Hi All,

Recently I needed a FLOOR function on my HP41. Easy enough, but I decided to generalize the solution. I wanted FLOOR and CEILING functions that did not effect the stack (Y, Z, and T are preserved) and that did not use any registers. LASTX does not have to be correct.

My solution uses 44 bytes on an HP41 including 19 bytes for the LBL FLOOR, LBL CEIL and the END instructions. My solution does not update LASTX correctly but it could if I were willing to use more memory. I use no SP tricks but they are not disallowed. Off the top of my head I can't see how SP would help but I am going to look into it in more detail to see if I am missing something.

I worked on this for longer than I expected when I started the exercise and would love to see other solutions. I just can't shake the feeling that I am missing something obvious--some trick that will shave the code to half the size and twice the speed. At one point I thought that I had a very elegant truly tiny solution but it did not work for certain cases. I'll share later.

Cheers,

-Marwan

How do you define FLOOR and CEIL for negative numbers?

FLOOR(-2.3) = -3.0

CEIL(-2.3) = -2.0

Therein lies the trick because otherwise (for a positive number) FLOOR is just INT or IP. INTG on the 33S or 35S is a FLOOR function but as far as I can tell there is no CEILING on those machines.

Cheers,

-Marwan

Edited: 29 Oct 2011, 8:10 p.m.

can we use alpha?

41 RPN:

  1 LBL FLOOR

  2 INT

  3 x<0?

  4 x<>0?

  5 RTN

  6 X<>L

  7 FRC 

  8 x<>0?

  9 ST- L

 10 X<>L

 11 INT

 12 RTN

Edited: 29 Oct 2011, 9:05 p.m. after one or more responses were posted

Extending Egan's code using CEIL(x) = -FLOOR(-x):

  1 LBL CEIL

  2 CHS

  3 XEQ 00

  4 CHS

  5 RTN

  6 LBL 00

  7 LBL FLOOR

  8 INT

  9 x<0?

 10 x<>0?

 11 RTN

 12 X<> L

 13 FRC 

 14 x<>0?

 15 ST- L

 16 X<> L

 17 INT

 18 END

Is this 38 bytes all up?

- Pauli

Edit: merged in Egan's improvements

Edited: 29 Oct 2011, 9:12 p.m. after one or more responses were posted

Doh! Concurrent updates. I changed it, shorter now.

Easily fixed :-)

- Pauli

No.

Don't think that this is right. For FLOOR(-2.3) it returns -2. FLOOR(-2.3) should be -3.00.

-2.3 return -2.0 not -3.0. INT would give you the same results. Solution not working correctly for negatives. Also, your solution to CEIL is exactly what I used in my solution. That is why LASTX does not work for CEIL in my solution.

Cheers,

-Marwan

Edited: 29 Oct 2011, 10:35 p.m. after one or more responses were posted

Blame Egan for the floor -- I just copied his code :-)

I checked the one byte longer version and I'm pretty sure floor(-2.3) was correct then.

- Pauli

I never got to see the 1 byte longer version. I count 39 bytes in the current version (?). If the original version was 1 byte longer and works it is 4 bytes shorter than mine (I think).

Cheers,

-Marwan

Edited: 29 Oct 2011, 10:41 p.m.

FLOOR(x) = -CEIL(-x) and vice versa. This should help on the 33/35S.

43 bytes, perfect function (YZT preserved, LASTX contains last X):

01*LBL"CEIL"

02 INT

03 X<>Y

04 X<> L

05 X>Y?

06 ISG Y

07 LBL 00      (NOP)

08 X<> L

09 X<>Y

10 RTN

11*LBL"FLOOR"

12 INT

13 X<>Y

14 X<> L

15 X<Y?

16 DSE Y

17 LBL 00      (NOP)

18 X<> L

19 X<>Y 

20 END

Nice. Actually I think it is pretty elegant--a single test, a single exit point (RTN), shorter, and you don't use a flag. Better than mine. But see Pauli's and Marcus's solution and comments above. If you are willing to give up LASTX (as I was) you can use:

LBL CEIL

CHS

XEQ 00

CHS

RTN

LBL FLOOR

LBL 00

...

