Create Your Own Jug Puzzle

[@oenrn]

1. Fill 5. 2. Move 5 to 9. 3. Fill 5. 4. Move 5 to 9, keeping remainder. 5. Empty 9. 6. Move 5 to 9. 7. Repeat steps 1-6. 8. Fill 5. 9. Move 5 to 9. 5,0 / 0,5 / 5,5 / 1,9 / 1,0 / 0,1 / 5,1 / 0,6 / 5,6 / 2,9 / 2,0 / 0,2 / 5,2 / 0,7

[@BrianStDenis-pj1tq]

The above is actually the solution, for all cases of this puzzle. The inverse is also, which is start with 9, pour into 5, and repeat. Dr Barker was fond of saying 7 was the hardest to get from this puzzle. But, that's only with a narrowly defined starting point. Fill both, pour 9 into 5 and you have 7, one move. This is a great dinner puzzle for kids and those who like to think. We call it "buckets".

[@bertsimpsan]

@@BrianStDenis-pj1tq "fill both, pour 9 into 5 and you have 7". Am*rican common core education system at its finest

[@user-wu8yq1rb9t]

In this video something (almost) new happened! Your eye contact with us! It was so good, do it more and more.

[@emanuellandeholm5657]

Nice video! Reminds me of the bouncing DVD logo problem.

[@wyattstevens8574]

For the most efficient solution, do the first thing on this list that doesn't make you backtrack: Transfer water from the jug with more water to the one with less Empty the full jug Fill the empty jug Mathologer has an old video about this type of problem.

[@ShivanshSharma]

I do these kind of problems in the following way: Find a (multiple of 5) and a (multiple of 9) whose difference is 7 25(multiple of 5) - 18(Multiple of 9) = 7(Target) [There are other combinations too] (5+5+5+5+5) - (9+9) = 7 This means that you have to fill the 5l jug 5 times and empty the 9l jug 2 times. The way you did in the video is: 27-20 i.e. (9 + 9 + 9) - (5 +5)

[@user-wu8yq1rb9t]

Hello Dr. Barker Honestly, I'm not familiar with this subject, but I want to watch it again and find what's going on! Thank you so much

[@jonathanmetcalfe4616]

0,0 0,9 Fill 9 5,4 Transfer 9 to 5 0,4 Dump 5 4,0 Transfer 9 to 5 4,9 Fill 9 5,8 Transfer 9 to 5 0,8 Dump 5 5,3 Transfer 9 to 5 0,3 Dump 5 3,0 Transfer 9 to 5 3,9 Fill 9 5,7 Transfer 9 to 5 The contents of 9L jug is 7L.

[@jursamaj]

The graph version of this I've seen before (and like a lot) is done on a equilateral triangular grid, instead of a square grid.

[@HoSza1]

What if you'd have 3 jugs? The diagram would need another dimension.

[@keithmasumoto9698]

Let (a,b) represent the number of liters in the 9 and 5 jugs respectively. Then (0,0)(0,5)(5,0)(5,5)(9,1)(1,0)(1,5)(6,0)(6,5)(9,2)(0,2)(2,0)(2,5)(7,0).

[@HoSza1]

You missed a step, (0,1).

[@wyattstevens8574]

Do you always get to any volume within x+y moves (where x and y are the volumes)? Given x, y, and target volume z, is there a systematic way just by looking at x, y, and z to figure out whether to fill j_x or j_y to minimize the number of moves?

[@sethdurais2477]

So since 9 and 5 are co-prime, there would be infinitely many x-y integer pairs satisfying the equation 9x+5y=1 So then the base case would be 9*(1)+5*(-2)=1 i.e. (1,-2) is a solution and this can be extended to 9x+5y=n I'm not particularly clued up on Diophantine equations but hopefully that made sense 😅 Please do correct me if that explanation is off, and thanks in advance for doing so 💪

[@GhostyOcean]

Your solution gives -1 instead of 1.

[@sethdurais2477]

@@GhostyOcean Oops 😅 my bad hehe. I meant (-1,2) 🤦🏽‍♂️

[@Mercredi00Addams]

You can find it in the movie "die hard 3" they have to solve it very quickly to stop thé bomb Bruce Willis & Samuel L Jackson

[@tiborgrun6963]

My attempt: If there's 7 in the 9, we are done. If there's 2 in the 9 and we pour 5, we have 7 in the 9. If there's 2 in the 5, we pour 2 in the 9. If there's room for exactly 3 in the 9 and we pour 5, there's 2 in the 5. If there's 6 in the 9, there's room for exactly 3 in the 9. If there's 1 in the 9 and we pour 5, there's 6 in the 9. If there's 1 in the 5, we pour 1 in the 9. If there's room for exactly 4 in the 9 and we pour 5, there's 1 in the 5. If there's 5 in the 9, there's room for exactly 4 in the 9. We pour 5 in the 9.

[@DrBarker]

I like your "working backwards" kind of approach!

[@bash2357]

are you Oxbridge Dr B?

[@AlexLoveBananas]

What about (0,0)-(0,5)-(5,0)-(5,5)-(9,1)-(0,1)-(1-0)-(1,5)-(6,0)-(6,5)-(9,2)-(0,2)-(2,0)-(2,5)-(7,0)?

[@bscutajar]

10:35 I don't understand why there are 28 different points. Why does 28 come from 9 and 5? Surely if all points are possible then there would be 60.

[@DrBarker]

I'm not sure if I explained this very clearly, but the 28 points are just the points on the outside of the rectangle. So we don't include all 60 of the points inside the rectangle (e.g. (2, 2) wouldn't be counted). This is essentially because there's no way to know when to stop pouring other than if we run out of water, or completely fill one of the jugs, so we're restricted to the boundary.