What is my licence plate number?

Hint #1: The four-digit number is the smallest number expressible as the sum of two cubes in two different ways, that is, n = a^3 + b^3 = c^3 + d^3; a, b, c and d are positive integers;

Hint #2: n < 2000;

Hint #3: the three-letter pefix is CAB backwards :-)

My straightforward approach (not to call it 'brute-force' :-) could be used on the HP-41CV, but I have preferred to do it on my HP 50g. My RPL program is somewhat clumsy because I wrote a QBASIC program first and then converted it to RPL, but it does find n, a, b, c and d in 70 seconds on the real calculator when the input is 1999. When the input is 9999, another larger solution is found, but it takes about 10 minutes before the solutions are shown. The ten solutions less than 100,000 are found in about 9 minutes on the emulator. There is much room for improvement as no mathematical approaches have been tried-- neither could I. (I would have chosen API-3142, but it wasn't available anymore. On second thought I should have chosen something as simple as ABC-1234 :-)

Have fun!

So how long have you been driving a taxicab in a suburb of London?

Ah, you know the story! Well, most everyone here does, I think :-)

Quote:

---

Ah, you know the story! Well, most everyone here does, I think :-)

---

Ramanujan, perchance ?

Best regards from V.

This is just the kind of problem I like to try to solve in my head when I go to bed. Usually it's calculating the time to a star at various speeds and distances (or something along those lines). I may have mentioned before my (young) kids and I once tried to estimate the number of atoms in the universe and came up with 10^75. Recently I saw an estimate of 10^80 elemental particles so we actually came pretty close! This would have been more difficult just remembering all the possible cubes but I'm sure it would have put me to sleep quickly (which is the desired effect)! Anyway, just doodling I wrote down all the possible cubes and have an answer (BAC-1729) but I did cheat by using a biological computer (I will accept my disqualification in a manly fashion).

.
Hi, LHH:
.

Quote:

---

This is just the kind of problem I like to try to solve in my head when I go to bed.

---

Hehe, just in case you're too sleepy to give it a try, this short HP-71B code will find the numeric part of the answer in negligible time:

>CAT

workfile   BASIC   150 07/26/04 15:00

>LIST

10 DESTROY ALL @ INPUT "Max=";M @ R=INT(M^(1/3)) @ DIM S(2*R*R*R)

20 FOR I=1 TO R @ FOR J=I TO R @ N=I*I*I+J*J*J

30 IF S(N) THEN DISP N;"is the minimum" @ END ELSE S(N)=1

40 NEXT J @ NEXT I @ DISP "None found up to";M

>RUN

Max=1000

None found up to 1000

>RUN

Max=2000

 1729 is the minimum

It's a trivial brute-force approach and it certainly could be made faster by simply pre-computing a cubes table up to the required limit so that N=I*I*I+J*J*J would be computed instead as N=C(I)+C(J) and other such tricks but it's pretty pointless as the running time is already completely negligible.

Other trivial optimizations would include:

- taking out of the J loop the invariant computation of I*I*I or C(I) if using the cubes table

- using a string S$ of the correct length instead of a vector filling it up initially to all "0" characters, afterwards storing a "1" to mark each achieved sum in the correct position within the string after previously checking if the position is already marked so that a minimum solution has been found. This would save lots of memory as the string would take much less space than the array (which could have been declared INTEGER to save about 50% memory, by the way).

- if your HP model has the equivalent of flag or bit arrays, this would be ideal, as the S array or the S$ string are just mimicking them

etc, etc, etc, but as it's so trivial and runs instantly I didn't bother, too lazy today ... XD

Best regards from V.

Edited: 23 Aug 2012, 6:11 a.m.

The program is for the HP-42S but can be adapted easily to other models:

STO 00

LBL 00

RCL 00

3

Y^X

STO 02

1

STO 01

LBL 01

3

Y^X

RCL+ 02

STO 03

3

1/X

Y^X

IP

STO 04

RCL 00

1

+

STO 05

LBL 02

RCL 04

RCL 05

X>Y?

GTO 06

3

Y^X

STO 07

1

STO 06

LBL 03

RCL 06

3

Y^X

RCL+ 07

RCL 03

X<=Y?

