Presentation on theme: "19.01.2014 Computer-based problems Sergey Pozdnyakov RUSSIA Workshop on problem creation."— Presentation transcript:
19.01.2014 Computer-based problems Sergey Pozdnyakov RUSSIA email@example.com Workshop on problem creation
19.01.2014 Logical problems with parameters Let's consider a class of logarithmic functions which is determined by the formula y=log a k(x+b) and a predicate which restricts the domain of parameters - a>0 & a not equal to 1. Find conditions on a, b, c for function graph neither has points inside the III quarter of coordinate plane.
19.01.2014 Program for verification (a 0 ) | (a>1 & k<0) There are wrong solutions (pic. to the left - red). Eliminate them. (a 0 & k*b 0) | (a>1 & k 1) We have missed some solutions (pic. to the right - yellow) Find all solutions
19.01.2014 Generated tasks Consider the 6-digit numbers of tram billets: 000000,… 194527,….999999. The billet is named lucky if the sum of three first digits is equal to sum of next three. For example, 194527 is lucky number. Now I get the number 123456. Find the nearest lucky number (simplification: next to number under consideration). If I get number abcdef, write the formula for next nearest lucky number.
19.01.2014 Program idea It is websudoku type problem. It is possible to find many problems of such type and use them for such named oral contests when the main parameter for answer is speed of reply. From the other side it is crossword type problem and can be used in SMS Olympyad.
19.01.2014 Not fully determined problems On the photo you see the way how to do Sugar cotton wool in Thailand. If we have caramel cylinder with H=0.3 m and D=0.01 m, how many times we need to curtail it twice and then stretch it to same length to get strings less than human hair.
19.01.2014 Internet based problems There is good possibility to use such a problem for internet-search to determine all the parameters of the problem. Other idea is to find other applied problems which has the same structure.
19.01.2014 Discrete math problems One can prove that the voting function of 3 variables is a kind of basic voting function, i.e. a voting function of n variables can be expressed only in terms of voting function of 3 variables. The problem of CIE contest was to provide such expression for a voting function of 7 variables. It is possible, as it was already said, but the proof does not provides any way to construct the expression. Let us consider another problem, it concerns Boolean functions. Voting function of n variables (for odd n) is a function, that equals 1 if and only if the number of variables equal to 1 is greater than n/2. In other words there are n voters that say 1 (pro) or 0 (contra), and if more than a half of voters says 1 (pro), then the result is 1 (i.e. pro).
Your consent to our cookies if you continue to use this website.