In this challenge, you are to develop an RPN program for the HP-11c that correctly calculates days between two dates. It must handle dates from Oct. 15, 1582 to Nov. 25, 4046 (same as the 12c). The evaluation criteria are the same as in an earlier thread on this subject: score = number of lines + 7*number of registers used (not including lastx or stack) + 2*number of labels used. The lowest score wins. The evaluation criteria are designed to reward those whose programs make the most economical use of scarce resources on the 11c.

Each entrant should calculate their score and post it with their code.

Here's an accurate days-between-dates 11C implementation based on this algorithm

01 LBL A

02 0

03 STO 0

04 Rv

05 GSB A

06 CHS

07 STO 0

08 R^

09 LBL A

10 STO I

11 INT

12 9

13 +

14 1

15 2

16 1/x

17 X<>Y

18 x

19 LASTx

20 X<>Y

21 INT

22 1

23 2

24 x

25 -

26 X<>I

27 FRAC

28 EEX

29 2

30 x

31 INT

32 STO - 0

33 Rv

34 LASTx

35 FRAC

36 EEX

37 4

38 x

39 RCL I

40 1

41 0

42 /

43 INT

44 -

45 X<>I

46 3

47 0

48 6

49 x

50 5

51 +

52 1

53 0

54 /

55 INT

56 STO - 0

57 Rv

58 RCL I

59 3

60 6

61 5

62 x

63 STO - 0

64 Rv

65 RCL I

66 4

67 /

68 INT

69 STO - 0

70 Rv

71 RCL I

72 EEX

73 2

74 /

75 INT

76 STO + 0

77 Rv

78 RCL I

79 4

80 0

81 0

82 /

83 INT

84 STO - 0

85 1

86 STO + 0

87 RCL 0

88 RTN

score: 88 lines, 2 registers, 1 label = 103

*Edited: 15 Aug 2009, 2:28 a.m. *

The following is based on the method used in the TI-59 Master Library Module (page 76). Day of the Week has also been included. The program is far from optimized: 91 + 3*7 + 3*2 = 116.

001- LBL A 035- * 069- RCL 1

002- GSB 0 036- STO+ 0 070- EEX

003- x<>y 037- Rv 071- 2

004- GSB 0 038- ENTER 072- /

005- - 039- ENTER 073- INT

006- RTN 040- . 074- 1

007- LBL 0 041- 4 075- +

008- 0 042- * 076- .

009- STO 0 043- 2 077- 7

010- Rv 044- . 078- 5

011- EEX 045- 3 079- *

012- 2 046- + 080- INT

013- * 047- INT 081- STO- 0

014- ENTER 048- STO 2 082- RCL 1

015- FRAC 049- Rv 083- 4

016- EEX 050- 1 084- /

017- 4 051- - 085- INT

018- * 052- 3 086- STO+ 0

019- STO 1 053- 1 087- RCL 2

020- 3 054- x<>y 088- STO- 0

021- 6 055- * 089- R^

022- 5 056- STO+ 0 090- RCL 0

023- * 057- Rv 091- RTN

024- STO+ 0 058- 1 092- LBL B

025- Rv 059- LSTx 093- GSB 0

026- INT 060- x>y 094- ENTER

027- EEX 061- GTO 1 095- CHS

028- 2 062- x<>y 096- 7

029- / 063- STO- 1 097- /

030- INT 064- CLx 098- INT

031- LSTx 065- STO 2 099- 7

032- FRAC 066- LBL 1 100- *

033- EEX 067- Rv 101- +

034- 2 068- Rv 102- RTN

Date format is MM.DDYYYY

GSB A: works just like DeltaDYS on the 12C.

GSB B: similar to DATE on the 12C, but requires only one argument, the date, and returns 0=Sat, 1=Sun, ... 6=Fri.

