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Frequently Asked Questions (FAQS);faqs.477 Then Sum ai ci x = delta1 error on weighing 1 Sum bi ci x = delta2 error on weighing 2 Now the ratio of delta1 to delta2 will be rational regardless of the value of x, since x will factor out; let's call this ratio p over q (p and q relatively prime). We would like to choose { ai } and { bi } such that for any set of mints J, which will be a subset of { 1 , 2 , ... , n }, that Sum aj ( = Sum ai ci ) is relatively prime to Sum bj. If this is true then we can determine the error x; it will simply be delta1/p, which is equal to delta2/q. If the { ai } have been carefully chosen, we should be able to figure out the bogus mints from one of the weighings, provided that all subsets ( { { aj } over all J } ) have unique sums. This was the strategy proposed above, where is was suggested that ai = 3 ** (i-1) ; note that you can use base 2 instead of base 3 since all the errors have the same sign. Well, for the time being I'm stumped. This agrees with the analysis I've been fighting with. I actually came up with a pair of functions that "almost" works. So that the rest of you can save some time (in case you think the way I did): Weighing 1: 1 coin from each mint Weighing 2: 2^(k-1) coins from mint k, for 1...k...n (total 2^n - 1 coins) Consider the n mints to be one-bit-each -- bit set -> mint makes bogus coins. Then we can just state that we're trying to discover "K", where K is a number whose bit pattern _just_ describes the bogosity of each mint. OK - now, assuming we know 'x', and we only consider the *difference* of the weighing from what it should be, for weighing 1, the devaiation is just the Hamming weight of K -- that is the number of 1-bits in it -- that is, the number of bogosifying mints. For weighing 2, the deviation is just K! When the nth bit of K is set, then that mint contributes just 2^n to the deviation, and so the total deviation will just be K. So that set me in search of a lemma: given H(x) is the hamming weight of x, is f(x) = x / H(x) a 1-1 map integers into rationals? That is, if x/H(x) = y/H(y) can we conclude that x = y? The answer (weep) is NO. The lowest pair I could find are 402/603 (both give the ratio 100.5). Boy it sure looked like a good conjecture for a while! Sigh. There are two parts to the problem. First let us try to come up with a solution to finding the answer in 2 weighings - then worry about using the min. number of coins. Solutions are for GENERAL n. Let N = set of all mints, 1 to n. Card(N) = n. Let P = set of all bogus mints. Let Card(P) = p. Weighing I: Weigh n coins, 1 from each mint. Since each "good" coins weighs one ounce, let delta1 be the error in weighing. Since all bogus coins are identical, let delta1 be abs(error). If x is the weight by which one bogus coin differs from a good coin, delta1 = p * x. Weighing II: The coins to be weighed are composed thusly. Let a1 be the number of coins from mint 1, a2 # from mint2 .. and an from mint n. All ai's are distinct integers. Let A = Set of all ai's. Let delta2 = (abs.) error in weighing 2 = x * k where k is the number of coins that are bogus in weighing two. Or more formally k = sigma(ai) (over all i in P) Assuming p is not zero (from Weighing I - in that case go back and get beheaded for giving the king BAAAAAD advice), Let ratio = delta1/delta2 = p/k. Let IR = delta2/delta1 = k/p = inverse-ratio (for later proof). Let S(i) be the bag of all numbers generated by summing i distinct elements from A. Clearly there will be nCi (that n comb. i) elements in S(i). [A bag is a set that can have the same element occur more than once.] So S(1) = A and S(n) will have one element that is the sum of all the elements of A. Let R(i) = {x : For-all y in S(i), x = i/y} expressed as p/q (no common factors). (R is a bag too). Let R-A = Bag-Union(R(i) for 1>= i >=n). (can include same element twice) Choose A, such that all elements of R-A are DISTINCT, i.e. Bag(R-A) = Set(R-A). Let the sequence a1, a2, .. an, be an L-sequence if the above property is true. Or more simply, A is in L. ********************************************************************** CONJECTURE: The bogus mint problem is solved in two weighings if A is in L. Sketchy proof: R(1) = all possible ratios (= delta1/delta2) when p=1. R(i) = all possible ratio's when p=i. Since all possible combinations of bogus mints are reflected in R, just match the actual ratio with the generated table for