Помогите переписать процедуру Spline с Pascal на C#
Формулировка задачи:
Помогите пожалуйста переписать процедуру Spline на C#
вот код паскаль:
Листинг программы
- Vector = Array [0..640] of Real;
- Vec = Array [0..30] of Real;
- Matr = Array [ 1..30,1..30] of Real; Ne,Ngr :Integer;
- Xe,Ye,Xg,Yg:Vector;
- M :Integer;
- al,bl:Real;
- K:Vec;
- Procedure Spline(Xe, Ye : Vector; Ne, Ngr : Integer; Var Xg, Yg : Vector);
- Var a, b, c, d : Vec;
- h, R : Real;
- i, j, L : Integer;
- //cubspln
- Procedure CubSpln(Xe,Ye:Vector;Ne:Integer;Var a,b,c,d:Vec);
- Var h, Y : Vec;
- Ma : Matr;
- i, j : Integer;
- R, Q : Real;
- //progonka
- Procedure Progonka (A:Matr; b:Vec; N:Integer; Var X:Vec);
- Var Alp, Bet : Vec;
- R : Real;
- i : Integer;
- Begin
- Alp[1] := - A[1, 2] / A[1, 1];
- Bet[1] := b[1] / A[1, 1];
- For i := 2 to N-1 do
- Begin
- R := A[i,i] + Alp[i - 1] * A[i, i - 1];
- Alp[i] := - A[i, i + 1] / R;
- Bet [i] := (b[i] - A[i, i - 1] * Bet[i - 1]) / R
- End;
- X[N] := (B[N] - A[N, N - 1] * Bet[N - 1]) / (A[N, N]
- + Alp[N - 1] * A[N, N - 1]);
- For i := N-1 downto 1 do
- X[i] := Alp [i] * X [i + 1] + Bet [i]
- End;
- //progonka
- Begin
- Ne := Ne - 1;
- For i := 1 to Ne do
- Begin
- H[i] := Xe[i] - Xe[i - 1];
- a[i] := Ye[i]
- End;
- For i := 1 to Ne - 1 do
- For j := 1 to Ne - 1 do
- If i=j then
- Begin
- R := H[i]; Q :=H[i + 1];
- Ma[i, i] := 2 * (R + Q);
- Y[i] := 6*((Ye[i+1]-Ye[i])/Q-(Ye[i]-Ye[i-1])/R);
- End Else
- Begin
- Q := H[i + 1];
- Ma[j, i] := 0;
- Ma[i + 1, i] := Q;
- Ma[i, i + 1] := Q
- End;
- Progonka (Ma,Y,Ne-1,c);
- c[0] := 0; c[Ne] := 0;
- For i:=1 to Ne do
- Begin
- R := h[i];
- d[i] := (c[i] - c[i-1]) / R;
- b[i] := R*c[i]/2 - Sqr(R)*d[i]/6+(Ye[i] - Ye[i-1])/R
- End;
- End;
- //cubspln
- Begin
- CubSpln (Xe, Ye, Ne, a, b, c, d);
- L := 1;
- h := (bl - al) / Ngr;
- Xg[0]:= al;
- For i := 0 to Ngr - 1 do
- Begin
- YG[i] := a[L] + b[L] * (Xg[i] - Xe[L])+ c[L] * Sqr(Xg[i] -
- Xe[L])/2 + d[L]*Sqr(Xg[i] - Xe[L])*(Xg[i] - Xe[L])/6;
- Xg[i + 1] := Xg[i] + h;
- If Xg[i] > Xe [L] then L := L + 1
- End
- END;
- //end spline
Решение задачи: «Помогите переписать процедуру Spline с Pascal на C#»
textual
Листинг программы
- private void Progonka(Double[,] A, Double[] B, Int32 N, Double[] X)
- {
- Double[] Alp = new Double[N];
- Double[] Bet = new Double[N];
- Double R;
- Int32 i;
- Alp[0] = -A[0, 1] / A[0, 0];
- Bet[0] = B[0] / A[0, 0];
- for (i = 1; i < N; i++)
- {
- R = A[i, i] + Alp[i - 1] * A[i, i - 1];
- Alp[i] = -A[i, i + 1] / R;
- Bet[i] = (B[i] - A[i, i - 1] * Bet[i - 1]) / R;
- }
- X[N] = (B[N] - A[N, N - 1] * Bet[N - 1]) / (A[N, N] + Alp[N - 1] * A[N, N - 1]);
- for (i = N - 1; i >= 0; i--)
- {
- X[i] = Alp[i] * X[i + 1] + Bet[i];
- }
- }
- private void CubSpline(Double[] Xe, Double[] Ye, Int32 Ne, Double[] a, Double[] b, Double[] c, Double[] d)
- {
- Double[,] Ma = new Double[50, 50];
- Double[] h = new Double[40];
- Double[] Y = new Double[40];
- Int32 i = 0, j = 0;
- Double R = 0, Q = 0;
- Ne--;
- for (i = 1; i <= Ne - 1; i++)
- {
- h[i] = Xe[i] = Xe[i - 1];
- a[i] = Ye[i];
- }
- for ( i = 1; i <= Ne - 1; i++)
- for ( j = 1; j <= Ne - 1; j++)
- {
- if (i == j)
- {
- R = h[i];
- Q = h[i + 1];
- Ma[i, i] = 2 * (R + Q);
- Y[i] = 6 * ((Ye[i + 1] - Ye[i]) / Q - (Ye[i] - Ye[i - 1]) / R);//Ye[i-1]???
- }
- else
- {
- Q = h[i + 1];//
- Ma[j, i] = 0;
- Ma[i + 1, i] = Q;
- Ma[i, i + 1] = Q;
- }
- }
- Progonka(Ma, Y, Ne-1, c);
- c[0] = 0;
- c[Ne] = 0;
- for (i = 1; i <= Ne; i++)
- {
- R = h[i];
- d[i] = (c[i] - c[i-1]) / R;
- b[i] = R * c[i] / 2 - R * R * d[i] / 6 + (Ye[i] - Ye[i - 1]) / R;
- }
- }
- private void TabSpline(Double[] Xe, Double[] Ye, Int32 Ne, Int32 Ngr ,Double[]Xg, Double[]Yg)
- {
- CubSpline(Xe, Ye, Ne, a, b, c, d);
- Int32 L = 1;
- Double h = (bl - al) / Ngr;
- Xg[0] = al;
- //Yg[0] = a[L] + b[L] * (Xg[0] - Xe[L]) + c[L] * Math.Pow(Xg[0] - Xe[L], 2) / 2 + d[L] * Math.Pow(Xg[0] - Xe[L], 2) * (Xg[0] - Xe[L]) / 6;
- for (int i = 0; i <= Ngr - 1 ; i++)
- {
- Yg[i] = a[L] + b[L] * (Xg[i] - Xe[L]) + c[L] * Math.Pow(Xg[i] - Xe[L], 2) / 2 + d[L] * Math.Pow(Xg[i] - Xe[L], 2) * (Xg[i] - Xe[L]) / 6;
- Xg[i + 1] = Xg[i] + h;
- if (Xg[i] > Xe[L]) L++;
- }
- }
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