_{
Equivalent TurboBCD program:
Program DaysBetweenDates;
var days, dt1, dt2: real;
function f(dt: real): real;
var d, m, y, t, x: real;
begin
y:=1e4*(Frac(100*dt));
m:=Int(dt);
d:=Int(100*(Frac(dt)));
x:=Int(0.4*m+2.3);
t:=365*y+d+31*(m-1);
if (m-1)<=1 then
begin
y:=y-1;
x:=0
end;
f:=t+Int(y/4)-Int(0.75*(Int(y/100)+1))-x
end;
begin
ReadLn(dt1,dt2);
days:=f(dt2)-f(dt1);
WriteLn(days:7:0)
end.
>
Running
6.011960 10.311976
5996
>
Running
2.281900 3.011900
1
>
}

_{
Edited to correct a typo in line 060}

*Edited: 16 Aug 2009, 7:19 p.m. *

Hi Gerson.

Don't you mean: 102+3*7+4*2=131?

Don:

Quote:

Don't you mean: 102+3*7+4*2=131?

He is not counting the LBL B and the last eleven steps as those are for the day of the week calculation and are not needed for days between dates

Recent correspondencc from Richard Nelson tells me that at one time there was a set of biorhythm programs for the SR-56. That should have some routines for a machine with a limited repertoire similar to that of the HP-11. So far I haven't got my hands on the programs. I'm working on it.

Palmer

Oh, OK thanks Palmer.

Don

Hello Don,

I should have mentioned the last few steps didn't have anything to do with your requirement. Sorry! (Thanks Palmer for taking the time to explain it). As I said, my implementation is far from optimized, so there's plenty of room for improvent.

This topic was discussed here some years ago. Thanks for bringing it back:

http://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/archv015.cgi?read=78807

Regards,

Gerson.

*Edited: 16 Aug 2009, 11:03 p.m. *

Thanks for that link, Gerson. It is absolutely fascinating. Reading that thread, I see some names we haven't seen here very much recently, like Valentin Albillo and James Prange. You can always count on Valentin to defend the 15c (and the 71, as I recall), as he did in that thread. If V or James Prange are still lurking out there, I for one would welcome their regular return to this forum. We would all gain from their participation.

Hi Katie,

That formula does look more "compact" than the JD formula I based my routines on. I cannot better the code, but I have a suggestion for optimising the maths.

As (365*y) is an integer, then I assume (365*y) + INT(y/4) = INT(365*y + y/4) = INT(365.25*y)

This would save 2 steps in your program.

Perhaps maths gurus could comment on the validity of my assumption?

Bart

In this particular case, on this platform, I think you could get away with it, but in the general case (even an emulated hp-11) you should not (and should expect to have a horrible bug to catch!)

My reasoning is as follows:

The first term we know is an integer say INT1, the second term is INT2.FRAC1, the integer of which is INT2, thus the sum is INT1+INT2.

If we did not take the integer of the second term before summation, the sum will be [INT1+INT2].FRAC1, if we then take the integer part it is INT1+INT2, as before.

Some rules apply

1) One term must be an integer to start with

2) Both must be of the same sign

As far as I see, both these rules are adhered to within the scope of the implementation of this formula.

The problem is in the nature of the actual implementation of floating point arithmetic. Values that are the results of the division of an integer by a power of 2 are "special", in that they can sometimes have exact representations in both decimal floating point and binary floating point. In general, this is not true; in this particular case it works. Change the divisor to, say, 53, and things won't be so neat.

Hmmm, so what you're saying is that it all looks good in the physical world but when going to & from 1's & 0's things get hairy.

Hairy? Oh my, yes. The entire Broadway musical (cast, orchestra, and crew) comes along, singing and dancing through your results, leaving you scratching your head and saying "What the F..."? If you're lucky you see it happen and eventually figure it out, if you don't, you swear at both the hardware and the gods to no effect.

Please read (it will take a while, sorry) What Every Computer Scientist Should Know About Floating Point **and Users need to know too.** (Handy .PDF version that has maybe a little too much white space; double side it when printing.)

An excellent observation!

If I'm not mistaken, it should save 3 lines.

And it should work on a real 11C because of the BCD math, and a simulated 11C because .25 is an inverse power of 2 as pointed out below.

