Is a primarily prime TM decidable? - theory

A language L over alphabet Σ is primarily prime if and only if for every length l, the majority of strings of length l do belong to L if l is a prime number, but do not belong to L if l is a composite number. Let PriPriTM = {〈M〉 : L(M) is primarily prime and M is a TM}.
Is PriPriTM Turing decidable?

This is a very complicated decision problem, but the answer is that no, it cannot possibly be decidable whether a TM accepts a primarily prime language. Why? Some TMs accept primarily prime languages (consider a TM that accepts exactly the strings of prime length) and some do not (consider the TM accepting the complement of the former's language). The property is semantic in that it deals with what strings are in the language - rather than syntactic, dealing with the form of the TM itself. In other words, two TMs accepting the same language would always be treated identically by a decider for our problem. By Rice's theorem, then, the problem of deciding whether a TM decides such a language is not computable.


What is the answer for this computational analysis problem?

Two algorithms have the same function, while algorithm A has computational complexity O(2^N) and algorithm B has computational complexity O(N^10). Suppose a real computer can continuously run 10^7seconds, performing 10^3 basic operations per second.
In this computer environment, please answer the following questions.
A) What is the approximate range of N for algorithms A and B, respectively?
B) Which algorithm is more suitable in the environment? Why?
The question is defective.
The fact that A has complexity O(2N) means the number of basic operations (presumably modeled as each basic operation taking the same amount of time) means A takes at most some constant times 2N steps for N at least some threshold N0. Similarly, the fact B has complexity O(N10) means B takes at most some constant times N10 steps for N at least some threshold N1. However, they may be different constants; the number of steps for A is at most C02N and the number of steps for B is at most C1N10, and they may have different thresholds N0 and N1.
In asking about a computer that can perform 103 basic operations for 107 seconds, the question asks for which N is the number of steps of A or B known to be at most 1010. In other words, it asks to solve for N in C02N ≤ 1010 and in C1N10 ≤ 1010.
These are clearly unsolvable without knowing C0 and C1, about which the question gives no information.
Further, we do not know the thresholds N0 and N1 where these bounds are known to apply. So even if we knew C0 and C1, we would not know any bound on how many steps the algorithms take for any particular N.
The question is also defective in that it neglects that the O notation puts only an upper bound on the algorithm. The algorithm may run in fewer steps than the values of the formulae. So it may be that, even with N for which C02N ≤ C1N10, algorithm B is better, or vice-versa.
Possibly it is intended that some simplifying assumptions are intended, such as C0 = C1 = 1, N0 = N1 = 0, and each algorithm takes exactly the number of steps of its formula. Then it is easy to solve 2N ≤ 1010 (N is at most about 33.22) and N10 = 1010 (N ≤ 10). However, if these assumptions are intended, then the author has missed the point of O notation; it characterizes a fundamental nature of an algorithm; it does not quantify its actual number of steps.

How can I scan a number having upto 10^18 digits in C [closed]

