In a reference manual (http://www.cse.unsw.edu.au/~se2011/DafnyDocumentation/Dafny%20-%20ValueTypes.pdf), we can find: two multisets are equal if they have exactly the same count of each element. However, there is no violation if I assert:
assert multiset({1,1}) == multiset{1};
So I am understanding something wrong.
Then, for instance, to prove this:
lemma seqSplit(s:seq<int>, c:int, p:int, f:int)
requires 0<=c<=p<=f+1<=|s|
ensures multiset(s[c..f+1]) == multiset(s[c..p])+multiset(s[p..f+1])
What is is necessary? I started with:
assert forall i :: c<=i<=f ==>
(s[i] in multiset(s[c..f+1]) <==> (s[i] in multiset(s[c..p]) || s[i] in multiset(s[p..f+1])));
It verifies, and I would say it is the same as in the ensures, but seems not. Any help?
multiset({1,1}) means "construct the set {1,1}, and then convert it to a multiset". Since the set {1,1} and the set {1} are the same, your assertion passes.
I think you want multiset{1,1}.
boolean lock = false;
// in thread 1
while (true) {
if (lock) {
lock = true;
criticalRegion1();
lock = false;
}
}
// in thread 2
while (true) {
if (!lock) {
lock = true;
criticalRegion(2);
lock = false;
}
}
Does this work correctly? If yes, explain how. If no, describe how the program can execute resulting in a race condition?
This is a homework question. To find the right answer for this (and for most things involving race conditions in general):
Break the code that each thread does into "atomic pieces". For example, something like x++; should be considered three steps (temp = x; temp++; x = temp; where temp is a register and not a variable). Forget about things that don't have globally visible state (e.g. accessing a local variable isn't important because a different thread won't care, but accessing a global variable is important because a different thread will see or effect it).
Next; imagine every possible order that these "important atomic pieces" could be done in; and see if there's a problem.
For example, if one thread has 2 atomic pieces A and B; and if the other thread has 2 atomic pieces C and D; then you would want to consider:
A, B, C then D
A, C, B then D
C, A, B then D
A, C, D then B
C, A, D then B
C, D, A then B
Note: This sounds like a lot of work at first; but it won't take much practice before you're able to quickly skip "not likely to be a problem" cases and focus on "more likely to be a problem" cases.
I'm new on golang and try to understand how the select statement work at
https://www.tutorialspoint.com/go/go_select_statement.htm
package main
import "fmt"
func main() {
var c1, c2, c3 chan int
var i1, i2 int
select {
case i1 = <-c1:
fmt.Printf("received ", i1, " from c1\n")
case c2 <- i2:
fmt.Printf("sent ", i2, " to c2\n")
case i3, ok := (<-c3): // same as: i3, ok := <-c3
if ok {
fmt.Printf("received ", i3, " from c3\n")
} else {
fmt.Printf("c3 is closed\n")
}
default:
fmt.Printf("no communication\n")
}
}
There was no explanation about channels at this point. Now I have no idea, how to trigger another output as "no communication".
Can anyone give me an example for each case?
The select statement chooses a case whose communication op would not block. If there are multiple cases whose comm. op would not block, one is chosen randomly.
Since in the example all communication ops would block, and since a default is provided, that will be executed.
To "trigger" another case, you have to make sure its comm. op does not block. In the example no one is sending or receiving anything from any of the channels that are used in the cases. Hell, they are not even initialized. And both sending to and receiving from a nil channel blocks forever (for details see How does a non initialized channel behave?). So you should do that: initialize a channel and send to / receive from it, the one whose case you want to trigger. For example:
c1 = make(chan int, 1)
c1 <- 1
This snippet initializes the c1 channel with a buffer of 1, and sends a value on it. So in the select statement after this the communication operation i1 = <-c1 would not block, so this will be chosen, and the output will be:
received 1 from c1
Try it on the Go Playground. (Note: I changed all Printf() calls to Println().)
Note that sending on / receiving from channels could happen concurrently, on other goroutines. I chose a buffered channel and the same goroutine for simplicity, and so that it behaves as you'd expect it even on the Go Playground.
More about channels: What are golang channels used for?
I am currently programming in BCPL for an OS course and wanted to write a simple is_digit() function for validation in a program of mine.
A code snippet of my current code follows:
let is_digit(n) be {
if ((n >= '0') /\ (n <= '9')) then
resultis true;
}
I am aware that BCPL has no notion of types, but how would I be able to accomplish this sort of thing in the language?
Passing in a number yields a false result instead of the expected true.
is_digit() is a function returning a value, rather than a routine, so should use = VALOF rather than BE. Otherwise, the code is OK.
let is_digit(n) = valof {
.....
resultis true
}
Functions that return values should be using valof rather than be, the latter (a routine rather than a function) can be called as a function but the return value you get back from it will be undefined(a).
In addition, you should ensure you return a valid value for every code path. At the moment, a non-digit will not execute a RESULTIS statement, and I'm not entirely certain what happens in that case (so best to be safe).
