Convolution Layer: Calculate output dimensions based on input and kernel dimensions - deep-learning

I am confused by the very last figure (figure 20) here:
https://towardsdatascience.com/a-comprehensible-explanation-of-the-dimensions-in-cnns-841dba49df5e
What I understood is that the output dimension of a convolution layer is calculated by
[(INPUT - KERNEL + 2xPADDING) / STRIDE] + 1
So in the first example where the input is 1 x 28 x 28, the kernel is 5 x 5 and valid-padding (which essentially means no padding at all as far as I know), stride is 1.
This would mean: [(28 - 5 + 0) / 1] + 1 = 24, so essentially 1 x 24 x 24 but instead the output is 16 x 13 x 13 - how can this be?

Related

idl radius of each pixel

I'm completely new to coding and have almost no idea what I'm doing. Currently I'm trying to find the radius of each pixel from the center of a galaxy in a fits file. I was told to try this by creating a blank array the same size as the fits file and I'm trying to use a for loop for each x value and another for each y. So far, I have the array and have attempted to create the for loops.
xcenter =249.8
ycenter =250.0
d=fltarr(500,500)
for i=0,499 do begin
d=d(i-xcenter,*)
endfor
for j=0,499 do begin
d=d(j-ycenter,*)
endfor
I know this look awful and I obviously have no idea what I'm doing. So if anyone can offer any help at all I'd be grateful.
Let's look at a simpler version first. Suppose you have 10 points on a line, and you want to calculate the distance of each point from some xref. IDL supports vector math, so you can do this:
xref = 5.4
x = dindgen(10)
distance = abs(x - xref)
I have used the DINDGEN command here, and you can look up the help. Now in a 2d case, you want 2-d arrays: one will be a 500 * 500 array containing the X coordinate of each pixel, the other containing the Y coordinate. So the arrays will have to be of the form,
0 1 2 3 ...
0 1 2 3 ...
0 1 2 3 ...
and
0 0 0 0 ...
1 1 1 1 ...
2 2 2 2 ...
We can generate them using the # operator. Note that IDL counts from 0 by default.
just_one = replicate(1d, 500) ; contains 1 1 1 1 ...
one_500 = dindgen(500) ; contains 0.0 1.0 2.0 ...
x = just_one # one_500
y = one_500 # just_one
Now you can calculate the distance as usual, d = sqrt(xx + yy), but using vector math again:
distance = sqrt( (x - xref) ^ 2 + (y - yref) ^ 2 )
This is a 500x500 array, which contains the distance of each pixel from your xref, yref points.

How to graph probability density function when given density function

I want to graph the pdf and verify the total area under the density curve is 1. I am given that
f(x) = .075x+.2 when 3 ≤ x ≤ 5 and f(x)=0 otherwise
Can you help me figure out what the code for this question is?
You can plot this function in Mathematica using:
Plot[Piecewise[{{0.075 x + 2, 3 <= x && x <= 5}}, 0], {x, 0, 5}]
And then verify area using definite integral (from 0 to infinity):
Integrate[Piecewise[{{0.075 x + 0.2, 3 <= x && x <= 5}}, 0], {x, 0, Infinity}]
(which results in 1).
You can of course calculate it by hand.
Integral represents area under the curve. When x is not in range [3,5] then area is always zero.
So your problem can be reduced to calculating integral from 0.075 x + 0.2, between 3 and 5.
Integral from (0.075 x + 0.2)dx equals to 0.2 x + 0.0375 x^2 + C. When calculated in range, equals to 1.9375 - 0.9375 = 1.

Converting decimal floating number to binary

Can anyone please help me convert number (for example) 143,625 to binary? I've been searching through net for quite a long time but didn't find anything with good explanation.
Thanks in advance!
The integer part can be done by dividing by 2 repeatedly and keeping track of the remainder:
143 / 2 = 71 remainder 1
71 / 2 = 35 remainder 1
35 / 2 = 17 remainder 1
17 / 2 = 8 remainder 1
8 / 2 = 4 remainder 0
4 / 2 = 2 remainder 0
2 / 2 = 1 remainder 0
1 / 2 = 0 remainder 1
So the integer part is 10001111
For the fractional part, multiply it by 2 repeatedly and look at the integer part of the result:
.625 x 2 = 1.25 - we need the integer part, which is 1 - so far we have 0.1, we then disregard the 1 and look at 0.25
0.25 x 2 = 0.5 - so far we have 0.10, we look at 0.5
0.5 x 2 = 1.0 - we have 0.101, no decimal part so we're good.
The whole number is the integer part + the decimal part, so 10001111.101

