I have seen number ranges represented as [first1,last1) and [first2,last2).
I would like to know what such a notation means.
A bracket - [ or ] - means that end of the range is inclusive -- it includes the element listed. A parenthesis - ( or ) - means that end is exclusive and doesn't contain the listed element. So for [first1, last1), the range starts with first1 (and includes it), but ends just before last1.
Assuming integers:
(0, 5) = 1, 2, 3, 4
(0, 5] = 1, 2, 3, 4, 5
[0, 5) = 0, 1, 2, 3, 4
[0, 5] = 0, 1, 2, 3, 4, 5
That's a half-open interval.
A closed interval [a,b] includes the end points.
An open interval (a,b) excludes them.
In your case the end-point at the start of the interval is included, but the end is excluded. So it means the interval "first1 <= x < last1".
Half-open intervals are useful in programming because they correspond to the common idiom for looping:
for (int i = 0; i < n; ++i) { ... }
Here i is in the range [0, n).
The concept of interval notation comes up in both Mathematics and Computer Science. The Mathematical notation [, ], (, ) denotes the domain (or range) of an interval.
The brackets [ and ] means:
The number is included,
This side of the interval is closed,
The parenthesis ( and ) means:
The number is excluded,
This side of the interval is open.
An interval with mixed states is called "half-open".
For example, the range of consecutive integers from 1 .. 10 (inclusive) would be notated as such:
[1,10]
Notice how the word inclusive was used. If we want to exclude the end point but "cover" the same range we need to move the end-point:
[1,11)
For both left and right edges of the interval there are actually 4 permutations:
(1,10) = 2,3,4,5,6,7,8,9 Set has 8 elements
(1,10] = 2,3,4,5,6,7,8,9,10 Set has 9 elements
[1,10) = 1,2,3,4,5,6,7,8,9 Set has 9 elements
[1,10] = 1,2,3,4,5,6,7,8,9,10 Set has 10 elements
How does this relate to Mathematics and Computer Science?
Array indexes tend to use a different offset depending on which field are you in:
Mathematics tends to be one-based.
Certain programming languages tends to be zero-based, such as C, C++, Javascript, Python, while other languages such as Mathematica, Fortran, Pascal are one-based.
These differences can lead to subtle fence post errors, aka, off-by-one bugs when implementing Mathematical algorithms such as for-loops.
Integers
If we have a set or array, say of the first few primes [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ], Mathematicians would refer to the first element as the 1st absolute element. i.e. Using subscript notation to denote the index:
a1 = 2
a2 = 3
:
a10 = 29
Some programming languages, in contradistinction, would refer to the first element as the zero'th relative element.
a[0] = 2
a[1] = 3
:
a[9] = 29
Since the array indexes are in the range [0,N-1] then for clarity purposes it would be "nice" to keep the same numerical value for the range 0 .. N instead of adding textual noise such as a -1 bias.
For example, in C or JavaScript, to iterate over an array of N elements a programmer would write the common idiom of i = 0, i < N with the interval [0,N) instead of the slightly more verbose [0,N-1]:
function main() {
var output = "";
var a = [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ];
for( var i = 0; i < 10; i++ ) // [0,10)
output += "[" + i + "]: " + a[i] + "\n";
if (typeof window === 'undefined') // Node command line
console.log( output )
else
document.getElementById('output1').innerHTML = output;
}
<html>
<body onload="main();">
<pre id="output1"></pre>
</body>
</html>
Mathematicians, since they start counting at 1, would instead use the i = 1, i <= N nomenclature but now we need to correct the array offset in a zero-based language.
e.g.
function main() {
var output = "";
var a = [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 ];
for( var i = 1; i <= 10; i++ ) // [1,10]
output += "[" + i + "]: " + a[i-1] + "\n";
if (typeof window === 'undefined') // Node command line
console.log( output )
else
document.getElementById( "output2" ).innerHTML = output;
}
<html>
<body onload="main()";>
<pre id="output2"></pre>
</body>
</html>
Aside:
In programming languages that are 0-based you might need a kludge of a dummy zero'th element to use a Mathematical 1-based algorithm. e.g. Python Index Start
Floating-Point
Interval notation is also important for floating-point numbers to avoid subtle bugs.