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No data type can store such large number. Using array
int a[pow(10,pow(10,18))] again won't do the job because pow() returns double and double can't store 10^(10^18).
Anyone having any idea?
I'm trying to solve the following problem:
Consider an integer with N
digits (in decimal notation, without leading zeroes) D1,D2,D3,…,DN. Here, D1 is the most significant digit and DN the least significant. The weight of this integer is defined as:
∑ i=2 -> N (Di−Di−1).
You are given integers N and W. Find the number of positive integers with N digits (without leading zeroes) and weight equal to W. Compute this number modulo 109+7.
The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first and only line of each test case contains two space-separated integers N and W denoting the number of digits and the required weight.
For each test case, print a single line containing one integer — the number of N-digit positive integers with weight W, modulo 109+7.
You don't need to store a number with 10^18 digits. Look at the definition of the weight:
∑ i=2 -> N (Di−Di−1)
Each element in the sum is the difference of two consecutive digits.
Let's take for example a 4 digit number whose digits are D1, D2, D3, D4. Then the sum is:
(D2 - D1) + (D3 - D2) + (D4 - D3)
Reording the operands:
D4 - D3 + D3 - D2 + D2 - D1
You'll see that all but the first and last digits cancel out! So the whole sum is D4 - D1. In fact, for any number of digits N, the sum is:
DN - D1
So only the first and the last digits are relevant. You should be able to figure out the rest from there.
There are libraries for handling these sorts of problems, like The GNU
Multiple Precision
Arithmetic Library:
What is GMP?
GMP is a free library for arbitrary precision arithmetic, operating on signed integers, rational numbers, and floating-point numbers. There is no practical limit to the precision except the ones implied by the available memory in the machine GMP runs on. GMP has a rich set of functions, and the functions have a regular interface.
The main target applications for GMP are cryptography applications and research, Internet security applications, algebra systems, computational algebra research, etc.
But 10^18 would take a huge (and effectively impossible) amount of memory (if my math is correct: 2.881 EiB).
The referenced problem takes input from 2 ≤ N ≤ 10^18. That isn't 10^18 digits (or 10^10^18 which is absurdly enormous) but 18 digits or 2 to 1,000,000,000,000,000,000. This will fit inside a 64 bit integer, signed or unsigned.
Use int64_t from stdint.h.
10^18 is pushing the limits of 64 bit integers, probably why they chose it. Anything larger should use an arbitrary precision math library such as GMP.
...but there are limits. Simply storing such a number would take about 1 million gigabytes. So while the problem is about solving for numbers with 10^18 digits, I strongly suspect you're not supposed to solve it by actually storing those numbers. There's some mathematical technique you're supposed to apply.

analphabetic strings decidable by DFA?

Is the language of panalphabetic strings decidable by DFA? If so, how can I prove it?
A string {a,…,z}* is said to be panalphabetic if it contains at least one occurrence of each letter.
Yes, it is decidable by a DFA, albeit one that needs 226 (that is, 67108864) states.
The easiest way to prove this is probably by the Myhill-Nerode theorem. To use that, you'll need to divide all strings into equivalence classes based on what can be added to the string to make it part of the language. (see the wikipedia article)
To do that, define a function f over strings in {a,...,z}* that is the set of all letters in the string. Obviously, for any two strings x and y, f(xy) is f(x) ⋃ f(y). (That is, the union of f(x) and f(y))
The language of panalphabetic strings is then all strings s such that f(s) is the set of all letters. That is, whether a string is panalphabetic can be determined just by what the value of f applied to that string is.
Now consider two strings x and y such that f(x) = f(y). Then, for any third string z, f(xz) = f(x) ⋃ f(z) = f(y) ⋃ f(z) = f(yz). Therefore, xz is panalphabetic if and only if yz is panalphabetic. Therefore, x and y are equivalent.
Therefore, there can only be as many different equivalence classes as there are possible values of f. Since f(s) is a subset of {a, b, ..., z}, there are only 226 possible values of f(s). This is finite, so the language of panalphabetic strings is recognizable by a DFA.
(To show that 226 is the smallest number of DFA states, you also need to show that if f(x) != f(y), then x and y are not equivalent, which therefore means that there are exactly as many equivalence classes as there are possible values for f(s). That's fairly straightforward, but I'll let you complete that bit)

Compiler does not recognise matching float values [duplicate]

