Resistor mesh: Difference between revisions
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B: zeromatrix(n, 1), |
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B[k: (bi - 1) * q + bj, 1]: 1, |
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M: lu_factor(A), |
M: lu_factor(A), |
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V: lu_backsub(M, B), |
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Revision as of 11:14, 28 August 2012
Given 10 × 10 grid nodes interconnected by 1Ω resistors as shown, find the resistance between point A and B.
See also [[1]]
Ada
<lang Ada>with Ada.Text_IO; use Ada.Text_IO; procedure ResistMesh is
H, W : constant Positive := 10; rowA, colA : constant Positive := 2; -- row/col indexed from 1 rowB : constant Positive := 7; colB : constant Positive := 8;
type Ntype is (A, B, Free);
type Vtype is digits 15;
type Node is record
volt : Vtype := 0.0;
name : Ntype := Free;
end record;
type NodeMesh is array (Positive range <>, Positive range <>) of Node;
package IIO is new Ada.Text_IO.Float_IO (Vtype);
mesh, dmesh : NodeMesh (1 .. H, 1 .. W);
curA, curB, diff : Vtype;
procedure set_AB (mesh : in out NodeMesh) is begin
mesh (rowA, colA).volt := 1.0; mesh (rowA, colA).name := A;
mesh (rowB, colB).volt := -1.0; mesh (rowB, colB).name := B;
end set_AB;
function sides (i : Positive; j : Positive) return Vtype is
s : Integer := 0;
begin
if i /= 1 and i /= H then s := s + 2; else s := s + 1; end if;
if j /= 1 and j /= W then s := s + 2; else s := s + 1; end if;
return Vtype (s);
end sides;
procedure calc_diff (mesh : NodeMesh; dmesh : out NodeMesh;
total : out Vtype) is
n : Natural;
v : Vtype := 0.0;
begin
total := 0.0;
for i in Positive range 1 .. H loop
for j in Positive range 1 .. W loop
n := 0; v := 0.0;
if i /= 1 then v := v + mesh (i - 1, j).volt; n := n + 1; end if;
if j /= 1 then v := v + mesh (i, j - 1).volt; n := n + 1; end if;
if i < H then v := v + mesh (i + 1, j).volt; n := n + 1; end if;
if j < W then v := v + mesh (i, j + 1).volt; n := n + 1; end if;
v := mesh (i, j).volt - v / Vtype (n);
dmesh (i, j).volt := v;
if mesh (i, j).name = Free then total := total + v ** 2; end if;
end loop;
end loop;
end calc_diff;
begin
loop
set_AB (mesh);
calc_diff (mesh, dmesh, diff);
exit when diff < 1.0e-40;
for i in Positive range 1 .. H loop
for j in Positive range 1 .. W loop
mesh (i, j).volt := mesh (i, j).volt - dmesh (i, j).volt;
end loop;
end loop;
end loop;
curA := dmesh (rowA, colA).volt * sides (rowA, colA); curB := dmesh (rowB, colB).volt * sides (rowB, colB); diff := 4.0 / (curA - curB); IIO.Put (diff, Exp => 0); New_Line;
end ResistMesh;</lang>
- Output:
1.60899124173073
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- define S 10
typedef struct { double v; int fixed; } node;
- define each(i, x) for(i = 0; i < x; i++)
node **alloc2(int w, int h) { int i; node **a = calloc(1, sizeof(node*)*h + sizeof(node)*w*h); each(i, h) a[i] = i ? a[i-1] + w : (node*)(a + h); return a; }
void set_boundary(node **m) { m[1][1].fixed = 1; m[1][1].v = 1; m[6][7].fixed = -1; m[6][7].v = -1; }
double calc_diff(node **m, node **d, int w, int h) { int i, j, n; double v, total = 0; each(i, h) each(j, w) { v = 0; n = 0; if (i) v += m[i-1][j].v, n++; if (j) v += m[i][j-1].v, n++; if (i+1 < h) v += m[i+1][j].v, n++; if (j+1 < w) v += m[i][j+1].v, n++;
