Approximate equality
Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the difference in floating point calculations between different language implementations becomes significant.
For example, a difference between 32 bit and 64 bit floating point calculations may appear by about the 8th significant digit in base 10 arithmetic.
- Task
Create a function which returns true if two floating point numbers are approximately equal.
The function should allow for differences in the magnitude of numbers, so that, for example,
100000000000000.01 may be approximately equal to 100000000000000.011,
even though 100.01 is not approximately equal to 100.011.
If the language has such a feature in its standard library, this may be used instead of a custom function.
Show the function results with comparisons on the following pairs of values:
- 100000000000000.01, 100000000000000.011 (note: should return true)
- 100.01, 100.011 (note: should return false)
- 10000000000000.001 / 10000.0, 1000000000.0000001000
- 0.001, 0.0010000001
- 0.000000000000000000000101, 0.0
- sqrt(2) * sqrt(2), 2.0
- -sqrt(2) * sqrt(2), -2.0
- 3.14159265358979323846, 3.14159265358979324
Answers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.
Factor
The ~ word takes three arguments: the two values to be compared, and an epsilon value representing the allowed distance between the two values. A positive epsilon performs an absolute distance test, an epsilon of zero performs an exact comparison, and a negative epsilon performs a relative distance test (as required by this task).
<lang factor>USING: formatting generalizations kernel math math.functions ;
100000000000000.01 100000000000000.011 100.01 100.011 10000000000000.001 10000.0 /f 1000000000.0000001000 0.001 0.0010000001 0.000000000000000000000101 0.0 2 sqrt dup * 2.0 2 sqrt dup neg * -2.0 3.14159265358979323846 3.14159265358979324
[ 2dup -1e-15 ~ "%+47.30f %+47.30f -1e-15 ~ : %u\n" printf ] 2 8 mnapply</lang>
- Output:
+100000000000000.015625000000000000000000000000 +100000000000000.015625000000000000000000000000 -1e-15 ~ : t
+100.010000000000005115907697472721 +100.010999999999995679900166578591 -1e-15 ~ : f
+1000000000.000000238418579101562500000000 +1000000000.000000119209289550781250000000 -1e-15 ~ : t
+0.001000000000000000020816681712 +0.001000000100000000054917270731 -1e-15 ~ : f
+0.000000000000000000000101000000 +0.000000000000000000000000000000 -1e-15 ~ : f
+2.000000000000000444089209850063 +2.000000000000000000000000000000 -1e-15 ~ : t
-2.000000000000000444089209850063 -2.000000000000000000000000000000 -1e-15 ~ : t
+3.141592653589793115997963468544 +3.141592653589793115997963468544 -1e-15 ~ : t
Go
Go's float64 type is limited to 15 or 16 digits of precision. As there are some numbers in this task which have more digits than this I've used big.Float instead. <lang go>package main
import (
"fmt" "log" "math/big"
)
func max(a, b *big.Float) *big.Float {
if a.Cmp(b) > 0 {
return a
}
return b
}
func isClose(a, b *big.Float) bool {
relTol := big.NewFloat(1e-9) // same as default for Python's math.isclose() function t := new(big.Float) t.Sub(a, b) t.Abs(t) u, v, w := new(big.Float), new(big.Float), new(big.Float) u.Mul(relTol, max(v.Abs(a), w.Abs(b))) return t.Cmp(u) <= 0
}
func nbf(s string) *big.Float {
n, ok := new(big.Float).SetString(s)
if !ok {
log.Fatal("invalid floating point number")
}
return n
}
func main() {
root2 := big.NewFloat(2.0)
root2.Sqrt(root2)
pairs := [][2]*big.Float{
{nbf("100000000000000.01"), nbf("100000000000000.011")},
{nbf("100.01"), nbf("100.011")},
{nbf("0").Quo(nbf("10000000000000.001"), nbf("10000.0")), nbf("1000000000.0000001000")},
{nbf("0.001"), nbf("0.0010000001")},
{nbf("0.000000000000000000000101"), nbf("0.0")},
{nbf("0").Mul(root2, root2), nbf("2.0")},
{nbf("0").Mul(nbf("0").Neg(root2), root2), nbf("-2.0")},
{nbf("100000000000000003.0"), nbf("100000000000000004.0")},
{nbf("3.14159265358979323846"), nbf("3.14159265358979324")},
}
for _, pair := range pairs {
s := "≉"
if isClose(pair[0], pair[1]) {
s = "≈"
}
fmt.Printf("% 21.19g %s %- 21.19g\n", pair[0], s, pair[1])
}
}</lang>
- Output:
100000000000000.01 ≈ 100000000000000.011
100.01 ≉ 100.011
1000000000.0000001 ≈ 1000000000.0000001
0.001 ≉ 0.0010000001
1.01e-22 ≉ 0
2.000000000000000273 ≈ 2
-2.000000000000000273 ≈ -2
100000000000000003 ≈ 100000000000000004
3.141592653589793239 ≈ 3.14159265358979324
Julia
Julia has an infix operator, ≈, which corresponds to Julia's buitin isapprox() function.
