December 2015

[lavalamp]

Haha, oops. I just submitted it to them, but now realised I only used 3 operations! I wonder if that will make me look like a fool or get me extra nerd points.

[Batalov]

Well, now that they explained that in their world sin() is also an "operation", it is not all that interesting.
Then, LaurV is exactly right, and precisely one operation is enough.

[lavalamp]

I interpret their clarification to mean any standard math operations a programming language could perform natively, or perhaps functions in a standard math library. That's still a tricky challenge, however I did not use any of these in my solution.

[R. Gerbicz]

The official solution is at: https://www.research.ibm.com/haifa/p...ember2015.html

My posted solution was similar to Dan Dima's solution, my another (not submitted) solution:
let d=sum(n=0,255,G(n)/256^n)+1/(256^256), where G() is the function for the EBCDIC letters into ASCII letters transformation, so G(129)=97 etc. then for h(n)=floor(d*256^n)%256 it is easy to see that h(n)=G(n) for every 0<=n<256 what was needed. In full form:

h(n)=floor(0.000000000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000000000000000000000000000000000000000\
0000000000000000000000000000000000000000000000000000000000000000000000000002116\
0871574023119343370726678609983331272809167795795371266002282812403796908468340\
1062120093394372785917997751852761389112682180034805416796769614756139395745340\
5316874418644225900820937735649889844185731317514412265218783789287435962695391\
6769618658126318692786005267883838614428630160146494263512345061469631963959879\
0847363623041992288026861287030630015653184034834413849332424159194251604138428\
0840024700654399175511236611888422369348515979897369815022618085414673227312285\
9200860264474643110935217054506660650588083317038524670978053386313535289037283\
0940299114102533990167415637892657794585932762593581402197984330138310070498970\
1147361623418615666834026027060449459530324938506075839327252338*256^n)%256

(note: added 1/(256^256) to avoid precision problems)

In some sense this is a much better solution than the posted since we can code every f:Z_+>Z_m function using at most 4 operations.

[lavalamp]

Mine was also similar to Dan's, but I don't feel as though his is quite so nice since it uses base 10.

The solution from Yan-Wu He is pretty beautiful though. Very nice 3 operation code using numbers that will handily fit inside a long int or even a double without loss of precision.

This was mine:
f(x) = n \ 128^x % 128

Where \ is integer division and n is 493 digits (1638 bits):

8660314991803751929461825923499426172658629134394367739779900947281057156471620670172581000271320265552857681513609896807258827213701656448003504939921977203999655544729545098561245769135132614624165711458986648644217231693992750110296972675461802376918477577190333909894402187783439723363380673978509841172906208789840686903659306840369831419292555756879108408590085051899189147701931633211690537069463072665259675689054778847243664513870245828869524606392755862278941061726515154178688417792

The code for generating n:

n=0;
l=65;

for(i=193,233,

	if(i<=201 || (209<=i && i<=217) || (226<=i && i<=233),
		n += 128^i * l;
		l++
	);

);

l=97;

for(i=129,169,

	if(i<=137 || (145<=i && i<=153) || (162<=i && i<=169),
		n += 128^i * l;
		l++
	);

);

This code could also be adapted to convert any character to any other.