Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

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Any commutative hash function will do.  Using RSA makes this relatively easy, I think. So Alice and Beth both establish their secret primes, and, in a twist, keep everything secret.  $% Make EA, EB, DA and DB look like functions; i.e., *not* italic: \def\EA{\operatorname{EA}} \def\EB{\operatorname{EB}} \def\DA{\operatorname{DA}} \def\DB{\operatorname{DB}}$...

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You only need Assume the genuine coins weigh $x$ grammes. Credit to Hexomino and Jaap for the corrections!

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Star Battle solution The first few steps: More progress: A bit more progress: Finishing off the puzzle: Cryptic solution The answers to the cryptic clues: Assembly To assemble the grid, notice that there are: This means that Here's how to do it:

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This works, although I'm not sure if it's unique: How I found it We have two 6s (red and black) with six cards between them that must be all the same colour (let's say red). Key fact: Now, to begin with, After that, Also,

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Alice takes 1 cardboard box, 10 envelopes, 9 matching coins or disk, and 1 coin/disk of the same size/shape/weight, with a different colour or pattern. This difference may be disclosed in advance - see the final step. Alice places 1 coin in each of the envelopes and seals them all (keeping track of the "different" coin) then stacks the envelopes inside the ...

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Let's see.. Since most of the information is in point 4, it seems sensible to start the reasoning there. 4(a) If clubs are the trump suit, then both the queen with the fruits and one with the wheat ears can beat the jack holding a scythe. From this we get that the suits of the fruit and wheat queens must be Combining this with The queen of the ...

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Okay, this was worth persisting with! It's hard to spot exactly where to start without some trial and error, and it's very hard to explain the logical steps retrospectively since a lot of it is a case of "Well, that didn't work... how about this instead...?" but the final solution is very satisfying... Note that in the diagrams that follow I have converted ...

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There are ${30 \choose 2} = 435$ ways of selecting 2 coins out of 30. Since the fakes can be heavy or light, that yields $435 \times 2 = 870$ possible combinations. Since each test yeilds one of 3 outcomes, and $3^6 = 729 < 870 < 2187 = 3^7$, at least 7 tests are needed. Below is proof that Update with more details

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These steps detail a full algorithm to figuring out which two coins are fake. Worst case, this takes $10$ comparisons. Edit: The algorithm was fixed to account for the error noted in the comments. This didn't affect the worst case scenario. Step 1: Split the $30$ coins into $3$ piles of $10$, which we will call $A$, $B$, and $C$. Compare $A$ and $B$. If ...

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Here is a solution that takes 1.5 minutes.

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A brute force solution: We start with the two sixes with six empty places in between. We must fill those six places with exactly one of each number 1 to 5, as well as 7, while following the rules of spacing. Now, filling in the 3: The 4: When placing the 5, we can easily reject any solution that would leave more than 1 empty place in each suit, since we ...

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Alice takes a basic calculator (a simple one that doesn't show history) and covers up the screen with something (paper, non-transparent tape, etc). Alice types in her number and hits the divide key. Beth takes the calculator, types in her number and hits the equals key. They take the cover off the screen and if the screen shows any number other than 1, they ...

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Update: Added a simplification of Hubble07's answer for variations 0 and 1. Variation 0 (original question) The goal is to get 2/3 correct guesses as often as possible. To do that, A and B should ensure that exactly one of them is correct if C is correct, and that either both or neither of them is correct if C is incorrect. One possible strategy is ...

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The number of weighings will be: What will you do is to: Mathematically speaking: So the next step is:

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Are you Because More specifically,

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What a great first puzzle and nicely related to real life! I can solve it in at most Like so

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I have an algorithm to achieve the task with no more than There are some interesting solutions here, most of them do not appear to be complete, however, so I have decided to add mine as well. First, divide the 30 coins into 5 groups of 6 coins each. Label them A, B, C, D, and E. The gist of the algorithm is to find out which of these groups have a fake and ...

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Here is my simple methodology; Let's call then Step 12v34 and the possible outcomes become; \begin{array}{c|c|c|c|} & \text{12} & \text{34}& \text{Result} \\ \hline \text{I} & GG & GF& NE\\ \hline \text{II} & GG & FF& NE\\ \hline \text{III} & GG & GG& E\\ \hline \text{IV} & GF & GF &E\\ \...

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Assuming Alice and Beth can be assigned the same number, then no such algorithm exists, since the algorithm would sometimes reveal that the numbers are not distinct, which means revealing their numbers to each other. (If Alice and Beth cannot be assigned the same number, then that needs to be specified)

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