# Deconvolution/1D

Deconvolution/1D
You are encouraged to solve this task according to the task description, using any language you may know.

The convolution of two functions ${\displaystyle {\mathit {F}}}$ and ${\displaystyle {\mathit {H}}}$ of an integer variable is defined as the function ${\displaystyle {\mathit {G}}}$ satisfying

${\displaystyle G(n)=\sum _{m=-\infty }^{\infty }F(m)H(n-m)}$

for all integers ${\displaystyle {\mathit {n}}}$. Assume ${\displaystyle F(n)}$ can be non-zero only for ${\displaystyle 0}$${\displaystyle {\mathit {n}}}$${\displaystyle |{\mathit {F}}|}$, where ${\displaystyle |{\mathit {F}}|}$ is the "length" of ${\displaystyle {\mathit {F}}}$, and similarly for ${\displaystyle {\mathit {G}}}$ and ${\displaystyle {\mathit {H}}}$, so that the functions can be modeled as finite sequences by identifying ${\displaystyle f_{0},f_{1},f_{2},\dots }$ with ${\displaystyle F(0),F(1),F(2),\dots }$, etc. Then for example, values of ${\displaystyle |{\mathit {F}}|=6}$ and ${\displaystyle |{\mathit {H}}|=5}$ would determine the following value of ${\displaystyle {\mathit {g}}}$ by definition.

${\displaystyle {\begin{array}{lllllllllll}g_{0}&=&f_{0}h_{0}\\g_{1}&=&f_{1}h_{0}&+&f_{0}h_{1}\\g_{2}&=&f_{2}h_{0}&+&f_{1}h_{1}&+&f_{0}h_{2}\\g_{3}&=&f_{3}h_{0}&+&f_{2}h_{1}&+&f_{1}h_{2}&+&f_{0}h_{3}\\g_{4}&=&f_{4}h_{0}&+&f_{3}h_{1}&+&f_{2}h_{2}&+&f_{1}h_{3}&+&f_{0}h_{4}\\g_{5}&=&f_{5}h_{0}&+&f_{4}h_{1}&+&f_{3}h_{2}&+&f_{2}h_{3}&+&f_{1}h_{4}\\g_{6}&=&&&f_{5}h_{1}&+&f_{4}h_{2}&+&f_{3}h_{3}&+&f_{2}h_{4}\\g_{7}&=&&&&&f_{5}h_{2}&+&f_{4}h_{3}&+&f_{3}h_{4}\\g_{8}&=&&&&&&&f_{5}h_{3}&+&f_{4}h_{4}\\g_{9}&=&&&&&&&&&f_{5}h_{4}\end{array}}}$

We can write this in matrix form as:

${\displaystyle \left({\begin{array}{l}g_{0}\\g_{1}\\g_{2}\\g_{3}\\g_{4}\\g_{5}\\g_{6}\\g_{7}\\g_{8}\\g_{9}\\\end{array}}\right)=\left({\begin{array}{lllll}f_{0}\\f_{1}&f_{0}\\f_{2}&f_{1}&f_{0}\\f_{3}&f_{2}&f_{1}&f_{0}\\f_{4}&f_{3}&f_{2}&f_{1}&f_{0}\\f_{5}&f_{4}&f_{3}&f_{2}&f_{1}\\&f_{5}&f_{4}&f_{3}&f_{2}\\&&f_{5}&f_{4}&f_{3}\\&&&f_{5}&f_{4}\\&&&&f_{5}\end{array}}\right)\;\left({\begin{array}{l}h_{0}\\h_{1}\\h_{2}\\h_{3}\\h_{4}\\\end{array}}\right)}$

or

${\displaystyle g=A\;h}$

For this task, implement a function (or method, procedure, subroutine, etc.) deconv to perform deconvolution (i.e., the inverse of convolution) by constructing and solving such a system of equations represented by the above matrix ${\displaystyle A}$ for ${\displaystyle {\mathit {h}}}$ given ${\displaystyle {\mathit {f}}}$ and ${\displaystyle {\mathit {g}}}$.

• The function should work for ${\displaystyle {\mathit {G}}}$ of arbitrary length (i.e., not hard coded or constant) and ${\displaystyle {\mathit {F}}}$ of any length up to that of ${\displaystyle {\mathit {G}}}$. Note that ${\displaystyle |{\mathit {H}}|}$ will be given by ${\displaystyle |{\mathit {G}}|-|{\mathit {F}}|+1}$.
• There may be more equations than unknowns. If convenient, use a function from a library that finds the best fitting solution to an overdetermined system of linear equations (as in the Multiple regression task). Otherwise, prune the set of equations as needed and solve as in the Reduced row echelon form task.
• Test your solution on the following data. Be sure to verify both that deconv${\displaystyle (g,f)=h}$ and deconv${\displaystyle (g,h)=f}$ and display the results in a human readable form.

 h = [-8,-9,-3,-1,-6,7] f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1] g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7] 

## BBC BASIC

As several others, this is a translation of the D solution.

