What are some programming problems like "Hello, world" and "FizzBuzz"?

TheUltimatum

onlyNones wrote:
Polyglot: Write a program that works in a language and also works in another language.

works in any language that supports interpreting or compiling an empty program.

solve the cube

badatprogrammingibe

when green flag clicked
repeat ((number) - (2)) 
  change [iter v] by (1)
  if <((number) mod (iter)) = [0]> then 
    say [comp]
    stop [this script v]
  end
end
say [prime]

I believe this is the shortest primality test (least amount of blocks) written in scratch.
Iter starts at 1, 13 blocks.

Last edited by badatprogrammingibe (Sept. 21, 2018 02:58:45)

onlyNones

TheUltimatum wrote:
onlyNones wrote:
Polyglot: Write a program that works in a language and also works in another language.

works in any language that supports interpreting or compiling an empty program.

Ok, added that it needs to print

Last edited by onlyNones (Sept. 25, 2018 12:25:14)

########################## E ^ ^       ^               
    | | v  v |v|  |||v   | N | | ^     |            ^  
    v | |  | | v  ||v    v C | ^ |  ^  |            |  
      | |  | |    ||     v O ^ | |  |  |  ^         |  
      | |  v |    ||       D | | | ^|  |  |         |  
      | v    v    ||       E | | ^ ||  |  |         |  
      v           vv       D ##########################

ShinigamiBlacky

The shortest way to find whether a number is a prime needs only 5 Blocks
I cheated around needing to reset the itterator by using custom Blocks :D

define Is (number1) prime?     || counter (number2)
if <<(number1)=[1]>or<<not <[2] = (number1)>> and <[0] = ((number1) mod ((1) + (number2)))>>> then 
  say [NOT PRIME] for (2) secs
else 
  if <([sqrt v] of (number1)) > (number2)> then 
    Is (number1) prime?     || counter ((1) + (number2)) :: custom
  else 
    say [PRIME] for (2) secs
  end
end

Last edited by ShinigamiBlacky (Sept. 25, 2018 18:00:55)

ateesdalejr

If you have iOS 12 and Shortcuts installed you can run my version of Fizz Buzz.

https://nofile.io/f/oTRRAyZCzsB/fizzbuzz.shortcut

No idea about the security of nofile so… run at your own risk.

…

gp469-sensei

ShinigamiBlacky wrote:

The shortest way to find whether a number is a prime needs only 5 Blocks
I cheated around needing to reset the itterator by using custom Blocks :D

define Is (number1) prime?     || counter (number2)
if <<(number1)=[1]>or<<not <[2] = (number1)>> and <[0] = ((number1) mod ((1) + (number2)))>>> then 
  say [NOT PRIME] for (2) secs
else 
  if <([sqrt v] of (number1)) > (number2)> then 
    Is (number1) prime?     || counter ((1) + (number2)) :: custom
  else 
    say [PRIME] for (2) secs
  end
end

Here's one with only 1 (or 2) block:
https://scratch.mit.edu/projects/241134479/

when green flag clicked
forever 
  add (letter (((<[data v] contains [p] ?> * (8)) + (<[data v] contains [c] ?> * (4))) + ((<(item (1 v) of [data v] :: list) = ((length of [data v] :: list) + (1))> * (2)) + <((item (1 v) of [data v] :: list) mod ((length of [data v] :: list) + (1))) = [0]>)) of [cppccccpppppppp]) to [data v]
end

ShinigamiBlacky

sure if you use a list as counter and output you can save blocks….
but you have to reset the list which means adding more 2 Blocks (which is still better than mine…)
and even this can made a lot better and even a bit smaller (only 3 Blocks and its resetting itself)

define Is (number1) prime?
delete (all v) of [data2 v]
repeat until <<([sqrt v] of (number1)) < ((2) + (length of [data2 v] :: list))> or <[0] = (item (last v) of [data2 v] :: list)>> 
  add ((number1) mod ((2) + (length of [data2 v] :: list))) to [data2 v]
end

<not <[0] = (item (last v) of [data2 v] :: list)>> // this is the output

Last edited by ShinigamiBlacky (Sept. 28, 2018 18:24:17)

badatprogrammingibe

Smallest fibonacci:

define f(x) (y)
f (y)((x)+(y))

onlyNones

Counter:
Given input n, output all numbers from 1 to n.
This challenge introduces for loops.

Example:

Input: 5
Output:
1
2
3
4
5

########################## E ^ ^       ^               
    | | v  v |v|  |||v   | N | | ^     |            ^  
    v | |  | | v  ||v    v C | ^ |  ^  |            |  
      | |  | |    ||     v O ^ | |  |  |  ^         |  
      | |  v |    ||       D | | | ^|  |  |         |  
      | v    v    ||       E | | ^ ||  |  |         |  
      v           vv       D ##########################

bybb

onlyNones wrote:
Counter:
Given input n, output all numbers from 1 to n.
This challenge introduces for loops.

Example:
Input: 5
Output:
1
2
3
4
5

(lambda x:print("".join(str(i)+("\n"if not i==x else"")for i in range(1, x+1))))(5)

Recursion:

def c(n,m):
    return str(n)+"\n"+c(n+1,m)if n!=m else str(n)
print(c(0,5))

Last edited by bybb (Oct. 1, 2018 11:56:58)

Game Over
You'll find me on @LastContinue from now on.

onlyNones

bybb wrote:

Recursion:

def c(n,m):
    return str(n)+"\n"+c(n+1,m)if n!=m else str(n)
print(c(0,5))

…your recursion uses 2 input?

