I am trying to print coordinate outputs of a program to a text file in order to use it in another program but I don't really know anything about GWBASIC and its my first time using MS-DOS. I need it to open a text file named plot.txt and print output there and save it without actually plotting on GWBASIC. Here is the program which I found in an old magazine.
810 REM MAKE A GLOBULAR
12 REM
14 R0=20: R2=R0*R0: R3=R2*R0
16 P1=3.14159265#
18 C0=P1*P1*R3/4
20 R1=R0/SQR(2)
22 XM=512: YM=512
24 X2=XM/2: Y2=YM/2: S=5
26 INPUT "HOW MANY STARS ";T
27 RANDOMIZE TIMER
28 CLS: REM CLEAR SCREEN
30 FOR I=1 TO T
32 C=C0*RND: R=R1
34 REM
36 REM NOW FIND R
38 FOR K=1 TO 5
40 GOSUB 100
42 R=R+(C-C1)/D
44 NEXT K
46 REM 3-DIMENSIONAL PLACE
48 X=RND-.5
50 Y=RND-.5
52 Z=RND-.5
54 S1=SQR(X*X+Y*Y+Z*Z)
56 IF S1>.5 THEN GOTO 48
58 REM POINT IS NOW IN SPHERE
60 R=R*S1: X=X*R: Y=Y*R: Z=Z*R
62 GOSUB 200
64 NEXT I
66 END
68 REM
100 REM NEWTON-RAPHSON ITERATION
105 A=R/R0
110 C1=ATN(A)*.5*R3
115 A=1+A*A
120 C1=C1+R*.5*R2/A
125 C1=P1*(C1-R*R2/(A*A))
130 D=4*P1*R*R/(A*A*A)
135 RETURN
140 REM
200 REM 2-DIMENSIONAL PLOT
203 SCREEN 9
205 X=X*S+X2: Y=Y*S+Y2
210 IF X<0 OR Y<0 THEN 225
215 IF X>=XM OR Y>=YM THEN 225
220 PSET(X,Y)
225 RETURN
230 REM ------------------------
240 REM APPEARED IN ASTRONOMICAL
250 REM COMPUTING, SKY & TELE-
260 REM SCOPE, APRIL, 1986
270 REM ------------------------
Here is a Python 3 paraphrase:
#globular.py
#Python paraphrase of model.bas from
#http://www.skyandtelescope.com/wp-content/uploads/model.bas
from math import pi, sqrt, atan
from random import uniform, random
#Global variables:
r0 = 20.0
r2 = r0**2
r3 = r0**3
c0 = pi**2*r3/4
r1 = r0/sqrt(2)
def NRI(c,r):
#Newton-Raphson Iteration
a = r/r0
c1 = atan(a)*0.5*r3
a = 1+a**2
c1 += r*0.5*r2/a
c1 = pi*(c1-r*r2/a**2)
d = 4*pi*r**2/a**3
return (c1,d)
def makeStars(t):
stars = []
for i in range(t):
c = c0*random()
r = r1
for k in range(5):
c1,d = NRI(c,r)
r += (c-c1)/d
while True:
x = uniform(-0.5,0.5)
y = uniform(-0.5,0.5)
z = uniform(-0.5,0.5)
s1 = sqrt(x**2 + y**2 + z**2)
if s1 <= 0.5: break
r *= s1
x *= r
y *= r
z *= r
stars.append((x,y,z))
return stars
def starsToFile(t,fname):
stars = makeStars(t)
f = open(fname,'w')
for star in stars:
print(*star, sep = ', ',file = f)
f.close()
I skipped the part about printing x and y and instead wrote a function makeStars to return a list of (x,y,z) tuples, as well as a related function which takes such an output and sends it to a text file. This last function is the only thing that used Python 3 instead of Python 2. If you are using Python 2 you can import Python 3's print function from the future.
Typing starsToFile(100,'stars.txt') in the Python shell gave me a text file which begins:
-0.32838465248713156, -0.3294895266926551, -1.2963580524762535
14.20224408569865, 1.4434961933043464, 6.450969593697097
1.6525937589658193, -0.24447292610082685, 1.0543647986350608
1.5707528567123823, 5.190972598268825, -2.0054790217091134
I don't have good 3-d scatter-plot graphing at my finger tips, but here is a screen shot of 50 points generated by the function and plotted using a computer algebra system called Derive:
Final remark: I wonder if there is a typo in the source code. The line
C0=P1*P1*R3/4
strikes me as suspicious since it is fairly rare in mathematics for pi to appear squared -- though it does happen. Maybe there should be only 1 factor of pi there (which would then have the effect of setting C0 proportional to the volume of the sphere of radius R0). On the other hand, I don't know exactly what is happening here, so I left it in. If the results seem problematic, you could maybe experiment with that line.
