared   library,   which   contains   necessary   function   calls   such   as   ‘system()’     has   to   be   executable.   By   using   return-­‐into-­‐libc   attack,   the   attacker   jumps   to   existing   executable   code,   which   circumvents   the   limitation  of  non-­‐executable  stack       Problem  4  (4  pts)     In   earlier   days   web   service   programs   were   often   run   as   root   user.   Later   on   those   programs   are   run   by   default   as   a   user   with   limited   privilege   on   the   system,   e.g.,   it   can  only  access  a  portion  of  the  file  system  that  contains  the  website’s  data.  From  a   security  perspective,  explain  why  this  change  happened.     Answer:   Root  user  is  the  most  powerful  user  account  in  a  system.  When  web  service   programs   run   as   root,   if   there   are   vulnerabilities,   such   as   buffer   overflow,   in   the   program,   attacker   can   take   advantage   of   such   vulnerabilities   to   alter   the   normal   control   flow   and   execute   arbitrary   code   on   behalf   of   the   root   user,   which   basically

gives  the  attacker  all  privileges  on  the  system.  By  running  web  service  programs  as  a   user   with   limited   privilege,   even   if   such   attacks   happen,   the   attacker   can   only   get   as   much   privilege   as   the   limited   user.   The   attacker   has   to   find   more   vulnerabilities   in   order  to  take  total  control  of  the  whole  system,  which  makes  his/her  job  more  difficult,   hence  the  system  is  more  secure.       Problem  5  (10  pts)     Why   are   password   hashes   stored   on   the   system?   Explain   both   the   need   of   storing   the  hashes,  and  the  security  of  storing  hashes.  What  properties  of  hash  functions  are   important  for  each  of  the  aspects?       Answer:   System  stores  password  hashes,  because  the  system  needs  evidence  to  verify   user   provided   password   in   order   to   authenticate   user.   The   reason   why   storing   password  hashes  rather  than  plaintexts  or  encryption  is  that  the  hashing  process  is  not   reversible.  By  storing  password  hashes  even  the  system  is  compromised  it  is  hard  to  get   the   clear-­‐text   passwords.   This   is   due   to   the   pre-­‐image   resistance   property   of   hash.   Hash’s   collision   resistance   property   ensures   that   if   the   calculated   hash   matches   the   stored  one,  the  user  input  must  be  identical  to  the  correct  password.       Problem  6  (10  pts)     In  order  to  ensure  data  transmitted  through  Internet  is  not  tampered  with,  Alice  and   Bob  conduct  the  following  protocol  in  communication.     1. Alice   sends   a   message   m   along   with   its   digest   c   =   H(m)   where   H   is   a   cryptographic  hash  function.   2. Upon  receiving  m  and  c,  Bob  verifies  that  c  =  H(m).       Can   this   protocol   protect   the   message’s   integrity   during   transmission?   Explain.   If   you  think  this  protocol  does  not  achieve  the  intended  objective,  can  you  modify  it  to   make  it  work?     Answer:   This   protocol   cannot   protect   the   message’s   integrity   during   transmission.   Cryptographic  hash  function  is  not  encryption.  Those  hash  algorithms  are  well  known   and   publicly   available.   Suppose   there   is   a   man   in   middle   capable   of   intercepting   messages   between   Alice   and   Bob.   The   attacker   can   send   arbitrary   massage   m’   and   using  the  same  hash  function  c’  =  H(m’)  and  send  (m’,  c’)  to  Bob.  When  Bob  receives  the   message,  there  is  no  way  for  him  to  be  sure  whether  the  message  is  from  Alice  or  not.     In  order  to  guarantee  the  message’s  integrity,  Alice  and  Bob  can  exchange  their  public   keys  offline  (or  through  other  secure  protocols).  Alice  and  Bob  can  sign  messages  using   his/her   own   private   key   and   send   the   plaintext   and   the   digital   signature   together

through   the   network.   When   the   other   side   receives   the   message   he/she   can   verify   signature  on  the  message  using  the  sender’s  public  key.       Alternatively,   since   we   only   care   about   the   message’s   integrity,   Alice   and   Bob   can   pick   a  shared  key  K  offline  (or  through  other  secure  protocols).  To  communication,  they  can   send   message   m   with   h,   where   h   =   MAC(m,   K),   when   the   other   side   receives   the   message,   he/she   can   verify   the   integrity   by   applying   the   same   MAC   calculation   with   message  m:  h’  =  MAC(m,  K),  if  h  =  h’  then  the  receiver  can  be  assured  the  integrity  of   message  is  not  compromised.