Related
Alternatively, is there a way to force a re-evaluation of a single watch expression?
Say I have the following watch expression:
> Random.splitmix 123 '(Random.natIn 0 100)
When I run this, I might see a result like:
Now evaluating any watch expressions (lines starting with `>`)... Ctrl+C cancels.
5 | > Random.splitmix 123 '(Random.natIn 0 100)
⧩
56
Saving the file again will show the same result every time (it is cached).
I'm not sure if Random results should never be cached (maybe that's still good default behavior to save on computation time), but just wondering what the best workarounds are for this.
debug.clear-cache doesn't work either in this situation, since each time the RNG (Random.splitmix) starts over with the same seed.
Of course, we can manually change the random seed, but this also may not always be desired behavior (and a minor nitpick would be that it involves unnecessary keystrokes and creates additional caching - one cached result per seed, so you have to recall which seeds you already used).
You can clear the expression cache with debug.clear-cache in UCM.
That said, re-evaluating your expression is actually going to give the same result every time! The splitMix function is completely deterministic, so the result you get depends on the seed you provide and on nothing else.
So you could clear the cache here, but it’s not going to do anything.
To get a really random value, you need to use IO which is not allowed in watch expressions. You’d need to provide I/O to your program using run in UCM.
Since the watch expression would somehow need to maintain the random state, which is likely more trouble than it is worth, manually editing the random seed is likely the best compromise. Just re-evaluating will always start from the initial value produced from the given random seed.
Alternatively or conjunctively, evaluating a list of random values may be useful.
This has been bugging me since a long time.
Suppose I have a boolean function F defined as follows:
Now, it can be expressed in its SOP form as:
F = bar(X)Ybar(Z)+ XYZ
But I fail to understand why we always complement the 0s to express them as 1. Is it assumed that the inputs X, Y and Z will always be 1?
What is the practical application of that? All the youtube videos I watched on this topic, how to express a function in SOP form or as sum of minterms but none of them explained why we need this thing? Why do we need minterms in the first place?
As of now, I believe that we design circuits to yield and take only 1 and that's where minterms come in handy. But I couldn't get any confirmation of this thing anywhere so I am not sure I am right.
Maxterms are even more confusing. Do we design circuits that would yield and take only 0s? Is that the purpose of maxterms?
Why do we need minterms in the first place?
We do not need minterms, we need a way to solve a logic design problem, i.e. given a truth table, find a logic circuit able to reproduce this truth table.
Obviously, this requires a methodology. Minterm and sum-of-products is mean to realize that. Maxterms and product-of-sums is another one. In either case, you get an algebraic representation of your truth table and you can either implement it directly or try to apply standard theorems of boolean algebra to find an equivalent, but simpler, representation.
But these are not the only tools. For instance, with Karnaugh maps, you rewrite your truth table with some rules and you can simultaneously find an algebraic representation and reduce its complexity, and it does not consider minterms. Its main drawback is that it becomes unworkable if the number of inputs rises and it cannot be considered as a general way to solve the problem of logic design.
It happens that minterms (or maxterms) do not have this drawback, and can be used to solve any problem. We get a trut table and we can directly convert it in an equation with ands, ors and nots. Indeed minterms are somehow simpler to human beings than maxterms, but it is just a matter of taste or of a reduced number of parenthesis, they are actually equivalent.
But I fail to understand why we always complement the 0s to express them as 1. Is it assumed that the inputs X, Y and Z will always be 1?
Assume that we have a truth table, with only a given output at 1. For instance, as line 3 of your table. It means that when x=0, y=1 and z=0 , the output will be zero. So, can I express that in boolean logic? With the SOP methodology, we say that we want a solution for this problem that is an "and" of entries or of their complement. And obviously the solution is "x must be false and y must be true and z must be false" or "(not x) must be true and y must be true and (not z) must be true", hence the minterm /x.y./z. So complementing when we have a 0 and leaving unchanged when we have a 1 is way to find the equation that will be true when xyz=010
If I have another table with only one output at 1 (for instance line 8 of your table), we can find similarly that I can implement this TT with x.y.z.
Now if I have a TT with 2 lines at 1, one can use the property of OR gates and do the OR of the previous circuits. when the output of the first one is 1, it will force this behavior and ditto for the second. And we directly get the solution for your table /xy/z+xyz
This can be extended to any number of ones in the TT and gives a systematic way to find an equation equivalent to a truth table.
