How to resolve the algorithm Resistor mesh step by step in the jq programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Resistor mesh step by step in the jq programming language
Table of Contents
Problem Statement
Given 10×10 grid nodes (as shown in the image) interconnected by 1Ω resistors as shown, find the resistance between points A and B.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Resistor mesh step by step in the jq programming language
Source code in the jq programming language
# 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)
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