<lang julia>testvalues = [[100000000000000.01, 100000000000000.011],
[100.01, 100.011],
[10000000000000.001 / 10000.0, 1000000000.0000001000],
[0.001, 0.0010000001],
[0.000000000000000000000101, 0.0],
[sqrt(2) * sqrt(2), 2.0],
[-sqrt(2) * sqrt(2), -2.0],
[3.14159265358979323846, 3.14159265358979324]]
for (x, y) in testvalues
println(rpad(x, 21), " ≈ ", lpad(y, 22), ": ", x ≈ y)
end
</lang>
- Output:
1.0000000000000002e14 ≈ 1.0000000000000002e14: true 100.01 ≈ 100.011: false 1.0000000000000002e9 ≈ 1.0000000000000001e9: true 0.001 ≈ 0.0010000001: false 1.01e-22 ≈ 0.0: false 2.0000000000000004 ≈ 2.0: true -2.0000000000000004 ≈ -2.0: true 3.141592653589793 ≈ 3.141592653589793: true
Perl 6
Is approximately equal to is a built-in operator in Perl 6. Unicode ≅, or the ASCII equivalent: =~=. By default it uses a tolerance of 1e-15 times the order of magnitude of the larger comparand, though that is adjustable by setting the dynamic variable $*TOLERANCE to the desired value. Probably a good idea to localize the changed $*TOLERANCE as it will affect all comparisons within its scope.
Most of the following tests are somewhat pointless in Perl 6. To a large extent, when dealing with Rational values, you don't really need to worry about "approximately equal to", and all of the test values below, with the exception of sqrt(2), are Rats by default, and exact. You would have specifically coerce them to Nums (floating point) to lose the precision.
For example, in Perl 6, the sum of .1, .2, .3, & .4 is identically equal to 1.
<lang perl6>say 0.1 + 0.2 + 0.3 + 0.4 === 1.0000000000000000000000000000000000000000000000000000000000000000000000000; # True</lang>
It's also approximately equal to 1 but... ¯\_(ツ)_/¯
<lang perl6>for
100000000000000.01, 100000000000000.011, 100.01, 100.011, 10000000000000.001 / 10000.0, 1000000000.0000001000, 0.001, 0.0010000001, 0.000000000000000000000101, 0.0, sqrt(2) * sqrt(2), 2.0, -sqrt(2) * sqrt(2), -2.0, 100000000000000003.0, 100000000000000004.0, 3.14159265358979323846, 3.14159265358979324
-> $a, $b {
say "$a ≅ $b: ", $a ≅ $b;
}
say "\nTolerance may be adjusted.";
say 22/7, " ≅ ", π, ": ", 22/7 ≅ π; { # Localize the tolerance to only this block
my $*TOLERANCE = .001; say 22/7, " ≅ ", π, ": ", 22/7 ≅ π;
}</lang>
- Output:
100000000000000.01 ≅ 100000000000000.011: True 100.01 ≅ 100.011: False 1000000000.0000001 ≅ 1000000000.0000001: True 0.001 ≅ 0.0010000001: False 0.000000000000000000000101 ≅ 0: True 2.0000000000000004 ≅ 2: True -2.0000000000000004 ≅ -2: True 100000000000000003 ≅ 100000000000000004: True 3.141592653589793226752 ≅ 3.14159265358979324: True Tolerance may be adjusted. 3.142857 ≅ 3.141592653589793: False 3.142857 ≅ 3.141592653589793: True
Phix
Traditionally I have always just used sprintf() to compare floating point atoms in phix.