GTO 04

1

STO+ 06

GTO 03

LBL 04

X#Y?

GTO 05

CLA

AIP

|-"LF"

ARCL 00

|-","

ARCL 01

|-" "

ARCL 05

|-","

ARCL 06

AVIEW

LBL 05

1

STO+ 05

GTO 02

LBL 06

RCL 00

RCL 01

1

+

STO 01

X<Y?

GTO 01

1

STO+ 00

GTO 00

As I don't have the real calculater at hand I can't tell you how fast it is to find the smallest solution. However the result is displayed immediately on my iPhone. Only for the next solutions it takes some time to consider.

Usage:

1 R/S

R/S

...

Thanks for the nice challenge: I had some fun.

Kind regards

Thomas

_{Edit: corrected usage}

Edited: 23 Aug 2012, 4:13 p.m. after one or more responses were posted

There's a nice BBC program on this number (part of their Five Numbers series):

The First Taxicab Number

Bill

Hi gerson

Here is a program for HP50 wich use GoferList Library
http://www.hpcalc.org/details.php?id=6529

«

 3 INV ^ IP -> Nm

 « 

  'n^3' 'n' 1 Nm 1 SEQ -> Li

  «

   Li 2 « DROP Li  NSUB 1 + Nm SUB ADD » DOSUBS 

   Concat DUP Nub Diff SORT  

  » 

 » 

»

Just type
9999 Plate? (or 99999 ...)

1999 or 9999 takes no time with emu48
100.000 takes 60 sec for the 10 solutions

All solutions are displayed sorted
If ypu want only the smallest, add HEAD after the SORT or << MIN >> STREAM in place of SORT

Edited: 23 Aug 2012, 10:29 a.m.

HI

10 DESTROY ALL @ INPUT "Max=";M @ R=INT(M^(1/3)) @ DIM S(2*R*R*R)

20 FOR I=1 TO R @ FOR J=I TO R @ N=I*I*I+J*J*J

30 IF S(N) THEN DISP N;"is the minimum" @ END ELSE S(N)=1

40 NEXT J @ NEXT I @ DISP "None found up to";M

Interesting. But how are sure that N is the minimum ?

or:

20 FOR I=1 TO R-1 @ FOR J=I+1 TO R @ N=I*I*I+J*J*J
?

Hi.

I used the same algorithm as you on my Commodore C128D.
In your code, you use the aray S() to "flag" with bi-cubic positions have already bee encounted.
That was my first approach idea. But, using a whole integer array to only record '0' and '1' appears to me as a too binary.
Instead of the '1' flag, I put the a value (a>0), that makes possible a more explicit display of the results n = a,b = c ,d :

LIST

10 A=0:B%=0:N=0:INPUT "Max:";M%:R%=(M%/2)^(1/3)

20 DIM P%(M%),C(2*R%)

30 FOR N=1 TO 2*R%:C(N)=N*N*N:NEXT N 

40 FOR A=1 TO R:B%=A:DO:B%=B%+1:N=C(A)+C(B%):IF N>M% THEN EXIT

50 IF P%(N) THEN PRINT N,"="A;B%"="P%(N);INT((N-P%(N))^(1/3)):ELSE P%(N)=A

60 LOOP:NEXT A:END

READY.

RUN

Max:? 32000

 1729 = 9  10 = 1  12 

 4104 = 9  15 = 2  16 

 13832 = 18  20 = 2  24 

 20663 = 19  24 = 10  27 

READY.

Edited: 23 Aug 2012, 11:40 a.m.

When I see this and other approaches here, mine looks quite silly! It could be done within two loops only once, instead I used two nested loops twice. Anyway, here it is, just for the records:

%%HP: T(3)A(R)F(,);

