I am reading Editorial about a problem on Codefoces but still not able to understand it beacuse as it is using PigeonHole principle , I am not getting how to apply pigeonhole priniciple on this problem
Here's problem Editorial:
In this problem we use the septimal number system. It is a very important limitation. Let's count how many digits are showed on the watch display and call it cnt. If cnt more than 7, the answer is clearly 0 (because of pigeonhole principle). If cnt is not greater than 7, then you can just bruteforces all cases.
Here' Problem Statement
http://codeforces.com/contest/686/problem/C
Robbers, who attacked the Gerda's cab, are very successful in covering from the kingdom police. To make the goal of catching them even harder, they use their own watches.
First, as they know that kingdom police is bad at math, robbers use the positional numeral system with base 7. Second, they divide one day in n hours, and each hour in m minutes. Personal watches of each robber are divided in two parts: first of them has the smallest possible number of places that is necessary to display any integer from 0 to n - 1, while the second has the smallest possible number of places that is necessary to display any integer from 0 to m - 1. Finally, if some value of hours or minutes can be displayed using less number of places in base 7 than this watches have, the required number of zeroes is added at the beginning of notation.
Note that to display number 0 section of the watches is required to have at least one place.
Little robber wants to know the number of moments of time (particular values of hours and minutes), such that all digits displayed on the watches are distinct. Help her calculate this number.
The "Pigeonhole Principle" is just this: if I try to put more than seven pigeons into seven pigeonholes, one of the holes has more than one pigeon in it.
Here, the septimal digits are the pigeonholes and the places in a time representation are the pigeons. So, if you can determine (quickly) for a given case that more than seven places would be needed to represent the time, the pigeonhole principle tells you there can be no representations having distinct digits, and you can skip the really slow work of counting such representations by brute-force and go on to the next case.
Related
I have been learning suffix arrays creation, & i understand that We first sort all suffixes according to first character, then according to first 2 characters, then first 4 characters and so on while the number of characters to be considered is smaller than 2n.
But my doubt is why don't we choose the first 3 characters, then 9... and so on. Why only 2 characters are taken into account since the strings are a part of same strings and not different random strings?
I haven't analyzed the suffix array construction algorithm thoroughly, but still would like to share my thoughts.
In my humble opinion, your question is similar to the following ones:
Why do computers use binary encoding of information instead of ternary?
Why does binary search bisect the range instead of trisecting it?
Why are there two sexes rather than three?
The reason is that the number 2 is special - it is the smallest plural number. The difference between 1 and 2 is qualitative, whereas the difference between 2 and 3 (as well as any other positive integer) is quantitative and therefore not as drastic.
As a result, binary formulation of many algorithms and data structures turns out to be the simplest one, though some of them may be generalized, with various degrees of added complexity, for an arbitrary base.
Answer is given from the post you linked. And as #Leon answered, the algorithm work because it use a dichotomous approach to solve the sorting problem. if you correctly read the answer, the main purpose is to divide word be small 2 character fragments. So that 4 characters can be easily sort base on the arrangement of the 2 pair of characters, 6 characters with 4-2 or 2-4 or 2-2-2 and so one. Thus have a word of 3 letters in the table is non-sense since word of 3 characters may be seen has 2 characters + the position in the alphabet of the last character.
I think you are considering only the speed of 2^x versus 3^x where you obviously would prefer the latter.
But you have to consider the effort you need for each step.
Since 3^x needs about 1.58 less steps than 2^x you would need to be able to compute a single step for the 3^x growth in less than 1.58 times what you need for a single step in the 2^x growth to perform better.
Generally the problems will get much more complex when you have to handle three elements in each step instead of two.
Also if you could expand it to 3^x you could also do it for a bigger n^x and then with big n your algorithm is suddenly not exponential but effectively linear.
I'm writing a present value calculator in java where the user enters the future amount, the inflation rate, and the term in months. I can get it to calculate correctly, but I need to display the full term including zero, so if they enter 24 months, I need to display 0 - 24 not 0 - 23. How do I account for the zero value?
Create an array with a size of 25. That will give you 0-24.
That is a good question. When you are calculating future value, you need to remember that you have a starting point that is always at time zero. And then at the end of the first time period, you have accumulated some interested, which is posted at the end of time 1 (etc) and that becomes the new beginning about. The program's array notation begins with zero, so that is an advantage for your calculation. If you are calculating with repeated iterations, then you will have to use an array size that begins with the beginning amount at time zero, and ends with time period 25. That would be a 0 to 26 array. I've done that calculation, by the way, and I don't recommend doing it in iterations that sum again and again to create a cumulative sum, because you may run into rounding errors. Therefore, what you want to do is use one of the standard formulas and calculate each period individually, according to the formula. If you use the formula correct, that will be faster and more correct. Be sure to divide the annual interest rate by the number of periods to get the per-period rate, (i.e.: monthly is rate/12) and be sure to multiply the number of periods by the number of periods per year (i.e.: Periods = years*12 if and only if your monthly rate is rate/12) There is an easier way to say that, but if you are looking at the formula and an explanation of the formula you will see what I am talking about.
I need a value as close to 0 as possible. I need to be able to divide through this value, but it should be effectively 0.
Does Java provide an easy way of generating a double with only the least significant bit set? Or do I have to calculate it myself?
//EDIT: A little background information, because someone requested it. I know that my soultion is not a particularly clean one, but here you are:
I am writing a program for homework. It calculates the resistance of a circuit consisting of multiple resistors in parallel and serial circuits.
