< Earlier Kibitzing · PAGE 244 OF 244 ·
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Oct2414   Schwartz: Oh.. ..oops!! Embarrassing.. Seems I need to relearn to count. My apologies, <al wazir>. Okay.. so.. 90, not 170, is the number to beat. 

Oct2414
  Shams: <al wazir> My ears prick up whenever you say a problem is "really easy". Thanks for the Mondaylevel puzzle. 

Oct2414
  Sneaky: I had a devil of a time just figuring out the decimal calendar cubes problem. I've seen them in banks (etc.) but never played with one to see how it works. Here's what I finally came up with: cube 1 = 1, 2, 3, 4, 5, 6
cube 2 = 0, 1, 2, 7, 8, 9
It's worth mentioning that there will be situations where you can't actually use both cubes, for example the number 8 cannot be expressed as "08", you have to just show the cube with 8 on its face, and hide the other one until it's needed again. I thought of one other method which eliminates the problem of not using two blocks, but it has a kind of a cheat: cube 1 = 0, 1, 2, 3, 4, 5
cube 2 = 0, 1, 2, 7, 8, 9
The cheat is, there is no face that reads "6" but you can just turn the 9 upside down. 

Oct2414
  Sneaky: Just to make sure we're asking the same question on the binarycubes problem: #1  <Are we allowed to flip the cubes to use the same face for two different numbers?> In other words, can I use the same face as both 1101 and 1011? #2  It's perfectly OK to only use one of the two blocks, right? So if I have a 001 on one block, I can use that for my number one, right? I don't need to also have a 0 on the other block to stick in front of it, or do I? 

Oct2414
  al wazir: <Sneaky: Here's what I finally came up with: cube 1 = 1, 2, 3, 4, 5, 6
cube 2 = 0, 1, 2, 7, 8, 9
It's worth mentioning that there will be situations where you can't actually use both cubes, for example the number 8 cannot be expressed as "08", you have to just show the cube with 8 on its face, and hide the other one until it's needed again.> By saying "every <pair> of numbers from 0 1 to 3 1," I hinted that you have to use both cubes (if you used only one in the bank, where would the other cube be, in a vault?), but maybe I should have made that condition explicit. <I thought of one other method which eliminates the problem of not using two blocks, but it has a kind of a cheat: cube 1 = 0, 1, 2, 3, 4, 5
cube 2 = 0, 1, 2, 7, 8, 9
The cheat is, there is no face that reads "6" but you can just turn the 9 upside down.> Bingo! 

Oct2414
  al wazir: For the record, in the decimal problem both cubes have to have 1 and 2 so that you can make 11 and 22. There has to be a 0 to allow for 10 and 20, and the 3 has be on the other block to allow for 30. So now you have 0, 1, 2, _, _, _ and 1, 2, 3, _, _, _.
That leaves six spaces for the six remaining digits: for example, 4, 5, and 6 on the first one and 7, 8, and 9 on the other. But you're required to use both cubes, which means there's no way to make 04, 05, and 06. There has to be a 0 on the second cube too. Now you don't have enough places for all the digits. That's where the trick of inverting the 9 to make a 6 comes in. So the two cubes contain the numbers 0, 1, 2, 3, 4, 5 and 0, 1, 2, 6/9, 7, 8, respectively. This is what I found on the calendar cubes I saw yesterday in a doctor's office. But they could just as well have had 3, 5, and 7 on one and 4, 6, and 8 on the other, or other arrangements. And of course the order in which they're placed on the faces doesn't matter, in contrast to ordinary dice, where the numbers on opposite faces add up to 7 (1 and 6, 2 and 5, 3 and 4). On to the binary problem! 

Oct2514
  Sneaky: So based on your interpretation of the normal calendar cubes puzzle, I gather that it's perfectly OK to invert numbers (110 can become 011, etc.) but it's NOT permitted to only use one of the two blocks. An interesting puzzle and I can't think of a way to solve it other that seeing a computer program churn through endless variations. I could write such a program, with great effort, but it seems that there should be a more intuitive solution. Very good puzzle. I haven't given up, just not ready to make a serious contribution. 