      *FLOAT 64      DIM h(5), f(15), g(20)      h() = -8,-9,-3,-1,-6,7      f() = -3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1      g() = 24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7       PROCdeconv(g(), f(), x())      PRINT "deconv(g,f) = " FNprintarray(x())      x() -= h() : IF SUM(x()) <> 0 PRINT "Error!"       PROCdeconv(g(), h(), y())      PRINT "deconv(g,h) = " FNprintarray(y())      y() -= f() : IF SUM(y()) <> 0 PRINT "Error!"      END       DEF PROCdeconv(g(), f(), RETURN h())      LOCAL f%, g%, i%, l%, n%      f% = DIM(f(),1) + 1      g% = DIM(g(),1) + 1      DIM h(g% - f%)      FOR n% = 0 TO g% - f%        h(n%) = g(n%)        IF n% < f% THEN l% = 0 ELSE l% = n% - f% + 1        IF n% THEN          FOR i% = l% TO n% - 1            h(n%) -= h(i%) * f(n% - i%)          NEXT        ENDIF        h(n%) /= f(0)      NEXT n%      ENDPROC       DEF FNprintarray(a())      LOCAL i%, a$FOR i% = 0 TO DIM(a(),1) a$ += STR$(a(i%)) + ", " NEXT = LEFT$(LEFT$(a$))
Output:
deconv(g,f) = -8, -9, -3, -1, -6, 7
deconv(g,h) = -3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1


## C

Using FFT:

#include <stdio.h>#include <stdlib.h>#include <math.h>#include <complex.h> double PI;typedef double complex cplx; void _fft(cplx buf[], cplx out[], int n, int step){	if (step < n) {		_fft(out, buf, n, step * 2);		_fft(out + step, buf + step, n, step * 2); 		for (int i = 0; i < n; i += 2 * step) {			cplx t = cexp(-I * PI * i / n) * out[i + step];			buf[i / 2]     = out[i] + t;			buf[(i + n)/2] = out[i] - t;		}	}} void fft(cplx buf[], int n){	cplx out[n];	for (int i = 0; i < n; i++) out[i] = buf[i];	_fft(buf, out, n, 1);} /* pad array length to power of two */cplx *pad_two(double g[], int len, int *ns){	int n = 1;	if (*ns) n = *ns;	else while (n < len) n *= 2; 	cplx *buf = calloc(sizeof(cplx), n);	for (int i = 0; i < len; i++) buf[i] = g[i];	*ns = n;	return buf;} void deconv(double g[], int lg, double f[], int lf, double out[]) {	int ns = 0;	cplx *g2 = pad_two(g, lg, &ns);	cplx *f2 = pad_two(f, lf, &ns); 	fft(g2, ns);	fft(f2, ns); 	cplx h[ns];	for (int i = 0; i < ns; i++) h[i] = g2[i] / f2[i];	fft(h, ns); 	for (int i = 0; i >= lf - lg; i--)		out[-i] = h[(i + ns) % ns]/32;	free(g2);	free(f2);} int main(){	PI = atan2(1,1) * 4;	double g[] = {24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7};	double f[] = { -3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1 };	double h[] = { -8,-9,-3,-1,-6,7 }; 	int lg = sizeof(g)/sizeof(double);	int lf = sizeof(f)/sizeof(double);	int lh = sizeof(h)/sizeof(double); 	double h2[lh];	double f2[lf]; 	printf("f[] data is : ");	for (int i = 0; i < lf; i++) printf(" %g", f[i]);	printf("\n"); 	printf("deconv(g, h): ");	deconv(g, lg, h, lh, f2);	for (int i = 0; i < lf; i++) printf(" %g", f2[i]);	printf("\n"); 	printf("h[] data is : ");	for (int i = 0; i < lh; i++) printf(" %g", h[i]);	printf("\n"); 	printf("deconv(g, f): ");	deconv(g, lg, f, lf, h2);	for (int i = 0; i < lh; i++) printf(" %g", h2[i]);	printf("\n");}
Output:
f[] data is :  -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1
deconv(g, h):  -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1
h[] data is :  -8 -9 -3 -1 -6 7

deconv(g, f):  -8 -9 -3 -1 -6 7

## Common Lisp

Uses the routine (lsqr A b) from Multiple regression and (mtp A) from Matrix transposition.

