My RPN entry as submitted was:
HHC 2013 RPN Programming Contest entry
Eric Smith <spacewar@gmail.com>
September 21, 2013WP34S version:
01: CONST #000
02: CONST #001
03: LBL 01
04: RCL Z
05: x = 0?
06: GTO 02
07: IP
08: STO T
09: RCL* Y
10: STO+ Z
11: RDN
12: CONST #010
13: STO* T
14: CLx
15: CONST #008
16: /
18: GTO 01
18: LBL 02
19: RCL Z
HP42S and HP41: replace CONST lines with ordinary numeric entry linesHP41C: replace "RCL* Y" with "x<>y STO* Y x<>y"
Congrats Bill and Eric.
Now that the contest is over I guess we can share our results.
I selected to use the 12C because it was designed to solve this problem (indirectly of course :).
Embedded in the 12C is the following function:
To use: set CF_{0} to 0, then set CF_{j (j=1 to n number of digits)} to each octaldecimal digit, then set your base (1 then 100x), and then press NPV.
Code:
01  43 25 INTG ; LSTx = x, x = int(x) (0 or 1)
02  43 35 x=0 ; if x=0
03  43,33 05 GTO 05 ; then LSTx < 1, continue with x = 0
04  43,33 00 GTO 00 ; else end program and leave x = 1
05  43 13 CFo ; x = 0, CFo = x, n = 1
06  43 36 LSTx ; x = LSTx
07  43 35 x=0 ; if x=0
08  43,33 17 GTO 17 ; then no digits left, continue with NPV setup and NPV
09  1 1 ; x = x * 10
10  0 0 ;
11  20 x ;
12  36 ENTER ; y = x
13  43 24 FRAC ; x = value right of decimal
14  30  ; x = y  x, i.e. value left of decimal, LSTx = right of decimal
15  43 14 CFj ; CFn = int(x), n = n + 1
16  43,33 06 GTO 06 ; loop up for next digit
17  7 7 ; sto 700 in i
18  0 0 ; NOTE: i = i / 100
19  0 0 ;
20  12 i ;
21  42 13 NPV ; do it, x = answer
Thanks again Gene for the challenge. It was fun to finally use NPV.
WP34S program. Seventeen steps. Executition time three ticks for any input.
# 099 // Scaling factor
STO K // Into loop counting register
# 000 // Result
RCL Z // Stack rearrangement to get the input in X
SDL[>]K // Scale the input
// Add IP here to avoid problems for 10^100 arguments
RCL X // Begin loop
# 010 // Extract lowest digit
MOD
RCL+ T // Accumulate digit into result
# 008 // Push result one octal digit
/
x[<>] Y // Get original number back
SDR 001 // Shift it and trim final digit off
IP
DSZ K // Loop until done
BACK 010
+ // Could use x[<>] Y here but that doesn't handle 1 as input
Some results:
Input Result True Delta
0.5 0.625 0.625 0
0.75 0.953125 0.953125 0
0.1 0.125 0.125 0
0.777777777 0.9999999925494194 0.9999999925494194 0
0 0 0 0
1 1 1 0
1e95 1.60861174670876E86 1.608611746708759E86 1e101
This routine handles inputs from 1e99 through 1 plus 0. The range can be altered by changing the constant on the first line, decreasing this reduces execution time. E.g. if the lower limit is restricted to 1e16, change this constant to 16 and the execution time will be essentially zero.
Take the LBLs out of Eric's solution using BACK and SKIP and his program is one step shorter than this and a bit faster.
 Pauli
Edited: 22 Sept 2013, 5:30 p.m.
Here are mine:
WP34S:001 # 000
002 # 008
003 RCL Z
004 SDL 001
005 IP
006 STO+ Z
007 8
008 STO* T
009 STO* Z
010 Rv
011 x<> L
012 FP
013 x#0?
014 BACK 010
015 Rv
016 /
017 x>1?
018 IP
HP15C:001 f MATRIX 1
002 f LBL 0
003 8
004 STO* 1
005 x<>y
006 1
007 0
008 *
009 g INT
010 RCL/ 1
011 STO 0
012 LSTx
013 f FRAC
014 g TEST 0
015 GTO 0
016 1
017 RCL 0
018 g TEST 2
019 g INT
020 
One of the things I like about the programming challenges is that I always learn new techniques from the other participants. Even though I had the winning RPN entry this year, and used a WP34S, I wasn't familiar with the SDL, SDR, SKIP, and BACK instructions.
