| Bytes | Lang | Time | Link |
|---|---|---|---|
| 037 | AWK | 250321T190621Z | xrs |
| 069 | SAKO | 250321T180950Z | Acrimori |
| 004 | Thunno 2 | 230614T170500Z | The Thon |
| 7271 | Scala | 230416T070359Z | 138 Aspe |
| 084 | PowerShell | 230404T204530Z | James Fl |
| 006 | Stax | 230331T170224Z | emirps |
| 018 | APL Dyalog/GNU | 230303T153404Z | Mark Ree |
| 005 | Pyt | 230303T015252Z | Kip the |
| nan | Qi | 230303T002535Z | gantz gi |
| 024 | Arturo | 230302T193938Z | chunes |
| 016 | Pip | 230302T201443Z | The Thon |
| nan | 230302T200928Z | The Thon | |
| nan | Fig | 220926T175840Z | Seggan |
| 041 | Regex Perl/PCRE+?^=RME | 210308T031739Z | Deadcode |
| 005 | MATL | 190316T133555Z | Suever |
| 063 | Prolog SWI | 220903T080150Z | Aiden Ch |
| 017 | Alice | 220808T211615Z | Julian |
| 045 | Desmos | 220323T034111Z | Aiden Ch |
| 4812 | MMIX | 210502T193347Z | NoLonger |
| 021 | Machine Language x86 32bit | 210313T030201Z | l4m2 |
| 023 | Machine Language x86 32bit | 210308T074048Z | Deadcode |
| 024 | Raku | 210308T225227Z | Sean |
| 055 | C++ clang | 210308T222748Z | Deadcode |
| 047 | Qbasic | 190313T151913Z | steenber |
| 003 | Vyxal | 210308T102447Z | lyxal |
| 040 | Retina 0.8.2 | 190312T103358Z | Neil |
| 004 | Gaia | 190328T203636Z | Giuseppe |
| 101 | VDMSL | 190328T152838Z | Expired |
| 080 | Labyrinth | 190315T230956Z | Jonathan |
| 038 | MACHINE LANGUAGEX86 | 190314T200144Z | user5898 |
| 043 | x86 Assembly | 190314T134903Z | Fayti170 |
| 009 | Pyth | 190312T074438Z | JPeroute |
| 004 | Japt ! | 190312T025347Z | Luis fel |
| 054 | Java JDK | 190312T092803Z | Olivier |
| 023 | TIBASIC TI84 | 190312T133649Z | absolute |
| 022 | APLNARS | 190313T155312Z | user5898 |
| 004 | MathGolf | 190313T132135Z | maxb |
| 038 | Kotlin | 190313T143748Z | Adam |
| 054 | Racket | 190313T131246Z | Galen Iv |
| 057 | Factor | 190313T124730Z | Galen Iv |
| 045 | Forth gforth | 190312T123547Z | reffu |
| 005 | Actually | 190313T082035Z | Jonathan |
| 005 | Husk | 190313T074427Z | Jonathan |
| 046 | C# Visual C# Interactive Compiler | 190312T035341Z | Gymhgy |
| 034 | Smalltalk | 190313T024450Z | Leandro |
| 108 | Twig | 190313T013247Z | Ismael M |
| 035 | Haskell | 190313T005244Z | Joseph S |
| 041 | C gcc | 190312T075848Z | Marcos |
| 056 | Bash | 190312T193631Z | Jonathan |
| 025 | Octave | 190312T093629Z | Luis Men |
| 008 | cQuents | 190312T153640Z | Stephen |
| 029 | R | 190312T031323Z | Giuseppe |
| 066 | Java 8 | 190312T131159Z | Benjamin |
| 043 | Powershell | 190312T121316Z | J. Bergm |
| 045 | Python | 190312T100525Z | Jonathan |
| 008 | Pyth | 190312T102825Z | Sok |
| 014 | Wolfram Language Mathematica | 190312T101906Z | ZaMoC |
| 038 | JavaScript | 190312T063337Z | tsh |
| 013 | Charcoal | 190312T100608Z | Neil |
| 081 | Batch | 190312T094019Z | Neil |
| 033 | Ruby | 190312T081031Z | Kirill L |
| 004 | 05AB1E | 190312T070714Z | Emigna |
| 003 | Neim | 190312T042759Z | Unrelate |
| 003 | Jelly | 190312T042138Z | Mr. Xcod |
| 004 | Brachylog | 190312T040049Z | Unrelate |
| 046 | Python 3 | 190312T035745Z | Neil A. |
| 062 | Javascript | 190312T025521Z | vityavv |
| 017 | CJam | 190312T025905Z | Esolangi |
SAKO, 69 bytes
PODPROGRAM:F(N)
W=0
*)W=W+I×0*MOD(N,I)
POWTORZ:I=1(1)N
F()=W-N×2
WROC
Returns 0 if the input is good enough for me (perfect), returns a negative or positive number if it isn't.
Full programme version, 79 bytes
CZYTAJ:N
W=0
*1)W=W+I×0*MOD(N,I)
POWTORZ:I=1(1)N
DRUKUJ(9,0):W-N×2
STOP1
KONIEC
Thunno 2, 4 bytes
FṫS=
Explanation
FṫS= # Implicit input -> 28
F # Factors of input -> [1,2,4,7,14,28]
ṫ # Without the tail -> [1,2,4,7,14]
S # Sum the list -> 28
= # Equals the input? -> 1
# Implicit output
Scala, 72 71 bytes
saved 1 byte thanks to the comment.
def f(n:Int):Boolean={val F=for(i<-1 until n if n%i<1)yield i;F.sum==n}
PowerShell, 84 bytes
param($n)1..[math]::sqrt($n)|%{$a+=$_,($n/$_)*!($n%$_)};$n-eq(($a|gu)-join"+"|iex)/2
Stax, 6 bytes
ôxⁿ♪σ▀
This is PackedStax, which unpacks to the following 7 bytes:
:dNs|+=
Explanation
:d # divisors
N # minus the last item (which is equal to the input)
|+ # summed
= # is equal to
s # the input
Times out for larger test cases on the online interpreter.
APL (Dyalog/GNU), 18 bytes, ⎕IO=0
{⍵=+/(0=⍵|⍨⍳⍵)/⍳⍵}
Try it online! (The in-browser version of APL has too small a workspace to handle the larger test cases.)