If you want a copy of the calculated coordinates simply add these lines:
1 OPEN "PLOT.TXT" FOR OUTPUT AS #1
65 CLOSE #1
221 PRINT #1, X + "," + Y
The program will work as before but in addition to this it outputs the coordinate to a file named plot.txt
Put them in an image with 640x350 size (that size is demanded by SCREEN 9) and you get the same result.
Related
The problem is how do we compute the integer value of floor(log2(5^x)) without floating point arithmetic or long integer computation? I'm looking for a simple, efficient and mathematically elegant way.
Observations:
The formula is just the number of bits in 5**x (plus 1)
Attempts:
I tried to simplify it to:
floor(x*log2(5))
In my use case, x is not extremely large, probably just 1-100. While an elegant formula that works for small values would suffice me, I would be interested in a formula/algorithm that works for any value of x
I'm making a reference software implementation of universal numbers (type III). I want to make everything easily convertible to microcode by purely using bitwise and basic operations. This is one of the formulas i need to simplify.
As you correctly note, log2(5**x) == x * log2(5). log2(5) is a constant, which can be approximated to 2.3219281.
However, floats aren't allowed per the question. Not an issue!
log2_5 = 23219281;
scale = 10000000; // note log2_5/scale is the actual value
result = x * log2_5;
output = (result - (result % scale)) / scale;
By reducing result by result % scale, dividing it by scale will be an integer division, not a float.
for a simple, efficient and mathematically elegant way... floor(x*log2(5))
Since x has integer values 1 to 100, various tests can to made to find the "best" that uses an integer multiply and a divide by power_of_2.
f(x) = x*a integer_divide power_of_2
For
f(x) = floor(x*log2(5)) = floor(x*some_float_c) the value of some_float_c is bounded by 100 minimum and maximums below.
x f(x) mn mx
f(x)/x (f(x) + 1)/x
1 2 2.00000 3.00000
2 4 2.00000 2.50000
3 6 2.00000 2.33333
...
59 136 2.30508 2.32203
...
87 202 2.32184 2.33333
...
98 227 2.31633 2.32653
99 229 2.31313 2.32323
100 232 2.32000 2.33000
The maximum min is 2.32184 and the minimum max is 2.32203, :
2.32184... <= some_float_c < 2.32203...
Since we cannot use float, find some_integer/some_power_of_2
2.32184... <= some_integer/some_power_of_2 < 2.32203...
ceil(some_power_of_2 * 2.32184...) <= some_integer < floor(some_power_of_2 * 2.32203...)
min max
2.32184 2.32203
2 5 4
4 10 9
8 19 18
...
1024 2378 2377
2048 4756 4755
4096 9511 9511 < first one where min <= max
8192 19021 19022
So 9511/4096 is the "simplest" and is a "best" candidate.
f(x) = (x*9511) integer_divide_by_power_of_2 4096
// In C
unsigned y = (x*9511u) >> 12;
Here is a very rough approximation, but it can help if you want to obtain it mentally
5^3 = 125
2^7 = 128
So for raising to the power of n:
5^n ~~ 2^(7n/3)
So 5^12 is near 2^28 might require up to 29 bits.
It's a bit overestimated because 2^7 > 5^3, so 28 bits are enough, a good usage is to simply round the fraction upper.
If I evaluate in Smalltalk:
(1 to: 50) reject: [:i | (5 raisedTo: i) highBit = (i*7/3) ceiling].
I get:
#(31 34 37 40 43 46 49)
You see that the very simple formulation works up to 5^30 which is not that bad.
I'm so lost on how to fix this, it should be so simple. I'm using a pic16F1526 and trying to scale the analog to digital reading from 0-255 to 50-100 roughly. I am using this equation
result = ((user_input + 200) * 200) / 800;
In code blocks and on my calculator it works at all numbers from 0-255 and it works perfectly whether I use 8 bit, 16 bit variables in code bloacks.
I've already verified that the AtoD input is working correctly sending the data to the UART. Even if I enter static numbers in place of the sample I get weird results.
When the acd reads a 255 or I enter a 255 the equation gives me a 31 in decimal instead of 100 like it's supposed to. The only thing I can think of is something is getting messed up in the way an 8 bit PIC does it's math since it's an a bit micro.
Sounds like you are getting the correct results on in codeblack because of integer promotion and getting the incorrect results in the hardware because of variable overflow.
uint8_t Can contain 0 to 255
int8_t Can contain -126 125
uint16_t Can contain 0 to 65635
...
Assuming you have uint16_t, the micro's math will go as follows:
((255 + 200) * 200) / 800
(455 * 200) / 800 : 455 * 200 Overflows the 16 bit variable!