d[i][j].v = v = m[i][j].v - v / n; if (!m[i][j].fixed) total += v * v; } return total; }
double iter(node **m, int w, int h) { node **d = alloc2(w, h); int i, j; double diff = 1e10; double cur[] = {0, 0, 0};
while (diff > 1e-24) { set_boundary(m); diff = calc_diff(m, d, w, h); each(i,h) each(j, w) m[i][j].v -= d[i][j].v; }
each(i, h) each(j, w) cur[ m[i][j].fixed + 1 ] += d[i][j].v * (!!i + !!j + (i < h-1) + (j < w -1));
free(d); return (cur[2] - cur[0])/2; }
int main() { node **mesh = alloc2(S, S); printf("R = %g\n", 2 / iter(mesh, S, S)); return 0; }</lang>
D
<lang d>import std.stdio, std.traits;
enum Node.FP differenceThreshold = 1e-40;
struct Node {
alias real FP;
enum Fixed : size_t { free = 0, A = 1, B = 2 }
FP voltage = 0.0; private Fixed fixed__ = Fixed.free;
this(in FP v, in Fixed f) pure nothrow {
this.voltage = v;
this.fixed__ = f;
}
@property Fixed fixed() const pure nothrow { return fixed__; }
}
Node.FP iter(size_t w, size_t h)(ref Node[w][h] m) pure nothrow {
static void enforceBoundaryConditions(size_t w, size_t h)
(ref Node[w][h] m)
pure nothrow {
m[1][1].voltage = 1.0;
m[6][7].voltage = -1.0;
}
static Node.FP calcDifference(size_t w, size_t h)
(in ref Node[w][h] m,
ref Node[w][h] d) pure nothrow {
Node.FP total = 0.0;
foreach (i; 0 .. h)
foreach (j; 0 .. w) {
Node.FP v = 0.0;
size_t n;
if (i != 0) { v += m[i-1][j].voltage; n++; }
if (j != 0) { v += m[i][j-1].voltage; n++; }
if (i < h-1) { v += m[i+1][j].voltage; n++; }
if (j < w-1) { v += m[i][j+1].voltage; n++; }
v = m[i][j].voltage - v / n;
d[i][j].voltage = v;
if (m[i][j].fixed == Node.Fixed.free)
total += v ^^ 2;
}
return total; }
Node[w][h] difference;
while (true) {
enforceBoundaryConditions(m);
if (calcDifference(m, difference) < differenceThreshold)
break;
foreach (i, const(Node[]) di; difference)
foreach (j, ref const(Node) dij; di)
m[i][j].voltage -= dij.voltage;
}
Node.FP[EnumMembers!(Node.Fixed).length] cur = 0.0;
foreach (i, const(Node[]) di; difference)
foreach (j, ref const(Node) dij; di)
cur[m[i][j].fixed] += dij.voltage *
(!!i + !!j + (i < h-1) + (j < w-1));
return (cur[Node.Fixed.A] - cur[Node.Fixed.B]) / 2.0;
}
void main() {
enum size_t w = 10,
h = w;
Node[w][h] mesh;
// Set A and B Nodes. mesh[1][1] = Node( 1.0, Node.Fixed.A); mesh[6][7] = Node(-1.0, Node.Fixed.B);
writefln("R = %.19f", 2 / iter(mesh));
}</lang>
- Output:
R = 1.6089912417307296556
Maxima
<lang maxima>/* Place a current souce between A and B, providing 1 A. Then we are really looking
for the potential at A and B, since I = R (V(B) - V(A)) where I is given and we want R. Atually, we will compute potential at each node, except A where we assume it's 0. Without with assumption, there would be infinitely many solutions since potential is known up to a constant. For A we will simply write the equation V(A) = 0, to keep the program simple. Hence, for a general grid of p rows and q columns, there are n = p * q nodes, so n unknowns, and n equations. Write Kirchhoff's current law at each node. Be careful with the node A (equation A = 0) and the node B (there is a constant current to add, from the source, that will go in the constant terms of the system). Finally, we have a n x n linear system of equations to solve. Simply use Maxima's builtin LU decomposition. Since all computations are exact, the result will be also exact, written as a fraction. Also, the program can work with any grid, and any two nodes on the grid.