So just think of minterms and maxterms as a tool to translate a TT into equations. What is important is the truth table (that describes the behaviour of what you want to do) and the equations (that give you a way to realize it).
Is there a way to ensure that:
if a==b then devfun(a)==devfun(b);
where devfun() is a device function involves some floating point maths ops (e.g. polynomials) and returns floating point results, a and b are floating point variables.
I don't care about cross-implentation consistence (e.g. different compiler/different OS/different driver versions or different compiler options), I only care about, within the same building/program, at runtime, can it ensure that during each function call, the result returned by devfun() are consistent in a way such that as long as a==b, devfun(a)==devfun(b)?
I am talking about SM2.0+ hardware and CUDA 5.0+, just in case being relevant.
Let's assume that your numbers a and b represent properly normalized IEEE-754 representation floating point numbers and that niether a nor b is a NaN value. Let's also assume a and b are both 32-bit, or else a and b are both 64-bit (IEEE-754 floating point representations).
In that case, I believe the (ISO C/C++, or CUDA C/C++) floating point test for equality (==) will return TRUE when the two numbers a and b are bitwise identical (and FALSE otherwise).
Under the TRUE case, with one exception, I believe it is safe to assume that devfun(a) == devfun(b) without any additional conditions except the obvious ones: there is no difference in the behavior of devfun on either side of the == operation, that is, it's the same code, compiled in the same way, executed under the same conditions (e.g. other variables that may be taking part in devfun, same GPU type, etc.), just as you've indicated in your question: "same building/program".
The one exception is if the result of devfun(a) is NaN, since (IEEE-754) NaN != NaN.
It would be interesting (to me) if you think you have a piece of code that disproves this assertion.
Perhaps floating point ninjas will come along and correct me.
Perhaps also I would be remiss if I did not say something about the hazards of floating point comparisons. If you're not familiar with this (most folks would never recommend performing a test a==b on two floating point numbers) you can find many questions about it on SO.
For the same reasons that floating point equality comparison (==) in general is unwise, I think relying on the above assertion, even if it's true, is unwise. Let me give you one example.
Suppose you compile code for architecture sm_20. Now you run the code on an sm_21 device. This one simple variation could result in a JIT-compile at runtime. Now you are no longer running the same code, and all bets are off.
So, again, even if the above is true, I think it's unwise for you to rely on such a statement:
if a==b, then devfun(a) == devfun(b)
Reading this question got me thinking: For a given function f, how can we know that a loop of this form:
while (x > 2)
x = f(x)
will stop for any value x? Is there some simple criterion?
(The fact that f(x) < x for x > 2 doesn't seem to help since the series may converge).
Specifically, can we prove this for sqrt and for log?
For these functions, a proof that ceil(f(x))<x for x > 2 would suffice. You could do one iteration -- to arrive at an integer number, and then proceed by simple induction.
For the general case, probably the best idea is to use well-founded induction to prove this property. However, as Moron pointed out in the comments, this could be impossible in the general case and the right ordering is, in many cases, quite hard to find.
Edit, in reply to Amnon's comment:
If you wanted to use well-founded induction, you would have to define another strict order, that would be well-founded. In case of the functions you mentioned this is not hard: you can take x << y if and only if ceil(x) < ceil(y), where << is a symbol for this new order. This order is of course well-founded on numbers greater then 2, and both sqrt and log are decreasing with respect to it -- so you can apply well-founded induction.
Of course, in general case such an order is much more difficult to find. This is also related, in some way, to total correctness assertions in Hoare logic, where you need to guarantee similar obligations on each loop construct.
There's a general theorem for when then sequence of iterations will converge. (A convergent sequence may not stop in a finite number of steps, but it is getting closer to a target. You can get as close to the target as you like by going far enough out in the sequence.)
The sequence x, f(x), f(f(x)), ... will converge if f is a contraction mapping. That is, there exists a positive constant k < 1 such that for all x and y, |f(x) - f(y)| <= k |x-y|.
(The fact that f(x) < x for x > 2 doesn't seem to help since the series may converge).
If we're talking about floats here, that's not true. If for all x > n f(x) is strictly less than x, it will reach n at some point (because there's only a limited number of floating point values between any two numbers).
Of course this means you need to prove that f(x) is actually less than x using floating point arithmetic (i.e. proving it is less than x mathematically does not suffice, because then f(x) = x may still be true with floats when the difference is not enough).
