ganeshie8
  • ganeshie8
Q4 and Q5 my answer is NOTA for both, looking for a better justification http://assets.openstudy.com/updates/attachments/55ac929ee4b0d48ca8ed17ac-imqwerty-1437374350288-5.jpg
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
#4 let \(i\) be the number of operations required excluding swaps, then : \[(15-3i) + (12+2i) = 10\] 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
  • dan815
actually the min is even more than 17 :), u can only exchange a total of 5 times before u exhaust all red :
ganeshie8
  • ganeshie8
exactly, i was being pessimistic actually it seems there is no way to get a state where \(red = yellow\) but the proof seems hard

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dan815
  • dan815
really
ganeshie8
  • ganeshie8
yes i checked for all the sequences upto a length of 50 operations (2^50 sequences)
dan815
  • dan815
how did u check so many
dan815
  • dan815
thats not too big for the computer ?
ganeshie8
  • ganeshie8
each 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
  • dan815
not quite i think since there is that one restrition where if u dont have enough red u cannot give away anymore
dan815
  • dan815
its slightly varying problem, the second u exchange u have stepped into a lower domain of numbers
ganeshie8
  • ganeshie8
when the red becomes negative, just break the loop
dan815
  • dan815
when u are on 27 u can use swap to acheive all combinations, then when u are on 26, use swap
ganeshie8
  • ganeshie8
here is the script https://jsfiddle.net/0keuvouv/
dan815
  • dan815
what does push method do
ganeshie8
  • ganeshie8
push adds an element to the array
ganeshie8
  • ganeshie8
//declare the array subset ``` subset = [ ]; ``` //add "2, 3" to the array ``` subset.push(2); subset.push(3); ```
ganeshie8
  • ganeshie8
after above commands, subset[0] contains 2 and subset[1] contains 3
dan815
  • dan815
for some reason i find it confusing still, can i ask u a simpler question
dan815
  • dan815
lets say we start with 5 Red and 4 Yellow
ganeshie8
  • ganeshie8
this spits out all the sequences https://jsfiddle.net/0keuvouv/1/
dan815
  • dan815
was is the exchange option double way
dan815
  • dan815
like u can exchange 2 yellow for 3 red and vice versa
ganeshie8
  • ganeshie8
if 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
  • ganeshie8
idk programming wise it looks easy but math wise it looks ahrd
ganeshie8
  • ganeshie8
*hard
dan815
  • dan815
okay so we can exhcange both ways?
dan815
  • dan815
Red to Yellow and Yellow to Red right
ganeshie8
  • ganeshie8
nope, i think the problem changes based on the assumtion
dan815
  • dan815
i think its red to yellow only
dan815
  • dan815
if u can go other way its infinite
ganeshie8
  • ganeshie8
yeah i got u
dan815
  • dan815
is it wrong to think this way okay wait
dan815
  • dan815
lets 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
  • dan815
oh i see the problem is that we cannot get to all the combinaions with that many swaps
dan815
  • dan815
wait i got an idea its not that many too i think
dan815
  • dan815
|dw:1437489711171:dw|
dan815
  • dan815
then ud do the same pattern to cover all the possibilities
dan815
  • dan815
isnt this way a lot less options and u still get all the possible combinations
dan815
  • dan815
the 2^50 stuff is when arrangement matters
dan815
  • dan815
im 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
  • dan815
for 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
  • dan815
i think finding a closer form answer for this pattern isnt anything too non trivial
dan815
  • dan815
am i just not making any sense lol, i dont know
dan815
  • dan815
findint the most efficient way to do these swaps looks like a very interesting problem though
anonymous
  • anonymous
heh, cute code @ganeshie8
anonymous
  • anonymous
dan815 here's better annotated code
anonymous
  • anonymous
https://jsfiddle.net/uL2ns14L/2/
anonymous
  • anonymous
hmm, presume we can represent a state as a binomial \(r+yX\) so then move-equivalence 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 $$10=-3i+2j\\7=2i-3j$$ so consider: $$20=-6i+4j\\21=6i-9j\\\implies 1=-5j$$so it follows \(5+5X\) is not move-equivalent to \(15+12X\)
anonymous
  • anonymous
and 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
  • anonymous
the 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
  • dan815
here there is atleast one solution for red = yellow when u have 5 red and u exchange ull have 2 red and 2 yellow
dan815
  • dan815
@oldrin.bataku oh thank you for that annotations!
dan815
  • dan815
@Empty
dan815
  • dan815
omg i misunderstood the swap operation xD, i thought swap only changed 1 red to 1 yellow
dan815
  • dan815
now everything makes sense again
anonymous
  • anonymous
well i gave a mathematical proof that 5 reds and 5 yellows from 15 and 12 is not possible no matter how many steps
ganeshie8
  • ganeshie8
That looks so neat @oldrin.bataku ! hope i can mimic the same to show that red=yellow is impossible
ganeshie8
  • ganeshie8
\(\cdots\) 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 $$15=a-3i+2j\\12=a+2i-3j$$ subtract and get $$3=5(j-i)$$ since \(5\nmid 3\), it follows \(a+aX\) is not move-equivalent to \(15+12X\)
anonymous
  • anonymous
yep, that works well :)
anonymous
  • anonymous
for counting all the possible reachable configurations, consider the system of inequalities: $$i\ge 0\\j\ge 0\\15-3i+2j\ge 0\\12+2i-3j\ge 0$$
anonymous
  • anonymous
this 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)
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=integer+solutions+to+x+%3E%3D+0%2C+y+%3E%3D+0%2C+15+-+3x+%2B+2y+%3E%3D+0%2C+12+%2B+2x+-+3y+%3E%3D+0

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