Generating Roman numerals with dice

Question

This puzzle is closely based on this one: Generating numbers with cubes

Now we want to generate Roman numerals by placing up to three 6-sided dice side by side. We are allowed to write multiple numerals on each side of a dice, such as "III" for 3 or "XX" for 20.

What is the largest contiguous range of positive numbers you can make?

Remember that:

I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1000

If you need larger numerals, then feel free to use this notation. Good luck!

UPDATE: I made a Java program that you can use to check your score. Simply pass your 3 dice as Strings (space or comma separated) to the score() method. Note that currently it does not support the flipping operation, but perhaps someone can add that.

Stealthy update from @Bass: I also made a program for checking the dice, this one's in Python, and you can try it online.

Answer 1

UPDATE (after a pretty sturdy hint from OP):

Found a 170:

 I  II III IV IX  LX
 V  X  XV  XL XLV XCV
 XX L  LXX C  CXX CL

(Again, one more number can be constructed, if flipping the dice is allowed.)

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Original answer:

I got all the way up to

160

with these dice:

 1: I  II III IV  IX C
 2: V  X  XV  XCV CX C
 3: XX L  LV  LXX C  CXX

Here's how to construct the numbers

 I              = 1
 II             = 1
 III            = 1
 IV             = 1
 V              = 2
 VI    = V.I    = 2+1
 VII   = V.II   = 2+1
 VIII  = V.III  = 2+1
 IX             = 1
 X              = 2
 XI    = X.I    = 2+1
 XII   = X.II   = 2+1
 XIII  = X.III  = 2+1
 XIV   = X.IV   = 2+1
 XV             = 2
 XVI   = XV.I   = 2+1
 XVII  = XV.II  = 2+1
 XVIII = XV.III = 2+1
 XIX   = X.IX   = 2+1

 In short: I to XIX can be made with dice 1 and 2.

 XX to XXXIX = XX(3) + I to XIX (1 and 2)

 XL     = X.L      = 2+3
 XLI    = X.L.I    = 2+3+1
 XLII   = X.L.II   = 2+3+1
 XLIII  = X.L.III  = 2+3+1
 XLIV   = X.L.IV   = 2+3+1
 XLV    = X.LV     = 2+3
 XLVI   = X.LV.I   = 2+3+1
 XLVII  = X.LV.II  = 2+3+1
 XLVIII = X.LV.III = 2+3+1
 XLIX   = X.L.IX   = 2+3+1

 L to LXIX = L (3) + I to XIX (2+1)

 LXX to LXXXIX = LXX (3) + I to XIX (2+1)
 
 XC     = X.C     = 2+3
 XCI    = X.C.I   = 2+3+1
 XCII   = X.C.II  = 2+3+1
 XCIII  = X.C.III = 2+3+1
 XCIV   = X.C.IV  = 2+3+1
 XCV              = 2
 XCVI   = XCV.I   = 2+1
 XCVII  = XCV.II  = 2+1
 XCVIII = XCV.III = 2+1
 XCIX   = X.C.IX  = 2+3+1
 
 C to CXIX = C(3) + I to XIX (2+1)

 CXX to CXXXIX = CXX (3) + I to XIX (2+1)
 
 CXL to CXLIX: like XL to XLIX, but CX on die 2 instead of X

 CL                = 2+3
 CLI    = C.L.I    = 2+3+1 
 CLII   = C.L.II   = 2+3+1 
 CLIII  = C.L.III  = 2+3+1 
 CLIV   = C.L.IV   = 2+3+1 
 CLV    = C.LV     = 2+3
 CLVI   = C.LV.I   = 2+3+1 
 CLVII  = C.LV.II  = 2+3+1 
 CLVIII = C.LV.III = 2+3+1 
 CLIX   = C.L.IX   = 2+3+1 

 and finally
 
 CLX = C.L.X = 1+3+2 (This could also be its own number on the sixth side of the first die.)
 

If I can flip the dice, there's also

161 = CLXI = C.L.XI = 2 + 3 + flipped 1

Answer 2

Partial answer, which goes up to

89

Solution:

1. Consider the following 2 dice: I,II,III,IV,VI,VII and X,XX,XXX,XL,LX,LXX. These can display any 2-digit number whose digits are in the set of {0,1,2,3,4,6,7} (0 = just don't place the corresponding dice).
2. Let's add L and V to the 1st and 2nd faces of dice 3 (leaving the others blank). This allows to additionally express any 2-digit number which has a 5 it it (except 55). By combining V and III, or L and XXX, you can also get a number which has an 8 in it (except 58, 85, 88).
3. Add IX to the 3rd face of dice 3. This can express any number ending with 9, except 89 (note that 59 can be expressed by using L+I+X).
4. Add LV to the 4th face of dice 3. This allows to express 55 (just as LV) and 58 (as LV+III).
5. Add XV to the 5th face of dice 3. This allows to express 85 (as LXX+XV) and 88 (as LXX+XV+III). 6. To get 89, you can for example add XIX to the remaining 6th face of dice 3 (getting LXX+XIX).

