# Fast Topswops calculation

From AZSPCS:

Suppose you have a deck containing n cards. Each card contains a number from 1 to n, and each number appears on exactly one card. You look at the number on the top card -- let's says it's k -- and then reverse the order of the top k cards. You continue this procedure -- reading the top number and then reversing the corresponding number of cards -- until the top card is 1.

Write the fastest program to compute the number of reversals for a given deck. Note that if you are participating in the contest you are not allowed to post your code (and thus I will not post my code yet).

• What is the input/output model? Any language restrictions? How will you determine how fast each entry is? – aaaaaaaaaaaa Jan 28 '11 at 2:24
• There could be a dedicated stackexchange for azspcs ;) – Eelvex Jan 30 '11 at 23:22
• So are we allowed to post solutions or not? – AShelly Feb 9 '11 at 16:25
• Yes. The contest has finished. – Alexandru Feb 13 '11 at 11:42
• The link to azspcs links to a page which is out of order. And it seems a meta-tag, which doesn't describe the puzzle. The tag should, perhaps, be removed. – user unknown May 1 '11 at 1:42

## JavaScript

function(d){for(t=0;x=(n=d[0])-1;t++)for(i=0;i<n/2;i++){m=d[x-i];d[x-i]=d[i];d[i]=m}return t}


You pass it the deck, like so:

f([3, 2, 1]) // 1
f([2, 3, 1]) // 2
f([1, 2, 3]) // 0

• So you're the winner! :) – user unknown May 1 '11 at 1:33

### Scala: (This isn't a golf - is it?)

def transform (l: List[Int], sofar: Int = 0) : Int =
if (l(0) == 1) sofar else transform (l.take (l(0)).reverse ::: l.drop (l(0)), sofar + 1)


Complete application with testcase and stopwatch, including the shuffling of the Deck:

object DeckReverse extends Application {

def transform (l: List[Int], sofar: Int = 0) : Int =
if (l(0) == 1) sofar else transform (l.take (l(0)).reverse ::: l.drop (l(0)), sofar + 1)

def stopwatch (count: Int, size: Int) = {
val li = (1 until size).toList
val r = util.Random

val start = System.currentTimeMillis ()
(0 until count).foreach (_ => transform (r.shuffle (li)))
val stop = System.currentTimeMillis ()

println ("count: " + count + "\tsize: " + size + "\tduration: " + (stop - start) + " msecs")
}

stopwatch (1000, 100)
}


count: 1000 size: 100 duration: 1614 msecs machine: Single Pentium M 2Ghz

# Python, 84 Chars

Golfing anyway... I'm using the numbers 0 through n-1. Assuming the array is stored in a variable x, it takes me 84 chars of Python.

while x[0]:x[:x[0]+1]=x[x[0]::-1]


However, performance is pretty bad due to memory abuse.

### C

int revno(int* deck, int n) {
int k,j,r=0;
for(;;) {
k=*deck;
if (k==1) {
return r;
}
for (j=0; j<k/2; j++) {
int tmp=deck[j];
deck[j]=deck[k-j];
deck[k-j]=tmp;
}
r++;
}
}


deck is a pointer to an integer array representing the decks. n is the number of the cards. Obviously memory safety is the task of the caller.

It is probably nearing the fastest algorithm on recent computers and on a high-level language. Only with asm-level tricks could it be made faster, but not heavily even with them.

# Perl 5, 58 + 2 (-ap) = 60 bytes

++$\&&(@F[0..$F[0]-1]=reverse@F[0..$F[0]-1])while$F[0]-1}{


Try it online!