CareerCup

A(5L) B(3L) C(Big)
---------------------------
0 0 9.5
0 3 6.5
3 3 3.5
5 1 3.5
0 1 3.5 (5L waste)
1 3 0.5
4 0 0.5

In SIX Steps with No Water Wastage..
Initially.. 0,0,9.5
step1: 5,0,4.5
step2: 2,3,4.5
step3: 2,0,7.5
step4: 0,2,7.5
step5: 5,2,2.5
step6: 4,3,2.5

But you cannot use 3 jars

From the problem statement, wastage can only happen if the user is allowed to use two jars only.

According to that assumption, I could think of following solution;
Fill the ---
5L 0
2L 3L
2L 0 (3L wastage)
0 2L
5L 2L
4L 3L => (6 attempts)

5L 3L 9.5L

0 0 9.5
0 3 6.5
3 0 6.5
3 3 3.5
5 1 3.5
0 1 3.5 (-5L)
1 0 3.5
1 3 0.5
4 0 0.5

Let the water be distributed in the 3 jars(separated by commas) as below.

Initially:
0L,0L,9.5L, Distribute them into 5L and 3L jars as below:
5L,3L,1.5L Now throw away the 1.5 (wastage = 1.5L) so the distribution becomes

5,3,0. From here...
5,0,3
2,3,3
2,0,6
0,2,6
5,2,1
4,3,1 (Four is captured inside the 5L can in a total of 8 steps.)

4 steps :
1.First fill 3L and pour it into 5L and now 5L container has 3L and 2L empty
2.another 3L into 5L, so 1L will be left in 3L container
3.clear the 5L and fill it with 1L that was left in 3L
4.fill 3L into the 5L , hence 4L.

See the water which has been thrown out is only 3L which is in step 4th that is 2, 0, 3

It can be done in following way but I don't know its minimal wastage or not...
I am assuming we can only use two jar not 3
5,3,9.5
-------------------
0,3,6.5
3,0,6.5
3,3,3.5
5,1,3.5
1,0,3.5
1,3,0.5
4,0,0.5
done
I understand wastage is 9L..., but this the only way I could think of using two jars

5 steps, no waste.
0,0,9.5
1) 3,0,6.5
2)0,3,6.5
3)3,3,3.5
4)1,5,3.5
5)4,2,3.5

0 Ltrs, 0 Ltrs, 9.5 Ltrs (Given)

5 Ltrs, 2 Ltrs, 2.5 Ltrs (Now given changes to)

4 Ltrs, 3 Ltrs, 2.5 Ltrs (Given remains the same)

Everyone is over thinking the problem. Assuming you have 5L and 3L beaker, and 9.5L of water....

1) Put the 3L beaker inside the 5L beaker, fill the remaining space inside the 5L beaker to the rim with water. You should now have 2L of water in the 5L beaker, transfer that water into the 3L beaker.

2) Repeat the same motion, you should now have 2L of water inside the 3L beaker, and another 2L of water inside the 5L beaker, which equals a total of 4L!!!

two steps, no water wasted. Don't over think life.

Pour half to 5 ltr i,e 2.5 ltr and full to 3 ltr , remaining water is 9.5-2.5-3 = 4ltr

9.5L 5L 3L
9.5 0 0
1.5 5 3
0 5 3
3 2 3
6 0 2
1 5 2
1 4 3

1. Fill 5L can , pour 3L to 3L can.
2. put remaining 2L in 5L can to 3L can.
3. Fill 5L can and pour 1L of the 5L to 3L can.
Now you have 4L in the 5L can

0 0 9.5
5 0 4.5
2 3 4.5
2 0 7.5
0 2 7.5
5 2 2.5
4 3 2.5

fill 5l transfer 5 to 3 remaining 2 l
do same process 2l exists
2l+2l=4l

Step1 : Pour from 3L to 5L.
Step2 : Now 5L has 3L and 3L flask empty. Fill the 3L flask and pour into 5L flask.
Step3: Now 5L flask is full and 3L flask has 1L remaining.
Step4: Throw the water from 5L flask and get the 1L from 3L flask.
Step5: Fill the 3L flask with water and pour into 5L which already has 1L in it. Altogether it makes 4L in flask 5L.

3 5 9.5l
0 0 9.5
3 5 1.5
3 5 0 (1.5 - waste)
3 0 5
0 3 5
3 3 2
0 5 2 (1l - waste)
3 2 2
3 4 0

step1: pour 5L in 3L (you will have 2L in 5L container)
step2: empty 3L container and now pour remaining 2L (from 5L container ) in 3L
Step3: fill 5L container again
step4: pour 5L container now in 3L container which already has 2L water, so it can only contain extra 1L.

So in 5L container you will have exact 4L

step1: pour 5L in 3L (you will have 2L in 5L container)
step2: empty 3L container and now pour remaining 2L (from 5L container ) in 3L
Step3: fill 5L container again
step4: pour 5L container now in 3L container which already has 2L water, so it can only contain extra 1L.

So in 5L container you will have exact 4L

first fill the 3L and pour it in the 5L and again do the same by filling the 5L completely now there is 1L of water is remaining in the 3L and pour it to the 5L and now fill the 3L and pour it again in the 5L now there is exactly 4L of water in the 5L bottle with the wastage of 5L of water

Put the 3L container into the 5L (supposed more large).
Fill the 5L (2L).
Put the content into the 3L.
Repeat :
Put the 3L container into the 5L (supposed more large).
Fill the 5L (2L).
Pour the 3L container into the 5L.

No waste

9.5;0;0
1.5;5;3 (1.5 waste)
0;5;3
3;2;3
6;2;0
6;0;2
1;5;2
1;4;3

#include<studio.h>
#include<conio.h>
return();
int;
char[ ] [ ]

5 3
---------------------------
5 0
2 3
2 0
0 2
4.5 2
4 2.5

vol(A) - 5, vol(B) - 3, water thrown/wasted
0, 0, 0
5, 0, 0
2, 3, 0
2, 0, 3
0, 2, 3
5, 2, 3
4, 3, 3
in A I get 4L with wastage of 3L

Fill the 5. Pour slowly into the 3 until the bottom of the 5 container is almost exposed. That is 2.5. Using the water already in it, do the same with the 3, that is 1.5. Pout that into the big container 2.5+1.5 = 4
total wastage 2.5-1.5 = 1 litre.
4 steps


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