Thursday, May 18, 2017
Playing with Memo in Delphi
I change the font in memo (tMemo) into courier. With this change, it's easy to do string manipulation with old Pascal style.
Below, I create small program with read input from edit into n variable. The input can only contain number 1 to 9.
The output is displayed on Memo. It's just number 123456789 (yeah, it has type string, but its number).
It's that all? No. The number's forming a 'cross' centered on number specified in edit. :)
Palindrom Checker
This simple program is using an edit as input, a button to trigger processing and a memo for output.
I used edit.text as s value and then reverse its value using for command and saved to rs variable.
We compared s and rs to determine that s is palindrom or not.
Here's the code
unit Unit1; interface uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, StdCtrls; type TForm1 = class(TForm) Edit1: TEdit; Memo1: TMemo; Button1: TButton; procedure proses; procedure FormCreate(Sender: TObject); procedure Button1Click(Sender: TObject); private { Private declarations } public { Public declarations } end; var Form1: TForm1; implementation {$R *.dfm} procedure tform1.proses; var s,rs:string; i:integer; palindrom:boolean; begin memo1.Text:=''; palindrom:=true; s:=edit1.Text; for i:=length(s) downto 1 do begin rs:=rs+s[i]; end; for i:=1 to length(s) do begin if s[i]<>rs[i] then begin palindrom:=false; break; end; end; memo1.Lines.Append(s); memo1.Lines.Append(rs); if palindrom=true then memo1.Lines.Append('is palindrom') else memo1.Lines.Append('is not palindrom'); end; procedure TForm1.FormCreate(Sender: TObject); begin memo1.Text:=''; end; procedure TForm1.Button1Click(Sender: TObject); begin proses; end; end.
Tuesday, May 16, 2017
Walking Star in Delphi's Stringgrid.
This code moves the star from cell to cell on string grid.
I use variable s, an array of string type variable.
For delay, or controlling the speed, I use application.processmessages and sleep() combo command.
This code fills cell with blank (space) value that corresponds with s, except one cell. This one cell then "moves" to the right, into the cell next to it.
For this purpose I declare two integer type variable, sx and sy. This variable add itself by one every step. Based on this two variable, the cell that should be filled with star is decided.
I use variable s, an array of string type variable.
For delay, or controlling the speed, I use application.processmessages and sleep() combo command.
This code fills cell with blank (space) value that corresponds with s, except one cell. This one cell then "moves" to the right, into the cell next to it.
For this purpose I declare two integer type variable, sx and sy. This variable add itself by one every step. Based on this two variable, the cell that should be filled with star is decided.
Monday, May 15, 2017
Digit Word.
A digit word is a word where, after possibly removing some letters, you are left with one of the single digits:
ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT or NINE.
For example:
• BOUNCE and ANNOUNCE are digit words, since they contain the digit ONE.
• ENCODE is not a digit word, even though it contains an O, N and E, since they are not in order.
Here's my code on Delphi
.
ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT or NINE.
For example:
• BOUNCE and ANNOUNCE are digit words, since they contain the digit ONE.
• ENCODE is not a digit word, even though it contains an O, N and E, since they are not in order.
Here's my code on Delphi
unit Unit1; interface uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, StdCtrls; type TForm1 = class(TForm) Edit1: TEdit; Button1: TButton; Memo1: TMemo; procedure proses; procedure FormCreate(Sender: TObject); procedure Button1Click(Sender: TObject); private { Private declarations } public { Public declarations } end; var Form1: TForm1; digit,cdigit:array[1..9] of string; implementation {$R *.dfm} procedure TForm1.FormCreate(Sender: TObject); begin memo1.Text:=''; digit[1]:='one'; digit[2]:='two'; digit[3]:='three'; digit[4]:='four'; digit[5]:='five'; digit[6]:='six'; digit[7]:='seven'; digit[8]:='eight'; digit[9]:='nine'; end; procedure tform1.proses; var s:string; i,j,k,n:integer; c:array[1..9]of integer; ck:array[1..9]of boolean; begin memo1.Text:=''; s:=edit1.Text; memo1.Lines.Append(s); memo1.Lines.Append(''); n:=length(s); for i:=1 to 9 do begin cdigit[i]:=''; c[i]:=1; ck[i]:=true; end; //looking for char for i:=1 to 9 do begin for j:=1 to length(digit[i]) do begin if ck[i]=true then begin ck[i]:=false; for k:=c[i] to n do begin if s[k]=digit[i][j] then begin ck[i]:=true; cdigit[i]:=cdigit[i]+s[k]; c[i]:=c[i]+1; break; end; end; end; end; end; //compare for i:=1 to 9 do begin memo1.Lines.Append(cdigit[i]); end; end; procedure TForm1.Button1Click(Sender: TObject); begin proses; end; end.
Wednesday, May 10, 2017
Animation using Matplotlib
Suppossed we want to animate our plot, say f(x) = (x-c)^(2) to see the effect of various c value, we could do it in Python using Matplotlib module.
As we could see at the code below that the animation part is in
As we could see at the code below that the animation part is in
ani = animation.FuncAnimation(fig, animate, np.arange(-10,10), interval = 25, blit=False)
What about our own def? We could call it inside animate and use variable i (defined in ani, the np.arange(-10,10) part) to whatever treatment on our self define function f(x). In this case, I use i as c parameter value. I like the result, :)
Sunday, May 7, 2017
What About Unbounded End?
Yeah, what about it? The previous code have the both end bounded.
If we want a free/unbound end, we could set the condition at the with this properties (or we could choose whatever we like)
dy/dx=0
So we will have
y[1]-y[0]=0
y[0] = y[1]
if we want both free ends, we could set the other end as well
y[n] = y[n-1]
So, we just have to modify the original just a bit.
Beware though, with both ends free, we could lost the strings, :)
Saturday, May 6, 2017
Waves Equation Animation in Python
I use matplotlib module to do the animation.
The main code is in def waves(y0,y1,cb) that use finite difference that solved initial value problem and boundary value problem simultaneously.
.
The main code is in def waves(y0,y1,cb) that use finite difference that solved initial value problem and boundary value problem simultaneously.
code from pylab import * import matplotlib.animation as animation fig,ax = subplots() def waves(y0,y1,cb): y2 = y0 for i in range(1,len(y0)-1): y2[i] = 2*y1[i]-y0[i]+cb*(y1[i+1]-2*y1[i]+y1[i-1]) return y2 x = linspace(0.,1.,20) dx = 1./(len(x)) y0 = sin(2*pi*x) vy0 = 12. b = 1./32. #dt2/dx2 dt = sqrt(b*dx*dx) print dt c = 1. cb = c*b y1 = y0 + vy0*dt print y0 print y1 line, = ax.plot(x,y0) def animate(i): global y0,y1,cb y2 = waves(y0,y1,cb) y0 = y1 y1 = y2 line.set_ydata(y0) return line, #plot (x,y0) ani = animation.FuncAnimation(fig, animate, np.arange(1,200), interval = 25, blit=False) grid(True) ylim(-10,10) show()
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