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From: stephen on 27 Mar 2010 13:13
Thank you very much, Dave and Han, for the encouragement and advice!

I've got it down to about 7.5 minutes now.

I created this little program to help me to time it:

Just put the name of the program on the stack and run this one:

TimIt:
<< TIME -> X << EVAL TIME HMS-> X HMS-> - 3600 * "dt(s)" ->TAG >> >>

Thanks,

Stephen
From: TW on 27 Mar 2010 14:58
> TimIt:
> << TIME -> X << EVAL TIME HMS-> X HMS-> - 3600 * "dt(s)" ->TAG >> >>

Good. Now look at the command TEVAL. Input is an item to
evaluate. . . :-)

TW
From: stephen on 27 Mar 2010 15:24
On 27 Mar, 18:58, TW <timwess...(a)gmail.com> wrote:
> > TimIt:
> > << TIME -> X << EVAL TIME HMS-> X HMS-> - 3600 * "dt(s)" ->TAG >> >>
>
> Good. Now look at the command TEVAL. Input is an item to
> evaluate. . . :-)
>
> TW

Oh great! I'm going to assign a user key for that one! :)

Thanks,

Stephen
From: Dave Hayden on 28 Mar 2010 11:27
Here's my solution. TEVAL on my 50g says it takes 26.3 seconds.
This program skips candidates ending in odd digits. It further
restricts
he candidates by considering only those ranges of X that can result
in X^4 beginning with a digit between 5 and 9.

« 1 1. 0.  N OK ANSWER«
@ N is the number of digits in X^4
@ OK is a temporary
@ ANSWER is the final answer

WHILE ANSWER NOT REPEAT
@ If X^4 has N digits then
@ 5x10^N < X^4 < 10^(N+1)
@ Also, it can be shown that numbers ending
@ in an odd digit always result in a small
@ value for the least significant digit (or
@ the second digit when last digit=5). Numbers
@ ending in zero are also bad.

@ Compute low bound for X
5. LOG N + 4. / ALOG CEIL
IF DUP 2. MOD THEN 1. + END

@ Compute high bound for X
N 1. + 4. / ALOG FLOOR
IF DUP 2. MOD THEN 1. - END

FOR X
X 1. DISP
X RI 4 ^ STR
1. 'OK' STO
WHILE DUP SIZE OK AND REPEAT
DUP HEAD "4" > 'OK' STO
TAIL
END
DROP
IF OK THEN
X 'ANSWER' STO
1.E499 'X' STO @ stop the loop
END
2.
STEP @ even numbers only
1. 'N' STO+
END
ANSWER
»
»
From: Dave Hayden on 28 Mar 2010 20:48
Here is a solution in HPGCC. On the 50g with TEVAL it takes 0.076
seconds.

Note that the original post specified User RPL, so this only qualifies
as an "interesting cheat." :)

Interestingly, calling sys_slowOff()/sys_slowOn() made no difference.
I suspect that most of the runtime is spent just loading the program.

Dave
---------------------

#include <hpgcc49.h>
#include "hpobjects.h"

int main (int argc, char **argv)
{
unsigned rpl_stack_bias = sat_stack_init();

int N;
long long X=0;
long long X4; /* X^4 */
int answer = 0;

for (N=1; answer == 0; ++N) {
int low, high;

low = ceil(pow((log10(5) + N) / 4, 10));
if (low % 2) ++low;

high = floor(pow( (N+1)/4.0, 10));
if (high % 2) --high;

for (X=low; X <= high; X += 2) {
char buf[50];
int i;
int len;
int ok = 1;
X4 = X*X*X*X;
sprintf(buf, "%L", X4);
len = strlen(buf);
for (i=1; ok && i < len; ++i) {
if (buf[i] < '5') ok = 0;
}
if (ok) {
// You found it!!
answer = X;
break;
}
}
}
STACKpush(REALencode((double)X, 0));
sat_stack_exit(rpl_stack_bias);
return 0;
}
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