Unicode polynomial equation: Difference between revisions
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x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ |
x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ |
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</pre> |
</pre> |
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=={{header|Nim}}== |
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{{trans|Go}} |
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<lang Nim>import re, sequtils, strutils, tables |
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from unicode import toRunes |
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const |
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Powers = [("0", "⁰"), ("1", "¹"), ("2", "²"), ("3", "³"), ("4", "⁴"), |
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("5", "⁵"), ("6", "⁶"), ("7", "⁷"), ("8", "⁸"), ("9", "⁹")] |
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Fractions = [(".25", "¼"), (".5", "½"), (".75", "¾"), (".14285714285714", "⅐"), |
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(".11111111111111", "⅑"), (".1", "⅒"), (".33333333333333", "⅓"), |
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(".66666666666667", "⅔"), (".2", "⅕"), (".4", "⅖"), (".6", "⅗"), |
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(".8", "⅘"), (".16666666666667", "⅙"), (".83333333333333", "⅚"), |
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(".125", "⅛"), (".375", "⅜"), (".625", "⅝"), (".875", "⅞")] |
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Reps1 = [(",", ""), (" ", ""), ("¼", ".25"), ("½", ".5"), ("¾", ".75"), |
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("⅐", ".14285714285714"), ("⅑", ".11111111111111"), ("⅒", ".1"), |
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("⅓", ".33333333333333"), ("⅔", ".66666666666667"), ("⅕", ".2"), |
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("⅖", ".4"), ("⅗", ".6"), ("⅘", ".8"), ("⅙", ".16666666666667"), |
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("⅚", ".83333333333333"), ("⅛", ".125"), ("⅜", ".375"), ("⅝", ".625"), |
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("⅞", ".875"), ("↉", ".0"), ("⏨", "e"), ("⁄", "/")] |
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Reps2 = [("⁰", "0"), ("¹", "1"), ("²", "2"), ("³", "3"), ("⁴", "4"), ("⁵", "5"), |
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("⁶", "6"), ("⁷", "7"), ("⁸", "8"), ("⁹", "9"), ("⁻⁻", ""), ("⁻", "-"), |
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("⁺", ""), ("**", ""), ("^", ""), ("↑", ""), ("⁄", "/")] |
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Eqs = [ |
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"-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹", |
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"x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1", |
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"0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵", |
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"1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰", |
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"+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1", |
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"x^5 - 2x**4 + 42x^3 + 40x + 1", |
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"x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1", |
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"x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰", |
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"x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1", |
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"x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1", |
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"0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1", |
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"1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1", |
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"x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2", |
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"x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5", |
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"x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x", |
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"⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝"] |
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type Coefs = Table[int, float] |
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proc printEquation(coefs: Coefs) = |
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stdout.write "=> " |
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if coefs.len == 0: |
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echo "0\n" |
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return |
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var max = int.low |
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var min = int.high |
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for k in coefs.keys: |
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if k > max: max = k |
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if k < min: min = k |
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for p in countdown(max, min): |
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var c = coefs.getOrDefault(p) |
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if c != 0: |
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if p < max: |
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var sign = '+' |
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if c < 0: |
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sign = '-' |
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c = -c |
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stdout.write ' ', sign, ' ' |
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if c != 1 or (c == 1 and p == 0): |
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var cs = $c |
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cs.trimZeros() |
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let ix = cs.find('.') |
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if ix >= 0: |
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let d = cs[ix..^1] |
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for frac in Fractions: |
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if d == frac[0]: |
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cs = cs.replace(d, frac[1]) |
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break |
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if cs[0] == '0' and cs.len > 1 and cs[1] != '.': |
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cs = cs[1..^1] |
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stdout.write cs |
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if p != 0: |
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var ps = ($p).multiReplace(Powers) |
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if ps == "¹": ps = "" |
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stdout.write 'x', ps |
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echo '\n' |
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template getFloat(s: string): float = |
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try: |
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s.parseFloat() |
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except ValueError: |
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quit "Expected float, got: " & s, QuitFailure |
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let rgx = re"\s+(\+|-)\s+" |
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for eq in Eqs: |
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echo eq |
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let terms = eq.split(rgx) |
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let ops = toSeq(eq.findAll(rgx)).mapIt(it.strip()) |
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var coefs: Coefs |
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for i, term in terms: |
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let s = term.split("x") |
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var t = s[0] |
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t = unicode.strip(t, runes = "·× ".toRunes) |
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t = t.multiReplace(Reps1) |
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var c = 1.0 |
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var inverse = false |
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if t == "" or t == "+" or t == "-": t &= '1' |
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let ix = t.find('/') |
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if ix == t.high: |
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inverse = true |
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t.setLen(t.high) |
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c = t.getFloat() |
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elif ix >= 0: |
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let u = t.split('/') |
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let m = u[0].getFloat() |
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let n = u[1].getFloat() |
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c = m / n |
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else: |
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c = t.getFloat() |
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if i > 0 and ops[i - 1] == "-": c = -c |
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if c == -0.0: c = 0.0 |
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if s.len == 1: |
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coefs[0] = coefs.getOrDefault(0) + c |
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continue |
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var u = s[1].strip() |
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if u.len == 0: |
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let p = if inverse: -1 else: 1 |
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if c != 0: |
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coefs[p] = coefs.getOrDefault(p) + c |
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continue |
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u = u.multiReplace(Reps2) |
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let jx = u.find('/') |
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var p: int |
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if jx >= 0: |
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let v = u.split('/') |
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p = try: v[0].parseInt() except ValueError: 1 |
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let d = v[1].getFloat() |
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c /= d |
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else: |
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p = try: u.strip().parseInt() except ValueError: 1 |
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if inverse: p = -p |
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if c != 0: coefs[p] = coefs.getOrDefault(p) + c |
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printEquation(coefs)</lang> |
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{{out}} |
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<pre>-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹ |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ |
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=> 0 |
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1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x^5 - 2x**4 + 42x^3 + 40x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ |
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=> 1 + 40x-¹ + 42x-³ - 2x-⁴ + x-⁵ |
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x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 |
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=> x⁵ - 2x⁴ + 42x³ + 40x + 1 |
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x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 |
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=> x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ |
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x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5 |
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=> x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ |
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x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x |
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=> -40x-¹ + 42x-³ - 2x-⁴ + x-⁵ |
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⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝ |
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=> ⅐x⁵ - x⁴ + 42⅕x³ + ⅑x - 40¾</pre> |
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=={{header|Phix}}== |
=={{header|Phix}}== |
Revision as of 19:54, 19 June 2021
The objective of this task is to parse in a difficult polynomial, and generate a "pretty" representation of the polynomial in Unicode.
