Resistor mesh
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
Given 10×10 grid nodes (as shown in the image) interconnected by 1Ω resistors as shown,
find the resistance between points A and B.
- See also
- (humor, nerd sniping) xkcd.com cartoon (you can solve that for extra credits)
- An article on how to calculate this and an implementation in Mathematica
11l
-V DIFF_THRESHOLD = 1e-40
T.enum Fixed
FREE
A
B
T Node
Float voltage
Fixed fixed
F (v = 0.0, f = Fixed.FREE)
.voltage = v
.fixed = f
F set_boundary(&m)
m[1][1] = Node( 1.0, Fixed.A)
m[6][7] = Node(-1.0, Fixed.B)
F calc_difference(m, &d)
V h = m.len
V w = m[0].len
V total = 0.0
L(i) 0 .< h
L(j) 0 .< w
V v = 0.0
V n = 0
I i != 0 {v += m[i - 1][j].voltage; n++}
I j != 0 {v += m[i][j - 1].voltage; n++}
I i < h-1 {v += m[i + 1][j].voltage; n++}
I j < w-1 {v += m[i][j + 1].voltage; n++}
v = m[i][j].voltage - v / n
d[i][j].voltage = v
I m[i][j].fixed == FREE
total += v ^ 2
R total
F iter(&m)
V h = m.len
V w = m[0].len
V difference = [[Node()] * w] * h
L
set_boundary(&m)
I calc_difference(m, &difference) < :DIFF_THRESHOLD
L.break
L(di) difference
V i = L.index
L(dij) di
V j = L.index
m[i][j].voltage -= dij.voltage
V cur = [0.0] * 3
L(di) difference
V i = L.index
L(dij) di
V j = L.index
cur[Int(m[i][j].fixed)] += (dij.voltage *
(Int(i != 0) + Int(j != 0) + (i < h - 1) + (j < w - 1)))
R (cur[Int(Fixed.A)] - cur[Int(Fixed.B)]) / 2.0
V w = 10
V h = 10
V mesh = [[Node()] * w] * h
print(‘R = #.16’.format(2 / iter(&mesh)))- Output:
R = 1.6089912417307285
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;
- Output:
1.60899124173073
BBC BASIC
INSTALL @lib$+"ARRAYLIB"
*FLOAT 64
@% = &F0F
PRINT "Resistance = "; FNresistormesh(10, 10, 1, 1, 7, 6) " ohms"
END
DEF FNresistormesh(ni%, nj%, ai%, aj%, bi%, bj%)
LOCAL c%, i%, j%, k%, n%, A(), B()
n% = ni% * nj%
DIM A(n%-1, n%-1), B(n%-1, 0)
FOR i% = 0 TO ni%-1
FOR j% = 0 TO nj%-1
k% = i% * nj% + j%
IF i% = ai% AND j% = aj% THEN
A(k%, k%) = 1
ELSE
c% = 0
IF (i% + 1) < ni% c% += 1 : A(k%, k% + nj%) = -1
IF i% > 0 c% += 1 : A(k%, k% - nj%) = -1
IF (j% + 1) < nj% c% += 1 : A(k%, k% + 1) = -1
IF j% > 0 c% += 1 : A(k%, k% - 1) = -1
A(k%, k%) = c%
ENDIF
NEXT
NEXT
k% = bi% * nj% + bj%
B(k%, 0) = 1
PROC_invert(A())
B() = A().B()
= B(k%, 0)
- Output:
Resistance = 1.60899124173071 ohms
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;
}
C#
using System;
using System.Collections.Generic;
namespace ResistorMesh {
class Node {
public Node(double v, int fixed_) {
V = v;
Fixed = fixed_;
}
public double V { get; set; }
public int Fixed { get; set; }
}
class Program {
static void SetBoundary(List<List<Node>> m) {
m[1][1].V = 1.0;
m[1][1].Fixed = 1;
m[6][7].V = -1.0;
m[6][7].Fixed = -1;
}
static double CalcuateDifference(List<List<Node>> m, List<List<Node>> d, int w, int h) {
double total = 0.0;
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
double v = 0.0;
int n = 0;
if (i > 0) {
v += m[i - 1][j].V;
n++;
}
if (j > 0) {
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++;
}
v = m[i][j].V - v / n;
d[i][j].V = v;
if (m[i][j].Fixed == 0) {
total += v * v;
}
}
}
return total;
}
static double Iter(List<List<Node>> m, int w, int h) {
List<List<Node>> d = new List<List<Node>>(h);
for (int i = 0; i < h; i++) {
List<Node> t = new List<Node>(w);
for (int j = 0; j < w; j++) {
t.Add(new Node(0.0, 0));
}
d.Add(t);
}
double[] curr = new double[3];
double diff = 1e10;
while (diff > 1e-24) {
SetBoundary(m);
diff = CalcuateDifference(m, d, w, h);
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
m[i][j].V -= d[i][j].V;
}
}
}
for (int i = 0; i < h; i++) {
for (int j = 0; j < w; j++) {
int k = 0;
if (i != 0) k++;
if (j != 0) k++;
if (i < h - 1) k++;
if (j < w - 1) k++;
curr[m[i][j].Fixed + 1] += d[i][j].V * k;
}
}
return (curr[2] - curr[0]) / 2.0;
}
const int S = 10;
static void Main(string[] args) {
List<List<Node>> mesh = new List<List<Node>>(S);
for (int i = 0; i < S; i++) {
List<Node> t = new List<Node>(S);
for (int j = 0; j < S; j++) {
t.Add(new Node(0.0, 0));
}
mesh.Add(t);
}
double r = 2.0 / Iter(mesh, S, S);
Console.WriteLine("R = {0:F15}", r);
}
}
}
- Output:
R = 1.608991241729890
C++
#include <iomanip>
#include <iostream>
#include <vector>
class Node {
private:
double v;
int fixed;
public:
Node() : v(0.0), fixed(0) {
// empty
}
Node(double v, int fixed) : v(v), fixed(fixed) {
// empty
}
double getV() const {
return v;
}
void setV(double nv) {
v = nv;
}
int getFixed() const {
return fixed;
}
void setFixed(int nf) {
fixed = nf;
}
};
void setBoundary(std::vector<std::vector<Node>>& m) {
m[1][1].setV(1.0);
m[1][1].setFixed(1);
m[6][7].setV(-1.0);
m[6][7].setFixed(-1);
}
double calculateDifference(const std::vector<std::vector<Node>>& m, std::vector<std::vector<Node>>& d, const int w, const int h) {
double total = 0.0;
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
double v = 0.0;
int n = 0;
if (i > 0) {
v += m[i - 1][j].getV();
n++;
}
if (j > 0) {
v += m[i][j - 1].getV();
n++;
}
if (i + 1 < h) {
v += m[i + 1][j].getV();
n++;
}
if (j + 1 < w) {
v += m[i][j + 1].getV();
n++;
}
v = m[i][j].getV() - v / n;
d[i][j].setV(v);
if (m[i][j].getFixed() == 0) {
total += v * v;
}
}
}
return total;
}
double iter(std::vector<std::vector<Node>>& m, const int w, const int h) {
using namespace std;
vector<vector<Node>> d;
for (int i = 0; i < h; ++i) {
vector<Node> t(w);
d.push_back(t);
}
double curr[] = { 0.0, 0.0, 0.0 };
double diff = 1e10;
while (diff > 1e-24) {
setBoundary(m);
diff = calculateDifference(m, d, w, h);
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
m[i][j].setV(m[i][j].getV() - d[i][j].getV());
}
}
}
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
int k = 0;
if (i != 0) ++k;
if (j != 0) ++k;
if (i < h - 1) ++k;
if (j < w - 1) ++k;
curr[m[i][j].getFixed() + 1] += d[i][j].getV()*k;
}
}
return (curr[2] - curr[0]) / 2.0;
}
const int S = 10;
int main() {
using namespace std;
vector<vector<Node>> mesh;
for (int i = 0; i < S; ++i) {
vector<Node> t(S);
mesh.push_back(t);
}
double r = 2.0 / iter(mesh, S, S);
cout << "R = " << setprecision(15) << r << '\n';
return 0;
}
- Output:
R = 1.60899124172989
D
import std.stdio, std.traits;
enum Node.FP differenceThreshold = 1e-40;
struct Node {
alias FP = real;
enum Kind : size_t { free, A, B }
FP voltage = 0.0;
/*const*/ private Kind kind = Kind.free;
// Remove kindGet once kind is const.
