Question

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

Answer 1

#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.

Answer 2

actually the min is even more than 17 :), u can only exchange a total of 5 times before u exhaust all red :

Answer 3

exactly, i was being pessimistic
actually it seems there is no way to get a state where \(red = yellow\)
but the proof seems hard

Answer 4

really

Answer 5

yes i checked for all the sequences upto a length of 50 operations
(2^50 sequences)

Answer 6

how did u check so many

Answer 7

thats not too big for the computer ?

Answer 8

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

Answer 9

not quite i think since there is that one restrition where if u dont have enough red u cannot give away anymore

Answer 10

its slightly varying problem, the second u exchange u have stepped into a lower domain of numbers

Answer 11

when the red becomes negative, just break the loop

Answer 12

when u are on 27 u can use swap to acheive all combinations, then when u are on 26, use swap

Answer 13

here is the script
https://jsfiddle.net/0keuvouv/

Answer 14

what does push method do

Answer 15

push adds an element to the array

Answer 16

//declare the array subset
```
subset = [ ];
```

//add "2, 3" to the array
```
subset.push(2);
subset.push(3);
```

Answer 17

after above commands, subset[0] contains 2 and subset[1] contains 3

Answer 18

for some reason i find it confusing still, can i ask u a simpler question

Answer 19

lets say we start with 5 Red and 4 Yellow

Answer 20

this spits out all the sequences
https://jsfiddle.net/0keuvouv/1/

Answer 21

was is the exchange option double way

Answer 22

like u can exchange 2 yellow for 3 red and vice versa

Answer 23

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

Answer 24

idk programming wise it looks easy
but math wise it looks ahrd

Answer 25

*hard

Answer 26

okay so we can exhcange both ways?

Answer 27

Red to Yellow
and Yellow to Red right

Answer 28

nope, i think the problem changes based on the assumtion

Answer 29

i think its red to yellow only

Answer 30

if u can go other way its infinite

Answer 31

yeah i got u

Answer 32

is it wrong to think this way okay wait

Answer 33

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
.
.

Answer 34

oh i see the problem is that we cannot get to all the combinaions with that many swaps

Answer 35

wait i got an idea its not that many too i think