257 lines
8.1 KiB
Plaintext
257 lines
8.1 KiB
Plaintext
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(* Copyright (c) 2011-2020 Stefan Krah. All rights reserved. *)
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The Matrix Fourier Transform:
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=============================
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In libmpdec, the Matrix Fourier Transform [1] is called four-step transform
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after a variant that appears in [2]. The algorithm requires that the input
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array can be viewed as an R*C matrix.
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All operations are done modulo p. For readability, the proofs drop all
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instances of (mod p).
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Algorithm four-step (forward transform):
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----------------------------------------
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a := input array
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d := len(a) = R * C
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p := prime
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w := primitive root of unity of the prime field
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r := w**((p-1)/d)
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A := output array
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1) Apply a length R FNT to each column.
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2) Multiply each matrix element (addressed by j*C+m) by r**(j*m).
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3) Apply a length C FNT to each row.
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4) Transpose the matrix.
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Proof (forward transform):
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--------------------------
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The algorithm can be derived starting from the regular definition of
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the finite-field transform of length d:
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d-1
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,----
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\
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A[k] = | a[l] * r**(k * l)
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/
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`----
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l = 0
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The sum can be rearranged into the sum of the sums of columns:
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C-1 R-1
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,---- ,----
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\ \
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= | | a[i * C + j] * r**(k * (i * C + j))
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/ /
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`---- `----
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j = 0 i = 0
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Extracting a constant from the inner sum:
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C-1 R-1
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,---- ,----
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\ \
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= | r**k*j * | a[i * C + j] * r**(k * i * C)
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/ /
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`---- `----
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j = 0 i = 0
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Without any loss of generality, let k = n * R + m,
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where n < C and m < R:
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C-1 R-1
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,---- ,----
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\ \
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A[n*R+m] = | r**(R*n*j) * r**(m*j) * | a[i*C+j] * r**(R*C*n*i) * r**(C*m*i)
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/ /
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`---- `----
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j = 0 i = 0
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Since r = w ** ((p-1) / (R*C)):
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a) r**(R*C*n*i) = w**((p-1)*n*i) = 1
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b) r**(C*m*i) = w**((p-1) / R) ** (m*i) = r_R ** (m*i)
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c) r**(R*n*j) = w**((p-1) / C) ** (n*j) = r_C ** (n*j)
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r_R := root of the subfield of length R.
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r_C := root of the subfield of length C.
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C-1 R-1
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,---- ,----
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\ \
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A[n*R+m] = | r_C**(n*j) * [ r**(m*j) * | a[i*C+j] * r_R**(m*i) ]
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/ ^ /
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`---- | `---- 1) transform the columns
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j = 0 | i = 0
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^ |
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| `-- 2) multiply
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`-- 3) transform the rows
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Note that the entire RHS is a function of n and m and that the results
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for each pair (n, m) are stored in Fortran order.
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Let the term in square brackets be f(m, j). Step 1) and 2) precalculate
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the term for all (m, j). After that, the original matrix is now a lookup
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table with the mth element in the jth column at location m * C + j.
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Let the complete RHS be g(m, n). Step 3) does an in-place transform of
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length n on all rows. After that, the original matrix is now a lookup
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table with the mth element in the nth column at location m * C + n.
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But each (m, n) pair should be written to location n * R + m. Therefore,
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step 4) transposes the result of step 3).
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Algorithm four-step (inverse transform):
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----------------------------------------
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A := input array
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d := len(A) = R * C
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p := prime
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d' := d**(p-2) # inverse of d
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w := primitive root of unity of the prime field
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r := w**((p-1)/d) # root of the subfield
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r' := w**((p-1) - (p-1)/d) # inverse of r
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a := output array
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0) View the matrix as a C*R matrix.
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1) Transpose the matrix, producing an R*C matrix.
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2) Apply a length C FNT to each row.
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3) Multiply each matrix element (addressed by i*C+n) by r**(i*n).
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4) Apply a length R FNT to each column.
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Proof (inverse transform):
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--------------------------
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The algorithm can be derived starting from the regular definition of
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the finite-field inverse transform of length d:
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d-1
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,----
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\
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a[k] = d' * | A[l] * r' ** (k * l)
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/
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`----
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l = 0
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The sum can be rearranged into the sum of the sums of columns. Note
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that at this stage we still have a C*R matrix, so C denotes the number
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of rows:
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R-1 C-1
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,---- ,----
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\ \
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= d' * | | a[j * R + i] * r' ** (k * (j * R + i))
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/ /
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`---- `----
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i = 0 j = 0
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Extracting a constant from the inner sum:
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R-1 C-1
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,---- ,----
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\ \
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= d' * | r' ** (k*i) * | a[j * R + i] * r' ** (k * j * R)
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/ /
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`---- `----
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i = 0 j = 0
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Without any loss of generality, let k = m * C + n,
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where m < R and n < C:
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R-1 C-1
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,---- ,----
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\ \
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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)
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/ /
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`---- `----
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i = 0 j = 0
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Since r' = w**((p-1) - (p-1)/d) and d = R*C:
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a) r' ** (R*C*m*j) = w**((p-1)*R*C*m*j - (p-1)*m*j) = 1
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b) r' ** (C*m*i) = w**((p-1)*C - (p-1)/R) ** (m*i) = r_R' ** (m*i)
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c) r' ** (R*n*j) = r_C' ** (n*j)
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d) d' = d**(p-2) = (R*C) ** (p-2) = R**(p-2) * C**(p-2) = R' * C'
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r_R' := inverse of the root of the subfield of length R.
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r_C' := inverse of the root of the subfield of length C.
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R' := inverse of R
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C' := inverse of C
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R-1 C-1
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,---- ,---- 2) transform the rows of a^T
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\ \
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A[m*C+n] = R' * | r_R' ** (m*i) * [ r' ** (n*i) * C' * | a[j*R+i] * r_C' ** (n*j) ]
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/ ^ / ^
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`---- | `---- |
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i = 0 | j = 0 |
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^ | `-- 1) Transpose input matrix
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| `-- 3) multiply to address elements by
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| i * C + j
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`-- 3) transform the columns
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Note that the entire RHS is a function of m and n and that the results
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for each pair (m, n) are stored in C order.
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Let the term in square brackets be f(n, i). Without step 1), the sum
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would perform a length C transform on the columns of the input matrix.
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This is a) inefficient and b) the results are needed in C order, so
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step 1) exchanges rows and columns.
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Step 2) and 3) precalculate f(n, i) for all (n, i). After that, the
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original matrix is now a lookup table with the ith element in the nth
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column at location i * C + n.
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Let the complete RHS be g(m, n). Step 4) does an in-place transform of
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length m on all columns. After that, the original matrix is now a lookup
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table with the mth element in the nth column at location m * C + n,
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which means that all A[k] = A[m * C + n] are in the correct order.
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--
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[1] Joerg Arndt: "Matters Computational"
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http://www.jjj.de/fxt/
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[2] David H. Bailey: FFTs in External or Hierarchical Memory
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http://crd.lbl.gov/~dhbailey/dhbpapers/
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