Re: A floating-point puzzle
Jarrett Billingsley wrote: On Wed, Aug 5, 2009 at 10:16 PM, Donnos...@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. 10T = T + T IF ( (1.+T)-T .EQ. 1. ) GOTO 10 B = 0. 20B = B + 1 IF ( T+B .EQ. T ) GOTO 20 IF ( T+2.*B .GT. T+B ) GOTO 30 B = B + B 30Q = 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.
A floating-point puzzle
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. 10T = T + T IF ( (1.+T)-T .EQ. 1. ) GOTO 10 B = 0. 20B = B + 1 IF ( T+B .EQ. T ) GOTO 20 IF ( T+2.*B .GT. T+B ) GOTO 30 B = B + B 30Q = 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
Re: A floating-point puzzle
Lars T. Kyllingstad pub...@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
Re: A floating-point puzzle
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. 10T = T + T IF ( (1.+T)-T .EQ. 1. ) GOTO 10 B = 0. 20B = B + 1 IF ( T+B .EQ. T ) GOTO 20 IF ( T+2.*B .GT. T+B ) GOTO 30 B = B + B 30Q = 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
Re: A floating-point puzzle
On Wed, Aug 5, 2009 at 10:16 PM, Donnos...@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.