n. ************************************************************************ A brief example. Say n=3. Skip to next line if you want. Let A=(2,3,7). p=1 possible ratios = 1/2 1/3 1/7 p=2 possible ratios = 2/5 2/9 1/5(2/10) p=3 possible ratios = 1/4(3/12) (lots of blood in Scandalvania). As all outcomes are distinct, and the actual ratio MUST be one of these, match it to the answer, and start sharpening the axe. Note that the minimum for n=3 is A=(0,1,3) possible ratios are p=1 infinity (delta2=0),1,1/3 p=2 2/1,2/3,1/2 p=3 3/4 ************************************************************************ All those with the determination to get this far are saying OK, OK how do we get A. I propose a solution that will generate A, to give you the answer in two weighings, but will not give you the optimal number of coins. Let a1=0 For i>=2 >=n ai = i*(a1 + a2 + ... + ai-1) + 1 ***************************************** * i-1 * * ai = i* [Sigma(aj)] + 1 * ****Generator function G***** * j=1 * ***************************************** If A is L, all RATIO's are unique. Also all inverse-ratio's (1/ratio) are unique. I will prove that all inverse-ratio's (or IR's) are unique. Let A(k), be the series generated by the first k elements from eqn. G.(above) ************************************************************************ PROOF BY INDUCTION. A(1) = {0} is in L. A(2) = {0,1} is in L. ASSUME A(k) = {0,1, ..., ak} is in L. T.P.T. A(k+1) = {0,1, ..., ak, D) is in L where D is generated from G. We know that all IR's(inverse ratio's) from A(k) are distinct. Let K = set of all IR's of A(k). Since A(k+1) contains A(k), all IR's of A(k) will also be IR's of A(K+1). So for all P, such that (k+1) is not in P, we get a distinct IR. So consider cases when (k+1) is in P. p=1 (i.e. (k+1) = only bogus mint), IR = D ______________________________________________________________________ CONJECTURE: Highest IR for A(k) = max(K) = ak Proof: Since max[A(k)] = ak, for p'= 1, max IR = ak/1 = ak for p'= 2, max IR (max sum of 2 ai's)/2 = (ak + ak-1)/2 < ak (as ak>ak-1). for p'= i max IR sum of largest i elements of A(k) -------------------------------- i < i * ak/i = ak. So max. IR for A(k) is ak. ______________________________________________________________________ D > ak So for p=1 IR is distinct. Let Xim be the IR formed by choosing i elements from A(k+1). Note: We are choosing D and (i-1) elements from A(k). m is just an index to denote each distinct combination of (i-1) elemnts of A(i). ______________________________________________________________________ CONJECTURE : For p=j, all new IR's Xjm are limited to the range D/(j-1) > Xjm > D/j. Proof: Xjm = (D + {j-1 elements of A(k)})/j Clearly Xjm > D/j. To show: max[Xjm] < D/(j-1) Note: a1 + a2 .. + ak < D/(k+1) max[Xjm] = (D + ak + ak-1 + ... + a(k-j+1))/j < (D + D/(k+1))/j = D (k+2)/(k+1)j = [D/(j-1)] * alpha. alpha = (j-1)/(j) * (k+2)/(k+1) Since j <= k, (j-1)/j <= (k-1)/k < (k+1)/(k+2) IMPLIES alpha < 1. Conjecture proved. ______________________________________________________________________ CONJECTURE : For a given p, all newly generated IR's are distinct. Proof by contradiction: Assume this is not so. Implies (D + (p-1) elements of A(k))/p = (D + some other (p-1) elements of A(k))/p Implies SUM[(p-1) elements of A(k)] = SUM[ some other (p-1) elements of A(k)] Implies SUM[(p-1) elements of A(k)]/(p-1) = SUM[some other (p-1) elements]/(p-1) Implies A(k) is NOT in L. Contra. Hence conjecture. ______________________________________________________________________ CONJECTURE: A(k+1) is in L. Since all newly generated IR's are distinct from each other, and all newly generated IR's are greater than previous IR's, A(k+1) is in L. ==> logic/zoo.p <== I took some nephews and nieces to the Zoo, and we halted at a cage marked Tovus Slithius, male and female. Beregovus Mimsius, male and female. Rathus Momus, male and female. Jabberwockius Vulgaris, male and female. The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no! It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right. As the four species were easily distinguished, no mistake would arise in pairing the animals; naturally a child who identified one animal as Mr Tove identified the other animal of the same species as Mrs Tove. The keeper, who consented to judge the lists, scrutinised them carefully. "Here's a queer thing. I take two of the lists, say, John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs Tove. It is just the same for every pair of lists, and for all four species. "Curiouser