MMM #16: Mario and the Problematic Yoshi
29 Sep 2008 Quan Quach 13 comments 511 views
The Problem Statement
In one day, Mario has a 10 day showdown with Bowser at the Castle of Doom. In order to get there, Mario must mount Yoshi to traverse the Donut Plains. Unfortuantely, there is one slight problem.
Mario has a stable filled with 650 Yoshi Dinosaurs. Out of all these Yoshis, there is one Yoshi that always causes Mario problems. After riding this particular Yoshi (even for a second), Mario is too sore to do his usual jumps, sprints, and spin moves for 10 days. This debilitating injury makes it difficult to battle the goombas, koopa troopas, and other various baddies in the Mushroom Kingdom.
There’s a good chance that Mario will not ride the Problematic Yoshi but he doesn’t want to leave anything to chance. Luckily, Mario is a popular guy and has an infinite number of friends who are willing to help him out. But being the nice guy that he is, Mario wants to involve as few of his friends as possible in this ordeal.
What is the minimum number of friends that Mario will have to call upon to determine with 100% certainty which Yoshi is the Problematic Yoshi?
Assume the following:
1) There are 650 Yoshi Dinosaurs in the Stable.
2) All Yoshis are identical in appearance and behavior.
3) Riding the Problematic Yoshi for even a second will cause severe soreness for 10 days.
4) Soreness is not felt until after one day has passed.
5) Riding “Normal” Yoshis for any duration of time causes NO soreness.
6) Mario has an infinite number of friends of whom will help him test these Yoshis, i.e. Samus, Kirby, Link, Pikachu, Luigi, etc.
7) The answer is not 650 (it is obvious that Mario can call upon 650 of his friends and have them each ride a different Yoshi to determine the Problematic Yoshi).
The Bounty
The winner will receive a 10 dollar gift certificate to Amazon.com!
The Rules
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Email your answers to mondaymathmadness at gmail dot com.
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Depending on how many submissions we receive, we will be giving out additional prizes, giving you yet more chances to win. We’d also like to give a prize out to the “best” or “most creative” answer.
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Inelgible for any one person to win more than once per year. But you should still submit your answer!
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Answer must be explained. You must show your work! We will accept answers in the form of a MATLAB script as well. We will be the final judge on whether an answer was properly explained or not.
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The deadline to submit answers is October 6th 2008, Monday 11:59 PM Pacific Standard Time
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The winners will be chosen randomly from all the submittals using a random number generator.
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The winner will be announced at 9:00 AM PST October 10th, 2008.
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Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
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13 Responses to “MMM #16: Mario and the Problematic Yoshi”
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So 365 days are the maximum time allowed.?
Mario needs to do battle in one day. Therein lies the problem.
That’s 24 hours then?
Riding the Problematic Yoshi for even a second will cause severe soreness after a period of 23-25 hours.
even if there is infinite people riding the problematic yoshi, if all of them feel soreness only after a period of 25 hours, it still doesn’t meet the requirement…
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so the time you have to find the problamatic Yoshi is 24 hours?? but the sourness can start after 25 hours?? HUH??
Mario needs just two friends.
10 are sufficient.
9 are not sufficient.
10 is the minimum.
Only one friend is necessary.
One friend rides one Yoshi; if that friend is sore after a day, then that Yoshi is the problematic one and he must pick from one of the other 649.
If that friend is not sore after a day, then he can just ride that Yoshi and be certain it is not the problematic Yoshi.
If the time allowed is 10 days.
Minimum is 6.
He has 9 days to find the bad yoshi.
The 10th day is reserved for testing.
Brands his yoshis to make sure this never happens again(or butchers the bad one).
Each day of testing results in one sore tester.
He splits his herd into 6 groups, one person rides each in that group. They are kept separate. The group that injured its rider is then split into 5, etc.
650/6! 1
You do not give us enough information to come to an exact solution.
If we assume that the problematic Yoshi must be known with 100% certainty within around one day, Mario needs at least 649 friends to know with 100% certainty 24 hours from now, which Yoshi is the problematic yoshi.
if we have 2 days, Mario needs around 26 friends. Each of them sit on 26 Yoshi’s for a few seconds each, then one day later, one friend will be sore. Have each of the non-sore friends sit on one of the Yoshi’s that the sore friend sat on. If none of them get sore a day later, the problematic Yoshi is the one that wasn’t retested. If one of the friends does get sore, then we have our problematic yoshi with 100% certainty.
Then just follow that step until you get to what the previous poster on 30 Oct 2008 said.
There’s plenty of information to solve the problem. Mgccl’s just being an ass and the rest of the people complaining aren’t bothering to carefully read the problem.
The best answer I can come up with is he needs 325 friends.
The first friend rides Yoshis 1-325
2nd friend rides 2-326
…
325th friend rides yoshis 325 - 649
The friends that are sore determine which Yoshi is the bad one. For instance, if the 4th friend is sore (and the 3rd isn’t) it’s the 328th Yoshi, since that’s the one Yoshi that friend 4 rode that friend 3 didn’t. If friends 1 - 12 are sore and friend 13 isn’t, the 12th Yoshi is the bad one. And of course, if no friends are sore, the last (650th) Yoshi is the bad one.
An easy way to visualize this is with a number line (representing the Yoshi’s) going from 1 - 650. The 325 friends are overlapping segments on the number line, 325 numbers long and mostly overlapping, with each subsequent friend being one spot to the right of the previous one. If you highlighted all the lines of the friends that are sore, you’d see that they only all overlap for a single Yoshi.
I have no proof that this is the absolute minimum number of friends.
the minimum is 10
it involves knowledge of binary
in binary every number is represented by a series of 1s and 0s
1 = 001
2 = 010
3 = 011
4 = 100
…
650=1010001010
each yoshi is number off from 1-650
that number is then converted into binary
mario calls 10 friends
each friend then rides every yoshi that has a 1 in their specific place in the number
ex
0000000001 friend 1 rides
0000000010 friend 2 rides
0000000011 friend 1 + 2 rides
0000000100 friend 3 rides
…
0010010110 friend 2 + 3 + 5 + 8
afterward mario just has to compare which friends get sore to the yoshis binary numbers to find out which it is.
i dont understand why he needs to know which one is the problematic yoshi when obviously he just needs to find a single yoshi that is not the problematic one. with two friends, one testing 1-325 and the other testing 326-650, mario can just choose a yoshi from the friend who does not get sore.