• Lars T. Kyllingstad (22/22) Aug 05 2009 Here's a puzzle for you floating-point wizards out there. I have to
• bearophile (5/7) Aug 05 2009 Generally double.inf+1 == double.inf
• Lars T. Kyllingstad (13/39) Aug 05 2009 I finally solved the puzzle by digging through ancient scientific
• Bill Baxter (15/54) Aug 05 2009 T,
• Jouko Koski (7/10) Aug 05 2009 Well, there is certain likelihood that decimal floating point will reapp...
• Don (4/49) Aug 05 2009 Not quite. They just used exponents which were powers of 10 or 16,
• Jarrett Billingsley (10/57) Aug 05 2009 ,
• Don (2/50) Aug 06 2009 Then Q is 0.5*ln(0.5). Dunno what use that is.

"Lars T. Kyllingstad" <public kyllingen.NOSPAMnet> writes:

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

bearophile <bearophileHUGS lycos.com> writes:

Lars T. Kyllingstad:
 (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?)

Generally double.inf+1 == double.inf
Inspecting the run time values inside a direct translation may give you clues.

Bye,
bearophile

"Lars T. Kyllingstad" <public kyllingen.NOSPAMnet> writes:

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.

I guess I can just drop this part from my code, then. ;)

-Lars

Bill Baxter <wbaxter gmail.com> writes:

On Wed, Aug 5, 2009 at 8:07 AM, Lars T.
Kyllingstad<public kyllingen.nospamnet> 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:

 =A0 =A0 =A0REAL B,Q,T
 C =A0 =A0 ------------------------------
 C =A0 =A0 |*** COMPUTE MACHINE BASE ***|
 C =A0 =A0 ------------------------------
 =A0 =A0 =A0T =3D 1.
 10 =A0 =A0T =3D T + T
 =A0 =A0 =A0IF ( (1.+T)-T .EQ. 1. ) GOTO 10
 =A0 =A0 =A0B =3D 0.
 20 =A0 =A0B =3D B + 1
 =A0 =A0 =A0IF ( T+B .EQ. T ) GOTO 20
 =A0 =A0 =A0IF ( T+2.*B .GT. T+B ) GOTO 30
 =A0 =A0 =A0B =3D B + B
 30 =A0 =A0Q =3D ALOG(B)
 =A0 =A0 =A0Q =3D .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 fort=


h. 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 b=

e
 an interesting piece of computer history trivia.

 After the above code has finished, the variable B contains the radix of t=

he
 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 system=

s
 as well. Instead of just power on/off, they had 10 or 16 separate voltage
 levels to differentiate between bit values.

 I guess I can just drop this part from my code, then. ;)

Very interesting!   Thanks for sharing that.
But I'm not so sure about the 16 separate voltage levels part.  I
think they were probably binary underneath.  Like the BCD (Binary
Coded Decimal) system.

--bb

"Jouko Koski" <joukokoskispam101 netti.fi> writes:

"Lars T. Kyllingstad" <public kyllingen.NOSPAMnet> wrote:

 After the above code has finished, the variable B contains the radix of 
 the computer's numerical system.

 I guess I can just drop this part from my code, then. ;)

Well, there is certain likelihood that decimal floating point will reappear 
in the future. However, determining the radix may be unnecessary, because 
decimal representations may get types of their own. Run-time computation of 
radix is bad anyway.

-- 
Jouko 

Don <nospam nospam.com> writes:

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.

 I guess I can just drop this part from my code, then. ;)

 
 -Lars

Jarrett Billingsley <jarrett.billingsley gmail.com> writes:

On Wed, Aug 5, 2009 at 10:16 PM, Don<nospam 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:

 =A0 =A0 =A0REAL B,Q,T
 C =A0 =A0 ------------------------------
 C =A0 =A0 |*** COMPUTE MACHINE BASE ***|
 C =A0 =A0 ------------------------------
 =A0 =A0 =A0T =3D 1.
 10 =A0 =A0T =3D T + T
 =A0 =A0 =A0IF ( (1.+T)-T .EQ. 1. ) GOTO 10
 =A0 =A0 =A0B =3D 0.
 20 =A0 =A0B =3D B + 1
 =A0 =A0 =A0IF ( T+B .EQ. T ) GOTO 20
 =A0 =A0 =A0IF ( T+2.*B .GT. T+B ) GOTO 30
 =A0 =A0 =A0B =3D B + B
 30 =A0 =A0Q =3D ALOG(B)
 =A0 =A0 =A0Q =3D .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 significa=



nt?)
 -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 syste=


ms
 as well. Instead of just power on/off, they had 10 or 16 separate voltag=


e
 levels to differentiate between bit values.

 Not quite. They just used exponents which were powers of 10 or 16, rather
 than 2. BTW, T =3D=3D 1/real.epsilon. I don't know what ALOG does, so I'v=

e no
 idea what Q is.

Apparently ALOG is just an old name for LOG.  At least that's what
Google tells me.

Don <nospam nospam.com> writes:

Jarrett Billingsley wrote:
 On Wed, Aug 5, 2009 at 10:16 PM, Don<nospam 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.