Everybody knows the

"12 Days of Christmas" Song (Partridge in a Pear Tree etc). How many gifts has my true love given to me altogether by the end of Day 3? Well, the sum of today's 6 gifts (3+2+1) plus yesterday's 3 gifts (2+1) plus the first day's 1 gift comes to 10 gifts in all. So the

Total

Gifts through Day 3 =

TG(3) = 10.

What is the

smallest possible (byte-count) User RPL program that calculates TG(x)?

To check your program, here are a few sample inputs and outputs:

TG(3)=10

TG(5)=35

TG(12)=364

TG(365)=8171255

N.B. This is called a

micro-challenge instead of a

mini-challenge because a REALLY SMALL solution exists. If it's obvious to you, please don't post it right away; let others try to find it first. "We only appreciate the answer if we struggle with the question." -- Ignatius O'Brien, O.P.

I still know the formula by heart, and its shorter form.

4 commands, 2 of which are single digit numbers ;-)

That would then be 20 bytes including the Rpl markers << >>

Cheers, Werner

(07-17-2017 02:53 PM)Werner Wrote: [ -> ]I still know the formula by heart, and its shorter form.

4 commands, 2 of which are single digit numbers ;-)

Cheers, Werner

I think I got the idea, but more than 4 commands ....

32.5 bytes. Only first attempt, though.

More like it: 27.5 bytes.

PS: Even more like it: 20 bytes.

Great question, Joe! Finite differences quickly gave me the formula; factoring it gave a simplified form (also 4 operations and two single digit constants). Several attempts keep giving me 9 step solutions. Do you have a shorter one? (RPN form used; LBL and RTN not counted)

(07-17-2017 04:32 PM)Gerson W. Barbosa Wrote: [ -> ]More like it: 27.5 bytes.

PS: Even more like it: 20 bytes.

I was happy with my 38 bytes solution but i have to search another way

(07-17-2017 06:52 PM)Gerald H Wrote: [ -> ]32.5 & stuck.

Just factor the expression you already have. This might give you an insight towards a shorter solution.

(07-17-2017 08:17 PM)Gerson W. Barbosa Wrote: [ -> ] (07-17-2017 06:52 PM)Gerald H Wrote: [ -> ]32.5 & stuck.

Just factor the expression you already have. This might give you an insight towards a shorter solution.

Yes but I cannot imagine how you arrive at 20 bytes ... Did you have a division in your expression ?

[EDIT] Eureka! I get it!
(07-17-2017 08:31 PM)Gilles59 Wrote: [ -> ][EDIT] Eureka! I get it!

Congratulations!

Apparently the 20-byte record cannot be broken.

Gerson.

(07-17-2017 04:54 PM)Jim Horn Wrote: [ -> ]Great question, Joe! Finite differences quickly gave me the formula; factoring it gave a simplified form (also 4 operations and two single digit constants). Several attempts keep giving me 9 step solutions. Do you have a shorter one? (RPN form used; LBL and RTN not counted)

This is an RPL challenge, but please feel free to implement it for the RPN machine of your choice.

I wonder whether the polynomial form or the shorter form would be faster on a machine that doesn't have the secret command that's needed (such as the bare-bones HP-41).

(07-17-2017 09:33 PM)Gerson W. Barbosa Wrote: [ -> ]Apparently the 20-byte record cannot be broken.

Yes, I'm sure that we're all arriving at the same 20-byte solution.

Final hint to those still hunting: the formula can be expressed this way, which should remind you of a shorter way of expressing it:

\(\frac { x(x+1)(x+2) }{ 1\cdot 2\cdot 3 } \)
7 steps (not including LBL and RTN) is the shortest I can get it so far on the 20S.

Four steps on the 34S.

Pauli

Ausgezeichnet! I had "6" in the denominator so didn't recognize the now-obvious simplification as per Paul's note. Many thanks, Joe - a clever insight on your part!

(07-17-2017 05:58 PM)Gilles59 Wrote: [ -> ] (07-17-2017 04:32 PM)Gerson W. Barbosa Wrote: [ -> ]More like it: 27.5 bytes.

PS: Even more like it: 20 bytes.

I was happy with my 38 bytes solution but i have to search another way

Now that you've found it I should tell you my next program yesterday was 38.5 bytes long.

An exotic memory-consuming solution:

« x x x IDN →DIAG x x x x »

I know there's a better equivalent of the two middle commands, but I don't remember which.

Don't anyone be misled by this one. Joe's optimum solution is simply « y y y y » :-)

All well & good saying you have a solution with so many bytes but eventually you lay your cards on the table.

Below a 32.5 Byte solution to the problem:

Code:

« DUPDUP 3 + * 2 +

* 6 /

»

A 38 bytes solution

Code:

`<< [ 1 3 2 0 ] SWAP PEVAL 6 / >>`

(07-19-2017 09:41 AM)Gerald H Wrote: [ -> ]All well & good saying you have a solution with so many bytes but eventually you lay your cards on the table.

Ok, we are far enough down the page to reveal the 20-byte solution:

« 2 + 3 COMB »
Why it works: The formula for TG(x) is \(\frac { x(x+1)(x+2) }{ 1\cdot 2\cdot 3 }\), but that looks very similar to the formula for COMB(x,3) which is \(\frac { x(x-1)(x-2) }{ 1\cdot 2\cdot 3 }\). All you have to do to turn the latter into the former is add 2 to x, and voila, TG(x) = COMB(x+2,3).

This is an example of program optimization being obtained not by code optimization but by math optimization.

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.

Oh, duh, I was doing COMB(x+2, x-1), completely forgetting about the symmetric nature of combinations.