\<< 3, XROOT IP \-> r1

  \<< { } 1, r1

    FOR a a r1

      FOR b a SQ a * b SQ b * + +

      NEXT

    NEXT OBJ\-> \->ARRY r1 SQ r1 + 2, / 1, r1 1, - \-> n e1 s1

    \<< 1, 1, n 1, -

      FOR i GETI UNROT i r1 1, + ==

        IF

        THEN s1 'r1' STO+ 1, 'e1' STO+ 's1' 1, STO-

        END e1 r1 s1 \-> e2 r2 s2

        \<< OVER i 1, + DUP n

          FOR j GETI j r2 1, + ==

            IF

            THEN s2 'r2' STO+ 1, 'e2' STO+ 's2' 1, STO-

            END 6, PICK ==

            IF

            THEN 5, PICK 1, IQUOT \->STR "=" + e1 1, IQUOT \->STR + "^3+" + 6, PICK e1 3, ^ - 3, XROOT 1, IQUOT \->STR
              + "^3=" + e2 1,IQUOT \->STR + "^3+" + 6, PICK e2 3, ^ - 3, XROOT 1, IQUOT \->STR + "^3" + 6, ROLLD

            END

          NEXT

        \>> DROP2 ROT DROP

      NEXT DROP

    \>>

  \>> DROP

\>>

40000 TN -->  7: "1729=1^3+12^3=9^3+10^3"

              6: "4104=2^3+16^3=9^3+15^3"

              5: "13832=2^3+24^3=18^3+20^3"

              4: "39312=2^3+34^3=15^3+33^3"

              3: "32832=4^3+32^3=18^3+30^3"

              2: "40033=9^3+34^3=16^3+33^3"

              1: "20683=10^3+27^3=19^3+24^3" 

(after whopping 154 seconds on the emulator!)

Quote:

---

Anyway, just doodling I wrote down all the possible cubes and have an answer (BAC-1729) but I did cheat by using a biological computer

---

Your superior carbon-based computer is absolutely right! I ought to have tried using mine as well :-)

Hello Valentin,

Quote:

---

Ramanujan, perchance ?

---

Yes, as we are aware of:

http://en.wikipedia.org/wiki/Taxicab_number

http://mathworld.wolfram.com/TaxicabNumber.html

Best regards,

Gerson.

Gilles you use a list to organize the seed of solutions. This gives me the basements of my HP-39gII implementation.

The HP39gII is easy to program and have a really large memory. So I use two list (L0 and L1) to store n=a³+b³ and a respectively.
To discover (a b) (c d ) couples I use the convenient POS function which return the position in a list (or 0 if nothing found !).

EXPORT GERSON(m)

BEGIN

LOCAL a,b,c,d,n,r,p;

r:=3NTHROOT(m/2);

L0:={};

L1:={};

FOR a FROM 1 TO r DO

  b:=a;

  REPEAT

    b:=b+1;

    n:= a^3+b^3;

    p:= POS(L0,n);

    IF p>0 THEN

      c:=L1(p);

      d:=3 NTHROOT (n-c^3);

      PRINT(L0(p)+"="+{a,b}+"="+{c,d});

    ELSE

      L0:=CONCAT(L0,{n});

      L1:=CONCAT(L1,{a});

    END;

  UNTIL n>m ;

END;

END;

Please note the elegant and powerful NTHROOT function of this HP-39gII !

EDIT : Sorry for this confusing PRINTing :
A better solution is :

PRINT(n+"="+a+"³+"+b+"³="+c+"³+"+d+"³"); 

Edited: 23 Aug 2012, 1:38 p.m.

Hello Thomas,

Quote:

---

Thanks for the nice challenge: I had some fun.

---

I'm glad you have liked it. This is not a challenge however, this is just an exercise. I would pose a challenge only if I had a hard to beat solution, which would be very unlikely here :-)

I tried your program in my HP-42S but I can't get it to work. It's a 107-byte program. I'll double-check the listing later.

Cheers,

Gerson.

Bill, thanks for the link. Although I can listen to BBC live and some recorded programs, to that particular one I cannot. Any other link?

Gerson.

Your program running on the real hardware 39gII takes ~7.3s for an input of 100000 returning 10 results.

Was that time for emu48 running real calc speed?

TW

Edited: 23 Aug 2012, 2:06 p.m.

Hi Gerson

After adding an END to the program I get the following listing of the labels:

00 { 111-Byte Prgm }

02 LBL 00

09 LBL 01

23 LBL 02

33 LBL 03

44 LBL 04

58 LBL 05

62 LBL 06

73 END

This might help to cross-check with what you have already entered.