@property Kind kindGet() const pure nothrow @nogc {return kind; }
}
Node.FP iter(size_t w, size_t h)(ref Node[w][h] m) pure nothrow @nogc {
static void enforceBoundaryConditions(ref Node[w][h] m)
pure nothrow @nogc {
m[1][1].voltage = 1.0;
m[6][7].voltage = -1.0;
}
static Node.FP calcDifference(in ref Node[w][h] m,
ref Node[w][h] d) pure nothrow @nogc {
Node.FP total = 0.0;
foreach (immutable i; 0 .. h) {
foreach (immutable j; 0 .. w) {
Node.FP v = 0.0;
{
size_t n = 0;
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].kindGet == Node.Kind.free)
total += v ^^ 2;
}
}
return total;
}
Node[w][h] difference;
while (true) {
enforceBoundaryConditions(m);
if (calcDifference(m, difference) < differenceThreshold)
break;
foreach (immutable i, const di; difference)
foreach (immutable j, const ref dij; di)
m[i][j].voltage -= dij.voltage;
}
Node.FP[EnumMembers!(Node.Kind).length] cur = 0.0;
foreach (immutable i, const di; difference)
foreach (immutable j, const ref dij; di)
cur[m[i][j].kindGet] += dij.voltage *
(!!i + !!j + (i < h-1) + (j < w-1));
return (cur[Node.Kind.A] - cur[Node.Kind.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.Kind.A);
mesh[6][7] = Node(-1.0, Node.Kind.B);
writefln("R = %.19f", 2 / mesh.iter);
}
- Output:
R = 1.6089912417307296556
ERRE
We'll solve the linear system. We'll write Kirchhoff's circuit laws at each node and search for a voltage distribution that creates a 1A current coming from A exiting in B. The difference of voltage between B and A is then the resistance.
PROGRAM RESISTENCE_MESH
!$BASE=1
!$DYNAMIC
DIM A[0,0]
BEGIN
N=10
NN=N*N
!$DIM A[NN,NN+1]
PRINT(CHR$(12);) !CLS
! generate matrix data
NODE=0
FOR ROW=1 TO N DO
FOR COL=1 TO N DO
NODE=NODE+1
IF ROW>1 THEN
A[NODE,NODE]=A[NODE,NODE]+1
A[NODE,NODE-N]=-1
END IF
IF ROW<N THEN
A[NODE,NODE]=A[NODE,NODE]+1
A[NODE,NODE+N]=-1
END IF
IF COL>1 THEN
A[NODE,NODE]=A[NODE,NODE]+1
A[NODE,NODE-1]=-1
END IF
IF COL<N THEN
A[NODE,NODE]=A[NODE,NODE]+1
A[NODE,NODE+1]=-1
END IF
END FOR
END FOR
AR=2 AC=2 A=AC+N*(AR-1)
BR=7 BC=8 B=BC+N*(BR-1)
A[A,NN+1]=-1
A[B,NN+1]=1
PRINT("Nodes ";A,B)
! solve linear system
! using Gauss-Seidel method
! with pivoting
R=NN
FOR J=1 TO R DO
FOR I=J TO R DO
EXIT IF A[I,J]<>0
END FOR
IF I=R+1 THEN
PRINT("No solution!")
!$STOP
END IF
FOR K=1 TO R+1 DO
SWAP(A[J,K],A[I,K])
END FOR
Y=1/A[J,J]
FOR K=1 TO R+1 DO
A[J,K]=Y*A[J,K]
END FOR
FOR I=1 TO R DO
IF I<>J THEN
Y=-A[I,J]
FOR K=1 TO R+1 DO
A[I,K]=A[I,K]+Y*A[J,K]
END FOR
END IF
END FOR
END FOR
PRINT("Resistence=";ABS(A[A,NN+1]-A[B,NN+1]))
END PROGRAM- Output:
Nodes 12 68 Resistence=1.608993
Euler Math Toolbox
The functions for this have been implemented in Euler already. Thus the following commands solve this problem.
>load incidence;
>{u,r}=solvePotentialX(makeRectangleX(10,10),12,68); r,
1.60899124173The necessary functions in the file incidence.e are as follows. There are versions with full matrices. But the functions listed here use compressed matrices and the conjugate gradient method.
function makeRectangleX (n:index,m:index)
## Make the incidence matrix of a rectangle grid in compact form.
## see: makeRectangleIncidence
K=zeros(n*(m-1)+m*(n-1),3);
k=1;
for i=1 to n;
for j=1 to m-1;
K[k,1]=(i-1)*m+j; K[k,2]=(i-1)*m+j+1; K[k,3]=1;
k=k+1;
end;
end;
for i=1 to n-1;
for j=1 to m;
K[k,1]=(i-1)*m+j; K[k,2]=i*m+j; K[k,3]=1;
k=k+1;
end;
end;
H=cpxzeros([n*m,n*m]);
H=cpxset(H,K);
H=cpxset(H,K[:,[2,1,3]]);
return H;
endfunction
function solvePotentialX (A:cpx, i:index ,j:index)
## Solve the potential problem of resistance in a graph.
## This functions uses the conjugate gradient method.
## A is a compressed incidence matrix.
## Return the potential u for the nodes in A,
## such that u[i]=1, u[j]=-1, and the flow
## to each knot is equal to the flow from the knot,
## and the flow from i to j is (u[i]-u[j])*A[i,j].
## see: makeIncidence
n=size(A)[1];
b=ones(n,1); f=-cpxmult(A,b);
h=1:n; B=cpxset(A,h'|h'|f);
B=cpxset(B,i|h'|0);
B=cpxset(B,[i,i,1]);
B=cpxset(B,j|h'|0);
B=cpxset(B,[j,j,1]);
v=zeros(n,1); v[i]=1; v[j]=-1;
u=cpxfit(B,v);
f=(-f[i])*u[i]-cpxmult(A,u)[i];
return {u,2/f}
endfunction
Here is the code for the conjugate gradient method for compressed, sparse matrices from cpx.e.
