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hoppingbunny123
Posts: **723** Trinity

binary

make path using binary, steps of binary power = path

= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0), (1 or 0), (1 or 0), (1 or 0)]]

learn path = one binary at a time completes binary power.

= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1), (1), (1), (1), (1)]]

redo binary power correcting binary errors in subsequent versions to match the number made in the binary power

if previous binary path = true, new binary path may also be true.

changing the binary power to be true for a lower binary power, 2 to the 0 power for example = 1,

greater binary power which does not know this binary value of 2 to the power of 1 = 0 incorporates the value into the number their binary power makes

= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1)]]

= [2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0)]]

in creating the value the binary power equals, you incorporate lower binary power values into the greater binary power value which sorts out errors = blend binary path

you branch out the various nouns into their own verb binary path, a verb binary path merges greater binary power paths with lower binary power paths

to make the greater binary path true.

the greater the binary power, the more smaller binary values have to be made to fit the greater binary power, the harder to complete the harder to merge.

for example the number pi, you can merge pi to math but not all of the number pi.

and for things that are a rational number you need to have the entire binary power path, this is considerate.

so fitting a irrational number to a rational number is corrupt. you need to take only a part of the irrational number to a rational value = proportion.

therefore, if a path = a binary to some power, made up of smaller binary power that are themselves made up of smaller binary power,

reversing the largest binary value down each power in the path to sort out its origin might fix the one largest path to sort out the smaller paths to their value,

but the smaller value or path is tied to the larger value so it holds the smaller values in relation to the largest path value.

therefore in sorting out the return to the smallest value the largest value having sorting out the smaller values to their position,

keeps the largest value open as reference for that value or power till that power has completed its return to the smallest value.

= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1), (1)]]

= [2 ^ 4) = [(1 or 0)]]

therefore, to remove smaller power values inaccessible to the larger path power, the larger path power creates a path intermediate that removes the small power value for the larger power path.

= 2 ^ 3

so thats my theory. in practice if you jimmy up an algorithm using my theory the bot would see it needs the intermediate to generate the result wanted.

the binary goes up to the problem, but not down to the solution.

make path using binary, steps of binary power = path

= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0), (1 or 0), (1 or 0), (1 or 0)]]

learn path = one binary at a time completes binary power.

= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1), (1), (1), (1), (1)]]

redo binary power correcting binary errors in subsequent versions to match the number made in the binary power

if previous binary path = true, new binary path may also be true.

changing the binary power to be true for a lower binary power, 2 to the 0 power for example = 1,

greater binary power which does not know this binary value of 2 to the power of 1 = 0 incorporates the value into the number their binary power makes

= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1)]]

= [2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0)]]

in creating the value the binary power equals, you incorporate lower binary power values into the greater binary power value which sorts out errors = blend binary path

you branch out the various nouns into their own verb binary path, a verb binary path merges greater binary power paths with lower binary power paths

to make the greater binary path true.

the greater the binary power, the more smaller binary values have to be made to fit the greater binary power, the harder to complete the harder to merge.

for example the number pi, you can merge pi to math but not all of the number pi.

and for things that are a rational number you need to have the entire binary power path, this is considerate.

so fitting a irrational number to a rational number is corrupt. you need to take only a part of the irrational number to a rational value = proportion.

therefore, if a path = a binary to some power, made up of smaller binary power that are themselves made up of smaller binary power,

reversing the largest binary value down each power in the path to sort out its origin might fix the one largest path to sort out the smaller paths to their value,

but the smaller value or path is tied to the larger value so it holds the smaller values in relation to the largest path value.

therefore in sorting out the return to the smallest value the largest value having sorting out the smaller values to their position,

keeps the largest value open as reference for that value or power till that power has completed its return to the smallest value.

= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1), (1)]]

= [2 ^ 4) = [(1 or 0)]]

therefore, to remove smaller power values inaccessible to the larger path power, the larger path power creates a path intermediate that removes the small power value for the larger power path.

