I'm a newbie at Objective-C and I'm looking for the best way to handle primitive floats and double when implementing the -hash method in an Objective-C class. I've found some good advise on isEqual and hash in general in this question:
Best practices for overriding isEqual: and hash
but it doesn't say anything on how to deal with floats and doubles.
My best attempt:
...
long lat = [[NSNumber numberWithDouble:self.latitude] longValue];
result = prime * result + (int) (lat ^ (lat >>> 32));
...
but I'm not sure this is the correct way. Any ideas ?
On the assumption that Apple's implementation of -hash is adequate, what is wrong with
result = [[NSNumber numberWithDouble: [self latitude]] hash];
Or using the modern syntax
result = [#([self latitude]) hash];
Advice of #JeremyP about using NSNumber was pretty good, but I followed this more into depth. Since CoreFoundation is open source, I found how does CFNumber implement hashing function and this is what I found:
Split the number to integral and fraction part, hash both parts and sum them. Integral part is hashed by modulo and a hash-factor, fraction part is hashed only by multiplication.
NSUInteger HashDouble(double value) {
double absolute = ABS(value);
double integral = round(absolute);
double fragment = absolute - integral;
NSUInteger integralHash = 2654435761U * fmod(integral, NSUIntegerMax);
NSUInteger fragmentHash = fragment * NSUIntegerMax;
return integralHash + fragmentHash;
}
Original source code from CoreFoundation wth comments:
/* For use by NSNumber and CFNumber.
Hashing algorithm for CFNumber:
M = Max CFHashCode (assumed to be unsigned)
For positive integral values: (N * HASHFACTOR) mod M
For negative integral values: ((-N) * HASHFACTOR) mod M
For floating point numbers that are not integral: hash(integral part) + hash(float part * M)
HASHFACTOR is 2654435761, from Knuth's multiplicative method
*/
#define HASHFACTOR 2654435761U
CF_INLINE CFHashCode _CFHashInt(long i) {
return ((i > 0) ? (CFHashCode)(i) : (CFHashCode)(-i)) * HASHFACTOR;
}
CF_INLINE CFHashCode _CFHashDouble(double d) {
double dInt;
if (d < 0) d = -d;
dInt = floor(d+0.5);
CFHashCode integralHash = HASHFACTOR * (CFHashCode)fmod(dInt, (double)ULONG_MAX);
return (CFHashCode)(integralHash + (CFHashCode)((d - dInt) * ULONG_MAX));
}
From file ForFoundationOnly.h on opensource.apple.com.
Related
I am not sure how to do achieve this logic.
I have an array of numbers:
num = # [0,0.5,1,1.5,2...,12.5,13,13.5,14,14.5,15,15.5.....75];
These numbers are increased with a difference of .5
and I have a result number 'x'.
I want to find out which number in the array is the close match to the 'x' .
Which mathematical operator should I use ? much less << ? or approximately equal ~~ operator in objective-c ?
I can do if (x<<y) in xcode but I cannot do x ~~ y.
At the moment I'm using the following way suggested in here:
static bool CloseEnoughForMe(double value1, double value2, double acceptableDifference)
{
return fabs(value1 - value2) <= acceptableDifference;
}
Is there any simpler method that objective-c provides ?
There are no such operators in Objective-C (or C). You will have to do your own checking. Since each number in your array is 0.5 from the next value, you know the number you wish to check will be closest to a value in the array +/- 0.25.
Something like this perhaps:
double x = ... // The number to check
NSArray *numbers = #[ #0, #0.5, #1, #1.5, ..., 75 ]; // Your array of numbers
for (NSNumber *number in numbers) {
if (x >= [number doubleValue] - 0.25 || x < [number doubleValue] + 0.25) {
NSLog(#"%f is closest to %#", x, number);
break;
}
}
You'll need to add a little extra logic if x is less than -0.5 or greater than 75.5.
So my friend asked me this question as interview practice:
Using Objective-C & Foundation Kit, Write a method that takes a single digit int, and logs out to the console the precise result of that int being raised to the power of 100.
Initially I thought it sounded easy, but then I realized that even a single digit number raised to the power of 100 would quickly come close to 100 digits, which would overflow.
So I tried tackling this problem by creating an NSArray w/ NSNumbers (for reflection), where each object in the array is a place in the final result number. Then I perform the multiplication math (including factoring in carries), and then print out a string formed by concatenating the objects in the array. Here is my implementation w/ input 3:
NSNumber *firstNum = [NSNumber numberWithInteger:3];
NSMutableArray *numArray = [NSMutableArray arrayWithArray:#[firstNum]];
for( int i=0; i<99; i++)
{
int previousCarry = 0;
for( int j=0; j<[numArray count]; j++)
{
int newInt = [firstNum intValue] * [[numArray objectAtIndex:j] intValue] + previousCarry;
NSNumber *calculation = [NSNumber numberWithInteger:newInt];
previousCarry = [calculation intValue]/10;
NSNumber *newValue = [NSNumber numberWithInteger:(newInt % 10)];
[numArray replaceObjectAtIndex:j withObject:newValue];
}
if(previousCarry > 0)
{
[numArray addObject:[NSNumber numberWithInteger:previousCarry]];
}
}
NSArray* reversedArray = [[numArray reverseObjectEnumerator] allObjects];
NSString *finalNumber = [reversedArray componentsJoinedByString:#""];
NSLog(#"%#", finalNumber);
This isn't a problem out of a textbook or anything so I don't have any reference to double check my work. How does this solution sound to you guys? I'm a little worried that it's pretty naive even though the complexity is O(N), I can't help but feel like I'm not utilizing a type/class or method unique to Objective-C or Foundation Kit that would maybe produce a more optimal solution-- or at the very least make the algorithm cleaner and look more impressive
Write a method that takes a single digit int, and logs out to the console the precise result of that int being raised to the power of 100.
