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Topic: Calendar Cubes with Roman Numerals (Read 3984 times) |
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mlowry
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Calendar Cubes with Roman Numerals
« on: Oct 16th, 2002, 9:05am » |
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If you've solved the Calendar Cubes puzzle or if you couldn't solve it and want to get a different perspective on the problem, try solving it using Roman numerals. Because the Roman numerals do not include a zero, one cube face must be blank.
I actually solved the puzzle first using Roman numerals. When trying to solve the Calendar Cubes problem, for some reason I lept to the assumption that there were too few faces to solve the problem using Arabic numerals. As far as I can tell, the same "trick" required to solve the puzzle with Arabic numerals is also required with Roman numerals. This trick is even more obvious when using Roman numerals -- that's how I discovered it.
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Jeremiah Smith
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Re: Calendar Cubes with Roman Numerals
« Reply #1 on: Oct 16th, 2002, 10:14am » |
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on Oct 16th, 2002, 9:05am, mlowry wrote:If you've solved the Calendar Cubes puzzle or if you couldn't solve it and want to get a different perspective on the problem, try solving it using Roman numerals. Because the Roman numerals do not include a zero, one cube face must be blank.
I actually solved the puzzle first using Roman numerals. When trying to solve the Calendar Cubes problem, for some reason I lept to the assumption that there were too few faces to solve the problem using Arabic numerals. As far as I can tell, the same "trick" required to solve the puzzle with Arabic numerals is also required with Roman numerals. This trick is even more obvious when using Roman numerals -- that's how I discovered it. |
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Well, if I recall, the trick for Arabic numerals was to flip the 6 over to get 9. But that doesn't really apply here, since I V and X don't flip over into each other..
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| « Last Edit: Oct 16th, 2002, 10:15am by Jeremiah Smith » |
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mlowry
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Re: Calendar Cubes with Roman Numerals
« Reply #2 on: Oct 16th, 2002, 12:00pm » |
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I think you've overlooked something:
XI is IX upside-down.
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Garzahd
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Re: Calendar Cubes with Roman Numerals
« Reply #3 on: Oct 16th, 2002, 12:59pm » |
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But XI isn't a one-digit number.
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mlowry
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Re: Calendar Cubes with Roman Numerals
« Reply #4 on: Oct 16th, 2002, 1:40pm » |
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There's no way to write many of the numbers between 1 and 31 in Roman numerals without using more than two numerals. For example: 3 (III), 7 (VII), 8 (VIII), 12 (XII), 13 (XIII), 14 (XIV), ..., 19 (XIX), 21 (XXI), and so on.
The Calendar Cubes puzzle specifies exacly 2 cubes, and does not specifically proscribe having more than one numeral on a face.
Given these facts, I solved the puzzle operating from the assumption that it was allowed to have as few (0) or as many (3 or more) numerals on a single face.
Enjoy!
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James Fingas
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Re: Calendar Cubes with Roman Numerals
« Reply #5 on: Oct 17th, 2002, 10:03am » |
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This is a really good point about Roman Numerals, which many people (but of course not our good puzzlers here!) don't realize.
1) To correctly use Roman Numerals, you transform the decimal digits one-at-a-time into their Roman Numeral equivalents:
1 9 9 9
M CM XC IX
2) Because of this, putting a smaller sign before a big sign can only be used in certain circumstances: a single I before a single V, a single I before a single X, a single X before a single L, a single X before a single C, etc.
49 is not IL, it's XLIX
also, 8 cannot be IIX, it has to be VIII (silly Romans)
29 is written XXIX, not IXXX or XIXX ("20" is XX, and "9" is IX)
So this problem is actually very much in the spirit of Roman Numerals, because each decimal digit corresponds to a single Roman Numeral grouping, which of course may have more than a single letter in it!
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william wu
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Re: Calendar Cubes with Roman Numerals
« Reply #6 on: Oct 21st, 2002, 4:18am » |
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Wow this is great. Now we have four versions of the Calendar Cubes problem floating around! (There's a Calendar Cubes 0.5 in the easy section.) I'll add this one in next week's batch of new problems. Thanks!
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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That Don Guy
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Re: Calendar Cubes with Roman Numerals
« Reply #7 on: Oct 30th, 2002, 12:51pm » |
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Either I read this problem wrong, or it seems a little easy for the "hard" section...
Solution Spoiler:
One die: i, ii, iii, iv, ix
The other: V, X, XV, XX, XXV, XXX
(Actually, both use uppercase letters - the lowercase ones are there just to show which die is which in the solution)
1-10: i, ii, iii, iv, V, Vi, Vii, Viii, ix, X
11-20: Xi, Xii, Xiii, Xiv, XV, XVi, XVii, XViii, Xix, XX
21-25: XXi, XXii, XXiii, XXiv, XXV
26-31: XXVi, XXVii, XXViii, XXix, XXX, XXXi
(For calendars with more days:
32-34: XXXii, XXXiii, XXXiv
35 is possible if the remaining face on the first die is "v")
Note that no face needs to be turned over
-- Don
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James Fingas
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Re: Calendar Cubes with Roman Numerals
« Reply #8 on: Oct 31st, 2002, 12:57pm » |
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Don,
You don't use both cubes for all of your days. Note that you don't have a blank face on the capital letter cube, so you can't form the numbers from i to iv.
