Algebraic Attack via Boolean Polynomial Ring (Gröbner Basis Key Recovery)

Updated 13 Sep 2026

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  What it is

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    Model unknown key bits as Boolean variables, propagate the generator symbolically, and constrain its outputs to the observed bits. Gröbner bases or SAT can recover keys for tractable systems. A polynomial representation alone guarantees neither efficient solving nor a unique solution; every candidate needs independent output verification.

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  When to apply

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    Known operations and observed output yield a manageable system in a bounded number of unknown bits. Evaluate degree, variable count, structure, solver runtime, and whether the data actually distinguish candidates.

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  Symbols and assumptions

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    Unknown key bits are $k_0,\ldots,k_{n-1}$. Work in a Boolean polynomial ring over $GF(2)$ with $k_i^2=k_i$. Observed bit $o_t$ constrains symbolic output $f_t(k)$.

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  Math — step by step

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    1. Translate XOR to +, AND to multiplication, NOT x to $1+x$, and OR(x,y) to $x+y+xy$. Rebuild every state transition and output at the correct clock index.
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    2. Add equations $f_t(k)+o_t=0$. A solution is an assignment consistent with all supplied observations. More equations can remove candidates, but count alone proves neither uniqueness nor efficient solvability.
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    3. Gröbner bases or SAT can solve some such systems. Degree growth, number of variables and equation structure control cost; merely expressing a cipher as polynomials does not make it easy to break.
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    4. Enumerate or otherwise account for multiple solutions and regenerate output with each candidate. Do not assume the first returned assignment is the unique key.
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  Worked example

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    Let the observed equations be $k_0+k_1=1$ and $k_1k_2=1$. The second forces $k_1=k_2=1$ because a product of bits is 1 only when both are 1.

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    Then $k_0+1=1$ forces $k_0=0$. Candidate key: $(0,1,1)$. Substitute: $0\oplus1=1$ and $1\cdot1=1$, so both observations are satisfied.

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    If only $k_0+k_1=1$ were observed, there would be four keys: $(0,1,0)$, $(0,1,1)$, $(1,0,0)$, $(1,0,1)$. An equation system need not determine every bit.

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  SageMath

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    B = BooleanPolynomialRing(69, "k")
    key_vars = list(B.gens())
    # Build symbolic_stream using the exact generator and clock convention.
    equations = [f + bit for f, bit in zip(symbolic_stream, real_stream)]
    solutions = B.ideal(equations).variety()
    # There may be zero, one, or several solutions.
    # For each candidate, rebuild the real generator and compare held-out output.
    

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  Related

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    Geffe Generator (Nonlinear LFSR Combination via Multiplexer)

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    Linear Feedback Shift Register (LFSR)

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  Cards

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    What three GF(2) operations do Boolean AND, XOR, and NOT translate into?

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      AND → multiplication, XOR → addition, NOT(x) → 1+x.

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    Why does setting "symbolic_output + known_bit = 0" as an equation correctly encode "these two bits are equal"?

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      In GF(2) arithmetic, x + y = 0 exactly when x = y (since 1+1=0 and 0+0=0), so this is the natural way to express an equality constraint as a polynomial equation.

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