Explanation:
{...}creates an anonymous function (dfn), whose argument becomes the parameter⍵. We'll examine the expression inside the braces from right to left.⍳Ngenerates a vector of the first N numbers starting with the index origin. Since that is set to 0, it goes up to N-1. For this function to work we only need to go up to N/2, but⍳only accepts integers, and making the division result an integer would take more characters./with two vector operands produces a new vector with each element of the right vector repeated a number of times equal to the corresponding element of the left vector. In this case we're going to build a vector of N 0s and 1s with 1s only at the positions corresponding to the proper divisors of N; the result will therefore be a vector containing only those divisors.- The expression inside parentheses builds the selector; going right to left again, we start with another
⍳⍵. (We could have built a DRYer version that didn't need to repeat that, but it would have taken more characters.) |is the divide-and-take-remainder operator (mod). By itself it takes its operands on the opposite order from e.g. C's%, but we are transposing it with⍨to flip that, which removes the need for another pair of parentheses.⍵|⍨⍳⍵divides each number from 0 to⍵-1into⍵and returns a vector of the remainders. (Fortunately for this use,0|Njust returns N rather than throwing a division-by-zero error.)0=is what it looks like: a simple equality test with 0. Applied to a vector, it returns a vector with 1's wherever the original vector had a 0, and 0's everywhere else - exactly what we want for our selector.- Having built our vector of the proper divisors, we then compute the sum of its elements with
+/(reduction via addition). - Finally,
⍵=returns our desired result: 1 if the sum is equal to⍵itself, 0 otheriwse.
Pyt, 5 bytes
ĐðƩ₂=
Returns True if perfect, False otherwise
Đ implicit input; Đuplicate
ð get ðivisors
Ʃ Ʃum
₂ divide by two
= is it equal to the input; implicit print
Qi, 72 bytes (45 characters + 22 brackets)
(λ(n)((☯(~> △(pass(~>(/ n _)integer?))+(= n)))(cdr(range n 0 -1))))
Pip, 16 bytes
a=$+Y{!\a%a}FI,a
There's no factors builtin so we have to implement it ourselves
a=$+Y{!\a%a}FI,a ; input on command line
,a ; range(input)
{ }FI ; filtered by:
\a%a ; the input mod this number
! ; logical not
$+Y ; sum of the list
a= ; equals the input?
Thunno D, \$ 5 \log_{256}(96) \approx \$ 4.12 bytes
fZHS=
Explanation
fZHS= # D flag duplicates input
f # Factors, including 1 and input
ZH # Without last item (input)
S # Sum this list
= # Equal to the input?
# Implicit output
Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes
=Sk
=Sk
S # Does the sum
k # Of the divisors
= # Equal the input?
Regex (Perl/PCRE+(?^=)RME), 44 41 bytes
((\2x+?|^x\B)(?^=\2+$))++$(?^!(\2x+)\3+$)
This was a port of my 48 byte .NET regex below, using lookinto instead of variable-length lookbehind. This makes it far faster, and able to scan all numbers from 0 to 9000 in about 7 seconds on my PC. It also enables somes golfs that aren't possible in the .NET version.
# No anchor needed, because \2 cannot be initially captured if
# matching is started anywhere besides the beginning.
# Sum as many divisors of N as we can (using the current position as the sum),
# without skipping any of them:
(
(
\2x+? # \2 = the smallest number larger than the previous value of \2
# for which the following is true; add \2 to the running total
# sum of divisors.
|
^x # This must always be the first step. Add 1 to the sum of
# divisors, thus setting \2 = 1 so the next iteration can keep
# going.
\B # Exclude 1 as a divisor if N == 1. We don't have to do this for
# N > 1, because this loop will never be able to add N to the
# already summed divisors (as 1 will always be one of them).
)
(?^=\2+$) # Assert that \2 is a divisor of N
)++ # Iterate the above at least once, with a possessive quantifier
# to lock in the match (i.e. prevent any iteration from being
# backtracked into).
$ # Require that the sum equal N exactly. If this fails, the loop
# won't be able to find a match by backtracking, because
# everything inside the loop is done atomically.
# Assert that there are no more proper divisors of N that haven't already been summed:
(?^! # Lookinto – Assert that the following cannot match:
(\2x+) # \3 = any number greater than \2
\3+$ # Assert \3 is a proper divisor of N
)
Regex (Perl/PCRE+(?^=)RME), 45 bytes
^(?=(x(?=(x*))(?^=(?(?=\2+$)(\3?+\2))))+$)\3$
This is a port of the largest-to-smallest 47 byte .NET regex below, using lookinto instead of variable-length lookbehind.
Regex (Perl/PCRE+(?^=)RME), 47 bytes
^(?=((?^0=(x*))(?^=(?(?=\2+$)(\3?+\2)))x)+$)\3$
This is a port of the smallest-to-largest 47 byte .NET regex below, using lookinto instead of variable-length lookbehind. This one is less conducive to being ported, and I had to use (?^0=), lookinto accessing the in-progress current match. I've not decided how to finalize the current behavior of this, so it's likely this version will not work with a future version of lookinto.
Regex (.NET), 47 bytes
^(?=((?<=(?=(?(\3+$)((?>\2?)\3)))(^x*))x)+$)\2$
^(?=(x(?=(x*$)(?<=(?(^\2+)(\2(?>\3?))))))+$)\3$
This is based on my 44 byte abundant numbers regex, with is in turn based on Martin Ender's 45 byte abundant numbers regex. It was trivial to adapt to matching perfect numbers; just two small changes were needed, with a third change made for aesthetic purposes.
# For the purpose of these comments, the input number will be referred to as N.
^(?= # Attempt to add up all the divisors of N.
(x # Cycle through all positive values of tail that are less than N,
# testing each one to see if it is a divisor of N. Start at N-1.
(?= # Do the below operations in a lookahead, so that upon popping back
# out of it our position will remaing the same as it is here.
(x*$) # \2 = tail, a potential divisor; go to end to that the following
# lookbehind can operate on N as a whole.
(?<= # Switch to right-to-left evaluation so that we can operate both
# on N and the potential divisor \2. This requires variable-length
# lookbehind, a .NET feature. Please read these comments in the
# order indicated, from [Step 1] to [Step 4].
(?(^\2+) # [Step 1] If \2 is a divisor of N, then...