It makes me wonder if we could include the other year terms in the constant on a real (or correctly emulated) HP-11C?

INT (y * 365.2425)

Certainly worth a try, right?

*Edited: 18 Aug 2009, 3:34 p.m. *

Edit: This program has been shown to be incorrect. Don't try this at home.

Here's my modifications of Katie's example:

01 LBL A

02 0

03 STO 0

04 Rv

05 GSB A

06 CHS

07 STO 0

08 R^

09 LBL A

10 STO I

11 INT

12 9

13 +

14 1

15 2

16 /

17 FRAC

18 1

19 2

20 .

21 1

22 x

23 INT

24 X<>I

25 FRAC

26 EEX

27 2

28 x

29 INT

30 STO - 0

31 Rv

32 LASTx

33 FRAC

34 EEX

35 4

36 x

37 RCL I

38 1

39 0

40 /

41 INT

42 -

43 X<>I

44 3

45 0

46 6

47 x

48 5

49 +

50 1

51 0

52 /

53 INT

54 STO - 0

55 Rv

56 RCL I

57 3

58 6

59 5

60 .

61 2

62 4

63 2

64 5

65 x

66 INT

67 STO - 0

68 1

69 STO + 0

70 RCL 0

71 RTN

Score 71 lines, 2 regs, 1 label = 87

*Edited: 19 Aug 2009, 4:20 p.m. *

i := -4004

while oldmethod(i) = newmethod(i) do i := i+1 endwhile;

display ("error at", i)

Yes, upon reflection (and testing) my idea was silly.

It doesn't work in many cases.

But the original y * 365.25 is good.

Here's my corrected version. I have more confidence in this one.

I don't need to recall the year to divide by 400, I use the year divided by 100 already there and just divide that by 4!

I also shortened the (m*306 + 5)/10 part by 1 line.

I tested these improvements to be the same. If someone finds a counterexample, I'd like to know.

Score 79 lines, 2 regs, 1 label = 95

01 LBL A

02 0

03 STO 0

04 Rv

05 GSB A

06 CHS

07 STO 0

08 R^

09 LBL A

10 STO I

11 INT

12 9

13 +

14 1

15 2

16 /

17 FRAC

18 1

19 2

20 .

21 1

22 x

23 INT

24 X<>I

25 FRAC

26 EEX

27 2

28 x

29 INT

30 STO - 0

31 Rv

32 LASTx

33 FRAC

34 EEX

35 4

36 x

37 RCL I

38 1

39 0

40 /

41 INT

42 -

43 X<>I

44 3

45 0

46 .

47 6

48 x

49 .

50 5

51 +

52 INT

53 STO - 0

54 Rv

55 RCL I

56 3

57 6

58 5

59 .

60 2

61 5

62 x

63 INT

64 STO - 0

65 Rv

66 RCL I

67 EEX

68 2

69 /

70 INT

71 STO + 0

72 4

73 /

74 INT

75 STO - 0

76 1

77 STO + 0

78 RCL 0

79 RTN

*Edited: 18 Aug 2009, 6:50 p.m. *

Quote:

Yes, upon reflection (and testing) my idea was silly.

It implicitly breaks my rule #1 :-)

*edit: and #2 as the y/100 is opposite in sign*
*Edited: 19 Aug 2009, 6:36 a.m. *

Quote:

I don't need to recall the year to divide by 400, I use the year divided by 100 already there and just divide that by 4!
I also shortened the (m*306 + 5)/10 part by 1 line.

Now why didn't I think of that!!

One caution though, you seem to be taking INT[INT[y/100]/4], rather than INT[[y/100]/4], which may lead to errors in some cases.

Bart

You are right, I am taking INT[INT[y/100]/4] intentionally. In my tests it was the same, and saved a step.

Let me know if you still think it's different, especially if you can show an example.

I wrote a 'C' program to check all year values and I think I'm OK.

A purist would have used a calculator program, but I was lazy I guess.

As it is we are interested in removing the leap days of whole centuries but counting every fouth one of those.

So your formula should indeed work fine.