For those who want more speed and less space, notice the system is sparse and necessarily symmetric, so one can use conjugate gradient or any other sparse symmetric solver. */
/* Auxiliary function to get rid of the borders */ ongrid(i, j, p, q) := is(i >= 1 and i <= p and j >= 1 and j <= q)$
grid_resistor(p, q, ai, aj, bi, bj) := block(
[n: p * q, A, B, M, k, c, V],
A: zeromatrix(n, n),
for i thru p do
for j thru q do (
k: (i - 1) * q + j,
if i = ai and j = aj then
A[k, k]: 1
else (
c: 0,
if ongrid(i + 1, j, p, q) then (c: c + 1, A[k, k + q]: -1),
if ongrid(i - 1, j, p, q) then (c: c + 1, A[k, k - q]: -1),
if ongrid(i, j + 1, p, q) then (c: c + 1, A[k, k + 1]: -1),
if ongrid(i, j - 1, p, q) then (c: c + 1, A[k, k - 1]: -1),
A[k, k]: c
)
),
B: zeromatrix(n, 1),
B[k: (bi - 1) * q + bj, 1]: 1,
M: lu_factor(A),
V: lu_backsub(M, B),
V[k, 1]
)$
grid_resistor(10, 10, 2, 2, 8, 7); 455859137025721 / 283319837425200
bfloat(%), fpprec = 40; 1.608991241730729655954495520510088761201b0</lang>
Perl
<lang perl>use strict;
my ($w, $h) = (9, 9); my @v = map([ (0) x ($w + 1) ], 0 .. $h); # voltage my @f = map([ (0) x ($w + 1) ], 0 .. $h); # fixed condition my @d = map([ (0) x ($w + 1) ], 0 .. $h); # diff
my @n; # neighbors for my $i (0 .. $h) { push @{$n[$i][$_]}, [$i, $_ - 1] for 1 .. $w; push @{$n[$i][$_]}, [$i, $_ + 1] for 0 .. $w - 1; } for my $j (0 .. $w) { push @{$n[$_][$j]}, [$_ - 1, $j] for 1 .. $h; push @{$n[$_][$j]}, [$_ + 1, $j] for 0 .. $h - 1; }
sub set_boundary { $f[1][1] = 1; $f[6][7] = -1; $v[1][1] = 1; $v[6][7] = -1; }
sub calc_diff { my $total_diff; for my $i (0 .. $h) { for my $j (0 .. $w) { my ($p, $v) = $n[$i][$j]; $v += $v[$_->[0]][$_->[1]] for @$p; $d[$i][$j] = $v = $v[$i][$j] - $v / scalar(@$p); $total_diff += $v * $v unless $f[$i][$j]; } } $total_diff; }
sub iter { my $diff = 1; while ($diff > 1e-24) { # 1e-24 is overkill (12 digits of precision) set_boundary(); $diff = calc_diff(); print "error^2: $diff\r"; for my $i (0 .. $h) { for my $j (0 .. $w) { $v[$i][$j] -= $d[$i][$j]; } } } print "\n";
my @current = (0) x 3; for my $i (0 .. $h) { for my $j (0 .. $w) { $current[ $f[$i][$j] ] += $d[$i][$j] * scalar(@{$n[$i][$j]}); } } return ($current[1] - $current[-1]) / 2; }
print "R = @{[2 / iter()]}\n";</lang>
Perl 6
<lang perl6>my $S = 10;
my @fixed;
sub allocmesh ($w, $h) {
gather for ^$h {
take [0 xx $w];
}
}
sub force-fixed(@f) {
@f[1][1] = 1; @f[6][7] = -1;
}
sub force-v(@v) {
@v[1][1] = 1; @v[6][7] = -1;
}
sub calc_diff(@v, @d, Int $w, Int $h) {
my $total = 0;
for ^$h X ^$w -> $i, $j {
my @neighbors = grep *.defined, @v[$i-1][$j], @v[$i][$j-1], @v[$i+1][$j], @v[$i][$j+1];
my $v = [+] @neighbors;
@d[$i][$j] = $v = @v[$i][$j] - $v / +@neighbors;
$total += $v * $v unless @fixed[$i][$j];
}
return $total;
}
sub iter(@v, Int $w, Int $h) {
my @d = allocmesh($w, $h); my $diff = 1e10; my @cur = 0, 0, 0;
while $diff > 1e-24 {
force-v(@v);
$diff = calc_diff(@v, @d, $w, $h);
for ^$h X ^$w -> $i, $j {