This is what I did in my original solution and why I said that LAST X is correct for FLOOR but not for CEIL. It shortens the code considerably.

Cheers,

-Marwan

Edited: 30 Oct 2011, 8:47 a.m.

This will not work. First, the inital test makes no sense. It uses a well-known technique from the good old days: use two consecutive tests, and you will get a "composite test" for (NOT condition1) OR condition2.

So x<0? followed by x<>0? logically equals "X>=0 OR X<>0?", or in other words: "is x greater than zero, or equal to zero, or not equal to zero". Which is the case for any number. So the program will always stop after this test, simply returning INT(x). #-)

There also is not much sense in the sequence "X<>0? ST-L" as the test for non-zero X is obsolete: if X actually is zero subtracting it from L will not change anything.

Here's my (also Q&D) first idea for the '41:

01 LBL"FLOOR"

02 INT

03 X<> L

04 X<0?

05 DSE L

06 LBL 00

07 X<> L

08 END

Dieter

Edited: 30 Oct 2011, 10:12 a.m.

*clap* *clap*

I see the idea with DSE/ISG is quite obvious. And it's sufficiently elegant as well. ;-)

Dieter

I regard the STO X command is the best NOP for the HP-41C and HP-42s.

Namir

But STO X requires two bytes, while LBL 00 only takes one. ;-)

Dieter

The Zenrom module offers a NOP instruction that is basically a text byte (I think) but I am not sure of the length. Also X <> X works well but is also two bytes as is STO X.

Cheers,

-Marwan

Here is my original solution. But as noted elsewhere Werner's is better (which was partially what drove the challenge--find a better solution).

LBL CEIL

CHS

XEQ 00

CHS

RTN

LBL FLOOR

LBL 00

CF 20

FRC

X=0?

SF 20

CLX          These two instruction could also be X <> L 

LAST X       which is the same number of bytes

INT 

FS?C 20

RTN

X>0?

RTN

DSE X

RTN

END

44 bytes total preserves Y, Z, and T. Preserves LASTX for FLOOR but not for CEIL.

I don't see any way to do this without using a DSE instruction so I would love to see Egan's original solution it works. The later solution does not.

Cheers,

-Marwan

Edited: 30 Oct 2011, 11:06 a.m.

Quote:

---

LBL CEIL

CHS

XEQ 00

CHS

RTN

---

I found this relationship yesterday, at Wikipedia:

FLOOR(x) can be defined in terms of INT(x) and FRC(x), but I don't see this being mentioned there:

FLOOR(x) = INT(x) + INT(FRC(x) + 1) - 1

On the HP-15C, the following avoids any test, but preserves Y only. Perhaps it can be optimized on the HP-41.

001- f LBL C      ; CEIL

002-   CHS

003-   GSB A

004-   CHS

005- g RTN

006- f LBL A      ; FLOOR

007- g INT

008- g LSTx

009- f FRAC

010-   1

011-   +

012- g INT

013-   1

014-   -

015-   +

016- g RTN

Regards,

Gerson.

Hi Gerson,

I found the same relationship. The fly in the ointment in my original challenge, so to speak, was the requirement that the stack be preserved. I am trying to think of ways to optimize your solution on the 41 to meet that requirement.

Cheers,

-Marwan

Quote:

---

The fly in the ointment in my original challenge, so to speak, was the requirement that the stack be preserved.

---

Yes, I am aware of it. That's not an easy task, congratulations for your and other's accomplishments!

Cheers,

Gerson.

Edited: 30 Oct 2011, 12:55 p.m.

Thanks. I have to say that it took me longer than Werner to come up with a less elegant solution <g>.

Cheers,

-Marwan

Edited: 30 Oct 2011, 1:24 p.m.

