1. ## Puzzle!!!

Hi all got a puzzle for you!

Find the next two lines of the following sequence:

11
21
12-11
11-12-21
31-22-11
13-11-22-21
11-13-21-32-11
...
... 2. They had something like this on the Google test and the python challenge...

11
21
12-11
11-12-21
31-22-11
13-11-22-21
11-13-21-32-11

31-13-12-11-13-12-21

13-21-13-11-13-31-13-11-22-11 3. Nice one :P 4. I don't get it 5. Say the code out loud and notice how what you say relates to the code before it. 6. i still don't get it 7. I finally get it!  8. TELL ME 9. If you're having trouble with this, I'll white out a hint for you all.

Read out the numbers one digit at a time:

11
21
12-11
11-12-21
31-22-11
13-11-22-21
11-13-21-32-11

One One
Two One
One Two One One
One One One Two Two One
Three One Two Two One One
One Three One One Two Two Two One
One One One Three Two One Three Two One One

Now compare each string to the previous line and look for a connection.

...and also if you're looking at Rouss' solution and not seeing a connection, note that he was slightly off. It should be:

11
21
12-11
11-12-21
31-22-11
13-11-22-21
11-13-21-32-11

31-13-12-11-13-12-21

13-21-13-11-12-31-13-11-22-11 10. I Still Don't Get It. Ive Been Reading It Over For Half An Hour  11. Want more math? If you think that the sequence is "non-mathematical", I derived this mathematical expression that gives the sequence... have fun! (D is a recursive function and t is the term number.) It's a lot easier if you think verbally, isn't it?

By the way, % here is a certain non-integer remainder function. 2.1%0.1 would be 0, 2.1%0.2 would be 0.1, 2.1%0.3 would be 0 since 0.3 fits evenly into 2.1, etc.) If you really want conventional operators, you could define % with limits and modular arithmetic...

D(t+1) = (sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,((D(t)-D(t)%10^(LOG(D(t))-
LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)
-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*
10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10
^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)
-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,
LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)
*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)
-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+1)%(((
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*
10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*
10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S
-1))%10^S))%10^(K-1))%10^K/10^(K-1)*100^(2*sigma(N=1,K,(((D(t)-D(t)%
10^(LOG(D(t))-LOG(D(t))%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-
(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%
1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)
*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))
%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))
%1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)
%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^
(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^
S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^
(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=1,
LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/
10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%
10^(S-1))%10^S))%10^(N-1))%10^N+1)%(((D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%
1)+sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*
10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^
(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-
D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-
(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(S-1))%10^S+.5)*2*
(D(t)-D(t)%10^(S-1))%10^S))-(D(t)-D(t)%10^(LOG(D(t))-LOG(D(t))%1)+
sigma(S=1,LOG(D(t))-LOG(D(t))%1,(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10
)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^
R)%10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/
10)%10^(S-1))%10^S+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,
ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%
10^(R+1)))/10)-(sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*
10-D(t)*10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))
/10)%10^(S-1))%10^S+.5)*2*(D(t)-D(t)%10^(S-1))%10^S))%10^(N-1))%10^
N+.5))))/100)+(sigma(K=1,LOG(D(t)*10)-LOG(D(t)*10)%1,100^(1+sigma(N=
1,K-1,2*((((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*
10%10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(
sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%
10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-
1)+1)%(((sigma(R=1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%
10^(R+1))%10^(R+2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)-(sigma(R=
1,LOG(D(t)*10)-LOG(D(t)*10)%1-1,ABS((D(t)*10-D(t)*10%10^(R+1))%10^(R+
2)/10-(D(t)*10-D(t)*10%10^R)%10^(R+1)))/10)%10^(N-1))%10^N/10^(N-1)+
.5)))))/10)
Erm, they say this gives you the pattern  12. Well, thanks for the crystal clear code. 13.   14. Originally Posted by h_howee  11

There are 2 1's here so we then write:

21

There is 1 2 followed by 1 1 so we write:

12-11

There is 1 1 followed by 1 2 followed by 2 1's so we write:

11-12-21

And so on and so on... 15. Nearly identical (well, same idea) to an ACM problem for this year's competition.

http://acm.fit.edu/icpc/ser2005/problems/K_number.pdf Popular pages Recent additions 