( # [Step 2] Add it to \3, the running total sum of divisors:
# \3 = \3 + \2
\2 # [Step 4] Iff we run out of space here, i.e. iff the sum would
# exceed N at this point, the match will fail, and the
# summing of divisors will be halted. It can't backtrack,
# because everything in the loop is done atomically.
(?>\3?) # [Step 3] Since \3 is a nested backref, it will fail to match on
# the first iteration. The "?" accounts for this, making
# it add zero to itself on the first iteration. This must
# be done before adding \2, to ensure there is enough room
# for the "?" not to cause the match to become zero-length
# even if \3 has a value.
)
)
)
)
)+ # Using "+" instead of "*" here is an aesthetic choice, to make it
# clear we don't want to match zero. It doesn't actually make a
# difference (besides making the regex ever so slightly more
# efficient) because \3 would be unset anyway for N=0.
$ # We can only reach this point if all proper divisors of N, all the
# way down to \2 = 1, been successfully summed into \3.
)
\3$ # Require the sum of divisors to be exactly equal to N.
Regex (.NET), 48 bytes
((?=.*(\1x+$)(?<=^\2+))\2|^x\B)+$(?<!^\3+(x+\2))
This is based on my 43 byte abundant numbers regex. It was interesting and somewhat tricky to adapt into a fully golfed perfect numbers regex.
# No anchor needed, because \1 cannot be initially captured if
# matching is started anywhere besides the beginning.
# Sum as many divisors of N as we can (using the current position as the sum),
# without skipping any of them:
(
(?= # Atomic lookahead:
.*(\1x+$) # \2 = the smallest number larger than the previous value of \1
# (which is identical to the previous value of \2) for which the
# following is true. We can avoid using "x+?" because the ".*"
# before this implicitly forces values to be tested in increasing
# order from the smallest.
(?<=^\2+) # Assert that \2 is a divisor of N
)
\2 # Add \2 to the running total sum of divisors
|
^x # This must always be the first step. Add 1 to the sum of divisors,
# thus setting \1 = 1 so the next iteration can keep going
\B # Exclude 1 as a divisor if N == 1. We don't have to do this for
# N > 1, because this loop will never be able to add N to the
# already summed divisors (as 1 will always be one of them).
)+$ # Require that the sum equal N exactly. If this fails, the loop
# won't be able to find a match by backtracking, because everything
# inside the loop is done atomically.
# Assert that there are no more proper divisors of N that haven't already been summed:
(?<! # Assert that the following, evaluated right-to-left, cannot match:
^\3+ # [Step 2] is a proper divisor of N
(x+\2) # [Step 1] Any value of \3 > \2
)
Regex (Perl / PCRE2 v10.35+), 64 60 57 bytes
This is a port of the 41 byte regex, emulating variable-length lookbehind using a recursive subroutine call.
(?=((\2x+?|^x\B)((?<=(?=^\2+$|(?3)).)))++$)(?!(\2x+)\4+$)
Try it online! - Perl
Attempt This Online! - PCRE2 v10.40+
# No anchor needed, because \2 cannot be initially captured if
# matching is started anywhere besides the beginning.
(?=
# Sum as many divisors of N as we can (using the current position as the sum),
# without skipping any of them:
(
(
\2x+? # \2 = the smallest number larger than the previous value of \2
# for which the following is true; add \2 to the running total
# sum of divisors.
|
^x # This must always be the first step. Add 1 to the sum of
# divisors, thus setting \2 = 1 so the next iteration can keep
# going.
\B # Exclude 1 as a divisor if N == 1. We don't have to do this for
# N > 1, because this loop will never be able to add N to the
# already summed divisors (as 1 will always be one of them).
)
((?<= # Define subroutine (?3); lookbehind
(?= # Circumvent the constant-width limitation of lookbehinds
# in PCRE by using a lookahead inside the lookbehind
^\2+$ # This is the payload of the emulated variable-length
# lookbehind: Assert that \2 is a divisor of N
|
(?3) # Recursive call - this is the only alternative that can
# match until we reach the beginning of the string
)
. # Go back one character at a time, trying the above
# lookahead for a match each time
))
)++ # Iterate the above at least once, with a possessive quantifier
# to lock in the match (i.e. prevent any iteration from being
# backtracked into).
$ # Require that the sum equal N exactly. If this fails, the loop
# won't be able to find a match by backtracking, because
# everything inside the loop is done atomically.
)
# Assert that there are no more proper divisors of N that haven't already been summed:
(?! # Assert that the following cannot match:
(\2x+) # \4 = any number greater than \2
\4+$ # Assert \4 is a proper divisor of N
)
It isn't practical to directly port the 47 byte regex to PCRE, because it changes the value of a capture group inside the lookbehind, and upon the return of any subroutine in PCRE, all capture groups are reset to the values they had upon entering the subroutine. It is possible, however, and I did this for the corresponding Abundant numbers regex by changing the main loop from iteration into recursion.
Regex (ECMAScript), 53 bytes – Even perfect numbers
If it is ever proved that there are no odd perfect numbers, the following would be a robust answer, but for now it is noncompeting, as it only matches even perfect numbers. It was written by Grimmy on 2019-03-15.
^(?=(x(x*?))(\1\1)+$)((x*)(?=\5$))+(?!(xx+)\6+$)\1\2$
This uses the fact that every even perfect number is of the form \$2^{n-1} (2^n-1)\$ where \$2^n-1\$ is prime. For \$2^n-1\$ to be prime, it is necessary that \$n\$ itself be prime, so the regex does not need to do a \$log_2\$ test on \$2^n\$ to verify \$n\$ is prime. (I have used \$n\$ here instead of \$p\$ to make that clear.)
^ # N = tail = input
(?=(x(x*?))(\1\1)+$) # Assert that the largest odd divisor of N is >= 3 (due to
# having a "+" here instead of "*"; this is not necessary, but
# speeds up the regex's non-match when N is a a power of 2);
# \1 = N / {largest odd divisor of N}
# == {largest power of 2 divisor of N}; \2 = \1 - 1
((x*)(?=\5$))+ # tail = N / {largest power of 2 divisor of N}
# == {largest odd divisor of N}
(?!(xx+)\6+$) # Assert tail is prime
\1\2$ # Assert tail == \1*2 - 1; if this fails to match, it will
# backtrack into the "((x*)(?=\5$))+" loop (effectively
# multiplying by 2 repeatedly), but this will always fail
# to match because every subsequent match attempt will be
# an even number, and "\1\2$" can only be odd.
MATL, 5 bytes
EGZ\s=
Try it out at MATL Online
Explanation
% Implicitly grab the input as an integer
% STACK: { 6 }
E % Multiply by two
% STACK: { 12 }
G % Grab the input again
% STACK: { 12, 6 }
Z\ % Compute all divisors (including itself)
% STACK: { 12, [1, 2, 3, 6] }
s % Sum up these divisors
% STACK: { 12, 12 }
= % Check that the two elements on the stack are equal
% STACK: { 1 }
% Implicitly display the result
Prolog (SWI), 70 67 63 bytes
-3 bytes thanks to @Jo King
-4 bytes thanks to @Steffan
$N:-bagof(M,(between(1,N,M),N mod M<1),L),sumlist(L,S),S=:=2*N.