Oct2514
  Sneaky: Until we tackle the complicated binary cubes problem, here is a less serious contribution for amusement. I will read your mind. First, I want you to think of any country that starts with a "D". Got it? OK, now look at the second letter in that country's name. Got it? OK, now think of an animal that starts with that letter. Got it? OK, now think of the color of that animal.
Got it? OK, now I read your mind. Behold the great Sneakdini: http://pastebin.com/56fXz8Tb 

Oct2514
  al wazir: <Sneaky: OK, now I read your mind.> No, from Deutschland. Sorry about that. 

Oct2514   tbentley: I suppose nobody came up with the jabiru. 

Oct2514
  Sneaky: Hey, at least I got the animal and color correct. Two out of three ain't bad, right? There's always one wise guy who comes up with a white ostrich from the Dominican Republic. 

Oct2514
  TheFocus: What do you get when you combine barium, cobalt, and nitrogen? 

Oct2614
  Sneaky: BaCoN!
I am contemplating writing a program to solve al wazir's "binary calendar cube problem", and even thinking about it gives me a headache. So how can we kill more time? I know! On the topic of jocular chemistry, the mineral magnesium iron silicate hydroxide is has a very humorous name to geologists ... it's known as "cummingtonite". No, seriously. It was discovered in a town called Cummington. 

Oct2614
  TheFocus: <Sneaky> BaCoN?
The other, (or only), good meat? 

Oct2614
  al wazir: Let me start by answering the bonus part of the first question I asked: How many different numbers can be displayed altogether the two cubes with face numbers 0, 1, 2, 3, 4, 5, 6 and 0, 1, 2, 6, 7, 8? One cube can show six different numbers and the other can show seven; 6 x 7 = 42. Since you can interchange the order of the two cubes, that makes 42 + 42 = 84. But nine of those numbers (00, 01, 02, 10, 11, 12, 20, 21, and 22) are generated twice, and you have to correct for that: 84  9 = 75. Another way to count is to note that you can make every number from 00 to 95, a total of 96, with the exception of these: 33, 34, 35; 43, 44, 45; 53, 54, 55; 66, 67, 68, 69; 76, 77, 78, 79; 86, 87, 88, 89. Subtracting 21 from 96 give 75 again. Check. Now for the binary problem. I haven't found a way to go about it *intuitively*, as suggested by <Sneaky>, but I can at least be systematic. First I'll prove my assertion that the largest number in the sequence 1, 2, 3, … that can be reached must be less than 288. Suppose that the 12 faces of the cubes all display different numbers, and that inverting them results in 12 more, with no duplication, for a total of 24 distinct numbers. Pairing them yields 12 x 12 = 144 different concatenated numbers (the term I use for the number formed by concatenating the two face numbers that are displayed); interchanging the two cubes yields 144 more, for a total of 288. But if the set of concatenated numbers thus produced starts with 01, then at least one of the faces has to be ǀ or 0ǀ or 00ǀ, etc., and the number on at least one face of the other cube has to consist of one or two or three, etc., zeroes. Whatever it is, it reads the same way when it's inverted, so that cube can display at most 11 different numbers. Hence the size of the set of different concatenated numbers that can be displayed by the two cubes together is at most 2 x 11 x 12 = 264. Since you're required to find a sequence of *consecutive numbers*, without skipping any, this means that the last one in the sequence can be no bigger than 264. I'll start by developing a notation. My notation may not be the best possible one, but it's adequate for my purposes. I'm going to compile a list of the various binary face numbers that are palindromic, i.e., that are the same when read backwards, and another list of the ones that are nonpalindromic, in both cases labeling them in order of increasing complexity, using capital letters for the palindromic ones and small letters for the nonpalindromic: A ǀ
B ǀǀ
C 0ǀ0
D ǀ0ǀ
E ǀǀǀ
F 0ǀǀ0
G ǀ00ǀ
H ǀǀǀǀ
I 00ǀ00
J 0ǀ0ǀ0
K ǀ000ǀ
L ǀ0ǀ0ǀ
M ǀǀǀǀǀ
N 00ǀǀ00
O 0ǀ00ǀ0
P ǀ0000ǀ
Q 0ǀǀǀǀ0
R ǀ0ǀǀ0ǀ
S ǀǀ00ǀǀ
T ǀǀǀǀǀǀ
Z 0...
I haven't decided yet how many zeroes the allzero face number will have (it doesn't matter when it's displayed by the lefthand cube), so for now I'll just represent it by the letter Z. If I have to be specific when it's displayed by the righthand cube, I will write the zeroes explicitly, e.g., 0, 00, 000, etc. a 0ǀ
b 00ǀ
c 0ǀǀ
d 000ǀ
e 00ǀ0
f 00ǀǀ
g 0ǀ0ǀ
h 0ǀǀǀ
i ǀ0ǀǀ
j 0000ǀ
k 000ǀ0
l 000ǀǀ
m 00ǀ0ǀ
n 00ǀǀ0
o 0ǀ00ǀ
p 00ǀǀǀ
q 0ǀ0ǀǀ
r 0ǀǀ0ǀ
s ǀ00ǀǀ
t 0ǀǀǀǀ
u ǀ0ǀǀǀ
(I'll leave off here; otherwise the notation becomes unwieldy. There are 26 different distinct 6bit nonpalindromic face numbers.) Note that this second list *excludes any face number that can be formed by inverting another face number* on the list. I will denote the inverted forms of the nonpalindromic face numbers with a prime, e.g., a' ǀ0
b' ǀ00
c' ǀǀ0, etc.
(cont'd) 

