I am not good at probability and I know it's not a coding problem directly. But I wish you would help me with this. While I was solving a computation problem I found this difficulty:
Problem definition:
The Little Elephant from the Zoo of Lviv is going to the Birthday
Party of the Big Hippo tomorrow. Now he wants to prepare a gift for
the Big Hippo. He has N balloons, numbered from 1 to N. The i-th
balloon has the color Ci and it costs Pi dollars. The gift for the Big
Hippo will be any subset (chosen randomly, possibly empty) of the
balloons such that the number of different colors in that subset is at
least M. Help Little Elephant to find the expected cost of the gift.
Input
The first line of the input contains a single integer T - the number
of test cases. T test cases follow. The first line of each test case
contains a pair of integers N and M. The next N lines contain N pairs
of integers Ci and Pi, one pair per line.
Output
In T lines print T real numbers - the answers for the corresponding test cases. Your answer will considered correct if it has at most 10^-6 absolute or relative error.
Example
Input:
2
2 2
1 4
2 7
2 1
1 4
2 7
Output:
11.000000000
7.333333333
So, Here I don't understand why the expected cost of the gift for the second case is 7.333333333, because the expected cost equals Summation[xP(x)] and according to this formula it should be 33/2?
Yes, it is a codechef question. But, I am not asking for the solution or the algorithm( because if I take the algo from other than it would not increase my coding potentiality). I just don't understand their example. And hence, I am not being able to start thinking about the algo.
Please help. Thanks in advance!
There are three possible choices, 1, 2, 1+2, with costs 4, 7 and 11. Each is equally likely, so the expected cost is (4 + 7 + 11) / 3 = 22 / 3 = 7.33333.
My computer uses ASCII, American Standard Code for Information Exchange.
It is my understanding that this uses a 7 bit code to represent all the letters, symbols, and numbers needed for the english language.
It is my understanding that these 7 bits can be represented with hexadecimal codes.
I thought that hexadecimal needed 8 bits. 4 bits per number.
Can some one explain to me how the hexadecimal system can be used to represent the codes in the 7 bit ASCII system.
Thanks in advance.
Hexadecimal numbers don't need eight bits, each hex digit can represent four bits but there's no upper limit, since you can just use more digits:
0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffffff...
For representing seven-bit values, you can just use the lower half of the eight-bit hex numbers, 0x00 through 0x7f.
That gives you the binary numbers 0000000 through 1111111.
With 7 bits you can represent every number from 0 to 2^7 = 127 (decimal) = 7f (hexadecimal).
Hexadecimal doesn't need 7 bits, it is just another way to write numbers.
You can combine any combination of up to 4 bits into one hexadecimal character.
1 bit: 0 to 1
2 bits: 0 to 3
3 bits: 0 to 7
4 bits: 0 to F
For 7 bits you need 2 hex digit, one coding 3 bits and the other 4 bits so you get a code of 00 to 7F.
Or you use 8 bits, but the most significant bit is always 0.
This is more of a maths question than programming but I figure a lot of people here are pretty good at maths! :)
My question is: Given a 9 x 9 grid (81 cells) that must contain the numbers 1 to 9 each exactly 9 times, how many different grids can be produced. The order of the numbers doesn't matter, for example the first row could contain nine 1's etc. This is related to Sudoku and we know the number of valid Sudoku grids is 6.67×10^21, so since my problem isn't constrained like Sudoku by having to have each of the 9 numbers in each row, column and box then the answer should be greater than 6.67×10^21.
My first thought was that the answer is 81! however on further reflection this assumes that the 81 numbers possible for each cell are different, distinct number. They are not, there are 81 possible numbers for each cell but only 9 possible different numbers.
My next thought was then that each of the cells in the first row can be any number between 1 and 9. If by chance the first row happened to be all the same number, say all 1s, then each cell in the second row could only have 8 possibilites, 2-9. If this continued down until the last row then number of different permutations could be calculated by 9^2 * 8^2 * 7^2 ..... * 1^2. However this doesn't work if each row doesn't contain 9 of the same number.
It's been quite a while since I studied this stuff and I can't think of a way to work it out, I'd appreciate any help anyone can offer.
Imagine taking 81 blank slips of paper and writing a number from 1 to 9 on each slip (nine of each number). Shuffle the deck, and start placing the slips on the 9x9 grid.