( 25464 ) / 800: Note that 91000 & 0xFFFF == 25464
31
You can work around this issue by simplifying your equation :
(user_input + 200) / 4 is equivalent to ((user_input + 200) * 200) / 800 and will not overflow at 16 bits although your accuracy is not very high as ImaginaryHuman072889 pointed out.
If I understand your question correctly, you want to linearly map the numbers 0-255 to the numbers 50-100.
Back to good old y = mx + b algebra.
When x = 0, y = 50. Therefore:
y = mx + b
50 = m*0 + b
b = 50
When x = 255, y = 100. Therefore:
y = mx + 50
100 = m*255 +50
m*255 = 50
m = 50/255 = 10/51
Therefore, the precise answer is:
y = (10/51)*x + 50
On a side note, I have no idea how you got the result of 31 when plugging in 100 into your formula. See below.
(255+200)*200/800 = 113.75
I am trying to fit 3 numbers inside 1 number.But numbers will be only between 0 and 11.So their (base) is 12.For example i have 7,5,2 numbers.I come up with something like this:
Three numbers into One number :
7x12=84
84x5=420
420+2=422
Now getting back Three numbers from One number :
422 MOD 12 = 2 (the third number)
422 - 2 = 420
420 / 12 = 35
And i understanded that 35 is multiplication of first and the second number (i.e 7 and 5)
And now i cant get that 7 and 5 anyone knows how could i ???
(I started typing this answer before the other one got posted, but this one is more specific to Arduino then the other one, so I'm leaving it)
The code
You can use bit shifting to get multiple small numbers into one big number, in code it would look like this:
int a, b, c;
//putting then together
int big = (a << 8) + (b << 4) + c;
//separating them again
a = (big >> 8) & 15;
b = (big >> 4) & 15;
c = big & 15;
This code only works when a, b and c are all in the range [0, 15] witch appears to be enough for you case.
How it works
The >> and << operators are the bitshift operators, in short a << n shifts every bit in a by n places to the left, this is equivalent to multiplying by 2^n. Similarly, a >> n shifts to to the right. An example:
11 << 3 == 120 //0000 1011 -> 0101 1000
The & operator performs a bitwise and on the two operands:
6 & 5 == 4 // 0110
// & 0101
//-> 0100
These two operators are combined to "pack" and "unpack" the three numbers. For the packing every small number is shifted a bit to the left and they are all added together. This is how the bits of big now look (there are 16 of them because ints in Arduino are 16 bits wide):
0000aaaabbbbcccc
When unpacking, the bits are shifted to the right again, and they are bitwise anded together with 15 to filter out any excess bits. This is what that last operation looks like to get b out again:
00000000aaaabbbb //big shifted 4 bits to the right
& 0000000000001111 //anded together with 15
-> 000000000000bbbb //gives the original number b
All is working exactly like in base 10 (or 16). Here after your corrected example.
Three numbers into One number :
7x12^2=1008
5*12^1=60
2*12^0=2
1008+60+2=1070
Now getting back Three numbers from One number :
1070 MOD 12 = 2 (the third number)
1070/12 = 89 (integer division) => 89 MOD 12 = 5
89 / 12 = 7
Note also that the maximum value will be 11*12*12+11*12+11=1727.
If this is really programming related, you will be using 16bits instead of 3*8 bits so sparing one byte. An easyer method not using base 12 would be fit each number into half a byte (better code efficiency and same transmission length):
7<<(4+4) + 5<<4 + 2 = 1874
1874 & 0x000F = 2
1874>>4 & 0x000F = 5
1874>>8 & 0x0F = 7
Because MOD(12) and division by 12 is much less efficient than working with powers of 2
you can use the principle of the positional notation to change from one or the other in any base
Treat yours numbers (n0,n1,...,nm) as a digit of a big number in the base B of your choosing so the new number is
N = n0*B^0 + n1*B^1 + ... + nm*B^m
to revert the process is also simple, while your number is greater than 0 find its modulo in respect to the base to get to get the first digit, then subtracts that digit and divide for the base, repeat until finish while saving each digit along the way
digit_list = []
while N > 0 do:
d = N mod B
N = (N - d) / B
digit_list.append( d )
then if N is N = n0*B^0 + n1*B^1 + ... + nm*B^m doing N mod B give you n0, then subtract it leaving you with n1*B^1 + ... + nm*B^m and divide by B to reduce the exponents of all B and that is the new N, N = n1*B^0 + ... + nm*B^(m-1) repetition of that give you all the digit you start with
here is a working example in python
def compact_num( num_list, base=12 ):
return sum( n*pow(base,i) for i,n in enumerate(num_list) )
def decompact_num( n, base=12):
if n==0:
return [0]
result = []
while n:
n,d = divmod(n,base)
result.append(d)
return result
example
>>> compact_num([2,5,7])
1070
>>> decompact_num(1070)
[2, 5, 7]
>>> compact_num([10,2],16)
42
>>> decompact_num(42,16)
[10, 2]
>>>
e.g. given the following number modulo 55
74627282173621618272362 % 55 = 47
why does splitting the number; calculate first part modulo 55; add result in front of the second part and use modulo 55 again; yield the same result again?