;; Assemble the mxn matrix A from the 2D row vector x.(defun make-conv-matrix (x m n)  (let ((lx (cadr (array-dimensions x)))        (A  (make-array (,m ,n) :initial-element 0)))     (loop for j from 0 to (- n 1) do         (loop for i from 0 to (- m 1) do              (setf (aref A i j)                    (cond ((or (< i j) (>= i (+ j lx)))                           0)                          ((and (>= i j) (< i (+ j lx)))                           (aref x 0 (- i j)))))))    A)) ;; Solve the overdetermined system A(f)*h=g by linear least squares.(defun deconv (g f)  (let* ((lg (cadr (array-dimensions g)))         (lf (cadr (array-dimensions f)))         (lh (+ (- lg lf) 1))         (A  (make-conv-matrix f lg lh)))     (lsqr A (mtp g))))

Example:

(setf f #2A((-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1)))(setf h #2A((-8 -9 -3 -1 -6 7)))(setf g #2A((24 75 71 -34 3 22 -45 23 245 25 52 25 -67 -96 96 31 55 36 29 -43 -7))) (deconv g f)#2A((-8.0)    (-9.000000000000002)    (-2.999999999999999)    (-0.9999999999999997)    (-6.0)    (7.000000000000002)) (deconv g h)#2A((-2.999999999999999)    (-6.000000000000001)    (-1.0000000000000002)    (8.0)    (-5.999999999999999)    (3.0000000000000004)    (-1.0000000000000004)    (-9.000000000000002)    (-9.0)    (2.9999999999999996)    (-1.9999999999999991)    (5.0)    (1.9999999999999996)    (-2.0000000000000004)    (-7.000000000000001)    (-0.9999999999999994))

## D

T[] deconv(T)(in T[] g, in T[] f) pure nothrow {    int flen = f.length;    int glen = g.length;    auto result = new T[glen - flen + 1];    foreach (int n, ref e; result) {        e = g[n];        immutable lowerBound = (n >= flen) ? n - flen + 1 : 0;        foreach (i; lowerBound .. n)                e -= result[i] * f[n - i];        e /= f[0];    }    return result;} void main() {    import std.stdio;    immutable h = [-8,-9,-3,-1,-6,7];    immutable f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1];    immutable g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,                   -96,96,31,55,36,29,-43,-7];    writeln(deconv(g, f) == h, " ", deconv(g, f));    writeln(deconv(g, h) == f, " ", deconv(g, h));}
Output:
true [-8, -9, -3, -1, -6, 7]
true [-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1]

## Fortran

This solution uses the LAPACK95 library.

 ! Build! Windows: ifort /I "%IFORT_COMPILER11%\mkl\include\ia32" deconv1d.f90 "%IFORT_COMPILER11%\mkl\ia32\lib\*.lib"! Linux: program deconv  ! Use gelsd from LAPACK95.  use mkl95_lapack, only : gelsd   implicit none  real(8), allocatable :: g(:), href(:), A(:,:), f(:)  real(8), pointer     :: h(:), r(:)  integer              :: N  character(len=16)    :: cbuff  integer              :: i  intrinsic            :: nint   ! Allocate data arrays  allocate(g(21),f(16))  g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7]  f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1]   ! Calculate deconvolution  h => deco(f, g)   ! Check result against reference  N = size(h)  allocate(href(N))  href = [-8,-9,-3,-1,-6,7]  cbuff = ' '  write(cbuff,'(a,i0,a)') '(a,',N,'(i0,a),i0)'  if (any(abs(h-href) > 1.0d-4)) then     write(*,'(a)') 'deconv(f, g) - FAILED'  else     write(*,cbuff) 'deconv(f, g) = ',(nint(h(i)),', ',i=1,N-1),nint(h(N))  end if   ! Calculate deconvolution  r => deco(h, g)   cbuff = ' '  N = size(r)  write(cbuff,'(a,i0,a)') '(a,',N,'(i0,a),i0)'  if (any(abs(r-f) > 1.0d-4)) then     write(*,'(a)') 'deconv(h, g) - FAILED'  else     write(*,cbuff) 'deconv(h, g) = ',(nint(r(i)),', ',i=1,N-1),nint(r(N))  end if contains  function deco(p, q)    real(8), pointer    :: deco(:)    real(8), intent(in) :: p(:), q(:)     real(8), allocatable, target :: r(:)    real(8), allocatable         :: A(:,:)    integer :: N     ! Construct derived arrays    N = size(q) - size(p) + 1    allocate(A(size(q),N),r(size(q)))    A = 0.0d0    do i=1,N       A(i:i+size(p)-1,i) = p    end do     ! Invoke the LAPACK routine to do the work    r = q    call gelsd(A, r)     deco => r(1:N)  end function deco end program deconv 

Results:

 deconv(f, g) = -8, -9, -3, -1, -6, 7deconv(h, g) = -3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1 

## Go

Translation of: D
package main import "fmt" func main() {    h := []float64{-8, -9, -3, -1, -6, 7}    f := []float64{-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1}    g := []float64{24, 75, 71, -34, 3, 22, -45, 23, 245, 25, 52, 25, -67, -96,        96, 31, 55, 36, 29, -43, -7}    fmt.Println(h)    fmt.Println(deconv(g, f))    fmt.Println(f)    fmt.Println(deconv(g, h))} func deconv(g, f []float64) []float64 {    h := make([]float64, len(g)-len(f)+1)    for n := range h {        h[n] = g[n]        var lower int        if n >= len(f) {            lower = n - len(f) + 1        }        for i := lower; i < n; i++ {            h[n] -= h[i] * f[n-i]        }        h[n] /= f[0]    }    return h}
Output:
[-8 -9 -3 -1 -6 7]
[-8 -9 -3 -1 -6 7]
[-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1]
[-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1]