There is no general algorithm to determine whether a function f and a variable x will end or not in that loop. The Halting problem is reducible to that problem.
For sqrt and log, we could safely do that because we happen to know the mathematical properties of those functions. Say, sqrt approaches 1, log eventually goes negative. So the condition x < 2 has to be false at some point.
Hope that helps.
In the general case, all that can be said is that the loop will terminate when it encounters xi≤2. That doesn't mean that the sequence will converge, nor does it even mean that it is bounded below 2. It only means that the sequence contains a value that is not greater than 2.
That said, any sequence containing a subsequence that converges to a value strictly less than two will (eventually) halt. That is the case for the sequence xi+1 = sqrt(xi), since x converges to 1. In the case of yi+1 = log(yi), it will contain a value less than 2 before becoming undefined for elements of R (though it is well defined on the extended complex plane, C*, but I don't think it will, in general converge except at any stable points that may exist (i.e. where z = log(z)). Ultimately what this means is that you need to perform some upfront analysis on the sequence to better understand its behavior.
The standard test for convergence of a sequence xi to a point z is that give ε > 0, there is an n such that for all i > n, |xi - z| < ε.
As an aside, consider the Mandelbrot Set, M. The test for a particular point c in C for an element in M is whether the sequence zi+1 = zi2 + c is unbounded, which occurs whenever there is a |zi| > 2. Some elements of M may converge (such as 0), but many do not (such as -1).
Sure. For all positive numbers x, the following inequality holds:
log(x) <= x - 1
(this is a pretty basic result from real analysis; it suffices to observe that the second derivative of log is always negative for all positive x, so the function is concave down, and that x-1 is tangent to the function at x = 1). From this it follows essentially immediately that your while loop must terminate within the first ceil(x) - 2 steps -- though in actuality it terminates much, much faster than that.
A similar argument will establish your result for f(x) = sqrt(x); specifically, you can use the fact that:
sqrt(x) <= x/(2 sqrt(2)) + 1/sqrt(2)
for all positive x.
If you're asking whether this result holds for actual programs, instead of mathematically, the answer is a little bit more nuanced, but not much. Basically, many languages don't actually have hard accuracy requirements for the log function, so if your particular language implementation had an absolutely terrible math library this property might fail to hold. That said, it would need to be a really, really terrible library; this property will hold for any reasonable implementation of log.
I suggest reading this wikipedia entry which provides useful pointers. Without additional knowledge about f, nothing can be said.
The word seems to get used in a number of contexts. The best I can figure is that they mean a variable that can't change. Isn't that what constants/finals (darn you Java!) are for?
An invariant is more "conceptual" than a variable. In general, it's a property of the program state that is always true. A function or method that ensures that the invariant holds is said to maintain the invariant.
For instance, a binary search tree might have the invariant that for every node, the key of the node's left child is less than the node's own key. A correctly written insertion function for this tree will maintain that invariant.
As you can tell, that's not the sort of thing you can store in a variable: it's more a statement about the program. By figuring out what sort of invariants your program should maintain, then reviewing your code to make sure that it actually maintains those invariants, you can avoid logical errors in your code.
It is a condition you know to always be true at a particular place in your logic and can check for when debugging to work out what has gone wrong.
The magic of wikipedia: Invariant (computer science)
In computer science, a predicate that,
if true, will remain true throughout a
specific sequence of operations, is
called (an) invariant to that
sequence.
This answer is for my 5 year old kid. Do not think of an invariant as a constant or fixed numerical value. But it can be. However, it is more than that.
Rather, an invariant is something like of a fixed relationship between varying entities. For example, your age will always be less than that compared to your biological parents. Both your age, and your parent's age changes in the passage of time, but the relationship that i mentioned above is an invariant.
An invariant can also be a numerical constant. For example, the value of pi is an invariant ratio between the circle's circumference over its diameter. No matter how big or small the circle is, that ratio will always be pi.
I usually view them more in terms of algorithms or structures.
For example, you could have a loop invariant that could be asserted--always true at the beginning or end of each iteration. That is, if your loop was supposed to process a collection of objects from one stack to another, you could say that |stack1|+|stack2|=c, at the top or bottom of the loop.
If the invariant check failed, it would indicate something went wrong. In this example, it could mean that you forgot to push the processed element onto the final stack, etc.
As this line states:
In computer science, a predicate that, if true, will remain true throughout a specific sequence of operations, is called (an) invariant to that sequence.