In the target language define a "polynomial" object (or structure or record). Using this object also define the routines for parsing a polynomial as input, and generating a normalised Unicode representation of the polynomial as output.
Task details:
Given a string containing an untidy Unicode polynomial, e.g.
-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹
Coerce (or convert) the string into the "polynomial" object, at the same time normalise the polynomial to a canonical form. The ideal normalised output (in this example) would be:
x⁵ - 2x⁴ + 42x³ + 40x + 1
Description | Input example test cases |
---|---|
"Zero" coefficients are removed | x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 |
The "0" polynomial case | 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ |
"One" coefficients are normalised | 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ |
Signs are normalised | +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 |
ASCII representations are parsed | x^5 - 2x**4 + 42x^3 + 40x + 1 |
Non-ASCII representations are parsed | x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 (c.f. ↑ & ·) |
Specifically permit non-polynomials where terms have negative exponents | x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ (n.b. Unicode Fraction) |
Spaces in numbers and between operators are ignored | x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 |
Single commas are ignored in numbers | x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 |
A coefficient may be duplicated, zero, or missing | 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 |
Support Scientific notation and optionally support Unicode Decimal Exponent Symbol U+23E8/⏨ |
1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 |
Unicode characters that must be specifically supported are: | ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁻ ⁺ · × ⁄ ↑ ⏨. Where · & × are multiplication, and ⁄ is Unicode Fraction. |
Support fractions for both input and output. | x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 On output round the decimal to appropriate fraction. |
Optionally support Unicode Vulgar fractions for both input and output. ¼ ½ ¾ ⅐ ⅑ ⅒ ⅓ ⅔ ⅕ ⅖ ⅗ ⅘ ⅙ ⅚ ⅛ ⅜ ⅝ ⅞ ↉ |
x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ On output round the decimal to appropriate fraction. |
There are (at least) three possible ways of achieving this task.
- Using an external parsing library.
- Using a built-in parsing/formatting library.
- Coding a custom polynomial parsing routing.
Either one, or all of these approaches are accepted and appear as a subtitle.
Go
Although this program provides full support for Unicode vulgar fractions, note that there is no guarantee that they (or arithmetic on them) will successfully 'round trip' due to floating point arithmetic being used in the underlying calculations and some of them being recurring rather than exact decimals in any case. <lang go>package main
import (
"fmt" "log" "math" "regexp" "strconv" "strings"
)
var powers = strings.NewReplacer(
"0", "⁰", "1", "¹", "2", "²", "3", "³", "4", "⁴", "5", "⁵", "6", "⁶", "7", "⁷", "8", "⁸", "9", "⁹", "-", "⁻",
)
var fractions = [][2]string{
{".25", "¼"}, {".5", "½"}, {".75", "¾"}, {".14285714285714285", "⅐"}, {".1111111111111111", "⅑"}, {".1", "⅒"}, {".3333333333333333", "⅓"}, {".6666666666666666", "⅔"}, {".2", "⅕"}, {".4", "⅖"}, {".6", "⅗"}, {".8", "⅘"}, {".16666666666666666", "⅙"}, {".8333333333333334", "⅚"}, {".125", "⅛"}, {".375", "⅜"}, {".625", "⅝"}, {".875", "⅞"},
}
func printEquation(coefs map[int]float64) {
fmt.Print("=> ") if len(coefs) == 0 { fmt.Println("0\n") return } max, min := math.MinInt32, math.MaxInt32 for k := range coefs { if k > max { max = k } if k < min { min = k } } for p := max; p >= min; p-- { if c := coefs[p]; c != 0 { if p < max { sign := "+" if c < 0 { sign = "-" c = -c } fmt.Printf(" %s ", sign) } if c != 1 || (c == 1 && p == 0) { cs := fmt.Sprintf("%v", c) ix := strings.Index(cs, ".") if ix >= 0 { dec := cs[ix:] for _, frac := range fractions { if dec == frac[0] { cs = strings.Replace(cs, dec, frac[1], 1) break } } } if cs[0] == '0' && len(cs) > 1 && cs[1] != '.' { cs = cs[1:] } fmt.Print(cs) } if p != 0 { ps := strconv.Itoa(p) ps = powers.Replace(ps) if ps == "¹" { ps = "" } fmt.Printf("x%s", ps) } } } fmt.Println("\n")
}
func check(err error) {
if err != nil { log.Fatal(err) }
}
func main() {
equs := []string{ `-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹`, `x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1`, `0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵`, `1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰`, `+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1`, `x^5 - 2x**4 + 42x^3 + 40x + 1`, `x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1`, `x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰`, `x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1`, `x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1`, `0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1`, `1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1`, `x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2`, `x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5`, `x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x`, `⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝`, } rgx := regexp.MustCompile(`\s+(\+|-)\s+`) rep := strings.NewReplacer( ",", "", " ", "", "¼", ".25", "½", ".5", "¾", ".75", "⅐", ".14285714285714285", "⅑", ".1111111111111111", "⅒", ".1", "⅓", ".3333333333333333", "⅔", ".6666666666666666", "⅕", ".2", "⅖", ".4", "⅗", ".6", "⅘", ".8", "⅙", ".16666666666666666", "⅚", ".8333333333333334", "⅛", ".125", "⅜", ".375", "⅝", ".625", "⅞", ".875", "↉", ".0", "⏨", "e", "⁄", "/", ) rep2 := strings.NewReplacer( "⁰", "0", "¹", "1", "²", "2", "³", "3", "⁴", "4", "⁵", "5", "⁶", "6", "⁷", "7", "⁸", "8", "⁹", "9", "⁻⁻", "", "⁻", "-", "⁺", "", "**", "", "^", "", "↑", "", "⁄", "/", ) var err error for _, equ := range equs { fmt.Println(equ) terms := rgx.Split(equ, -1) ops := rgx.FindAllString(equ, -1) for i := 0; i < len(ops); i++ { ops[i] = strings.TrimSpace(ops[i]) } coefs := make(map[int]float64) for i, term := range terms { s := strings.Split(term, "x") t := s[0] t = strings.TrimRight(t, "·× ") t = rep.Replace(t) c := 1.0 inverse := false if t == "" || t == "+" || t == "-" { t += "1" } ix := strings.Index(t, "/") if ix == len(t)-1 { inverse = true t = t[0 : len(t)-1] c, err = strconv.ParseFloat(t, 64) check(err) } else if ix >= 0 { u := strings.Split(t, "/") m, err := strconv.ParseFloat(u[0], 64) check(err) n, err := strconv.ParseFloat(u[1], 64) check(err) c = m / n } else { c, err = strconv.ParseFloat(t, 64) check(err) } if i > 0 && ops[i-1] == "-" { c = -c } if c == -0.0 { c = 0 } if len(s) == 1 { coefs[0] += c continue } u := s[1] u = strings.TrimSpace(u) if u == "" { p := 1 if inverse { p = -1 } if c != 0 { coefs[p] += c } continue } u = rep2.Replace(u) jx := strings.Index(u, "/") p := 1 if jx >= 0 { v := strings.Split(u, "/") p, err = strconv.Atoi(v[0]) if (err != nil) { p = 1 } d, err := strconv.ParseFloat(v[1], 64) check(err) c /= d } else { p, err = strconv.Atoi(strings.TrimSpace(u)) if (err != nil) { p = 1 } } if inverse { p = -p } if c != 0 { coefs[p] += c } } printEquation(coefs) }
}</lang>
- Output:
-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹ => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ => 0 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ => x⁵ - 2x⁴ + 42x³ + 40x + 1 +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x**4 + 42x^3 + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ => 1 + 40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x => -40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ ⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝ => ⅐x⁵ - x⁴ + 42⅕x³ + ⅑x - 40¾
Julia
The task allows the "polynomials" to be parsed to have negative exponents. This makes them Laurent polynomials, not ordinary polynomials. <lang julia>import Base.print
struct LaurentPolynomial{T}
powtocoef::Dict{Int,T} varname::Char
end
function tosuper(i)
s = "" if i != 1 if i < 0 s = "\u207b" i = -i end s *= prod([reverseudi[x] for x in reverse(digits(i))]) end return s
end
function print(io::IO, lp::LaurentPolynomial)
if isempty(lp.powtocoef) || all(iszero, values(lp.powtocoef)) print("0") return end firstterm = true dorev = minimum(keys(lp.powtocoef)) >= 0 for p in sort!(collect(lp.powtocoef), lt =(a,b)->a[1]<b[1], rev=dorev) e, c = p[1], p[2] if c != 0 if !firstterm print(c < 0 ? " - " : " + ") c = abs(c) end sfrac = get(tofracs, c - trunc(c), "") sint = string(Int(trunc(c))) s = round(c) == c ? sint : sfrac == "" ? string(c) : sint == "0" ? sfrac : sint * sfrac print(c == 1 && e != 0 ? "" : s) e != 0 && print(lp.varname, tosuper(e)) firstterm = false end end
end
const uniexp = [['\u2070', '\u00b9', '\u00b2', '\u00b3']; collect('\u2074':'\u2079')] const allsuper = String(vcat(uniexp, ['\u207a', '\u207b'])) const udi = Dict([u => i-1 for (i, u) in enumerate(uniexp)]) const reverseudi = Dict([v => k for (k, v) in udi]) const uch = Dict(['\u207a' => '+', '\u207b' => '-', '\u00b7' => ' ',
'\u00d7' => ' ', '\u2091' => '^', '\u23e8' => 'E'])
const fracs = Dict(['¼' => "$(1/4)", '½' => "$(1/2)", '¾' => "$(3/4)", '⅐' => "$(1/7)",
'⅑' => "$(1/9)", '⅒' => "$(1/10)", '⅓' => "$(1/3)", '⅔' => "$(2/3)", '⅕' => "$(1/5)", '⅖' => "$(2/5)", '⅗' => "$(3/5)", '⅘' => "$(4/5)", '⅙' => "(1/6)", '⅚' => "$(5/6)", '⅛' => "$(1/8)", '⅜' => "$(3/8)", '⅝' => "$(5/8)", '⅞' => "$(7/8)", '↉' => "$(0/3)"])
const tofracs = Dict(0.25 => '¼', 0.5 => '½', 0.75 => '¾', 0.2 => '⅕') allfrac = join(vcat(collect(keys(fracs))), "|")
utoascii(c) = (x = haskey(udi, c) ? Char('0' + udi[c]) : haskey(uch, c) ? uch[c] : c)
function fcoef(termstring, varname)
m = match(r"(\d+)⁄(\d+)", termstring).captures return "$(parse(Float64, m[1])/parse(Float64, m[2]))$varname^0"
end
function xcoef(termstring, varname)
m = match(Regex("(\\d+)$varname([^⁄\\+]*)⁄(\\d+)"), termstring).captures coe = parse(Float64, m[1])/parse(Float64, m[3]) return "$(coe)$varname$(m[2])"
end
function fromvfrac(termstring)
m = match(r"(\d*)(\D+)", termstring).captures s = string((m[1] == "" ? 0 : parse(Float64, m[1])) + parse(Float64, fracs[m[2][1]]))
end
function normalizeexpression(s, varname='x')