Alice, 17 bytes
/O\B!d v
@I/-?+&<
Outputs 0 if the number is perfect, non 0 if imperfect
Flattened
/I\B!d&+?-/O@
/I\ Reads the input
B Get all the divisors in ascending order
! Remove the input number and save it on the tape
d&+ Sum all the remaining divisors of the stack
? Get the input from the tape
- Subtract from the sum
/O@ Output and exit
Desmos, 45 bytes
f(n)=\{\sum_{N=1}^nN\{\mod(n,N)=0,0\}=2n,0\}
\$f(n)\$ returns \$1\$ if \$n\$ is a perfect number, \$0\$ is \$n\$ is an imperfect number.
MMIX, 48 bytes (12 instrs)
(jxd¹ -T)
00000000: 27010001 f7010000 e3020000 1eff0001 '¢¡¢ẋ¢¡¡ẉ£¡¡œ”¡¢
00000010: feff0006 72ffff01 220202ff 27010101 “”¡©r””¢"££”'¢¢¢
00000020: 5b01fffb 32000002 73000001 f8010000 [¢”»2¡¡£s¡¡¢ẏ¢¡¡
Disassembly:
perf SUBU $1,$0,1 // i = n - 1
PUT rD,0 // zero out hidiv register
SETL $2,0 // s = 0
0H DIVU $255,$0,$1 // loop: t:rR = rD:n /% i
GET $255,rR // t = rR
ZSZ $255,$255,$1 // t = t? 0 : i
ADDU $2,$2,$255 // s += i
SUBU $1,$1,1 // i--
PBNZ $1,0B // iflikely(i) goto loop
CMPU $0,$0,$2 // r = n <=> s
ZSZ $0,$0,1 // r = !r
POP 1,0 // return r
The best part of this is that the only branch in the whole thing is to do a loop. Testing the division is handled by the ZSZ instruction (which is also used later to handle a negation).
I could save an instruction if I were allowed to return different results for abundant and deficient numbers.
- jxd is a program I wrote, available on request; it is like xxd, but with support for only jelly and 05AB1E codepages (and no
-por-rsupport, it would be redundant).
Machine Language (x86 32-bit), 21 bytes
is_perfect:
mov ecx, esi ; copy input number into ECX
inc esi
mov ebx, esi ; copy input number + 1 into EBX
fact_loop:
xor edx, edx ; clear EDX (high word for dividend)
mov eax, esi
div ecx ; divide (n+1) / ECX
dec edx ; if remainder = 1, n/ECX is zero and ECX is not 1
; In case EAX = 2^32-1, this doesn't work as intended
; (get edx=0 for all ECXs)
; According to test 2^(2^n)-1 up to 2^128-1 are all
; deficient so it happens to be correct
jnz cont_loop ; if not, continue loop
sub ebx, eax ; Subtract divisor from EBX; this simultaneously
; changes the running total, and sets the flags for
; the return value.
jc done ; If CF=1, the sum of divisors subtracted from EBX has
; exceeded our input number, meaning it is abundant,
; and we need to exit the loop now to return this
; result in the flags.
cont_loop:
loop fact_loop
dec ebx ; Subtract the 1 added into esi for ZF
done: ; return value:
; (JZ) = ZF=1 = perfect
; (JNZ) = ZF=0 = imperfect
end_is_perfect:
Machine Language (x86 32-bit), 22 bytes edited from here with also abundant? output
is_perfect:
mov ecx, esi ; copy input number into ECX
mov ebx, esi ; copy input number into EBX
fact_loop:
xor edx, edx ; clear EDX (high word for dividend)
lea eax, [esi+1]
div ecx ; divide (n+1) / ECX
dec edx ; if remainder = 1, n/ECX is zero and ECX is not 1
; In case EAX = 2^32-1, this doesn't work as intended
; (get edx=0 for all ECXs)
; According to test 2^(2^n)-1 up to 2^128-1 are all
; deficient so it happens to be correct
jnz cont_loop ; if not, continue loop
sub ebx, eax ; Subtract divisor from EBX; this simultaneously
; changes the running total, and sets the flags for
; the return value.
jc done ; If CF=1, the sum of divisors subtracted from EBX has
; exceeded our input number, meaning it is abundant,
; and we need to exit the loop now to return this
; result in the flags.
cont_loop:
loop fact_loop
test ebx, ebx ; Not SUBed in last cycle
done: ; return value:
; (JZ) = ZF=1 = perfect
; (JNZ) = ZF=0 = imperfect
; (JC) = CF=1 = abundant
; (JNC) = CF=0 = non-abundant
; (JA) = CF=0 and ZF=0 = deficient
; (JNA) = CF=1 or ZF=1 = non-deficient
end_is_perfect:
Bad luck in last round sub is not runned and a test is needed
Machine Language (x86 32-bit), 24 23 bytes
This is a macro that takes an unsigned integer in EAX as input (N), and outputs the result in the Zero Flag (set=perfect, clear=imperfect). It also outputs a result in the Carry Flag (set=abundant, clear=non-abundant). As such, it's actually an answer for both this question and An abundance of integers!, requiring no modification.
It's based on 640KB's abundant numbers answer, with the two additional optimizations from l4m2's answer, where the divisors are subtracted from N instead of added to it (thus simultaneously setting the flags for the return value). An input of 1 is correctly handled, costing 3 2 bytes.
89 C1 89 C3 9E EB 0E 31 D2 50 F7 F1 58 85 D2 75 04 29 CB 72 02 E2 F0
89 C1 mov ecx, eax ; copy input number into ECX
89 C3 mov ebx, eax ; copy input number into EBX
9E sahf ; in case input number is 1, let ZF=0 and CF=0
EB 0E jmp cont_loop ; exclude original number as a divisor
fact_loop:
31 D2 xor edx, edx ; clear EDX (high word for dividend)
50 push eax ; save original dividend
F7 F1 div ecx ; divide EDX:EAX / ECX
58 pop eax ; restore dividend
85 D2 test edx, edx ; if remainder = 0, it divides evenly
75 04 jnz cont_loop ; if not, continue loop
29 CB sub ebx, ecx ; Subtract divisor from EBX; this simultaneously
; changes the running total, and sets the flags for
; the return value.