My program produces correct results over the domain [0 to 7.777...]
My (nonwinning) RPL entry was:
148 bytes (not including a global name)« UNROT
« { } SWAP 1 12 START
10 DUP2 MOD UNROT / IP UNROT + SWAP
NEXT DROP »
ROT OVER EVAL UNROT EVAL
ADD SWAP MOD REVLIST
« SWAP 10 * + » STREAM
»
An earlier version used a logarithm to determine how many digits to use, but I made the tradeoff of assuming 12 digits for a slightly shorter program. Bill Butler's winning program is shorter than mine ever though it does compute the digit count with a logarithm.
Edited: 22 Sept 2013, 6:33 p.m.
My RPL entry was (didn't win, I got it down to 168.5 bytes)  HP 50g
<< 'N' STO 0 UNROT
DUP2 XPON SWAP XPON
MAX 1 + 1 SWAP FOR
K DUP2 K ALOG / FP
1 ALOG * IP SWAP K
ALOG / FP 1 ALOG *
IP + N MOD K 1  ALOG
4 ROLL + UNROT NEXT
DROP2 { N K } PURGE >>
An RPL solution (HP 50g) to the RPN problem. The handling of 0 and 1 cases is left as an exercise to the reader.
%%HP: T(3)A(R)F(.);
\<< \>STR TAIL DUP SIZE SWAP "o" "# " ROT + SWAP + OBJ\> B\>R 8. ROT ^ /
\>>
(63.5 bytes)
Input Result
0.5 0.625
0.75 0.953125
0.1 0.125
0.777777777 0.9999999925494
Gerson.
Does this handle 1e50 ?
 Pauli
Only decimal numbers from 0.000000000001 through 0.777777777777, in this format only (the first zeroes may be omitted).
Gerson.
Edited: 22 Sept 2013, 11:04 p.m.
Although my stab at this uses 3 registers, it did come in at 15 steps on the WP34S. It handles values down to 1e99
CLREGS
INC 00
STO 01
x=0?
SKIP 009
IP
RCL/ 00
STO+ 02
8
STOx 00
RCL 01
FP
SDL 001
BACK 011
RCL 02
Quote:
Take the LBLs out of Eric's solution using BACK and SKIP and his program is one step shorter than this and a bit faster.
One more step can be save by replacing steps 001 and 002 with
01: ^{C}# 001
No numbered register was killed during the execution of this program, but down to 17 steps only:
001 ^{C}# 001
002 RCL Z
003 SDL 001
004 IP
005 8
006 STO* Z
007 Rv
008 t<> L
009 RCL/ Y
010 STO+ Z
011 <> TYZT
012 FP
013 x#0?
014 BACK 011
015 RCL Z
016 x>1?
017 IP
0.11037552421 > 0.14159265358
I played with a few line noniterated RPN approach which would use a built in octal to decimal conversion:
Multiply by 10^n (with n=number of digits for that calculator's precision)
convert that integer from octal to decimal
divide by 8^n
But, to my surprise, I don't see any RPN machines with base conversion except for display values only. So step 2 isn't available, even though the actual math is in many of them.
My congradulations to the winners  and all who submitted entries!
That's what I have in my HP41C with the Advantage Module right now:
01 OCTIN
02 ENTER^
03 LN
04 8
05 LN
06 /
07 1
08 +
09 INT
10 8
11 X<>Y
12 Y^X
13 /R/S 5 R/S > 0.625000000
R/S 75 R/S > 0.953125000
R/S 777777777 R/S > 0.999999993
Obviously 0 and 1 are not possible here.
Gerson.
Edited: 23 Sept 2013, 12:33 p.m.
Isn't this the basis for the five step 41 program? Using the OCT and DEC functions?
If not, I'd like to know how they were done.
 Pauli
The baton has passed. Others are exploiting the 34S function set better than I :)
Congratulations to all,
Pauli
You can have on HP41 (naked) and also on HP65 (but in 20 steps)
01 1E10
02 *
03 DEC
04 1073741824
05 /
But that's cheating as told previously
(to have 1 working, use 1E9 and 134217728)
Edited: 24 Sept 2013, 3:28 a.m. after one or more responses were posted
Thanks, it is as I thought.
 Pauli
Quote:
But that's cheating as told previously
Why? Rule #4 says:
"No custom built ROM or machine code can be built and used for this problem. Any already existing functionality in the machine is ok."