and curiouser! Each boy supposes Mr Tove to be the animal which he supposes to be Mr Tove; but each girl supposes Mr Tove to be the animal which she supposes to be Mrs Tove. And similarly for the oth- er animals. I mean, for instance, that the animal Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs Tove." "It seems a little involved," I said, "but I suppose it is a remarkable coincidence." "Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children." How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right? [by Sir Arthur Eddington] ==> logic/zoo.s <== Given that there is at least one boy and one girl (John and Mary are mentioned) then the answer is that there were 3 nephews and 2 nieces, the winner was a boy who got 4 right. Number the animals 1 through 8, such that the females are even and the males are odd, with members of the same species consecutive; i.e. 1 is Mr. Tove, 2 Mrs. Tove, etc. Then each childs guesses can be represented by a permutation. I use the standard notation of a permutation as a set of orbits. For example: (1 3 5)(6 8) means 1 -> 3, 3 -> 5, 5 -> 1, 6 -> 8, 8 -> 6 and 2,4,7 are unchanged. [1] Let P be any childs guesses. Then P(mate(i)) = mate(P(i)). [2] If Q is another childs guesses, then [P,Q] = T, where [P,Q] is the commutator of P and Q (P composed with Q composed with P inverse composed with Q inverse) and T is the special permutation (1 2) (3 4) (5 6) (7 8) that just swaps each animal with its spouse. [3] If P represents a boy, then P*P = I (I use * for composition, and I for the identity permutation: (1)(2)(3)(4)(5)(6)(7)(8) [4] If P represents a girl, then P*P = T. [1] and [4] together mean that all girl's guesses must be of the form: (A B C D) (E F G H) where A and C are mates, as are B & D, E & F G & H. So without loss of generality let Mary = (1 3 2 4) (5 7 6 8) Without to much effort we see that the only possibilities for other girls "compatible" with Mary (I use compatible to mean the relation expressed in [2]) are: g1: (1 5 2 6) (3 8 4 7) g2: (1 6 2 5) (3 7 4 8) g3: (1 7 2 8) (3 5 4 6) g4: (1 8 2 7) (3 6 4 5) Note that g1 is incompatible with g2 and g3 is incompatible with g4. Thus no 4 of Mary and g1-4 are mutually compatible. Thus there are at most three girls: Mary, g1 and g3 (without loss of generality) By [1] and [3], each boy must be represented as a product of transpostions and/or singletons: e.g. (1 3) (2 4) (5) (6) (7) (8) or (1) (2) (3 4) (5 8) (6 7). Let J represent John's guesses and consider J(1). If J(1) = 1, then J(2) = 2 (by [1]) using [2] and Mary J(3) = 4, J(4) = 3, and g1 & J => J(5) = 6, J(6) = 5, & g3 & J => J(8) = 7 J(7) = 8 i.e. J = (1)(2)(3 4)(5 6)(7 8). But the [J,Mary] <> T. In fact, we can see that J must have no fixed points, J(i) <> i for all i, since there is nothing special about i = 1. If J(1) = 2, then we get from Mary that J(3) = 3. contradiction. If J(1) = 3, then J(2) = 4, J(3) = 1, J(4) = 2 (from Mary) => J(5) = 7, J(6) = 8, J(7) = 5, J(8) = 6 => J = (1 3)(2 4)(5 7)(6 8) (from g1) But then J is incompatible with g3. A similar analysis shows that J(1) cannot be 4,5,6,7 or 8; i.e. no J can be compatible with all three girls. So without loss of generality, throw away g3. We have Mary = (1 3 2 4) (5 7 6 8) g1 = (1 5 2 6) (3 8 4 7) The following are the only possible boy guesses which are compatible with both of these: B1: (1)(2)(3 4)(5 6)(7)(8) B2: (1 2)(3)(4)(5)(6)(7 8) B3: (1 3)(2 4)(5 7)(6 8) B4: (1 4)(2 3)(5 8)(6 7) B5: (1 5)(2 6)(3 8)(4 7) B6: (1 6)(2 5)(3 7)(4 8) Note that B1 & B2 are incombatible, as are B3 & B4, B5 & B6, so at most three of them are mutually compatible. In fact, Mary, g1, B1, B3 and B5 are all mutually compatible (as are all the other possibilities you can get by choosing either B1 or B2, B3 or B4, B5 or B6. So if there are 2 girls there can be 3 boys, but no more, and we have already eliminated the case of 3 girls and 1 boy. The only other possibility to consider is whether there can be 4 or more boys and 1 girl. Suppose there are Mary and 4 boys. Each boy must map 1 to a different digit or they would not be mutually compatible. For example if b1 and b2 both map 1 to 3, then they both map 3 to 1 (since a boy's map consists of transpositions), so both b1*b2 and b2*b1 map 1 to 1. Furthermore, b1 and b2 cannot map 1 onto spouses. For example, if b1(1) = a and b is the spouse of a, then b1(2) = b. If b2(1) = b, then b2(2) = a. Then b1*b2(1) = b1(b) = 2 and b2*b1(1) = b2(a) = 2 (again using the fact that boys are all transpostions). Thus