Oct2614
  al wazir: Now for the systematic part. Using the binary numbers in the above lists, and only these, extending them as necessary with face numbers having more bits, I will compile a table of all the possible ways to represent each number in the sequence 1, 2, 3, …, 264. (I don't expect to get that far, but that's the plan.) Here is how it starts out: 1: Z a; Z b; Z d; Z j; Z n
2: A 0; Z a'; Z C; Z e; Z k
3: a A; b A; d A; j A; Z B; Z f; Z l
4: A 00; a' 0; C 0; e 0; Z b'; Z e'; Z I
5: A a; a' A; b a; C A; e A; k A; Z D; Z g; Z m
6: A a'; B 0; b a'; c 0; d a'; f 0; j a'; l 0; Z c'; Z n 7: A B; a B; b B; c A; d B; f A; j B; l A; Z E; Z h; Z p 8: A 000; a 000; a' 00; b 000; b' 0; C 00; d 000; e 00; j 000; Z d' 9: A b; a b; b' A; C a; d b; e a; e' A; I A; j b; k a; Z G; Z o 10: A C; a C; b C; C a'; D 0; d C; e a'; g 0; j C; k a'; m 0; Z g'; Z J I may have missed some, but this sample is enough to show that the higher the number we're trying to generate, the more ways there are to do it. (Obviously as we go to bigger and bigger concatenated numbers we'll need to augment the two lists, both palindromic and nonpalindromic, with face numbers that have more bits, but the trend is clear.) The trick now is to pick out the face numbers that occur most often, i.e., the ones that generate the biggest selection of concatenated numbers. Off hand it looks advantageous to have a 0 on one cube and 00 on the other., and to use one more palindromic number, E. Of course that cuts down the size of the largest set of numbers that can be generated (in principle) to 2 x 10 x 11 = 220; but I'm sure that even that is an overestimate. The above table shows that it is possible to generate the first 10 natural numbers with cubes printed with 0, a _, _, _, _ and 00, b, E, f, g, _, where the blanks are available for additional assignments in order to generate more numbers. Using just the face numbers already chosen we can generate the following additional concatenated numbers (I have inserted the binary representations for those that can't be generated with the available face numbers): 11: a f
12: 0 f'
13: f a
14: f a'
15: a E
16: (10000)
17: b' a
18: b' a'
19: a f
20: g 0
21: a g
22: g a'
23: a' E
24: f' 0
25: (11001)
26: a g'
27: (11011)
28: a f'
29: E a
30: E a'
31: (11111)
32: (100000)
I can fill in the remaining blanks so as to supply the missing numbers. The simplest way is to assign the binary expressions for 16, 25, 27, 31, and 32 to the five blanks, but that is far from optimum. A better choice is to use just two of the blanks, making the faces read 0, a, c, _, _, _ and 00, b, E, f, g, j; but it may be possible to improve on this. Now those five missing numbers can be represented as follows: 16: 0 j'
25: c b
27: f c
31: c E
32: j' 0
and we can go on:
33: a j
34: j' a'
35: a' f
But I will stop here, at least for now. 