I wasn't aware of that function, but DEC (octal to decimal conversion) is part of the HP41 instruction set. So, that should be ok, except for the 1 case. Likewise, the equivalent HP42S program should be ok too:
01 1E12
02 *
03 >DEC
04 68719476736
05 /
I am always very careful with INT commands following transcendetal functions like the logarithm here. Unexpected behaviour may occur due to roundoff errors. Your program evaluates the base8 log  which may cause such a problem:
8 ENTER 7 y^{x}I do not know if this will also happen with the program you posted (my HP41 does not have an Advantage ROM, and I do not even know what exactly the OCTIN command does), but maybe you can try some integer powers of 8. If roundoff errors occur, using [LOG] instead of [LN] may give better results since lg 8 rounds much better to ten digits than ln 8.
=> 2097152
[LN]
=> 14,56609079 ; rounded down from ...79 17588 ...
8 [LN]
=> 2,079441542 ; rounded up from ...41 67983 ...
[/]
=> 6,999999998 ; so the quotient is below the true value
and INT(x) becomes 6 instead of 7
Dieter
True, but that is why at the conference I had to put up some rule clarifications...such as
No base conversion functions allowed.
Wlodek came up with the large constant similar to what you indicated, but to keep things "fair" between machines, no base conversion functions were allowed.
We had entries for these machines:
HP 65 !
HP 29C
HP 41CX
HP 42S
WP 34S
HP 15c LE (by Bill Carter!) oops.
I especially like Egan's 12c approach. Fascinating! Wish you had been at the conference this year!
Thanks again all. I hope these problems were interesting.
Edited: 25 Sept 2013, 8:23 a.m. after one or more responses were posted
Quote:
at the conference I had to put up some rule clarifications...such asNo base conversion functions allowed.
Eric Smith said in another thread: "There was another rules clarification at the conference that no base conversion functions can be used." I'd forgotten about that, sorry!
Quote:
Wlodek came up with the large constant similar to what you indicated,
That's 8^12, as Jim Horn has suggested. I kind of found the trick but I dumbly used different constants, according to the number of significant digits rather than the machine's number of digits.
Quote:
I especially like Egan's 12c approach. Fascinating!
Really fascinating, as nonconventional approaches usually are!
Quote:
Wish you had been at the conference this year!
I visited a forum member last month and I asked him about going to the next year's conference. He told me his English was not good enough to fully enjoy it. I told him not to worry much about that, but on second thought I realize I would need to go at least one month in advance to get used to spoken language. I still watch American movies with English subtitles on :)
Thanks again for the interesting problems. They have the right difficulty level, the real challenge being size (or speed) optimization.
Gerson.
OCTIN allows for the input of integer octal constants. For instance:
keystrokes display
XEQ ALPHA OCTIN ALPHA _ O
75301 ENTER^ 31425 ; 75301_{8} = 31425_{10}
However a better program (better if DEC were not natively available) would be:
01 OCTIN
02 1073741824 ; 8^10
03 /R/S 5000000000 R/S > 0.625000000
R/S 7500000000 R/S > 0.953125000
R/S 7777777770 R/S > 0.999999993
R/3 7777777777 R/S > 0.999999999
Gerson.
Edited: 24 Sept 2013, 10:25 a.m.
Hi Eric,
There may be an error on the transcription of your solution. When I try it as written in your post, the value in T is lost after 12: and the program returns always 1.
If, for example, I make the following change, then the program runs correctly :
01: c# 001 (saves one step)
02: LBL 01
03: RCL Z
04: x = 0?
05: GTO 02
06: IP
07: STO T
08: RCL* Y
09: STO+ Z
10: RDN
11: x <> T
12: SDL 001
13: STO T
14: CLx
15: CONST #008
16: /
17: GTO 01
18: LBL 02
19: RCL Z
Or perhaps there is something I am not seeing? Forgive me if am mistaken.
Miguel
CLx after STO* T is not disabling the stack lift as it should. Same here (v3.2r3382), but it does work on the emulator and on the HP42S. Probably an old wp34s bug that's already been fixed. What is the version of your wp34s?
Best regards,
Gerson
PS.: This is Eric's program optimized for the wp34s:
01: CPX CONST #001 ; CPX 1
02: RCL Z
03: x = 0?
04: SKIP 11
05: IP
06: STO T
07: RCL* Y
08: STO+ Z
09: RDN
10: CONST #010
11: STO* T
12: CLx
13: CONST #008
14: /
15: BACK 013
16: RCL Z
Edited: 24 Sept 2013, 1:57 p.m.
