## digitalmars.D.learn - A floating-point puzzle

- 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.

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

Aug 05 2009

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

Aug 05 2009

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?) -LarsI 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

Aug 05 2009

On Wed, Aug 5, 2009 at 8:07 AM, Lars T. Kyllingstad<public kyllingen.nospamnet> wrote:Lars T. Kyllingstad wrote:T,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 =h. IB, and Q can be expressed in terms of real.epsilon, real.min and so fort=behave no idea how, though. Any ideas? (I am especially puzzled by the line after l.20. How can this test ever =astrue? Is the fact that the 1 in l.20 is an integer literal significant?) -LarsI finally solved the puzzle by digging through ancient scientific papers,=well as some old FORTRAN and ALGOL code, and the solution turned out to b=ean interesting piece of computer history trivia. After the above code has finished, the variable B contains the radix of t=hecomputer'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 =50sand 60s there were computers that used the decimal and hexadecimal system=sas 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

Aug 05 2009

"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

Aug 05 2009

Lars T. Kyllingstad wrote:Lars T. Kyllingstad wrote: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.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?) -LarsI 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

Aug 05 2009

On Wed, Aug 5, 2009 at 10:16 PM, Don<nospam nospam.com> wrote:Lars T. Kyllingstad wrote:nt?)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=,-LarsI finally solved the puzzle by digging through ancient scientific papers=toas well as some old FORTRAN and ALGOL code, and the solution turned out =50sbe 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=msand 60s there were computers that used the decimal and hexadecimal syste=eas well. Instead of just power on/off, they had 10 or 16 separate voltag=e nolevels 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=idea what Q is.Apparently ALOG is just an old name for LOG. At least that's what Google tells me.

Aug 05 2009

Jarrett Billingsley wrote:On Wed, Aug 5, 2009 at 10:16 PM, Don<nospam nospam.com> wrote:Then Q is 0.5*ln(0.5). Dunno what use that is.Lars T. Kyllingstad wrote:Apparently ALOG is just an old name for LOG. At least that's what Google tells me.Lars T. Kyllingstad wrote: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.

Aug 06 2009