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ganeshie8
 one year ago
Q4 and Q5
my answer is NOTA for both, looking for a better justification
http://assets.openstudy.com/updates/attachments/55ac929ee4b0d48ca8ed17acimqwerty14373743502885.jpg
ganeshie8
 one year ago
Q4 and Q5 my answer is NOTA for both, looking for a better justification http://assets.openstudy.com/updates/attachments/55ac929ee4b0d48ca8ed17acimqwerty14373743502885.jpg

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4#4 let \(i\) be the number of operations required excluding swaps, then : \[(153i) + (12+2i) = 10\] solving this gives me \(i=17\) so a minimum of \(17\) operations is required to get a sum of \(10\). since all the options are less than \(17\), i ticked last option.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1actually the min is even more than 17 :), u can only exchange a total of 5 times before u exhaust all red :

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4exactly, i was being pessimistic actually it seems there is no way to get a state where \(red = yellow\) but the proof seems hard

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4yes i checked for all the sequences upto a length of 50 operations (2^50 sequences)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1how did u check so many

dan815
 one year ago
Best ResponseYou've already chosen the best response.1thats not too big for the computer ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4each move consists of choosing between two operations : 1) swap 2) give away 3 red, get 2 yellow for any sequence of length n, there are exactly 2^n possibilities

dan815
 one year ago
Best ResponseYou've already chosen the best response.1not quite i think since there is that one restrition where if u dont have enough red u cannot give away anymore

dan815
 one year ago
Best ResponseYou've already chosen the best response.1its slightly varying problem, the second u exchange u have stepped into a lower domain of numbers

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4when the red becomes negative, just break the loop

dan815
 one year ago
Best ResponseYou've already chosen the best response.1when u are on 27 u can use swap to acheive all combinations, then when u are on 26, use swap

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4here is the script https://jsfiddle.net/0keuvouv/

dan815
 one year ago
Best ResponseYou've already chosen the best response.1what does push method do

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4push adds an element to the array

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4//declare the array subset ``` subset = [ ]; ``` //add "2, 3" to the array ``` subset.push(2); subset.push(3); ```

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4after above commands, subset[0] contains 2 and subset[1] contains 3

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for some reason i find it confusing still, can i ask u a simpler question

dan815
 one year ago
Best ResponseYou've already chosen the best response.1lets say we start with 5 Red and 4 Yellow

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4this spits out all the sequences https://jsfiddle.net/0keuvouv/1/

dan815
 one year ago
Best ResponseYou've already chosen the best response.1was is the exchange option double way

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like u can exchange 2 yellow for 3 red and vice versa

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4if we start with 5 red and 5 yellow, then we can give away 2 yellow, get 3 yellow and see if we ever arrive at 15 red and 12 yellow

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4idk programming wise it looks easy but math wise it looks ahrd

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay so we can exhcange both ways?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1Red to Yellow and Yellow to Red right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nope, i think the problem changes based on the assumtion

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i think its red to yellow only

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if u can go other way its infinite

dan815
 one year ago
Best ResponseYou've already chosen the best response.1is it wrong to think this way okay wait

dan815
 one year ago
Best ResponseYou've already chosen the best response.1lets say I start at 5  Red 4 Yellow i currently have 9 flowers i can have 0 red to 9 red a total of 9 options i will swap in a way that i end up with 3 red and 6 yellow in the end so i can go to 8 flow option with no problem therefore the total nubmer of combinations must be 10 swaps 1 exchange 9 swaps 1 exchange 8 1 7 . .

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh i see the problem is that we cannot get to all the combinaions with that many swaps

dan815
 one year ago
Best ResponseYou've already chosen the best response.1wait i got an idea its not that many too i think

dan815
 one year ago
Best ResponseYou've already chosen the best response.1then ud do the same pattern to cover all the possibilities

dan815
 one year ago
Best ResponseYou've already chosen the best response.1isnt this way a lot less options and u still get all the possible combinations

dan815
 one year ago
Best ResponseYou've already chosen the best response.1the 2^50 stuff is when arrangement matters

dan815
 one year ago
Best ResponseYou've already chosen the best response.1im not saying what im doing is optimal, but it is much less and shows that u can get to all the possiblites with less doesnt it

dan815
 one year ago
Best ResponseYou've already chosen the best response.1for example lets say with that method defined above, inefficient maybe but still lets say starting with 5 Red and 4 Yellow (5,4) (4,5) (3,6) . . (0,9) . . (9,0) Now exchnges stars the 8 flower case with (6,2) (5,3) . . (0,8) . . (8,0) Now exhange starts the 7 floser case with (5,2) . . and so on

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i think finding a closer form answer for this pattern isnt anything too non trivial

dan815
 one year ago
Best ResponseYou've already chosen the best response.1am i just not making any sense lol, i dont know

dan815
 one year ago
Best ResponseYou've already chosen the best response.1findint the most efficient way to do these swaps looks like a very interesting problem though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0heh, cute code @ganeshie8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dan815 here's better annotated code

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, presume we can represent a state as a binomial \(r+yX\) so then moveequivalence is represented by: $$r+yX \sim (r+yX)+ i(3 + 2X) + j(2  3X)$$ so we're interested where \(i\ge 0\) represents exchanges and \(j\ge 0\) swapped exchanges (since swapping only exists to let us exchange in the reverse direction). so it follows if we want \(15+12X\sim 5+5X\) we'd need to show that there is a pair of nonnegative integer solutions to $$10=3i+2j\\7=2i3j$$ so consider: $$20=6i+4j\\21=6i9j\\\implies 1=5j$$so it follows \(5+5X\) is not moveequivalent to \(15+12X\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if we can never reach \(5+5X\) from \(15+12X\) then it follows that not every combination of flowers is reachable from \(15+12X\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the set of reachable combinations (R,Y) is given here by Xs where the row gives Y and column gives R http://pastebin.com/d8eT8WbJ

dan815
 one year ago
Best ResponseYou've already chosen the best response.1here there is atleast one solution for red = yellow when u have 5 red and u exchange ull have 2 red and 2 yellow

dan815
 one year ago
Best ResponseYou've already chosen the best response.1@oldrin.bataku oh thank you for that annotations!

dan815
 one year ago
Best ResponseYou've already chosen the best response.1omg i misunderstood the swap operation xD, i thought swap only changed 1 red to 1 yellow

dan815
 one year ago
Best ResponseYou've already chosen the best response.1now everything makes sense again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well i gave a mathematical proof that 5 reds and 5 yellows from 15 and 12 is not possible no matter how many steps

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That looks so neat @oldrin.bataku ! hope i can mimic the same to show that red=yellow is impossible

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(\cdots\) so it follows if we want \(15+12X\sim a+aX\) we'd need to show that there is a pair of nonnegative integer solutions to $$15=a3i+2j\\12=a+2i3j$$ subtract and get $$3=5(ji)$$ since \(5\nmid 3\), it follows \(a+aX\) is not moveequivalent to \(15+12X\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, that works well :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for counting all the possible reachable configurations, consider the system of inequalities: $$i\ge 0\\j\ge 0\\153i+2j\ge 0\\12+2i3j\ge 0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this bounds a quadrilateral in the plane of \((i,j)\) so we can count the number of possible such configurations (each has a unique sequence of moves I think)
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