Oct2614
  john barleycorn: Fancy algebra often comes with fancy algepanty. 

Oct2614
  TheFocus: What makes it harder is algegirdle. 

Oct2714
  al wazir: To summarize: With two cubes whose face numbers are 0, a, c, _, _, _ and 00, b, E, f, g, j, where the notation is the same as before and the blanks stand for faces not yet filled in, we can represent the natural numbers though 33 as follows: 1: 00 a
2: 00 a'
3: 0 f / 00 c
4: a 00
5: b a / 0 g
6: b a' / j a'
7: 0 E
8: a' 00 / b' 0
9: a' b
10: g 0 / 0 g'
11: a f
12: 0 f'
13: f a
14: f a'
15: a E
16: 0 j'
17: b' a
18: b' a'
19: a f
20: g' 0
21: a g / g a
22: g a'
23: a' E
24: f' 0
25: c b
26: a g'
27: f c
28: a f'
29: E a
30: E a'
31: c E
32: j' 0
33: a j
(I have corrected some mistakes in my previous posts.) There is no way to represent 34 with the existing face numbers. But by reshuffling them a little and some trial and error, I find that the choices 0, a, B, c, E, g (0, 0ǀ, ǀǀ, 0ǀǀ, ǀǀǀ, 0ǀ0ǀ) and 00, b, e, f, i, j (00, 00ǀ, 00ǀ0, 00ǀǀ, ǀ0ǀǀ, 0000ǀ) work fairly well: 1: 00 a
2: 00 a'
3: 0 f / 00 c
4: a 00
5: b a / e a'
6: b a' / j a' / 00 c'
7: 00 E
8: a' 00 / b' 0
9: a' b
10: 00 g'
11: a f
12: 0 f'
13: 0 i' / f a
14: f a'
15: b E / j E
16: 0 j'
17: b' a
18: b' a'
19: a f
20: g' 00
21: b g
22: i 0
23: e E
24: f' 0
25: c b
26: b g' / i' 0
27: f c
28: E 00
29: a i'
30: f c'
31: f E
32: j' 0
33: a j
34: a' e
35: a' f
36: a' e'
37: e g
38: b' c'
39: e' E
40. g' 00
41: g b
42: g' a' / a' g' / e g'
43: g c
44: a' f'
45: i a / a' i'
46: i a'
47: i B
48: a j'
49: f' a
50: f' a'
51: c f
52: c' b'
53: i' a
54: i' a'
55: i' B
56: (ǀǀǀ000) ???
57: E b
58: (ǀǀǀ0ǀ0) ???
59: B i
60: h b'
61: ǀǀǀǀ0ǀ B i'
62: ǀǀǀǀǀ0 E c'
63: ǀǀǀǀǀǀ h E
64: (ǀ000000) ???
The sequence stops after 55, which is the best I can do at this point. I don't see any pattern emerging, so this approach isn't likely to be the best one. An exhaustive search by computer would probably improve on it considerably. 

Oct2714
  Sneaky: Just playing around, I constructed dice like these:
die 1 = ("00","001","11","101","1101","111")
die 2 = ("1","01","010","1000","1100","001")
These can count up to 37 (100101) but there's no way to make 38 (100110). That's the best I could come up with just with trial and error. Every time I try to change something to allow 100110 one of the previous numbers is unattainable. So that's a pretty poor starting point, but that's the best I could find so far. Maybe it will help shed light on why it can be tricky to get to higher numbers. Think: if we just want to make it up to 63, we'll have to have 32 numbers with 6 binary digits, which implies that a whole bunch of those faces will have to have a lot of significant digits. But they can't all be long, or we'll never get numbers like 01 or 11. (I know that's an oversimplification, but I think you know what I'm getting at.) 