Add 25 & 30 as binary number

Using 8 bit registers and signed magnitude representation.
I thought 25 in BCD is 010 0101 but my text book says it as 001 1001. Can somebody explain?
25 / 2 = 12r1 (12 with a remainder of 1)
12 / 2 = 6r0 (6 with a remainder of 0)
6 / 2 = 3r0 (3 with a remainder of 0)
3 / 2 = 1r1 (1 with a remainder of 0)
1 / 2 = 0r1 (0 with a remainder of 0)
So 11001 (working backward up the tree) is the binary equivalent to 25.
Another way to think about it is with powers of 2:
(1*16) + (1*8) + (0*4) + (0*2) + (1*1) = 25
And it's worth noting, just as in base 10, leading zeros do not change the value of a number. (00025 == 25) (0011001 == 11001).
The leading zeros are there in your case because your needing to populate an 8 bit register (there needs to be 8 binary digits regardless of their value).

Code Golf: Calculate Orthodox Easter date

Locked. This question and its answers are locked because the question is off-topic but has historical significance. It is not currently accepting new answers or interactions.
The Challenge
Calculate the Date of the Greek Orthodox Easter (http://www.timeanddate.com/holidays/us/orthodox-easter-day) Sunday in a given Year (1900-2100) using the least amount of characters.
Input is just a year in the form '2010'. It's not relevant where you get it (Input, CommandLineArgs etc.) but it must be dynamic!
Output should be in the form day-month-year (say dd/mm/yyyy or d/m/yyyy)
Restrictions No standard functions, such as Mathematica's EasterSundayGreekOrthodox or PHP's easter_date(), which return the (not applicable gregorian) date automatic must be used!
Examples
2005 returns 1/5/2005
2006 returns 23/4/2006
2007 returns 8/4/2007
2008 returns 27/4/2008
2009 returns 19/4/2009
2010 returns 4/4/2010
2011 returns 24/4/2011
2012 returns 15/4/2012
2013 returns 5/5/2013
2014 returns 20/4/2014
2015 returns 12/4/2015
Code count includes input/output (i.e full program).
Edit:
I mean the Eastern Easter Date.
Reference: http://en.wikipedia.org/wiki/Computus
Python (101 140 132 115 chars)
y=input()
d=(y%19*19+15)%30
e=(y%4*2+y%7*4-d+34)%7+d+127
m=e/31
a=e%31+1+(m>4)
if a>30:a,m=1,5
print a,'/',m,'/',y
This one uses the Meeus Julian algorithm but since this one only works between 1900 and 2099, an implementation using Anonymous Gregorian algorithm is coming right up.
Edit: Now 2005 is properly handled. Thanks to Mark for pointing it out.
Edit 2: Better handling of some years, thanks for all the input!
Edit 3: Should work for all years in range. (Sorry for hijacking it Juan.)
PHP CLI, no easter_date(), 125 characters
Valid for dates from 13 March 1900 to 13 March 2100, now works for Easters that fall in May
Code:
<?=date("d/m/Y",mktime(0,0,0,floor(($b=($a=(19*(($y=$argv[1])%19)+15)%30)+(2*($y%4)+4*$y%7-$a+34)%7+114)/31),($b%31)+14,$y));
Invocation:
$ php codegolf.php 2010
$ php codegolf.php 2005
Output:
04/04/2010
01/05/2005
With whitespace:
