Ladder

Modular lift — the ladder

3 distinct arguments against one idea, from the one that introduces it to the one that assumes the rest.
  1. 34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

    How many primes the answer needs

    Work modulo a word-sized prime and no intermediate can exceed twenty-six bits, whatever the matrix does. The catch is that the answer must be reassembled from several such computations, and the number of them has to be fixed before the first one runs — by a theorem about how large a determinant can be, not by trying more until it settles.

    rung 1 · exact
  2. det mod p, as a fraction of p3713a prime that divides the answerdet A3·10⁴primes swept25unlucky5rate0.2det, in bits15singular mod p is not singularand one residue cannot tell them apart

    A prime that divides the answer

    A modular elimination reports a singular matrix and is telling the truth — over the field with p elements the matrix is singular. Over the rationals it is not. Nothing in the residue distinguishes the two cases, no quantity is small enough to be suspicious, and the wrong answer is a correct computation of a different question.

    rung 2 · exact
  3. 2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

    A fraction recovered from one remainder

    A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

    rung 3 · exact

All ladders