To better understand this hope this example in C++ helps.
Consider a scenario where you have to get some values and get the total count of them in a variable called as count and add them in a variable called as sum
The invariant (again it's more like a concept):
// invariant:
// we have read count grades so far, and
// sum is the sum of the first count grades
The code for the above would be something like this,
int count=0;
double sum=0,x=0;
while (cin >> x) {
++count;
sum+=x;
}
What the above code does?
1) Reads the input from cin and puts them in x
2) After one successful read, increment count and sum = sum + x
3) Repeat 1-2 until read stops ( i.e ctrl+D)
Loop invariant:
The invariant must be True ALWAYS. So initially you start out your code with just this
while(cin>>x){
}
This loop reads data from standard input and stores in x. Well and good. But the invariant becomes false because the first part of our invariant wasn't followed (or kept true).
// we have read count grades so far, and
How to keep the invariant true?
Simple! increment count.
So ++count; would do good!. Now our code becomes something like this,
while(cin>>x){
++count;
}
But
Even now our invariant (a concept which must be TRUE) is False because now we didn't satisfy the second part of our invariant.
// sum is the sum of the first count grades
So what to do now?
Add x to sum and store it in sum ( sum+=x) and the next time
cin>>x will read a new value into x.
Now our code becomes something like this,
while(cin>>x){
++count;
sum+=x;
}
Let's check
Whether code matches our invariant
// invariant:
// we have read count grades so far, and
// sum is the sum of the first count grades
code:
while(cin>>x){
++count;
sum+=x;
}
Ah!. Now the loop invariant is True always and code works fine.
The above example was taken and modified from the book Accelerated C++ by Andrew-koening and Barbara-E
Something that doesn't change within a block of code
All the answers here are great, but i felt that i can shed more light on the matter:
Invariant from a language point of view means something that never changes. The concept though comes actually from math, it's one of the popular proof techniques when combined with induction.
Here is how a proof goes, If you can find an invariant that is in the initial state, And that this invariant persists regardless of any [legal] transformation applied to the state, then you can prove that If a certain state does not have this invariant then it can never occur, no matter what sequence of transformations are applied to the initial state.
Now the previous way of thinking (again combined with induction) makes it possible to predicate the logic of computer software. Especially important when the execution goes in loops, in which an invariant can be used to prove that a certain loop will yield a certain result or that it will never change the state of a program in a certain way.
When invariant is used to predicate a loop logic its called loop invariant. It can be used outside loops, but for loops it is really important, because you often have a lot of possibilities, or an infinite number of possibilities.
Notice that i use the word "predicate" the logic of a computer software, and not prove. And that's because while in math invariant can be used as a proof, it can never prove that the computer software when executed will yield what is expected, due to the fact that the software is executed on top of many abstractions, that can never be proved that they will yield what is expected (think of the hardware abstraction for example).
Finally while theoretically and rigorously predicting software logic is only important for high critical applications like Medical, and Military ones. Invariant can still be used to aid the typical programmer when debugging. It can be used to know where at a certain location The program failed because it has failed to maintain a certain invariant - many of us use it anyway without giving a thought about it.
Class Invariant
Class Invariant is a condition which should be always true before and after calling relevant function
For example balanced tree has an Invariant which is called isBalanced. When you modify your tree through some methods (e.g. addNode, removeNode...) - isBalanced should be always true before and after modifying the tree
Following on from what it is, invariants are quite useful in writing clean code, since knowing conceptually what invariants should be present in your code allows you to easily decide how to organize your code to reach those aims. As mentioned ealier, they're also useful in debugging, as checking to see if the invariant's being maintained is often a good way of seeing if whatever manipulation you're attempting to perform is actually doing what you want it to.
It's typically a quantity that does not change under certain mathematical operations.
An example is a scalar, which does not change under rotations. In magnetic resonance imaging, for example, it is useful to characterize a tissue property by a rotational invariant, because then its estimation ideally does not depend on the orientation of the body in the scanner.
The ADT invariant specifes relationships
among the data fields (instance variables)
that must always be true before and after
the execution of any instance method.
There is an excellent example of an invariant and why it matters in the book Java Concurrency in Practice.
Although Java-centric, the example describes some code that is responsible for calculating the factors of a provided integer. The example code attempts to cache the last number provided, and the factors that were calculated to improve performance. In this scenario there is an invariant that was not accounted for in the example code which has left the code susceptible to race conditions in a concurrent scenario.