72 02 jc done ; If CF=1, the sum of divisors subtracted from EBX has
; exceeded our input number, meaning it is abundant,
; and we need to exit the loop now to return this
; result in the flags.
cont_loop:
E2 F0 loop fact_loop
done: ; return value:
; (JZ) = ZF=1 = perfect
; (JNZ) = ZF=0 = imperfect
; (JC) = CF=1 = abundant
; (JNC) = CF=0 = non-abundant
; (JA) = CF=0 and ZF=0 = deficient
; (JNA) = CF=1 or ZF=1 = non-deficient
C++ (clang), 55 bytes
[](long n){long i,s;for(i=s=n;--i;n%i?:s-=i);return!s;}
Can handle 8589869056 just fine thanks to the use of long, but this input is not included in the TIO as it takes more than 60 seconds to process. (On my i7-4930K @ 4.1 GHz it takes 65 seconds.)
This has a golf optimization that can't be done in C – n%i?:s-=i, using not only the GNU extension of allowing the true or false field of a ternary to be empty, but taking advantage of the fact that in C++, ?: and -= have the same operator precedence with right-to-left associativity – whereas in C ?: has a higher precedence than -=, so the only way to make it work would be as n%i?:(s-=i) and then it'd cost 1 byte over s-=n%i?0:i rather than saving 1 byte.
Qbasic, 53 47 bytes
Thank you @Deadcode, -6 bytes
INPUT n
FOR i=1TO n-1
s=s-i*(n\i=n/i)
NEXT
?n=s
Literally quite Basic... Prints -1 for perfect numbers, 0 otherwise.
Vyxal, 3+1 3 bytes
∆K=
-1 byte thanks to @Deadcode for telling me to not be an idiot and remember that I added a sum of proper divisors built-in. I actually forgot that I did
This is: does the sum of the proper divisors of the input (∆K) equal the input itself (=).
Retina 0.8.2, 44 40 bytes
crossed out 44 is still regular 44
.+
$*
M!&`(.+)$(?<=^\1+)
1>`¶
^(1+)¶\1$
Try it online! Edit: Saved 4 bytes thanks to @Deadcode. Still somewhat slow, so link excludes largest test cases. Explanation:
.+
$*
Convert to unary.
M!&`(.+)$(?<=^\1+)
Match all factors of the input. This uses overlapping mode, which in Retina 0.8.2 requires all of the matches to start at different positions, so the matches are actually returned in descending order, starting with the original input.
1>`¶
Delete all of the newlines except for the first. This sums together the proper factors, leaving the original input and sum of factors.
^(1+)¶\1$
Test whether they are the same.
Gaia, 4 bytes
dΣ⁻=
1 for perfect, 0 for imperfect.
d | implicit input, n, push divisors
Σ | take the sum
⁻ | subtract n
= | equal to n?VDM-SL, 101 bytes
f(i)==s([x|x in set {1,...,i-1}&i mod x=0])=i;s:seq of nat+>nat
s(x)==if x=[]then 0 else hd x+s(tl x)
Summing in VDM is not built in, so I need to define a function to do this across the sequences, this ends up taking up the majority of bytes
A full program to run might look like this:
functions
f:nat+>bool
f(i)==s([x|x in set {1,...,i-1}&i mod x=0])=i;s:seq of nat+>nat
s(x)==if x=[]then 0 else hd x+s(tl x)
Labyrinth, 80 bytes
?::`}:("(!@
perfect:
{:{:;%"}
+puts; "
}zero: "
}else{(:
"negI" _~
""""""{{{"!@
The Latin characters perfect puts zero else neg I are actually just comments*.
i.e. if the input is perfect a 0 is printed, otherwise -1 is.
?::`}:("(!@ ?::`}:("(!@
: BEWARE :
{:{:;%"} {:{:;%"}
+ ; " +LAIR; "
} : " } OF : "
} {(: }MINO{(:
" " _~ "TAUR" _~
""""""{{{"!@ """"""{{{"!@
How?
Takes as an input a positive integer n and places an accumulator variable of -n onto the auxiliary stack, then performs a divisibility test for each integer from n-1 down to, and including, 1, adding any which do divide n to the accumulator. Once this is complete if the accumulator variable is non-zero a -1 is output, otherwise a 0 is.
The ?::`}:( is only executed once, at the beginning of execution:
?::`}:( Main,Aux
? - take an integer from STDIN and place it onto Main [[n],[]]
: - duplicate top of Main [[n,n],[]]
: - duplicate top of Main [[n,n,n],[]]
` - negate top of Main [[n,n,-n],[]]
} - place top of Main onto Aux [[n,n],[-n]]
: - duplicate top of Main [[n,n,n],[-n]]
( - decrement top of Main [[n,n,n-1],[-n]]
The next instruction, ", is a no-op, but we have three neighbouring instructions so we branch according to the value at the top of Main, zero takes us forward, while non-zero takes us right.
If the input was 1 we go forward because the top of Main is zero:
(!@ Main,Aux
( - decrement top of Main [[1,1,-1],[-1]]
! - print top of Main, a -1
@ - exit the labyrinth
But if the input was greater than 1 we turn right because the top of Main is non-zero:
:} Main,Aux
: - duplicate top of Main [[n,n,n-1,n-1],[-n]]
} - place top of Main onto Aux [[n,n,n-1],[-n,n-1]]
At this point we have a three-neighbour branch, but we know n-1 is non-zero, so we turn right...
"% Main,Aux
" - no-op [[n,n,n-1],[-n,n-1]]
% - place modulo result onto Main [[n,n%(n-1)],[-n,n-1]]
- ...i.e we've got our first divisibility indicator n%(n-1), an
- accumulator, a=-n, and our potential divisor p=n-1:
- [[n,n%(n-1)],[a,p]]
We are now at another three-neighbour branch at %.
