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kier007/README.md

MAKARU

"The mathematics of perception exists between the quantum states of observation and transformation."

NEURAL FRAMEWORK v3.7.25

graph TD
    A[Perceptual Input] -->|Quantum Transform| B{Feature Extraction}
    B -->|Primary Path| C[Neural Encoding φ]
    B -->|Secondary Path| D[Eigenface Matrix Ω]
    C --> E[Temporal Integration]
    D --> E
    E -->|Cryptographic Hash| F[Identity Manifold]
    F -->|Schrödinger Transform| G[Perceptual Output]
    G --> H{Decision Boundary}
    H -->|Rejection| I[Recalibration Loop]
    I --> A
    H -->|Acceptance| J[Quantum Entanglement]
    J -->|Dimensional Collapse| K[Facial Transformation]
    
    classDef default fill:#0d1117,stroke:#6A11CB,color:#2575FC,stroke-width:2px;
    classDef main fill:#0d1117,stroke:#6A11CB,color:#FFFFFF,stroke-width:4px;
    classDef crypto fill:#0d1117,stroke:#2575FC,color:#6A11CB,stroke-width:3px;
    
    class A,G,K main;
    class F,J crypto;
Loading

First Riddle Key

Click to reveal riddle...
I travel at incalculable speeds yet never move. I process but do not think.

When $f'(x) = 0$ and $f''(x) > 0$, the local minima of the function:

$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$

reveals the first access key.

QUANTUM PERCEPTUAL MANIFOLD

sequenceDiagram
    participant Observer
    participant QuantumState
    participant NeuralNetwork
    participant FacialMatrix
    
    Observer->>QuantumState: Initiate Observation
    Note over QuantumState: Superposition Collapse
    QuantumState->>NeuralNetwork: Transmit Eigenvalues
    
    loop Transformation Sequence
        NeuralNetwork->>NeuralNetwork: Self-optimization
        NeuralNetwork->>FacialMatrix: Calculate Projections
        FacialMatrix-->>NeuralNetwork: Return Eigenvectors
    end
    
    FacialMatrix->>Observer: Project Transformed Image
    Note over Observer,FacialMatrix: Facial Recognition Uncertainty: ΔP·ΔQ ≥ ħ/2
    
    NeuralNetwork->>QuantumState: Reset Quantum State
    QuantumState-->>Observer: Ready for Next Observation
Loading

The transform of visual information across the neural-quantum boundary follows:

$$\int_{\Omega} \nabla f(x,y) \cdot \nabla \phi(x,y) , dA = \oint_{\partial \Omega} \phi(x,y) \nabla f(x,y) \cdot \hat{n} , ds - \int_{\Omega} \phi(x,y) \Delta f(x,y) , dA$$

Where $f(x,y)$ maps the domain of visual perception to the range of computational understanding, and $\phi(x,y)$ represents the cognitive field across manifold $\Omega$.

Second Riddle Key

Click to reveal riddle...
I have no weight, yet you can see me. I have no voice, yet I speak volumes.

For coefficient $a_n$ defined by:

$a_n = \frac{1}{n!} \frac{d^n}{dx^n} \left[ \prod_{k=1}^{n-1} (f(x) - a_k x^k) \right]_{x=0}$

When n=42, $a_n$ encodes the second access key in hexadecimal.

NEUROMORPHIC TRANSFORMATION PROTOCOL

classDiagram
    class FacialTransformation {
        +int dimensionality
        +float quantumCoupling
        +Matrix eigenfaces
        +performTransform(Face source, Face target)
        +calculateEigenvalues()
        -normalizeFeatureSpace()
        -applySchrödingerEquation()
    }
    
    class NeuralEncoder {
        +int[] layerStructure
        +float learningRate
        +Matrix weights
        +encode(Image input)
        +backpropagate(Error delta)
        -activationFunction(float x)
        -calculateGradient()
    }
    
    class QuantumObserver {
        +float uncertaintyPrinciple
        +int observationState
        +observe(QuantumState state)
        +collapseWavefunction()
        -calculateProbabilityMatrix()
        -entangleStates()
    }
    
    class FacialFeatures {
        +Vector landmarks
        +float[] vectorSpace
        +extractFeatures(Image face)
        +normalizeCoordinates()
        -calculateDistances()
        -createFeatureVector()
    }
    
    FacialTransformation --> NeuralEncoder : uses
    FacialTransformation --> QuantumObserver : triggers
    NeuralEncoder --> FacialFeatures : processes
    QuantumObserver --> FacialFeatures : observes
Loading

The central theorem of facial transposition can be formulated as:

$$\lim_{n \to \infty} \int_0^1 \frac{e^{-x^2} \sin(nx)}{1+x^2} , dx = \frac{\pi}{2e}$$

This applies iteratively to the convergence sequence:

$$S_n = \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} \cdot \prod_{j=1}^{k-1} \frac{4j^2}{4j^2-1}$$

Third Riddle Key

Click to reveal riddle...
I am present in every exchange but cannot be seen directly. 
Remove me and communication fails.