the four boys must be: B1: (1)(2)... or (1 2).... B2: (1 3)... or (1 4) ... B3: (1 5) ... or (1 6) ... B4: (1 7) ... or (1 8) ... Consider B4. The only permutation of the form (1 7)... which is compatible with Mary ( (1 3 2 4) (5 7 6 8) ) is: (1 7)(2 8)(3 5)(4 6) The only (1 8)... possibility is: (1 8)(2 7)(3 6)(4 5) Suppose B4 = (1 7)(2 8)(3 5)(4 6) If B3 starts (1 5), it must be (1 5)(2 6)(3 8)(4 7) to be compatible with B4. This is compatible with Mary also. Assuming this and B2 starts with (1 3) we get B2 = (1 3)(2 4)(5 8)(6 7) in order to be compatible with B4. But then B2*B3 and B3*B2 moth map 1 to 8. I.e. no B2 is mutually compatible with B3 & B4. Similarly if B2 starts with (1 4) it must be (1 4)(2 3)(5 7)(6 8) to work with B4, but this doesn't work with B3. Likewise B3 starting with (1 6) leads to no possible B2 and the identical reasoning eliminates B4 = (1 8)... So no B4 is possible! I.e at most 3 boys are mutually compatiblw with Mary, so 2 girls & 3 boys is optimal. Thus: Mary = (1 3 2 4) (5 7 6 8) Sue = (1 5 2 6) (3 8 4 7) John = (1)(2)(3 4)(5 6)(7)(8) Bob = (1 3)(2 4)(5 7)(6 8) Jim = (1 5)(2 6)(3 8)(4 7) is one optimal solution, with the winner being John (4 right: 1 2 7 & 8) ==> physics/balloon.p <== A helium-filled balloon is tied to the floor of a car that makes a sharp right turn. Does the balloon tilt while the turn is made? If so, which way? The windows are closed so there is no connection with the outside air. ==> physics/balloon.s <== Because of buoyancy, the helium balloon on the string will want to move in the direction opposite the effective gravitational field existing in the car. Thus, when the car turns the corner, the balloon will deflect towards the inside of the turn. ==> physics/bicycle.p <== A boy, a girl and a dog go for a 10 mile walk. The boy and girl can walk 2 mph and the dog can trot at 4 mph. They also have bicycle which only one of them can use at a time. When riding, the boy and girl can travel at 12 mph while the dog can peddle at 16 mph. What is the shortest time in which all three can complete the trip? ==> physics/bicycle.s <== First note that there's no apparent way to benefit from letting either the boy or girl ride the bike longer than the other. Any solution which gets the boy there faster, must involve him using the bike (forward) more; similarly for the girl. Thus the bike must go backwards more for it to remain within the 10-mile route. Thus the dog won't make it there in time. So the solution assumes they ride the bike for the same amount of time. Also note that there's no apparent way to benefit from letting any of the three arrive at the finish ahead of the others. If they do, they can probably take time out to help the others. So the solution assumes they all finish at the same time. The boy starts off on the bike, and travels 5.4 miles. At this point, he drops the bike and completes the rest of the trip on foot. The dog eventually reaches the bike, and takes it *backward* .8 miles (so the girl gets to it sooner) and then returns to trotting. Finally, the girl makes it to the bike and rides it to the end. The answer is 2.75 hours. The puzzle is in Vasek Chvatal, Linear Programming, W. H. Freeman & Co. The generalized problem (n people, 1 bike, different walking and riding speeds) is known as "The Bicycle Problem". A couple references are Masuda, S. (1970). "The bicycle problem," University of California, Berkeley: Operations Research Center Technical Report ORC 70-35. Chvatal, V. (1983). "On the bicycle problem," Discrete Applied Mathematics 5: pp. 165 - 173. As for the linear program which gives the lower bound of 2.75 hours, let t[person, mode, direction] by the amount of time "person" (boy, girl or dog) is travelling by "mode" (walk or bike) in "direction" (forward or backwards). Define Time[person] to be the total time spent by person doing each of these four activities. The objective is to minimize the maximum of T[person], for person = boy, girl, dog, e.g. minimize T subject to T >= T[boy], T >= T[girl], T >= T[dog]. Now just think of all the other linear constraints on the variables t[x,y,z], such as everyone has to travel 10 miles, etc. In all, there are 8 contraints in 18 variables (including slack variables). Solving this program yields the lower bound. ==> physics/boy.girl.dog.p <== A boy, a girl and a dog are standing together on a long, straight road. Simulataneously, they all start walking in the same direction: The boy at 4 mph, the girl at 3 mph, and the dog trots