Quote:Indeed. Everything works fine on my hardware 34s with v. 3.2 3405.
CLx after STO* T is not disabling the stack lift as it should. Same here (v3.2r3382), but it does work on the emulator and on the HP42S. Probably an old wp34s bug that's already been fixed.
Dieter
Hello Gerson,
I am using V 3.2 3450. Physical device (not the emulator).
Regards,
Miguel
Strange, that's more recent than the emulator version (3448).
Regards,
Gerson.
P.S.: Eric Smith's solution works for arguments up to 7.77777777777
3.11037552421 > 3.14159265358
7.77777777777 > 7.99999999988
Edited: 24 Sept 2013, 3:07 p.m.
Well, I really do not get it. Could someone try it with version 3.2 3450 ? I think it is behaving correctly:
Suppose x = 0.5t:   0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0 < value lost after 10: const #10
z:  0.5 0 0 0 0 0 0 0 0 1
y:  0 1 1 1 1 1 1 1 1 0
x: 0.5 1 0.5 0.5 0.5 0 0 0 0 0 10
steps:  01 02 03 04 05 06 07 08 09 10
I do not think it is because of the CLx not disabling the stack lift. The version on post #2, works because Clx disable the stack lift
Edited: 24 Sept 2013, 3:45 p.m.
Programs using DEC or >DEC aren't going to correctly deal with really small numbers in the [0, 1] interval. Try 1e50 and see what happens.
Actually, they might be able to with some extra steps  get the exponent via LOG and IP. Then use these to calculate the required constants instead of hard coding them. I suspect this will still be shorter than the digit by digit approaches.
 Pauli
Edited: 24 Sept 2013, 4:19 p.m.
Hello Miguel
Quote:
I do not think it is because of the CLx not disabling the stack lift.
You are right. It turns out I was getting wrong results on the wp34c because my last line was RCL 2 instead of RCL Z. Sorry about the confusion. I will check it again step by step.
Gerson.
P.S.:
Quote:
The version on post #2, works because Clx disable the stack lift
I meant CLx disabled the stack lift (so that 8 replaces the 10 that was previously there).
Edited: 24 Sept 2013, 4:32 p.m.
Aren't you confusing RDN with RND? In Eric Smith's notation RDN means Rv (Roll stack down). That's the first thing I had checked previously, but I didn't detect it because I had introduced an error myself.
Regards,
Gerson.
Edited: 24 Sept 2013, 4:29 p.m.
Or you could use DROP instead of Rv in this instance. It will also be imperceptibly faster.
At one point we had a display of program checksums in the 34S firmware which would help avoid this kind of thing. I don't remember exactly why they were removed  I suspect due to opcodes changing but aren't certain anymore.
 Pauli
For the record: this one is even one step shorter.
01: CPX CONST #001Yes, I like these 34s stack shuffle commands. :)
02: RCL Z
03: x = 0?
04: SKIP 10
05: IP
06: STO T
07: RCL* Y
08: STO+ Z
09: X<> T
10: SDL 001
11: CONST #008
12: STO/ Z
13: <> ZTYY
14: BACK 012
15: RCL Z
Dieter
Quote:
001 # 000
002 # 008
These can of course be replaced with
Quote:
001 ^{C}# 008
QBASIC equivalent of the first program:
DEFDBL AZ
DEFINT D
DEF FNFRAC (NN) = NN  INT(NN)
CLS
INPUT N
P = 8
S = 0
DO
N = 10 * N
D = INT(N)
P = P * 8
S = 8 * (S + D)
N = FNFRAC(N)
LOOP UNTIL N = 0
R = S / P
IF R > 1 THEN R = 1
PRINT R? 0.11037552421
.141592653584667
Yes. I was confusing those two and confused myself :)
Thank you Gerson.
Regards,
Miguel
Quote:Or use EXPT and MANT to get the exponent and mantissa directly.
get the exponent via LOG and IP.
By the way, I noticed an inconsistency between the 34S manual and the software. On the last page of my manual (3.18 3242) it says that in version 3.0, "returned >BIN, >HEX, and >OCT." These commands were removed in v2.1. But I can't find these commands in any of the menus or in the manual. Were they removed again?
Dave
I don't remember what happened to these commands. I think we removed them because they didn't achieve anything useful  it is easier to switch integer mode and switch back when done rather than introducing extra complexity in the display code. Walter might remember better.