It appears the usage I gave was wrong. Instead of 0 start with 1. Apologies for that lapsus. However you should get:

1729

10,9 12,1

4104

15,9 16,2

13832

20,18 24,2

...

Though I've double checked the listing I might still have made some typos. Please forgive me as I'm typing all this on an iPhone which is a pain.

Then once more thanks for the exercise and for taking the time to test my solution. For me it was still a challenge. Part of it was the limitation of the available memory.

Cheers

Thomas

Just for the record: a solution for the HP-48GX without additional libraries:

\<< 3 XROOT IP \-> n

  \<< { } DUP 1 n

    FOR i 1 i

      FOR j

        i SQ i *

        j SQ j *

        + DUP2

        IF POS 0 >

        THEN

          ROT SWAP + SWAP

        ELSE

          +

        END

      NEXT

    NEXT

    DROP

  \>>

\>>

It's reasonable fast for 1999 or 9999 but gets rather slow for bigger values like 10⁵. My emulator m48 just tells me: DON'T PANIC. But I wasn't patient enough to wait for the result.

Cheers

Thomas

Hi Tim,

the algoritm is not the same but on real 50G hdw (the EMU48 is not accurate at all for 50G program) i get for all solutions :

1999 : 3.3 s

9999 : 16.3 s

99999 : 1504 s :O

So the 39gII is ~200 times faster here !. I will try to translate the CRET prog on my 50G for an accurate benchmark

I estimate in general the 39gII language 40 to 80 times faster than the UserRPL/50 depending what you do (and even far more quick with recursive program wich are incredibly slow in user RPL)

For sure the 39gII speed is impressive (and use almost no battery !)

Edited: 23 Aug 2012, 5:14 p.m.

Another way, use the Valentin Albillo algorithm but all with stack and modify to show all solutions :

« 

 3 XROOT IP DUP DUPDUP * * 2 * O {}  -> r t n s

 «

  0. t NDUPN DROP

  1. r FOR a

   a  r FOR b

    a DUPDUP * * b DUPDUP * * + DUP 'n' STO

    IF PICK THEN n 's' STO+ ELSE 1. n UNPICK END

  NEXT

 NEXT

 t DROPN s

 »

»

9999 takes 7,3 s on emulator at real speed (probably less with a real 50G cal)

99999 results 'out of memory'

Use a string instead the stack could be interesting

Edited: 23 Aug 2012, 6:07 p.m.

Shades of Ramanujan and Hardy?

Hello Thomas,

Quote:

---

It appears the usage I gave was wrong. Instead of 0 start with 1. Apologies for that lapsus. However you should get:

1729

10,9 12,1

4104

15,9 16,2

13832

20,18 24,2

...

---

The lapsus was mine, sorry! I had forgotten to insert the concatenation characters (shame on me!). Starting with 0 also works. The first solution is found after 32 seconds. The second one is shown about one minute later. Very nice!

Quote:

---

For me it was still a challenge. Part of it was the limitation of the available memory.

---

Doing it fast and with little memory is really challenging. When I said it was not a challenge I meant I didn't intend to challenge no one here, rather invite you do do a programming exercise and find better solutions than the clumsy one I had found. My naive method consisted of finding all the possible sums of two cubes and storing them in a list, a second list would store a. When this was done I would search in the first list for repeated sums and take the respective a and c in the second list; b and d would be simply calculated. Because I intended to port the program to the HP-41CV, I replaced the second array with a scheme to keep track of a and b. The QBASIC program below is easier to follow than the RPL version elsewhere, but it still looks awkward :-)