s = replace(s, Regex("\\d*(" * allfrac * ")") => (x) -> fromvfrac(x)) s = replace(replace(s, r"[^\+].+" => (x) -> "+$x"), r".+[^\+]$" => (x) -> "$x+") s = replace(replace(replace(s, r"\s+" => ""), r"\*\*|↑" => "^"), "\u23e8" => "e") s = replace(s, Regex(varname * "(?=[$allsuper]+)") => "$varname^") s = replace(prod([utoascii(c) for c in s]), r"\s+" => "") s = replace(s, r"[\-\+]+" => (x) -> isodd(count(y -> y == '-', x)) ? "+-" : "+") s = replace(replace(s, r"\^\+\-" => "^-"), r"\^\+" => "^") s = replace(s, Regex("(?<=[\\-\\+])$varname") => "1$varname") s = replace(s, r"(\d+)x([^⁄\+]*)⁄(\d+)" => (x) -> xcoef(x, varname)) s = replace(s, r"(?<=[0-9])(,)(?=[0-9])" => "") s = replace(s, r"\d+⁄\d+" => (x) -> fcoef(x, varname)) s = replace(s, r"\+([\d\.]+)\+" => SubstitutionString("+\\1$varname^0+")) return replace(s, Regex("(⁄|/)(" * varname * ")\\^?") => (x) -> varname * (x[end] == '^' ? "^-" : "^-1"))
end
function topoly(s::String, varname='x')
allcoef = Dict{Int,Float64}(0 => 0.0) s = normalizeexpression(s, varname) reg = Regex("[\\.\\-\\+\\deE\\/]*" * varname * "(?:\\^[\\d\\-]+)?(?=(?:[^\\+\\-]?(?:\\+|\\-)))") matched = collect(eachmatch(reg, s)) pairs = [split(x.match, varname * "^") for x in matched] for p in pairs p[1] = replace(p[1], "+-" => "-") if length(p) == 1 push!(p, "1") p[1] = replace(p[1], Regex(varname * "\$") => "") end coef, expo = tryparse(Float64, p[1]), tryparse(Int, p[2]) coef = coef == nothing ? 1.0 : coef expo = expo == nothing ? 1 : expo if haskey(allcoef, expo) coef += allcoef[expo] end allcoef[expo] = coef end return LaurentPolynomial(allcoef, varname)
end
testcases = [ "1x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1", "0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵", "1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰", "+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1", "x^5 - 2x**4 + 42x^3 + 40x + 1", "x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1", "x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰", "x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1", "x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1", "0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1", "1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1", "x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2", "x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½", ]
for s in testcases
println(lpad(s, 48), " => ", topoly(s, 'x'))
end
</lang>
- Output:
1x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ => 0 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ => x⁵ - 2x⁴ + 42x³ + 40x + 1 +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x**4 + 42x^3 + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ => x⁻⁵ - 2x⁻⁴ + 42x⁻³ + 40x⁻¹ + 1 x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½
Nim
<lang Nim>import re, sequtils, strutils, tables from unicode import toRunes
const
Powers = [("0", "⁰"), ("1", "¹"), ("2", "²"), ("3", "³"), ("4", "⁴"), ("5", "⁵"), ("6", "⁶"), ("7", "⁷"), ("8", "⁸"), ("9", "⁹")]
Fractions = [(".25", "¼"), (".5", "½"), (".75", "¾"), (".14285714285714", "⅐"), (".11111111111111", "⅑"), (".1", "⅒"), (".33333333333333", "⅓"), (".66666666666667", "⅔"), (".2", "⅕"), (".4", "⅖"), (".6", "⅗"), (".8", "⅘"), (".16666666666667", "⅙"), (".83333333333333", "⅚"), (".125", "⅛"), (".375", "⅜"), (".625", "⅝"), (".875", "⅞")]
Reps1 = [(",", ""), (" ", ""), ("¼", ".25"), ("½", ".5"), ("¾", ".75"), ("⅐", ".14285714285714"), ("⅑", ".11111111111111"), ("⅒", ".1"), ("⅓", ".33333333333333"), ("⅔", ".66666666666667"), ("⅕", ".2"), ("⅖", ".4"), ("⅗", ".6"), ("⅘", ".8"), ("⅙", ".16666666666667"), ("⅚", ".83333333333333"), ("⅛", ".125"), ("⅜", ".375"), ("⅝", ".625"), ("⅞", ".875"), ("↉", ".0"), ("⏨", "e"), ("⁄", "/")]
Reps2 = [("⁰", "0"), ("¹", "1"), ("²", "2"), ("³", "3"), ("⁴", "4"), ("⁵", "5"), ("⁶", "6"), ("⁷", "7"), ("⁸", "8"), ("⁹", "9"), ("⁻⁻", ""), ("⁻", "-"), ("⁺", ""), ("**", ""), ("^", ""), ("↑", ""), ("⁄", "/")]
Eqs = [ "-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹", "x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1", "0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵", "1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰", "+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1", "x^5 - 2x**4 + 42x^3 + 40x + 1", "x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1", "x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰", "x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1", "x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1", "0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1", "1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1", "x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2", "x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5", "x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x", "⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝"]
type Coefs = Table[int, float]
proc printEquation(coefs: Coefs) =
stdout.write "=> " if coefs.len == 0: echo "0\n" return
var max = int.low var min = int.high for k in coefs.keys: if k > max: max = k if k < min: min = k
for p in countdown(max, min): var c = coefs.getOrDefault(p) if c != 0: if p < max: var sign = '+' if c < 0: sign = '-' c = -c stdout.write ' ', sign, ' '
if c != 1 or (c == 1 and p == 0): var cs = $c cs.trimZeros() let ix = cs.find('.') if ix >= 0: let d = cs[ix..^1] for frac in Fractions: if d == frac[0]: cs = cs.replace(d, frac[1]) break if cs[0] == '0' and cs.len > 1 and cs[1] != '.': cs = cs[1..^1] stdout.write cs
if p != 0: var ps = ($p).multiReplace(Powers) if ps == "¹": ps = "" stdout.write 'x', ps
echo '\n'