When dealing with floating-point numbers especially in Computer Graphics (color conversion, computational geometry, animation easing/blending, etc.) often times normalized numbers are used. That is, numbers between 0.0 and 1.0.
It is important to know the edge cases if the endpoints are inclusive or exclusive:
(0,1) = 1e-M .. 0.999...
(0,1] = 1e-M .. 1.0
[0,1) = 0.0 .. 0.999...
[0,1] = 0.0 .. 1.0
Where M is some machine epsilon. This is why you might sometimes see const float EPSILON = 1e-# idiom in C code (such as 1e-6) for a 32-bit floating point number. This SO question Does EPSILON guarantee anything? has some preliminary details. For a more comprehensive answer see FLT_EPSILON and David Goldberg's What Every Computer Scientist Should Know About Floating-Point Arithmetic
Some implementations of a random number generator, random() may produce values in the range 0.0 .. 0.999... instead of the more convenient 0.0 .. 1.0. Proper comments in the code will document this as [0.0,1.0) or [0.0,1.0] so there is no ambiguity as to the usage.
Example:
You want to generate random() colors. You convert three floating-point values to unsigned 8-bit values to generate a 24-bit pixel with red, green, and blue channels respectively. Depending on the interval output by random() you may end up with near-white (254,254,254) or white (255,255,255).
+--------+-----+
|random()|Byte |
|--------|-----|
|0.999...| 254 | <-- error introduced
|1.0 | 255 |
+--------+-----+
For more details about floating-point precision and robustness with intervals see Christer Ericson's Real-Time Collision Detection, Chapter 11 Numerical Robustness, Section 11.3 Robust Floating-Point Usage.
It can be a mathematical convention in the definition of an interval where square brackets mean "extremal inclusive" and round brackets "extremal exclusive".
Related
In a simple range I try to get the amount of successive assignments for a variable. The values should be between 6-12 or should be 0. For example in the case a hospital has 24 shifts and an employee should work between 6 and 12 hours or not at all.
# Build shifts
shifts = {}
for n in all_nurses:
for d in all_days:
for s in all_shifts:
shifts[(n, d, s)] = model.NewBoolVar('shift_n%id%is%i' % (n, d, s))
# Count successive occurrences
for e_count in all_nurses:
s_count = 0
while s_count < len(all_shifts):
model.Add(sum(shifts[e_count, s_count] for s in range(e_count, e_count + 6 == 6) #min
model.Add(sum(shifts[e_count, s_count] for s in range(e_count, e_count + 12 <= 12) #min
Unfortunately this doesn't work since it increases the value with only one, what would be the best approach to check if how many hours have been assigned and increase s_count with that value?
If you just want to constrain the sum, you should use this method
model.AddLinearExpressionInDomain(sum(bool_vars), cp_model.Domain.FromIntervals([0, 0], [6, 12]))
If you want to constrain the length of a sequence, you should look at the shift_scheduling example
In particular, the soft sequence constraint.
The idea is the following, for every starting point, you want to forbid 010, 0110, 01110, ..., 0111110 and 01111111111110 (0110 means work[start] is false, work[start + 1] is true, work[start + 2] is true, work[start + 3] is false.
To forbid a sequence, just add a nogood, that is a clause (or AddBoolOr containing the negation of the pattern.
in my example bool_or(work[start], work[start + 1].Not(), work[start + 2].Not(), work[start + 3]).
Loop over all starting points and all patterns. And pay attention to the boundary conditions.