<?=date("d/m/Y", mktime(0, 0, 0, floor(($b = ($a = (19 * (($y = $argv[1]) % 19) + 15) % 30) + (2 * ($y % 4) + 4 * $y % 7 - $a + 34) % 7 + 114) / 31), ($b % 31) + 14, $y));
This iteration is no longer readable thanks to PHP's handling of assignments. It's almost a functional language!
For completeness, here's the previous, 127 character solution that does not rely on short tags:
Code:
echo date("d/m/Y",mktime(0,0,0,floor(($b=($a=(19*(($y=$argv[1])%19)+15)%30)+(2*($y%4)+4*$y%7-$a+34)%7+114)/31),($b%31)+14,$y));
Invocation:
$ php -r 'echo date("d/m/Y",mktime(0,0,0,floor(($b=($a=(19*(($y=$argv[1])%19)+15)%30)+(2*($y%4)+4*$y%7-$a+34)%7+114)/31),($b%31)+14,$y));' 2010
$ php -r 'echo date("d/m/Y",mktime(0,0,0,floor(($b=($a=(19*(($y=$argv[1])%19)+15)%30)+(2*($y%4)+4*$y%7-$a+34)%7+114)/31),($b%31)+14,$y));' 2005
C#, 155 157 182 209 212 characters
class P{static void Main(string[]i){int y=int.Parse(i[0]),c=(y%19*19+15)%30,d=c+(y%4*2+y%7*4-c+34)%7+128;System.Console.Write(d%31+d/155+"/"+d/31+"/"+y);}}
Python 2.3, 97 characters
y=int(input())
c=(y%19*19+15)%30
d=c+(y%4*2+y%7*4-c+34)%7+128
print"%d/%d/%d"%(d%31+d/155,d/31,y)
This also uses the Meeus Julian algorithm (and should work for dates in May).
removed no longer necessary check for modern years and zero-padding in output
don't expect Easters in March anymore because there are none between 1800-2100
included Python 2.3 version (shortest so far)
Mathematica
<<Calendar`;a=Print[#3,"/",#2,"/",#]&##EasterSundayGreekOrthodox##&
Invoke with
a[2010]
Output
4/4/2010
Me too: I don't see the point in not using built-in functions.
Java - 252 196 190 chars
Update 1: The first algo was for Western Gregorian Easter. Fixed to Eastern Julian Easter now. Saved 56 chars :)
Update 2: Zero padding seem to not be required. Saved 4 chars.
class E{public static void main(String[]a){long y=new Long(a[0]),b=(y%19*19+15)%30,c=b+(y%4*2+y%7*4-b+34)%7+(y>1899&y<2100?128:115),m=c/31;System.out.printf("%d/%d/%d",c%31+(m<5?0:1),m,y);}}
With newlines
class E{
public static void main(String[]a){
long y=new Long(a[0]),
b=(y%19*19+15)%30,
c=b+(y%4*2+y%7*4-b+34)%7+(y>1899&y<2100?128:115),
m=c/31;
System.out.printf("%d/%d/%d",c%31+(m<5?0:1),m,y);
}
}
JavaScript (196 characters)
Using the Meeus Julian algorithm. This implementation assumes that a valid four-digit year was given.
y=~~prompt();d=(19*(y%19)+15)%30;x=d+(2*(y%4)+4*(y%7)-d+34)%7+114;m=~~(x/31);d=x%31+1;if(y>1899&&y<2100){d+=13;if(m==3&&d>31){d-=31;m++}if(m==4&&d>30){d-=30;m++}}alert((d<10?"0"+d:d)+"/0"+m+"/"+y)
Delphi 377 335 317 characters
Single line:
var y,c,n,i,j,m:integer;begin Val(ParamStr(1),y,n);c:=y div 100;n:=y-19*(y div 19);i:=c-c div 4-(c-((c-17)div 25))div 3+19*n+15;i:=i-30*(i div 30);i:=i-(i div 28 )*(1-(i div 28)*(29 div(i+1))*((21 -n)div 11));j:=y+y div 4 +i+2-c+c div 4;j:=j-7*(j div 7);m:=3+(i-j+40 )div 44;Write(i-j+28-31*(m div 4),'/',m,'/',y)end.
Formatted:
var
y,c,n,i,j,m:integer;
begin
Val(ParamStr(1),y,n);
c:=y div 100;
n:=y-19*(y div 19);
i:=c-c div 4-(c-((c-17)div 25))div 3+19*n+15;
i:=i-30*(i div 30);
