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Perfect numbers: Difference between revisions

m
→‎traditional method: added/changed comments and whitespace.
m (→‎Lucas-Lehmer + other optimizations: add/changed comments and whitespace.)
m (→‎traditional method: added/changed comments and whitespace.)
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:::* a ''corresponding factor'' is used when a factor is found
:::* testing is stopped if the sum of the factors exceeds '''X'''
<lang rexx>/*REXX program tests if a number (or a range of numbers) is/are perfect. */
parse arg low high . /*obtain the specified number(s). */
if high=='' & low=='' then high=34000000 /*if no argsarguments, then use a range. */
if low=='' then low=1 /*if no LOW, then assume unity. */
if high=='' then high=low /*if no HIGH, then assume LOW. */
w=length(high) /*use W for formatting the output. */
numeric digits max(9,w+2) /*ensure enough digits to handle# number*/
 
do i=low to high /*process the single #number or a range. */
if isPerfect(i) then say right(i,w) 'is a perfect number.'
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
/*──────────────────────────────────ISPERFECT subroutine────────────────*/
isPerfect: procedure; parse arg x 1 y /*getobtain the number to be tested. */
if x<==6 then return 01 /*perfecthandle numbersthe can'tspecial becase of < six. */
s=1 /*the first factor of X. _ /*[↓] perfect number's digitalRoot = 1*/
do until y<10 do j=2 while j*j<=x /*startingfind the digital root of Y. at 2, find factors ≤√X*/
parse var y r 2; ifdo x//j\=k=02 thenfor iteratelength(y)-1; /*J isn't ar=r+substr(y,k,1); factor of X,end so skip./*k*/
y=r s = s + j + x%j /*···find adddigital itroot andof the otherdigit root. factor*/
end /*until*/ if s>x then return 0 /*Sumwash, rinse, repeat ··· too big? It ain't perfect.*/
 
end /*j*/ /*(above) is marginally faster. */
returnif sr\==x1 then return 0 /*ifDigital theroot sum¬ matches1? X, Then ¬ perfect!. */</lang>
s=1 /*the first factor of X. ___*/
do j=2 while j*j<=x /*starting at 2, find the factors ≤√ X */
if x//j\==0 then iterate /*J isn't a factor of X, so skip it. */
s = s + j + x%j /*··· add it and the other factor. */
if s>x then return 0 /*Is the sum too big? It ain't perfect*/
end /*j*/ /*(above) is marginally faster. */
return s==x /*if the sum matches X, it's perfect! *//</lang>
'''output''' &nbsp; when using the default inputs:
6 is a perfect number.
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