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+--- Thread: RPL Micro-Challenge: Christmas in July (/thread-8692.html)
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RE: RPL Micro-Challenge: Christmas in July - Gerson W. Barbosa - 07-19-2017 02:07 PM
(07-19-2017 01:42 PM)Dave Britten Wrote: Oh, duh, I was doing COMB(x+2, x-1), completely forgetting about the symmetric nature of combinations.
I did that as well.
(x+2)!(x-1)!/6. (37.5 bytes)
COMB(x+2, x-1) (27.5 bytes)
COMB(x+2, 3) (20 bytes)
PS: Next, this "masterpiece" :-)
« DUP 2. + IDN →DIAG PCOEF SWAP GET ABS
» (38.5 bytes)
RE: RPL Micro-Challenge: Christmas in July - rprosperi - 07-19-2017 03:24 PM
(07-19-2017 12:56 PM)Joe Horn Wrote: This is an example of program optimization being obtained not by code optimization but by math optimization.
Clever, and good eye to recognize the pattern, which I did not. Though, I have become re-acquainted with the AUR as a result of this challenge.
I was looking for some form of the PI (Product) function, since this is also PI, for i, 0 to 2 of (x + i) / (1 + i), though even if I found it, the formula still needed more bytes than 20.
Still, educational and fun, so thanks for that Joe.
For those interested, John Meyer wrote-up a very interesting PI solution, here.
RE: RPL Micro-Challenge: Christmas in July - Gerald H - 07-19-2017 03:28 PM
If you're interested in the series much info here:
https://oeis.org/A000292
RE: RPL Micro-Challenge: Christmas in July - Claudio L. - 07-19-2017 03:51 PM
(07-19-2017 02:07 PM)Gerson W. Barbosa Wrote: PS: Next, this "masterpiece" :-)
« DUP 2. + IDN →DIAG PCOEF SWAP GET ABS
» (38.5 bytes)
This is what I had:
Code:
« [ 1 3 2 0 ] SWAP PEVAL 6. / »Code:
« { 0 1 2 } ADD ΠLIST 6. / »Needs to be compiled in exact mode (no dot in the integer vector) and you get 38 bytes and 38.5 bytes respectively.
RE: RPL Micro-Challenge: Christmas in July - Gerson W. Barbosa - 07-19-2017 04:14 PM
82 bytes and counting :-)
« → n
« '∑(j=1,n,∑(k=1,j,k))' EVAL
»
»
RE: RPL Micro-Challenge: Christmas in July - Gerson W. Barbosa - 07-19-2017 04:37 PM
(07-19-2017 12:56 PM)Joe Horn Wrote: EDIT: I just noticed something cool. If you run the program above on an input of 'X' (undefined), and then EVAL the resulting mess, you get this:
\[\frac { { X }^{ 3 }+3{ X }^{ 2 }+2X }{ 6 }\]
Now press FACTOR (or COLLECT) and see this:
\[\frac { X\cdot (X+1)\cdot (X+2) }{ 3\cdot 2 }\]
It almost does all the thinking for you.
It works with most programs:
X « 1 + DUP ACOSH SINH SQ * 6 / » EVAL LIN FACTOR --> 'X*(X+1)*(X+2)/(3*2)'
RE: RPL Micro-Challenge: Christmas in July - Gilles59 - 07-19-2017 06:22 PM
(07-19-2017 04:14 PM)Gerson W. Barbosa Wrote:Code:
∑(j=1,n,∑(k=1,j,k))
It was what I did in the equation writer to find the expression.
After evalution, just do FACTOR ;D