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Dinesman's multiple-dwelling problem: Difference between revisions

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Dinesman's multiple-dwelling problem (view source)

Revision as of 05:08, 2 March 2021

8 bytes removed ,  3 years ago
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m (→‎{{header|Perl}}: revised where 'warnings' pragma used)
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names= 'Baker Cooper Fletcher Miller Smith' /*names of multiple─dwelling tenants. */
#tenants= words(names) /*the number of tenants in the building*/
floors= 5; top= floors; bottom= 1 /*floor 1 is the ground (bottom) floor.*/
sols#= 0 /*the number of solutions found so far.*/
do @.1=1 for floors /*iterate through all floors for rules.*/
do @.2=1 for floors /* " " " " " " */
do @.3=1 for floors /* " " " " " " */
do @.4=1 for floors /* " " " " " " */
do @.5=1 for floors /* " " " " " " */
call set
do j=1 for floors-1; a= @.j /* [↓] people don't live on same floor*/
do k=j+1 to floors /*see if any people live on same floor.*/
if a==@.k then iterate @.5 /*Is anyone cohabiting? Then not valid*/
end /*k*/
end /*j*/
call Waldo /* ◄══ where the rubber meets the road.*/
end /*@.5*/
end /*@.4*/
end /*@.3*/
end /*@.2*/
end /*@.1*/
 
say 'found ' sols # " solution"s(sols#). /*display the number of solutions found*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
set: do p=1 for #tenants; call value word(names, p), @.p; end; return
s: if arg(1)=1 then return ''; return "s" /*a simple pluralizer function.*/
th: arg x; x=abs(x); return word('th st nd rd', 1 +x// 10* (x//100%10\==1)*(x//10<4))
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if Smith == Fletcher - 1 | Smith == Fletcher + 1 then return
if Fletcher == Cooper - 1 | Fletcher == Cooper + 1 then return
sols#= sols# + 1 /* [↑] "|" is REXX's "or" comparator.*/
say; do p=1 for #tenants; tenant= word(names, p)
say right(tenant, 35) 'lives on the' @.p || th(@.p) "floor."
end /*p*/ /* [↑] "||" is REXX's concatenation. */
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