xref: /third_party/python/Modules/_decimal/libmpdec/literature/matrix-transform.txt

(* Copyright (c) 2011-2020 Stefan Krah. All rights reserved. *)


The Matrix Fourier Transform:
=============================

In libmpdec, the Matrix Fourier Transform [1] is called four-step transform
after a variant that appears in [2]. The algorithm requires that the input
array can be viewed as an R*C matrix.

All operations are done modulo p. For readability, the proofs drop all
instances of (mod p).


Algorithm four-step (forward transform):
----------------------------------------

  a := input array
  d := len(a) = R * C
  p := prime
  w := primitive root of unity of the prime field
  r := w**((p-1)/d)
  A := output array

  1) Apply a length R FNT to each column.

  2) Multiply each matrix element (addressed by j*C+m) by r**(j*m).

  3) Apply a length C FNT to each row.

  4) Transpose the matrix.


Proof (forward transform):
--------------------------

  The algorithm can be derived starting from the regular definition of
  the finite-field transform of length d:

            d-1
           ,----
           \
    A[k] =  |  a[l]  * r**(k * l)
           /
           `----
           l = 0


  The sum can be rearranged into the sum of the sums of columns:

            C-1     R-1
           ,----   ,----
           \       \
         =  |       |  a[i * C + j] * r**(k * (i * C + j))
           /       /
           `----   `----
           j = 0   i = 0


  Extracting a constant from the inner sum:

            C-1           R-1
           ,----         ,----
           \             \
         =  |  r**k*j  *  |  a[i * C + j] * r**(k * i * C)
           /             /
           `----         `----
           j = 0         i = 0


  Without any loss of generality, let k = n * R + m,
  where n < C and m < R:

                C-1                          R-1
               ,----                        ,----
               \                            \
    A[n*R+m] =  |  r**(R*n*j) * r**(m*j)  *  |  a[i*C+j] * r**(R*C*n*i) * r**(C*m*i)
               /                            /
               `----                        `----
               j = 0                        i = 0


  Since r = w ** ((p-1) / (R*C)):

     a) r**(R*C*n*i) = w**((p-1)*n*i) = 1

     b) r**(C*m*i) = w**((p-1) / R) ** (m*i) = r_R ** (m*i)

     c) r**(R*n*j) = w**((p-1) / C) ** (n*j) = r_C ** (n*j)

     r_R := root of the subfield of length R.
     r_C := root of the subfield of length C.


                C-1                             R-1
               ,----                           ,----
               \                               \
    A[n*R+m] =  |  r_C**(n*j) * [ r**(m*j)  *   |  a[i*C+j] * r_R**(m*i) ]
               /                     ^         /
               `----                 |         `----    1) transform the columns
               j = 0                 |         i = 0
                 ^                   |
                 |                   `-- 2) multiply
                 |
                 `-- 3) transform the rows


   Note that the entire RHS is a function of n and m and that the results
   for each pair (n, m) are stored in Fortran order.

   Let the term in square brackets be f(m, j). Step 1) and 2) precalculate
   the term for all (m, j). After that, the original matrix is now a lookup
   table with the mth element in the jth column at location m * C + j.

   Let the complete RHS be g(m, n). Step 3) does an in-place transform of
   length n on all rows. After that, the original matrix is now a lookup
   table with the mth element in the nth column at location m * C + n.

   But each (m, n) pair should be written to location n * R + m. Therefore,
   step 4) transposes the result of step 3).



Algorithm four-step (inverse transform):
----------------------------------------

  A  := input array
  d  := len(A) = R * C
  p  := prime
  d' := d**(p-2)             # inverse of d
  w  := primitive root of unity of the prime field
  r  := w**((p-1)/d)         # root of the subfield
  r' := w**((p-1) - (p-1)/d) # inverse of r
  a  := output array

  0) View the matrix as a C*R matrix.

  1) Transpose the matrix, producing an R*C matrix.

  2) Apply a length C FNT to each row.

  3) Multiply each matrix element (addressed by i*C+n) by r**(i*n).

  4) Apply a length R FNT to each column.


Proof (inverse transform):
--------------------------

  The algorithm can be derived starting from the regular definition of
  the finite-field inverse transform of length d:

                  d-1
                 ,----
                 \
    a[k] =  d' *  |  A[l]  * r' ** (k * l)
                 /
                 `----
                 l = 0


  The sum can be rearranged into the sum of the sums of columns. Note
  that at this stage we still have a C*R matrix, so C denotes the number
  of rows:

                  R-1     C-1
                 ,----   ,----
                 \       \
         =  d' *  |       |  a[j * R + i] * r' ** (k * (j * R + i))
                 /       /
                 `----   `----
                 i = 0   j = 0


  Extracting a constant from the inner sum:

                  R-1                C-1
                 ,----              ,----
                 \                  \
         =  d' *  |  r' ** (k*i)  *  |  a[j * R + i] * r' ** (k * j * R)
                 /                  /
                 `----              `----
                 i = 0              j = 0


  Without any loss of generality, let k = m * C + n,
  where m < R and n < C:

                     R-1                                  C-1
                    ,----                                ,----
                    \                                    \
    A[m*C+n] = d' *  |  r' ** (C*m*i) *  r' ** (n*i)   *  |  a[j*R+i] * r' ** (R*C*m*j) * r' ** (R*n*j)
                    /                                    /
                    `----                                `----
                    i = 0                                j = 0


  Since r' = w**((p-1) - (p-1)/d) and d = R*C:

     a) r' ** (R*C*m*j) = w**((p-1)*R*C*m*j - (p-1)*m*j) = 1

     b) r' ** (C*m*i) = w**((p-1)*C - (p-1)/R) ** (m*i) = r_R' ** (m*i)

     c) r' ** (R*n*j) = r_C' ** (n*j)

     d) d' = d**(p-2) = (R*C) ** (p-2) = R**(p-2) * C**(p-2) = R' * C'

     r_R' := inverse of the root of the subfield of length R.
     r_C' := inverse of the root of the subfield of length C.
     R'   := inverse of R
     C'   := inverse of C


                     R-1                                      C-1
                    ,----                                    ,----  2) transform the rows of a^T
                    \                                        \
    A[m*C+n] = R' *  |  r_R' ** (m*i) * [ r' ** (n*i) * C' *  |  a[j*R+i] * r_C' ** (n*j) ]
                    /                           ^            /       ^
                    `----                       |            `----   |
                    i = 0                       |            j = 0   |
                      ^                         |                    `-- 1) Transpose input matrix
                      |                         `-- 3) multiply             to address elements by
                      |                                                     i * C + j
                      `-- 3) transform the columns



   Note that the entire RHS is a function of m and n and that the results
   for each pair (m, n) are stored in C order.

   Let the term in square brackets be f(n, i). Without step 1), the sum
   would perform a length C transform on the columns of the input matrix.
   This is a) inefficient and b) the results are needed in C order, so
   step 1) exchanges rows and columns.

   Step 2) and 3) precalculate f(n, i) for all (n, i). After that, the
   original matrix is now a lookup table with the ith element in the nth
   column at location i * C + n.

   Let the complete RHS be g(m, n). Step 4) does an in-place transform of
   length m on all columns. After that, the original matrix is now a lookup
   table with the mth element in the nth column at location m * C + n,
   which means that all A[k] = A[m * C + n] are in the correct order.


--

  [1] Joerg Arndt: "Matters Computational"
      http://www.jjj.de/fxt/
  [2] David H. Bailey: FFTs in External or Hierarchical Memory
      http://crd.lbl.gov/~dhbailey/dhbpapers/