I know UIKit uses CGFloat because of the resolution independent coordinate system.
But every time I want to check if for example frame.origin.x is 0 it makes me feel sick:
if (theView.frame.origin.x == 0) {
// do important operation
Isn't CGFloat vulnerable to false positives when comparing with ==, <=, >=, <, >?
It is a floating point and they have unprecision problems: 0.0000000000041 for example.
Is Objective-C handling this internally when comparing or can it happen that a origin.x which reads as zero does not compare to 0 as true?
First of all, floating point values are not "random" in their behavior. Exact comparison can and does make sense in plenty of real-world usages. But if you're going to use floating point you need to be aware of how it works. Erring on the side of assuming floating point works like real numbers will get you code that quickly breaks. Erring on the side of assuming floating point results have large random fuzz associated with them (like most of the answers here suggest) will get you code that appears to work at first but ends up having large-magnitude errors and broken corner cases.
First of all, if you want to program with floating point, you should read this:
What Every Computer Scientist Should Know About Floating-Point Arithmetic
Yes, read all of it. If that's too much of a burden, you should use integers/fixed point for your calculations until you have time to read it. :-)
Now, with that said, the biggest issues with exact floating point comparisons come down to:
The fact that lots of values you may write in the source, or read in with scanf or strtod, do not exist as floating point values and get silently converted to the nearest approximation. This is what demon9733's answer was talking about.
The fact that many results get rounded due to not having enough precision to represent the actual result. An easy example where you can see this is adding x = 0x1fffffe and y = 1 as floats. Here, x has 24 bits of precision in the mantissa (ok) and y has just 1 bit, but when you add them, their bits are not in overlapping places, and the result would need 25 bits of precision. Instead, it gets rounded (to 0x2000000 in the default rounding mode).
The fact that many results get rounded due to needing infinitely many places for the correct value. This includes both rational results like 1/3 (which you're familiar with from decimal where it takes infinitely many places) but also 1/10 (which also takes infinitely many places in binary, since 5 is not a power of 2), as well as irrational results like the square root of anything that's not a perfect square.
Double rounding. On some systems (particularly x86), floating point expressions are evaluated in higher precision than their nominal types. This means that when one of the above types of rounding happens, you'll get two rounding steps, first a rounding of the result to the higher-precision type, then a rounding to the final type. As an example, consider what happens in decimal if you round 1.49 to an integer (1), versus what happens if you first round it to one decimal place (1.5) then round that result to an integer (2). This is actually one of the nastiest areas to deal with in floating point, since the behaviour of the compiler (especially for buggy, non-conforming compilers like GCC) is unpredictable.
Transcendental functions (trig, exp, log, etc.) are not specified to have correctly rounded results; the result is just specified to be correct within one unit in the last place of precision (usually referred to as 1ulp).
When you're writing floating point code, you need to keep in mind what you're doing with the numbers that could cause the results to be inexact, and make comparisons accordingly. Often times it will make sense to compare with an "epsilon", but that epsilon should be based on the magnitude of the numbers you are comparing, not an absolute constant. (In cases where an absolute constant epsilon would work, that's strongly indicative that fixed point, not floating point, is the right tool for the job!)
Edit: In particular, a magnitude-relative epsilon check should look something like:
if (fabs(x-y) < K * FLT_EPSILON * fabs(x+y))
Where FLT_EPSILON is the constant from float.h (replace it with DBL_EPSILON fordoubles or LDBL_EPSILON for long doubles) and K is a constant you choose such that the accumulated error of your computations is definitely bounded by K units in the last place (and if you're not sure you got the error bound calculation right, make K a few times bigger than what your calculations say it should be).
Finally, note that if you use this, some special care may be needed near zero, since FLT_EPSILON does not make sense for denormals. A quick fix would be to make it:
if (fabs(x-y) < K * FLT_EPSILON * fabs(x+y) || fabs(x-y) < FLT_MIN)
and likewise substitute DBL_MIN if using doubles.
Since 0 is exactly representable as an IEEE754 floating-point number (or using any other implementation of f-p numbers I've ever worked with) comparison with 0 is probably safe. You might get bitten, however, if your program computes a value (such as theView.frame.origin.x) which you have reason to believe ought to be 0 but which your computation cannot guarantee to be 0.
To clarify a little, a computation such as :
areal = 0.0
will (unless your language or system is broken) create a value such that (areal==0.0) returns true but another computation such as
areal = 1.386 - 2.1*(0.66)
may not.