i:=i-(i div 28 )*(1-(i div 28)*(29 div(i+1))*((21 -n)div 11));
j:=y+y div 4 +i+2-c+c div 4;j:=j-7*(j div 7);
m:=3+(i-j+40 )div 44;
Write(i-j+28-31*(m div 4),'/',m,'/',y)
end.
Tcl
Eastern Easter
(116 chars)
puts [expr 1+[incr d [expr ([set y $argv]%4*2+$y%7*4-[
set d [expr ($y%19*19+15)%30]]+34)%7+123]]%30]/[expr $d/30]/$y
Uses the Meeus algorithm. Takes the year as a command line argument, produces Eastern easter. Could be a one-liner, but it's slightly more readable when split...
Western Easter
(220 chars before splitting over lines)
interp alias {} tcl::mathfunc::s {} set;puts [expr [incr 3 [expr {
s(2,(s(4,$argv)%100/4*2-s(3,(19*s(0,$4%19)+s(1,$4/100)-$1/4-($1-($1+8)/25+46)
/3)%30)+$1%4*2-$4%4+4)%7)-($0+11*$3+22*$2)/451*7+114}]]%31+1]/[expr $3/31]/$4
Uses the Anonymous algorithm.
COBOL, 1262 chars
WORKING-STORAGE SECTION.
01 V-YEAR PIC S9(04) VALUE 2010.
01 V-DAY PIC S9(02) VALUE ZERO.
01 V-EASTERDAY PIC S9(04) VALUE ZERO.
01 V-CENTURY PIC S9(02) VALUE ZERO.
01 V-GOLDEN PIC S9(04) VALUE ZERO.
01 V-GREGORIAN PIC S9(04) VALUE ZERO.
01 V-CLAVIAN PIC S9(04) VALUE ZERO.
01 V-FACTOR PIC S9(06) VALUE ZERO.
01 V-EPACT PIC S9(06) VALUE ZERO.
PROCEDURE DIVISION
XX-CALCULATE EASTERDAY.
COMPUTE V-CENTURY = (V-YEAR / 100) + 1
COMPUTE V-GOLDEN= FUNCTION MOD(V-YEAR, 19) + 1
COMPUTE V-GREGORIAN = (V-CENTURY * 3) / 4 - 12
COMPUTE V-CLAVIAN
= (V-CENTURY * 8 + 5) / 25 - 5 - V-GREGORIAN
COMPUTE V-FACTOR
= (V-YEAR * 5) / 4 - V-GREGORIAN - 10
COMPUTE V-EPACT
= FUNCTION MOD((V-GOLDEN * 11 + 20 + V-CLAVIAN), 30)
IF V-EPACT = 24
ADD 1 TO V-EPACT
ELSE
IF V-EPACT = 25
IF V-GOLDEN > 11
ADD 1 TO V-EPACT
END-IF
END-IF
END-IF
COMPUTE V-DAY = 44 - V-EPACT
IF V-DAY < 21
ADD 30 TO V-DAY
END-IF
COMPUTE V-DAY
= V-DAY + 7 - (FUNCTION MOD((V-DAY + V-FACTOR), 7))
IF V-DAY <= 31
ADD 300 TO V-DAY GIVING V-EASTERDAY
ELSE
SUBTRACT 31 FROM V-DAY
ADD 400 TO V-DAY GIVING V-EASTERDAY
END-IF
.
XX-EXIT.
EXIT.
Note: Not mine, but I like it
EDIT: I added a char count with spaces but I don't know how spacing works in COBOL so I didn't change anything from original. ~vlad003
UPDATE: I've found where the OP got this code: http://www.tek-tips.com/viewthread.cfm?qid=31746&page=112. I'm just putting this here because the author deserves it. ~vlad003
C, 128 121 98 characters
Back to Meeus' algorithm. Computing the day in Julian, but adjusting for Gregorian (this still seems naive to me, but I cannot find a shorter alternative).
main(y,v){int d=(y%19*19+15)%30;d+=(y%4*2+y%7*4-d+34)%7+128;printf("%d/%d/%d",d%31+d/155,d/31,y);}
I have not found a case where floor(d/31) would actually be needed. Also, to account for dates in May, the m in Meeus' algorithm must be at least 5, therefore the DoM is greater than 154, hence the division.
The year is supplied as the number of program invocation arguments plus one, ie. for 1996 you must provide 1995 arguments. The range of ARG_MAX on modern systems is more than enough for this.
PS. I see Gabe has come to the same implementation in Python 2.3, surpassing me by one character. Aw. :(
