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Perfect numbers: Difference between revisions

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Perfect numbers (view source)

Revision as of 06:27, 6 April 2016

549 bytes removed ,  8 years ago
→‎traditional method: replaced with the appropriate REXX program.
m (→‎optimized using digital root: add/changed comments and whitespace.)
(→‎traditional method: replaced with the appropriate REXX program.)
Line 1,718:
:::* 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). optional arguments from the CL*/
if high=='' & low=='' then high=34000000 /*if no arguments, 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: procedure; parse arg x 1 y /*obtain the number to be tested. */
if x==6 then return 1 if x<6 then return 0 /*handleperfect thenumbers specialcan't casebe of < six. */
s=1 /*the first factor /*[↓]of perfectX. number's digitalRoot = 1 ___*/
do until y<10 do j=2 while j*j<=x /*starting at 2, /*find the digital root of Y. factors ≤√ X */
parse var y r 2; do k=2 for length(y)-1; r=r+substr(y,k,1); endif x//j\==0 then iterate /*kJ isn't a factor of X, so skip it.*/
y=r s = s + j + x%j /* ··· add it and /*find digital root of the digitother rootfactor. */
end /*until*/ if s>x then return 0 /*wash, rinse, repeat ··· Is the sum too big? It ain't perfect*/
end /*j*/ end /*j*/ /*(above) is marginally faster. */▼
 
if r\==1 then return 0s==x /*Digitalif rootthe ¬sum 1?matches X, Then ¬it's 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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