function cgX (H:cpx, b:real column, x0:real column=none, f:index=10)
## Conjugate gradient method to solve Hx=b for compressed H.
##
## This is the method of choice for large, sparse matrices. In most
## cases, it will work well, fast, and accurate.
##
## H must be positive definite. Use cpxfit, if it is not.
##
## The accuarcy can be controlled with an additional parameter
## eps. The algorithm stops, when the error gets smaller then eps, or
## after f*n iterations, if the error gets larger. x0 is an optional
## start vector.
##
## H : compressed matrix (nxm)
## b : column vector (mx1)
## x0 : optional start point (mx1)
## f : number of steps, when the method should be restarted
##
## See: cpxfit, cg, cgXnormal
if isvar("eps") then localepsilon(eps); endif;
n=cols(H);
if x0==none then x=zeros(size(b));
else; x=x0;
endif;
loop 1 to 10
r=b-cpxmult(H,x); p=r; fehler=r'.r;
loop 1 to f*n
if sqrt(fehler)~=0 then return x; endif;
Hp=cpxmult(H,p);
a=fehler/(p'.Hp);
x=x+a*p;
rn=r-a*Hp;
fehlerneu=rn'.rn;
p=rn+fehlerneu/fehler*p;
r=rn; fehler=fehlerneu;
end;
end;
return x;
endfunction
FreeBASIC
' version 01-07-2018
' compile with: fbc -s console
#Define n 10
Dim As UInteger nn = n * n
Dim As Double g(-nn To nn +1, -nn To nn +1)
Dim As UInteger node, row, col
For row = 1 To n
For col = 1 To n
node += 1
If row > 1 Then
g(node, node) += 1
g(node, node - n) = -1
End If
If row < n Then
g(node, node) += 1
g(node, node + n) = -1
End If
If col > 1 Then
g(node, node) += 1
g(node, node -1) = -1
End If
If col < n Then
g(node, node) += 1
g(node, node +1) = -1
End If
Next
Next
Dim As UInteger ar = 2, ac = 2
Dim As UInteger br = 7, bc = 8
Dim As UInteger a = ac + n * (ar -1)
Dim As UInteger b = bc + n * (br -1)
g(a, nn +1) = -1
g(b, nn +1) = 1
Print : Print "Nodes a: "; a, " b: "; b
' solve linear system using Gauss-Seidel method with pivoting
Dim As UInteger i, j, k
Dim As Double y
Do
For j = 1 To nn
For i = j To nn
If g(i, j) <> 0 Then Exit For
Next
If i = nn +1 Then
Print : Print "No solution"
Exit Do
End If
For k = 1 To nn +1
Swap g(j, k), g(i, k)
Next
y = g(j, j)
For k = 1 To nn +1
g(j, k) = g(j, k) / y
Next
For i = 1 To nn
If i <> j Then
y = -g(i, j)
For k = 1 To nn +1
g(i, k) = g(i, k) + y * g(j, k)
Next
End If
Next
Next
Print
Print "Resistance ="; Abs(g(a, nn +1) - g(b, nn +1)); " Ohm"
Exit Do
Loop
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
Nodes a: 12 b: 68 Resistance = 1.60899124173073 Ohm
FutureBasic
// Rosetta Code Resistor Mesh task
//https://rosettacode.org/wiki/Resistor_mesh
// Translation of Yabasic to FutureBasic
window 1,@"Resistor Mesh Task"
#build ShowMoreWarnings NO
short N=10
short NN=100
double A(100,101)
short NODE=0
short ROW,COL
FOR ROW=1 TO N
FOR COL=1 TO N
NODE=NODE+1
IF ROW>1
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE-N)=-1
END IF
IF ROW<N
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE+N)=-1
END IF
IF COL>1
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE-1)=-1
END IF
IF COL<N
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE+1)=-1
END IF
NEXT
NEXT
short AR,AC,ANode,BNode,BR,BC
// These are columns and rows in the matrix to calculate the position of A and B
//A is row 2 column 2
//B is row 7 column 8
AR=2 : AC=2 : ANode=AC+N*(AR-1)
BR=7 : BC=8 : BNode=BC+N*(BR-1)
A(ANode,NN+1)=-1
A(BNode,NN+1)=1
// solve linear system
// using Gauss-Seidel method
// with pivoting
short R
R=NN
short J,I,K
double T
double Y
FOR J=1 TO R
FOR I=J TO R
IF A(I,J)<>0 then break
NEXT
IF I=R+1
PRINT "No solution!"
END
END IF
FOR K=1 TO R+1
T = A(J,K)
A(J,K) = A(I,K)
A(I,K) = T
NEXT
Y= (1 / A(J,J))
FOR K=1 TO R+1
A(J,K) = Y * A(J,K)
NEXT
FOR I=1 TO R
IF I<>J
Y=-A(I,J)
FOR K=1 TO R+1
A(I,K)=A(I,K)+ Y*A(J,K)
NEXT
END IF
NEXT
NEXT
double Resistance
Resistance = A(ANode,101)-A(BNode,101)
PRINT
PRINT "Resistance between Nodes" str$(ANode) + " and" + str$(BNode) + " is" + str$(abs(Resistance)) + " Ohms"
handleevents- Output:
Resistance between Nodes 12 and 68 is 1.6089912417 Ohms
Go
package main
import "fmt"
const (
S = 10
)
type node struct {
v float64
fixed int
}
func alloc2(w, h int) [][]node {
a := make([][]node, h)
for i := range a {
a[i] = make([]node, w)
}
return a
}
func set_boundary(m [][]node) {
m[1][1].fixed = 1
m[1][1].v = 1
m[6][7].fixed = -1
m[6][7].v = -1
}
func calc_diff(m [][]node, d [][]node, w, h int) float64 {
total := 0.0
for i := 0; i < h; i++ {
for j := 0; j < w; j++ {
v := 0.0
n := 0
if i != 0 {
v += m[i-1][j].v
n++
}
if j != 0 {
v += m[i][j-1].v
n++
}
if i+1 < h {
v += m[i+1][j].v
n++
}
if j+1 < w {
v += m[i][j+1].v
n++
}
v = m[i][j].v - v/float64(n)
d[i][j].v = v
if m[i][j].fixed == 0 {
total += v * v
}
}
}
return total
}
func iter(m [][]node, w, h int) float64 {
d := alloc2(w, h)
diff := 1.0e10
cur := []float64{0, 0, 0}
for diff > 1e-24 {
set_boundary(m)
diff = calc_diff(m, d, w, h)
for i := 0; i < h; i++ {
for j := 0; j < w; j++ {
m[i][j].v -= d[i][j].v
}
}
}
for i := 0; i < h; i++ {
for j := 0; j < w; j++ {
t := 0
if i != 0 {
t += 1
}
if j != 0 {
t += 1
}
if i < h-1 {
t += 1
}
if j < w-1 {
t += 1
}
cur[m[i][j].fixed+1] += d[i][j].v * float64(t)
}
}
return (cur[2] - cur[0]) / 2
}
func main() {
mesh := alloc2(S, S)
fmt.Printf("R = %g\n", 2/iter(mesh, S, S))
}
Haskell
All mutations are expressed as monoidal operations.
{-# LANGUAGE ParallelListComp #-}
import Numeric.LinearAlgebra (linearSolve, toDense, (!), flatten)
import Data.Monoid ((<>), Sum(..))
rMesh n (ar, ac) (br, bc)
| n < 2 = Nothing
| any (\x -> x < 1 || x > n) [ar, ac, br, bc] = Nothing
| otherwise = between a b <$> voltage
where
a = (ac - 1) + n*(ar - 1)
b = (bc - 1) + n*(br - 1)
between x y v = abs (v ! a - v ! b)
voltage = flatten <$> linearSolve matrixG current
matrixG = toDense $ concat [ element row col node
| row <- [1..n], col <- [1..n]
| node <- [0..] ]
element row col node =
let (Sum c, elements) =
(Sum 1, [((node, node-n), -1)]) `when` (row > 1) <>
(Sum 1, [((node, node+n), -1)]) `when` (row < n) <>
(Sum 1, [((node, node-1), -1)]) `when` (col > 1) <>
(Sum 1, [((node, node+1), -1)]) `when` (col < n)
in [((node, node), c)] <> elements
x `when` p = if p then x else mempty
current = toDense [ ((a, 0), -1) , ((b, 0), 1) , ((n^2-1, 0), 0) ]
- Output:
λ> rMesh 10 (2,2) (7,8) Just 1.6089912417307304
J
We represent the mesh as a Ybus matrix with B as the reference node and A as the first node and invert it to find the Z bus (which represents resistance). The first element of the first row of this Z matrix is the resistance between nodes A and B. (It has to be the first element because A was the first node. And we can "ignore" B because we made everything be relative to B.) Most of the work is defining Y which represents the Ybus.