I encountered the following in a leetcode article to determine whether an integer is a palindrome
Now let's think about how to revert the last half of the number. For number 1221, if we do 1221 % 10, we get the last digit 1, to get the second to the last digit, we need to remove the last digit from 1221, we could do so by dividing it by 10, 1221 / 10 = 122. Then we can get the last digit again by doing a modulus by 10, 122 % 10 = 2, and if we multiply the last digit by 10 and add the second last digit, 1 * 10 + 2 = 12, it gives us the reverted number we want. Continuing this process would give us the reverted number with more digits.
Now the question is, how do we know that we've reached the half of the number?
Since we divided the number by 10, and multiplied the reversed number by 10, when the original number is less than the reversed number, it means we've processed half of the number digits.
Can someone explain the last two sentences please?! Thank you!
Here is the enclosed C# code:
public class Solution {
public bool IsPalindrome(int x) {
// Special cases:
// As discussed above, when x < 0, x is not a palindrome.
// Also if the last digit of the number is 0, in order to be a palindrome,
// the first digit of the number also needs to be 0.
// Only 0 satisfy this property.
if(x < 0 || (x % 10 == 0 && x != 0)) {
return false;
}
int revertedNumber = 0;
while(x > revertedNumber) {
revertedNumber = revertedNumber * 10 + x % 10;
x /= 10;
}
// When the length is an odd number, we can get rid of the middle digit by revertedNumber/10
// For example when the input is 12321, at the end of the while loop we get x = 12, revertedNumber = 123,
// since the middle digit doesn't matter in palidrome(it will always equal to itself), we can simply get rid of it.
return x == revertedNumber || x == revertedNumber/10;
}
}
The reason is due to the original given input, x will be decreasing by 1 digit while the reverted string increases by 1 digit at the same time. The process keeps on going until x is less than or equal to the reverted string. Hence due to the change length changes, when it terminates, we would approximately reach half of the string.
Let's visit a few examples with positive numbers to understand the process. I would write (x,y) as the (original number, reverted string). The third example is purposely designed to show that it need not be exactly half though but the code would still work.
The first example is 1221, where there are even number of digits. It will go from (1221, 0) to (122, 1) to (12, 12), at this stage, the two terms are equal and hence the process terminates and we can conclude that it is a palindrome.
The next example is 1223, where there are even number of digits. It will go from (1223, 0) to (122, 3) to (12, 32), at this stage, the termination condition holds and hence the process terminates and we can conclude that it is not a palindrome.
Now, the third example is 1211,then the sequence is (1211,0), (121, 1), (12,11), (1,112), after which we terminates from the string and it would conclude that it is not a palindrome
Now, let's make the number consists of odd number of digits:
For 12321. It will go from (12321, 0) to (1232, 1) to (123, 12) to (12, 123) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,12) and we can conclude that it is a palindrome.
For 12323. It will go from (12323, 0) to (1232, 3) to (123, 32) to (12, 323) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,32) and we can conclude that it is not a palindrome.
For 12311. It will go from (12311, 0) to (1231, 1) to (123, 11) to (12, 113) and at this point, the condition breaks. We then divide the reverted string by 10 and we end up with (12,11) and we can conclude that it is not a palindrome.
I hope these examples would help you to understand what the post mean.
I want to convert the decimal number 27 into binary such a way that , first the digit 2 is converted and its binary value is placed in an array and then the digit 7 is converted and its binary number is placed in that array. what should I do?
thanks in advance
That's called binary-coded decimal. It's easiest to work right-to-left. Take the value modulo 10 (% operator in C/C++/ObjC) and put it in the array. Then integer-divide the value by 10 (/ operator in C/C++/ObjC). Continue until your value is zero. Then reverse the array if you need most-significant digit first.
If I understand your question correctly, you want to go from 27 to an array that looks like {0010, 0111}.
If you understand how base systems work (specifically the decimal system), this should be simple.
First, you find the remainder of your number when divided by 10. Your number 27 in this case would result with 7.