The third key is hidden in the alternating pattern of prime indices in $S_n$.

HYPERSPACE RESEARCH VECTORS

erDiagram
    NEURAL-FRAMEWORK ||--o{ PERCEPTUAL-DOMAIN : processes
    NEURAL-FRAMEWORK ||--|| QUANTUM-LAYER : integrates
    PERCEPTUAL-DOMAIN ||--o{ FEATURE-VECTOR : contains
    FEATURE-VECTOR }o--|| EIGENFACE-MATRIX : maps_to
    EIGENFACE-MATRIX ||--|| TRANSFORMATION-MATRIX : generates
    TRANSFORMATION-MATRIX ||--o{ FACIAL-OUTPUT : produces
    QUANTUM-LAYER ||--|| UNCERTAINTY-PRINCIPLE : adheres_to
    UNCERTAINTY-PRINCIPLE ||--o{ PROBABILITY-DISTRIBUTION : defines
    PROBABILITY-DISTRIBUTION }o--|| FACIAL-OUTPUT : constrains
    
    NEURAL-FRAMEWORK {
        float learning_rate
        int[] layer_structure
        function activation_type
    }
    
    QUANTUM-LAYER {
        complex[] eigenstates
        float uncertainty
        function wave_function
    }
    
    TRANSFORMATION-MATRIX {
        complex[][] coefficients
        float determinant
        boolean invertible
    }
    
    EIGENFACE-MATRIX {
        float[][] principal_components
        int dimensionality
        float variance_explained
    }
Loading

The multidimensional framework operates through parallel computations:

$$R_1 = \mathcal{L}{t^n e^{at}} = \frac{n!}{(s-a)^{n+1}}$$

$$R_2 = \mathcal{F}{f(t)} = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} , dt$$

$$R_3 = \prod_{n=1}^{\infty} \frac{1}{1-q^n} = \sum_{k=0}^{\infty} p(k)q^k$$

Where $p(k)$ represents the partition function from quantum information theory.

Fourth Riddle Key

I am the bridge between past and future. I am determined yet unpredictable.
Quantum State Eigenvalue Probability Amplitude Complex Phase Transformation Coefficient
Ψ₁ λ = 0x7F 0.732 + 0.214i π/7 Φ₁ = 0xD1A3B7
Ψ₂ λ = 0xA2 0.446 - 0.389i π/3 Φ₂ = 0xC4F102
Ψ₃ λ = 0xC4 0.911 + 0.000i 0 Φ₃ = 0xE30591
Ψ₄ λ = 0xD1 0.205 - 0.673i 5π/9 Φ₄ = 0x8F3A6C
Ψ₅ λ = 0xE3 0.558 + 0.461i π/4 Φ₅ = 0xF72E14

Plotting $R_3$ on the complex plane reveals a QR code when q=0.618.

CORE SYNTHESIS ALGORITHM

stateDiagram-v2
    [*] --> InitialState
    InitialState --> QuantumObservation
    QuantumObservation --> FeatureExtraction
    
    state FeatureExtraction {
        [*] --> LandmarkDetection
        LandmarkDetection --> GeometricAnalysis
        GeometricAnalysis --> FeatureVector
        FeatureVector --> [*]
    }
    
    FeatureExtraction --> DimensionalReduction
    DimensionalReduction --> EigenfaceCalculation
    EigenfaceCalculation --> TransformationMatrix
    
    state TransformationMatrix {
        [*] --> MatrixInitialization
        MatrixInitialization --> SingularValueDecomposition
        SingularValueDecomposition --> OptimalTransform
        OptimalTransform --> MatrixInversion
        MatrixInversion --> [*]
    }
    
    TransformationMatrix --> FacialWarping
    FacialWarping --> ColorCorrection
    ColorCorrection --> PostProcessing
    PostProcessing --> FinalOutput
    FinalOutput --> [*]
Loading

The principal facial transformation is governed by:

$$T(f) = \lim_{\epsilon \to 0} \frac{1}{2\epsilon} \int_{-\epsilon}^{\epsilon} f(x+t) , dt = f(x)$$

This evolves through the neural network backpropagation:

$$f_{n+1}(x) = f_n(x) - \frac{f_n(x)}{f_n'(x)} \cdot \left(1 - \frac{f_n(x) \cdot f_n''(x)}{2(f_n'(x))^2}\right)$$

Fifth Riddle Key

Born of light, yet I cast shadows. I exist only when observed.