You'd be able to create 81! different patterns if you considered each slip to be unique.
But instead you want to consider all the 1's to be equivalent.
For any particular configuration, how many times will that configuration be repeated
due to the 1's all being equivalent? The answer is 9!, the number of ways you can permute the nine slips with 1 written on them.
So that cuts the total number of permutations down to 81!/9!. (You divide by the number of indistinguishable permutations. Instead of 9! indistinguishable permutations, imagine there were just 2 indistinguishable permutations. You would divide the count by 2, right? So the rule is, you divide by the number of indistinguishable permutations.)
Ah, but you also want the 2's to be equivalent, and the 3's, and so forth.
By the same reasoning, that cuts down the number of permutations to
81!/(9!)^9
By Stirling's approximation, that is roughly 5.8 * 10^70.
First, let's start with 81 numbers, 1 through 81. The number of permutations for that is 81P81, or 81!. Simple enough.
However, we have nine 1s, which can be arranged in 9! indistinguishable permutations. Same with 2, 3, etc.
So what we have is the total number of board permutations divided by all the indistinguishable permutations of all numbers, or 81! / (9! ** 9).
>>> reduce(operator.mul, range(1,82))/(reduce(operator.mul, range(1, 10))**9)
53130688706387569792052442448845648519471103327391407016237760000000000L
I'm trying to learn C and have come across the inability to work with REALLY big numbers (i.e., 100 digits, 1000 digits, etc.). I am aware that there exist libraries to do this, but I want to attempt to implement it myself.
I just want to know if anyone has or can provide a very detailed, dumbed down explanation of arbitrary-precision arithmetic.
It's all a matter of adequate storage and algorithms to treat numbers as smaller parts. Let's assume you have a compiler in which an int can only be 0 through 99 and you want to handle numbers up to 999999 (we'll only worry about positive numbers here to keep it simple).
You do that by giving each number three ints and using the same rules you (should have) learned back in primary school for addition, subtraction and the other basic operations.
In an arbitrary precision library, there's no fixed limit on the number of base types used to represent our numbers, just whatever memory can hold.
Addition for example: 123456 + 78:
12 34 56
78
-- -- --
12 35 34
Working from the least significant end:
initial carry = 0.
56 + 78 + 0 carry = 134 = 34 with 1 carry
34 + 00 + 1 carry = 35 = 35 with 0 carry
12 + 00 + 0 carry = 12 = 12 with 0 carry
This is, in fact, how addition generally works at the bit level inside your CPU.
Subtraction is similar (using subtraction of the base type and borrow instead of carry), multiplication can be done with repeated additions (very slow) or cross-products (faster) and division is trickier but can be done by shifting and subtraction of the numbers involved (the long division you would have learned as a kid).
I've actually written libraries to do this sort of stuff using the maximum powers of ten that can be fit into an integer when squared (to prevent overflow when multiplying two ints together, such as a 16-bit int being limited to 0 through 99 to generate 9,801 (<32,768) when squared, or 32-bit int using 0 through 9,999 to generate 99,980,001 (<2,147,483,648)) which greatly eased the algorithms.
Some tricks to watch out for.
1/ When adding or multiplying numbers, pre-allocate the maximum space needed then reduce later if you find it's too much. For example, adding two 100-"digit" (where digit is an int) numbers will never give you more than 101 digits. Multiply a 12-digit number by a 3 digit number will never generate more than 15 digits (add the digit counts).
2/ For added speed, normalise (reduce the storage required for) the numbers only if absolutely necessary - my library had this as a separate call so the user can decide between speed and storage concerns.
3/ Addition of a positive and negative number is subtraction, and subtracting a negative number is the same as adding the equivalent positive. You can save quite a bit of code by having the add and subtract methods call each other after adjusting signs.
4/ Avoid subtracting big numbers from small ones since you invariably end up with numbers like:
10
11-
-- -- -- --
99 99 99 99 (and you still have a borrow).
Instead, subtract 10 from 11, then negate it:
11
10-
--
1 (then negate to get -1).
Here are the comments (turned into text) from one of the libraries I had to do this for. The code itself is, unfortunately, copyrighted, but you may be able to pick out enough information to handle the four basic operations. Assume in the following that -a and -b represent negative numbers and a and b are zero or positive numbers.
