About

About CSE Blog

CSE Blog is a collection of around 300 mathematics, quantitative aptitude and computer science puzzles for interview preparation and brain teasing.

Started in 2009, the archive covers probability, algorithms, number theory, logic, strategy and programming. Posts are arranged by difficulty and include solutions or reader credits where available.

About the author

I am Pratik Poddar, an IIT Bombay computer science alumnus and technology investor. I write and curate CSE Blog in my personal capacity. The views here are my own.

How to use the site

Start with Easy Puzzles, then Medium Puzzles and Difficult Puzzles. Try each problem before revealing the solution.

Privacy Policy

CSE Blog uses Blogger, Google Analytics and Google AdSense. These services may use cookies, device identifiers and similar technologies to operate the site, measure traffic and serve advertising.

Third-party vendors, including Google, use cookies to serve ads based on prior visits. Google's advertising cookies enable Google and its partners to serve personalised ads.

Opt out at Google Ads Settings or review other choices at AboutAds.

Traffic information may be processed for analytics, security, advertising measurement and service improvement. CSE Blog does not sell personal information. Email addresses and messages are used only to respond or manage submissions, attribution and corrections. You can control cookies in your browser.

Disclaimer

This is a personal weblog. The views expressed here are the author's own and not those of his employer or affiliated organisations.

Information is provided for educational purposes. Although the site is prepared with care, no representation is made about its accuracy, completeness, currency, suitability or reliability. The author is not liable for errors, omissions, delays or losses arising from its use.

Content and this policy may change without prior notice.

Contact

For privacy questions, corrections or attribution requests, email pratik.phodu@gmail.com.

Comments

Post a Comment

Popular posts from this blog

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

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

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