template getFloat(s: string): float =
try: s.parseFloat() except ValueError: quit "Expected float, got: " & s, QuitFailure
let rgx = re"\s+(\+|-)\s+"
for eq in Eqs:
echo eq let terms = eq.split(rgx) let ops = toSeq(eq.findAll(rgx)).mapIt(it.strip()) var coefs: Coefs for i, term in terms:
let s = term.split("x") var t = s[0] t = unicode.strip(t, runes = "·× ".toRunes) t = t.multiReplace(Reps1) var c = 1.0 var inverse = false if t == "" or t == "+" or t == "-": t &= '1'
let ix = t.find('/') if ix == t.high: inverse = true t.setLen(t.high) c = t.getFloat() elif ix >= 0: let u = t.split('/') let m = u[0].getFloat() let n = u[1].getFloat() c = m / n else: c = t.getFloat() if i > 0 and ops[i - 1] == "-": c = -c if c == -0.0: c = 0.0
if s.len == 1: coefs[0] = coefs.getOrDefault(0) + c continue
var u = s[1].strip() if u.len == 0: let p = if inverse: -1 else: 1 if c != 0: coefs[p] = coefs.getOrDefault(p) + c continue
u = u.multiReplace(Reps2) let jx = u.find('/') var p: int if jx >= 0: let v = u.split('/') p = try: v[0].parseInt() except ValueError: 1 let d = v[1].getFloat() c /= d else: p = try: u.strip().parseInt() except ValueError: 1 if inverse: p = -p if c != 0: coefs[p] = coefs.getOrDefault(p) + c
printEquation(coefs)</lang>
- Output:
-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹ => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ => 0 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ => x⁵ - 2x⁴ + 42x³ + 40x + 1 +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x**4 + 42x^3 + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ => 1 + 40x-¹ + 42x-³ - 2x-⁴ + x-⁵ x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x => -40x-¹ + 42x-³ - 2x-⁴ + x-⁵ ⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝ => ⅐x⁵ - x⁴ + 42⅕x³ + ⅑x - 40¾
Phix
To simplify this task I first created a test file (save as utf8, Unicode_polynomial_equation.txt):
-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹ ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 ===> x^5 - 2x^4 + 42x^3 + 40x + 1 x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ ==> 0 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x**4 + 42x^3 + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ ==> 1 + 40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ ===> 1 + 40x^-1 + 42x^-3 - 2x^-4 + x^-5 x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 ==> x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ ===> x^5 - 0.5x^4 + 101.25x^3 + 100.75x + 2.5 x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ ==> x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x<sup>5</sup> - 2x<sup>4</sup> + 42x<sup>3</sup> + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x^4 + 42x^3 + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 42x³ + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1 x5 - 2x4 + 42x3 + 40x + 1 ==> x⁵ - 2x⁴ + 42x³ + 40x + 1
Explanation:
On finding a line beginning with "==> ", parse the previous line and check that the output matches the rest of the line.
Lines beginning "===> " contain alternative (valid) representations of the previous line, for use when uni_frac is false. Note these equivalences can then be used further on, eg the result on line 5 maps to line 3 via 2, not that it matters should the same equivalent be defined twice, and also note that "==> " should always be the unicode version and "===> " the ascii one.
Obviously this is not well tested and //will// fail on the next thing thrown at it, but it shows the basic approach. Error handling omitted using that good ol' standby excuse "for clarity".
Note that space is not skipped when getting exponents, otherwise it might wrongly treat say "x + 2" as "x^2", since we /are/ allowing exponentiation symbols to be omitted. At the start of parse() we strip spaces around '^' (etc) to help with that.
Processing the input in utf32 form is a complete no-brainer; whereas constructing utf8 output is a little more subtle, but avoids creating a utf32 "string" just because all bytes happen to be less than 255, which would not necessarily be valid utf8/32.
<lang Phix>-- demo\rosetta\Unicode_polynomial_equation.exw constant uni_frac = false -- if true output unicode superscripts and vulgar fractions
constant UTF8BOM = {#EF,#BB,#BF},
SPTWO = #00B2, -- superscript two STHRE = #00B3, -- superscript three MIDOT = #00B7, -- middle dot SPONE = #00B9, -- superscript one QUART = #00BC, -- one quarter AHALF = #00BD, -- one half THQTR = #00BE, -- three quarters MULTY = #00D7, -- multiplication sign DIVDE = #2044, -- division sign SZERO = #2070, -- superscript zero SFOUR = #2074, -- superscript four SFIVE = #2075, -- superscript five SPSIX = #2076, -- superscript six SSEVN = #2077, -- superscript seven SEGHT = #2078, -- superscript eight SNINE = #2079, -- superscript nine SPLUS = #207A, -- superscript plus SMNUS = #207B, -- superscript minus SVNTH = #2150, -- one seventh NINTH = #2151, -- one ninth TENTH = #2152, -- one tenth THIRD = #2153, -- one third TWTHD = #2154, -- two thirds FIFTH = #2155, -- one fifth TWFTH = #2156, -- two fifths THFTH = #2157, -- three fifths FRFTH = #2158, -- four fifths SIXTH = #2159, -- one sixth FVSIX = #215A, -- five sixths EIGTH = #215B, -- one eigth THEGH = #215C, -- three eigths FVEGH = #215D, -- five eigths
-- ZTHRD = #2189, -- zero thirds[??]