That means something like this is what you're after, keeping in mind there can be implementation variations, such as & and /\ for and, or {...} and $(...$) for the block delimiters - I've used the ones documented in Martin's latest manual:
LET is_digit(n) = VALOF {
RESULTIS (n >= '0') & (n <= '9')
}
(a) Since Martin Richards is still doing stuff with BCPL, this manual may help in any future questions (or see his home page for a large selection of goodies).
Given an arbitrary list of booleans, what is the most elegant way of determining that exactly one of them is true?
The most obvious hack is type conversion: converting them to 0 for false and 1 for true and then summing them, and returning sum == 1.
I'd like to know if there is a way to do this without converting them to ints, actually using boolean logic.
(This seems like it should be trivial, idk, long week)
Edit: In case it wasn't obvious, this is more of a code-golf / theoretical question. I'm not fussed about using type conversion / int addition in PROD code, I'm just interested if there is way of doing it without that.
Edit2: Sorry folks it's a long week and I'm not explaining myself well. Let me try this:
In boolean logic, ANDing a collection of booleans is true if all of the booleans are true, ORing the collection is true if least one of them is true. Is there a logical construct that will be true if exactly one boolean is true? XOR is this for a collection of two booleans for example, but any more than that and it falls over.
You can actually accomplish this using only boolean logic, although there's perhaps no practical value of that in your example. The boolean version is much more involved than simply counting the number of true values.
Anyway, for the sake of satisfying intellectual curiosity, here goes. First, the idea of using a series of XORs is good, but it only gets us half way. For any two variables x and y,
x ⊻ y
is true whenever exactly one of them is true. However, this does not continue to be true if you add a third variable z,
x ⊻ y ⊻ z
The first part, x ⊻ y, is still true if exactly one of x and y is true. If either x or y is true, then z needs to be false for the whole expression to be true, which is what we want. But consider what happens if both x and y are true. Then x ⊻ y is false, yet the whole expression can become true if z is true as well. So either one variable or all three must be true. In general, if you have a statement that is a chain of XORs, it will be true if an uneven number of variables are true.
Since one is an uneven number, this might prove useful. Of course, checking for an uneven number of truths is not enough. We additionally need to ensure that no more than one variable is true. This can be done in a pairwise fashion by taking all pairs of two variables and checking that they are not both true. Taken together these two conditions ensure that exactly one if the variables are true.
Below is a small Python script to illustrate the approach.
from itertools import product
print("x|y|z|only_one_is_true")
print("======================")
for x, y, z in product([True, False], repeat=3):
uneven_number_is_true = x ^ y ^ z
max_one_is_true = (not (x and y)) and (not (x and z)) and (not (y and z))
only_one_is_true = uneven_number_is_true and max_one_is_true
print(int(x), int(y), int(z), only_one_is_true)
And here's the output.
x|y|z|only_one_is_true
======================
1 1 1 False
1 1 0 False
1 0 1 False
1 0 0 True
0 1 1 False
0 1 0 True
0 0 1 True
0 0 0 False
Sure, you could do something like this (pseudocode, since you didn't mention language):
found = false;
alreadyFound = false;
for (boolean in booleans):
if (boolean):
found = true;
if (alreadyFound):
found = false;
break;
else:
alreadyFound = true;
return found;
After your clarification, here it is with no integers.
bool IsExactlyOneBooleanTrue( bool *boolAry, int size )
{
bool areAnyTrue = false;
bool areTwoTrue = false;
for(int i = 0; (!areTwoTrue) && (i < size); i++) {
areTwoTrue = (areAnyTrue && boolAry[i]);
areAnyTrue |= boolAry[i];
}
return ((areAnyTrue) && (!areTwoTrue));
}
No-one mentioned that this "operation" we're looking for is shortcut-able similarly to boolean AND and OR in most languages. Here's an implementation in Java:
public static boolean exactlyOneOf(boolean... inputs) {
boolean foundAtLeastOne = false;
for (boolean bool : inputs) {
if (bool) {
if (foundAtLeastOne) {
// found a second one that's also true, shortcut like && and ||
return false;
}
foundAtLeastOne = true;
}
}
// we're happy if we found one, but if none found that's less than one
return foundAtLeastOne;
}
With plain boolean logic, it may not be possible to achieve what you want. Because what you are asking for is a truth evaluation not just based on the truth values but also on additional information(count in this case). But boolean evaluation is binary logic, it cannot depend on anything else but on the operands themselves. And there is no way to reverse engineer to find the operands given a truth value because there can be four possible combinations of operands but only two results. Given a false, can you tell if it is because of F ^ F or T ^ T in your case, so that the next evaluation can be determined based on that?.
booleanList.Where(y => y).Count() == 1;
Due to the large number of reads by now, here comes a quick clean up and additional information.
Option 1:
Ask if only the first variable is true, or only the second one, ..., or only the n-th variable.
x1 & !x2 & ... & !xn |
!x1 & x2 & ... & !xn |
...
!x1 & !x2 & ... & xn
This approach scales in O(n^2), the evaluation stops after the first positive match is found. Hence, preferred if it is likely that there is a positive match.
