[Community Puzzle] Kiss the girls

Puzzles
#1

Coding Games and Programming Challenges to Code Better

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Created by @Westicles,validated by @emh,@FredericLocquet and @pluieciel.
If you have any issues, feel free to ping them.

#2

Hi @Westicles
I must be missing something in the probability function.
How a number divided by its own square + 1 can be greater than 0.4 ?

#3

Hi @LudovicB, that is the odds for one girl. So if it is less than 0.4 he might be able to kiss some more.

#4

The statement is not clear.
Should we have to find the best dot from where we can have the most possible G’s?

1 Like
#5

You want to kiss as many girls as possible without having more than a 40% chance of catching monkeypox, like it says.

#6

But how?
We must be next to a girl to kiss her.

#7

I think my program is good, but differences appears with tests results as the number of G increases.
Could we have here or in the description an example of intermediate calculus for the first test ?
Thx for the Puzzle

#8

This is basic probability calculation.

Given two dice.
One die has one face painted 1.
Another die has two faces painted 1 (obviously this is a non-regular die).

Throwing these two dice together, what is the P of getting “at least one 1”?

“Getting 1” means getting infected in the problem.

#9

I’m sorry, but I don’t get the link with the “Kiss the girls” problem, no dices here.

#10

getting 1 on die k means infected by girl k. (Pk)
so you calculate the odd of not infected. (1-Pk)
you go through different dice, and keep not having any 1. and the complement of that is getting “at least one 1”.

1 Like
#11

For the first test i obtain this sorted results for each girl ,it is ok or i’m wrong?
0,
0,
0,
0.05555555555555555,
0.05555555555555555,
0.09803921568627451,
0.12121212121212122,
0.14285714285714285,
0.14285714285714285,
0.15789473684210525,
0.2222222222222222,
0.3333333333333333

1 Like
#12

As far as I remember, your results are right.

#13

Thx all, I was misreading the statement.

#14

P(A+B)=P(A)+P(B)-P(A)*P(B)

5 Likes
#15

Not all problems are best solved by programming/math.
I’m sure the girls on the edges (where x = 0 or y = 0) are all perfectly nice, fine girls… but that’s not enough for Bob… Bob has to have more.
That is all.

Nice puzzle. I learned from it. Thanks.

2 Likes
#16

@nicola (or anyone),

I am calculating those same values that @frayth posted earlier, but I am getting 7 as the most girls to kiss. And when I do the math using this list and starting at the top, when I get to the 8th value and add the first value that starts with .14, it is then adding up to .47, so I don’t count that girl. How do you get 8 from that list of values? I’m clearly missing something, but I just can’t figure out what it could be.

#17

You can’t just add them, otherwise the odds of getting at least one head in two coin flips would be 100%.

1 Like
#18

Thanks @Westicles. That finally put it in place for me. Just passed all test cases and validators.

Statistics and probability were never one of my strongpoints.

And I’ve never been to a rave.

1 Like
#19

You should definitely go to one sometime :rofl:

#20

:wave: Hi to All, and thanks @xujun117 for your quick “probability lesson” :wink: …
→ got it too :+1: !

=> Thanks a lot for your help, and to the author @Westicles too :handshake:.
As @Lisa-Has-Ideas said :

all have been said ;-).

1 Like