Translation of: C
package main import (    "fmt"    "math"    "math/cmplx") func main() {    h := []float64{-8, -9, -3, -1, -6, 7}    f := []float64{-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1}    g := []float64{24, 75, 71, -34, 3, 22, -45, 23, 245, 25, 52, 25, -67, -96,        96, 31, 55, 36, 29, -43, -7}    fmt.Printf("%.1f\n", h)    fmt.Printf("%.1f\n", deconv(g, f))    fmt.Printf("%.1f\n", f)    fmt.Printf("%.1f\n", deconv(g, h))} func deconv(g, f []float64) []float64 {    n := 1    for n < len(g) {        n *= 2    }    g2 := make([]complex128, n)    for i, x := range g {        g2[i] = complex(x, 0)    }    f2 := make([]complex128, n)    for i, x := range f {        f2[i] = complex(x, 0)    }    gt := fft(g2)    ft := fft(f2)    for i := range gt {        gt[i] /= ft[i]    }    ht := fft(gt)    it := 1 / float64(n)    out := make([]float64, len(g)-len(f)+1)    out[0] = real(ht[0]) * it    for i := 1; i < len(out); i++ {        out[i] = real(ht[n-i]) * it    }    return out} func fft(in []complex128) []complex128 {    out := make([]complex128, len(in))    ditfft2(in, out, len(in), 1)    return out} func ditfft2(x, y []complex128, n, s int) {    if n == 1 {        y[0] = x[0]        return    }    ditfft2(x, y, n/2, 2*s)    ditfft2(x[s:], y[n/2:], n/2, 2*s)    for k := 0; k < n/2; k++ {        tf := cmplx.Rect(1, -2*math.Pi*float64(k)/float64(n)) * y[k+n/2]        y[k], y[k+n/2] = y[k]+tf, y[k]-tf    }}
Output:

Some results have errors out in the last decimal place or so. Only one decimal place shown here to let results fit in 80 columns.

[-8.0 -9.0 -3.0 -1.0 -6.0 7.0]
[-8.0 -9.0 -3.0 -1.0 -6.0 7.0]
[-3.0 -6.0 -1.0 8.0 -6.0 3.0 -1.0 -9.0 -9.0 3.0 -2.0 5.0 2.0 -2.0 -7.0 -1.0]
[-3.0 -6.0 -1.0 8.0 -6.0 3.0 -1.0 -9.0 -9.0 3.0 -2.0 5.0 2.0 -2.0 -7.0 -1.0]