Neural Iteration Results for f_7(3.14159):

Iteration Function Value Derivative
f₁(π) -0.9999999999999991 -3.739192358E-16
f₂(π) -0.9999999999999738 -7.522736845E-15
f₃(π) -0.9999999999982447 -3.902345671E-13
f₄(π) -0.9999999998546225 -6.782349012E-12
f₅(π) -0.9999999834754892 -7.812938471E-10
f₆(π) -0.9999973458103589 -5.293847192E-08
f₇(π) CODE:F7A31E28D0B5 KEY:19375AF82

The seventh iteration of $f_n(x)$ when x=3.14159 reveals the fifth key.

QUANTUM COMPUTATIONAL MATRIX

pie
    title Component Analysis of Neural Framework
    "Computer Vision" : 95.2
    "Neural Networks" : 89.7
    "Facial Analysis" : 97.8
    "Quantum Algorithms" : 84.3
    "Feature Extraction" : 92.1
    "Transformation Matrix" : 91.6
Loading

The quantum entanglement matrix uses the following coefficients:

$$M_{jk} = \frac{1}{2\pi} \int_0^{2\pi} e^{i(j-k)x} , dx = \delta_{jk}$$

Sixth Riddle Key

I am the face that isn't there, yet you know my features.
I change your appearance, yet leave you untouched.

Quantum Computing Status:

Algorithm Status Completion
Face Detection [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
Feature Extraction [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
Eigenface Calculation [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
Transformation Matrix [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
Quantum Entanglement [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
Neural Calibration [ACTIVE] 🟩🟩🟩🟩🟩🟩🟩🟩
M_173,37 Computation [CALCULATING] 🟩🟩🟩🟩🟩🟩⬜⬜
Euler Multiplication [PENDING] ⬜⬜⬜⬜⬜⬜⬜⬜

Compute $M_{jk}$ when j=173 and k=37, then multiply by Euler's number.

QUANTUM CONNECTION PROTOCOL

graph TB
    subgraph "Connection Layer"
        A1[Email Protocol] --- B1[Hash: 47A1F3]
        A2[Repository Access] --- B2[Hash: 89C2E5]
        A3[Professional Network] --- B3[Hash: D306B7]
    end
    
    subgraph "Encryption Layer"
        C1[RSA-4096] --- D1[Quantum Key]
        C2[AES-256] --- D2[Time-Based Token]
        C3[Homomorphic] --- D3[Zero-Knowledge Proof]
    end
    
    subgraph "Integration Layer"
        E1[Mathematical Solution] --- F1[Access Control]
        E2[Riddle Verification] --- F2[Authentication]
        E3[Quantum Entanglement] --- F3[Trust Establishment]
    end
    
    B1 --> C1
    B2 --> C2
    B3 --> C3
    
    D1 --> E1
    D2 --> E2
    D3 --> E3
    
    F1 --> G[Secure Communication Channel]
    F2 --> G
    F3 --> G
    
    G --> H{Decoded Access Point}
    
    classDef blue fill:#0d1117,stroke:#2575FC,color:#FFFFFF,stroke-width:2px;
    classDef purple fill:#0d1117,stroke:#6A11CB,color:#FFFFFF,stroke-width:2px;
    classDef green fill:#0d1117,stroke:#30D158,color:#FFFFFF,stroke-width:2px;
    
    class A1,A2,A3,B1,B2,B3 blue;
    class C1,C2,C3,D1,D2,D3 purple;
    class E1,E2,E3,F1,F2,F3,G,H green;
Loading

The following temporal integral reveals the communication vector:

$$\int_{0}^{\infty} \frac{\sin(ax)}{x} , dx = \begin{cases} \frac{\pi}{2}, & \text{if } a > 0 \\ 0, & \text{if } a = 0 \\ -\frac{\pi}{2}, & \text{if } a < 0 \end{cases}$$

Final Riddle

The limit of A_n as n approaches the atomic number of gold
reveals how to reach me across the digital void.

Where:

$$A_n = \sum_{k=1}^{n} \binom{n}{k} (-1)^{k+1} \frac{1}{k}$$

n A_n Value Partial Sum Convergence Distance
1 1.0000000000 1.0000000000 0.3068528195
2 0.5000000000 1.5000000000 0.1931471805
10 0.0976420412 1.9757837325 0.0242162675
25 0.0400953365 1.9983050203 0.0016949797
50 0.0200481125 1.9999232654 0.0000767346
79 0.0126682279 1.9999997182 0.0000002818
196 0.0051010392 2.0000000000 0.0000000000

The limit of A_n as n approaches 79 (the atomic number of gold) reveals how to reach me across the digital void.

Pinned Loading

  1. Deep-Live-Cam Deep-Live-Cam Public

    Forked from hacksider/Deep-Live-Cam

    real time face swap and one-click video deepfake with only a single image

    Python