If the result of % was non-zero we go left to decrement our potential divisor, p=p-1, and leave the accumulator, a, as it is:
;:{(:""}" Main,Aux
; - drop top of Main [[n],[a,p]]
: - duplicate top of Main [[n,n],[a,p]]
{ - place top of Aux onto Main [[n,n,p],[a]]
- three-neighbour branch but n-1 is non-zero so we turn left
( - decrement top of Main [[n,n,p-1],[a]]
: - duplicate top of Main [[n,n,p-1,p-1],[a]]
"" - no-ops [[n,n,p-1,p-1],[a]]
} - place top of Main onto Aux [[n,n,p-1],[a,p-1]]
" - no-op [[n,n,p-1],[a,p-1]]
% - place modulo result onto Main [[n,n%(p-1)],[a,p-1]]
- ...and we branch again according to the divisibility
- of n by our new potential divisor, p-1
...but if the result of % was zero (for the first pass only when n=2) we go straight on to BOTH add the divisor to our accumulator, a=a+p, AND decrement our potential divisor, p=p-1:
;:{:{+}}""""""""{(:""} Main,Aux
; - drop top of Main [[n],[a,p]]
: - duplicate top of Main [[n,n],[a,p]]
{ - place top of Aux onto Main [[n,n,p],[a]]
: - duplicate top of Main [[n,n,p,p],[a]]
{ - place top of Aux onto Main [[n,n,p,p,a],[]]
+ - perform addition [[n,n,p,a+p],[]]
} - place top of Main onto Aux [[n,n,p],[a+p]]
} - place top of Main onto Aux [[n,n],[a+p,p]]
""""""" - no-ops [[n,n],[a+p,p]]
- a branch, but n is non-zero so we turn left
" - no-op [[n,n],[a+p,p]]
{ - place top of Aux onto Main [[n,n,p],[a+p]]
- we branch, but p is non-zero so we turn right
( - decrement top of Main [[n,n,p-1],[a+p]]
: - duplicate top of Main [[n,n,p-1,p-1],[a+p]]
"" - no-ops [[n,n,p-1,p-1],[a+p]]
} - place top of Main onto Aux [[n,n,p-1],[a+p,p-1]]
At this point if p-1 is still non-zero we turn left:
"% Main,Aux
" - no-op [[n,n,p-1],[a+p,p-1]]
% - modulo [[n,n%(p-1)],[a+p,p-1]]
- ...and we branch again according to the divisibility
- of n by our new potential divisor, p-1
...but if p-1 hit zero we go straight up to the : on the second line of the labyrinth (you've seen all the instructions before, so I'm leaving their descriptions out and just giving their effect):
:":}"":({):""}"%;:{:{+}}"""""""{{{ Main,Aux
: - [[n,n,0,0],[a,0]]
" - [[n,n,0,0],[a,0]]
- top of Main is zero so we go straight
- ...but we hit the wall and so turn around
: - [[n,n,0,0,0],[a,0]]
} - [[n,n,0,0],[a,0,0]]
- top of Main is zero so we go straight
"" - [[n,n,0,0],[a,0,0]]
: - [[n,n,0,0,0],[a,0,0]]
( - [[n,n,0,0,-1],[a,0,0]]
{ - [[n,n,0,0,-1,0],[a,0]]
- top of Main is zero so we go straight
- ...but we hit the wall and so turn around
( - [[n,n,0,0,-1,-1],[a,0]]
: - [[n,n,0,0,-1,-1,-1],[a,0]]
"" - [[n,n,0,0,-1,-1,-1],[a,0]]
} - [[n,n,0,0,-1,-1],[a,0,-1]]
- top of Main is non-zero so we turn left
" - [[n,n,0,0,-1,-1],[a,0,-1]]
% - (-1)%(-1)=0 [[n,n,0,0,0],[a,0,-1]]
; - [[n,n,0,0],[a,0,-1]]
: - [[n,n,0,0,0],[a,0,-1]]
{ - [[n,n,0,0,0,-1],[a,0]]
: - [[n,n,0,0,0,-1,-1],[a,0]]
{ - [[n,n,0,0,0,-1,-1,0],[a]]
+ - [[n,n,0,0,0,-1,-1],[a]]
} - [[n,n,0,0,0,-1],[a,-1]]
} - [[n,n,0,0,0],[a,-1,-1]]
""""""" - [[n,n,0,0,0],[a,-1,-1]]
- top of Main is zero so we go straight
{ - [[n,n,0,0,0,-1],[a,-1]]
{ - [[n,n,0,0,0,-1,-1],[a]]
{ - [[n,n,0,0,0,-1,-1,a],[]]
Now this { has three neighbouring instructions, so...
...if a is zero, which it will be for perfect n, then we go straight:
"!@ Main,Aux
" - [[n,n,0,0,0,-1,-1,a],[]]
- top of Main is a, which is zero, so we go straight
! - print top of Main, which is a, which is a 0
@ - exit the labyrinth
...if a is non-zero, which it will be for non-perfect n, then we turn left:
_~"!@ Main,Aux
_ - place a zero onto Main [[n,n,0,0,0,-1,-1,a,0],[]]
~ - bitwise NOT top of Main (=-1-x) [[n,n,0,0,0,-1,-1,a,-1],[]]
" - [[n,n,0,0,0,-1,-1,a,-1],[]]
- top of Main is NEGATIVE so we turn left
! - print top of Main, which is -1
@ - exit the labyrinth
MACHINE LANGUAGE(X86, 32 bit), 38 bytes
00000788 53 push ebx
00000789 8B442408 mov eax,[esp+0x8]
0000078D 31DB xor ebx,ebx
0000078F 31C9 xor ecx,ecx
00000791 43 inc ebx
00000792 39C3 cmp ebx,eax
00000794 730E jnc 0x7a4
00000796 31D2 xor edx,edx
00000798 50 push eax
00000799 F7F3 div ebx
0000079B 58 pop eax
0000079C 09D2 or edx,edx
0000079E 75F1 jnz 0x791
000007A0 01D9 add ecx,ebx
000007A2 EBED jmp short 0x791
000007A4 29C8 sub eax,ecx
000007A6 0F94C0 setz al
000007A9 0FB6C0 movzx eax,al
000007AC 5B pop ebx
000007AD C3 ret
000007AE
Function lenght: 7AEh-788h=26h=38d; below assembly file that generate obj for linking:
; nasmw -fobj this.asm
; bcc32 -v file.c this.obj
section _DATA use32 public class=DATA
global _sumd
section _TEXT use32 public class=CODE
_sumd:
push ebx
mov eax, dword[esp+ 8]
xor ebx, ebx
xor ecx, ecx
.1: inc ebx
cmp ebx, eax
jae .2
xor edx, edx
push eax
div ebx
pop eax
or edx, edx
jnz .1
add ecx, ebx
jmp short .1
.2: sub eax, ecx
setz al
movzx eax, al
pop ebx
ret
below the C file for link and test that function:
#include <stdio.h>
unsigned es[]={1,12,13,18,20,1000,33550335,6,28,496,8128,33550336,0};
int sumd(unsigned);
main(void)
{int i;
for(i=0;es[i];++i)
printf("f(%u)=%u\n",es[i],sumd(es[i]));
return 0;
}
below the macro assembly file source of the .asm one:
; nasmw -fobj this.asm
; bcc32 -v file.c this.obj
section _DATA use32 public class=DATA
global _sumd
section _TEXT use32 public class=CODE
_sumd:
<b |a=^8|b^=b|c^=c
.1: ++b |b>=a#.2|r^=r|<a|div b|>a|r|=r|jnz .1|c+=b|#.1
.2: a-=c|setz al|movzx eax, al
>b
ret
below the 113 bytes golfing code of above:
_sumd:<b|a=^8|b^=b|c^=c|.1:++b|b>=a#.2|r^=r|<a|div b|>a|r|=r|jnz .1|c+=b|#.1|.2:a-=c|setz al|movzx eax, al|>b|ret
the results:
f(1)=0
f(12)=0
f(13)=0
f(18)=0
f(20)=0
f(1000)=0
f(33550335)=0
f(6)=1
f(28)=1
f(496)=1
f(8128)=1
f(33550336)=1
I got the trick of use push eax|div | pop eax from https://codegolf.stackexchange.com/a/181515/58988
x86 Assembly, 45 43 Bytes.