For this task, it proved much harder to get decent-looking output, than it did to perform the tests, and to that end I
decided to allow the display format (dfmt) to be overidden, when needed, and for the tricker/ambiguous test 5, I also allow the compare format (cfmt) to be overidden, getting both a true and false result. Likewise I have a different result for test 4 to everyone else, but simply setting the cfmt to "%.8f" would get it the NOT.
<lang Phix>procedure test(atom a,b, string dfmt="%g", cfmt="%g")
bool eq = sprintf(cfmt,a)==sprintf(cfmt,b)
string eqs = iff(eq?"":"NOT "),
sa = sprintf(dfmt,a),
sb = sprintf(dfmt,b)
printf(1,"%30s is %sapproximately equal to %s\n",{sa,eqs,sb})
end procedure
test(100000000000000.01,100000000000000.011,"%.3f") test(100.01,100.011) test(10000000000000.001/10000.0,1000000000.0000001000,"%.10f") test(0.001,0.0010000001,"%.8f") test(0.000000000000000000000101,0.0,"%f") test(0.000000000000000000000101,0.0,"%f","%6f") test(sqrt(2)*sqrt(2),2.0) test(-sqrt(2)*sqrt(2),-2.0) test(3.14159265358979323846,3.14159265358979324,"%.20f")</lang>
- Output:
64 bit (implied by some of the accuracies specified for this task):
100000000000000.010 is approximately equal to 100000000000000.011
100.01 is NOT approximately equal to 100.011
1000000000.0000001001 is approximately equal to 1000000000.0000001000
0.00100000000 is approximately equal to 0.0010000001
0.000000000000000000000101000 is NOT approximately equal to 0.000000
0.000000000000000000000101000 is approximately equal to 0.000000
2 is approximately equal to 2
-2 is approximately equal to -2
3.14159265358979323851 is approximately equal to 3.14159265358979324003
32 bit (in fact a couple of them, the first and last pairs, are actually genuinely identical):
100000000000000.016 is approximately equal to 100000000000000.016
100.01 is NOT approximately equal to 100.011
1000000000.0000002384 is approximately equal to 1000000000.0000001192
0.0010000000 is approximately equal to 0.0010000001
0.000000000000000000000101000 is NOT approximately equal to 0.000000
0.000000000000000000000101000 is approximately equal to 0.000000
2 is approximately equal to 2
-2 is approximately equal to -2
3.1415926535897931 is approximately equal to 3.1415926535897931
Python
The Python source documentation states:
math.isclose -> bool
a: double
b: double
*
rel_tol: double = 1e-09
maximum difference for being considered "close", relative to the
magnitude of the input values
abs_tol: double = 0.0
maximum difference for being considered "close", regardless of the
magnitude of the input values
Determine whether two floating point numbers are close in value.
Return True if a is close in value to b, and False otherwise.
For the values to be considered close, the difference between them
must be smaller than at least one of the tolerances.
-inf, inf and NaN behave similarly to the IEEE 754 Standard. That
is, NaN is not close to anything, even itself. inf and -inf are
only close to themselves.
<lang python>from numpy import sqrt from math import isclose
testvalues = [[100000000000000.01, 100000000000000.011],
[100.01, 100.011],
[10000000000000.001 / 10000.0, 1000000000.0000001000],
[0.001, 0.0010000001],
[0.000000000000000000000101, 0.0],
[sqrt(2) * sqrt(2), 2.0],
[-sqrt(2) * sqrt(2), -2.0],
[3.14159265358979323846, 3.14159265358979324]]
for (x, y) in testvalues:
maybenot = "is" if isclose(x, y) else "is NOT" print(x, maybenot, "approximately equal to ", y)
</lang>
- Output:
100000000000000.02 is approximately equal to 100000000000000.02 100.01 is NOT approximately equal to 100.011 1000000000.0000002 is approximately equal to 1000000000.0000001 0.001 is NOT approximately equal to 0.0010000001 1.01e-22 is NOT approximately equal to 0.0 2.0 is approximately equal to 2.0 -2.0 is approximately equal to -2.0 3.141592653589793 is approximately equal to 3.141592653589793
REXX
Since the REXX language uses decimal digits for floating point numbers (and integers), it's just a matter of choosing
the number of decimal digits for the precision to be used for arithmetic (in this case, fifteen decimal digits).
