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@lastforkbender
Created July 19, 2025 09:18
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Collapsed LUYH Equation Complex Resolve
# Collapsed LUYH Equation Complex Set Solution:
# The mind bending modulus solve luyh equation done with slots & list comprehensions
# and+ collapsed forced n-resolves. Use of paired numbers @iks more probable returns
# of -1. If you need the specific luyh series number found set returnLuyhSeries=True.
# Is computational intense and needs to be placed in a multi-processing distinction.
# (Solid use for top-grade trignometric encryption solutions involving seq-erroring)
#_______________________________________________________________________________________
import cmath
import math
#_______________________________________________________________________________________
class COLLAPSED_LUYH():
slots = ['_lstA', '_lstB', '_lstC', '_lstD', '_lstE', '_lstDrd1', '_lstDrd2',
'_lstLe1', '_iks1', '_intA', '_intB', '_intC', '_strA', '_blnA']
def __init__(self):
pass
#_______________________________________________________________________________________
def rotate_pair(self, p: tuple, k_chr: str) -> tuple:
if k_chr == 'L':
return (p[1], p[0])
elif k_chr == 'U':
return (p[0]+1, p[1]+1)
elif k_chr == 'Y':
return p
elif k_chr == 'H':
return (p[0]-1, p[1]-1)
else:
return p
#_______________________________________________________________________________________
def drhl_distribute(self, isRtn: bool, n: int) -> tuple:
if n == 0:
return [[]]
else:
self._lstA = self.drhl_distribute(isRtn, n-1)
self._lstB = [(complex(math.floor(math.sqrt((self._intB**self._intC)/(math.pi**3))), self._intB+1), complex(math.floor(math.sqrt((self._intB+self._intC)/(self._intC*math.pi))), self._intC-1)) for self._intB in range(1, self._intA+1) for self._intC in range(1, self._intA+1)]
self._lstC = [self.rotate_pair(self._lstD, self._iks1[self._intB%len(self._iks1)]) for self._intB, self._lstD in enumerate(self._lstB)]
if isRtn:
return self._lstA+self._lstC
else:
return self._lstB+self._lstA
#_______________________________________________________________________________________
def luyh_resolve(self, isReassemble: bool, halfLoopMode: int):
if isReassemble:
self._lstLe1 = list(set([(cmath.sqrt(self._intB[0]*cmath.pi**self._intB[1])-self._intC[1])/cmath.acos(self._intC[0]*self._intB[1]*math.pi) for self._intB in self._lstDrd1 for self._intC in self._lstDrd2]))
else:
if halfLoopMode == 2:
self._lstLe1 = [(abs(self._intB.real)-abs(self._intB.imag))*(math.cos(self._intB.real+math.pi)**math.sqrt(self._intB.imag+math.pi)) for self._intB in self._lstLe1]
else:
self._lstLe1 = [(((math.sin(self._intB.real+self._intB.imag-math.pi)/math.sqrt(math.pi*4))*math.pi)%(self._intA-self._intB.imag-self._intB.real))/math.pi for self._intB in self._lstLe1]
#_______________________________________________________________________________________
def check_iks_and_n(self, iks: str, n: int):
self._intA = n
if isinstance(iks, str) and isinstance(n, int):
if n > 5:
raise Exception(f'<luyh err> invalid param @n={n}, impossible without a quantum computer; 1-5 valid @n')
else:
raise Exception('<luyh err> invalid luyh equation set params; iks<str> & n<int> are valid params only')
#_______________________________________________________________________________________
def set_iks(self):
self._lstA = []
for self._strA in iks:
if self._strA == '1': self._lstA.append('L')
elif self._strA == '2': self._lstA.append('U')
elif self._strA == '3': self._lstA.append('Y')
elif self._strA == '4': self._lstA.append('H')
else:
raise Exception(f'<luyh err> invalid iks sequence, ({self._strA}) not a 1-4 digit')
self._iks1 = ''.join(self._lstA)
#_______________________________________________________________________________________
def luyh_half_loops_operator(self, returnLuyhSeries: bool, iks: str, n: int):
self.check_iks_and_n(iks, n)
try:
self._lstE, self._iks1, self._bln = [], iks, True
while 1:
if self._bln: self._bln = False
else:
if len(self._iks1) > 1:
self._iks1 = self._iks1[1:len(self._iks1)-1]
else:
break
self._lstDrd1 = self.drhl_distribute(True, self._intA)
self._lstDrd1.pop(0)
self._lstDrd2 = self.drhl_distribute(False, self._intA)
self._lstDrd2.pop(len(self._lstDrd2)-1)
self.luyh_resolve(True, 1)
self.luyh_resolve(False, 2)
self.luyh_resolve(False, 4)
self._lstE.append(math.trunc(sum(self._lstLe1)))
self._intB = sum(self._lstE)
self._intC = math.floor((self._intB*max(self._lstE)*min(self._lstE)*(self._intA+min(self._lstE)))*((self._intA*max(self._lstE))*self._intA))
if self._intC%self._intB == 0:
if not returnLuyhSeries:
return self._intB
else:
return (self._intC, self._intB)
else:
return -1
except Exception as lattice_luyh_iks_pair:
return -1
#_______________________________________________________________________________________
def luyh_resolve(returnLuyhSeries: bool, iks: str, n: int):
cls = COLLAPSED_LUYH()
return cls.luyh_half_loops_operator(returnLuyhSeries, iks, n)
#_______________________________________________________________________________________
def test():
# luyh keys are 1-4 just as LtoR or RtoL count on a Menorah center; also yo hands
print(luyh_resolve(True, '434123421444312321421', 5))
test()
@lastforkbender

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Okay. First, the LUYH Equation is a very hard trigonometric problem to solve with large iks keys, both this version that doesn't need the key to be 1-4 specifically chars or the non-forced real rotational based LUYH Equation resolve that must be 1-4 chars. Of sequence erroring use you would typically splice in end to ends or begin to begins with a known standard LUYH Equation solve of rotation erroring. Thus, you can use this version hidden of it's specific use yet still have a length to length solve. Both variations are based on Menorah half-loop complex rotational solves.

@lastforkbender

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Secondly, the quantum version of the LUYH Equation, n being greater than 5. That requires triple recursion inside a specific type of vector bracketing. Once at n=6 then is all half-loops whole and a classical computer cannot perform.

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