nodes=: 10 10 #: i. 100
nodeA=: 1 1
nodeB=: 6 7
NB. verb to pair up coordinates along a specific offset
conn =: [: (#~ e.~/@|:~&0 2) ([ ,: +)"1
ref =: ~. nodeA,nodes-.nodeB NB. all nodes, with A first and B omitted
wiring=: /:~ ref i. ,/ nodes conn"2 1 (,-)=i.2 NB. connected pairs (indices into ref)
Yii=: (* =@i.@#) #/.~ {."1 wiring NB. diagonal of Y represents connections to B
Yij=: -1:`(<"1@[)`]}&(+/~ 0*i.1+#ref) wiring NB. off diagonal of Y represents wiring
Y=: _1 _1 }. Yii+Yij
Here, the result of nodes conn offset represents all pairs of nodes where we can connect the argument nodes to neighboring nodes at the specified offset, and wiring is a list of index pairs representing all connections made by all resistors (note that each connection is represented twice -- node e is connected to node f AND node f is connected to node e). Yii contains the values for the diagonal elements of the Y bus while Yij contains the values for the off diagonal elements of the Y bus.
So:
{.{. %. Y
1.60899
Or, if we want an exact answer (this is slow), we can assume our resistors are perfect:
{.{.%. x:Y
455859137025721r283319837425200
(here, the letter 'r' separates the numerator from the denominator)
To get a better feel for what the conn operation is doing, here is a small illustration:
3 3 #: i.9
0 0
0 1
0 2
1 0
1 1
1 2
2 0
2 1
2 2
(3 3 #: i.9) conn 0 1
0 0
0 1
0 1
0 2
1 0
1 1
1 1
1 2
2 0
2 1
2 1
2 2
In other words, each coordinate pair is matched up with the coordinate pair that you would get by adding the offset to the first of the pair. In actual use, we use this four times, with four offsets (two horizontal and two vertical) to get our complete mesh.
Java
import java.util.ArrayList;
import java.util.List;
public class ResistorMesh {
private static final int S = 10;
private static class Node {
double v;
int fixed;
Node(double v, int fixed) {
this.v = v;
this.fixed = fixed;
}
}
private static void setBoundary(List<List<Node>> m) {
m.get(1).get(1).v = 1.0;
m.get(1).get(1).fixed = 1;
m.get(6).get(7).v = -1.0;
m.get(6).get(7).fixed = -1;
}
private static double calcDiff(List<List<Node>> m, List<List<Node>> d, int w, int h) {
double total = 0.0;
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
double v = 0.0;
int n = 0;
if (i > 0) {
v += m.get(i - 1).get(j).v;
n++;
}
if (j > 0) {
v += m.get(i).get(j - 1).v;
n++;
}
if (i + 1 < h) {
v += m.get(i + 1).get(j).v;
n++;
}
if (j + 1 < w) {
v += m.get(i).get(j + 1).v;
n++;
}
v = m.get(i).get(j).v - v / n;
d.get(i).get(j).v = v;
if (m.get(i).get(j).fixed == 0) {
total += v * v;
}
}
}
return total;
}
private static double iter(List<List<Node>> m, int w, int h) {
List<List<Node>> d = new ArrayList<>(h);
for (int i = 0; i < h; ++i) {
List<Node> t = new ArrayList<>(w);
for (int j = 0; j < w; ++j) {
t.add(new Node(0.0, 0));
}
d.add(t);
}
double[] cur = new double[3];
double diff = 1e10;
while (diff > 1e-24) {
setBoundary(m);
diff = calcDiff(m, d, w, h);
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
m.get(i).get(j).v -= d.get(i).get(j).v;
}
}
}
for (int i = 0; i < h; ++i) {
for (int j = 0; j < w; ++j) {
int k = 0;
if (i != 0) k++;
if (j != 0) k++;
if (i < h - 1) k++;
if (j < w - 1) k++;
cur[m.get(i).get(j).fixed + 1] += d.get(i).get(j).v * k;
}
}
return (cur[2] - cur[0]) / 2.0;
}
public static void main(String[] args) {
List<List<Node>> mesh = new ArrayList<>(S);
for (int i = 0; i < S; ++i) {
List<Node> t = new ArrayList<>(S);
for (int j = 0; j < S; ++j) {
t.add(new Node(0.0, 0));
}
mesh.add(t);
}
double r = 2.0 / iter(mesh, S, S);
System.out.printf("R = %.15f", r);
}
}
- Output:
R = 1.608991241729889
JavaScript
Kirchhoff's circuit laws on the resistor mesh are represented as a linear equation system for the electric potential at each node of the grid. The linear equation is then solved using the conjugate gradient method:
// Vector addition, scalar multiplication & dot product:
const add = (u, v) => {let i = u.length; while(i--) u[i] += v[i]; return u;};
const sub = (u, v) => {let i = u.length; while(i--) u[i] -= v[i]; return u;};
const mul = (a, u) => {let i = u.length; while(i--) u[i] *= a; return u;};
const dot = (u, v) => {let s = 0, i = u.length; while(i--) s += u[i]*v[i]; return s;};
const W = 10, H = 10, A = 11, B = 67;
function getAdjacent(node){ // Adjacency lists for square grid
let list = [], x = node % W, y = Math.floor(node / W);
if (x > 0) list.push(node - 1);
if (y > 0) list.push(node - W);
if (x < W - 1) list.push(node + 1);
if (y < H - 1) list.push(node + W);
return list;
}
function linOp(u){ // LHS of the linear equation
let v = new Float64Array(W * H);
for(let i = 0; i < v.length; i++){
if ( i === A || i === B ) {
v[i] = u[i];
continue;
}
// For each node other then A, B calculate the net current flow:
for(let j of getAdjacent(i)){
v[i] += (j === A || j === B) ? u[i] : u[i] - u[j];
}
}
return v;
}
function getRHS(phiA = 1, phiB = 0){ // RHS of the linear equation
let b = new Float64Array(W * H);
// Setting boundary conditions (electric potential at A and B):
b[A] = phiA;
b[B] = phiB;
for(let j of getAdjacent(A)) b[j] = phiA;
for(let j of getAdjacent(B)) b[j] = phiB;
return b;
}
function init(phiA = 1, phiB = 0){ // initialize unknown vector
let u = new Float64Array(W * H);
u[A] = phiA;
u[B] = phiB;
return u;
}
function solveLinearSystem(err = 1e-20){ // conjugate gradient solver
let b = getRHS();
let u = init();
let r = sub(linOp(u), b);
let p = r;
let e = dot(r,r);
while(true){
let Ap = linOp(p);
let alpha = e / dot(p, Ap);
u = sub(u, mul(alpha, p.slice()));
r = sub(linOp(u), b);
let e_new = dot(r,r);
let beta = e_new / e;
if(e_new < err) return u;
e = e_new;
p = add(r, mul(beta, p));
}
}
function getResistance(u){
let curr = 0;
for(let j of getAdjacent(A)) curr += u[A] - u[j];
return 1 / curr;
}
let phi = solveLinearSystem();
let res = getResistance(phi);
console.log(`R = ${res} Ohm`);
- Output:
R = 1.608991241730955 Ohm
jq
Works with jq and gojq, that is, the C and Go implementations of jq.
