How to resolve the algorithm Solve the no connection puzzle step by step in the REXX programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Solve the no connection puzzle step by step in the REXX programming language
Table of Contents
Problem Statement
You are given a box with eight holes labelled A-to-H, connected by fifteen straight lines in the pattern as shown below: You are also given eight pegs numbered 1-to-8.
Place the eight pegs in the holes so that the (absolute) difference between any two numbers connected by any line is greater than one.
In this attempt: Note that 7 and 6 are connected and have a difference of 1, so it is not a solution.
Produce and show here one solution to the puzzle.
No Connection Puzzle (youtube).
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Solve the no connection puzzle step by step in the REXX programming language
Source code in the rexx programming language
/*REXX program solves the "no─connection" puzzle (the puzzle has eight pegs). */
parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */
if limit=='' | limit=="," then limit= 1 /* ║ A B ║ */
/* ║ /│\ /│\ ║ */
@. = /* ║ / │ \/ │ \ ║ */
@.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */
@.2 = 'B D E F' /* ║ / │/ \│ \ ║ */
@.3 = 'C A D G' /* ║ C────D────E────F ║ */
@.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */
@.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */
@.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */
@.7 = 'G C D E' /* ║ \│/ \│/ ║ */
@.8 = 'H D E F' /* ║ G H ║ */
cnt= 0 /* ╚═══════════════════════════╝ */
do pegs=1 while @.pegs\==''; _= word(@.pegs, 1)
subs= 0
do #=1 for words(@.pegs) -1 /*create list of node paths.*/
__= word(@.pegs, # + 1); if __>_ then iterate
subs= subs + 1; !._.subs= __
end /*#*/
!._.0= subs /*assign the number of the node paths. */
end /*pegs*/
pegs= pegs - 1 /*the number of pegs to be seated. */
_= ' ' /*_ is used for indenting the output.*/
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
say _ 'a='a _ "b="||b _ 'c='c _ "d="d _ 'e='e _ "f="f _ 'g='g _ "h="h
cnt= cnt + 1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/
s= left('s', cnt\==1) /*handle the case of plurals (or not).*/
say 'found ' cnt " solution"s'.' /*display the number of solutions found*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
?: parse arg node; nn= value(node)
nH= nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there any are duplicates.*/
nL= nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$= !.node.ch; fn= value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK.*/
/*REXX program solves the "no─connection" puzzle (the puzzle has eight pegs). */
@abc= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */
if limit=='' | limit=="," then limit= 1 /* ║ A B ║ */
oLimit= limit; limit= abs(limit) /* ║ /│\ /│\ ║ */
@. = /* ║ / │ \/ │ \ ║ */
@.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */
@.2 = 'B D E F' /* ║ / │/ \│ \ ║ */
@.3 = 'C A D G' /* ║ C────D────E────F ║ */
@.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */
@.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */
@.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */
@.7 = 'G C D E' /* ║ \│/ \│/ ║ */
@.8 = 'H D E F' /* ║ G H ║ */
cnt= 0 /* ╚═══════════════════════════╝ */
do pegs=1 while @.pegs\==''; _= word(@.pegs, 1)
subs= 0
do #=1 for words(@.pegs) -1 /*create list of node paths.*/
__= word(@.pegs, #+1); if __>_ then iterate
subs= subs + 1; !._.subs= __
end /*#*/
!._.0= subs /*assign the number of the node paths. */
end /*pegs*/
pegs= pegs - 1 /*the number of pegs to be seated. */
_= ' ' /*_ is used for indenting the output. */
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
call showNodes
cnt= cnt + 1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/
s= left('s', cnt\==1) /*handle the case of plurals (or not).*/
say 'found ' cnt " solution"s'.' /*display the number of solutions found*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
?: parse arg node; nn= value(node)
nH= nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there're any duplicates.*/
nL= nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$= !.node.ch; fn= value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showNodes: _= left('', 5) /*_ is used for padding the output. */
show= 0 /*indicates no graph has been found yet*/
do box=1 for sourceline() while oLimit<0 /*Negative? Then display the diagram. */
xw= sourceline(box) /*get a source line of this program. */
p2= lastpos('*', xw) /*the position of last asterisk.*/
p1= lastpos('*', xw, max(1, p2-1) ) /* " " " penultimate " */
if pos('╔', xw)\==0 then show= 1 /*Have found the top-left box corner ? */
if \show then iterate /*Not found? Then skip this line. */
xb= substr(xw, p1+1, p2-p1-2) /*extract the "box" part of line. */
xt= xb /*get a working copy of the box. */
do jx=1 for pegs /*do a substitution for all the pegs. */
@= substr(@abc, jx, 1) /*get the name of the peg (A ──► Z). */
xt= translate(xt, value(@), @) /*substitute the peg name with a value.*/
end /*jx*/ /* [↑] graph is limited to 26 nodes.*/
say _ xb _ _ xt /*display one line of the graph. */
if pos('╝', xw)\==0 then return /*Is this last line of graph? Then stop*/
end /*box*/
say _ 'a='a _ "b="||b _ 'c='c _ "d="d _ ' e='e _ "f="f _ 'g='g _ "h="h
return
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