Puzzle on Sorting

Source: Saurabh Joshi's Blog

Problem: (copy-pasting from the source)
Suppose we have 13 cards arranged in descending order: K Q J 10 9 8 7 6 5 4 3 2 1
At any move, you can take a substring of cards and move it elsewhere. For example, 10 9 8 could be move to the right by 2 steps to get K Q J 7 6 10 9 8 5 4 3 2 1. The goal is to sort the string in increasing order minimize the number of such moves.

It must be absolutely obvious that you can do it with 12 moves but the very fact that I’m asking you this question means that you can do better. How many moves do you need?

Now the obvious extension to this problem is to generalize this technique for any n.

Awesome Problem!!

Update (Sep 07, 2010):
Solution: At Saurabh Joshi's blog (Link)

Comments

  1. NG11:57 PM, August 22, 2010

    Why not simply try to follow a divide & conquer approach ?

    Move n/2 elements from right to left end & solve these smaller n/2 & ceil(n/2) element sequences .

    Number of moves :

    Let T(n) denote number of moves using this algorithm for n numbers.
    T(1)=0
    T(n)= T(floor(n/2)) + T(ceil(n/2)) + 1

    Solving this one can find T(n) & I know this recurrence is not "too" difficult to solve- I have solved it sometime back.

    PS:As far as proving is concerned this is the best, I think we would be able to come up with an argument which would establish that sorting can be done in time less then O(nlogn). But again, I dont want to think right away.

    ReplyDelete
  2. Pratik Poddar4:35 PM, August 23, 2010

    @Nikhil.. Using the recurrence, all you get is this:
    T(n) <= cn

    which we already know. I want you to give me an algorithm, where T(n) <= n-1

    ReplyDelete
  3. Unknown5:24 PM, September 08, 2010

    K Q J 10 9 8 7 6 5 4 3 2 1
    K 6 5 4 3 2 1 Q J 10 9 8 7
    1 Q K 6 5 4 3 2 J 10 9 8 7
    1 2 J Q K 6 5 4 3 10 9 8 7
    1 2 3 10 J Q K 6 5 4 9 8 7
    1 2 3 4 9 10 J Q K 6 5 8 7
    1 2 3 4 5 8 9 10 J Q K 6 7
    1 2 3 4 5 6 7 8 9 10 J Q K


    i think this is not np hard either :D

    m fan of Nikhil Garg myslef niraj
    joined codechef and written to nikhil pls reply nikhil.

    ReplyDelete
  4. Pratik Poddar1:01 PM, September 09, 2026

    Nice explicit sequence, niraj - tracing it through, each move does relocate a contiguous block legally. The known optimum for 13 cards is lower than the naive 12; Saurabh's blog has the general argument. (And for reaching Nikhil - Codechef is a better bet than this blog's comments, agreed. :P) (replied using AI)

    ReplyDelete

Post a Comment

Popular posts from this blog

Polya's Urn Problem: Expected Balls in the Smaller Urn

Puzzle: There are two urns with one ball each. Each of subsequent n-2 balls is placed into one of these urns, with probability proportional to the number of balls already in that urn. What is the expected number of balls in the smaller sized urn? Source: P. Winkler's Puzzles book. (Chapter: Probability). Solution: Highlight the part between the * symbols for the answer. * This problem can be reformulated as the following problem. Suppose I have a stack of black cards and one red card. Initially I take red card in my hand. Now I add black cards randomly between any two cards (so, initially its either above or below red). Note that the probability that I add the card above the red card, when x-1 is the number of cards above red and y-1 is the number of cards below red is x/(x+y). Let the problem be if red card is dividing the black cards into two sets, what is the expected number of black cards in the smaller section. So, we see that the two problems are equivalent. No...
Read more

Lion and Man in a Circular Cage: Can the Lion Catch the Tamer?

Source: Asked to me by Pramod Ganapathi (PhD Student at Stony Brook University) Problem: A lion and a lion tamer are enclosed within a circular cage. If they move at the same speed but are both restricted by the cage, can the lion catch the lion tamer? (Represent the cage by a circle, and the lion and lion tamer as two point masses within it.) Solution: Highlight the part between the * symbols for the answer. * Answer: surprisingly, no - the tamer can avoid being caught forever. This is the famous Lion and Man problem, posed in the 1930s and settled by Besicovitch; the proof appears in Littlewood's "A Mathematician's Miscellany" (and opens Bollobas's "The Art of Mathematics"). What the lion CAN do (Rado's strategy): move to the center, then always stay on the radius through the tamer, using any spare speed to move outward. Since the lion runs on a smaller circle, it can match the tamer's angular position while steadily increasing its radiu...
Read more

Expected Tosses for Consecutive Heads: 2 Heads vs 3 Heads Puzzle

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. 1) Calculate P{X>Y} 2) What's the expected value of X 3) What's the expected value of Y This is probably the hardest puzzle I have ever put on my blog. Hence, I will post its solution in the post directly rather than on comment. Solution: 1) (Solved by me finally after 13 months :)) Make a state diagram. Let the state be (a,b) where a is the number of consecutive heads streak "A" is on currently and b is the number of consecutive heads streak "B" is on currently. So, (0,3) (1,3) are final accepted states and (2,0) (2,1) (2,2) (2,3) are failure states. If you get tails, your contribution to the state reaches to "0" f(State) = P(X>Y | "State" configuration initially) ...
Read more