Roman Numeral Calibrated Gematria

Ever since I created Calibrated Gematria, I've been wanting to expand it to accommodate Roman numerals. This page is where you can watch me stumbling toward that goal. "Under construction" would be an extreme understatement.

Clearly, Roman Numeral Calibrated Gematria (RNCG) means more than just another set of letter-number mappings. If IV and VI are to return two different values, the gematria mechanism itself must be modified. It can't be as simple as just adding up all the letters in a word.

So here's my first attempt at a rule:

If a letter's value is lower than that of the letter which follows it, then it is subtracted from the word total; otherwise, it is added.

This works for Roman numerals. For example, IV is <1, 5>, which becomes -1 + 5 = 4; whereas VI is <5, 1>, which becomes 5 + 1 = 6. When I try to apply it to number words, though, I find it less than satisfactory.

Six is the natural number to begin with, since i and x are Roman numerals, leaving s as the only unknown quantity. Six = <s, 1, 10>, which must equal 6, so I need the letter s to represent subtracting 3. This, it turns out, is impossible. If I define s as -3, we get <-3, 1, 10>. Because -3 is less than 1, it must be subtracted from the total, giving us 3 - 1 + 10 = 12. If we instead define s as 3, the result is the same; 3 is now greater than 1 and must be added, which is the same thing as subtracting -3. No matter what value we assign to s, it's impossible to make six equal 6.

So here's my modified rule:

If a letter's absolute value is lower than that of the letter which follows it, then it is subtracted from the word total; otherwise, it is added.

This makes six doable. If s = -3, then six = <-3, 1, 10> = -3 - 1 + 10 = 6.

Here are the letters we have defined so far — the Roman numerals, plus s:

• c = 100
• d = 500
• i = 1
• l = 50
• m = 1000
• s = -3
• v = 5
• x = 10
• Undefined: {a, b, e, f, g, h, j, k, n, o, p, q, r, t, u, w, y, z}

And here are some basic number words to be used for calibration:

• one = <o, n, e> = 1
• two = <t, w, o> = 2
• three = <t, h, r, e, e> = 3
• four = <f, o, u, r> = 4
• five = <f, 1, 5, e> = 5
• six = <-3, 1, 10> = 6
• seven = <-3, e, 5, e, n> = 7
• eight = <e, 1, g, h, t> = 8
• nine = <n, 1, n, e> = 9
• ten = <t, e, n> = 10
• eleven = <e, 50, e, 5, e, n> = 11
• twelve = <t, w, e, 50, 5, e> = 12
• thirteen = <t, h, 1, r, t, e, e, n> = 13
• fourteen = <f, o, u, r, t, e, e, n> = 14
• fifteen = <f, 1, f, t, e, e, n> = 15
• sixteen = <-3, 1, 10, t, e, e, n> = 16
• seventeen = <-3, e, 5, e, n, t, e, e, n> = 17
• eighteen = <e, 1, g, h, t, e, e, n> = 18
• nineteen = <n, 1, n, e, t, e, e, n> = 19
• twenty = <t, w, e, n, t, y>

The two letters to tackle next are e and n, the only unknown quantities in seven, nine, and eleven.

Barring some error of mathematical reasoning, the following table includes all possible solutions for seven = 7.

e n seven nine eleven
-8 9 7 9 62
-7 9 7 10 61
-6 9 7 11 60
-5 9 7 12 69
-4 -9 7 -23 58
-4 -1 7 -3 58
-1 3 7 4 61
0 5 7 9 60
1 7 7 14 59
2 9 7 19 58
3 11 7 24 57
4 -1 7 5 50
4 7 7 17 50
6 -3 7 5 58
6 9 7 23 58
7 -5 7 6 47
7 9 7 24 47
8 -7 7 7 46
8 9 7 25 46
9 -9 7 -10 45

Eleven appears to be a lost cause, but we have two different solutions that work for both seven and nine. I opt for the latter of the two, because it has the useful feature of a zero value for e, making it possible for ten to equal -teen. Once e and n have been assigned, the values of a few other letters can be derived trivially. Here, then, are the updated code and calibration set:

• c = 100
• d = 500
• e = 0
• f = 1
• i = 1
• l = 50
• m = 1000
• n = 5
• o = 4
• s = -3
• t = 5
• v = 5
• x = 10
• Undefined: {a, b, g, h, j, k, p, q, r, u, w, y, z}
• Hits: one = 1, five = 5, six = 6, seven = 7, nine = 9, ten = 10, sixteen = 16, seventeen = 17, nineteen = 19
• Misses: eleven = 60, fifteen = 11
• two = <5, w, 4> = 2
• three = <5, h, r, 0, 0> = 3
• four = <1, 4, u, r> = 4
• eight = <0, 1, g, h, 5> = 8
• twelve = <5, w, 0, 50, 5, 0> = 12
• thirteen = <5, h, 1, r, 5, 0, 0, 5> = 13
• fourteen = <1, 4, u, r, 5, 0, 0, 5> = 14
• eighteen = <0, 1, g, h, 5, 0, 0, 5> = 18
• twenty = <5, w, 0, 5, 5, y>

Two is now impossible to solve, since every possible value for w will yield a number either greater than 3 or less than -6. We therefore use twelve to calibrate w and then twenty to calibrate y. There are infinitely many solutions for the set {three, four, eight}, none of which can deal with thirteen or eighteen, so of them I choose the one that happens also to yield the correct value for forty.

• c = 100
• d = 500
• e = 0
• f = 1
• g = 17
• h = 21
• i = 1
• l = 50
• m = 1000
• n = 5
• o = 4
• r = -13
• s = -3
• t = 5
• u = 22
• v = 5
• w = -38
• x = 10
• y = 63
• Undefined: {a, b, j, k, p, q, z}
• Hits: one = 1, three = 3, four = 4, five = 5, six = 6, seven = 7, eight = 8, nine = 9, ten = 10, twelve = 12, fourteen = 14, sixteen = 16, seventeen = 17, nineteen = 19, twenty = 20, twenty = 20, twenty-one = 21, twenty-three = 23, twenty-four = 24, twenty-five = 25, twenty-six = 26, twenty-seven = 27, twenty-eight = 28, twenty-nine = 29, forty = 40, forty-one = 41, forty-three = 43, forty-four = 44, forty-five = 45, forty-six = 46, forty-seven = 47, forty-eight = 48, forty-nine = 49
• Misses: two = -39, eleven = 60, thirteen = 12, fifteen = 11, eighteen = 13

This leaves only seven letters unassigned. Zero seems the most natural choice for calibrating z.

• z = 9
• zero = 0
page revision: 19, last edited: 08 Mar 2008 03:21