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Berger game
03-14-2016, 11:58 AM
Post: #1
Berger game
Hi,
In my studiying programs in library WP34s, in the Berger program, I think is mistake. But I am with prudency writting because it is a Marcus Von Cube's program.
It is not possible write "JMP". But it is said " Using a label and GTO doesn't save noticable time". Then me I puted "back" (as Dieter was said) instructions and I found answer is 2592.
2^5*9^2 = 2592.
I think it the alone solution.

I am wonder why this name, is it like sheapard ? In french a berger is a sheapard.

Am-I right in the number ?

Gérard.
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03-16-2016, 05:51 PM
Post: #2
RE: Berger game
(03-14-2016 11:58 AM)ggauny@live.fr Wrote:  In my studiying programs in library WP34s, in the Berger program, I think is mistake. But I am with prudency writting because it is a Marcus Von Cube's program.

I don't remember this being my program. Nevertheless, You are always free to critizise anybody's work if you find mistakes or bugs or have impovements.

The JMP instruction is translated by the PC based assembler into GTO, SKIP or BACK, whatever is needed. You cannot use it directly on the machine. In most cases it saves a label. It is most useful for library routines which are not meant to edited on the machine itself but on the PC and then recompiled and sent to the emulator and/or calculator.

Marcus von Cube
Wehrheim, Germany
http://www.mvcsys.de
http://wp34s.sf.net
http://mvcsys.de/doc/basic-compare.html
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03-16-2016, 07:42 PM
Post: #3
RE: Berger game
Marcus, I think the OP is referring to a thread I did a few years ago. Find a, b, c, and d such that a^b x c^d = abcd. I think the only solution was 2592 (2^5 * 9^2). It's rather easy to find using brute force.

Don
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03-16-2016, 11:04 PM
Post: #4
RE: Berger game
(03-16-2016 07:42 PM)Don Shepherd Wrote:  Marcus, I think the OP is referring to a thread I did a few years ago. Find a, b, c, and d such that a^b x c^d = abcd. I think the only solution was 2592 (2^5 * 9^2). It's rather easy to find using brute force.

Don

Are a, b, c, and d supposed to be single digit numbers? Or is something like 10^01*02^11 also permitted?
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03-16-2016, 11:08 PM (This post was last modified: 03-16-2016 11:16 PM by Don Shepherd.)
Post: #5
RE: Berger game
(03-16-2016 11:04 PM)Harald Wrote:  
(03-16-2016 07:42 PM)Don Shepherd Wrote:  Marcus, I think the OP is referring to a thread I did a few years ago. Find a, b, c, and d such that a^b x c^d = abcd. I think the only solution was 2592 (2^5 * 9^2). It's rather easy to find using brute force.

Don

Are a, b, c, and d supposed to be single digit numbers? Or is something like 10^01*02^11 also permitted?

I had in mind single digits for a, b, c, and d when I posed the challenge.
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03-16-2016, 11:20 PM
Post: #6
RE: Berger game
(03-16-2016 11:08 PM)Don Shepherd Wrote:  
(03-16-2016 11:04 PM)Harald Wrote:  Are a, b, c, and d supposed to be single digit numbers? Or is something like 10^01*02^11 also permitted?

I had in mind single digits for a, b, c, and d when I posed the challenge.

Ah, I see, reading that would have helped ;-) "4 digit number" Smile
Thanks!
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03-17-2016, 07:28 AM
Post: #7
RE: Berger game
Hi Marcus,
It was not a criticisum of course. Only a simple question. It is for that reason that I was impossible write "JMP" instruction.
And I was thinking you have do this programme it is because your name is
in library in front this program.

Thanks for all for respons and bravo to mister Shepherd.

Gérard.
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