@v[$i][$j] -= @d[$i][$j];
}
}
for ^$h X ^$w -> $i, $j {
@cur[ @fixed[$i][$j] + 1 ]
+= @d[$i][$j] * (?$i + ?$j + ($i < $h - 1) + ($j < $w - 1));
}
return (@cur[2] - @cur[0]) / 2;
}
my @mesh = allocmesh($S, $S);
@fixed = allocmesh($S, $S); force-fixed(@fixed);
say 2 / iter(@mesh, $S, $S);</lang>
- Output:
1.60899124172989
Python
<lang python>DIFF_THRESHOLD = 1e-40
class Fixed:
FREE = 0 A = 1 B = 2
class Node:
__slots__ = ["voltage", "fixed"]
def __init__(self, v=0.0, f=Fixed.FREE):
self.voltage = v
self.fixed = f
def set_boundary(m):
m[1][1] = Node( 1.0, Fixed.A) m[6][7] = Node(-1.0, Fixed.B)
def calc_difference(m, d):
h = len(m) w = len(m[0]) total = 0.0
for i in xrange(h):
for j in xrange(w):
v = 0.0
n = 0
if i != 0: v += m[i-1][j].voltage; n += 1
if j != 0: v += m[i][j-1].voltage; n += 1
if i < h-1: v += m[i+1][j].voltage; n += 1
if j < w-1: v += m[i][j+1].voltage; n += 1
v = m[i][j].voltage - v / n
d[i][j].voltage = v
if m[i][j].fixed == Fixed.FREE:
total += v ** 2
return total
def iter(m):
h = len(m) w = len(m[0]) difference = [[Node() for j in xrange(w)] for i in xrange(h)]
while True:
set_boundary(m) # Enforce boundary conditions.
if calc_difference(m, difference) < DIFF_THRESHOLD:
break
for i, di in enumerate(difference):
for j, dij in enumerate(di):
m[i][j].voltage -= dij.voltage
cur = [0.0] * 3
for i, di in enumerate(difference):
for j, dij in enumerate(di):
cur[m[i][j].fixed] += (dij.voltage *
(bool(i) + bool(j) + (i < h-1) + (j < w-1)))
return (cur[Fixed.A] - cur[Fixed.B]) / 2.0
def main():
w = h = 10 mesh = [[Node() for j in xrange(w)] for i in xrange(h)] print "R = %.16f" % (2 / iter(mesh))
main()</lang>
- Output:
R = 1.6089912417307286
REXX
This version allows specification of the grid size and the location of the A & B points. <lang rexx>/*REXX pgm calculates resistance between any 2 points on a resister grid*/ numeric digits 20 /*use moderate digits (precision)*/ minVal=(1'e-'||(digits()*2))/1 /*calculate the threshold min val*/ if 1=='f1'x then ohms='ohms' /*EBCDIC machine? Use 'ohms'. */
else ohms='ea'x /* ASCII machine? Use Greek Ω.*/
parse arg wide high Arow Acol Brow Bcol . say 'minVal = ' format(minVal,,,,0) ; say say 'resistor mesh is of size: ' wide "wide, " high 'high.' ; say say 'point A is at (row,col): ' Arow","Acol say 'point B is at (row,col): ' Brow","Bcol ; say @.=0; cell.=1
do until $<=minVal; $=0; v=0
@.Arow.Acol = +1 ; cell.Arow.Acol = 0
@.Brow.Bcol = -1 ; cell.Brow.Bcol = 0
do i =1 for high; im=i-1; ip=i+1
do j=1 for wide; jm=j-1; jp=j+1; n=0; v=0
if i\==1 then do; v=v+@.im.j; n=n+1; end
if j\==1 then do; v=v+@.i.jm; n=n+1; end
if i<high then do; v=v+@.ip.j; n=n+1; end
if j<wide then do; v=v+@.i.jp; n=n+1; end
v=@.i.j-v/n; #.i.j=v; if cell.i.j then $=$+v*v
end /*j*/
end /*i*/
do r =1 for High
do c=1 for Wide; @.r.c = @.r.c - #.r.c
end /*r*/
end /*c*/
end /*until $<=minVal*/
