I was looking through my box of relics that has old hardware, software, books, and unearthed this DOS 6.22 diskette I was looking for. I have another box in the attic with a Kaypro CP/M, MacIntosh SE, and a PET Commodore with a joystick I played Golf and other games. Is any of this old stuff have collectible value?
in 2012 (just checked my records and I also found the eBay listing photo!) I e-bayed a boxed and unused multiple floppy disk Windows v1.0 for an Olivetti M24. I netted £133 from an italian buyer.
Showcase your relics? Hell, I've got my picture as my avatar. How much more showcasing do we need? (Yes, I HAVE been accused of being a relic and a dinosaur.)
Back in the late 1960's, I used an English Electric Leo Marconi KDF8 computer which filled a room the size of a football pitch and probably had less power than a calculator.
Col
I also conserved the yellow colored punched paper tape that stores the first BASIC program I wrote in high school. A program for finding Perfect Numbers.
We had an old Teletype Model 33 terminal with a 300 baud dialup modem that connected to a GE Mark II Timesharing System in Dartmouth, MA.
The terminal had a paper tape reader/puncher which would read our programs to execute them.
Code:
100 REM FIND PERFECT NUMBERS BETWEEN 2 AND 1E+5
110 REM NORTHFIELD MOUNT HERMON SCHOOL, GILL, MASSACHUSETTS, 1973.
120 REM
130 LET START = 2
140 LET FINISH = 1E+5
150 REM
160 FOR N = START TO FINISH
170 LET SUM = 1
180 FOR I = 2 TO SQR(N)
190 IF N / I = INT(N / I) THEN GOTO 210
200 GOTO 230
210 LET SUM = SUM + I
220 IF N / I <> I THEN LET SUM = SUM + N / I
230 NEXT I
240 IF SUM = N THEN PRINT N;" IS A PERFECT NUMBER"
250 NEXT N
260 END
I rewrote the code in VBA and use it to benchmark CPU's. It has nested loops and is very CPU intensive.
How many seconds does it take for your computer to find perfect numbers between 2 and 100,000?
Keep in mind that the program has to find each number's proper divisors that are whole integers and then has to add those divisors up to see if the sum equals the number being evaluated. The sum of proper divisors of a number is called its aliquot sum, so a perfect number is one that is equal to its aliquot sum. As the number being checked gets higher, the more divisors it has to evaluate to see if they're whole numbers. You may want to shorten the range below 100,000, unless you have a super fast CPU. The first seven perfect numbers are 6, 28, 496, 8,128, 33,550,336, 8,589,869,056, and 137,438,691,328.
Code:
Option Compare Database
Option Explicit
' Function to check if a number is a perfect
Public Function IsPerfectNumber(ByVal NumberToCheck As Long) As Boolean
Dim SumOfDivisors As Long
Dim i As Long
' A perfect number must be a positive integer greater than 1
If NumberToCheck <= 1 Then
IsPerfectNumber = False
Exit Function
End If
SumOfDivisors = 1 ' Start with 1, as 1 is always a divisor
' Loop through potential divisors up to half of the number
For i = 2 To NumberToCheck / 2
If NumberToCheck Mod i = 0 Then
SumOfDivisors = SumOfDivisors + i
End If
Next i
' If the sum of proper divisors equals the number, it's a perfect number
If SumOfDivisors = NumberToCheck Then
IsPerfectNumber = True
Else
IsPerfectNumber = False
End If
End Function
' Sub-procedure to find and display perfect numbers
Public Sub FindPerfectNumbers()
Dim StartRange As Long
Dim EndRange As Long
Dim CurrentNumber As Long
Dim PerfectNumbersFound As String
' Get the range from the user (e.g., using InputBox)
StartRange = InputBox("Enter the starting number for the search:", "Perfect Number Search", 2)
EndRange = InputBox("Enter the ending number for the search:", "Perfect Number Search", 100000)
If StartRange > EndRange Then
MsgBox "Starting number cannot be greater than the ending number.", vbCritical
Exit Sub
End If
PerfectNumbersFound = "Perfect Numbers found in the range " & StartRange & " to " & EndRange & ":" & vbCrLf
' Iterate through the specified range
For CurrentNumber = StartRange To EndRange
If IsPerfectNumber(CurrentNumber) Then
PerfectNumbersFound = PerfectNumbersFound & CurrentNumber & vbCrLf
End If
Next CurrentNumber
' Display the results
If InStr(PerfectNumbersFound, vbCrLf & vbCrLf) > 0 Then ' Check if any perfect numbers were found
MsgBox PerfectNumbersFound, vbInformation, "Perfect Numbers"
Else
MsgBox "No perfect numbers found in the specified range.", vbInformation, "Perfect Numbers"
End If
End Sub
Any challenge takers to see how long it takes for your machine to run the above code?
