[11.1.4] Triggered to a state that will induce catastrophic failure - Syntax - 8/28/08

No one can run from the Machines...No one.

(( Despite my eventual conclusion about the security system having infinite solutions, that was the most fun I've had in a while in MXO. ))

((As a programmer, I kinda wanna see what you guys were seeing just out of curiosity))

((If I still have the screens I took at the time I'll PM them to you.))

I was frustrated about not being able to help, my coding skills are based around clothing and abilities, which are simple compared to the override code we were faced with, so I was very little help when attempting to solve the equation.

Despite our best efforts cracking the code, the success of this operation was down to those below in making the area safe and protecting that bluepill without whom the plant would have gone into meltdown. Not only did they distract the attention of the Ouroboros security away from the infiltration group during our attempts, but they also allowed the Ouroboros (ex?) employee to feel safe enough to show up and get through the security to the terminal. So she, and by extention everyone below, saved untold numbers of innocent lives.

I wonder where she went after security eventually attacked us... Hopefully we'll see her again and see if she knows any more details about the current situation.

Bayamos wrote:

(( Despite my eventual conclusion about the security system having infinite solutions, that was the most fun I've had in a while in MXO. ))

((As a programmer, I kinda wanna see what you guys were seeing just out of curiosity))

OUROBOR-OS

PROCESSING...
LOADING PARAMETERS VALUES...
PROCESSING...
PROCESSING...
ERROR!!!

EXECUTION ERROR:UNKNOWN PARAMETER VALUE(S10)
PRINTING EXECUTION TRACE
----------------------------


int s1,s2,s4,s6,s8;
boolean s3,s5,s7,s9;

s10; //unknown type

//S1
int s1val;
for (int i=1;i<11;i++){
   if (!si.isNaN){ //si = s1, s2, ..., s10
      s1val+= si;
   }
}
s1 = s1val;

//S2
int s2val;
for (int i=1;i<11;i++){
   if (si){
      s2val++;
   }
}
s2 = s2val;

//S3
s3 = true;
for (int i=1;i<11;i++){
   if (si > s1){
      s3 = false;
   }
}

//S4
int s4val = 0;
for (int i=1;i<11;i++){
   if (si == s4){
      s4val++;
   }
}
s4=s4val;

//S5
s5 = true;
for (int i=1;i<11;i++){
   if (!si.isNaN){
      if (si < 0){
         s5 = false;
      }
   }
}

//S6
int s6sum;
int s6count;
for (int i=1;i<11;i++){
   if (!si.isNaN){
      s6sum += si;
      s6count++;
   }
}
s6 = s6sum / s6count;

//S7
if (s4 > s2){
   s7 = true;
}else{
   s7 = false;
}

//S8
s8 = s1/s8;

//S9
if (s6 == ((s2-s4)-(s8*s4))){
   s9 = true;
}else{
   s9 = false;
}

//S10 = ?


Please input S10:
>

PRESS [CTRL-H] OR DIAL #4559 FOR HELP
[ESC] EXIT TO DOS [UP / DOWN ARROW] MOVE CURSOR [ENTER] RUN

As far as I can tell, there is no unique solution to the value of s10.

So, how many real life computer programmer's chose to belong to the machine faction? Like someone there said, I boggled. I do not speak math!

Great job and thanks for saving the City.

S1: if it equals S1 + S2 + .... S10, S2 - S10 must collectively equal zero.
S2: # true
S3: true unless # > S1, pretty much guaranteed to be true
S4: # = S4
S5: false if there's a negative number, has to be false
S6: average, but we know that the sum is s1, so we can re-write as s1/6.
S7: true if s4 > s2
S8: S1/S8, which can be rewritten as +/- sqrt(S1)
S9: true if S6 = (s2-s4) - (s8 * s4)
S10: must be a negative integer because when S1 > 36, a negative S8 would decrease more slowly than a positive S6 grew. For numbers < 36, S1 can't be both divisible by 6 and a square root of something.

Working through this we come to the equilibrium situation where:

S1:
S2: 2
S3: true
S4: 2
S5: false
S6:
S7: false
S8:
S9: true
S10:

Otherwise you get into a mess where you have s2 and s4 both equalling 1 which doesn't work, so on so forth. However in order for this situation to exist we need the unlikely constraint of s9 to be true. That means S6 = (S2-S4) - (S8*S4).

Rewriting that,

S1/6 = 0 - ( +/- 2root(S1) ). Because S6 is positive, S8 has to be negative. So, S1/6 = 2sqrt(S1), bring it over, S1/6 - 2sqrt(S1). There's a solution to this equation where S1 = 144. With this, everything falls into place.


S1: 144
S2: 2
S3: true
S4: 2
S5: false
S6: 24
S7: false
S8: -12
S9: true
S10: -16

S1: if it equals S1 + S2 + .... S10, S2 - S10 must collectively equal zero.
S2: # true
S3: true unless # > S1, pretty much guaranteed to be true
S4: # = S4
S5: false if there's a negative number, has to be false
S6: average, but we know that the sum is s1, so we can re-write as s1/6.
S7: true if s4 > s2
S8: S1/S8, which can be rewritten as +/- sqrt(S1)
S9: true if S6 = (s2-s4) - (s8 * s4)
S10: must be a negative integer because when S1 > 36, a negative S8 would decrease more slowly than a positive S6 grew. For numbers < 36, S1 can't be both divisible by 6 and a square root of something.

Working through this we come to the equilibrium situation where:

S1:
S2: 2
S3: true
S4: 2
S5: false
S6:
S7: false
S8:
S9: true
S10:

Otherwise you get into a mess where you have s2 and s4 both equalling 1 which doesn't work, so on so forth. However in order for this situation to exist we need the unlikely constraint of s9 to be true. That means S6 = (S2-S4) - (S8*S4).

Rewriting that,

S1/6 = 0 - ( +/- 2root(S1) ). Because S6 is positive, S8 has to be negative. So, S1/6 = 2sqrt(S1), bring it over, S1/6 - 2sqrt(S1). There's a solution to this equation where S1 = 144. With this, everything falls into place.


S1: 144
S2: 2
S3: true
S4: 2
S5: false
S6: 24
S7: false
S8: -12
S9: true
S10: -16