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Help with reversing a function

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  Help with reversing a function Tenjin <jdiscordbusiness@gmail.com> - 2021-08-30 05:13 -0700
    Re: Help with reversing a function Julio Di Egidio <julio@diegidio.name> - 2021-08-31 08:05 -0700

#40362 — Help with reversing a function

<?xml version="1.0"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html>
  <head>
    <title>Fun with && and ||</title>
    <script type="text/javascript">
      function g(s, C, m) {
        console.log(String.fromCharCode(96 + C));
        return (
          (!C && !s[m]) ||
          (C &&
            s[m] &&
            s[m++] == String.fromCharCode(96 + C) &&
            f(s, (++C * C++ + C++ * ++C) >> ++m, --m))
        );
      }
    </script>
  </head>
  <body style="text-align: center">
    <form
      onsubmit="alert((g(pass.value,1,0) ? 'Correct! The password is ' : 'Wrong! not ')+pass.value); return false;"
    >
      <div>
        <input type="text" name="pass" />
        <input type="submit" value="Login" />
      </div>
    </form>
  </body>
</html>

I need some help in finding what the correct password is. So far I understand that te first character must contain an "a"

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#40363

On Monday, 30 August 2021 at 14:14:45 UTC+2, Tenjin wrote:

> function g(s, C, m) { 
> console.log(String.fromCharCode(96 + C)); 
> return ( 
> (!C && !s[m]) || 
> (C && 
> s[m] && 
> s[m++] == String.fromCharCode(96 + C) && 
> f(s, (++C * C++ + C++ * ++C) >> ++m, --m))

I take that should read 'g', not 'f'.

> ); 
> } 

> onsubmit="alert((g(pass.value,1,0) ? 'Correct! The password is ' : 'Wrong! not ')+pass.value); return false;"

> I need some help in finding what the correct password is. So far I understand that te first character must contain an "a"

1) Firstly, few transformations [best viewed in monospace]:

1.1) Some clean-up and fixes:

(Fixed the exit conditions: note that 'false || undefined' is 'undefined', not 'false'.)

function g_1(s, c, m) {
    var L = s.length;
    return (
        (!c && m === L)
    ||  (c && m < L
        &&  s[m++] === String.fromCharCode(96+c)
        &&  g_1(s, (++c*c++ + c++*++c) >> ++m, --m))
    );
}

1.2) Resolve increments:

function g_2(s, c, m) {
    var L = s.length;
    return (
        (!c && m === L)
    ||  (c && m < L
        &&  s[m] === String.fromCharCode(96+c)
        &&  g_2(s, ((c+1)*(c+1) + (c+2)*(c+4)) >> (m+2), (m+1)))
    );
}

1.3) Unwind recursion and subsume 'm':

(We can subsume 'm' since it is the index into the string, bound to start from zero.)

function g(s, c) {
    var L = s.length;
    if (c === 0) {  // short-cut
        return L === 0;
    }
    for (var m = 0; m < L; ++m) {
        if(!(s[m] === String.fromCharCode(96+c))) {
            return false;
        }
        c = (c+1)*(c+1) + (c+2)*(c+4) >> m+2;
    }
    return c === 0;
}

2) Now from 1.3 we can extract the password generator:

(There is in 1.3 simply a comparison char by char with the expected password, so we can extrapolate the logic that just produces the password.)

function h(c) {
    var s = "";
    for (var m = 0; c !== 0; ++m) {
        s +=  String.fromCharCode(96+c);
        c = (c+1)*(c+1) + (c+2)*(c+4) >> m+2;
    }
    return s;
}

T) Some basic (and quick) testing in the console:

(Preliminarily, copy-paste the functions g_1, g, and h in the console.)

T.1) Check the code transformation:

-> h(1)
<- "adiorlc"

-> g_1("adiorlc", 1, 0)
<- true

-> g_1("adiorlC", 1, 0)
<- false

-> g_1("adiorl", 1, 0)
<- false

-> g("adiorlc", 1)
<- true

-> g("adiorlC", 1)
<- false

-> g("adiorl", 1)
<- false

T.2) Test the roundtrip:

-> g(h(100), 100)
<- true

-> g(h(100), 99)
<- false

-> var lo = -9, hi = 5; for (var c = lo; c <= hi; ++c) { var s = h(c), e_1 = g_1(s, c, 0), e = g(s, c); console.log(c, s, e_1, e); }
  -9 "Wxĉວ셭ﵗ黡繗目护눕뙉ꯡ찁ᗋȫa" true true
  -8 "XrÄմ鸾늲户抉韙哵많䮑䰶ⵐࡄ" true true
  -7 "Yl‘ƥ᩻侚碸酟풂툲㒂ጕ鮐벋ቘā" true true
  -6 "Zhy»ɼ␼ᨲ皏徕騐鼧弫Ⱪ빗ᩓ́c" true true
  -5 "[diorlc" true true
  -4 "\\bddb" true true
  -3 "]" true true
  -2 "^" true true
  -1 "_" true true
  0 "" true true
  1 "adiorlc" true true
  2 "bhy»ɼ␼ᨲ皏徕騐鼧弫Ⱪ빗ᩓ́c" true true
  3 "cl‘ƥ᩻侚碸酟풂툲㒂ጕ鮐벋ቘā" true true
  4 "drÄմ鸾늲户抉韙哵많䮑䰶ⵐࡄ" true true
  5 "exĉວ셭ﵗ黡繗目护눕뙉ꯡ찁ᗋȫa" true true
<- undefined

Seems to work fine for all 'c' integer, but I have neither *proved* that, nor I have proved that 'h' always terminates.

N.B.  It is definitely NOT a secure scheme...

Have fun,

Julio

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