Adapted from Wren
# Create a $rows * $columns matrix initialized with the input value
def matrix($rows; $columns):
. as $in
| [range(0;$columns)|$in] as $row
| [range(0;$rows)|$row];
def Node($v; $fixed):
{$v, $fixed};
# input: a suitable matrix of Nodes
def setBoundary:
.[1][1].v = 1
| .[1][1].fixed = 1
| .[6][7].v = -1
| .[6][7].fixed = -1 ;
# input: {d, m} where
# .d and .m are matrices (as produced by matrix(h; w)) of Nodes
# output: {d, m, diff} with d updated
def calcDiff($w; $h):
def adjust($cond; action): if $cond then action | .n += 1 else . end;
reduce range(0; $h) as $i (.diff = 0;
reduce range(0; $w) as $j (.;
.v = 0
| .n = 0
| adjust($i > 0; .v += .m[$i-1][$j].v)
| adjust($j > 0; .v += .m[$i][$j-1].v)
| adjust($i + 1 < $h; .v += .m[$i+1][$j].v)
| adjust($j + 1 < $w; .v += .m[$i][$j+1].v)
| .v = .m[$i][$j].v - .v/.n
| .d[$i][$j].v = .v
| if (.m[$i][$j].fixed == 0) then .diff += .v * .v else . end ) ) ;
# input: a mesh of width w and height h, i.e. a matrix as prodcued by matrix(h;w)
def iter:
length as $h
| (.[0]|length) as $w
| { m : .,
d : (Node(0;0) | matrix($h; $w)),
cur: [0,0,0],
diff: 1e10 }
| until (.diff <= 1e-24;
.m |= setBoundary
| calcDiff($w; $h)
| reduce range(0;$h) as $i (.;
reduce range(0;$w) as $j (.;
.m[$i][$j].v += (- .d[$i][$j].v) )) )
| reduce range(0; $h) as $i (.;
reduce range(0; $w) as $j (.;
.k = 0
| if ($i != 0) then .k += 1 else . end
| if ($j != 0) then .k += 1 else . end
| if ($i < $h - 1) then .k += 1 else . end
| if ($j < $w - 1) then .k += 1 else . end
| .cur[.m[$i][$j].fixed + 1] += .d[$i][$j].v * .k ))
| (.cur[2] - .cur[0]) / 2 ;
def task($S):
def mesh: Node(0; 0) | matrix($S; $S);
(2 / (mesh | iter)) as $r
| "R = \($r) ohms";
task(10)- Output:
R = 1.608991241729889 ohms
Julia
We construct the matrix A that relates voltage v on each node to the injected current b via Av=b, and then we simply solve the linear system to find the resulting voltages (from unit currents at the indicated nodes, i and j) and hence the resistance.
We can write A=DTD in terms of the incidence matrix D of the resistor graph (see e.g. Strang, Introduction to Linear Algebra, 4th ed., sec. 8.2).
Because the graph is a rectangular grid, we can in turn write the incidence matrix D in terms of Kronecker products ⊗ (kron in Julia) of "one-dimensional" D1 matrices (the incidence matrix of a 1d resistor network).
We use Julia's built-in sparse-matrix solvers (based on SuiteSparse) to solve the resulting sparse linear system efficiently
N = 10
D1 = speye(N-1,N) - spdiagm(ones(N-1),1,N-1,N)
D = [ kron(D1, speye(N)); kron(speye(N), D1) ]
i, j = N*1 + 2, N*7+7
b = zeros(N^2); b[i], b[j] = 1, -1
v = (D' * D) \ b
v[i] - v[j]
- Output:
1.6089912417307288
One advantage of this formulation is that it is easy to generalize to non-constant resistance.
If we have a vector y of admittance (1/resistance) values on each resistor, then one simply replaces D' * D with D' * spdiagm(y) * D.
Kotlin
// version 1.1.4-3
typealias List2D<T> = List<List<T>>
const val S = 10
class Node(var v: Double, var fixed: Int)
fun setBoundary(m: List2D<Node>) {
m[1][1].v = 1.0; m[1][1].fixed = 1
m[6][7].v = -1.0; m[6][7].fixed = -1
}
fun calcDiff(m: List2D<Node>, d: List2D<Node>, w: Int, h: Int): Double {
var total = 0.0
for (i in 0 until h) {
for (j in 0 until w) {
var v = 0.0
var n = 0
if (i > 0) { v += m[i - 1][j].v; n++ }
if (j > 0) { 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++ }
v = m[i][j].v - v / n
d[i][j].v = v
if (m[i][j].fixed == 0) total += v * v
}
}
return total
}
fun iter(m: List2D<Node>, w: Int, h: Int): Double {
val d = List(h) { List(w) { Node(0.0, 0) } }
val cur = DoubleArray(3)
var diff = 1e10
while (diff > 1e-24) {
setBoundary(m)
diff = calcDiff(m, d, w, h)
for (i in 0 until h) {
for (j in 0 until w) m[i][j].v -= d[i][j].v
}
}
for (i in 0 until h) {
for (j in 0 until w) {
var k = 0
if (i != 0) k++
if (j != 0) k++
if (i < h - 1) k++
if (j < w - 1) k++
cur[m[i][j].fixed + 1] += d[i][j].v * k
}
}
return (cur[2] - cur[0]) / 2.0
}
fun main(args: Array<String>) {
val mesh = List(S) { List(S) { Node(0.0, 0) } }
val r = 2.0 / iter(mesh, S, S)
println("R = $r")
}
- Output:
R = 1.608991241729889
Mathematica /Wolfram Language
Use KirchhoffMatrix and DrazinInverse to compute the effective resistance matrix of a graph:
ResistanceMatrix[g_Graph] := With[{n = VertexCount[g], km = KirchhoffMatrix[g]},
Table[ ReplacePart[ Diagonal[ DrazinInverse[ ReplacePart[km, k -> UnitVector[n, k]]]], k -> 0],
{k, n}]
]
rm = ResistanceMatrix[GridGraph[{10, 10}]];
N[rm[[12, 68]], 40]
- Output:
1.608991241730729655954495520510088761201
Use KirchhoffMatrix and PseudoInverse to compute the effective resistance matrix of a graph to the desired precision:
ResistanceMatrix[g_, prec_:$MachinePrecision]:= With[{m = PseudoInverse[N[KirchhoffMatrix[g], prec]]},
Outer[Plus, Diagonal[m], Diagonal[m]] - m - Transpose[m]
]
rm = ResistanceMatrix[GridGraph[{10, 10}], 40];
rm[[12, 68]]
- Output:
1.608991241730729655954495520510088761201
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 this 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
/* Some larger example */
grid_resistor(20, 20, 1, 1, 20, 20);
129548954101732562831760781545158173626645023 / 33283688571680493510612137844679320717594861
bfloat(%), fpprec = 40;
3.89226554090400912102670691601064387507b0
Modula-2
MODULE ResistorMesh;
FROM RConversions IMPORT RealToStringFixed;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
CONST S = 10;
TYPE Node = RECORD
v : LONGREAL;
fixed : INTEGER;
END;
PROCEDURE SetBoundary(VAR m : ARRAY OF ARRAY OF Node);
BEGIN
m[1][1].v := 1.0;
m[1][1].fixed := 1;
m[6][7].v := -1.0;
m[6][7].fixed := -1;
END SetBoundary;
PROCEDURE CalcDiff(VAR m,d : ARRAY OF ARRAY OF Node) : LONGREAL;
VAR
total,v : LONGREAL;
i,j,n : INTEGER;
BEGIN
total := 0.0;
FOR i:=0 TO S DO
FOR j:=0 TO S DO
v := 0.0;
n := 0;
IF i>0 THEN
v := v + m[i-1][j].v;
INC(n);
END;
IF j>0 THEN
v := v + m[i][j-1].v;
INC(n);
END;
IF i+1<S THEN
v := v + m[i+1][j].v;
INC(n);
END;
IF j+1<S THEN
v := v + m[i][j+1].v;
INC(n);
END;
v := m[i][j].v - v / LFLOAT(n);
d[i][j].v := v;
IF m[i][j].fixed=0 THEN
total := total + v*v;
END;
END;
END;
RETURN total;
END CalcDiff;
PROCEDURE Iter(m : ARRAY OF ARRAY OF Node) : LONGREAL;
VAR
d : ARRAY[0..S] OF ARRAY[0..S] OF Node;
i,j,k : INTEGER;
cur : ARRAY[0..2] OF LONGREAL;
diff : LONGREAL;
BEGIN
FOR i:=0 TO S DO
FOR j:=0 TO S DO
d[i][j] := Node{0.0,0};
END;
END;
diff := 1.0E10;
WHILE diff>1.0E-24 DO
SetBoundary(m);
diff := CalcDiff(m,d);
FOR i:=0 TO S DO
FOR j:=0 TO S DO
m[i][j].v := m[i][j].v - d[i][j].v;
END;
END;
END;
FOR i:=0 TO S DO
FOR j:=0 TO S DO
k:=0;
IF i#0 THEN INC(k) END;
IF j#0 THEN INC(k) END;
IF i<S-1 THEN INC(k) END;
IF j<S-1 THEN INC(k) END;
cur[m[i][j].fixed+1] := cur[m[i][j].fixed+1] + d[i][j].v*LFLOAT(k);
END;
END;
RETURN (cur[2]-cur[0]) / 2.0;
END Iter;
VAR
mesh : ARRAY[0..S] OF ARRAY[0..S] OF Node;
buf : ARRAY[0..32] OF CHAR;
r : LONGREAL;
pos : CARDINAL;
ok : BOOLEAN;
BEGIN
pos := 0;
r := 2.0 / Iter(mesh);
WriteString("R = ");
RealToStringFixed(r, 15,0, buf, pos, ok);
WriteString(buf);
WriteString(" ohms");
WriteLn;
ReadChar;
END ResistorMesh.