import Data.List h, f, g :: [Double]h = [-8,-9,-3,-1,-6,7]f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1]g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7] scale x ys = map (x*) ys deconv1d :: (Fractional a) => [a] -> [a] -> [a]deconv1d xs ys = takeWhile (/=0) $deconv xs ys where [] deconv _ = [] (0:xs) deconv (0:ys) = xs deconv ys (x:xs) deconv (y:ys) = q : zipWith (-) xs (scale q ys ++ repeat 0) deconv (y:ys) where q = x / y Check: *Main> h == deconv1d g fTrue *Main> f == deconv1d g hTrue ## J This solution borrowed from Formal power series: Ai=: (i.@] =/ i.@[ -/ i.@>:@-)&#divide=: [ +/ .*~ [:%.&.x: ] +/ .* Ai Sample data: h=: _8 _9 _3 _1 _6 7f=: _3 _6 _1 8 _6 3 _1 _9 _9 3 _2 5 2 _2 _7 _1g=: 24 75 71 _34 3 22 _45 23 245 25 52 25 _67 _96 96 31 55 36 29 Example use:  g divide f_8 _9 _3 _1 _6 7 g divide h_3 _6 _1 8 _6 3 _1 _9 _9 3 _2 5 2 _2 _7 _1 That said, note that this particular implementation is slow since it uses extended precision intermediate results. It will run quite a bit faster for this example with no notable loss of precision if floating point is used. In other words: divide=: [ +/ .*~ [:%. ] +/ .* Ai ## Java Translation of: Go import java.util.Arrays; public class Deconvolution1D { public static int[] deconv(int[] g, int[] f) { int[] h = new int[g.length - f.length + 1]; for (int n = 0; n < h.length; n++) { h[n] = g[n]; int lower = Math.max(n - f.length + 1, 0); for (int i = lower; i < n; i++) h[n] -= h[i] * f[n - i]; h[n] /= f[0]; } return h; } public static void main(String[] args) { int[] h = { -8, -9, -3, -1, -6, 7 }; int[] f = { -3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1 }; int[] g = { 24, 75, 71, -34, 3, 22, -45, 23, 245, 25, 52, 25, -67, -96, 96, 31, 55, 36, 29, -43, -7 }; StringBuilder sb = new StringBuilder(); sb.append("h = " + Arrays.toString(h) + "\n"); sb.append("deconv(g, f) = " + Arrays.toString(deconv(g, f)) + "\n"); sb.append("f = " + Arrays.toString(f) + "\n"); sb.append("deconv(g, h) = " + Arrays.toString(deconv(g, h)) + "\n"); System.out.println(sb.toString()); }} Output: h = [-8, -9, -3, -1, -6, 7] deconv(g, f) = [-8, -9, -3, -1, -6, 7] f = [-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1] deconv(g, h) = [-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1]  ## Kotlin Translation of: Go // version 1.1.3 fun deconv(g: DoubleArray, f: DoubleArray): DoubleArray { val fs = f.size val h = DoubleArray(g.size - fs + 1) for (n in h.indices) { h[n] = g[n] val lower = if (n >= fs) n - fs + 1 else 0 for (i in lower until n) h[n] -= h[i] * f[n -i] h[n] /= f[0] } return h} fun main(args: Array<String>) { val h = doubleArrayOf(-8.0, -9.0, -3.0, -1.0, -6.0, 7.0) val f = doubleArrayOf(-3.0, -6.0, -1.0, 8.0, -6.0, 3.0, -1.0, -9.0, -9.0, 3.0, -2.0, 5.0, 2.0, -2.0, -7.0, -1.0) val g = doubleArrayOf(24.0, 75.0, 71.0, -34.0, 3.0, 22.0, -45.0, 23.0, 245.0, 25.0, 52.0, 25.0, -67.0, -96.0, 96.0, 31.0, 55.0, 36.0, 29.0, -43.0, -7.0) println("${h.map { it.toInt() }}")    println("${deconv(g, f).map { it.toInt() }}") println() println("${f.map { it.toInt() }}")    println("${deconv(g, h).map { it.toInt() }}")} Output: [-8, -9, -3, -1, -6, 7] [-8, -9, -3, -1, -6, 7] [-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1] [-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1]  ## Lua Using metatables: function deconvolve(f, g) local h = setmetatable({}, {__index = function(self, n) if n == 1 then self[1] = g[1] / f[1] else self[n] = g[n] for i = 1, n - 1 do self[n] = self[n] - self[i] * (f[n - i + 1] or 0) end self[n] = self[n] / f[1] end return self[n] end}) local _ = h[#g - #f + 1] return setmetatable(h, nil)end Tests:  local f = {-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1}local g = {24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7}local h = {-8,-9,-3,-1,-6,7}print(unpack(deconvolve(f, g))) --> -8 -9 -3 -1 -6 7print(unpack(deconvolve(h, g))) --> -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1 ## Mathematica / Wolfram Language This function creates a sparse array for the A matrix and then solves it with a built-in function. It may fail for overdetermined systems, though. Fast approximate methods for deconvolution are also built into Mathematica. See Deconvolution/2D+  deconv[f_List, g_List] := Module[{A = SparseArray[ Table[Band[{n, 1}] -> f[[n]], {n, 1, Length[f]}], {Length[g], Length[f] - 1}]}, Take[LinearSolve[A, g], Length[g] - Length[f] + 1]]  Usage: f = {-3, -6, -1, 8, -6, 3, -1, -9, -9, 3, -2, 5, 2, -2, -7, -1}; g = {24, 75, 71, -34, 3, 22, -45, 23, 245, 25, 52, 25, -67, -96, 96, 31, 55, 36, 29, -43, -7}; deconv[f,g]  Output: {-8, -9, -3, -1, -6, 7} ## MATLAB The deconvolution function is built-in to MATLAB as the "deconv(a,b)" function, where "a" and "b" are vectors storing the convolved function values and the values of one of the deconvoluted vectors of "a". To test that this operates according to the task spec we can test the criteria above: >> h = [-8,-9,-3,-1,-6,7];>> g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7];>> f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1];>> deconv(g,f) ans = -8.0000 -9.0000 -3.0000 -1.0000 -6.0000 7.0000 >> deconv(g,h) ans = -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1 Therefore, "deconv(a,b)" behaves as expected. ## Perl 6 Works with: Rakudo 2015-09-07 Translation of Python, using a modified version of the Reduced Row Echelon Form subroutine rref() from here. sub deconvolve (@g, @f) { my$h = 1 + @g - @f;    my @m;    @m[^@g;^$h] >>+=>> 0; @m[^@g;$h] >>=<< @g;    for ^$h ->$j { for @f.kv -> $k,$v { @m[$j +$k][$j] =$v } }    return rref( @m )[^$h;$h];} sub convolve (@f, @h) {    my @g = 0 xx + @f + @h - 1;    @g[^@f X+ ^@h] >>+=<< (@f X* @h);    return @g;} # Reduced Row Echelon Form simultaneous equation solver.# Can handle over-specified systems of equations.# (n unknowns in n + m equations)sub rref ($m is copy) { return unless$m;    my ($lead,$rows, $cols) = 0, +$m, +$m[0]; # Trim off over specified rows if they exist. # Not strictly necessary, but can save a lot of # redundant calculations. [remove until debugged]# if$rows >= $cols {#$m = trim_system($m);#$rows = +$m;# } for ^$rows -> $r {$lead < $cols or return$m;        my $i =$r;        until $m[$i][$lead] { ++$i == $rows or next;$i = $r; ++$lead == $cols and return$m;        }        $m[$i, $r] =$m[$r,$i] if $r !=$i;        my $lv =$m[$r][$lead];        $m[$r] >>/=>> $lv; for ^$rows -> $n { next if$n == $r;$m[$n] >>-=>>$m[$r] >>*>> ($m[$n][$lead]//0);        }        ++$lead; } return$m;     # Reduce a system of equations to n equations with n unknowns.    # Looks for an equation with a true value for each position.    # If it can't find one, assumes that it has already taken one    # and pushes in the first equation it sees. This assumtion    # will alway be successful except in some cases where an    # under-specified system has been supplied, in which case,    # it would not have been able to reduce the system anyway.    sub trim_system ($m is rw) { my ($vars, @t) = +$m[0]-1, (); for ^$vars -> $lead { for ^$m -> $row { @t.push($m.splice( $row, 1 ) ) and last if$m[$row][$lead];            }        }        while (+@t < $vars) and +$m { @t.push( $m.splice( 0, 1 ) ) }; return @t; }} my @h = (-8,-9,-3,-1,-6,7);my @f = (-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1);my @g = (24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7); .say for ~@g, ~convolve(@f, @h),''; .say for ~@h, ~deconvolve(@g, @f),''; .say for ~@f, ~deconvolve(@g, @h),''; Output: 24 75 71 -34 3 22 -45 23 245 25 52 25 -67 -96 96 31 55 36 29 -43 -7 24 75 71 -34 3 22 -45 23 245 25 52 25 -67 -96 96 31 55 36 29 -43 -7 -8 -9 -3 -1 -6 7 -8 -9 -3 -1 -6 7 -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1 -3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1  ## Phix Translation of: D function deconv(sequence g, sequence f)integer lf = length(f)sequence h = repeat(0,length(g)-lf+1) for n = 1 to length(h) do atom e = g[n] for i=max(n-lf,0) to n-2 do e -= h[i+1] * f[n-i] end for h[n] = e/f[1] end for return hend function constant h = {-8,-9,-3,-1,-6,7}constant f = {-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1}constant g = {24,75,71,-34,3,22,-45,23,245,25,52,25,-67, -96,96,31,55,36,29,-43,-7}sequence rr = deconv(g, f) ?{r==h,r}r = deconv(g, h) ?{r==f,r} Output: {1,{-8,-9,-3,-1,-6,7}} {1,{-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1}}  ## PicoLisp (load "@lib/math.l") (de deconv (G F) (let A (pop 'F) (make (for (N . H) (head (- (length F)) G) (for (I . M) (made) (dec 'H (*/ M (get F (- N I)) 1.0) ) ) (link (*/ H 1.0 A)) ) ) ) ) Test: (setq F (-3. -6. -1. 8. -6. 3. -1. -9. -9. 3. -2. 5. 2. -2. -7. -1.) G (24. 75. 71. -34. 3. 22. -45. 23. 245. 25. 52. 25. -67. -96. 96. 31. 55. 36. 29. -43. -7.) H (-8. -9. -3. -1. -6. 7.) ) (test H (deconv G F))(test F (deconv G H)) ## Python Works with: Python version 3.x Inspired by the TCL solution, and using the ToReducedRowEchelonForm function to reduce to row echelon form from here def ToReducedRowEchelonForm( M ): if not M: return lead = 0 rowCount = len(M) columnCount = len(M[0]) for r in range(rowCount): if lead >= columnCount: return i = r while M[i][lead] == 0: i += 1 if i == rowCount: i = r lead += 1 if columnCount == lead: return M[i],M[r] = M[r],M[i] lv = M[r][lead] M[r] = [ mrx / lv for mrx in M[r]] for i in range(rowCount): if i != r: lv = M[i][lead] M[i] = [ iv - lv*rv for rv,iv in zip(M[r],M[i])] lead += 1 return M def pmtx(mtx): print ('\n'.join(''.join(' %4s' % col for col in row) for row in mtx)) def convolve(f, h): g = [0] * (len(f) + len(h) - 1) for hindex, hval in enumerate(h): for findex, fval in enumerate(f): g[hindex + findex] += fval * hval return g def deconvolve(g, f): lenh = len(g) - len(f) + 1 mtx = [[0 for x in range(lenh+1)] for y in g] for hindex in range(lenh): for findex, fval in enumerate(f): gindex = hindex + findex mtx[gindex][hindex] = fval for gindex, gval in enumerate(g): mtx[gindex][lenh] = gval ToReducedRowEchelonForm( mtx ) return [mtx[i][lenh] for i in range(lenh)] # h if __name__ == '__main__': h = [-8,-9,-3,-1,-6,7] f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1] g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7] assert convolve(f,h) == g assert deconvolve(g, f) == h ## R Here we won't solve the system but use the FFT instead. The method : • extend vector arguments so that they are the same length, a power of 2 larger than the length of the solution, • solution is ifft(fft(a)*fft(b)), truncated. conv <- function(a, b) { p <- length(a) q <- length(b) n <- p + q - 1 r <- nextn(n, f=2) y <- fft(fft(c(a, rep(0, r-p))) * fft(c(b, rep(0, r-q))), inverse=TRUE)/r y[1:n]} deconv <- function(a, b) { p <- length(a) q <- length(b) n <- p - q + 1 r <- nextn(max(p, q), f=2) y <- fft(fft(c(a, rep(0, r-p))) / fft(c(b, rep(0, r-q))), inverse=TRUE)/r return(y[1:n])}  To check :  h <- c(-8,-9,-3,-1,-6,7)f <- c(-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1)g <- c(24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7) max(abs(conv(f,h) - g))max(abs(deconv(g,f) - h))max(abs(deconv(g,h) - f))  This solution often introduces complex numbers, with null or tiny imaginary part. If it hurts in applications, type Re(conv(f,h)) and Re(deconv(g,h)) instead, to return only the real part. It's not hard-coded in the functions, since they may be used for complex arguments as well. R has also a function convolve,  conv(a, b) == convolve(a, rev(b), type="open")  ## Racket  #lang racket(require math/matrix)(define T matrix-transpose) (define (convolution-matrix f m n) (define l (matrix-num-rows f)) (for*/matrix m n ([i (in-range 0 m)] [j (in-range 0 n)]) (cond [(or (< i j) (>= i (+ j l))) 0] [(matrix-ref f (- i j) 0)]))) (define (least-square X y) (matrix-solve (matrix* (T X) X) (matrix* (T X) y))) (define (deconvolve g f) (define lg (matrix-num-rows g)) (define lf (matrix-num-rows f)) (define lh (+ (- lg lf) 1)) (least-square (convolution-matrix f lg lh) g))  Test:  (define f (col-matrix [-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1]))(define h (col-matrix [-8 -9 -3 -1 -6 7]))(define g (col-matrix [24 75 71 -34 3 22 -45 23 245 25 52 25 -67 -96 96 31 55 36 29 -43 -7])) (deconvolve g f)(deconvolve g h)  Output:  #<array '#(6 1) #[-8 -9 -3 -1 -6 7]>#<array '#(16 1) #[-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1]>  ## Tcl Works with: Tcl version 8.5 This builds the a command, 1D, with two subcommands (convolve and deconvolve) for performing convolution and deconvolution of these kinds of arrays. The deconvolution code is based on a reduction to reduced row echelon form. package require Tcl 8.5namespace eval 1D { namespace ensemble create; # Will be same name as namespace namespace export convolve deconvolve # Access core language math utility commands namespace path {::tcl::mathfunc ::tcl::mathop} # Utility for converting a matrix to Reduced Row Echelon Form # From http://rosettacode.org/wiki/Reduced_row_echelon_form#Tcl proc toRREF {m} { set lead 0 set rows [llength$m]	set cols [llength [lindex $m 0]] for {set r 0} {$r < $rows} {incr r} { if {$cols <= $lead} { break } set i$r	    while {[lindex $m$i $lead] == 0} { incr i if {$rows == $i} { set i$r		    incr lead		    if {$cols ==$lead} {			# Tcl can't break out of nested loops			return $m } } } # swap rows i and r foreach j [list$i $r] row [list [lindex$m $r] [lindex$m $i]] { lset m$j $row } # divide row r by m(r,lead) set val [lindex$m $r$lead]	    for {set j 0} {$j <$cols} {incr j} {		lset m $r$j [/ [double [lindex $m$r $j]]$val]	    } 	    for {set i 0} {$i <$rows} {incr i} {		if {$i !=$r} {		    # subtract m(i,lead) multiplied by row r from row i		    set val [lindex $m$i $lead] for {set j 0} {$j < $cols} {incr j} { lset m$i $j \ [- [lindex$m $i$j] [* $val [lindex$m $r$j]]]		    }		}	    }	    incr lead	}	return $m } # How to apply a 1D convolution of two "functions" proc convolve {f h} { set g [lrepeat [+ [llength$f] [llength $h] -1] 0] set fi -1 foreach fv$f {	    incr fi	    set hi -1	    foreach hv $h { set gi [+$fi [incr hi]]		lset g $gi [+ [lindex$g $gi] [*$fv $hv]] } } return$g    }     # How to apply a 1D deconvolution of two "functions"    proc deconvolve {g f} {	# Compute the length of the result vector	set hlen [- [llength $g] [llength$f] -1] 	# Build a matrix of equations to solve	set matrix {}	set i -1	foreach gv $g { lappend matrix [list {*}[lrepeat$hlen 0] $gv] set j [incr i] foreach fv$f {		if {$j < 0} { break } elseif {$j < $hlen} { lset matrix$i $j$fv		}		incr j -1	    }	} 	# Convert to RREF, solving the system of simultaneous equations	set reduced [toRREF $matrix] # Extract the deconvolution from the last column of the reduced matrix for {set i 0} {$i<$hlen} {incr i} { lappend result [lindex$reduced $i end] } return$result    }}