For addition, if signs are different, use subtraction of the negation:
-a + b becomes b - a
a + -b becomes a - b
For subtraction, if signs are different, use addition of the negation:
a - -b becomes a + b
-a - b becomes -(a + b)
Also special handling to ensure we're subtracting small numbers from large:
small - big becomes -(big - small)
Multiplication uses entry-level math as follows:
475(a) x 32(b) = 475 x (30 + 2)
= 475 x 30 + 475 x 2
= 4750 x 3 + 475 x 2
= 4750 + 4750 + 4750 + 475 + 475
The way in which this is achieved involves extracting each of the digits of 32 one at a time (backwards) then using add to calculate a value to be added to the result (initially zero).
ShiftLeft and ShiftRight operations are used to quickly multiply or divide a LongInt by the wrap value (10 for "real" math). In the example above, we add 475 to zero 2 times (the last digit of 32) to get 950 (result = 0 + 950 = 950).
Then we left shift 475 to get 4750 and right shift 32 to get 3. Add 4750 to zero 3 times to get 14250 then add to result of 950 to get 15200.
Left shift 4750 to get 47500, right shift 3 to get 0. Since the right shifted 32 is now zero, we're finished and, in fact 475 x 32 does equal 15200.
Division is also tricky but based on early arithmetic (the "gazinta" method for "goes into"). Consider the following long division for 12345 / 27:
457
+-------
27 | 12345 27 is larger than 1 or 12 so we first use 123.
108 27 goes into 123 4 times, 4 x 27 = 108, 123 - 108 = 15.
---
154 Bring down 4.
135 27 goes into 154 5 times, 5 x 27 = 135, 154 - 135 = 19.
---
195 Bring down 5.
189 27 goes into 195 7 times, 7 x 27 = 189, 195 - 189 = 6.
---
6 Nothing more to bring down, so stop.
Therefore 12345 / 27 is 457 with remainder 6. Verify:
457 x 27 + 6
= 12339 + 6
= 12345
This is implemented by using a draw-down variable (initially zero) to bring down the segments of 12345 one at a time until it's greater or equal to 27.
Then we simply subtract 27 from that until we get below 27 - the number of subtractions is the segment added to the top line.
When there are no more segments to bring down, we have our result.
Keep in mind these are pretty basic algorithms. There are far better ways to do complex arithmetic if your numbers are going to be particularly large. You can look into something like GNU Multiple Precision Arithmetic Library - it's substantially better and faster than my own libraries.
It does have the rather unfortunate misfeature in that it will simply exit if it runs out of memory (a rather fatal flaw for a general purpose library in my opinion) but, if you can look past that, it's pretty good at what it does.
If you cannot use it for licensing reasons (or because you don't want your application just exiting for no apparent reason), you could at least get the algorithms from there for integrating into your own code.
I've also found that the bods over at MPIR (a fork of GMP) are more amenable to discussions on potential changes - they seem a more developer-friendly bunch.
While re-inventing the wheel is extremely good for your personal edification and learning, its also an extremely large task. I don't want to dissuade you as its an important exercise and one that I've done myself, but you should be aware that there are subtle and complex issues at work that larger packages address.
For example, multiplication. Naively, you might think of the 'schoolboy' method, i.e. write one number above the other, then do long multiplication as you learned in school. example:
123
x 34
-----
492
+ 3690
---------
4182
but this method is extremely slow (O(n^2), n being the number of digits). Instead, modern bignum packages use either a discrete Fourier transform or a Numeric transform to turn this into an essentially O(n ln(n)) operation.
And this is just for integers. When you get into more complicated functions on some type of real representation of number (log, sqrt, exp, etc.) things get even more complicated.
If you'd like some theoretical background, I highly recommend reading the first chapter of Yap's book, "Fundamental Problems of Algorithmic Algebra". As already mentioned, the gmp bignum library is an excellent library. For real numbers, I've used MPFR and liked it.
Don't reinvent the wheel: it might turn out to be square!
Use a third party library, such as GNU MP, that is tried and tested.
You do it in basically the same way you do with pencil and paper...
The number is to be represented in a buffer (array) able to take on an arbitrary size (which means using malloc and realloc) as needed
you implement basic arithmetic as much as possible using language supported structures, and deal with carries and moving the radix-point manually
you scour numeric analysis texts to find efficient arguments for dealing by more complex function
you only implement as much as you need.