It is a 2nd year programming class. Our teacher still designs classes for us, we need to implement them according to his design.
Parallel circuits involve calculation of 1/*resistance*, therefore my program prohibits creation of resistors with 0 Ohm. Physics tells you that this is impossible anyway (you have just a tiny little resistance in every metal).
However, the example circuit we should use to test the program contains a 0 Ohm resistor. It is placed in a serial circuit, but resistors do not know where they are (the teacher designed it that way), so I cannot change my program to allow resistors with 0 Ohm resistance in serial circuits only.
Two solutions:
Allow 0 Ohm resistors in any case - if division by 0 occurs, well, bad luck
Set the resistor not to 0, but to a resistance one can neglect.
Both are not very good. The first one seemed not too good to me, and neither did the second, but I had to decide.
It was just a random choice that threw up the problem. I could not let go without solving it, so switching to the first one was not an option anymore ;-)
Use Double.MIN_VALUE:
A constant holding the smallest positive nonzero value of type double, 2-1074. It is equal to the hexadecimal floating-point literal 0x0.0000000000001P-1022 and also equal to Double.longBitsToDouble(0x1L).
If you would like to divide by "zero" you can actually just use Double.POSITIVE_INFINITY as the result.
I'm look for the "how do you find it" because I have no idea how to approach finding the algorithm complexity of my program.
I wrote a sudoku solver using java, without efficiency in mind (I wanted to try to make it work recursively, which i succeeded with!)
Some background:
my strategy employs backtracking to determine, for a given Sudoku puzzle, whether the puzzle only has one unique solution or not. So i basically read in a given puzzle, and solve it. Once i found one solution, i'm not necessarily done, need to continue to explore for further solutions. At the end, one of three possible outcomes happens: the puzzle is not solvable at all, the puzzle has a unique solution, or the puzzle has multiple solutions.
My program reads in the puzzle coordinates from a file that has one line for each given digit, consisting of the row, column, and digit. By my own convention, the upper left square of 7 is written as 007.
Implementation:
I load the values in, from the file, and stored them in a 2-D array
I go down the array until i find a Blank (unfilled value), and set it to 1. And check for any conflicts (whether the value i entered is valid or not).
If yes, I move onto the next value.
If no, I increment the value by 1, until I find a digit that works, or if none of them work (1 through 9), I go back 1 step to the last value that I adjusted and I increment that one (using recursion).
I am done solving when all 81 elements have been filled, without conflicts.
If any solutions are found, I print them to the terminal.
Otherwise, if I try to "go back one step" on the FIRST element that I initially modified, it means that there were no solutions.
How can my programs algorithm complexity? I thought it might be linear [ O(n) ], but I am accessing the array multiple times, so i'm not sure :(
Any help is appreciated
O(n ^ m) where n is the number of possibilities for each square (i.e., 9 in classic Sudoku) and m is the number of spaces that are blank.
This can be seen by working backwards from only a single blank. If there is only one blank, then you have n possibilities that you must work through in the worst case. If there are two blanks, then you must work through n possibilities for the first blank and n possibilities for the second blank for each of the possibilities for the first blank. If there are three blanks, then you must work through n possibilities for the first blank. Each of those possibilities will yield a puzzle with two blanks that has n^2 possibilities.
This algorithm performs a depth-first search through the possible solutions. Each level of the graph represents the choices for a single square. The depth of the graph is the number of squares that need to be filled. With a branching factor of n and a depth of m, finding a solution in the graph has a worst-case performance of O(n ^ m).
In many Sudokus, there will be a few numbers that can be placed directly with a bit of thought. By placing a number in the first empty cell, you give up on a lot of opportunities to reduce the possibilities. If the first ten empty cells have lots of possibilities, you get exponential growth. I'd ask the questions:
Where in the first line can the number 1 go?
Where in the first line can the number 2 go?
...
Where in the last line can the number 9 go?
Same but with nine columns?
Same but with the nine boxes?
Which number can go into the first cell?
Which number can go into the 81st cell?
That's 324 questions. If any question has exactly one answer, you pick that answer. If any question has no answer at all, you backtrack. If every question has two or more answers, you pick a question with the minimal number of answers.
You may get exponential growth, but only for problems that are really hard.
I’m working on a problem where its demanded to get the last 100 digits of 2^1001. The solution must be in java and without using BigInteger, only with int or Long. For the moment I think to create a class that handle 100 digits. So my question is, if there is another way the handle the overflow using int or long to get a number with 100 digits.
Thanks for all.
EDITED: I was off by a couple powers of 10 in my modulo operator (oops)
The last 100 digits of 2^1001 is the number 2^1001 (mod 10^100).
Note that 2^1001 (mod 10^100) = 2*(2^1000(mod 10^100)) (mod 10^100).
Look into properties of modulo: http://www.math.okstate.edu/~wrightd/crypt/lecnotes/node17.html
This is 99% math problem, 1% programming problem. :)
With this approach, though, you wouldn't be able to use just an integer, as 10^100 won't fit in an int or long.
However, you could use this to find, say, the last 10 digits of 2^1001, then use a separate routine to find the next 10 digits, etc, where each routine uses this feature...
The fastest way to do it (coding wise) is to create an array of 100 integers and then have a function which starts from the ones place and doubles every digit. Make sure to take the modulus of 10 and carry over 1s if needed. For the 100th digit, simply eliminate the carry over. Do it 1001 times and you're done.