If you can assure yourself that your computations produce values which are 0 (and not just that they produce values which ought to be 0) then you can go ahead and compare f-p values with 0. If you can't assure yourself to the required degree, best stick to the usual approach of 'toleranced equality'.
In the worst cases the careless comparison of f-p values can be extremely dangerous: think avionics, weapons-guidance, power-plant operations, vehicle navigation, almost any application in which computation meets the real world.
For Angry Birds, not so dangerous.
I want to give a bit of a different answer than the others. They are great for answering your question as stated but probably not for what you need to know or what your real problem is.
Floating point in graphics is fine! But there is almost no need to ever compare floats directly. Why would you need to do that? Graphics uses floats to define intervals. And comparing if a float is within an interval also defined by floats is always well defined and merely needs to be consistent, not accurate or precise! As long as a pixel (which is also an interval!) can be assigned that's all graphics needs.
So if you want to test if your point is outside a [0..width[ range this is just fine. Just make sure you define inclusion consistently. For example always define inside is (x>=0 && x < width). The same goes for intersection or hit tests.
However, if you are abusing a graphics coordinate as some kind of flag, like for example to see if a window is docked or not, you should not do this. Use a boolean flag that is separate from the graphics presentation layer instead.
Comparing to zero can be a safe operation, as long as the zero wasn't a calculated value (as noted in an above answer). The reason for this is that zero is a perfectly representable number in floating point.
Talking perfectly representable values, you get 24 bits of range in a power-of-two notion (single precision). So 1, 2, 4 are perfectly representable, as are .5, .25, and .125. As long as all your important bits are in 24-bits, you are golden. So 10.625 can be repsented precisely.
This is great, but will quickly fall apart under pressure. Two scenarios spring to mind:
1) When a calculation is involved. Don't trust that sqrt(3)*sqrt(3) == 3. It just won't be that way. And it probably won't be within an epsilon, as some of the other answers suggest.
2) When any non-power-of-2 (NPOT) is involved. So it may sound odd, but 0.1 is an infinite series in binary and therefore any calculation involving a number like this will be imprecise from the start.
(Oh and the original question mentioned comparisons to zero. Don't forget that -0.0 is also a perfectly valid floating-point value.)
[The 'right answer' glosses over selecting K. Selecting K ends up being just as ad-hoc as selecting VISIBLE_SHIFT but selecting K is less obvious because unlike VISIBLE_SHIFT it is not grounded on any display property. Thus pick your poison - select K or select VISIBLE_SHIFT. This answer advocates selecting VISIBLE_SHIFT and then demonstrates the difficulty in selecting K]
Precisely because of round errors, you should not use comparison of 'exact' values for logical operations. In your specific case of a position on a visual display, it can't possibly matter if the position is 0.0 or 0.0000000003 - the difference is invisible to the eye. So your logic should be something like:
#define VISIBLE_SHIFT 0.0001 // for example
if (fabs(theView.frame.origin.x) < VISIBLE_SHIFT) { /* ... */ }
However, in the end, 'invisible to the eye' will depend on your display properties. If you can upper bound the display (you should be able to); then choose VISIBLE_SHIFT to be a fraction of that upper bound.
Now, the 'right answer' rests upon K so let's explore picking K. The 'right answer' above says:
K is a constant you choose such that the accumulated error of your
computations is definitely bounded by K units in the last place (and
if you're not sure you got the error bound calculation right, make K a
few times bigger than what your calculations say it should be)
So we need K. If getting K is more difficult, less intuitive than selecting my VISIBLE_SHIFT then you'll decide what works for you. To find K we are going to write a test program that looks at a bunch of K values so we can see how it behaves. Ought to be obvious how to choose K, if the 'right answer' is usable. No?
We are going to use, as the 'right answer' details:
if (fabs(x-y) < K * DBL_EPSILON * fabs(x+y) || fabs(x-y) < DBL_MIN)
Let's just try all values of K:
#include <math.h>
#include <float.h>
#include <stdio.h>
void main (void)
double x = 1e-13;
double y = 0.0;
double K = 1e22;
int i = 0;
for (; i < 32; i++, K = K/10.0)
printf ("K:%40.16lf -> ", K);
if (fabs(x-y) < K * DBL_EPSILON * fabs(x+y) || fabs(x-y) < DBL_MIN)
printf ("YES\n");
printf ("NO\n");
ebg#ebg$ gcc -o test test.c
ebg#ebg$ ./test
K:10000000000000000000000.0000000000000000 -> YES
K: 1000000000000000000000.0000000000000000 -> YES
K: 100000000000000000000.0000000000000000 -> YES
K: 10000000000000000000.0000000000000000 -> YES
K: 1000000000000000000.0000000000000000 -> YES
K: 100000000000000000.0000000000000000 -> YES
K: 10000000000000000.0000000000000000 -> YES
K: 1000000000000000.0000000000000000 -> NO
K: 100000000000000.0000000000000000 -> NO
K: 10000000000000.0000000000000000 -> NO
K: 1000000000000.0000000000000000 -> NO
K: 100000000000.0000000000000000 -> NO
K: 10000000000.0000000000000000 -> NO
K: 1000000000.0000000000000000 -> NO
K: 100000000.0000000000000000 -> NO
K: 10000000.0000000000000000 -> NO
K: 1000000.0000000000000000 -> NO
K: 100000.0000000000000000 -> NO
K: 10000.0000000000000000 -> NO