6A 00 31 C9 31 D2 41 39 C1 7D 0B 50 F7 F9 58 85
D2 75 F1 51 EB EE 31 D2 59 01 CA 85 C9 75 F9 39
D0 75 05 31 C0 40 EB 02 31 C0 C3
Explaination (Intel Syntax):
PUSH $0 ; Terminator for later
XOR ECX, ECX ; Clear ECX
.factor:
XOR EDX, EDX ; Clear EDX
INC ECX
CMP ECX, EAX ; divisor >= input number?
JGE .factordone ; if so, exit loop.
PUSH EAX ; backup EAX
IDIV ECX ; divide EDX:EAX by ECX, store result in EAX and remainder in EDX
POP EAX ; restore EAX
TEST EDX, EDX ; remainder == 0?
JNZ .factor ; if not, jump back to loop start
PUSH ECX ; push factor
JMP .factor ; jump back to loop start
.factordone:
XOR EDX, EDX ; clear EDX
.sum:
POP ECX ; pop divisor
ADD EDX, ECX ; sum into EDX
TEST ECX, ECX ; divisor == 0?
JNZ .sum ; if not, loop.
CMP EAX, EDX ; input number == sum?
JNE .noteq ; if not, skip to .noteq
XOR EAX, EAX ; clear EAX
INC EAX ; increment EAX (sets to 1)
JMP .return ; skip to .return
.noteq:
XOR EAX, EAX ; clear EAX
.return:
RETN
Input should be provided in EAX.
Function sets EAX to 1 for perfect and to 0 for imperfect.
EDIT: Reduced Byte-Count by two by replacing MOV EAX, $1 with XOR EAX, EAX and INC EAX
Pyth, 9 13 bytes
qsf!%QTSt
Thank you to the commentors for the golf help
Finds all the factors of the input, sums them, and compares that to the original input.
Japt -!, 4 bytes
¥â¬x
-----------------
Implicit Input U
¥ Equal to
x Sum of
â Factors of U
¬ Without itself
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Java (JDK), 54 bytes
n->{int s=0,d=0;for(;++d<n;)s+=n%d<1?d:0;return s==n;}
Though for a strict number by number matching, the following will return the same values, but is only 40 bytes.
n->n==6|n==28|n==496|n==8128|n==33550336
TI-BASIC (TI-84), 30 23 bytes
:2Ans=sum(seq(Ans/Xnot(remainder(Ans,X)),X,1,Ans,1
Horribly inefficient, but it works.
Reducing the bytecount sped up the program by a lot.
Input is in Ans.
Output is in Ans and is automatically printed out when the program completes.
Explanation:
(TI-BASIC doesn't have comments, so just assume that ; makes a comment)
:2Ans=sum(seq(Ans/Xnot(remainder(Ans,X)),X,1,Ans ;Full program
2Ans ;double the input
seq( ;generate a list
X, ;using the variable X,
1, ;starting at 1,
Ans ;and ending at the input
;with an implied increment of 1
Ans/X ;from the input divided by X
not( ), ;multiplied by the negated result of
remainder(Ans,X) ;the input modulo X
;(result: 0 or 1)
sum( ;sum up the elements in the list
= ;equal?
Example:
6
6
prgmCDGF2
1
7
7
prgmCDGF2
0
Note: The byte count of a program is evaluated using the value in [MEM]>[2]>[7] (36 bytes) then subtracting the length of the program's name, CDGF2, (5 bytes) and an extra 8 bytes used for storing the program:
36 - 5 - 8 = 23 bytes
APL(NARS), chars 11, bytes 22
{⍵=2÷⍨11π⍵}
11π return the sum of all divisors, test:
f←{⍵=2÷⍨11π⍵}
f¨1 12 13 18 20 1000 33550335
0 0 0 0 0 0 0
f¨6 28 496 8128 33550336 8589869056
1 1 1 1 1 1
MathGolf, 5 4 bytes
─╡Σ=
Explanation
─ get divisors (includes the number itself)
╡ discard from right of list (removes the number itself)
Σ sum(list)
= pop(a, b), push(a==b)
Since MathGolf returns divisors rather than proper divisors, the solution is 1 byte longer than it would have been in that case.
Kotlin, 38 bytes
{i->(1..i-1).filter{i%it==0}.sum()==i}
Expanded:
{ i -> (1..i - 1).filter { i % it == 0 }.sum() == i }
Explanation:
{i-> start lambda with parameter i
(2..i-1) create a range from 2 until i - 1 (potential divisors)
.filter{i%it==0} it's a divisor if remainder = 0; filter divisors
.sum() add all divisors
==i compare to the original number
} end lambda
Factor, 57 bytes
: f ( x -- ? ) dup [1,b) [ dupd divisor? ] filter sum = ;
There is a shorter solution on the Rosetta Code that uses the divisors builtin:
: f ( n -- ? ) [ divisors sum ] [ 2 * ] bi = ;
but I wanted to come up with my own solution.