I'm curious to see how fast your machines CPU's can crunch those nested loops.
I scaled down the run to 29,999 numbers. Processing 100K might take too long.
CAVEAT: If you overclocked your cores, make sure you have adequate cooling when running this code.
3 Seconds for 2 to 30000 but I had to fix a bug and added a timer.
Code:
' Sub-procedure to find and display perfect numbers
Public Sub FindPerfectNumbers()
Dim StartRange As Long
Dim EndRange As Long
Dim CurrentNumber As Long
Dim PerfectNumbersFound As String
Dim NumbersFound As String
Dim StartTime As Date
Dim RunSec As Long
' Get the range from the user (e.g., using InputBox)
StartRange = InputBox("Enter the starting number for the search:", "Perfect Number Search", 2)
EndRange = InputBox("Enter the ending number for the search:", "Perfect Number Search", 30000)
If StartRange > EndRange Then
MsgBox "Starting number cannot be greater than the ending number.", vbCritical
Exit Sub
End If
PerfectNumbersFound = "Perfect Numbers found in the range " & StartRange & " to " & EndRange & ":" & vbCrLf
StartTime = Now
' Iterate through the specified range
For CurrentNumber = StartRange To EndRange
If IsPerfectNumber(CurrentNumber) Then
NumbersFound = NumbersFound & CurrentNumber & vbCrLf
' PerfectNumbersFound = PerfectNumbersFound & CurrentNumber & vbCrLf
End If
Next CurrentNumber
RunSec = DateDiff("s", StartTime, Now)
' Display the results
If NumbersFound <> vbNullString Then
' If InStr(PerfectNumbersFound, vbCrLf & vbCrLf) > 0 Then ' Check if any perfect numbers were found
' MsgBox PerfectNumbersFound, vbInformation, "Perfect Numbers"
MsgBox PerfectNumbersFound & vbCrLf & NumbersFound, vbInformation, "Perfect Numbers: " & RunSec & " Seconds"
Else
MsgBox "No perfect numbers found in the specified range.", vbInformation, "Perfect Numbers"
End If
End Sub
The commented out IF at the end checking if any were found did not work when found.
I tried 2 to 100000, it to 28 seconds but there must be another bug because the found list was the same.
Oops, I must have accidentally commented that line when posting the code. Your results for 2 to 100K are valid because there's no perfect numbers between 8,129 and 100K. The next perfect number is 33,550,336. That one has many whole number divisors. You can set min to 1 number below that num, and max to 1 after to see how long it takes to calc that one.
The first seven perfect numbers are 6, 28, 496, 8,128, 33,550,336, 8,589,869,056, and 137,438,691,328.
Yeah, as the loop increments the number being evaluated, the more divisors it has to examine to see which one's are integers, add them to the running total, and compare the final sum to the number being evaluated to see if it's a perfect num.
Try 33,550,336 to 33,550,336 to see if it displays that num as being perfect. The inner loop will eval 33,550,334 sequential numbers to see which one's are whole number divisors, add them up, and compare if the sum equals 33,550,336.
Yes, they're crunching in parallel. I don't see CPUØ. Does your machine have Intel Xeon or AMD Ryzen cores?
In task mgr you can set affinity to the MSACCESS.EXE process to make it run faster.