= 2 ^ 3

so thats my theory. in practice if you jimmy up an algorithm using my theory the bot would see it needs the intermediate to generate the result wanted.

the binary goes up to the problem, but not down to the solution.

so the intermediate 2 to some power, is held open binary to some power, made up of smaller binary power that are themselves made up of smaller binary power, eventually to 2 ^ 0. each time this happens a new pice of the puzzle is added until the entire puzzle can be solved, all the pieces are there.

in my theory, the bot would see it needs to sort out a value to its original spot, it cant, so it uses an intermediate paths to generate the result wanted.

im not able to test this theory practically yet, im still learning my binary theory and other maths and game programming but for future reference both to myself and others and people who could jimmy up an algorithm using my theory i put this here now.

i could have put this in the off topic forum but its useful for bots which might come in handy for oculus.

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## Comments

2,274Valuable Player723Trinity723Trinitya -> b

let a = b

a^ 2 = ab

a^2 - b^2 = ab - b^2

(a + b)(a - b) = b(a - b) // a - b = 0

a + b = b

b + b = b

2b = b

2 = 1

let (2 ^ 4) = (2 ^ 2 = 4)

(2 ^ 4)^ 2 = (2 ^ 4)(2 ^ 2 = 4)

(2 ^ 4)^2 - (2 ^ 2 = 4)^2 = (2 ^ 4)(2 ^ 2 = 4) - (2 ^ 2 = 4)^2

((2 ^ 4) + (2 ^ 2 = 4))((2 ^ 4) - (2 ^ 2 = 4)) = (2 ^ 2 = 4)((2 ^ 4) - (2 ^ 2 = 4)) // (2 ^ 4) - (2 ^ 2 = 4) = 0

(2 ^ 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)

(2 ^ 2 = 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)

2(2 ^ 2 = 4) = (2 ^ 2 = 4)

2 = 1

2,274Valuable Player723Trinityits very unpleasant trying to log in. i have a very difficult time trying to log in. its egregiously tedious and demanding work. so much so its distressful i cant log in for my username and password in multiple browsers for so many tries. by the time i finally get in im so fatigued by it all i have no joy in posting whatsoever.

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723Trinity2,274Valuable Player723Trinity2,274Valuable Player14,879Valuable PlayerAre you a fan of theMystgames? Check out my Mod at http://www.mystrock.com/Catch me on Twitter: twitter.com/zenbane

1,859Valuable PlayerI'm a software dev and I don't have any idea what you're talking about. When you explain something the most important thing is the high level non-technical description. Can you explain what the problem you're trying to solve is and just generally what your solution is?

Is this a path search algorithm like A* search algorithm?

723Trinity723Trinity14,879Valuable PlayerAre you a fan of theMystgames? Check out my Mod at http://www.mystrock.com/Catch me on Twitter: twitter.com/zenbane

723Trinity723Trinity723Trinity8073Jane723Trinity723Trinity([(->, <-), (->, <-)] + (->, <-)) = 1

becomes the factor in the [] brackets goes on the right side of the + sign = the true factor is offended:

([(->, <-), (->, <-)] + (->, <-)) = 1

([(->, <-), (->, <-)] + (->, <-) + ->) = 0

([(->, <-), (->, <-)] + (->, <-) + (->, <-)) = 1

becomes the factor on the right side of the + sigh goes back to be the factor in the [] brackets = the true factor becomes the offender;

([(->, <-), (->, <-)] + ->) = 0

([(->, <-), (->, <-)] + (->, <-)) = 1

([(->, <-), (->, <-), (->, <-)] + (->, <-) + ->) = 0

([(->, <-), (->, <-), (->, <-)] + (->, <-) + (->, <-)) = 1

([(->, <-), (->, <-), (->, <-), (->, <-)] + (->, <-) + (->, <-) + ->) = 0

...

then for every increment the false factor makes, there is a corresponding increment the true factor makes keeping the false factor perpetually false.

that is the error in multiplicative logic. its the proverbial left side or left hand that's culturally associated as a sign of evil or malpractice.

the converse of this is gud or addition, and that comes from 0, 1, and the left hand increasing the factor in the [] brackets is counted completely, and as the multiplication adds in more sets they're added in to the total number, which results in the addition moving left so that the addition pushes left on the offender.

the offender has then to do both being offended = (1, 0, 1), and being the offender = (0, 1, 0), while the addition has to do only the total sum of left pointing arrows to make, incrementing the value, and so pushing left as addition is drawn left.