Hi Marwan,

I didn't come up with this on the fly, it's been in my files for quite a while now ;-)

cheers, Werner

Equivalent formulas:

    CEIL(x)  = x - FRC(FRC(x) - 1)

    FLOOR(x) = x - FRC(FRC(x) + 1)

Cheers, Werner

You know, you guys are all too damn honest! :)

Cheers,

-Marwan

Well, I continued to think about this while out cycling today and I think I have a 37 byte solution. There are a couple of caveats:

1.  LASTX is not preserved and

2.  The solution is NOT symmetrical.  That is you can't write a solution for FLOOR on it's own it has to be done through a call to CEIL because 

the HP41 does not have symmetrical tests (that is there is an X<=0? but no X>=0? on the 41.  It is symetrical on the 42S which has both tests.

Here is my latest effort:

LBL FLOOR

CHS

XEQ 00

CHS

RTN

LBL CEIL

LBL 00

FRC

X<=0?

GTO 00

CLX

1

ST+ L

LBL 00  This is *NOT* a NO-OP

X<>L

INT

END

Cheers,

-Marwan

EDIT: Fixes issue with longer fraction (ss.eeeii) as pointed out by Werner but adds 2 bytes. 39 bytes now.

Edited: 31 Oct 2011, 7:49 a.m. after one or more responses were posted

Here's a 28 byte version, including 17 bytes for the alpha labels:
Using D. Knuth's definitions (Concrete Mathemetics, Ch. 3, p.67

FLOOR(x)="the greatest integer less than or equal to x." 

CEIL(x) ="the least integer greater than or equal to x."

For CIEL, we increment 1 for negative input, for FLOOR we decrement 1 for positive input, then return the IP of the result in all 4 cases (pos,neg input and CIEL,FLOOR selection).

00 { 28-Byte Prgm }

01>LBL "CIEL"

02 X>0?

03 ISG ST X

04 IP

05 GTO 00

06>LBL "FLOOR"

07 X<0?

08 DSE ST X

09>LBL 00

10 IP

11 .END.

Note: Does not account for integer inputs when ciel(x>0) or floor(x<0).

Edited: 30 Oct 2011, 11:29 p.m. after one or more responses were posted

I came up with this solution very early on in my efforts but it doesn't work. CEIL(2.0) is 2.0 not 3.0 as returned by this algorithm. Similarly FLOOR(-2.0) should return -2.0 and not -3.0.

Cheers,

-Marwan

Edited: 30 Oct 2011, 11:33 p.m. after one or more responses were posted

yep, just noted that in the comments.

Yes. So does not work as a generalized solution.

I like the trick you use to save a byte by:

ISG X

IP

GTO 00

Saves a byte over:

ISG X

LBL 00      NOP

IP 

RTN

Very clever. Can be used to save 3 bytes in Werner's solution (which is, IMHO, the best so far).

Cheers,

-Marwan

Edited: 30 Oct 2011, 11:52 p.m.

Hi Werner,

Did you see Alan's solution below? It does not work for integer values but he uses a trick that can shorten your version by 3 bytes.

Instead of:

ISG Y

LBL 00

X<> L

X<>Y

RTN

You can use:

ISG Y

LBL 00

GTO 00

A savings of 3 bytes.

Cheers,

-Marwan

Ok, last try for the night. Does NOT use synthetic programming or flags.

It works by hard-coding a 3-variable K-map for POS/NEG, INT/Non-INT, inputs and CIEL/FLOOR function. (8 cases)

1) If NEG CEIL then return IP(X)         (2 cases)

2) Else If POS FLOOR then return IP(X)   (2 cases)

3) Else If FP(X)=0 then return IP(X)     (2 cases)

4) Else If X>0 return IP(X+1)            (1 case)

5) Else If X<0 return IP(X-1)            (1 case)

00 { 43-Byte Prgm }

01>LBL "CIEL"

02 X<0?

03 GTO 02

04 GTO 00

05>LBL "FLOOR"

06 X>0?

07 GTO 02

08>LBL 00

09 STO ST L

10 FP

11 X=0?

12 GTO 01

13 X>0?

14 ISG ST L

15 X<0?

16 DSE ST L

17>LBL 01

18 X<> ST L

19>LBL 02

20 IP

21 .END.

Edited: 31 Oct 2011, 2:04 a.m.

Allen, all your solutions have something 'heavenly' ;-)

But that doesn't work.. ISG and DSE work on numbers in sss.eeeii format,
and will thus add/subtract ii, or 1 when ii=0.