So, the resulting dice are:

(I,II,III,IV,VI,VII), (X,XX,XXX,XL,LX,LXX) and (V,L,IX,LV,XV,XIX)

Note:

We still cannot get 90 or more in this configuration because it requires new C numeral (90=XC, 91=XCI etc.)

Answer 3

First, to give an upper bound on the solution:

The maximum potential is 342 (6^3+3*6^2+3*6) as this is the number of discrete representations available with three dice where not all dice need be used.

Second, the potential solution I'll be outlining here yields all numbers up to and including:

190

How we get there:

The first die is our ones die and will have I II III IV and IX, which also leaves one space for a maximum value that sadly won't be able to pair with I's
The second die is our tens die, but also needs a V so it looks like V X XX XXX XL which leaves a space for trickiness later, since XC will only be fully used once (Too few places to put Vs), there is no point putting it here
The third die requires a V and an L, which takes us to 89, an XC and C to get us through 149 and a CL (150) to reach 164, so V L XC C CL with one space remaining
Now for the tricky bit, CLX (160) on the third die gets us through to 174 and XXV on the second allows for combinations with both CL and CLX to get us to 189
Finally, we stick CLXC on the largely useless sixth spot from the first die because we can

Final Dice are:

(I,II,III,IV,IX,CLXC),(V,X,XX,XXX,XL,XXV),(V,L,XC,C,CL,CLX)

Compared to the upper bound, this solution rates a:

55.56%

EDIT: I just noticed that CXV (155) is not possible with this solution

replacing CLXC with CXLV, XXV with XV, CL with CX and CLX with CXX allows us to still hit 145, for a new percentage of 42.40%

Answer 4

I've found a handful of solutions that will get us up to 89:

89: [['I', 'II', 'III', 'IV', 'VI', 'VII'], ['X', 'XX', 'XXX', 'XL', 'LX', 'LXX'], ['V', 'IX', 'XI', 'XV', 'XIX', 'L', 'LV']]
89: [['I', 'II', 'IV', 'IX', 'X', 'XI', 'XXV'], ['III', 'V', 'VI', 'VII', 'VIII', 'X'], ['IIX', 'XII', 'XX', 'L', 'LX', 'LXX', 'LXXX']]
89: [['I', 'II', 'IV', 'IX', 'X', 'XI', 'XXV'], ['III', 'V', 'VI', 'VII', 'VIII', 'X'], ['II', 'IIX', 'XII', 'XX', 'L', 'LX', 'LXXX']]
89: [['I', 'II', 'III', 'V', 'VIII', 'X'], ['V', 'IX', 'XI', 'XVII', 'XX', 'L', 'LXX'], ['IV', 'VI', 'VII', 'IX', 'X', 'XI', 'LXX']]
89: [['I', 'II', 'III', 'IX', 'X', 'XI', 'XVI'], ['IV', 'V', 'XX', 'L', 'LX', 'LXX'], ['V', 'VI', 'VII', 'VIII', 'X', 'L']]

Can't find anything that beats it...

Answer 5

9..156 = 147 contiguous numbers

For the first die,

A1 I A2 II A3 III A4 IIII A5 CLV A6 CLVI

For the second,

B1 X B2 XX B3 XXX B4 XXXV B5 XXV B6 XV

For the third,

C1 L C2 LV C3 C C4 CV C5 CX C6 CL

Explanation:

A1B1 is 9. We can construct 10..19 using B1 or B6 plus A1..A4. Likewise 20..29, 30..39. For 40..49, we can make XL/XLV using B1C1 or B1C2 and A1..A4. 50 is C1; 51..54 is C1 plus A1..A4; 55..59 is C2 plus A1..A4. 60..89 proceeds as 10..39 with C1 prepended. 90 is B1C3; for 91..94 append A1..A4. Substitute C3 for C4 to get to 99. 100 is C3; from here 100..139 proceeds as 51..89 140 is C5B3 and append A1..A4 for 141..144, then substitute B4 to get to 149. 150 is C6 and append A1..A4. Hitherto-unused are A5 and A6