<lang rexx>═════════════════════════ processing pair 1 ══════════════════════════
A= 100000000000000.01
B= 100000000000000.011
A approximately equal to B? true
═════════════════════════ processing pair 2 ══════════════════════════ A= 100.01 B= 100.011
A approximately equal to B? false
═════════════════════════ processing pair 3 ══════════════════════════ A= 1000000000 B= 1000000000.0000001000
A approximately equal to B? true
═════════════════════════ processing pair 4 ══════════════════════════ A= 0.001 B= 0.0010000001
A approximately equal to B? false
═════════════════════════ processing pair 5 ══════════════════════════ A= 0.00000000000000000000101 B= 0.0
A approximately equal to B? false
═════════════════════════ processing pair 6 ══════════════════════════ A= 2.00000000000000 B= 2.0
A approximately equal to B? true
═════════════════════════ processing pair 7 ══════════════════════════ A= -2.00000000000000 B= -2.0
A approximately equal to B? true
═════════════════════════ processing pair 8 ══════════════════════════ A= 100000000000000003.0 B= 100000000000000004.0
A approximately equal to B? true
═════════════════════════ processing pair 9 ══════════════════════════ A= 3.14159265358979323846 B= 3.14159265358979324
A approximately equal to B? true</lang>
- output when using the internal default inputs:
═════════════════════════ processing pair 1 ══════════════════════════
A= 100000000000000.01
B= 100000000000000.011
A approximately equal to B? true
═════════════════════════ processing pair 2 ══════════════════════════
A= 100.01
B= 100.011
A approximately equal to B? false
═════════════════════════ processing pair 3 ══════════════════════════
A= 1000000000
B= 1000000000.0000001000
A approximately equal to B? true
═════════════════════════ processing pair 4 ══════════════════════════
A= 0.001
B= 0.0010000001
A approximately equal to B? false
═════════════════════════ processing pair 5 ══════════════════════════
A= 0.00000000000000000000101
B= 0.0
A approximately equal to B? false
═════════════════════════ processing pair 6 ══════════════════════════
A= 2.00000000000000
B= 2.0
A approximately equal to B? true
═════════════════════════ processing pair 7 ══════════════════════════
A= -2.00000000000000
B= -2.0
A approximately equal to B? true
═════════════════════════ processing pair 8 ══════════════════════════
A= 100000000000000003.0
B= 100000000000000004.0
A approximately equal to B? true
═════════════════════════ processing pair 9 ══════════════════════════
A= 3.14159265358979323846
B= 3.14159265358979324
A approximately equal to B? true
Sidef
Two values can be compared for approximate equality by using the built-in operator ≅, available in ASCII as =~=, which does approximate comparison by rounding both operands at (PREC>>2)-1 decimals. However, by default, Sidef uses a floating-point precision of 192 bits. <lang ruby>[
100000000000000.01, 100000000000000.011, 100.01, 100.011, 10000000000000.001 / 10000.0, 1000000000.0000001000, 0.001, 0.0010000001, 0.000000000000000000000101, 0.0, sqrt(2) * sqrt(2), 2.0, -sqrt(2) * sqrt(2), -2.0, sqrt(-2) * sqrt(-2), -2.0, cbrt(3)**3, 3, cbrt(-3)**3, -3, 100000000000000003.0, 100000000000000004.0, 3.14159265358979323846, 3.14159265358979324
].each_slice(2, {|a,b|
say ("#{a} ≅ #{b}: ", a ≅ b)
})</lang>
- Output:
100000000000000.01 ≅ 100000000000000.011: false 100.01 ≅ 100.011: false 1000000000.0000001 ≅ 1000000000.0000001: true 0.001 ≅ 0.0010000001: false 0.000000000000000000000101 ≅ 0: false 2 ≅ 2: true -2 ≅ -2: true -2 ≅ -2: true 3 ≅ 3: true -3-7.82914889268316957969274243345625157631318402415e-58i ≅ -3: true 100000000000000003 ≅ 100000000000000004: false 3.14159265358979323846 ≅ 3.14159265358979324: false
The Number n.round(-k) can be used for rounding the number n to k decimal places. A positive argument can be used for rounding before the decimal point.
