# Tag Info

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Answer: Explanation: Here are the functions for the output (in Java). Output 1 = !((!C)|C) Output 2 = (!C)|C Output 3 = !((!C)|C) Output 4 = (A|B)&C Output 5 = (!(A^B))&(!(B^C)) Output 6 = ((!A)&(B^C))|((A&C)&(!B)) Output 7 = ((A&C)&(!B))|(!((B|C)|A)) Output 8 = (!(B&C))|A Using these functions

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I built a logic gate structure based on supercat's textual explanation (given in user2357112's answer and Gamow's comment); trace it at your leisure. It's a good thing we are given an infinite supply of AND and OR gates.

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So this is a classic example of Markov chains. The only slight trick is that you only want the material that's coming in fresh. But that's no problem: we just create a separate container next to each container to absorb the rest of the liquid. [Aside: As @Bass points out in the comments, this is known as an absorbing Markov chain. However, we will just ...

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Unfortunately I don't know how to draw the diagram but I will try to explain the logic behind it and where you dropped the acid you clumsy, you... You build a And the 3 operations were: (let's say the input was X and Y to make it easier to explain. No need to involve letters and numbers also) Why did the initial circuit work? And you dropped acid on ...

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The circuit alone rigorously tests one ironic aphorism and another time-tested adage. A specific scientific hypothesis for the overall rebus: Interpretation The circuit has an input at the top whose components get copied and split to the right. The legend along the right of the picture identifies crossings as no-contact crossovers, square components ...

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PART 1 As Mordechai already found, the image filename holds a clue. PART 2 Step 1: Step 2: Step 3: PART 3 Step 1: Step 2: CONCLUSION I came back to this question more than a year later just to find another terrible pun.

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Here is a horrible attempt at ascii Explanation

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I can't draw a circuit diagram right now (see below), but I think I have a way for two fuses to test three boxes. Connect each box in series with a 1 ohm resistor. Connect fuse 1 in series with a 1 ohm resistor. Connect fuse 2 in series with a 2 ohm resistor. Now connect all of the preceding parts together in parallel. Finally make a circuit with that, a 2 ...

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It can be done with 63 48 44 fuses (22 at top and bottom). The basic idea is that there's a top and bottom circuit in parallel, and a fuse will blow in whichever circuit has lower resistance. This will increase the resistance of that circuit and the process will repeat. The bottom circuit therefore establishes roughly how many insulators there are, and the ...

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Here's DrXorile's excellent answer for the first question again, but redone with closed form matrix algebra instead of computational iteration. (Contains a program for solving the general case.) Start by tabulating the transitions in two tables: How much keeps moving (to where?): How much stops (in where?): Then, we look up the correct wikipedia page, ...

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I'm not entirely sure that my interpretation of how the backwards timer works is entirely correct, but if it works like this: Then I think I can get Here's the diagram (pardon the mess): Here's how we'd set the timers, and what would happen:

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PART TWO I won't present a detailed description of a solution, but will attempt to give some heuristics and ideas that answer the question not only in the case of multiples of 3, but for multiples of any $k$. (Though different circuits should be designed for different $k$s.) First, However, A construction that achieves this: Once we have this, PART ...

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Partial

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I think the configuration we are looking at is Here, courtesy of the OP, is a diagram showing how this works: Now, So diagram 1 is obtained by Then diagram 2

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EDIT: I have underestimated the complexity of the four container system. There are multiple loops and each container will have its own common factor when computing the sum of the geometric series. Alas, I was unable to determine what that factor is yet, because the loops are nesting and my final sum in all containers does not equal exactly 1. My solution for ...

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PART FOUR This can probably be done with fewer gates, but at least it's a solution

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This is a solution:

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Solved over at http://electronics.stackexchange.com in the link provided by Gamow in the comments. This is a direct quote of supercat's excellent answer; go upvote that. It is possible to construct a purely-combinatorial three-input circuit consisting of a number of AND and OR gates along with exactly two independent inverters, with three outputs ...

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I think this is how we can construct a solution with 501 fuses 32 fuses. (Previous 45-fuse solution turned out to be incorrect... sorry =( ...) 501 fuses This is the general diagram: Let $N = 1000$ be the number of boxes, $0 \leq m \leq N$ be the number of conductor boxes, $s = N - m$ be the number of insulator boxes, and define $A = \sum_{i=1}^{F}{a_i}$....

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If I am correctly understanding the problem statement (see my comment above) then I think this does it:

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The key thing seems to be that So perhaps I suppose I should say explicitly that It's not clear to me, though, As for the application in physics,

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PART TWO   (only, overhauled) – divisibility by 3 (also 4 and 8) Circuits for divisibility by 3, 4 and 8 are worth presenting, despite a general solution’s already being accepted, as they occupy a neat little middle-ground between the absolute minimality of division by 2 and the complex scaling required for other higher factors.   ...

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PART ONE PART THREE PART FOUR PART FIVE NOTE

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The biggest I can get for now that should fulfill all conditions: Only using the four pieces of the top row (and ordering them exactly like in the top row) we set the following on-states: Timer1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 24 Timer2: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 We start with an on-time for 10 hours. Afterwards timer2 is set to off and ...

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I don't have an answer, but I can provide limits. Obviously, the upper limit for the problem is 1000 fuses. Just wire 1000 simple circuits, each with a 10V cell, a fuse, a $5 \Omega$ resistor, and a box. Surprisingly, the lower limit is $5$. This represents the minimum number of fuses that can blow in at least $1001$ different orderings. The constraints of ...

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Taking the bottom output as bit 0 up to the top output as bit 7. You can easily see that bits 5, 6, and 7 are always 010. Given that I wrote the following Python code (without looking at any answers to make it more fun!): def bit0(a, b, c): return a or not (b and c) def bit1(a, b, c): return bit1_5(a, b, c) or not (a or b or c) def bit1_5(a, b, c): ...

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Well, I'm afraid to state the obvious, but I can do $D=11$, $N=1$, no extras: Rig the circuit up like in the first picture: Set one timer to all on and the other to all on except hours 12 and 24 This obviously works because one timer acts as a wire and the other as a 11-on, 1-off timer with 12-hour period. I claim that no period of ...

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As an example of how this puzzle isn't necessarily as convoluted as it might look, this solution produces a 48 hour lighting cycle with D = 41 hours on and N = 7 hours off. Another example produces a 48 hour lighting cycle with D = 46 hours on and N = 2 hours off, where timer A actually powers ...

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Here's my solution First, the three gates are The are arranged such that As a diagram, it looks like this: That satisfies the initial condition. As for what the acid burned away, it would be the wire

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