Activity for H_H
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| Edit | Post #290667 |
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| Edit | Post #290667 |
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| Comment | Post #290667 |
@trichoplax Thank you. Changed the »Explanation« section title and removed some parts of it (i don't think it needs subheadings now, do you?).
Yes there are probably a lot of language errors. I even do them in high German and even in my dialect, so no chance getting the English correct.
You c... (more) |
— | 10 months ago |
| Edit | Post #290667 |
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| Comment | Post #290667 |
Updated the challenge. Do you think it is better now? Any comment how it can be improved? (more) |
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| Edit | Post #290667 |
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| Comment | Post #290667 |
To 3:
A P-adic-number is one that has an infinite amount of digits to the left of the "decimal"-comma/point (instead of the right). Normally you would choose a prime number as base, such as base 2 or base 3 and not 10, this is why there is a P in P-adic. But lets make an example with base 10, a 10-a... (more) |
— | 10 months ago |
| Comment | Post #290667 |
To 1.: In this example, with `a==3`, you would have to call `f()` with the value `43691` or `0xAAAB`.
Because `3 * 0xAAAB` results in `0x20001`. And since we only return a 16 bit variable, the `0x2` at the beginning is droped. That means `f(0xAAAB)` will return 1.
Now think about which argument we ... (more) |
— | 10 months ago |
| Comment | Post #290667 |
The term virtual fraction is one that i created. But fractions with only a single number do exist in Mathematics.
Now there are multiple ways you can look at this challenge. I know at least 3:
1. Consider a C-function: `int a=3; uint16_t f(uint16_t x) { return a*x; }` With what value do you nee... (more) |
— | 10 months ago |
| Comment | Post #290667 |
Thank you for your feedback.
Ok, i should have used a `≡` instead of a `=`. For the rest, i probably explained it a bit too complicated. Maybe you didn't learn about that branch of mathematics and that is where the confusion comes from (which i also had in some challenges from other people) or i d... (more) |
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| Edit | Post #290667 |
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| Edit | Post #290667 | Initial revision | — | 10 months ago |
| Question | — |
Print the modular multiplicative inverse / virtual fractions Goal Print the modular multiplicative inverse with a modulus of 65536 (or a higher exponent of 2) for odd numbers 1-97. Example output Print these values (or an extension of them): ``` 1 43691 52429 28087 36409 35747 20165 61167 61681 51739 53053 14247 23593 55827 49717 3... (more) |
— | 10 months ago |
| Edit | Post #290612 |
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| Comment | Post #290626 |
`<body bgcolor=0>` (more) |
— | 10 months ago |
| Edit | Post #290612 |
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| Edit | Post #290612 | Initial revision | — | 10 months ago |
| Article | — |
Print virtual fractions / 2-adic fractions / Modular multiplicative inverse [FINALIZED] Now posted: Print virtual fractions / 2-adic fractions / Modular multiplicative inverse TL;DR Print this values (or an extension of them): ``` 1 43691 52429 28087 36409 35747 20165 61167 61681 51739 53053 14247 23593 55827 49717 31711 33761 44939 7085 28567 39961 4877... (more) |
— | 10 months ago |
| Comment | Post #290328 |
Is it valid to output a bool by saying output `N` means 1 and all other outputs means 0? (or vice versa). (more) |
— | 12 months ago |
| Edit | Post #289926 |
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— | about 1 year ago |
| Comment | Post #290206 |
@#65963d Yes, absolutely. But then it isn't a (vanilla) python solution anymore and should be mention in the title. For example you can say N Byte Python+itertoolz
You just count the bytes in your source code as normal. When you have:
```
import itertoolz as z
z.DoSomeStuff()
```
Then you ha... (more) |
— | about 1 year ago |
| Edit | Post #289926 |
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| Edit | Post #290109 |
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| Comment | Post #290104 |
`(4*(i%2==1)+1)` can be shorten to `(i%2*4+1)`
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— | about 1 year ago |
| Comment | Post #290104 |
`i%2` can only be 0 or 1, so the `==1` is not needed.
(There is more to improve, but before that the solution has to work in the first place.) (more) |
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| Edit | Post #290109 | Initial revision | — | about 1 year ago |
| Answer | — |
A: Single digit Roman numeral C, 59 byte ```c h(n){return n&4?n&2?5:n&1?1e3:n&8?50:500:n&2?100:n&1?1:10;} ``` Old version, 60 byte: ```c h(n){return n&4?n&2?5:n&1?1000:n&8?50:500:n&2?100:n&1?1:10;} ``` Not very creative, there is probably a smaller version. Takes a ASCII character as argument and returns the value... (more) |
— | about 1 year ago |
| Comment | Post #289996 |
Is `,.` as a solution allowed, then? (i would consider this a loophole for this particular challenge).
Even when you explicit forbid this form of output, it could still be made possible by using a special form of character encoding for the input. A encoding where `I` has the number 1, `V` the numb... (more) |
— | about 1 year ago |
| Comment | Post #289996 |
Do you have any restrictions on the output format? (more) |
— | about 1 year ago |
| Edit | Post #289926 |
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| Edit | Post #289926 |
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| Comment | Post #289926 |
Yes, you are correct, thank you. And `1or 2` is even shorter, but it requires to invert the logic. (more) |
— | about 1 year ago |
| Edit | Post #289926 |
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— | about 1 year ago |
| Comment | Post #289926 |
Wouldn't that need more bytes? I use `&` to save 2 byte over `and`. (more) |
— | about 1 year ago |
| Edit | Post #289926 |
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| Edit | Post #289926 | Initial revision | — | about 1 year ago |
| Answer | — |
A: Borromean coprimes SageMath, 68 66 64 Byte. 62 if you don't count the `m=` ``` g=gcd;m=lambda a,b,c:min(g(a,b),g(a,c),g(b,c))1 ``` Returns `False` for borromean coprimes and `True` for all other natural numbers >1. Use it like this `m(6,10,15)`. Using `min` to get the lowest `gcd` of all pairs. It is short... (more) |
— | about 1 year ago |