Typically you will use as you basic unit of computation
bytes containing with 0-99 or 0-255
16 bit words contaning wither 0-9999 or 0--65536
32 bit words containing...
...
as dictated by your architecture.
The choice of binary or decimal base depends on you desires for maximum space efficiency, human readability, and the presence of absence of Binary Coded Decimal (BCD) math support on your chip.
You can do it with high school level of mathematics. Though more advanced algorithms are used in reality. So for example to add two 1024-byte numbers :
unsigned char first[1024], second[1024], result[1025];
unsigned char carry = 0;
unsigned int sum = 0;
for(size_t i = 0; i < 1024; i++)
{
sum = first[i] + second[i] + carry;
carry = sum - 255;
}
result will have to be bigger by one place in case of addition to take care of maximum values. Look at this :
9
+
9
----
18
TTMath is a great library if you want to learn. It is built using C++. The above example was silly one, but this is how addition and subtraction is done in general!
A good reference about the subject is Computational complexity of mathematical operations. It tells you how much space is required for each operation you want to implement. For example, If you have two N-digit numbers, then you need 2N digits to store the result of multiplication.
As Mitch said, it is by far not an easy task to implement! I recommend you take a look at TTMath if you know C++.
One of the ultimate references (IMHO) is Knuth's TAOCP Volume II. It explains lots of algorithms for representing numbers and arithmetic operations on these representations.
#Book{Knuth:taocp:2,
author = {Knuth, Donald E.},
title = {The Art of Computer Programming},
volume = {2: Seminumerical Algorithms, second edition},
year = {1981},
publisher = {\Range{Addison}{Wesley}},
isbn = {0-201-03822-6},
}
Assuming that you wish to write a big integer code yourself, this can be surprisingly simple to do, spoken as someone who did it recently (though in MATLAB.) Here are a few of the tricks I used:
I stored each individual decimal digit as a double number. This makes many operations simple, especially output. While it does take up more storage than you might wish, memory is cheap here, and it makes multiplication very efficient if you can convolve a pair of vectors efficiently. Alternatively, you can store several decimal digits in a double, but beware then that convolution to do the multiplication can cause numerical problems on very large numbers.
Store a sign bit separately.
Addition of two numbers is mainly a matter of adding the digits, then check for a carry at each step.
Multiplication of a pair of numbers is best done as convolution followed by a carry step, at least if you have a fast convolution code on tap.
Even when you store the numbers as a string of individual decimal digits, division (also mod/rem ops) can be done to gain roughly 13 decimal digits at a time in the result. This is much more efficient than a divide that works on only 1 decimal digit at a time.
To compute an integer power of an integer, compute the binary representation of the exponent. Then use repeated squaring operations to compute the powers as needed.
Many operations (factoring, primality tests, etc.) will benefit from a powermod operation. That is, when you compute mod(a^p,N), reduce the result mod N at each step of the exponentiation where p has been expressed in a binary form. Do not compute a^p first, and then try to reduce it mod N.
Here's a simple ( naive ) example I did in PHP.
I implemented "Add" and "Multiply" and used that for an exponent example.
http://adevsoft.com/simple-php-arbitrary-precision-integer-big-num-example/
Code snip
// Add two big integers
function ba($a, $b)
{
if( $a === "0" ) return $b;
else if( $b === "0") return $a;
$aa = str_split(strrev(strlen($a)>1?ltrim($a,"0"):$a), 9);
$bb = str_split(strrev(strlen($b)>1?ltrim($b,"0"):$b), 9);
$rr = Array();
$maxC = max(Array(count($aa), count($bb)));
$aa = array_pad(array_map("strrev", $aa),$maxC+1,"0");
$bb = array_pad(array_map("strrev", $bb),$maxC+1,"0");
for( $i=0; $i<=$maxC; $i++ )
{
$t = str_pad((string) ($aa[$i] + $bb[$i]), 9, "0", STR_PAD_LEFT);
if( strlen($t) > 9 )
{
$aa[$i+1] = ba($aa[$i+1], substr($t,0,1));
$t = substr($t, 1);
}
array_unshift($rr, $t);
}
return implode($rr);
}