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172
Task/Numerical-integration/Go/numerical-integration.go
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172
Task/Numerical-integration/Go/numerical-integration.go
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package main
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import (
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"fmt"
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"math"
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)
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// specification for an integration
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type spec struct {
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lower, upper float64 // bounds for integration
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n int // number of parts
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exact float64 // expected answer
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fs string // mathematical description of function
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f func(float64) float64 // function to integrate
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}
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// test cases per task description
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var data = []spec{
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spec{0, 1, 100, .25, "x^3", func(x float64) float64 { return x * x * x }},
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spec{1, 100, 1000, float64(math.Log(100)), "1/x",
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func(x float64) float64 { return 1 / x }},
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spec{0, 5000, 5e5, 12.5e6, "x", func(x float64) float64 { return x }},
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spec{0, 6000, 6e6, 18e6, "x", func(x float64) float64 { return x }},
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}
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// object for associating a printable function name with an integration method
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type method struct {
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name string
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integrate func(spec) float64
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}
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// integration methods implemented per task description
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var methods = []method{
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method{"Rectangular (left) ", rectLeft},
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method{"Rectangular (right) ", rectRight},
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method{"Rectangular (midpoint)", rectMid},
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method{"Trapezium ", trap},
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method{"Simpson's ", simpson},
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}
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func rectLeft(t spec) float64 {
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parts := make([]float64, t.n)
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r := t.upper - t.lower
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nf := float64(t.n)
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x0 := t.lower
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for i := range parts {
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x1 := t.lower + float64(i+1)*r/nf
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// x1-x0 better than r/nf.
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// (with r/nf, the represenation error accumulates)
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parts[i] = t.f(x0) * (x1 - x0)
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x0 = x1
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}
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return sum(parts)
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}
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func rectRight(t spec) float64 {
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parts := make([]float64, t.n)
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r := t.upper - t.lower
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nf := float64(t.n)
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x0 := t.lower
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for i := range parts {
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x1 := t.lower + float64(i+1)*r/nf
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parts[i] = t.f(x1) * (x1 - x0)
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x0 = x1
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}
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return sum(parts)
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}
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func rectMid(t spec) float64 {
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parts := make([]float64, t.n)
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r := t.upper - t.lower
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nf := float64(t.n)
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// there's a tiny gloss in the x1-x0 trick here. the correct way
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// would be to compute x's at division boundaries, but we don't need
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// those x's for anything else. (the function is evaluated on x's
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// at division midpoints rather than division boundaries.) so, we
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// reuse the midpoint x's, knowing that they will average out just
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// as well. we just need one extra point, so we use lower-.5.
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x0 := t.lower - .5*r/nf
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for i := range parts {
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x1 := t.lower + (float64(i)+.5)*r/nf
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parts[i] = t.f(x1) * (x1 - x0)
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x0 = x1
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}
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return sum(parts)
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}
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func trap(t spec) float64 {
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parts := make([]float64, t.n)
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r := t.upper - t.lower
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nf := float64(t.n)
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x0 := t.lower
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f0 := t.f(x0)
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for i := range parts {
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x1 := t.lower + float64(i+1)*r/nf
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f1 := t.f(x1)
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parts[i] = (f0 + f1) * .5 * (x1 - x0)
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x0, f0 = x1, f1
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}
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return sum(parts)
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}
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func simpson(t spec) float64 {
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parts := make([]float64, 2*t.n+1)
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r := t.upper - t.lower
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nf := float64(t.n)
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// similar to the rectangle midpoint logic explained above,
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// we play a little loose with the values used for dx and dx0.
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dx0 := r / nf
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parts[0] = t.f(t.lower) * dx0
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parts[1] = t.f(t.lower+dx0*.5) * dx0 * 4
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x0 := t.lower + dx0
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for i := 1; i < t.n; i++ {
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x1 := t.lower + float64(i+1)*r/nf
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xmid := (x0 + x1) * .5
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dx := x1 - x0
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parts[2*i] = t.f(x0) * dx * 2
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parts[2*i+1] = t.f(xmid) * dx * 4
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x0 = x1
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}
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parts[2*t.n] = t.f(t.upper) * dx0
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return sum(parts) / 6
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}
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// sum a list of numbers avoiding loss of precision
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func sum(v []float64) float64 {
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if len(v) == 0 {
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return 0
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}
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var parts []float64
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for _, x := range v {
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var i int
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for _, p := range parts {
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sum := p + x
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var err float64
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if math.Abs(x) < math.Abs(p) {
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err = x - (sum - p)
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} else {
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err = p - (sum - x)
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}
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if err != 0 {
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parts[i] = err
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i++
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}
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x = sum
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}
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parts = append(parts[:i], x)
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}
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var sum float64
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for _, x := range parts {
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sum += x
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}
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return sum
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}
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func main() {
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for _, t := range data {
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fmt.Println("Test case: f(x) =", t.fs)
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fmt.Println("Integration from", t.lower, "to", t.upper,
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"in", t.n, "parts")
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fmt.Printf("Exact result %.7e Error\n", t.exact)
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for _, m := range methods {
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a := m.integrate(t)
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e := a - t.exact
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if e < 0 {
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e = -e
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}
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fmt.Printf("%s %.7e %.7e\n", m.name, a, e)
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}
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fmt.Println("")
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}
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}
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