Option 2:
Ask if there is at least one variable true in total. Additionally check every pair to contain at most one true variable (Anders Johannsen's answer)
(x1 | x2 | ... | xn) &
(!x1 | !x2) &
...
(!x1 | !xn) &
(!x2 | !x3) &
...
(!x2 | !xn) &
...
This option also scales in O(n^2) due to the number of possible pairs. Lazy evaluation stops the formula after the first counter example. Hence, it is preferred if its likely there is a negative match.
(Option 3):
This option involves a subtraction and is thus no valid answer for the restricted setting. Nevertheless, it argues how looping the values might not be the most beneficial solution in an unrestricted stetting.
Treat x1 ... xn as a binary number x. Subtract one, then AND the results. The output is zero <=> x1 ... xn contains at most one true value. (the old "check power of two" algorithm)
x 00010000
x-1 00001111
AND 00000000
If the bits are already stored in such a bitboard, this might be beneficial over looping. Though, keep in mind this kills the readability and is limited by the available board length.
A last note to raise awareness: by now there exists a stack exchange called computer science which is exactly intended for this type of algorithmic questions
It can be done quite nicely with recursion, e.g. in Haskell
-- there isn't exactly one true element in the empty list
oneTrue [] = False
-- if the list starts with False, discard it
oneTrue (False : xs) = oneTrue xs
-- if the list starts with True, all other elements must be False
oneTrue (True : xs) = not (or xs)
// Javascript
Use .filter() on array and check the length of the new array.
// Example using array
isExactly1BooleanTrue(boolean:boolean[]) {
return booleans.filter(value => value === true).length === 1;
}
// Example using ...booleans
isExactly1BooleanTrue(...booleans) {
return booleans.filter(value => value === true).length === 1;
}
One way to do it is to perform pairwise AND and then check if any of the pairwise comparisons returned true with chained OR. In python I would implement it using
from itertools import combinations
def one_true(bools):
pairwise_comp = [comb[0] and comb[1] for comb in combinations(bools, 2)]
return not any(pairwise_comp)
This approach easily generalizes to lists of arbitrary length, although for very long lists, the number of possible pairs grows very quickly.
Python:
boolean_list.count(True) == 1
OK, another try. Call the different booleans b[i], and call a slice of them (a range of the array) b[i .. j]. Define functions none(b[i .. j]) and just_one(b[i .. j]) (can substitute the recursive definitions to get explicit formulas if required). We have, using C notation for logical operations (&& is and, || is or, ^ for xor (not really in C), ! is not):
none(b[i .. i + 1]) ~~> !b[i] && !b[i + 1]
just_one(b[i .. i + 1]) ~~> b[i] ^ b[i + 1]
And then recursively:
none(b[i .. j + 1]) ~~> none(b[i .. j]) && !b[j + 1]
just_one(b[i .. j + 1] ~~> (just_one(b[i .. j]) && !b[j + 1]) ^ (none(b[i .. j]) && b[j + 1])
And you are interested in just_one(b[1 .. n]).
The expressions will turn out horrible.
Have fun!
That python script does the job nicely. Here's the one-liner it uses:
((x ∨ (y ∨ z)) ∧ (¬(x ∧ y) ∧ (¬(z ∧ x) ∧ ¬(y ∧ z))))
Retracted for Privacy and Anders Johannsen provided already correct and simple answers. But both solutions do not scale very well (O(n^2)). If performance is important you can stick to the following solution, which performs in O(n):
def exact_one_of(array_of_bool):
exact_one = more_than_one = False
for array_elem in array_of_bool:
more_than_one = (exact_one and array_elem) or more_than_one
exact_one = (exact_one ^ array_elem) and (not more_than_one)
return exact_one
(I used python and a for loop for simplicity. But of course this loop could be unrolled to a sequence of NOT, AND, OR and XOR operations)
It works by tracking two states per boolean variable/list entry:
is there exactly one "True" from the beginning of the list until this entry?
are there more than one "True" from the beginning of the list until this entry?
The states of a list entry can be simply derived from the previous states and corresponding list entry/boolean variable.
Python:
let see using example...
steps:
below function exactly_one_topping takes three parameter
stores their values in the list as True, False
Check whether there exists only one true value by checking the count to be exact 1.
def exactly_one_topping(ketchup, mustard, onion):
args = [ketchup,mustard,onion]
if args.count(True) == 1: # check if Exactly one value is True
return True
else:
return False
How do you want to count how many are true without, you know, counting? Sure, you could do something messy like (C syntax, my Python is horrible):
for(i = 0; i < last && !booleans[i]; i++)
;
if(i == last)
return 0; /* No true one found */
/* We have a true one, check there isn't another */
for(i++; i < last && !booleans[i]; i++)
;
if(i == last)
return 1; /* No more true ones */
else
return 0; /* Found another true */
I'm sure you'll agree that the win (if any) is slight, and the readability is bad.
It is not possible without looping. Check BitSet cardinality() in java implementation.
http://fuseyism.com/classpath/doc/java/util/BitSet-source.html
We can do it this way:-
if (A=true or B=true)and(not(A=true and B=true)) then
<enter statements>
end if