Output Hello, World! ... sometimes
code-golf
On PPCG, we typically do not allow programs to work only part of the time. That is, all programs must work with 100% probability unless otherwise stated.
This challenge is going to state otherwise. Your program must, with nonzero probability, output...
There is this challenge in the sandbox, which I can understand, but other people may find hard to understand, and I don't know how to reword it. Can anyone help?
personally, i prefer such a challenge to be stated in graph-theoretic terms. if not, i think "phone", "link" and "switch" should be defined first
when i first read the challenge, i thought a "switch" might be like a home router, with multiple ports where cables are plugged and one special uplink port
@user202729 what is part 1's input? is it a graph of pre-existing links, or just the number of phones, or nothing at all?
@user202729 one more thing: i think "just point" should be defined too, and it should be clear that it can't be part of more than one path in the matching
otherwise the trivial solution is: add a point and connect all phones to it
Play snake on a 50*50 grid. The snake will start at length 3, heading (2,0) with body (1,0) and (0,0). It shouldn't bump into wall or itself. There is always one food, which increases the length of snake by 1 when eaten.
Smallest amount of steps till there's no space to place food win. Flexible...
@flawr That would indeed rule out the direct use of trigonomentric functions so I think it's worth it. Plus, it'd make the challenge slightly more general so having variable bounda would be totally ok imo.
I'll Just Make A Snake
I noticed a post (now deleted) that pointed out that 05AB1E (legacy), when called with no arguments and non-empty input, will make a "snake" using the input:
Not sure what to do now with the input, so I'll just make a snake:
*****
*
*
*
*****
*
*
* ...
Int main ()
{
Int a , b, sum;
cout << "Enter number one";
cin >> a;
cout << "Enter number two";
cin >> b;
sum = a + b ;
cout << sum;
return 0;
}
So for that case want after get sum the program ask user either to continue or quit
If continue ask user to enter again two number
CMC: We define a : nat -> nat by writing n = 2^v1 + 2^v2 + ... + 2^vk and defining a(n) = v1 | v2 | ... | vk where | is bitwise-or. Given n : nat, return n | a(n) | a(a(n)) | a(a(a(n))) | ...
cookies for whomever can figure out the relationship between this and the notion of the transitive closure of a set
CMC: We define a : nat -> nat by writing n = 2^v1 + 2^v2 + ... + 2^vk and defining a(n) = v1 | v2 | ... | vk where | is bitwise-or. Given n : nat, return n | a(n) | a(a(n)) | a(a(a(n))) | ...
@H.PWiz Thanks a bunch! |> and ≠ I've already been using, but don't think I've used @. anywhere yet--may need to look and see where it could be useful. The other . trick seems vaguely familiar, not sure if I've already used it.
Java (JDK), 162 bytes
static void m(){String n="AWSALILAND";for(int i=0;i<n.length();i++){char c=n.charAt(i);int v=c;for(int z=0;z<v-64;z++)System.out.print(c);System.out.println("");}
Try it online!
Play snake on a 50*50 grid. The snake will start at length 3, heading (2,0) with body (1,0) and (0,0). It shouldn't bump into wall or itself. There is always one food, which increases the length of snake by 1 when eaten.
Smallest amount of steps till there's no space to place food win. Flexible...
I'll Just Make A Snake
I noticed a post (now deleted) that pointed out that 05AB1E (legacy), when called with no arguments and non-empty input, will make a "snake" using the input:
Not sure what to do now with the input, so I'll just make a snake:
*****
*
*
*
*****
*
*
* ...
Int main ()
{
Int a , b, sum;
cout << "Enter number one";
cin >> a;
cout << "Enter number two";
cin >> b;
sum = a + b ;
cout << sum;
return 0;
}
So for that case want after get sum the program ask user either to continue or quit
If continue ask user to enter again two number