back and forth between them at 10 mph. Assume all reversals of direction instantaneous. In one hour, where is the dog and in which direction is he facing? ==> physics/boy.girl.dog.s <== The dog's position and direction are indeterminate, other than that the dog must be between the boy and girl (endpoints included). To see this, simply time reverse the problem. No matter where the dog starts out, the three of them wind up together in one hour. This argument is not quite adequate. It is possible to construct problems where the orientation changes an infinite number of times initially, but for which there can be a definite result. This would be the case if the positions at time t are uniformly continuous in the positions at time s, s small. But suppose that at time a the dog is with the girl. Then the boy is at 4a, and the time it takes the dog to reach the boy is a/6, because the relative speed is 6 mph. So the time b at which the dog reaches the boy is proportional to a. A similar argument shows that the time the dog next reaches the girl is b + b/13, and is hence proportional to b. This makes the position of the dog at time (t > a) a periodic function of the logarithm of a, and thus does not approach a limit as a -> 0. ==> physics/brick.p <== What is the maximum overhang you can create with an infinite supply of bricks? ==> physics/brick.s <== You can create an infinite overhang. Let us reverse the problem: how far can brick 1 be from brick 0? Let us assume that the brick is of length 1. To determine the place of the center of mass a(n): a(1)=1/2 a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n) Thus n 1 n 1 a(n)=Sum -- = 1/2 Sum - = 1/2 H(n) m=1 2m m=1 m Needless to say the limit for n->oo of half the Harmonic series is oo. ==> physics/cannonball.p <== A person in a boat drops a cannonball overboard; does the water level change? ==> physics/cannonball.s <== The cannonball in the boat displaces an amount of water equal to the MASS of the cannonball. The cannonball in the water displaces an amount of water equal to the VOLUME of the cannonball. Water is unable to support the level of salinity it would take to make it as dense as a cannonball, so the first amount is definitely more than the second amount, and the water level drops. ==> physics/dog.p <== A body of soldiers form a 50m-by-50m square ABCD on the parade ground. In a unit of time, they march forward 50m in formation to take up the position DCEF. The army's mascot, a small dog, is standing next to its handler at location A. When the B----C----E soldiers start marching, the dog | | | forward--> begins to run around the moving A----D----F body in a clockwise direction, keeping as close to it as possible. When one unit of time has elapsed, the dog has made one complete circuit and has got back to its handler, who is now at location D. (We can assume the dog runs at a constant speed and does not delay when turning the corners.) How far does the dog travel? ==> physics/dog.s <== Let L be the side of the square, 50m, and let D be the distance the dog travels. Let v1 be the soldiers' marching speed and v2 be the speed of the dog. Then v1 = L / (1 time unit) and v2 = v1*D/L. Let t1, t2, t3, t4 be the time the dog takes to traverse each side of the square, in order. Find t1 through t4 in terms of L and D and solve t1+t2+t3+t4 = 1 time unit. While the dog runs along the back edge of the square in time t1, the soldiers advance a distance d=t1*v1, so the dog has to cover a distance sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2. Solving for t1 gives t1=L/sqrt(v2^2 - v1^2). The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1). In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by using v1=L/(1 time unit), obtaining 2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1 which can be turned into D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0 which has a root D = 4.18113L = 209.056m. ==> physics/magnets.p <== You have two bars of iron. One is magnetic, the other is not. Without using any other instrument (thread, filings, other magnets, etc.), find out which is which. ==> physics/magnets.s <== Take the two bars, and put them together like a T, so that one bisects the other. ___________________ bar A ---> |___________________| | | | | | | | | bar B ------------> | | | | | | |_| If they stick together, then bar B is the magnet. If they don't, bar A is the magnet. (reasoning follows) Bar magnets are "dead" in their centers (ie there is no magnetic force, since the two