Then you integer divide your number by 10 and store it back in that variable. Your number 27 would result in 2.
How many times do you do this?
You do this until you have no more digits.
How many digits can you have?
Well, if you think about the number 100, it has 3 digits because the number needs to remember that one 10^2 exists in the number. On the other hand, 99 does not.
The answer to the previous question is 1 + floor of Log base 10 of the input number.
Log of 100 is 2, plus 1 is 3, which equals number of digits.
Log of 99 is a little less than 2, but flooring it is 1, plus 1 is 2.
In java it is like this:
int input = 27;
int number = 0;
int numDigits = Math.floor(Log(10, input)) + 1;
int[] digitArray = new int [numDigits];
for (int i = 0; i < numDigits; i++) {
number = input % 10;
digitArray[numDigits - i - 1] = number;
input = input / 10;
}
return digitArray;
Java doesn't have a Log function that is portable for any base (it has it for base e), but it is trivial to make a function for it.
double Log( double base, double value ) {
return Math.log(value)/Math.log(base);
}
Good luck.
I'm attempting to hash the values
10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0
I need a function that will map them to an array that has a size of 13 without causing any collisions.
I've spent several hours thinking this over and googling and can't figure this out. I haven't come close to a viable solution.
How would I go about finding a hash function of this sort? I've played with gperf, but I don't really understand it and I couldn't get the results I was looking for.
if you know the exact keys then it is trivial to produce a perfect hash function -
int hash (int n) {
switch (n) {
case 10: return 0;
case 100: return 1;
case 32: return 2;
// ...
default: return -1;
}
}
Found One
I tried a few things and found one semi-manually:
(n ^ 28) % 13
The semi-manual part was the following ruby script that I used to test candidate functions with a range of parameters:
t = [10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0]
(1..200).each do |i|
t2 = t.map { |e| (e ^ i) % 13 }
puts i if t2.uniq.length == t.length
end
On some platforms (e.g. embedded), modulo operation is expensive, so % 13 is better avoided. But AND operation of low-order bits is cheap, and equivalent to modulo of a power-of-2.
I tried writing a simple program (in Python) to search for a perfect hash of your 11 data points, using simple forms such as ((x << a) ^ (x << b)) & 0xF (where & 0xF is equivalent to % 16, giving a result in the range 0..15, for example). I was able to find the following collision-free hash which gives an index in the range 0..15 (expressed as a C macro):
#define HASH(x) ((((x) << 2) ^ ((x) >> 2)) & 0xF)
Here is the Python program I used:
data = [ 10, 100, 32, 45, 58, 126, 3, 29, 200, 400, 0 ]
def shift_right(value, shift_value):
"""Shift right that allows for negative values, which shift left
(Python shift operator doesn't allow negative shift values)"""
if shift_value == None:
return 0
if shift_value < 0:
return value << (-shift_value)
else:
return value >> shift_value
def find_hash():
def hashf(val, i, j = None, k = None):
return (shift_right(val, i) ^ shift_right(val, j) ^ shift_right(val, k)) & 0xF
for i in xrange(-7, 8):
for j in xrange(i, 8):
#for k in xrange(j, 8):
#j = None
k = None
outputs = set()
for val in data:
hash_val = hashf(val, i, j, k)
if hash_val >= 13:
pass
#break
if hash_val in outputs:
break
else:
outputs.add(hash_val)
else:
print i, j, k, outputs
if __name__ == '__main__':
find_hash()
Bob Jenkins has a program for this too: http://burtleburtle.net/bob/hash/perfect.html
Unless you're very lucky, there's no "nice" perfect hash function for a given dataset. Perfect hashing algorithms usually use a simple hashing function on the keys (using enough bits so it's collision-free) then use a table to finish it off.
Just some quasi-analytical ramblings:
In your set of numbers, eleven in all, three are odd and eight are even.