I'm not sure I'd call this an inconsistency. These aren't documented in the command listing and the change log is known to not be totally complete and accurate. There have been a lot of changes and refinements and not everything made the last page.
 Pauli
Gene
No love at all for the 15c entrant? I'm not sure my wife really believes my account of where I was all weekend as it is!
Bill
Edited: 25 Sept 2013, 12:05 a.m.
Quote:Glad you liked it. When I read the challenge and given your financial background I had assumed you had NPV in mind.
I especially like Egan's 12c approach.
Updated version that should support 0 through 7777777777 including fractional. I didn't have a lot of time to test or optimize it. Perhaps another day.
01  43 25 INTG
02  44 48 1 STO .1
03  43 35 x=0
04  43,33 21 GTO 21
05  43 36 LSTx
06  1 1
07  0 0
08  43 36 LSTx
09  43 23 LN
10  1 1
11  0 0
12  43 23 LN
13  10 /
14  43 25 INTG
15  1 1
16  40 +
17  44 48 1 STO .1
18  21 y^x
19  10 /
20  43 25 INTG
21  43 13 CFo
22  43 36 LSTx
23  43 35 x=0
24  43,33 33 GTO 33
25  1 1
26  0 0
27  20 x
28  36 ENTER
29  43 24 FRAC
30  30 
31  43 14 CFj
32  43,33 22 GTO 22
33  7 7
34  0 0
35  0 0
36  12 i
37  42 13 NPV
38  8 8
39  45 48 1 RCL .1
40  21 y^x
41  20 x
HP15C version as of last Sunday:
001 g x=0
002 g RTN
003 EEX
004 9
005 *
006 STO RAN#
007 g LOG
009 g INT
009 8
010 
011 8
012 x<>y
013 y^x
014 STO 1
015 g CLx
016 STO 0
017 RCL RAN#
018 f LBL 0
019 1
020 0
021 *
022 g INT
023 8
024 STO/ 1
025 x<>y
026 RCL* 1
027 STO+ 0
028 g LSTx
029 f FRAC
030 g TEST 0
031 GTO 0
032 RCL 0
7777777777 > 1073741823
1234567 > 342391.0000
3.110375524 > 3.141592651
Quote:
Yes, I like these 34s stack shuffle commands. :)
Me too, but they won't help here, it appears.
Gerson.
001 SDR 001
002 ^{C} #001
003 RCL Z
004 SDL 001
005 IP
006 t<> L
007 RCL* Y
008 STO+ Z
009 DROP
010 8
011 /
012 R^
013 FP
014 x#0?
015 BACK 011
016 RCL Z
_{Edited to correct misspelling.}
Edited: 25 Sept 2013, 2:41 p.m. after one or more responses were posted
This piece of code almost makes me wish I'd implemented the integer mode shifts in real mode too :)
 Pauli
SDL and SDR are great but binary shifts would be handy in a lot of situations as well. No more room or just a matter of design philosophy?
Gerson.
Hello,
Here is some HP Prime code for both contests.
Both are solved in 1 line of code (plus declarations) and no loops (at least that meets the eye).
note: replace SigmaList byt he appropriate symbol
export RplProgContest(Base, X, Y)
begin
SigmaList(((makelist(ip(fp(X:=X/10)*10),C,1,12)+
makelist(ip(fp(Y:=Y/10)*10),C,1,12)) mod Base)*
{1,10,100,1000,10000,100000,..,100000000000});
end;
note, you can replace the {1,10..} by makelist(10^A,A,0,11), smaller, but slower
export RpnProgContest(X)
begin
SigmaList(makelist((ip(X) mod 10)*1/8^C + (0*(X:=X*10)),C,0,12));
end;
Cyrille
A bit of both really. Flash is basically full so finding space would take lots of effort or lose something else.
We'd also have had to convince Walter that they were useful (well, Marcus and I would also need convincing too I suspect). Two of the shifts would map to *2^x and /2^x in real mode and the rest still do nothing introducing a small inconsistency.
The alternative is 2^x * or 2^x / so these would only save one step. There would only be a performance boost if the algorithm were coded to deal with integers only  either another inconsistency or more core. Both options are bad.
 Pauli
From my integer base conversion routine circa 1984, so some optimization is possible.