Oct2814
  johnlspouge: @<Sneaky>: I notice <al wazir> uses a special font for his 1, a font invariant to rotation through 180 degrees. Thus, "01" (in <al>'s font) is also "10" after the rotation, though not in yours. I am still catching up on the "Two envelopes problem" [ http://en.wikipedia.org/wiki/Two_en... ] Lurking is good for my productivity at work, but I don't want you two (especially <al> ;>) to think that I don't find your discussions endlessly fascinating :) 

Oct2914
  Sneaky: I'm assuming al is writing the ones as  and not 1 because he wants to emphasize that they are completely symmetrical from top to bottom. I'm not literally standing on my head to see these things upside down so I'm not worried about that. I find it mildly interesting that if you turn a binary number upside down, it's the same thing as reading the string backwards. In PHP there is a function called strrev that reverses a string, e.g. strrev("evian") = "naive". So if you want to turn 1011101010 upside down, just take the strrev(). <I am still catching up on the "Two envelopes problem"> The wikipedia article is lacking in some respects, but it gets to the core of the matter. The problem as originally stated is flawed, and some people point out those flaws and think they can the entire discussion. But if you want to get to the heart of it, you can carefully fix those flaws, effectively creating a new of paradox, one with true mathematical rigor. One that can no longer be dismissed so easily. Some people are inspired to invent even more clever versions of the paradox that are still harder to dismiss. In short, don't think of it as a single paradox, but rather a complex of related paradoxes. 

Oct2914
  al wazir: <Sneaky>: I've thought abut the twoenvelope problem so more and even looked at the wiki article. I have to concede that what I've posted so far hasn't come to grips with the essence of the problem. The simplest version, without any cointossing or powers of 3, is this one, from the wiki article: <You have two indistinguishable envelopes that each contain money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. You pick at random, but before you open the envelope, you are offered the chance to take the other envelope instead.> Since there is no way to tell which envelope is which, any argument in favor of a strategy that gives a preference to swapping or to keeping must be spurious. (A "paradox" is paradoxical only until it is explained away.) Smullyan formulates the dilemma this way: <1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss. 2. Let the amounts in the envelopes be X and 2X. Now by swapping, the player may gain X or lose X. So the potential gain is equal to the potential loss.> Argument #1 must be wrong. Why?
'A' and 'X' are different things, or rather, different *concepts*. X and 2X are the *actual* contents of the envelopes. A/2 and 2A are the possible contents of the second envelope *relative to the first*. If the first envelope happens to contain X, then A = X. If the first envelope happens to contain 2X, then A = 2X. So the calculation of the "expected" gain from switching is faulty, because what is meant by A changes during the calculation. I'm pretty sure that this point must be covered in the wiki article, and I'll come across it if I get around to reading the article all the way through. 

Oct3014
  al wazir: The point I'm making probably isn't obvious. I can try to clarify it by contrasting the twoenvelope problem with the *threeenvelope* problem. You are given an envelope containing a certain amount of money, say, $100. You can look in it and confirm that it does contain $100. On the table are two other envelopes, one containing $50 and the other containing $200. You can look in them too and confirm that they hold these amounts. (But it isn't essential that you know the amounts; you just have to be convinced that one of the two envelopes on the table contains half as much and the other twice as much as the one you start with.) Now you have a choice. You can either keep the first envelope, or you can choose one of the other envelopes *randomly*  for example, by a coin toss  and keep its contents. Which do you do? If you have even a rudimentary grasp of probability you go with the coin toss. That way you have an even chance of getting $50 or $200, so your expectation is 1/2 x $50 + 1/2 x $200 = $125, i.e., $25 more than if you simply hold onto the $100. What is the difference between this problem and the twoenvelope problem? In this problem you are calculating an expected payoff based on a 50:50 chance of getting either half or double a *specific* amount. In the twoenvelope problem the spurious "expectation" is based on a 50:50 chance of going from one amount (Smullyan's X) to 2X or from a *different* amount (2X) to X. 

Oct3014
  john barleycorn: There is a story about the two envelopes.
A manager lost his job and left two envelopes in his desk addressed "To my successor  first envelope to be opened in an emergency" and "To my successor  second envelope to be opened in extreme emergency". The new manager started his job but found himself in troubles shortly after.
So he opened the first envelope. It contained a card with "Put the blame on me" and so this manager did. He told to his boss that he was suffering from the bad management of his predecessor. The excuse worked for a while but results did not improve. He opened the second envelope. It had a card inside saying: "Prepare two envelopes." 


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