10 CLS

12 INPUT MX

15 R1 = INT(MX ^ (1 / 3))

17 DIM M((R1 * (R1 + 1)) / 2)

20 N = 0

25 FOR A = 1 TO R1

30   FOR B = A TO R1

32     N = N + 1

35     M(N) = A ^ 3 + B ^ 3

50   NEXT B

55 NEXT A

57 E1 = 1: S1 = R1 - 1

60 FOR I = 1 TO N - 1

65   T = M(I)

67   IF I = R1 + 1 THEN R1 = R1 + S1: E1 = E1 + 1: S1 = S1 - 1

68   E2 = E1: R2 = R1: S2 = S1

70   FOR J = I + 1 TO N

71   IF J = R2 + 1 THEN R2 = R2 + S2: E2 = E2 + 1: S2 = S2 - 1

72   IF M(J) <> T THEN 94

74   PRINT USING "#######"; T;

76   PRINT " = ";

78   PRINT USING "##"; E1;

80   PRINT "^3 + ";

82   PRINT USING "##"; (T - E1 ^ 3) ^ (1 / 3);

84   PRINT "^3 = ";

86   PRINT USING "##"; E2;

88   PRINT "^3 + ";

90   PRINT USING "##"; (T - E2 ^ 3) ^ (1 / 3);

92   PRINT "^3"

94   NEXT J

96 NEXT I

? 100000

   1729 =  1^3 + 12^3 =  9^3 + 10^3

   4104 =  2^3 + 16^3 =  9^3 + 15^3

  13832 =  2^3 + 24^3 = 18^3 + 20^3

  39312 =  2^3 + 34^3 = 15^3 + 33^3

  46683 =  3^3 + 36^3 = 27^3 + 30^3

  32832 =  4^3 + 32^3 = 18^3 + 30^3

  40033 =  9^3 + 34^3 = 16^3 + 33^3

  20683 = 10^3 + 27^3 = 19^3 + 24^3

  65728 = 12^3 + 40^3 = 31^3 + 33^3

  64232 = 17^3 + 39^3 = 26^3 + 36^3

The problem in using QBASIC is it much faster than our calculators, real or emulated ones. The ten solutions above are found in about three seconds in my notebook, which disguises a bad algorithm.

Cheers,

Gerson.

Yes! :-)

http://en.wikipedia.org/wiki/Taxicab_number

http://mathworld.wolfram.com/TaxicabNumber.html

I like Thomas Klemm's version.
Consise, good use of the stack and the lists.
It also work like a charm on my old HP28S !
I get the { 1729 } answer alter about 1 min on it.

I have to admit that I am a bit lost with Gille's version despite they are both written in user RPL and look to follow relatively close algorithms.

Edited: 24 Aug 2012, 4:11 a.m.

Here is a direct translation of the HP39gII version for HP28S.
You still have to found any tricks to make it run faster on HP50g.

As for the HP39gII, the two large lists are store in main (global) memory. The local variables are exactly the same { m r a b n p c d } but are only set in separate nested sub-structures:

« DUP 2 / 3 INV ^ -> m r

  « {} 'L0' STO

    {} 'L1' STO

    1 r FOR a

      a

      DO

        1 +

        DUP 3 ^ a 3 ^ +

        L0 OVER POS     -> b n p

        « IF p THEN

            'L1' p GET

            n OVER 3 ^ - 3 INV ^ -> c d

            « n a b c d 5 ->LIST DUP 1 DISP »

          ELSE

            L0 n + 'L0' STO

            L1 a + 'L1' STO

          END

          b n

        »

      UNTIL m > END

      DROP

    NEXT

    CLMF

  »

»

'GERSON' STO

2000  [GERSON]   ---->  1:{ 1729 9 10 1 12 }                in  ~1'15"

                        4:{ 1729 9 10 1 12 }

                        3:{ 4104 9 15 2 16 }

                        2:{ 13832 18 20 2 23.9999999999 }

20000 [GERSON]   ---->  1:{ 20683 19 24 10 26.9999999999 }  in  ~39'37" on real HP28S

Edited: 24 Aug 2012, 4:03 a.m.

Hi Gerson,

I couldn't get the program to play either.

I have the program in MP3 format somewhere.

Send me an email and I'll see if I can find it for you.
The whole set of BBC series, Five Numbers, Another Fire Numbers, and A Further Five Numbers are funb to listen to. No heavy math, just fun programs.

Bill

Cret, Some comments :

@ Get the cubic rooth of the max numbre

   3 INV ^ IP -> Nm 

@ Create Li : the list of all the cube < max number

   'n^3' 'n' 1 Nm 1 SEQ -> Li

@ For each element of the list

@   Get the tail of the list after the element position

@   Use ADD. 5 { 1 2 3 4 } ADD -> { 6 7 8 9 }

@   At the end of process, we get a list of list :

@   { { first add with all following}{ second add with...} ... { sum of the 2 last} }

@   The number 2 (Li 2) is to avoid an error for the last item.