Nim
const S = 10
type
NodeKind = enum nodeFree, nodeA, nodeB
Node = object
v: float
fixed: NodeKind
Mesh[H, W: static int] = array[H, array[W, Node]]
func setBoundary(m: var Mesh) =
m[1][1].v = 1.0
m[1][1].fixed = nodeA
m[6][7].v = -1.0
m[6][7].fixed = nodeB
func calcDiff[H, W: static int](m,: Mesh[H, W]; d: var Mesh[H, W]): float =
for i in 0..<H:
for j in 0..<W:
var v = 0.0
var n = 0
if i > 0:
v += m[i - 1][j].v
inc n
if j > 0:
v += m[i][j - 1].v
inc n
if i + 1 < m.H:
v += m[i + 1][j].v
inc n
if j + 1 < m.W:
v += m[i][j + 1].v
inc n
v = m[i][j].v - v / n.toFloat
d[i][j].v = v
if m[i][j].fixed == nodeFree:
result += v * v
func iter[H, W: static int](m: var Mesh[H, W]): float =
var
d: Mesh[H, W]
cur: array[NodeKind, float]
diff = 1e10
while diff > 1e-24:
m.setBoundary()
diff = calcDiff(m, d)
for i in 0..<H:
for j in 0..<W:
m[i][j].v -= d[i][j].v
for i in 0..<H:
for j in 0..<W:
var k = 0
if i != 0: inc k
if j != 0: inc k
if i < m.H - 1: inc k
if j < m.W - 1: inc k
cur[m[i][j].fixed] += d[i][j].v * k.toFloat
result = (cur[nodeA] - cur[nodeB]) / 2
when isMainModule:
var mesh: Mesh[S, S]
let r = 2 / mesh.iter()
echo "R = ", r
- Output:
R = 1.608991241729889
Octave
We'll solve the linear system. We'll write Kirchhoff's circuit laws at each node and search for a voltage distribution that creates a 1A current coming from A exiting in B. The difference of voltage between B and A is then the resistance.
N = 10;
NN = N*N;
G = sparse(NN, NN);
node = 0;
for row=1:N;
for col=1:N;
node++;
if row > 1
G(node, node)++;
G(node, node - N) = -1;
end
if row < N;
G(node, node)++;
G(node, node + N) = -1;
end
if col > 1
G(node, node)++;
G(node, node - 1) = -1;
end
if col < N;
G(node, node)++;
G(node, node + 1) = -1;
end
end
end
current = sparse(NN, 1);
Ar = 2; Ac = 2; A = Ac + N*( Ar - 1 );
Br = 7; Bc = 8; B = Bc + N*( Br - 1 );
current( A ) = -1;
current( B ) = +1;
voltage = G \ current;
VA = voltage( A );
VB = voltage( B );
full( abs( VA - VB ) )
- Output:
ans = 1.6090
Perl
use strict;
use warnings;
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-15) {
set_boundary();
$diff = calc_diff();
#print "error^2: $diff\n"; # un-comment to see slow convergence
for my $i (0 .. $h) {
for my $j (0 .. $w) {
$v[$i][$j] -= $d[$i][$j];
}
}
}
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;
}
printf "R = %.6f\n", 2 / iter();
- Output:
R = 1.608991
Phix
uses inverse() from Gauss-Jordan_matrix_inversion#Phix and matrix_mul() from Matrix_multiplication#Phix
function resistormesh(integer ni, nj, ai, aj, bi, bj) integer n = ni*nj, k, c sequence A = repeat(repeat(0,n),n), B = repeat({0},n) for i=1 to ni do for j=1 to nj do k = (i-1)*nj + j--1 if i=ai and j=aj then A[k,k] = 1 else c = 0 if i<ni then c += 1; A[k,k+nj] = -1 end if if i>1 then c += 1; A[k,k-nj] = -1 end if if j<nj then c += 1; A[k,k+1] = -1 end if if j>1 then c += 1; A[k,k-1] = -1 end if A[k,k] = c end if end for end for k = (bi-1)*nj +bj B[k,1] = 1 A = inverse(A) B = matrix_mul(A,B) return B[k,1] end function printf(1,"Resistance = %.13f ohms\n",resistormesh(10, 10, 2, 2, 8, 7))
- Output:
Resistance = 1.6089912417307 ohms
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()
- Output:
R = 1.6089912417307286
import sys, copy
from fractions import Fraction
def gauss(a, b):
n, p = len(a), len(a[0])
for i in range(n):
t = abs(a[i][i])
k = i
for j in range(i + 1, n):
if abs(a[j][i]) > t:
t = abs(a[j][i])
k = j
if k != i:
for j in range(i, n):
a[i][j], a[k][j] = a[k][j], a[i][j]
b[i], b[k] = b[k], b[i]
t = 1/a[i][i]
for j in range(i + 1, n):
a[i][j] *= t
b[i] *= t
for j in range(i + 1, n):
t = a[j][i]
for k in range(i + 1, n):
a[j][k] -= t*a[i][k]
b[j] -= t * b[i]
for i in range(n - 1, -1, -1):
for j in range(i):
b[j] -= a[j][i]*b[i]
return b
def resistor_grid(p, q, ai, aj, bi, bj):
n = p*q
I = Fraction(1, 1)
v = [0*I]*n
a = [copy.copy(v) for i in range(n)]
for i in range(p):
for j in range(q):
k = i*q + j
if i == ai and j == aj:
a[k][k] = I
else:
c = 0
if i + 1 < p:
c += 1
a[k][k + q] = -1
if i >= 1:
c += 1
a[k][k - q] = -1
if j + 1 < q:
c += 1
a[k][k + 1] = -1
if j >= 1:
c += 1
a[k][k - 1] = -1
a[k][k] = c*I
b = [0*I]*n
k = bi*q + bj
b[k] = 1
return gauss(a, b)[k]
def main(arg):
r = resistor_grid(int(arg[0]), int(arg[1]), int(arg[2]), int(arg[3]), int(arg[4]), int(arg[5]))
print(r)
print(float(r))
main(sys.argv[1:])
# Output:
# python grid.py 10 10 1 1 7 6
# 455859137025721/283319837425200
# 1.6089912417307297
Racket
This version avoids mutation... possibly a little more costly than C, but more functional.