That strikes me as a typical interview "trick"[*] question - "single digit", "logs out to console"...
Here goes:
NSString *singleDigitTo100(int d)
{
static NSString *powers[] =
{
#"0",
#"1",
#"1267650600228229401496703205376",
#"515377520732011331036461129765621272702107522001",
#"1606938044258990275541962092341162602522202993782792835301376",
#"7888609052210118054117285652827862296732064351090230047702789306640625",
#"653318623500070906096690267158057820537143710472954871543071966369497141477376",
#"3234476509624757991344647769100216810857203198904625400933895331391691459636928060001",
#"2037035976334486086268445688409378161051468393665936250636140449354381299763336706183397376",
#"265613988875874769338781322035779626829233452653394495974574961739092490901302182994384699044001"
};
return powers[d % 10]; // simple bounds check...
}
And the rest is easy :-)
And if you are wondering, those numbers came from bc - standard command line calculator in U*ix and hence OS X. You could of course invoke bc from Objective-C if you really want to calculate the answers on the fly.
[*] It is not really a "trick" question but asking if you understand that sometimes the best solution is a simple lookup table.
As you have correctly figured out, you will need to use some sort of big integer library. This is a nice example you can refer to: https://mattmccutchen.net/bigint/
Furthermore, you can calculate x^n in O(lg(n)) rather than in O(n), using divide and conquer:
f(x, n):
if n == 0: # Stopping condition
return 1
temp = f(n/2)
result = temp * temp
if n%2 == 1:
result *= x
return result
x = 5 # Or another one digit number.
n = 100
result = f(x, 100) # This is the result you are looking for.
Note that x represents your integer and n the power you are raising x to.
I am trying to floor a float value to the third decimal. For example, the value 2.56976 shall be 2.569 not 2.570. I searched and found answers like these:
floor double by decimal place
Such answers are not accurate. For example the code:
double value = (double)((unsigned int)(value * (double)placed)) / (double)placed
can return the value - 1 and this is not correct. The multiplication of value and placed value * (double)placed) could introduce something like: 2100.999999996. When changed to unsigned int, it becomes 2100 which is wrong (the correct value should be 2101). Other answers suffer from the same issue. In Java, you can use BigDecimal which saves all that hassels.
(Note: of course, rounding the 2100.9999 is not an option as it ruins the whole idea of flooring to "3 decimals correctly")
The following code should work:
#include <stdio.h>
#include <math.h>
int main(void) {
double value = 1.23456;
double val3;
val3 = floor(1000.0 * value + 0.0001) * 0.001; // add 0.0001 to "fix" binary representation problem
printf("val3 is %.8f; the error is %f\n", val3, 1.234 - val3);
}
this prints out
val3 is 1.23400000; the error is 0.000000
If there are any residual errors, it comes about from the fact that floating point numbers cannot necessarily be represented exactly - the idea behind BigDecimal and things like that is to work around that in a very explicit way (for example by representing a number as its digits, rather than a binary representation - it's less efficient, but maintains accuracy)
I had to consider a solution involving NSString and it worked like a charm. Here is the full method:
- (float) getFlooredPrice:(float) passedPrice {
NSString *floatPassedPriceString = [NSString stringWithFormat:#"%f", passedPrice];
NSArray *floatArray = [floatPassedPriceString componentsSeparatedByString:#"."];
NSString *fixedPart = [floatArray objectAtIndex:0];
NSString *decimalPart = #"";
if ([floatArray count] > 1) {
NSString *decimalPartWhole = [floatArray objectAtIndex:1];
if (decimalPartWhole.length > 3) {
decimalPart = [decimalPartWhole substringToIndex:3];
} else {
decimalPart = decimalPartWhole;
}
}
NSString *wholeNumber = [NSString stringWithFormat:#"%#.%#", fixedPart, decimalPart];
return [wholeNumber floatValue];
}
For example, the value 2.56976 shall be 2.569 not 2.570
Solution is has simple as that :
double result = floor(2.56976 * 1000.0) / 1000.0;
I don't know why you search complication... this works perfectly, doesn't need to pass by some unsigned int or other + 0.0001 or whatever.
Important note :
NSLog(#"%.4f", myDouble);
Actually do a round on your variable. So it's improper to believe you can floor with a %.Xf
I've got a really large decimal number in an NSString, which is too large to fit into any variable including NSDecimal. I was doing the math manually, but if I can't fit the number into a variable then I can't be dividing it. So what would be a good way to convert the string?