Looking at the simplest forms of hashing - %13 - will give you the following hash values:
10 - 3,
100 - 9,
32 - 6,
45 - 6,
58 - 6,
126 - 9,
3 - 3,
29 - 3,
200 - 5,
400 - 10,
0 - 0
Which, of course, is unusable due to the number of collisions. Something more elaborate is needed.
Why state the obvious?
Considering that the numbers are so few any elaborate - or rather, "less simple" - algorithm will likely be slower than either the switch statement or (which I prefer) simply searching through an unsigned short/long vector of size eleven positions and using the index of the match.
Why use a vector search?
You can fine-tune it by placing the most often occuring values towards the beginning of the vector.
I assume the purpose is to plug in the hash index into a switch with nice, sequential numbering. In that light it seems wasteful to first use a switch to find the index and then plug it into another switch. Maybe you should consider not using hashing at all and go directly to the final switch?
The switch version of hashing cannot be fine-tuned and, due to the widely differing values, will cause the compiler to generate a binary search tree which will result in a lot of comparisons and conditional/other jumps (especially costly) which take time (I've assumed you've turned to hashing for its speed) and require space.
If you want to speed up the vector search additionally and are using an x86-system you can implement a vector search based on the assembler instructions repne scasw (short)/repne scasd (long) which will be much faster. After a setup time of a few instructions you will find the first entry in one instruction and the last in eleven followed by a few instructions cleanup. This means 5-10 instructions best case and 15-20 worst. This should beat the switch-based hashing in all but maybe one or two cases.
I did a quick check and using the SHA256 hash function and then doing modular division by 13 worked when I tried it in Mathematica. For c++ this function should be in the openssl library. See this post.
If you were doing a lot of hashing and lookup though, modular division is a pretty expensive operation to do repeatedly. There is another way of mapping an n-bit hash function into a i-bit indices. See this post by Michael Mitzenmacher about how to do it with a bit shift operation in C. Hope that helps.
Try the following which maps your n values to unique indices between 0 and 12
(1369%(n+1))%13
the thing is that, the 1st number is already ORACLE LONG,
second one a Date (SQL DATE, no timestamp info extra), the last one being a Short value in the range 1000-100'000.
how can I create sort of hash value that will be unique for each combination optimally?
string concatenation and converting to long later:
I don't want this, for example.
Day Month
12 1 --> 121
1 12 --> 121
When you have a few numeric values and need to have a single "unique" (that is, statistically improbable duplicate) value out of them you can usually use a formula like:
h = (a*P1 + b)*P2 + c
where P1 and P2 are either well-chosen numbers (e.g. if you know 'a' is always in the 1-31 range, you can use P1=32) or, when you know nothing particular about the allowable ranges of a,b,c best approach is to have P1 and P2 as big prime numbers (they have the least chance to generate values that collide).
For an optimal solution the math is a bit more complex than that, but using prime numbers you can usually have a decent solution.
For example, Java implementation for .hashCode() for an array (or a String) is something like:
h = 0;
for (int i = 0; i < a.length; ++i)
h = h * 31 + a[i];
Even though personally, I would have chosen a prime bigger than 31 as values inside a String can easily collide, since a delta of 31 places can be quite common, e.g.:
"BB".hashCode() == "Aa".hashCode() == 2122
Your
12 1 --> 121
1 12 --> 121
problem is easily fixed by zero-padding your input numbers to the maximum width expected for each input field.
For example, if the first field can range from 0 to 10000 and the second field can range from 0 to 100, your example becomes:
00012 001 --> 00012001
00001 012 --> 00001012
In python, you can use this:
#pip install pairing
import pairing as pf
n = [12,6,20,19]
print(n)
key = pf.pair(pf.pair(n[0],n[1]),
pf.pair(n[2], n[3]))
print(key)
m = [pf.depair(pf.depair(key)[0]),
pf.depair(pf.depair(key)[1])]
print(m)
Output is:
[12, 6, 20, 19]
477575
[(12, 6), (20, 19)]