#lang racket
(require racket/flonum)
(define-syntax-rule (fi c t f) (if c f t))
(define (neighbours w h)
(define h-1 (sub1 h))
(define w-1 (sub1 w))
(lambda (i j)
(+ (fi (zero? i) 1 0)
(fi (zero? j) 1 0)
(if (< i h-1) 1 0)
(if (< j w-1) 1 0))))
(define (mesh-R probes w h)
(define h-1 (sub1 h))
(define w-1 (sub1 w))
(define-syntax-rule (v2ref v r c) ; 2D vector ref
(flvector-ref v (+ (* r w) c)))
(define w*h (* w h))
(define (alloc2 (v 0.))
(make-flvector w*h v))
(define nghbrs (neighbours w h))
(match-define `((,fix+r ,fix+c) (,fix-r ,fix-c)) probes)
(define fix+idx (+ fix+c (* fix+r w)))
(define fix-idx (+ fix-c (* fix-r w)))
(define fix-val
(match-lambda**
[((== fix+idx) _) 1.]
[((== fix-idx) _) -1.]
[(_ v) v]))
(define (calc-diff m)
(define d
(for*/flvector #:length w*h ((i (in-range h)) (j (in-range w)))
(define v
(+ (fi (zero? i) (v2ref m (- i 1) j) 0)
(fi (zero? j) (v2ref m i (- j 1)) 0)
(if (< i h-1) (v2ref m (+ i 1) j) 0)
(if (< j w-1) (v2ref m i (+ j 1)) 0)))
(- (v2ref m i j) (/ v (nghbrs i j)))))
(define Δ
(for/sum ((i (in-naturals)) (d.v (in-flvector d)) #:when (= (fix-val i 0.) 0.))
(sqr d.v)))
(values d Δ))
(define final-d
(let loop ((m (alloc2)) (d (alloc2)))
(define m+ ; do this first will get the boundaries on
(for/flvector #:length w*h ((j (in-naturals)) (m.v (in-flvector m)) (d.v (in-flvector d)))
(fix-val j (- m.v d.v))))
(define-values (d- Δ) (calc-diff m+))
(if (< Δ 1e-24) d (loop m+ d-))))
(/ 2
(/ (- (* (v2ref final-d fix+r fix+c) (nghbrs fix+r fix+c))
(* (v2ref final-d fix-r fix-c) (nghbrs fix-r fix-c)))
2)))
(module+ main
(printf "R = ~a~%" (mesh-R '((1 1) (6 7)) 10 10)))
- Output:
R = 1.6089912417301238
Raku
(formerly Perl 6)
my $*TOLERANCE = 1e-12;
sub set-boundary(@mesh,@p1,@p2) {
@mesh[ @p1[0] ; @p1[1] ] = 1;
@mesh[ @p2[0] ; @p2[1] ] = -1;
}
sub solve(@p1, @p2, Int \w, Int \h) {
my @d = [0 xx w] xx h;
my @V = [0 xx w] xx h;
my @fixed = [0 xx w] xx h;
set-boundary(@fixed,@p1,@p2);
loop {
set-boundary(@V,@p1,@p2);
my $diff = 0;
for (flat ^h X ^w) -> \i, \j {
my @neighbors = (@V[i-1;j], @V[i;j-1], @V[i+1;j], @V[i;j+1]).grep: *.defined;
@d[i;j] = my \v = @V[i;j] - @neighbors.sum / @neighbors;
$diff += v × v unless @fixed[i;j];
}
last if $diff =~= 0;
for (flat ^h X ^w) -> \i, \j {
@V[i;j] -= @d[i;j];
}
}
my @current;
for (flat ^h X ^w) -> \i, \j {
@current[ @fixed[i;j]+1 ] += @d[i;j] × (?i + ?j + (i < h-1) + (j < w-1) );
}
(@current[2] - @current[0]) / 2
}
say 2 / solve (1,1), (6,7), 10, 10;
- Output:
1.60899124172989
REXX
This version allows specification of the grid size, the locations of the A and B points, and the number of decimal digits precision.
Dropping the decimal digits precision (numeric digits) to 10 makes the execution 3 times faster.
/*REXX program calculates the resistance between any two points on a resistor grid.*/
if 2=='f2'x then ohms = "ohms" /*EBCDIC machine? Then use 'ohms'. */
else ohms = "Ω" /* ASCII " " " Greek Ω.*/
parse arg high wide Arow Acol Brow Bcol digs . /*obtain optional arguments from the CL*/
if high=='' | high=="," then high= 10 /*Not specified? Then use the default.*/
if wide=='' | wide=="," then wide= 10 /* " " " " " " */
if Arow=='' | Arow=="," then Arow= 2 /* " " " " " " */
if Acol=='' | Acol=="," then Acol= 2 /* " " " " " " */
if Brow=='' | Brow=="," then Brow= 7 /* " " " " " " */
if Bcol=='' | Bcol=="," then Bcol= 8 /* " " " " " " */
if digs=='' | digs=="," then digs= 20 /* " " " " " " */
numeric digits digs /*use moderate decimal digs (precision)*/
minVal = 1'e-' || (digs*2) /*calculate the threshold minimal value*/
say ' minimum value is ' format(minVal,,,,0) " using " digs ' decimal digits'; say
say ' resistor mesh size is: ' wide "wide, " high 'high' ; say
say ' point A is at (row,col): ' Arow"," Acol
say ' point B is at (row,col): ' Brow"," Bcol
@.=0; cell.= 1
do until $<=minVal; v= 0
@.Arow.Acol= 1 ; cell.Arow.Acol= 0
@.Brow.Bcol= '-1' ; cell.Brow.Bcol= 0
$=0
do i=1 for high; im= i-1; ip= i+1
do j=1 for wide; n= 0; v= 0
if i\==1 then do; v= v + @.im.j; n= n+1; end
if j\==1 then do; jm= j-1; 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; jp= j+1; 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 /*c*/
end /*r*/
end /*until*/
say
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 all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sides: parse arg row,col; z=0; if row\==1 & row\==high then z= z+2; else z= z+1
if col\==1 & col\==wide then z= z+2; else z= z+1
return z
- output when using the default inputs:
minimum value is 1E-40 using 20 decimal digits
resistor mesh size is: 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
Sidef
var (w, h) = (10, 10)
var v = h.of { w.of(0) } # voltage
var f = h.of { w.of(0) } # fixed condition
var d = h.of { w.of(0) } # diff
var n = [] # neighbors
for i in ^h {
for j in (1 ..^ w ) { n[i][j] := [] << [i, j-1] }
for j in (0 ..^ w-1) { n[i][j] := [] << [i, j+1] }
}
for j in ^w {
for i in (1 ..^ h ) { n[i][j] := [] << [i-1, j] }
for i in (0 ..^ h-1) { n[i][j] := [] << [i+1, j] }
}
func set_boundary {
f[1][1] = 1; f[6][7] = -1;
v[1][1] = 1; v[6][7] = -1;
}
func calc_diff {
var total_diff = 0
for i,j in (^h ~X ^w) {