UPARW = #2191, -- uparrow BASET = #23E8, -- base 10 SPACE = ' ', -- space T = 10, -- align nxt tbl
$
constant {vulgar_fractions,unicode_vulgar_fractions} = columnize({{{1,4},QUART},
{{1,2},AHALF}, {{3,4},THQTR}, {{1,7},SVNTH}, {{1,9},NINTH}, {{1,T},TENTH}, {{1,3},THIRD}, {{2,3},TWTHD}, {{1,5},FIFTH}, {{2,5},TWFTH}, {{3,5},THFTH}, {{4,5},FRFTH}, {{1,6},SIXTH}, {{5,6},FVSIX}, {{1,8},EIGTH}, {{3,8},THEGH}, {{5,8},FVEGH}})
constant EXPONENTS = {SZERO,SPONE,SPTWO,STHRE,SFOUR,SFIVE,SPSIX,SSEVN,SEGHT,SNINE}
function skip(sequence s, integer sdx, sequence set)
while sdx<=length(s) and find(s[sdx],set) do sdx += 1 end while return sdx
end function
function get_sign(sequence s, integer sdx, bool allow_superscripts) integer sgn = +1, ch
for sdx=sdx to length(s) do ch = s[sdx] if allow_superscripts then ch = iff(ch=SPLUS?'+': iff(ch=SMNUS?'-': iff(ch=SPACE?'?':ch))) -- (do not skip spaces, see note) end if if ch!='+' and ch!=' ' then if ch!='-' then exit end if sgn *= -1 end if end for return {sgn,sdx}
end function
function get_num(sequence s, integer sdx, atom n=0, bool allow_superscripts=false, as_fraction=false) integer sgn = +1, ch, f, e10, d = 1 atom p10 = 10 bool none = not as_fraction -- (cope with "x" == "1x^1" and != "0x^0")
-- (but not when processing the "34" of "12.34", obvs) if not as_fraction then {sgn,sdx} = get_sign(s,sdx,allow_superscripts) end if if not allow_superscripts then sdx = skip(s,sdx,{' '}) end if for sdx=sdx to length(s) do ch = s[sdx] if ch>='0' and ch<='9' then if as_fraction then n += (ch-'0')/p10 p10 *= 10 else n = n*10+ch-'0' end if none = false elsif allow_superscripts then f = find(ch,EXPONENTS) if f=0 then exit end if n = n*10+f-1 none = false elsif not find(ch," ,") then exit end if end for if not allow_superscripts then if find(ch,{'e','E',BASET}) then {e10,f,sdx} = get_num(s,sdx+1) if f!=1 then ?9/0 end if n *= power(10,e10) elsif ch='.' then if as_fraction then ?9/0 end if {n,f,sdx} = get_num(s,sdx+1,n,as_fraction:=true) if f!=1 then ?9/0 end if none = false else f = find(ch,unicode_vulgar_fractions) if f!=0 then if as_fraction then ?9/0 end if integer {vfn,vfd} = vulgar_fractions[f] if uni_frac then n = n*vfd + vfn if d!=1 then ?9/0 end if d = vfd else n += vfn/vfd end if sdx += 1 none = false end if end if end if if none then n = 1 end if n *= sgn return {n,d,sdx}
end function
function get_term(sequence s, integer sdx) integer last_sdx = sdx, -- (infinite loop check/prevent)
e = 0
atom c, d = 1, f bool sdiv = false
{c,d,sdx} = get_num(s,sdx) sdx = skip(s,sdx,{' ',MIDOT,MULTY}) if sdx<=length(s) and (s[sdx]='/' or s[sdx]=DIVDE) then sdx += 1 if sdx<=length(s) and s[sdx]!='x' then {d,f,sdx} = get_num(s,sdx) if f!=1 then ?9/0 end if if not uni_frac then c /= d d = 1 end if else sdiv = true end if end if if sdx<=length(s) and s[sdx]='x' then sdx = skip(s,sdx+1,{'^','*',SPLUS,UPARW}) {e,f,sdx} = get_num(s,sdx,allow_superscripts:=true) if f!=1 then ?9/0 end if if sdiv then e = -e end if else if sdiv then ?9/0 end if end if if sdx<=length(s) and (s[sdx]='/' or s[sdx]=DIVDE) then if d!=1 then ?9/0 end if {d,f,sdx} = get_num(s,sdx+1) if f!=1 then ?9/0 end if if not uni_frac then c /= d d = 1 end if end if if sdx=last_sdx then ?9/0 end if return {e,c,d,sdx}
end function
function unicode_superscripts(integer e) -- display helper string res = ""
if e>9 then res = unicode_superscripts(floor(e/10)) e = remainder(e,10) end if res &= utf32_to_utf8({EXPONENTS[e+1]}) return res
end function
enum EXP, COEF, FRAC -- contents of terms[i]
function poly(sequence terms) -- display helper string r = ""
for t=length(terms) to 1 by -1 do {integer e, atom c, integer f} = terms[t] if c!=0 then if c=1 and f=1 and e!=0 then r &= iff(r=""? "":" + ") elsif c=-1 and f=1 and e!=0 then r &= iff(r=""?"-":" - ") else if r!="" then r &= iff(c<0?" - ":" + ") c = abs(c) end if if f!=1 then -- (hence/only when uni_frac==true) integer k = find({remainder(c,f),f},vulgar_fractions) if k then c = floor(c/f) if c!=0 then r &= sprintf("%d",c) end if r &= utf32_to_utf8({unicode_vulgar_fractions[k]}) else r &= sprintf("%g",c/f) end if else r &= sprintf("%g",c) end if end if if e!=0 then r &= 'x' if e!=1 then if uni_frac then if e<0 then r &= utf32_to_utf8({SMNUS}) e = -e end if r &= unicode_superscripts(e) else r &= sprintf("^%d",e) end if end if end if end if end for if r="" then r="0" end if return r
end function
function parse(sequence s) sequence terms = {} integer sdx = 1, e, f atom c