K: 1000.0000000000000000 -> NO
K: 100.0000000000000000 -> NO
K: 10.0000000000000000 -> NO
K: 1.0000000000000000 -> NO
K: 0.1000000000000000 -> NO
K: 0.0100000000000000 -> NO
K: 0.0010000000000000 -> NO
K: 0.0001000000000000 -> NO
K: 0.0000100000000000 -> NO
K: 0.0000010000000000 -> NO
K: 0.0000001000000000 -> NO
K: 0.0000000100000000 -> NO
K: 0.0000000010000000 -> NO
Ah, so K should be 1e16 or larger if I want 1e-13 to be 'zero'.
So, I'd say you have two options:
Do a simple epsilon computation using your engineering judgement for the value of 'epsilon', as I've suggested. If you are doing graphics and 'zero' is meant to be a 'visible change' than examine your visual assets (images, etc) and judge what epsilon can be.
Don't attempt any floating point computations until you've read the non-cargo-cult answer's reference (and gotten your Ph.D in the process) and then use your non-intuitive judgement to select K.
The correct question: how does one compare points in Cocoa Touch?
The correct answer: CGPointEqualToPoint().
A different question: Are two calculated values are the same?
The answer posted here: They are not.
How to check if they are close? If you want to check if they are close, then don't use CGPointEqualToPoint(). But, don't check to see if they are close. Do something that makes sense in the real world, like checking to see if a point is beyond a line or if a point is inside a sphere.
The last time I checked the C standard, there was no requirement for floating point operations on doubles (64 bits total, 53 bit mantissa) to be accurate to more than that precision. However, some hardware might do the operations in registers of greater precision, and the requirement was interpreted to mean no requirement to clear lower order bits (beyond the precision of the numbers being loaded into the registers). So you could get unexpected results of comparisons like this depending on what was left over in the registers from whoever slept there last.
That said, and despite my efforts to expunge it whenever I see it, the outfit where I work has lots of C code that is compiled using gcc and run on linux, and we have not noticed any of these unexpected results in a very long time. I have no idea whether this is because gcc is clearing the low-order bits for us, the 80-bit registers are not used for these operations on modern computers, the standard has been changed, or what. I'd like to know if anyone can quote chapter and verse.
You can use such code for compare float with zero:
if ((int)(theView.frame.origin.x * 100) == 0) {
// do important operation
This will compare with 0.1 accuracy, that enough for CGFloat in this case.
Another issue that may need to be kept in mind is that different implementations do things differently. One example of this that I am very familiar with is the FP units on the Sony Playstation 2. They have significant discrepancies when compared to the IEEE FP hardware in any X86 device. The cited article mentions the complete lack of support for inf and NaN, and it gets worse.
Less well known is what I came to know as the "one bit multiply" error. For certain values of float x:
y = x * 1.0;
assert(y == x);
would fail the assert. In the general case, sometimes, but not always, the result of a FP multiply on the Playstation 2 had a mantissa that was a single bit less than the equivalent IEEE mantissa.
My point being that you should not assume that porting FP code from one platform to another will produce the same results. Any given platform is internally consistent, in that results don't change on that platform, it's just that they may not agree with a different platform. E.g. CPython on X86 uses 64 bit doubles to represent floats, while CircuitPython on a Cortex MO has to use software FP, and only uses 32 bit floats. Needless to say that will introduce discrepancies.
A quote I learned over 40 years ago is as true today as the day I learned it. "Doing floating point maths on a computer is like moving a pile of sand. Every time you do anything, you leave a little sand behind and pick up a little dirt."
Playstation is a registered trademark of Sony Corporation.
-(BOOL)isFloatEqual:(CGFloat)firstValue secondValue:(CGFloat)secondValue{
BOOL isEqual = NO;
NSNumber *firstValueNumber = [NSNumber numberWithDouble:firstValue];
NSNumber *secondValueNumber = [NSNumber numberWithDouble:secondValue];
isEqual = [firstValueNumber isEqualToNumber:secondValueNumber];
return isEqual;
I am using the following comparison function to compare a number of decimal places:
bool compare(const double value1, const double value2, const int precision)
int64_t magnitude = static_cast<int64_t>(std::pow(10, precision));
int64_t intValue1 = static_cast<int64_t>(value1 * magnitude);
int64_t intValue2 = static_cast<int64_t>(value2 * magnitude);
return intValue1 == intValue2;
// Compare 9 decimal places:
if (compare(theView.frame.origin.x, 0, 9)) {
// do important operation
I'd say the right thing is to declare each number as an object, and then define three things in that object: 1) an equality operator. 2) a setAcceptableDifference method. 3)the value itself. The equality operator returns true if the absolute difference of two values is less than the value set as acceptable.
You can subclass the object to suit the problem. For example, round bars of metal between 1 and 2 inches might be considered of equal diameter if their diameters differed by less than 0.0001 inches. So you'd call setAcceptableDifference with parameter 0.0001, and then use the equality operator with confidence.