########################## E ^ ^       ^               
    | | v  v |v|  |||v   | N | | ^     |            ^  
    v | |  | | v  ||v    v C | ^ |  ^  |            |  
      | |  | |    ||     v O ^ | |  |  |  ^         |  
      | |  v |    ||       D | | | ^|  |  |         |  
      | v    v    ||       E | | ^ ||  |  |         |  
      v           vv       D ##########################

bybb

onlyNones wrote:
bybb wrote:
Recursion:
def c(n,m):
    return str(n)+"\n"+c(n+1,m)if n!=m else str(n)
print(c(0,5))
…your recursion uses 2 input?

Yes

Game Over
You'll find me on @LastContinue from now on.

1a3c5e7g9i

ShinigamiBlacky wrote:

The shortest way to find whether a number is a prime needs only 5 Blocks
I cheated around needing to reset the itterator by using custom Blocks :D

define Is (number1) prime?     || counter (number2)
if <<(number1)=[1]>or<<not <[2] = (number1)>> and <[0] = ((number1) mod ((1) + (number2)))>>> then 
  say [NOT PRIME] for (2) secs
else 
  if <([sqrt v] of (number1)) > (number2)> then 
    Is (number1) prime?     || counter ((1) + (number2)) :: custom
  else 
    say [PRIME] for (2) secs
  end
end

That's twenty three blocks. Reporters and booleans are blocks, too.

Wao how is that possible

Uh Squidward? Yes Spongebob? You might want to look at this… HOLY SHRIMP! If we don't act now, well, think about how many clams I'll have to cough up JUST to see an extra portion of nick.com… or worse, to see ALL of it in general! *** Reader! This is not a fictional sketch! It actually is, but that's not the point. Learn more at the link that Spongebob left at the first instance of “this”.

novice27b

Write a program to find the nth fibonacci number in O(log n) time.

i use arch btw

TheLogFather

novice27b wrote:
Write a program to find the nth fibonacci number in O(log n) time.

Are you aware that the nth Fib.no. is just A^n – B^n, where A is (1+sqrt(5))/2 and B is (1-sqrt(5))/2…?

Since |B|<1, that means B^n goes to zero as n increases, which means simply using round(A^n) works pretty well.
See https://scratch.mit.edu/discuss/post/1568178/ for a simple Scratch version

EDIT: oops! –Forgot the division by sqrt(5) in above equations. See the post linked above for correct ones…

Or perhaps you're asking for a solution that's not limited to only ~15 digits (or whatever precision you're using) – i.e. uses some bignum arithmetic to avoid integer precision limits during the recursion…?

Last edited by TheLogFather (Oct. 9, 2018 11:20:37)

Siggy the Kumquat slayer:

Main account: DadOfMrLog –– Frameworks for basic pen-rendered 3D in scratch (see studio). Examples:

- - - - 3D Text - - - - - - Simple shapes - - - Controllable structures - - - On the ground - - - - - - In space - - - -

goldfish678

1a3c5e7g9i wrote:
That's twenty three blocks. Reporters and booleans are blocks, too.

yup. counting reporters/booleans, the shortest is either @badatprogrammingibe's method or my method (both 13 blocks) ((i think)).

can we still get it shorter?? :thinking:

i'm planning on making a program that you can't use more than once in scratch, going to have to wait until tomorrow though bc it's getting super late here

TheAspiringHacker

Would writing a rudimentary compiler or interpreter be a task appropriate for this thread?

Long live Kyoto Animation!

XayCraft360

1a3c5e7g9i wrote:
Make a program that you can't use more than once.

Example:
#This program uses the 'os' module to remove itself.
import os
os.remove(__file__)

shorter:

import os; os.remove(__file__)

<Insert copy of NilsTheBest's OLD uncreative copy of adsuri's uncreative copy of Wahsp's uncreative copy of -ShadowOfTheFuture's- uncreative signature here>

I AM NO LONGER ON SCRATCH

jokebookservice1

XayCraft360 wrote:
1a3c5e7g9i wrote:
Make a program that you can't use more than once.

Example:
#This program uses the 'os' module to remove itself.
import os
os.remove(__file__)
shorter:
import os; os.remove(__file__)

Well, depends how you count.

If you're talking about the number of lines, sure, it's shorter. But by number of characters, on a Unix-like machine, the first one is actually shorter, and on a Windows-like machine, they are the same length. You're replacing a newline character with a semicolon and a space.

And sure, the semicolon is optional, but then it's still a tie.

goldfish678

I have found a technically shorter Scratch primality test method. Since it is, in essence, a primality test and not a compositry test, we can argue that if the program says nothing when you give it a number, you can simply assume it is composite (or know that it is because I just told you this). This allows us to eliminate one of the “say” blocks and bring our test down to a grand total of 12 blocks.

The sad news is that unfortunately, I believe that's the end of the road, at least for this method. Eliminating the other say block would give us two identical outputs for two separate outcomes.

I have also found a way to create a “one-use” program in Scratch 1.4. When fed an imaginary or undefined math operator like (x / 0) or (sqrt of -1), Scratch 1.x doesn't just return an error, it stops the entire project, essentially breaking it and rendering it dysfunctional. This allows us to create a very straightforward one-time-use program in 5 blocks, with the variable “divisor” starting at any number that isn't 0. Note: This could also be done in a similar fashion with the square root block, but I didn't feel like taking another screenshot and it isn't any shorter anyway. It yields the same results.

But wait, there's more! There's an even shorter way to do this in just 4 blocks. Without further ado…

Check these in 1.4 if you don't believe me

That's all for now! Feel free to debate my questionable definition of output to your heart's content.
-Goldfish

Last edited by goldfish678 (Dec. 9, 2018 21:55:25)

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