s = match_replace("",s,"^") s = match_replace("",s,"") s = match_replace("**",s,"^") s = match_replace(" ^",s,"^") s = match_replace("^ ",s,"^") s = match_replace({' ',SPLUS},s,{SPLUS}) s = match_replace({' ',SMNUS},s,{SMNUS}) for i=1 to length(EXPONENTS) do e = EXPONENTS[i] s = match_replace({' ',e},s,{e}) end for while sdx<=length(s) do {e,c,f,sdx} = get_term(s,sdx) if c!=0 then -- (aside: +5 -5 may yet leave c==0) terms = append(terms,{e,c,f}) end if end while terms = sort(terms) -- merge, eg "10x^2 + 10x^2" -> 20x^2 for i=length(terms) to 2 by -1 do if terms[i][EXP] = terms[i-1][EXP] then if terms[i-1][FRAC]!=terms[i][FRAC] then ?9/0 -- placeholder for more code else terms[i-1][COEF] += terms[i][COEF] end if terms[i..i] = {} -- (delete 2nd) end if end for return poly(terms)
end function
sequence alts = {}, -- (unicode versions)
altn = {}, -- (idx of ascii equivalents) lines = read_lines("Unicode_polynomial_equation.txt")
if lines[1][1..3] = UTF8BOM then
-- remove/ignore any utf8 byte order mark lines[1] = lines[1][4..$]
end if
for i=2 to length(lines) do
if length(lines[i])>5 and lines[i][1..5] = "===> " then alts = append(alts,lines[i-1][5..$]) altn = append(altn,i) end if
end for
for i=2 to length(lines) do
if length(lines[i])>4 and lines[i][1..4] = "==> " then sequence line = utf8_to_utf32(lines[i-1]) sequence res = parse(line) sequence expected = lines[i][5..$] if res=expected then -- (res is the unicode version) if platform()!=WINDOWS or res="0" then printf(1,"%2d: %40s ok\n",{i-1,res}) else -- (unicode output on windows consoles is fiddly...) printf(1,"%2d: ok\n",i-1) end if else integer k = find(expected,alts) if k and res=lines[altn[k]][6..$] then -- (res is the ascii equivalent) printf(1,"%2d: %40s ok\n",{i-1,res}) else printf(1,"%d: error - %s\n",{i-1,res}) end if end if end if
end for</lang>
- Output:
uni_frac = false
1: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 4: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 6: 0 ok 8: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 10: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 12: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 14: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 16: 1 + 40x^-1 + 42x^-3 - 2x^-4 + x^-5 ok 19: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 21: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 23: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 25: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 27: x^5 - 0.5x^4 + 101.25x^3 + 100.75x + 2.5 ok 30: x^5 - 0.5x^4 + 101.25x^3 + 100.75x + 2.5 ok 32: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 34: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 36: x^5 - 2x^4 + 42x^3 + 40x + 1 ok 38: x^5 - 2x^4 + 42x^3 + 40x + 1 ok
uni_frac = true (linux only, unless you have managed to find and install a decent windows console unicode font, which I haven't)
1: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 4: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 6: 0 ok 8: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 10: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 12: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 14: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 16: 1 + 40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ ok 19: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 21: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 23: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 25: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 27: x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ ok 30: x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ ok 32: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 34: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 36: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok 38: x⁵ - 2x⁴ + 42x³ + 40x + 1 ok
Wren
<lang ecmascript>import "/pattern" for Pattern
var powers = [
["0", "⁰"], ["1", "¹"], ["2", "²"], ["3", "³"], ["4", "⁴"], ["5", "⁵"], ["6", "⁶"], ["7", "⁷"], ["8", "⁸"], ["9", "⁹"], ["-", "⁻"]
]
var fractions = [
[".25", "¼"], [".5", "½"], [".75", "¾"], [".14285714285714", "⅐"], [".11111111111111", "⅑"], [".1", "⅒"], [".33333333333333", "⅓"], [".66666666666667", "⅔"], [".2", "⅕"], [".4", "⅖"], [".6", "⅗"], [".8", "⅘"], [".16666666666667", "⅙"], [".83333333333333", "⅚"], [".125", "⅛"], [".375", "⅜"], [".625", "⅝"], [".875", "⅞"]
]
var printEquation = Fn.new { |coefs|