Try your example with PI for instance ...

Cheers, Werner

Only two bytes ;-)

Werner

Perhaps we can use a 'test' suite to guarantee that a program returns the correct solution in all cases.

So, how about (just for CEIL):

CEIL(0)      :   0 

CEIL(1e-50)  :   1 

CEIL(-1e-50) :   0 

CEIL(2e-5)   :   1 

CEIL(-2e-5)  :   0 

CEIL(1)      :   1 

CEIL(-1)     :  -1 

CEIL(PI)     :   4 

CEIL(-PI)    :  -3 

CEIL(10)     :  10 

CEIL(-10)    : -10 

CEIL(1e50)   : 1e50

Cheers, Werner

Right, sorry, GTO 00 is 2 bytes not 1 byte.

Cheers,

-Marwan

Good idea. There are so many cases where true decrement / increment operators would have been useful.

Cheers,

-Marwan

That's why the 34S has them :-)

I was also tempted to do the equivalent of ISG and DSE but without the skips.

- Pauli

This looks like a 42S solution. I count 44 bytes for a 41. Also see Werner's comment below. That got me as well in my second solution (now fixed but 2 bytes longer).

Cheers,

-Marwan

It always seemed to me that they should have been able to squeeze them into the 41.

Wow- you're right, I get so used to using ISG/DSE on integers and forget there is a seldom-used increment value on the right!

Following is a test driver based on Werner's test scenarios. It is not small but it does test all the suggested cases automatically. Since I currently have a number of FLOOR solutions in memory I have set it up so that the user enters the name of the CEIL and FLOOR functions. Run it by "XEQ FLTST" or if from within the program area "XEQ A" or simply A in user mode and then follow the prompts. You will be asked first for the CEIL FN name and then the FLOOR FN name. The calculator is already in ALPHA mode so enter the name and press R/S. Total size: 270 bytes.

LBL FLTST

LBL A

SF 21

CEIL FN

AON

AVIEW

AOFF

CLD

ASTO 01

0

ENTER^

XEQ IND 01

X<>Y?

GTO 01

1

ENTER^

1E-50

XEQ IND 01

X<>Y?

GTO 01

0

-1E-50

XEQ IND 01

X<>Y?

GTO 01

1

2E-5

XEQ IND 01

X<>Y?

GTO 01

1

ENTER^

XEQ IND 01

X<>Y?

GTO 01

-1

ENTER^

XEQ IND 01

X<>Y?

GTO 01

4

PI

XEQ IND 01

X<>Y?

GTO 01

-3

PI

CHS

XEQ IND 01

X<>Y?

GTO 01

10

ENTER^

XEQ IND 01

X<>Y?

GTO 01

-10

ENTER^

XEQ IND 01

X<>Y?

GTO 01

1E50

ENTER^

XEQ IND 01

X<>Y?

GTO 01

FLOOR FN

AON

AVIEW

AOFF

CLD

ASTO 01

0

ENTER^

XEQ IND 01

X<>Y?

GTO 01

0

1E-50

XEQ IND 01

X<>Y?

GTO 01

-1

-1E-50

XEQ IND 01

X<>Y?

GTO 01

0

2E-5

XEQ IND 01

X<>Y?

GTO 01

-1

-2E-5

XEQ IND 01

X<>Y?

GTO 01

1

ENTER^

XEQ IND 01

X<>Y?

GTO 01

-1

ENTER^

XEQ IND 01

X<>Y?

GTO 01

3

PI

XEQ IND 01

X<>Y?

GTO 01

-4

PI

CHS

XEQ IND 01

X<>Y?

GTO 01

10

ENTER^

XEQ IND 01

X<>Y?

GTO 01

-10

ENTER^

XEQ IND 01

X<>Y?

GTO 01

1E50

ENTER^

XEQ IND 01

X<>Y?

GTO 01

SUCCESS

AVIEW

RTN

LBL 01

FAIL

AVIEW

STOP

END

I hope this helps a little with testing these routines.

Cheers,

-Marwan