| Can such a program be written? |
| Yes |
|
100% |
[ 10 ] |
| No |
|
0% |
[ 0 ] |
|
| Total Votes : 10 |
|
Collatz Conjecture
[Pick a number ~ if it is odd, then [*3, +1]. If it is even, then [Divide by 2]. Keep doing this. You will eventually reach a 4-2-1 loop.]
Picture a chart with 0-100 on the bottom & 0-100 on the left side. The bottom row will be for the starting number. The left side will be for the highest number reached before dropping to the 4-2-1 Loop for that starting number. Can such a program be written?
Someone asked what language I wanted. I am not sure of the language, but I wuld like to program it into my TI-~83 Plus.
What makes you think such a program couldn't be written? I don't see anything crazy going on here.
Edit: I made the program both for the number of iterations and for the max value reached.
Code: #include <stdlib.h>
#include <graphx.h>
#include <tice.h>
int main(void)
{
int num, max, iterations = 0;
char title1[] = "Collatz Iterations";
char title2[] = "Collatz Max Value";
gfx_Begin();
//setup
gfx_SetDrawScreen();
gfx_ZeroScreen();
gfx_SetColor(255);
gfx_SetTextFGColor(255);
gfx_SetTextBGColor(5);
gfx_SetTextTransparentColor(5);
gfx_SetTextScale(2,2);
gfx_PrintStringXY(title1,(LCD_WIDTH - gfx_GetStringWidth(title1)) / 2, 3);
for (int i=1;i<LCD_WIDTH - 1;i++){
num = i;
iterations = 1;
do{
if(num % 2 == 0){
num = num / 2;
} else {
num = num * 3 + 1;
}
iterations++;
} while (num != 4);
gfx_FillRectangle_NoClip(i, LCD_HEIGHT - iterations, 2, 2);
}
while (!os_GetCSC());
gfx_ZeroScreen();
gfx_PrintStringXY(title2,(LCD_WIDTH - gfx_GetStringWidth(title2)) / 2, 3);
for (int i=1;i<LCD_WIDTH - 1;i++){
num = i;
max = 1;
do{
if(num % 2 == 0){
num = num / 2;
} else {
num = num * 3 + 1;
}
if (max < num){
max = num;
}
} while (num != 4);
gfx_FillRectangle_NoClip(i, LCD_HEIGHT - (max / 55) , 2, 2);
}
while (!os_GetCSC());
gfx_End();
return 0;
}
Download
Yes, that's a relatively easy program. Without actually making it show a table, you could throw it together like this:
Code: Input "STARTING NUMBER:",X
Repeat getKey
If remainder(X,2):Then
3X+1->X
Else
.5X->X
End
Disp X
End
You can stop this version by pressing any key. You could also store the three most recent values in a list and stop when the last three are [4, 2, 1].
Code: def callitz():
def cg(n):
yield n
while n != 1:
n = 3 * n + 1 if n % 2 else n // 2
yield n
for n in range(1, 101):
print(n, max(list(cg(n)))
You didn't specify a language, but the bottom line is that it is definitely possible (as womp said, why wouldn't it be?).
Also, 0 as a starting point is invalid as posed, since 0 never reaches the 4-2-1 loop. Note additionally that many values <100 reach a rather large maximum.
Edit: thought I'd golf this in TI-BASIC, cause that's what we do here after all.
Display code could be adjusted as desired; current form is 50 49 bytes.
Code: :For(I,1,|E2
:I->M
:While ln(Ans
:.5Ans
:If fPart(Ans:6Ans+1
:If Ans>M:Ans->M
:End
:Disp toString(I
:Pause M
:End
For what it's worth, :While Ans!=1 -> :While Ans-1.
KermMartian wrote:
For what it's worth, :While Ans!=1 -> :While Ans-1.
That doesn't offer any size benefits, correct? Both ≠ and - are single byte tokens, so is there any reason to use - besides being easier to type?
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