@   With '2' DOSUBS works with 2 elements each time and then stop to the antepenultimate element

  Li 2 « DROP Li  NSUB 1 + Nm SUB ADD » DOSUBS

@ 'Concat' the list so that { { a b c }{ d e }} -> { a b c d e }

   Concat  @ here we have a list with _all_ the a^3+b^3 combinaison

   DUP     @ but we want only the duplicates items ! so :   

@ Delete the duplicate of one list. The initial list is kept on the stack

  Nub    @ Ex { 1 4 5 4 1 } Nub -> {5}

@ Get all the duplicates by the difference of the 2 lists 

@ (there if no command "get the duplicate of a list" in Gofer

@ but the sequance <DUP Nub Diff> does the job 

  Diff     @ Ex { 1 4 5 4 1 }  { 5 } Diff -> { 1 4 }

  SORT     @ Here we get all the solutions sorted

I use a lot of this kind of programmation with Eulers projects.

in fact with global variables and few change, it s just 3 simple lines of code and few bytes

3 XROOT IP 'Nm' STO 

1 Nm 1 Seq 3 ^ 'Li' STO

Li 2 « DROP Li  NSUB 1 + Nm SUB ADD » DOSUBS

Concat DUP Nub Diff SORT

and i think this 3 lines could work :

1 3 XROOT IP 1 Seq 3 ^ 'Li' STO

Li 2 « DROP Li  NSUB 1 + MAXR SUB ADD » DOSUBS

Concat DUP Nub Diff SORT

Edited: 24 Aug 2012, 12:46 p.m.

Quote:

---

Use a string instead the stack could be interesting

---

Not exactly a string but the HP-42S allows for bit operations. So we can squeeze 36 bits into one register which is enough to find the smallest solutions.

00 { 65 Byte Prgm }

   CLRG

   LBL 00

   RCL 00

   STO 01

05 ENTER

   X^2

   *

   STO 02

   LBL 01

10 RCL 01

   ENTER

   X^2

   *

   RCL+ 02

15 CLA

   AIP

   36

   RCL ST Y

   RCL ST Y

20 BASE/

   3

   +

   RCL IND ST X

   RDN

25 RDN

   MOD

   BIT?

   GTO 02

   +/-

30 1

   X<>Y

   ROTXY

   OR

   STO IND ST Y

35 GTO 03

   LBL 02

   AVIEW

   LBL 03

   DSE 01

40 GTO 01

   1

   STO+ 00

   GTO 00

44 END

Since Free42 allows to set the size to 9999 it's possible to find even more. But of course it's still rather limited compared to other calculators.

Cheers

Thomas

Thomas,

I tried another approach on Emu71 using no matrices at all. Is it similar to what you have used on Free42? The timing is not accurate (sometimes I get about twice the actual seconds). Thanks!

Cheers,

Gerson.