var w = n[i][j].map { |a| v.dig(a...) }.sum
d[i][j] = (w = (v[i][j] - w/n[i][j].len))
f[i][j] || (total_diff += w*w)
}
total_diff
}
func iter {
var diff = 1
while (diff > 1e-24) {
set_boundary()
diff = calc_diff()
for i,j in (^h ~X ^w) {
v[i][j] -= d[i][j]
}
}
var current = 3.of(0)
for i,j in (^h ~X ^w) {
current[ f[i][j] ] += (d[i][j] * n[i][j].len)
}
(current[1] - current[-1]) / 2
}
say "R = #{2 / iter()}"
- Output:
R = 1.60899124172988902191367295307304
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 {{epsilon 1e-24}} {
set voltageDifference [expr {
[lindex $forcePoints 2] - [lindex $forcePoints 5]}]
expr {$voltageDifference / [my FindCurrentFixpoint $epsilon]}
}
}
Setting up and solving this particular problem:
ResistorMesh create mesh {
size {10 10}
fixed {1 1 1.0}
fixed {6 7 -1.0}
}
puts [format "R = %.12g" [mesh solveForResistance]]
- Output:
R = 1.60899124173
Wren
class Node {
construct new(v, fixed) {
_v = v
_fixed = fixed
}
v { _v }
v=(value) { _v = value }
fixed { _fixed }
fixed=(value) { _fixed = value }
}
var setBoundary = Fn.new { |m|
m[1][1].v = 1
m[1][1].fixed = 1
m[6][7].v = -1
m[6][7].fixed = -1
}
var calcDiff = Fn.new { |m, d, w, h|
var total = 0
for (i in 0...h) {
for (j in 0...w) {
var v = 0
var n = 0
if (i > 0) {
v = v + m[i-1][j].v
n = n + 1
}
if (j > 0) {
v = v + m[i][j-1].v
n = n + 1
}
if (i + 1 < h) {
v = v + m[i+1][j].v
n = n + 1
}
if (j + 1 < w) {
v = v + m[i][j+1].v
n = n + 1
}
v = m[i][j].v - v/n
d[i][j].v = v
if (m[i][j].fixed == 0) total = total + v*v
}
}
return total
}
var iter = Fn.new { |m, w, h|
var d = List.filled(h, null)
for (i in 0...h) {
d[i] = List.filled(w, null)
for (j in 0...w) d[i][j] = Node.new(0, 0)
}
var cur = [0] * 3
var diff = 1e10
while (diff > 1e-24) {
setBoundary.call(m)
diff = calcDiff.call(m, d, w, h)
for (i in 0...h) {
for (j in 0...w) m[i][j].v = m[i][j].v - d[i][j].v
}
}
for (i in 0...h) {
for (j in 0...w) {
var k = 0
if (i != 0) k = k + 1
if (j != 0) k = k + 1
if (i < h - 1) k = k + 1
if (j < w - 1) k = k + 1
cur[m[i][j].fixed + 1] = cur[m[i][j].fixed + 1] + d[i][j].v * k
}
}
return (cur[2] - cur[0]) / 2
}
var S = 10
var mesh = List.filled(S, null)
for (i in 0...S) {
mesh[i] = List.filled(S, null)
for (j in 0...S) mesh[i][j] = Node.new(0, 0)
}
var r = 2 / iter.call(mesh, S, S)
System.print("R = %(r)")
- Output:
R = 1.6089912417299
XPL0
code real RlRes=46, RlOut=48;
def S = 10;
proc SetBoundary(MV, MF);
real MV; int MF;
[MF(1,1):= 1; MV(1,1):= 1.0;
MF(6,7):= -1; MV(6,7):= -1.0;
];
func real CalcDiff(MV, MF, DV, W, H);
real MV; int MF; real DV; int W, H;
int I, J, N; real V, Total;
[Total:= 0.0;
for I:= 0 to H-1 do
for J:= 0 to W-1 do
[V:= 0.0; N:= 0;
if I then [V:= V + MV(I-1,J); N:= N+1];
if J then [V:= V + MV(I,J-1); N:= N+1];
if I+1 < H then [V:= V + MV(I+1,J); N:= N+1];
if J+1 < W then [V:= V + MV(I,J+1); N:= N+1];
V:= MV(I,J) - V/float(N); DV(I,J):= V;
if MF(I,J) = 0 then Total:= Total + V*V;
];
return Total;
];
func real Iter(MV, MF, W, H);
real MV; int MF, W, H;
real DV, Diff, Cur; int I, J;
[DV:= RlRes(W); for I:= 0 to W-1 do DV(I):= RlRes(H);
Diff:= 1E10;
Cur:= [0.0, 0.0, 0.0];
while Diff > 1E-24 do
[SetBoundary(MV, MF);
Diff:= CalcDiff(MV, MF, DV, W, H);
for I:= 0 to H-1 do
for J:= 0 to W-1 do
MV(I,J):= MV(I,J) - DV(I,J);
];
for I:= 0 to H-1 do
for J:= 0 to W-1 do
Cur(MF(I,J)+1):= Cur(MF(I,J)+1) +
DV(I,J) * float(-(I>0) - (J>0) - (I<H-1) - (J<W-1));
\middle=4; side=3; corner=2
return (Cur(2)-Cur(0))/2.0;
];
real MeshV(S,S); int MeshF(S,S);
RlOut(0, 2.0 / Iter(MeshV, MeshF, S, S))- Output:
1.60899
Yabasic
N=10
NN=N*N
DIM A(NN,NN+1)
NODE=0
FOR ROW=1 TO N
FOR COL=1 TO N
NODE=NODE+1
IF ROW>1 THEN
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE-N)=-1
END IF
IF ROW<N THEN
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE+N)=-1
END IF
IF COL>1 THEN
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE-1)=-1
END IF
IF COL<N THEN
A(NODE,NODE)=A(NODE,NODE)+1
A(NODE,NODE+1)=-1
END IF
NEXT
NEXT
AR=2 : AC=2 : A=AC+N*(AR-1)
BR=7 : BC=8 : B=BC+N*(BR-1)
A(A,NN+1)=-1
A(B,NN+1)=1
PRINT "Nodes ",A,B
// solve linear system
// using Gauss-Seidel method
// with pivoting
R=NN
FOR J=1 TO R
FOR I=J TO R
IF A(I,J)<>0 BREAK
NEXT
IF I=R+1 THEN
PRINT "No solution!"
END
END IF
FOR K=1 TO R+1
T = A(J,K)
A(J,K) = A(I,K)
A(I,K) = T
NEXT
Y=1/A(J,J)
FOR K=1 TO R+1
A(J,K)=Y*A(J,K)
NEXT
FOR I=1 TO R
IF I<>J THEN
Y=-A(I,J)
FOR K=1 TO R+1
A(I,K)=A(I,K)+Y*A(J,K)
NEXT
END IF
NEXT
NEXT
PRINT "Resistence = "; : PRINT ABS(A(A,NN+1)-A(B,NN+1)) USING "%1.13f"zkl
Uses the GNU Scientific Library.
var [const] GSL=Import("zklGSL"); // libGSL (GNU Scientific Library)
fcn onGrid(i,j,p,q){ ((0<=i<p) and (0<=j<q)) }
fcn gridResistor(p,q, ai,aj, bi,bj){
n,A := p*q, GSL.Matrix(n,n); // zero filled
foreach i,j in (p,q){
k:=i*q + j;
if(i==ai and j==aj) A[k,k]=1;
else{
c:=0;
if(onGrid(i+1,j, p,q)){ c+=1; A[k, k+q]=-1 }
if(onGrid(i-1,j, p,q)){ c+=1; A[k, k-q]=-1 }
if(onGrid(i, j+1, p,q)){ c+=1; A[k, k+1]=-1 }
if(onGrid(i, j-1, p,q)){ c+=1; A[k, k-1]=-1 }
A[k,k]=c;
}
}
b:=GSL.Vector(n); // zero filled
b[k:=bi*q + bj]=1;
A.AxEQb(b)[k];
}gridResistor(10,10, 1,1, 7,6).println();- Output:
1.60899