To use the above code, a simple demonstration driver (which solves the specific task):

# Simple pretty-printerproc pp {name nlist} {    set sep ""    puts -nonewline "$name = $" foreach n nlist { puts -nonewline [format %s%g sep n] set sep , } puts "$"} set h {-8 -9 -3 -1 -6 7}set f {-3 -6 -1 8 -6 3 -1 -9 -9 3 -2 5 2 -2 -7 -1}set g {24 75 71 -34 3 22 -45 23 245 25 52 25 -67 -96 96 31 55 36 29 -43 -7} pp "deconv(g,f) = h" [1D deconvolve$g $f]pp "deconv(g,h) = f" [1D deconvolve$g $h]pp " conv(f,h) = g" [1D convolve$f \$h]
Output:
deconv(g,f) = h = [-8,-9,-3,-1,-6,7]
deconv(g,h) = f = [-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1]
conv(f,h) = g = [24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7]


## Ursala

The user defined function band constructs the required matrix as a list of lists given the pair of sequences to be deconvolved, and the lapack..dgelsd function solves the system. Some other library functions used are zipt (zipping two unequal length lists by truncating the longer one) zipp0 (zipping unequal length lists by padding the shorter with zeros) and pad0 (making a list of lists all the same length by appending zeros to the short ones).

#import std#import nat band = pad0+ ~&rSS+ zipt^*D(~&r,^lrrSPT/~&ltK33tx zipt^/~&r ~&lSNyCK33+ zipp0)^/~&rx ~&B->NlNSPC ~&bt deconv = lapack..dgelsd^\~&l ~&||0.!**+ band 

test program:

h = <-8.,-9.,-3.,-1.,-6.,7.>f = <-3.,-6.,-1.,8.,-6.,3.,-1.,-9.,-9.,3.,-2.,5.,2.,-2.,-7.,-1.>g = <24.,75.,71.,-34.,3.,22.,-45.,23.,245.,25.,52.,25.,-67.,-96.,96.,31.,55.,36.,29.,-43.,-7.> #cast %eLm test = <   'h': deconv(g,f),   'f': deconv(g,h)> 
Output:
<
'h': <
-8.000000e+00,
-9.000000e+00,
-3.000000e+00,
-1.000000e+00,
-6.000000e+00,
7.000000e+00>,
'f': <
-3.000000e+00,
-6.000000e+00,
-1.000000e+00,
8.000000e+00,
-6.000000e+00,
3.000000e+00,
-1.000000e+00,
-9.000000e+00,
-9.000000e+00,
3.000000e+00,
-2.000000e+00,
5.000000e+00,
2.000000e+00,
-2.000000e+00,
-7.000000e+00,
-1.000000e+00>>


## zkl

Using GNU Scientific Library:

var [const] GSL=Import("zklGSL");	// libGSL (GNU Scientific Library)fcn dconv1D(f,g){   fsz,hsz:=f.len(), g.len() - fsz +1;   A:=GSL.Matrix(g.len(),hsz);   foreach n,fn in ([0..].zip(f)){ foreach rc in (hsz){ A[rc+n,rc]=fn } }   h:=A.AxEQb(g);   h}
f:=GSL.VectorFromData(-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1);g:=GSL.VectorFromData(24,75,71,-34,3,22,-45,23,245,25,52,25,-67,-96,96,31,55,36,29,-43,-7);h:=dconv1D(f,g);h.format().println(); f:=dconv1D(h,g);f.format().println();
Output:
-8.00,-9.00,-3.00,-1.00,-6.00,7.00
-3.00,-6.00,-1.00,8.00,-6.00,3.00,-1.00,-9.00,-9.00,3.00,-2.00,5.00,2.00,-2.00,-7.00,-1.00


Or, using lists:

Translation of: D
fcn deconv(g,f){   flen, glen, delta:=f.len(), g.len(), glen - flen + 1;   result:=List.createLong(delta); // allocate list with space for items   foreach n in (delta){      e:=g[n];      lowerBound:=(if (n>=flen) n - flen + 1 else 0);      foreach i in ([lowerBound .. n-1]){ e-=result[i]*f[n - i]; }      result.append(e/f[0]);    }    result;}
h:=T(-8,-9,-3,-1,-6,7);f:=T(-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1);g:=T(24,75,71,-34,3,22,-45,23,245,25,52,25,-67,                   -96,96,31,55,36,29,-43,-7);println(deconv(g, f) == h, " ", deconv(g, f));println(deconv(g, h) == f, " ", deconv(g, h));
Output:
True L(-8,-9,-3,-1,-6,7)
True L(-3,-6,-1,8,-6,3,-1,-9,-9,3,-2,5,2,-2,-7,-1)
`