PPS. Anybody looking at a tabular method for 1800-2099?
Edit - Shortened Gabe's answer to 88 characters:
y=input()
d=(y%19*19+15)%30
d+=(y%4*2+y%7*4-d+34)%7+128
print"%d/%d/%d"%(d%31+d/155,d/31,y)
BASIC, 973 chars
Sub EasterDate (d, m, y)
Dim FirstDig, Remain19, temp 'intermediate results
Dim tA, tB, tC, tD, tE 'table A to E results
FirstDig = y \ 100 'first 2 digits of year
Remain19 = y Mod 19 'remainder of year / 19
' calculate PFM date
temp = (FirstDig - 15) \ 2 + 202 - 11 * Remain19
Select Case FirstDig
Case 21, 24, 25, 27 To 32, 34, 35, 38
temp = temp - 1
Case 33, 36, 37, 39, 40
temp = temp - 2
End Select
temp = temp Mod 30
tA = temp + 21
If temp = 29 Then tA = tA - 1
If (temp = 28 And Remain19 > 10) Then tA = tA - 1
'find the next Sunday
tB = (tA - 19) Mod 7
tC = (40 - FirstDig) Mod 4
If tC = 3 Then tC = tC + 1
If tC > 1 Then tC = tC + 1
temp = y Mod 100
tD = (temp + temp \ 4) Mod 7
tE = ((20 - tB - tC - tD) Mod 7) + 1
d = tA + tE
'return the date
If d > 31 Then
d = d - 31
m = 4
Else
m = 3
End If
End Sub
Credit: Astronomical Society of South Australia
EDIT: I added a char count but I think many spaces could be removed; I don't know BASIC so I didn't make any changes to the code. ~vlad003
I'm not going to implement it, but I'd like to see one where the code e-mails the Pope, scans any answer that comes back for a date, and returns that.
Admittedly, the calling process may be blocked for a while.
Javascript 125 characters
This will handle years 1900 - 2199. Some of the other implementations cannot handle the year 2100 correctly.
y=prompt();k=(y%19*19+15)%30;e=(y%4*2+y%7*4-k+34)%7+k+127;m=~~(e/31);d=e%31+m-4+(y>2099);alert((d+=d<30||++m-34)+"/"+m+"/"+y)
Ungolfed..ish
// get the year to check.
y=prompt();
// do something crazy.
k=(y%19*19+15)%30;
// do some more crazy...
e=(y%4*2+y%7*4-k+34)%7+k+127;
// estimate the month. p.s. The "~~" is like Math.floor
m=~~(e/31);
// e % 31 => get the day
d=e%31;
if(m>4){
d += 1;
}
if(y > 2099){
d += 1;
}
// if d is less than 30 days add 1
if(d<30){
d += 1;
}
// otherwise, change month to May
// and adjusts the days to match up with May.
// e.g., 32nd of April is 2nd of May
else{
m += 1;
d = m - 34 + d;
}
// alert the result!
alert(d + "/" + m + "/" + y);
A fix for dates up to 2399.
I'm sure there is a way to algorithmically calculate dates beyond this but I don't want to figure it out.
y=prompt();k=(y%19*19+15)%30;e=(y%4*2+y%7*4-k+34)%7+k+127;m=~~(e/31);d=e%31+m-4+(y<2200?0:~~((y-2000)/100));alert((d+=d<30||++m-34)+"/"+m+"/"+y)
'VB .Net implementation of:
'http://aa.usno.navy.mil/faq/docs/easter.php
Dim y As Integer = 2010
Dim c, d, i, j, k, l, m, n As Integer
c = y \ 100
n = y - 19 * (y \ 19)
k = (c - 17) \ 25
i = c - c \ 4 - (c - k) \ 3 + 19 * n + 15
i = i - 30 * (i \ 30)
i = i - (i \ 28) * (1 - (i \ 28) * (29 \ (i + 1)) * ((21 - n) \ 11))
j = y + y \ 4 + i + 2 - c + c \ 4
j = j - 7 * (j \ 7)
l = i - j
m = 3 + (l + 40) \ 44
d = l + 28 - 31 * (m \ 4)
Easter = DateSerial(y, m, d)