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Showing posts from February, 2012

Maximum Family of Subsets Intersecting in Exactly One Element

Source: Asked to me by  Santosh Ananthakrishnan (EE IITB Fifth year undergraduate, To be Worldquant Analyst) Problem: At most, how many subsets can you find of the set A = {1, 2, ..., n} such that any two intersect in exactly one element? Solution: Highlight the part between the * symbols for the answer. * Answer: n. Construction (n subsets): {1}, {1,2}, {1,3}, ..., {1,n} - every pair intersects in exactly {1}. Upper bound: let A_1, ..., A_K be such a family and v_1, ..., v_K their 0-1 incidence vectors in R^n; the condition says v_i . v_j = 1 for i != j. The Gram matrix G (entries v_i . v_j) has |A_i| on the diagonal and 1 everywhere off it. If every |A_i| >= 2 then G = diag(|A_i| - 1) + J, a positive diagonal matrix plus a positive semidefinite one, hence invertible - so the v_i are linearly independent in an n-dimensional space and K <= n. If instead some set is a singleton, say A_1 = {x}: every other set must contain x (to meet A_1 in one element), and no two other...

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...

Algorithm Puzzle: Triplets in Array

Source: Asked to me by Anuj Jain (EE IITB 2010 Graduate, MFE Student at Baruch College NY) Problem: Given an array of n integers, find an algorithm to find triplets in the array such that sum of the three numbers is zero. What is the order of your algorithm? Make sure its quadratic in size of array. :-) Solution: Highlight the part between the * symbols for the answer. * O(n^2) algorithm: sort the array (O(n log n)). For each i from 1 to n, look for a pair (j, k) with i < j < k and a[j] + a[k] = -a[i] using two pointers: set j = i+1, k = n; if a[j] + a[k] equals the target, record the triplet and move both; if the sum is too small, j++; if too big, k--. Each pair-search is O(n) because the pointers only move toward each other, so the total is O(n^2) plus the one-time sort. Correctness: when a[j] + a[k] < target, no k' < k can work with this j (the array is sorted, sums only get smaller), so k is never needed again - and symmetrically for j - so no...

Original Problem : ATM Money Withdrawal Puzzle

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Source: One of the very few original problems on the blog. Remember ' Mathematics of Housie '? Please share if you like! Problem: Assumptions: a) I have infinite money in my account b) The daily limit of amount of money that can be withdrawn from an ATM is finite c) You can login into an ATM Machine only once a day d) If you login into the machine and enter a large number to withdraw, you will not get anything. And hence, you will not be able to withdraw anything from the ATM for the day. e) I do not know what the daily limit is. What strategy should I choose so that I can withdraw N rupees in minimum number of days? Of course, you can do it in N days (withdrawing only one rupee daily) you could do it in N-limit + ceiling(N/limit)-1 days (check N, check N-1, .. check limit. Once you know the limit, and you have already withdrawn 'limit' rupees, you will take ceiling ((N-limit)/limit) days more. Can you do better? Update: (19-07-2012) This is e...