001 f LBL B ; define base
002 STO .0
003 g LOG
004 g INT
005 1
006 +
007 STO .1
008 RCL .0
009 g RTN010 f LBL E ; convert from decimal to defined base
011 STO 1
012 0
013 STO I
014 STO 2
015 f LBL 4
016 RCL 1
017 g INT
018 RCL/ .0
019 STO 1
020 f FRAC
021 RCL* .0
022 RCL I
023 10^x
024 *
025 STO+ 2
026 RCL .1
027 STO+ I
028 RCL 1
029 GTO 4
030 RCL 2
031 g RTN8 GSB B > 8.000000000 ; octal base
342391 GSB E > 1234567.000
1073741823 GSB E > 7777777778 ; the last digit should be a 7
25894 GSB E > 62446.0000016 GSB B > 16.00000000 ; hexadecimal base
99774 GSB E > 108051114 ; 1.08.05.11.14, that is, 185BE
257 GSB B > 257.0000000 ; base 257
12345678 GSB E > 186235169.0 ; 186.235.169_{257}
123456789 GSB E > 7070043158 ; 7.070.043.158_{257} ; the last digit should be a 7
This is subject to rounding error for large inputs, as we can see from the examples. I used it mainly for decimal to hexadecimal conversions  no problem in the short range I used. Someone ought to have told me there was something called HP16C :)
Edited: 25 Sept 2013, 6:12 p.m.
While bit shifts for reals would be nice, I personally do not see a big advantage for real life applications. With one exception: doubling or halving a number is a common operation. A dedicated command (if possible, without affecting the stack) would add very nicely to the existing INC and DEC commands. Yes, we already had that discussion some time ago, but I still consider such commands very useful for everydayprogramming. So please forgive me if I dig out this idea once again.
Dieter
Over the past couple of days I've put together a solution to the RPL problem that takes a different approach. This converts the digits to binary and then does carryfree addition with bit twiddling.
At 281 bytes, it's huge, but because there are NO LOOPS it's very fast.
For base 2 you just XOR the numbers.
For base 8, you covert the octal digits as a hex number (e.g., 77. to #77h), thus there is one extra bit to the left of each octet. Now you can add the numbers and any carry goes into the extra bit, which is then masked out.
Base 10... oh you silly base 10.... As with base 8, I convert the digits in hex mode (e.g., 94. to #94h). So each digit occupies 4 bits. The trick is to add them while subtracting 10 from the sum in such a way that the sum is never greater than 15, which would cause it to spill over to the next digit's bits. If two digits are A and B, the code says "if A is greater than 5 then subtract 5 from it and add 5 to B. If B is now greater than 10 then subtract 10 from it. Now A is in the range [0..5] and B is [0..9], so you can add them and the result is [0..14] which still fits in 4 bits. If the sum is greater than 10 then subtract 10." All of this is done with bit manipulation on all digits simultaneously.
I may write this up in a Datafile article.
Replace the ">" characters with the right arrow. Sorry, I keep forgetting now to convert a listing to ascii. :(
«
{ { R>I >STR "#" SWAP + OBJ> }
SWAP OVER EVAL UNROT EVAL } @ You now have binaries with the same digitsSWAP CASE
DUP 2. == THEN
@ This is easy. Convert to binary and XOR together
BIN DROP EVAL
XOREND
DUP 8. == THEN
HEX DROP EVAL
ADD #7777777777777777h ANDEND
@ It better be 10
HEX DROP EVALDUPDUP SL AND @ a & (a<<1)
OVER SR OR #4444444444444444h AND @ x.4 = a.8  (a.4 & a.2)
@ level 1 now contains 4 if a>5. subtract from a and add
@ to b
SWAP OVER 
UNROT +
@ b a
{ DUPDUP SL OR @ b  (b<<1)
SL OVER AND @ b.8 & (b.4  b.2)
#8888888888888888h AND
DUP SR SR OR

}
SWAP OVER EVAL @ sub10(b)
@ b Sub10 a
ROT + @ a+b Sub10
SWAP EVAL @ here is the answer, but in hexEND
@ Level 1 contains a binary number reprenting the answer.
@ convert the digits to a real.
>STR
DUP SIZE 1.  2. SWAP SUB @ take off the h
OBJ>»
That's a heck of a single line of code Cyrille!
:)
How are the renovations going?
Geoff
Quote:
export RpnProgContest(X) begin SigmaList(makelist((ip(X) mod 10)*1/8^C + (0*(X:=X*10)),C,0,12)); end;
Reminds me this can be done on the HP17BII as well, in one line of code so to say:
0*L(X:X/10)+SIGMA(C:0:11:1:0*L(T:X)+L(X:(IP(10*X)))/8^C+0*L(X:FP(10*G(T))))=D
Gerson.