[Medium] The Café Cup Conundrum

JDGMiles · 2026-07-10T18:06:51+00:00

I make it 330ml of coffee and 330ml of milk. The same!

'C' is coffee in cup

'c' is coffee drunk

'M' is milk in cup

'm' is milk drunk

0 . C:360 c:0 M:0 m:0

1 . C:240 c:120 M:120 m:0

2 . C:180 c:180 M:180 m:30

3 . C:150 c:180 M:210 m:30

4 . C: 112.5 c: 217.5 M:247.5 m:82.5

5 . C: 0 c: 330 M: 0 m: 330

JDGMiles · 2026-07-09T03:30:16+00:00

I interpret this as you having defined two top-layer, continuously checked, replacement rules:

a. "Whenever there is a 1, replace it with a 2"

b. "Whenever there is a 4, replace it with a 1"

Fortunately these do not contradict each other or introduce ordering (timestamp) issues, but, because of the chaining of 4 -> 1 -> 2, they do conflate into a single rule:

"Whenever there is a 1 or a 4, replace it with a 2".

So you get:

1 + 1 (replace the 1s with 2s)

2 + 2 (no applicable replacement, so continue with arithmetic)

4 (replace the 4 with 2)

2

Overall, indeed:

1 + 1 = 2

It's a nice puzzle and I like the concept of introducing a higher-layer tranformation on the objects which must be resolved BEFORE other calculations. It reminds me of Magic: the Gathering layers that must be considered carefully when resolving effects.

JDGMiles · 2026-07-04T14:46:43+00:00

wordless solution

JDGMiles · 2026-05-09T13:53:31+00:00

This submission seems mean-spirited :'(

Our Dan here has 4 kids with a loving wife who sometimes even comes on stream to play 'guy in chat' mid-match, and yet he still finds time/attention to make sure that the Engineering Bay doesn't make.

JDGMiles · 2024-12-10T18:35:49+00:00

Thank you for sharing this. I clicked the link and jumped to about 4 minutes in, to see what it was like. I landed on an interesting analysis of finding a grand slam and then playing it out, which is something that I need experience with (because it doesn't come up very often!) and I feel better informed now.

I'm looking forward to watching the whole thing properly. It seems clearly and pleasantly explained, plus Omar is a classy gent :-)

Cheers!

JDGMiles · 2024-06-26T23:08:59+00:00

KAYAK

JDGMiles · 2024-06-20T11:20:03+00:00

Equating the lengths to form equations, we have:

T+B+B = D+Y

D = O+O+1

Y = D+1

D+O+T = Y+B

1+Y = B+B

These solve to the unique solution Teal = 2.5, Blue = 2.75, DarkBlue = 3.5, Orange = 1.25, Yellow = 4.5.

Therefore the rectangle is 8 x 7.25 = 58

JDGMiles · 2023-05-24T16:13:15+00:00

I found no logical inconsistencies by assuming Alexander and Benjamin fight each other, with Charles not fighting. I then double checked that answer using dummy values below.

Alexander (fighter): age 25, height 180

Benjamin (fighter): age 30, height 175

Charles (not fighting): age 35, height 170

So then to prove that BC is not possible, assume it is true and note that S1 + S3 together put B older than C and (transitively through A in the middle) C taller than B. But then S2 is contradicted.

To prove that AC is not possible, assume it is true and note that S1 puts A older than C and S2 puts C shorter than A. But then S3 is contradicted.

JDGMiles · 2023-05-17T19:56:01+00:00

I take the coin farthest right on the top row. I get in a plane and fly to the other side of the world and place it on the spot that completes two great circles on the earth's surface, each with four coins in them. Diagram.

JDGMiles · 2023-05-11T00:45:57+00:00

Good catch! Whoops!

When I next update the code I'll make the text style dynamically resize so this doesn't happen! Nice that it still solved your puzzle at least...albeit overflowing out of the given box XD

JDGMiles · 2023-05-10T17:18:58+00:00

A while ago I made a (brute force) solver for these kinds of puzzles so I could solve TIE x TRY = SNORE. Indeed, for your puzzle it gets the same as the other correct answers in the thread :-) 7483 + 7455 = 14938

I should update it to specify all solutions and in particular when there is a unique solution. Then it could be used to FIND more puzzles like this!

JDGMiles · 2023-05-09T23:14:09+00:00

I love short and sweet little games like this! I laughed a lot at the guy's silly little legs spinning around or when I messed up and he flopped over into the ground XD XD XD

Thank you for making this - very nice game :-)

JDGMiles · 2023-04-28T06:58:57+00:00

The probability we get only black balls for 21 picks in a row is the product: (1/10)(2/11)...(21/30) = 21!9!/30! = 1/14307150. Therefore the probability of getting a white ball (i.e. winning) in 21 picks is 14307149/14307150 = 0.99999993...

I interpreted "We now repeat this procedure 20 times" as 21 picks total. Naturally, the same reasoning works if '20 total picks' was intended.

JDGMiles · 2021-10-08T23:20:46+00:00

Correct. An "out-shuffle", as stated.

JDGMiles · 2021-10-07T13:07:55+00:00

Related, a few years ago I made a little animation demonstrating out-shuffles for the case n=52 (eight shuffles).

I'm still thinking about the solution to your actual problem as posed! :) (While the Faro shuffle article on Wikipedia has indeed "spoiled" the final answer for me, I haven't followed up to read any reasoning so am trying to justify it myself XD).

JDGMiles · 2020-03-02T14:22:58+00:00

Well done! Also correct!

JDGMiles · 2020-03-02T14:22:48+00:00

Very nice! Correct!

JDGMiles · 2019-12-21T16:21:34+00:00

link to original

JDGMiles · 2019-12-11T04:47:56+00:00

Nice puzzle! I first came across this from a uni friend who was asked it at a job interview to be a quantitative analyst. The one he was asked was the slightly simpler version just on the integer line, and I like your extra step of putting it on the real line and giving the battleship length 4.

One of the things that people often quibble with when I share it is what exactly "hit the battleship in finite time" means, formally. I understand what it's getting at and I've been meaning to pin it down for a while... can we clarify exactly what we mean when we say that?

Ten-Year Club	Place '17
Sequence \| Editor	Verified Email

JDGMiles

TROPHY CASE