System.write("=> ") if (coefs.isEmpty) { System.print("0\n") return } var keys = coefs.keys.toList var max = keys[0] var min = keys[0] for (k in keys.skip(1)) { if (k > max) max = k if (k < min) min = k } for (p in max..min) { var c = coefs[p] ? coefs[p] : 0 if (c != 0) { if (p < max) { var sign = "+" if (c < 0) { sign = "-" c = -c } System.write(" %(sign) ") } if (c != 1 || (c == 1 && p == 0)) { var cs = c.toString var ix = cs.indexOf(".") if (ix >= 0) { var dec = cs[ix..-1] for (frac in fractions) { if (dec == frac[0]) { cs = cs[0...ix] + frac[1] break } } } if (cs[0] == "0" && cs.count > 1 && cs[1] != ".") cs = cs[1..-1] System.write(cs) } if (p != 0) { var ps = p.toString for (power in powers) ps = ps.replace(power[0], power[1]) if (ps == "¹") ps = "" System.write("x%(ps)") } } } System.print("\n")
}
var equs = [
"-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹", "x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1", "0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵", "1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰", "+x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1", "x^5 - 2x**4 + 42x^3 + 40x + 1", "x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1", "x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰", "x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1", "x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1", "0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1", "1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1", "x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2", "x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5", "x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x", "⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝"
]
var pattern = Pattern.new("+1/s/n+1/s")
var reps = [
[",", ""], [" ", ""], ["¼", ".25"], ["½", ".5"], ["¾", ".75"], ["⅐", ".14285714285714"], ["⅑", ".11111111111111"], ["⅒", ".1"], ["⅓", ".33333333333333"], ["⅔", ".66666666666667"], ["⅕", ".2"], ["⅖", ".4"], ["⅗", ".6"], ["⅘", ".8"], ["⅙", ".16666666666667"], ["⅚", ".83333333333333"], ["⅛", ".125"], ["⅜", ".375"], ["⅝", ".625"], ["⅞", ".875"], ["↉", ".0"], ["⏨", "e"], ["⁄", "/"]
]
var reps2 = [
["⁰", "0"], ["¹", "1"], ["²", "2"], ["³", "3"], ["⁴", "4"], ["⁵", "5"], ["⁶", "6"], ["⁷", "7"], ["⁸", "8"], ["⁹", "9"], ["⁻⁻", ""], ["⁻", "-"], ["⁺", ""], ["**", ""], ["^", ""], ["↑", ""], ["⁄", "/"]
]
for (equ in equs) {
System.print(equ) var terms = pattern.splitAll(equ) var ops = pattern.findAll(equ).map { |m| m.text.trim() }.toList var coefs = {} var i = 0 for (term in terms) { var s = term.split("x") var t = s[0].trimEnd("·× ") for (rep in reps) t = t.replace(rep[0], rep[1]) var c = 1 var inverse = false if (t == "" || t == "+" || t == "-") t = t + "1" var ix = t.indexOf("/") if (ix == t.count - 1) { inverse = true t = t[0..-2] c = Num.fromString(t) } else if (ix >= 0) { var u = t.split("/") var m = Num.fromString(u[0]) var n = Num.fromString(u[1]) c = m / n } else { c = Num.fromString(t) } if (i > 0 && ops[i-1] == "-") c = -c if (c == -0) c = 0 var cont = false if (s.count == 1) { coefs[0] = coefs[0] ? coefs[0] + c : c cont = true } if (!cont) { var u = s[1].trim() if (u == "") { var p = 1 if (inverse) p = -1 if (c != 0) coefs[p] = coefs[p] ? coefs[p] + c : c cont = true } if (!cont) { for (rep in reps2) u = u.replace(rep[0], rep[1]) var jx = u.indexOf("/") var p = 1 if (jx >= 0) { var v = u.split("/") p = Num.fromString(v[0]) if (p == null) p = 1 var d = Num.fromString(v[1]) c = c / d } else { p = Num.fromString(u.trim()) if (p == null) p = 1 } if (inverse) p = -p if (c != 0) coefs[p] = coefs[p] ? coefs[p] + c : c } } i = i + 1 } printEquation.call(coefs)
}</lang>
- Output:
-0.00x⁺¹⁰ + 1.0·x ** 5 + -2e0x^4 + +0,042.00 × x ⁺³ + +.0x² + 20.000 000 000x¹ - -1x⁺⁰ + .0x⁻¹ + 20.x¹ => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 42x³ + 0x² + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0e+0x⁰⁰⁷ + 00e-00x + 0x + .0x⁰⁵ - 0.x⁴ + 0×x³ + 0x⁻⁰ + 0/x + 0/x³ + 0x⁻⁵ => 0 1x⁵ - 2x⁴ + 42x³ + 40x + 1x⁰ => x⁵ - 2x⁴ + 42x³ + 40x + 1 +x⁺⁵ + -2x⁻⁻⁴ + 42x⁺⁺³ + +40x - -1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x^5 - 2x**4 + 42x^3 + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x↑5 - 2.00·x⁴ + 42.00·x³ + 40.00·x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁻⁵ - 2⁄x⁴ + 42x⁻³ + 40/x + 1x⁻⁰ => 1 + 40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ x⁵ - 2x⁴ + 42.000 000x³ + 40x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - 2x⁴ + 0,042x³ + 40.000,000x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 0x⁷ + 10x + 10x + x⁵ - 2x⁴ + 42x³ + 20x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 1E0x⁵ - 2,000,000.e-6x⁴ + 4.2⏨1x³ + .40e+2x + 1 => x⁵ - 2x⁴ + 42x³ + 40x + 1 x⁵ - x⁴⁄2 + 405x³⁄4 + 403x⁄4 + 5⁄2 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁵ - 0.5x⁴ + 101.25x³ + 100.75x + 2.5 => x⁵ - ½x⁴ + 101¼x³ + 100¾x + 2½ x⁻⁵ - 2⁄x⁴ + 42x⁻³ - 40/x => -40x⁻¹ + 42x⁻³ - 2x⁻⁴ + x⁻⁵ ⅐x⁵ - ⅓x⁴ - ⅔x⁴ + 42⅕x³ + ⅑x - 40⅛ - ⅝ => ⅐x⁵ - x⁴ + 42⅕x³ + ⅑x - 40¾