6,735Volunteer Moderator5,465Volunteer Moderatorbut only on a tuesday

723Trinitywhat there is is there's two mechanisms at odds, and the two fighters fight to decide who gets what mechanism.

- the first mechanism is the back and forth or starting over after a flub.

- the second mechanism is taking the flub and building on it in a sweep.

- mechanism 1 = back and forth

- mechanism 2 = sweep

the whole thing works like opening a door, and then walking through it.

the opponent is the door, they are opened, then they have to close again and this opening and closing is the back and forth to fix a flub.

as the door is opened, the person who opened the door walks through the door and this is a sweeping motion of first opening the door then walking through it.

to get the door open the door opener has to physically take the closed door, and make it have a flub that the door tries to fix by closing.

this door opening is a physical interaction from the door opener to the door, and that is different martial arts strategies of striking or grappling.

if the door is too heavy, you can try to pull the door open but it wont open, so there you have the body strength difference at play. a woman might have the know how to open a door like a fire fighter ramming a door to open it, running into the door, but the door is too strong for the light body of the woman.

if guns, they say the first to land a shot usually wins the fight, so here the first hit is the door being opened. and so if you have a fight with weapons and they can land a shot then you have to take cover, making the door heavy by manoeuvring yourself to cover. then taking opportunity to open them like a door.

and this leads to juking having a place where juking can keep you from getting moved as a door.

juking has two mechanisms;

- circle parameter - the fighter is the door trying not to be opened hides its degree on the circle, and then hides its 3d attributes.

- circle center - the fighter trying to open the door is the circle centre moving to the circle degree to open the door, and has to account for the 3d attributes too.

if the target can juke successfully, their circle degree is hidden, and the 3d attributes help hide the door.

now to problem solve this you need to find the degree the opponent is on, then target it this means the circle will be filled at that 3d space so that as the opponent is moving on that 3d space at various degrees the degree wont matter so much because the entire 3d space for the circle plane is being filled.

as seen in the clip from the movie serenity where rivers brother uses the flash stun tool to free river.

this has a time limit, because they can hit you and you have to take this into account to keep the door too heavy to move.

the problem happens when juking works then the circle centre is compromised and transferred to the opponent and then your the target on the circle parameter. you had a time to become movable in return for hitting them but missed now your exposed.

in boxing you have the movement of moving to the side when its obvious to fight commentators you used up your time to go forward. so you create a obstacle by moving to the side so there's a re-balancing that happens much like tight rope walking.

how does math of addition and multiplication fit into this?factor in morals and that adds another re-balancing to be done of when to move to the side and become passive and so stop the offence.

if you have the skill to not have to move to the side its still technically wrong for the reason of having to move to the side if your using the whole scope of people who have to do it not just the strongest fighter, the weaker fighter would have to step off and re-balance so the strongest fighter has to as well. hence morals of re-balancing to stop from being oppressive come into play and so morals actually fit in with the fighting mechanism. the group consists of both multiplication and addition, hence if multiplication is true addition also has to be true to have the group true.

i watched a Bloomberg documentary of using brain implants to help a paralyzed man regain the use of his hands. he had some mobility regained but not enough to be useful. showing me that there are sets or movements involved in moving the hand functionally. looping the (multiplication to addition) to have fine movements.

the first loop is the most crude set, set 0, then set = set + 1, have increasingly refined movements. this is what must stop the multiplication from looping forever, because the set has multiplication tied to addition. and while multiplication might be immoral, its still tied to the set that includes addition hence whats wrong isn't based on strength but math. and the math to refined movements in increment the loop counting the set. if the multiplication ignores addition then the loop incrementing finer movement stops and your paralyzed, or crippled. so absolute strength must include addition and so have fine meowvement.

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723Trinity2,274Valuable Player6,932Valuable Player6,001Volunteer Moderator723Trinity