Example Input: 423723487924398723478243789243879243978234
Output: 4DD361F5A772159224CE9EB0C215D2915FA
I was looking at the first answer here, but it's in C# and I don't know it's objective C equivalent.
Does anyone have any ideas that don't involve using an external library?
If this is all you need, it's not too hard to implement, especially if you're willing to use Objective-C++. By using Objective-C++, you can use a vector to manage memory, which simplifies the code.
Here's the interface we'll implement:
// NSString+BigDecimalToHex.h
#interface NSString (BigDecimalToHex)
- (NSString *)hexStringFromDecimalString;
#end
To implement it, we'll represent an arbitrary-precision non-negative integer as a vector of base-65536 digits:
// NSString+BigDecimalToHex.mm
#import "NSString+BigDecimalToHex.h"
#import <vector>
// index 0 is the least significant digit
typedef std::vector<uint16_t> BigInt;
The "hard" part is to multiply a BigInt by 10 and add a single decimal digit to it. We can very easily implement this as long multiplication with a preloaded carry:
static void insertDecimalDigit(BigInt &b, uint16_t decimalDigit) {
uint32_t carry = decimalDigit;
for (size_t i = 0; i < b.size(); ++i) {
uint32_t product = b[i] * (uint32_t)10 + carry;
b[i] = (uint16_t)product;
carry = product >> 16;
}
if (carry > 0) {
b.push_back(carry);
}
}
With that helper method, we're ready to implement the interface. First, we need to convert the decimal digit string to a BigInt by calling the helper method once for each decimal digit:
- (NSString *)hexStringFromDecimalString {
NSUInteger length = self.length;
unichar decimalCharacters[length];
[self getCharacters:decimalCharacters range:NSMakeRange(0, length)];
BigInt b;
for (NSUInteger i = 0; i < length; ++i) {
insertDecimalDigit(b, decimalCharacters[i] - '0');
}
If the input string is empty, or all zeros, then b is empty. We need to check for that:
if (b.size() == 0) {
return #"0";
}
Now we need to convert b to a hex digit string. The most significant digit of b is at the highest index. To avoid leading zeros, we'll handle that digit specially:
NSMutableString *hexString = [NSMutableString stringWithFormat:#"%X", b.back()];
Then we convert each remaining base-65536 digit to four hex digits, in order from most significant to least significant:
for (ssize_t i = b.size() - 2; i >= 0; --i) {
[hexString appendFormat:#"%04X", b[i]];
}
And then we're done:
return hexString;
}
You can find my full test program (to run as a Mac command-line program) in this gist.
tl;dr: How should I be dealing with numbers like 20! * 20! in Objective-C?
I'm learning Objective-C by working through Project Euler. It's been quite fun, but one problem I've been running in to is working with arbitrarily large numbers. I'm still pretty green on these things, so I don't know why something like, say, Python, handles large numbers with ease compared to Obj-C.
Take for example Problem 15:
Starting in the top left corner of a 2 x 2 grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a 20 x 20 grid?
That's easy. Using combinatorics:
(20+20)! / 20!(20!)
-> 815915283247897734345611269596115894272000000000 / 5919012181389927685417441689600000000
-> 137846528820
In Python:
import math
print math.factorial(40) / (math.factorial(20) * math.factorial(20))
In Objective-C, though? I have't yet found a way to force such a large numbers through. Using the 2 x 2 example works fine. I can get 9C4 = 126, as it should be. But how should I be dealing with numbers like 20!?
I've dallied with trying to use NSDecimalNumber, which appears to support more numerals per number, assuming you can convert it to Mantissa x Exponent and don't mind loss of precision, but that didn't prove to be too useful, as I couldn't figure out how to have Obj-C create a Mantissa from a %llu and I do mind loss of precision.
The code I have so far properly generates the factorials, as it appears unsigned long long handles values so large, but chokes up on x * y, and thus getCombinatoricOf:20 and:20 returns 1.
#import "Problem15.h"
#implementation Problem15
- (unsigned long long)factorial:(NSNumber *)number {
unsigned long long temp = 1;
for (int i = [number intValue]; i > 0; i--) {
temp *= i;
}
return temp;
}
- (unsigned long long)getCombinatorcOf:(NSNumber *)x and:(NSNumber *)y {
NSNumber *n = #([x intValue] + [y intValue]);
NSNumber *n_factorial = #([self factorial:n]);
NSNumber *x_factorial = #([self factorial:x]);
NSNumber *y_factorial = #([self factorial:y]);
return ([n_factorial unsignedLongLongValue] / ([x_factorial unsignedLongLongValue] * [y_factorial unsignedLongLongValue]));
}
- (NSString *)answer {
NSNumber *x = #5;
NSNumber *y = #4;
unsigned long long answer = [self getCombinatoricOf:x and:y];
return [NSString stringWithFormat:#"\n\nProblem 15: \nHow many routes are there through a 20 x 20 grid? \n%llu", answer];
}
#end
It's not Objective-C, but you could just use GMP just as an ordinary C library.
There are also Objective-C wrappers fo GMP, like GMPInt.