<lang ruby>var a = 100000000000000.01 var b = 100000000000000.011
- Rounding at 2 and 3 decimal places, respectively
say (round(a, -2) == round(b, -2)) # true say (round(a, -3) == round(b, -3)) # false</lang>
There is also the built-in approx_cmp(a, b, k) method, which is equivalent with a.round(k) <=> b.round(k).
<lang ruby>var a = 22/7 var b = Num.pi
say ("22/7 ≅ π at 2 decimals: ", approx_cmp(a, b, -2) == 0) say ("22/7 ≅ π at 3 decimals: ", approx_cmp(a, b, -3) == 0)</lang>
- Output:
22/7 ≅ π at 2 decimals: true 22/7 ≅ π at 3 decimals: false
Additionally, the rat_approx method can be used for computing a very good rational approximation to a given real value:
<lang ruby>say (1.33333333.rat_approx == 4/3) # true say (zeta(-5).rat_approx == -1/252) # true</lang>
Rational approximations illustrated for substrings of PI: <lang ruby>for k in (3..19) {
var r = Str(Num.pi).first(k)
say ("rat_approx(#{r}) = ", Num(r).rat_approx.as_frac)
}</lang>
- Output:
rat_approx(3.1) = 31/10 rat_approx(3.14) = 22/7 rat_approx(3.141) = 245/78 rat_approx(3.1415) = 333/106 rat_approx(3.14159) = 355/113 rat_approx(3.141592) = 355/113 rat_approx(3.1415926) = 86953/27678 rat_approx(3.14159265) = 102928/32763 rat_approx(3.141592653) = 103993/33102 rat_approx(3.1415926535) = 1354394/431117 rat_approx(3.14159265358) = 833719/265381 rat_approx(3.141592653589) = 17925491/5705861 rat_approx(3.1415926535897) = 126312511/40206521 rat_approx(3.14159265358979) = 144029661/45846065 rat_approx(3.141592653589793) = 325994779/103767361 rat_approx(3.1415926535897932) = 903259831/287516534 rat_approx(3.14159265358979323) = 1726375805/549522486
zkl
Floats are 64 bit and have the closeTo method, which takes a comparison value and tolerance. If the tolerance is >=0, comparison is absolute. If tolerance is <0 (and x!=0 and y!=0), the comparison is relative. <lang zkl>testValues:=T(
T(100000000000000.01,100000000000000.011), T(100.01, 100.011), T(10000000000000.001 / 10000.0, 1000000000.0000001), T(0.001, 0.0010000001), T(0.00000000000000000101, 0.0), T( (2.0).sqrt()*(2.0).sqrt(), 2.0), T( -(2.0).sqrt()*(2.0).sqrt(), -2.0), T(100000000000000003.0, 100000000000000004.0), T(3.14159265358979323846, 3.14159265358979324)
);
tolerance:=-1e-9; // <0 for relative comparison foreach x,y in (testValues){
maybeNot:=( if(x.closeTo(y,tolerance)) " \u2248" else "!\u2248" );
println("% 25.19g %s %- 25.19g %g".fmt(x,maybeNot,y, (x-y).abs()));
}</lang>
- Output:
100000000000000.0156 ≈ 100000000000000.0156 0
100.0100000000000051 !≈ 100.0109999999999957 0.001
1000000000.000000238 ≈ 1000000000.000000119 1.19209e-07
0.001000000000000000021 !≈ 0.001000000100000000055 1e-10
1.010000000000000018e-18 !≈ 0 1.01e-18
2.000000000000000444 ≈ 2 4.44089e-16
-2.000000000000000444 ≈ -2 4.44089e-16
100000000000000000 ≈ 100000000000000000 0
3.141592653589793116 ≈ 3.141592653589793116 0