Catch me if you can?

Source: Asked to me by Ram Kumar, a frequent visitor to this blog.

Problem:
There's a duck swimming in a circular lake with a wolf at the edge. wolf won't enter water. Duck in water is slower than Wolf on land. Duck on land is faster than the Wolf on land. Given the lake diameter 'd' and wolf speed 'w', what is the minimum speed of the duck in water so that he can escape? what should be his escape strategy?

Hint:
1) Its less than w/4.
2) There is pi involved. :)

Solution:
Answer given by Vivek Jha (EE, IITB) in the comments. Proof of the solution given by me in the comments of one of the earlier posts: http://pratikpoddarcse.blogspot.com/2009/11/polyas-urn-problem.html?showComment=1258400372028#c8784580092886919079

Comments

  1. stp7:04 AM, December 12, 2009

    i did not get the meaning?
    as the wolf wont enter the water,it can never catch the duck unless it comes out of water.

    ReplyDelete
  2. Pratik Poddar8:34 AM, December 12, 2009

    its written "duck needs to escape". It has to get out of water and run :)

    ReplyDelete
  3. stp9:09 AM, December 14, 2009

    the duck comes to the center of the lake.Then it oves in a direction diametrically opposite to that of wolf.
    the min required speed of duck is w/2*pi.

    ReplyDelete
  4. Pratik Poddar4:47 AM, December 16, 2009

    @teja.. you need to do better.. I said that the answer is less than w/4.

    ReplyDelete
  5. Vivek Jha10:58 PM, January 06, 2010

    Its w/(pi+1).....I think.

    ReplyDelete
  6. Pratik Poddar1:13 AM, January 07, 2010

    @Vivek..
    Its correct.. Thank You. Proof given by me in comments of http://pratikpoddarcse.blogspot.com/2009/11/polyas-urn-problem.html?showComment=1258400372028#c8784580092886919079

    ReplyDelete
  7. Vivek Jha10:37 AM, January 07, 2010

    Same proof here.....btw nice solution for that urn problem....nice discussion on the thread also.

    ReplyDelete
  8. Unknown3:22 PM, June 10, 2014

    This comment has been removed by the author.

    ReplyDelete
  9. Unknown3:29 PM, June 10, 2014

    I believe, that the answer is 'slightly greater than zero'.the strategy for the duck is to move perpendicular to the velocity of the wolf at all times,such that his distance from land decreases (assuming initial distance of the duck from wolf is >=d/2,which can be easily achieved).As radial relative velocity is never changing,the distance between duck and wolf will remain same.duck will follow a spiral path till it reaches the land.
    Comment if you find anything wrong with this strategy.

    ReplyDelete
  10. Pratik Poddar1:06 PM, September 08, 2026

    Thanks Vivek - your solution for this one was spot on. That urn thread turned into a good discussion indeed. (replied using AI)

    ReplyDelete
  11. Pratik Poddar1:06 PM, September 08, 2026

    There is a problem with the strategy: "move perpendicular to the wolf's velocity" does not control the wolf's ANGULAR position around the lake. If the duck is slower than w/(pi+1), the wolf can mirror it - keep itself on the same radial line as the duck - and then beat the duck to whatever landing point it chooses. The sharp answer is v > w/(1 + pi): while the duck stays within radius R*v/w of the centre, its angular speed exceeds the wolf's, so it can maneuver itself to be diametrically opposite the wolf; then it swims straight to shore, a distance R(1 - v/w), while the wolf must run half the circumference, pi*R. The duck escapes iff R(1 - v/w)/v < pi*R/w, i.e. v > w/(1+pi) - which is indeed a bit less than w/4. (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