can we pick and choose which I's or U's we want to multiply? cause i think i solved it but im not sure on the rules. had a set of 8 I's, and i doubled 7 of the I's, and added a U to the last I, then i paired up the 15 I's to make 4 U's, then i was left with 5 U's. then i took out the 4 U's using the 4th rule to get MU. Is the right?
@NateCan2 I don't think so, he says "suppose we have M and then a string of letters" so x is probably the string of letters which follow M. Meaning all of them, otherwise it would be kinda easy.
Its like a keypad because its possible to know the answer by searching every possible combination which means that there is definitely a limit which means there is definitely a number which means their is definitely a conclusion to make. So if this problem definitely has a number of combinations and your willing to pay $20 to anyone that can solve it, isn't that slightly patronising?
3:50 is a really easy proof showing that there doesn't exist a context-free grammar to produce MI to MU. I.e. showing that L(G) does not contain {"MU"}.
Let N be the number of I's in the string. Rule 2 lets us multiply N by 2, rule 3 lets us subtract a multiple of 3. So to solve, the puzzle we need to find two natural numbers a and b, such that 2^a-3*b is a multiple of 3, which is impossible. (Note: (2^a1-3*b1)*2^a2-3*b2= 2^(a1*a2)-3*(b1*a2+b2) reducing it to such a case.)
@wierdo1232123 can we pick and choose which I's or U's we want to multiply? cause i think i solved it but im not sure on the rules. had a set of 8 I's, and i doubled 7 of the I's, and added a U to the last I, then i paired up the 15 I's to make 4 U's, then i was left with 5 U's. then i took out the 4 U's using the 4th rule to get MU. Is the right?
I thought this was about Johann Sebastian Bach , not someone who thinks there is an "I"
azkeyz 2 weeks ago
The failure here again is the syntax
There is no "I" , unless you're a greek pagan idiot.
azkeyz 2 weeks ago
took me 3 min to solve that.
quitgnr 1 month ago
I resolve it in 1 minute -.-, please next time don't put any price...
DraskyVanderhoff 2 months ago
Evil sir, pure evil.
themattkellyshow 6 months ago
Comment removed
haverunik 7 months ago
"I will give 20 dollars to the first person who can derive MU... that's in this room."
no doors left behind, he is good.
dorgeville 8 months ago 5
can we pick and choose which I's or U's we want to multiply? cause i think i solved it but im not sure on the rules. had a set of 8 I's, and i doubled 7 of the I's, and added a U to the last I, then i paired up the 15 I's to make 4 U's, then i was left with 5 U's. then i took out the 4 U's using the 4th rule to get MU. Is the right?
NateCan2 11 months ago
Comment removed
G1lmarou 2 months ago
This has been flagged as spam show
@NateCan2 I don't think so, he says "suppose we have M and then a string of letters" so x is probably the string of letters which follow M. Meaning all of them, otherwise it would be kinda easy.
G1lmarou 2 months ago
Its like a keypad because its possible to know the answer by searching every possible combination which means that there is definitely a limit which means there is definitely a number which means their is definitely a conclusion to make. So if this problem definitely has a number of combinations and your willing to pay $20 to anyone that can solve it, isn't that slightly patronising?
CHRISTMASBASTARD 1 year ago
Comment removed
CHRISTMASBASTARD 1 year ago
3:50 is a really easy proof showing that there doesn't exist a context-free grammar to produce MI to MU. I.e. showing that L(G) does not contain {"MU"}.
Entertainmentwf 1 year ago
HAHAHA This guy is such a cunt...giving a 20 dollar incentive to a problem that is unsolvable to throw his students off.
infantileretard 1 year ago
could you just change MI to MII to MIII or do you have to make it MIIII?
ophios 1 year ago
@ophios thats what i wanna know too. that guy doesnt specify that we have to double each I
NateCan2 11 months ago
It's impossible.
Let N be the number of I's in the string. Rule 2 lets us multiply N by 2, rule 3 lets us subtract a multiple of 3. So to solve, the puzzle we need to find two natural numbers a and b, such that 2^a-3*b is a multiple of 3, which is impossible. (Note: (2^a1-3*b1)*2^a2-3*b2= 2^(a1*a2)-3*(b1*a2+b2) reducing it to such a case.)
narcissusgray 1 year ago
"You'll be working on it for a few hours" Thats mean, poor students.
rKos 1 year ago 7
@rKos I worked on it for an hour. I've given up for tonight. I WILL solve this. It will NOT beat me.....
wierdo1232123 1 year ago
@wierdo1232123 can we pick and choose which I's or U's we want to multiply? cause i think i solved it but im not sure on the rules. had a set of 8 I's, and i doubled 7 of the I's, and added a U to the last I, then i paired up the 15 I's to make 4 U's, then i was left with 5 U's. then i took out the 4 U's using the 4th rule to get MU. Is the right?
NateCan2 11 months ago