using the example above:
746272821736 % 55 = 46
'46' + '21618272362' = 4621618272362
4621618272362 % 55 = 47
same result if you calculate the number digit by digit using the way described above
7 % 55 = 7
'7' + '4' = 74 % 55 = 19
'19' + '6' = 196 % 55 = 31
'31' + '2' = 312 % 55 = 37
....
result = 47
could someone clarify WHY?
It comes so because the property which holds here is basic and is
Dividend = Divisor * Quotient + Remainder
This is all because of long-division method.
I am working out a question for you.
Ex :- 123456789 % 4
Here,
55 ) 123456789 ( 22...
-110_______
13456789 // here, you'd have replaced it as 13456789 which indeed comes from long-division
-110_____
2456789 // here, you'd have replaced it as 2456789 which indeed comes from long-division
It is so because the remainders are itself put down in the
long-division method exactly below the number so that the number gets
reduced and the number in next stage is substituted by remainder left
appended next by the rest-undivided number.
What you're quoting is none-different from this case.
So,you see that once you find a number which yields remainder 0 in
between, you can drop the digits upto which your remainder is 0.
And,then fresh start with the further digits assuming it as the given
number. This is surely a formula-type thing.
But,your hypothesis was correct and is what we use in the long-division method of formulating the remainders/modulo!
I'd like to draw a shape which has holes in it such that I can fill the shape it and not have the holes filled with that colour (leave them transparent).
According to the W3 path spec:
Compound paths (i.e., a path with multiple subpaths) are possible to allow effects such as "donut holes" in objects.
Can somebody please give a very simple example of how to perform this with a vector path in Raphael?
Thanks very much.
This turns out to be quite straightforward if you know the trick. For example this doesn't work:
paper.path("M 50 50 L 50 150 L 150 150 L 150 50 z" +
" M 75 75 L 75 125 L 125 125 L 125 75 z")
.attr("fill", "#f00");
But this does work*:
paper.path("M 50 50 L 50 150 L 150 150 L 150 50 z" +
" M 75 75 L 125 75 L 125 125 L 75 125 z")
.attr("fill", "#f00");
The difference is that for the donut to appear the the inner path has to have it's vertices drawn in reverse order to the outer path (ie. draw one clockwise, the other anti-clockwise). A tidbit I found on the text.xml.svg.devel archives.
(*) At least, it works in Chrome, Opera and Firefox 4.0 beta, but not in 3.6
To make this work in Firefox 3.6, you need to close the hole; i.e. make the coordinates join back to themselves when defining the inner boundary. Curiously, this doesn't appear necessary for the outer boundary.
paper.path("M 50 50 L 50 150 L 150 150 L 150 50 z" +
" M 75 75 L 125 75 L 125 125 L 75 125 L 75 75 z")
.attr("fill", "#f00");
Just a quick note to follow up on the comment - the clockwise/counter-clockwise concept might seem strange at first, but it's pretty standard throughout GIS / CAD technologies.
I think the correct way to do this is setting the attribute "fill-rule" to the value "evenodd". Take a look at the svg spec:
Don't try to set it with "Raphael.Element.attr()". It doesn't work. I use the jQuery.attr() function instead:
// assuming paper is a Raphael.Paper object
path = paper.path('Mx,y{some path commands for the main shape}Z'
+'Mx,y{some path commands for the hole}Z'
);
// this doesn't work
path.attr({'fill-rule': 'evenodd'});
// neither this
path.attr({fillRule: 'evenodd'});
// if you inspect the object returned by paper.path
// you can see it has a reference to the DOM element
console.debug(path)
// so a bit of jQuery and it's done
$(path[0]).attr('fill-rule', 'evenodd');
I have used this on complex paths with successful results.
For anyone looking to do circular donuts, great easy plugin Raphael-donut-plugin
Gist:
Raphael.fn.donut = function(x, y, innerRadius, outerRadius) {
y -= outerRadius;
return this.path('M'+x+' '+y+'a'+outerRadius+' '+outerRadius +
' 0 1 0 1 0m-1 '+
(outerRadius - innerRadius)+
'a'+innerRadius+' '+innerRadius+
' 0 1 1 -1 0');
};