Forth (gforth), 45 bytes
: f 0 over 1 ?do over i mod 0= i * - loop = ;
Explanation
Loops over every number from 1 to n-1, summing all values that divide n perfectly. Returns true if sum equals n
Code Explanation
: f \ start word definition
0 over 1 \ create a value to hold the sum and setup the bounds of the loop
?do \ start a counted loop from 1 to n. (?do skips if start = end)
over \ copy n to the top of the stack
i mod 0= \ check if i divides n perfectly
i * - \ if so, use the fact that -1 = true in forth to add i to the sum
loop \ end the counted loop
= \ check if the sum and n are equal
; \ end the word definition
Actually, 5 bytes
;÷Σ½=
Full program taking input from STDIN which prints 1 for perfect numbers and 0 otherwise.
How?
;÷Σ½= - full program, implicitly place input, n, onto the stack
; - duplicate top of stack
÷ - divisors (including n)
Σ - sum
½ - halve
= - equal?
Husk, 5 bytes
=¹ΣhḊ
Yields 1 for a perfect argument and 0 otherwise.
How?
=¹ΣhḊ - Function: integer, n e.g. 24
Ḋ - divisors [1,2,3,4,6,8,12,24]
h - initial items [1,2,3,4,6,8,12]
Σ - sum 36
¹ - first argument 24
= - equal? 0
C# (Visual C# Interactive Compiler), 46 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)^n*2
Returns 0 if perfect, otherwise returns a positive number. I don't know if outputting different types of integers are allowed in place of two distinct truthy and falsy values, and couldn't find any discussion on meta about it. If this is invalid, I will remove it.
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Smalltalk, 34 bytes
((1to:n-1)select:[:i|n\\i=0])sum=n
Twig, 108 bytes
Yeah, I managed to get something longer than Java :/
This creates a macro that you import into your own template and call the method a
{%macro a(n,x=0)%}{%if n>1%}{%for i in 1..n-1%}{%set x=x+i*(n%i==0)%}{%endfor%}{{x==n}}{%endif%}{%endmacro%}
Returns 1 for perfect numbers, nothing for imperfect.
You can try it on https://twigfiddle.com/0or03v (testcases included)
C (gcc), 41 bytes
f(n,i,s){for(i=s=n;--i;s-=n%i?0:i);n=!s;}
1: 0
12: 0
13: 0
18: 0
20: 0
1000: 0
33550335: 0
6: 1
28: 1
496: 1
8128: 1
33550336: 1
-65536: 0 <---- Unable to represent final test case with four bytes, fails
Let me know if that failure for the final case is an issue.
Bash, 56 bytes
Maybe a recursive function will be shorter? My Bash is certainly not strong.
s=0;for ((;v++<$1;)){ $[s+=($1%$v<1)*$v-1];};[ $s = $1 ]
A full program taking a command line argument which has a return code 0 if it was perfect and 1 otherwise.
Octave, 25 bytes
@(n)~mod(n,t=1:n)*t'==2*n
Explanation
@(n)~mod(n,t=1:n)*t'==2*n
@(n) % Define anonymous function with input n
1:n % Row vector [1,2,...,n]
t= % Store in variable t
mod(n, ) % n modulo [1,2,...,n], element-wise. Gives 0 for divisors
~ % Logical negate. Gives 1 for divisors
t' % t transposed. Gives column vector [1;2;...;n]
* % Matrix multiply
2*n % Input times 2
== % Equal? This is the output value
cQuents, 8 bytes
?#N=U\zN
Explanation
? Mode query: return whether or not input is in sequence
# Conditional: iterate N, add N to sequence if condition is true
N= Condition: N ==
U ) sum( )
\z ) proper_divisors( )
N N
)) implicit
R, 33 29 bytes
!2*(n=scan())-(x=1:n)%*%!n%%x
Returns TRUE for perfect numbers and FALSE for imperfect ones.
Java 8, 66 bytes
Someone has to use the stream API at some point, even if there's a shorter way to do it
n->java.util.stream.IntStream.range(1,n).filter(i->n%i<1).sum()==n
Powershell, 46 bytes 43 bytes
param($i)1..$i|%{$o+=$_*!($i%$_)};$o-eq2*$i
Edit: -3 bytes thanks to @AdmBorkBork
Python, 45 bytes
lambda n:sum(d*(n%d<1)for d in range(1,n))==n
True for perfect; False for others (switch this with == -> !=)
44 42 41 bytes (-2 thanks to ovs) if we may output using "truthy vs falsey":
f=lambda n,i=1:i/n or-~f(n,i+1)-(n%i<1)*i
(falsey (0)) for perfect; truthy (a non-zero integer) otherwise
Pyth, 8 bytes
qs{*MPyP
Try it online here.
qs{*MPyPQQ Implicit: Q=eval(input())
Trailing QQ inferred
PQ Prime factors of Q
y Powerset
P Remove last element - this will always be the full prime factorisation
*M Take product of each
{ Deduplicate
s Sum
q Q Is the above equal to Q? Implicit print
JavaScript, 38 bytes
n=>eval("for(i=s=n;i--;)n%i||!(s-=i)")
(Last testcase timeout on TIO.)
Charcoal, 13 bytes
Ｎθ⁼θΣΦθ∧ι¬﹪θι
Try it online! Link is to verbose version of code. Outputs - for perfect numbers. Uses brute force. Explanation:
Ｎθ Numeric input
Φθ Filter on implicit range
ι Current value (is non-zero)
∧ Logical And
θ Input value
﹪ Modulo
ι Current value
¬ Is zero
Σ Sum of matching values
⁼ Equals
θ Input value
Batch, 81 bytes
@set s=-%1
@for /l %%i in (1,1,%1)do @set/as+=%%i*!(%1%%%%i)
@if %s%==%1 echo 1
Takes n as a command-line parameter and outputs 1 if it is a perfect number. Brute force method, starts the sum at -n so that it can include n itself in the loop.
05AB1E, 4 bytes
ѨOQ
Explanation
O # the sum
Ñ # of the divisors of the input
¨ # with the last one removed
Q # equals the input
Neim, 3 bytes
𝐕𝐬𝔼
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
Brachylog, 4 bytes
fk+?
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Brute force, sums the factors and checks for equality.
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!