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(33s) OEIS A4186: Digits in Descending Order - Gerald H - 09-22-2022 03:33 AM
Takes a natural number from the stack & returns A4186(N) to stack.
Digits of index decreasing.
eg Input
10477
returns
77410.
I use [i] to represent the built in sign for use of i in indirect addressing to avoid confusion in say
IP(LOG((i)))
or
5*(i).
Suggestions for improvements most welcome.
https://oeis.org/A004186
Preserves stack.
Code:
1. LBL D
2. CLVARS
3. STO K
1. LBL F
2. x=0?
3. GTO G
4. CLx
5. RMDR(K:10)+1
6. STO i
7. SGN
8. STO+ [i]
9. CLx
10. IDIV(K:10)
11. STO K
12. GTO F
1. LBL G
2. CLx
3. 10
4. STO i
5. CLx
6. STO T
1. LBL H
2. CLx
3. RCL [i]
4. x=0?
5. GTO J
1. LBL I
2. CLx
3. 10* T+i-1
4. STO T
5. DSE [i]
6. GTO I
1. LBL J
2. DSE i
3. GTO H
4. CLx
5. RCL T
6. RTN
D: LN = 9 H: LN = 15
F: LN = 58 I: LN = 26
G: LN = 30 J: LN = 18RE: (33s) OEIS A4186: Digits in Descending Order - C.Ret - 09-24-2022 07:41 AM
Hello,
Since I still have no HP-35S, here is a version for an HP-41C without any extra RAM or ROM module:
Code:
001 LBL "4186 // real integer input N in register X:
002 CLRG
003 LBL 01 // LOOP1: strip integer N digit by digit
004 RCL X 10 ST/ Z MOD X<>Y ISG 00
010 LBL 02 // LOOP2: sort digits from register R10 to R01
011 X<> IND L X<=Y? X<>Y X<> IND L DSE L GTO 02 // R10:highest digit R01:smallest digits (or zero)
017 INT X>0? GTO 01 // loop until no more digit in N
020 10 X<>Y
022 LBL 03 // LOOP3: build-up result from digits in R10 down to R0#
023 10 * RCl IND Y + DSE Y DEG DSE 00 GTO 03
031 END // Result in X: N'= jih...dcba[ ][RTN]
1234 [R/S] ... goose ... goose ... goose ... 4321.
10477 [R/S] ... goose ... goose ... goose ... 77410.
This is a simplified version from a code published as one of the numerous solutions of the code challenge Re: Misez p'tit, Optimisez - N°85 (L'Algorithme de Kaprekar) on the French Forum Silicium (20 août 2018).
A close related code may produce both decreasing and increasing digits sorted integers from the input:
Code:
X: N=hdebcijadf
001 LBL "4186B 011 X<> IND L 021 SIGN 031 X<> IND L
002 CLRG 012 X<=Y? 022 X<> Z 032 ST+ Y
003 LBL 01 013 X<>Y 023 LBL 03 033 RDN
004 RCL X 014 X<> IND L 024 10 034 DSE L
005 10 015 DSE L 025 ST* Z 035 DEG
006 ST/ Z 016 GTO 02 026 X<> T 036 DSE 00
007 MOD 017 INT 027 ST* T 037 GTO 03
008 X<>Y 018 X>0? 028 X<> IND L 038 END
009 ISG 00 019 GTO 01 029 ST+ Z
010 LBL 02 020 10 030 ST* IND L X: abcdefghij Y: jihgfedcba[ ][RTN]
1342 [R/S] ... goose ... goose ... goose ... 1234. [X<>Y] 4321.
10477 [R/S] ... goose ... goose ... goose ... 1477. [X<>Y] 77410.
RE: (33s) OEIS A4186: Digits in Descending Order - Gerald H - 09-24-2022 11:32 AM
Interesting stuff, Ret, thank you.
The last programme you mention works nicely on a 42S - but I don't see the point of line 35?
RE: (33s) OEIS A4186: Digits in Descending Order - Gerald H - 09-24-2022 11:56 AM
OK, I've now replaced DEG with RAD & it works fine - usage clarified.
RE: (33s) OEIS A4186: Digits in Descending Order - C.Ret - 09-24-2022 12:44 PM
Yes, everyone can put whatever they want in step 035 as long as it can be used as a NOP.
Maybe I should have left the original LBL 04 which more clearly indicates a non-operation.
This also applies to step 028 of the first program.
RE: (33s) OEIS A4186: Digits in Descending Order - C.Ret - 09-25-2022 04:18 PM
Hello,
This weekend, I found a way to rent a friend's HP-35S along with its owner's manual. What a funny machine!
It is with great difficulty that I managed to adapt my second algorithm to it, the one that gives the integers with the numbers sorted in ascending and descending order.
Be indulgent, this is my first program on HP-35S; it is it is intended to be used in RPN mode and with the ALL display option.
Code:
O001 ►LBL O O011 x<>y O021 1 O031 × O041 1 O051▫ STO I
O002 CLVARS O012▪ x<>(i) O022 STO+ J O032 RCL+(i) O042 STO- I O052▫ 2
O003 STO N O013 FS? 1 O023 10¨x O033 x<>y O043 10¨x O053▫ -
O004 13 O014 CLx O024 INT÷ O034 STO×(i) O044 × O054▫ RTN
O005 STO I O015 x<y? O025 x>0? O035 R↑ O045 DSE J
O006 STO(I) O016 x<>y O026 GTO O009 O036 RCL+(i) O046 GTO O029
O007 SF 1 O017 x<>(i) O027▫ XEQ O049 O037 x<>y O047 R↓
O008 RCL N O018 DSE I O028 SGN O038 R↓ O048 RTN ■
O009▫ XEQ O049 O019 GTO O012 O029▪ x<>y O039 x<>y O049▪ REGX
O010 RMDR O020 CF 1 O030 10 O040 R↑ O050▫ 12
LBL O
CK=A4D8
LN=175It's a bit long code but it has the advantage of using only one label (LBL O). Unlike the HP-41C version, it handles integers up to 12 digits which it temporarily stores in twelve anonymous registers using indirect addressing via the I register.
It also uses flag #1 because I couldn't find a simple instruction to reset or reserve the anonymous register range. The flag is used to reset the anonymous registers on the first pass. At the beginning of the code, I put 13 in the anonymous register (13) in order to 'reserve' without surprise the range (1) to (12).
It's a bit of a hassle on this machine, isn't it?
The keyboard is far too overloaded, I prefer, by far, to use my more sober HP-41C and with better organized registers.
3142 XEQ O ENTER
run a few seconds and display
1234
4321
The initial value 3142 is kept in the register N
This version clears all the registers. To preserve memory contents, please alter the code with the starter sequence:
Code:
O001 LBL O
O002 STO N
O003 13
O004 STO I
O005 STO(I)
O006 CLx
O007 STO J
O008 SF 1
...