Acur = #.Arow.Acol * sides(Arow,Acol) Bcur = #.Brow.Bcol * sides(Brow,Bcol) say 'resistance between point A and point B is: ' 4/(Acur-Bcur) ohms exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────sides subroutine────────-───────────*/ sides: parse arg i,j; !=0; if i\==1 & i\==high then !=!+2
else !=!+1
if j\==1 & j\==wide then !=!+2
else !=!+1
return !</lang> output when using the input of: 10 10 2 2 7 8
minVal = 1E-40 resistor mesh is of size: 10 wide, 10 high. point A is at (row,col): 2,2 point B is at (row,col): 7,8 resistance between point A and point B is: 1.6089912417307296559 Ω
Tcl
<lang tcl>package require Tcl 8.6; # Or 8.5 with the TclOO package
- This code is structured as a class with a little trivial DSL parser so it is
- easy to change what problem is being worked on.
oo::class create ResistorMesh {
variable forcePoints V fixed w h
constructor {boundaryConditions} {
foreach {op condition} $boundaryConditions { switch $op { size { lassign $condition w h set fixed [lrepeat $h [lrepeat $w 0]] set V [lrepeat $h [lrepeat $w 0.0]] } fixed { lassign $condition j i v lset fixed $i $j [incr ctr] lappend forcePoints $j $i $v } } }
}
method CalculateDifferences {*dV} {
upvar 1 ${*dV} dV set error 0.0 for {set i 0} {$i < $h} {incr i} { for {set j 0} {$j < $w} {incr j} { set v 0.0 set n 0 if {$i} { set v [expr {$v + [lindex $V [expr {$i-1}] $j]}] incr n } if {$j} { set v [expr {$v + [lindex $V $i [expr {$j-1}]]}] incr n } if {$i+1 < $h} { set v [expr {$v + [lindex $V [expr {$i+1}] $j]}] incr n } if {$j+1 < $w} { set v [expr {$v + [lindex $V $i [expr {$j+1}]]}] incr n } lset dV $i $j [set v [expr {[lindex $V $i $j] - $v/$n}]] if {![lindex $fixed $i $j]} { set error [expr {$error + $v**2}] } } } return $error
}
method FindCurrentFixpoint {epsilon} {
set dV [lrepeat $h [lrepeat $w 0.0]] set current {0.0 0.0 0.0} while true { # Enforce the boundary conditions foreach {j i v} $forcePoints { lset V $i $j $v } # Compute the differences and the error set error [my CalculateDifferences dV] # Apply the differences for {set i 0} {$i < $h} {incr i} { for {set j 0} {$j < $w} {incr j} { lset V $i $j [expr { [lindex $V $i $j] - [lindex $dV $i $j]}] } } # Done if the error is small enough if {$error < $epsilon} break } # Compute the currents from the error for {set i 0} {$i < $h} {incr i} { for {set j 0} {$j < $w} {incr j} { lset current [lindex $fixed $i $j] [expr { [lindex $current [lindex $fixed $i $j]] + [lindex $dV $i $j] * (!!$i+!!$j+($i<$h-1)+($j<$w-1))}] } } # Compute the actual current flowing between source and sink return [expr {([lindex $current 1] - [lindex $current 2]) / 2.0}]
}
# Public entry point method solveForResistance Template:Epsilon 1e-24 {
set voltageDifference [expr { [lindex $forcePoints 2] - [lindex $forcePoints 5]}] expr {$voltageDifference / [my FindCurrentFixpoint $epsilon]}
}
}</lang> Setting up and solving this particular problem: <lang tcl>ResistorMesh create mesh {
size {10 10}
fixed {1 1 1.0}
fixed {6 7 -1.0}
} puts [format "R = %.12g" [mesh solveForResistance]]</lang>
- Output:
R = 1.60899124173