A floating-point puzzle

[Don]

Jarrett Billingsley wrote:
> On Wed, Aug 5, 2009 at 10:16 PM, Don<nospam at nospam.com> wrote:
>> Lars T. Kyllingstad wrote:
>>> Lars T. Kyllingstad wrote:
>>>> Here's a puzzle for you floating-point wizards out there. I have to
>>>> translate the following snippet of FORTRAN code to D:
>>>>
>>>>      REAL B,Q,T
>>>> C     ------------------------------
>>>> C     |*** COMPUTE MACHINE BASE ***|
>>>> C     ------------------------------
>>>>      T = 1.
>>>> 10    T = T + T
>>>>      IF ( (1.+T)-T .EQ. 1. ) GOTO 10
>>>>      B = 0.
>>>> 20    B = B + 1
>>>>      IF ( T+B .EQ. T ) GOTO 20
>>>>      IF ( T+2.*B .GT. T+B ) GOTO 30
>>>>      B = B + B
>>>> 30    Q = ALOG(B)
>>>>      Q = .5/Q
>>>>
>>>> Of course I could just do a direct translation, but I have a hunch that
>>>> T, B, and Q can be expressed in terms of real.epsilon, real.min and so
>>>> forth. I have no idea how, though. Any ideas?
>>>>
>>>> (I am especially puzzled by the line after l.20. How can this test ever
>>>> be true? Is the fact that the 1 in l.20 is an integer literal significant?)
>>>>
>>>> -Lars
>>>
>>> I finally solved the puzzle by digging through ancient scientific papers,
>>> as well as some old FORTRAN and ALGOL code, and the solution turned out to
>>> be an interesting piece of computer history trivia.
>>>
>>> After the above code has finished, the variable B contains the radix of
>>> the computer's numerical system.
>>>
>>> Perhaps the comment should have tipped me off, but I had no idea that
>>> computers had ever been anything but binary. But apparently, back in the 50s
>>> and 60s there were computers that used the decimal and hexadecimal systems
>>> as well. Instead of just power on/off, they had 10 or 16 separate voltage
>>> levels to differentiate between bit values.
>> Not quite. They just used exponents which were powers of 10 or 16, rather
>> than 2. BTW, T == 1/real.epsilon. I don't know what ALOG does, so I've no
>> idea what Q is.
> 
> Apparently ALOG is just an old name for LOG.  At least that's what
> Google tells me.
Then Q is 0.5*ln(0.5). Dunno what use that is.