Do the rules of arithmetic hold for addition in C?

I'm new to C, and I'm having such a hard time understanding this material. I really need help! Please someone help.
In arithmetic, the sum of any two positive integers is great than either:
(n+m) > n for n, m > 0
(n+m) > m for n, m > 0
C has an addition operator +. Does this arithmetic rule hold in C?
I know this is False. But can please someone explain to me why so, I can understand it? Please provide counter-example?
Thank you in advance.
(I won't solve this for you, but will provide some pointers.)
It is false for both integer and floating-point arithmetic, for different reasons.
Integers are susceptible to overflow.
Adding a very small floating-point number m to a very large number n returns n. Have a read of What Every Computer Scientist Should Know About Floating-Point Arithmetic.
It doesn't hold, since C's integers are not "abstract" infinititely-sized integers that the real integers (in mathematics) are.
In C, integers are discrete and digital, and implemented using a fixed number of bits. This leads to limited range, and problems when you go (try to) out of range. Typically integers will wrap, which is very "un-natural".
I brief search did not show up nice answers describing these, so I rather attempt to answer this nicely here, for beginners.
The answer is false, of course, but why so?
In C, or any programming language providing some kind of integer type, this type does not mean it in the mathematical sense. In mathematical sense non-negative integers range from 0 to infinity. A computer, however has only limited storage, so integers necessarily are constrained to something less than infinity.
This alone proves that a + b > a and a + b > b can not be true all the time, since it can be set up so both a and b is less than the largest number the computer can represent in it's storage, but a + b is larger than that.
What exactly happens here, depends. Some mentioned wraparound, but that's not necessarily the case. The C language the first place defines integer overflow to be an undefined behaviour, that whatever, including fire and smoke, may happen if the code happens to step on it (of course in the reality that won't happen, but interpreting the standard strictly it could, as well as the breach of the space-time continuum).
I won't describe how wraparound works here since it is beyond the scope of the problem itself.
Floating point
The case here is again just the same like for integers: the key to understand why mathematics don't fully apply here is that the computer has a limited storage.
Floating numbers in the computers memory are represented much like scientific notation: a mantissa, and an exponent. Both of these have a fixed limited range depending on the type of the floating point variable.
In base 10, you may conceive this like you have the exponent ranging from 10 ^ -10 to 10 ^ 10, and the mantissa having like 4 fraction digits after the decimal point, always normalized.
With this in mind, check these example additions:
1.2345 * (10 ^ 0) + 1.0237 * (10 ^ 5)
5.2345 * (10 ^ 10) + 6.7891 * (10 ^ 10)
The first is an example where the result will equal one of the input numbers while both were larger than zero. The second is an example where the result is out of range.
The floating point representation computers use however is capable to represent infinity, and two at that: positive infinity and negative infinity. So while the first example passes as a proof, the second does not, since that addition's result is positive infinity.
However with this in mind, you could produce an another proofing example:
3.1416 * (10 ^ 0) + (+ infinity)
Of course the result is positive infinity, no matter what you add it to. And of course positive infinity is not larger than positive infinity, so proved again.