>LIST

10 DESTROY ALL @ T=TIME @ INPUT N @ R=INT(N^(1/3))

15 FOR A=1 TO R @ A3=A*A*A

20 FOR B=A+1 TO R @ B3=B*B*B

25 FOR C=A+1 TO R @ C3=C*C*C

30 FOR D=C+1 TO R @ D3=D*D*D

35 IF B3-C3=D3-A3 THEN PRINT A3+B3;A;B;C;D

40 NEXT D @ NEXT C @ NEXT B @ NEXT A @ PRINT TIME-T



>RUN

? 1000000

 1729  1  12  9  10

 4104  2  16  9  15

 13832  2  24  18  20

 39312  2  34  15  33

 704977  2  89  41  86

 46683  3  36  27  30

 216027  3  60  22  59

 32832  4  32  18  30

 110656  4  48  36  40

 314496  4  68  30  66

 216125  5  60  45  50

 439101  5  76  48  69

 110808  6  48  27  45

 373464  6  72  54  60

 593047  7  84  63  70

 149389  8  53  29  50

 262656  8  64  36  60

 885248  8  96  72  80

 40033  9  34  16  33

 195841  9  58  22  57

 20683  10  27  19  24

 513000  10  80  45  75

 805688  11  93  30  92

 65728  12  40  31  33

 134379  12  51  38  43

 886464  12  96  54  90

 515375  15  80  54  71

 64232  17  39  26  36

 171288  17  55  24  54

 443889  17  76  38  73

 320264  18  68  32  66

 165464  20  54  38  48

 920673  20  97  33  96

 842751  23  94  63  84

 525824  24  80  62  66

 955016  24  98  63  89

 994688  29  99  60  92

 327763  30  67  51  58

 558441  30  81  57  72

 513856  34  78  52  72

 984067  35  98  59  92

 402597  42  69  56  61

 1016496  47  97  66  90

 1009736  50  96  59  93

 684019  51  82  64  75

 1024.84

Hi Gerson

Quote:

---

Is it similar to what you have used on Free42?

---

I don't think so. I'm using the following conditions:

1. b < a
   
2. c > a
   
3. c³ <= a³ + b³
   

These give additional restrictions on c and d.

Here's a Python program that does (more or less) the same:

n = 20

for a in range(n):

    A = a**3

    for b in range(1, a):

        S = A + b**3

        m = trunc(pow(S, 1.0/3))

        for c in range(a+1, m+1):

            C = c**3

            d = 1

            while True:

                T = C + d**3

                if S <= T:

                    break

                d += 1

            if T == S:

                print S, a, b, c, d

    a += 1

And here the HP-42S program with some additional comments:

STO 00             ; input a

LBL 00             ; for a in ...

  RCL 00           

  3                

  Y^X              

  STO 02           ; A = a^3

  1                

  STO 01           ; b = 1

  LBL 01           ; for b in range(1, a)

    3              

    Y^X            

    RCL+ 02        

    STO 03         ; S = A + b^3

    3              

    1/X            

    Y^X            

    IP             

    STO 04         ; m = trunc(pow(S, 1.0/3))

    RCL 00         

    1              

    +              

    STO 05         ; c = a + 1

    LBL 02         ; for c in range(a+1, m+1)

      RCL 04       

      RCL 05       

      X>Y?         

      GTO 06       ; break if c > m

      3            

      Y^X          

      STO 07       ; C = c^3

      1            

      STO 06       ; d = 1

      LBL 03       ; while True

        RCL 06     

        3          

        Y^X        

        RCL+ 07    ; T = C + d^3

        RCL 03     

        X<=Y?      ; break if S <= T

        GTO 04     

        1          

        STO+ 06    ; d += 1

      GTO 03       

      LBL 04       

      X#Y?         ; next c if T # S

      GTO 05       

      CLA          ; if T == S

      AIP          

      |-"LF"       

      ARCL 00      

      |-","        

      ARCL 01      

      |-" "        

      ARCL 05      

      |-","        

      ARCL 06      

      AVIEW        ; print S, a, b, c, d

      LBL 05       

      1            

      STO+ 05      ; c += 1

    GTO 02         

    LBL 06         

    RCL 00         

    RCL 01         

    1              

    +              

    STO 01         ; b += 1

    X<Y?           

  GTO 01           

  1                

  STO+ 00          ; a += 1

GTO 00             

HTH

Thomas

Hello Thomas,

Quote:

---

Quote:

---

Is it similar to what you have used on Free42?

---

I don't think so. I'm using the following conditions:

1. b < a
   
2. c > a
   
3. c³ <= a³ + b³
   

---

Yes, the only similarity might be both algorithms don't need a large amount of memory and thus can be implemented on many programmable calculators. Also, larger numbers can be found given enough time. I estimate I would need about two or three minutes to find the first number on the HP-42S using my algorithm. Way better than my first solution but at least four times slower than yours. Well, the problem is more difficult than I initially though. No wonder it's associated with Hardy, Ramanujan and the like.

It is stated here that

1729 is the smallest number which can be represented in two different ways as the sum of two cubes:


1729 = 13 + 123 

       93 + 103


The largest known similar number is:


885623890831 = 75113 + 77303 

               87593 + 59783 

Thanks for Python code and the comment to your HP-42S program.

Gerson.