poles cance out). So, if bar A is the magnet, then bar B won't stick to its center. However, bar magnets are quite "alive" at their edges (ie the magnetic force is concentrated). So, if bar B is the magnet, then bar A will stick nicely to its end. ==> physics/milk.and.coffee.p <== You are just served a hot cup of coffee and want it to be as hot as possible when you drink it some number of minutes later. Do you add milk when you get the cup or just before you drink it? ==> physics/milk.and.coffee.s <== Normalize your temperature scale so that 0 degrees = room temperature. Assume that the coffee cools at a rate proportional to the difference in temperature, and that the amount of milk is sufficiently small that the constant of proportinality is not changed when you add the milk. An early calculus homework problem is to compute that the temperature of the coffee decays exponentially with time, T(t) = exp(-ct) T0, where T0 = temperature at t=0. Let l = exp(-ct), where t is the duration of the experiment. Assume that the difference in specific heats of coffee and milk are negligible, so that if you add milk at temperature M to coffee at temperature C, you get a mix of temperature aM+bC, where a and b are constants between 0 and 1, with a+b=1. (Namely, a = the fraction of final volume that is milk, and b = fraction that is coffee.) If we let C denote the original coffee temperature and M the milk temperature, we see that Add milk later: aM + blC Add milk now: l(aM+bC) = laM+blC The difference is d=(1-l)aM. Since l<1 and a>0, we need to worry about whether M is positive or not. M>0: Warm milk. So d>0, and adding milk later is better. M=0: Room temp. So d=0, and it doesn't matter. M<0: Cold milk. So d<0, and adding milk now is better. Of course, if you wanted to be intuitive, the answer is obvious if you assume the coffee is already at room temperature and the milk is either scalding hot or subfreezing cold. Moral of the story: Always think of extreme cases when doing these puzzles. They are usually the key. Oh, by the way, if we are allowed to let the milk stand at room temperature, then let r = the corresponding exponential decay constant for your milk container. Add acclimated milk later: arM + blC We now have lots of cases, depending on whether r<l: The milk pot is larger than your coffee cup. (E.g, it really is a pot.) r>l: The milk pot is smaller than your coffee cup. (E.g., it's one of those tiny single-serving things.) M>0: The milk is warm. M<0: The milk is cold. Leaving out the analysis, I compute that you should... Add warm milk in large pots LATER. Add warm milk in small pots NOW. Add cold milk in large pots NOW. Add cold milk in small pots LATER. Of course, observe that the above summary holds for the case where the milk pot is allowed to acclimate; just treat the pot as of infinite size. ==> physics/mirror.p <== Why does a mirror appear to invert the left-right directions, but not up-down? ==> physics/mirror.s <== Mirrors invert front to back, not left to right. The popular misconception of the inversion is caused by the fact that a person when looking at another person expects him/her to face her/him, so with the left-hand side to the right. When facing oneself (in the mirror) one sees an 'uninverted' person. See Martin Gardner, ``Hexaflexagons and other mathematical diversions,'' University of Chicago Press 1988, Chapter 16. A letter by R.D. Tschigi and J.L. Taylor published in this book states that the fundamental reason is: ``Human beings are superficially and grossly bilaterally symmetrical, but subjectively and behaviorally they are relatively asymmetrical. The very fact that we can distinguish our right from our left side implies an asymettry of the perceiving system, as noted by Ernst Mach in 1900. We are thus, to a certain extent, an asymmetrical mind dwelling in a bilaterally symmetrical body, at least with respect to a casual visual inspection of our external form.'' Martin Gardner has also written the book ``The Amidextrous Universe.'' ==> physics/monkey.p <== Hanging over a pulley, there is a rope, with a weight at one end. At the other end hangs a monkey of equal weight. The rope weighs 4 ounces per foot. The combined ages of the monkey and it's mother is 4 years. The weight of the monkey is as many pounds as the mother is years old. The mother is twice as old as the monkey was when the mother was half as old as the monkey will be when the monkey is 3 times as old as the mother was when she was 3 times as old as the monkey.