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SWF
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Re: Calendar Cubes with Roman Numerals
« Reply #9 on: Nov 1st, 2002, 5:46pm » |
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After what I am pretty sure is an exhaustive search  , there appears to be no solution which uses no tricks. The best I can find is a pair of February blocks:
<blank>, V, X, XV, XX, XXV and <blank>, I, II, III, IV, XIX
With the reasonable trick of letting IX be flipped over to become XI results in at least 10 combinations that permit 30 of the 31 values (missing either the 24 or the 30). In the following pair, the last face can have either a 24 or a 30 on it, which makes the other number undoable:
<blank>, V, X, XV, XX, XXV and <blank>, I, II, III, IX, XXX
A desparate move lacking in aestetics can be taken to get the last number from the above pair of cubes, but I hope there is a cleaner solution:
_____________begin spoiler___________
To make the 24, prop up the IX behind the XX block and shifted slightly to the right, such that the XX cube obscures the bottom half of the IX making it IV.
_____________end spoiler_____________
Here is combination (allowing 29 of the 31 values) which is fairly interesting because getting several numbers is less obvious than when there is a boring cube with multiples of 5:
I, II, V, X, XVII, XXV and <blank>, I, VI, IX, X, XXI
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mlowry
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Re: Calendar Cubes with Roman Numerals
« Reply #10 on: Nov 4th, 2002, 4:51am » |
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Perhaps I should make one thing clear. My solution does not require that all of the Roman numerals representing a single Arabic numeral be on a single cube, but only that together the numerals form the correct sequence.
For example, the number 26 could be XXV on one cube and I on the other cube. Together, they form XXVI, and that's what matters. A simpler example is the number 6: V on one cube and I on the second cube.
This is a twist on the original Calendar Cubes puzzle, but that's what makes it interesting.
If there is the additional requirement that all Roman numbers representing each Arabic numeral fall on a single cube face, then there just aren't enough faces to solve the puzzle with only 2 cubes.
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That Don Guy
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Re: Calendar Cubes with Roman Numerals
« Reply #11 on: Nov 18th, 2002, 12:36pm » |
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This may be a bit of a cheat, but this is as close to a solution that I've gotten so far (pretty much every other "solution" has either 24 or 30 missing):
Cube 1 - I, II, III, V, X, XI/IX
Cube 2 - v, x, xv, xxi, xxv, (blank)
(Note that Cube 2 uses uppercase like Cube 1 - they're shown as lowercase here to differentiate between the two in my solution)
The solution:
* indicates the blank side of cube 2
1-10 - I*, II*, III*, Iv, V*, vI, vII, vIII, IX*, X*
11-20 - xI, xII, xIII, XIv, Xv, xvI, xvII, xvIII, xIX, xX
21-29 - xXI, xxiI, xxIII, xxiV, Xxv, xxvI, xxvII, xxvIII, xxiX
31 - Xxxi
Here's the cheat:
30 - put xxv on top of an upside-down V such that the two Vs form a giant X; you now have xx"X", or 30.
-- Don
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blah
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Re: Calendar Cubes with Roman Numerals
« Reply #12 on: Mar 20th, 2003, 2:49pm » |
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Well, I thought about it a bit, and since he said it was sly, all I did was this:
Let's label the cubes cube1 and cube2. Have the numbers 0, 1, and 2 on both cubes. Then, on cube1, put 3, 4, and 5. On cube2, put 6, 7, and 8. Now, we can almost get any number that is a calendar day (except for 09, 19, and 29 because of the 9). What do we do now? Simply flip the 6 around when we need a 9, and there you have it.
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SWF
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Re: Calendar Cubes with Roman Numerals
« Reply #13 on: Mar 20th, 2003, 5:25pm » |
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In Roman Numeral an upsidedown 6 does not look like a 9.
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mistysakura
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Re: Calendar Cubes with Roman Numerals
« Reply #14 on: Mar 24th, 2003, 11:17pm » |
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Please exlpain how on earth you get X with your solution.
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Margit Schubert-While
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Re: Calendar Cubes with Roman Numerals
« Reply #15 on: Jul 5th, 2003, 11:38am » |
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I'm wondering if we can get all the days from 1 to 365 (366) with 3 cubes.
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jonderry
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Re: Calendar Cubes with Roman Numerals
« Reply #16 on: Jan 10th, 2005, 1:41pm » |
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Is there a way to solve this without cheating (i.e. the cubes can be arranged side by side flush with each other and the desk so that any of the 31 first Roman numerals can be read from left to right using no tricks other than the XI/IX duality)?
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Nigel_Parsons
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Re: Calendar Cubes with Roman Numerals
« Reply #17 on: Jan 16th, 2005, 8:34am » |
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Give me two or three weeks and I'll be able to do it.
:: Of course, it will the be February, with only 28 days needed!::
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| « Last Edit: Jan 16th, 2005, 8:34am by Nigel_Parsons » |
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That Don Guy
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Re: Calendar Cubes with Roman Numerals
« Reply #18 on: Jul 27th, 2005, 12:24pm » |
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on Jul 5th, 2003, 11:38am, Margit Schubert-While wrote:| I'm wondering if we can get all the days from 1 to 365 (366) with 3 cubes. |
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No.
The numbers 111, 222, 333, 144, 155, 166, 177, and 188 require 19 separate digits (the 1 in 144 through 188 can be one of the 1s in 111), and 3 cubes can only have up to 18 digits.
-- Don (has it been that long since I've been here?)
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