lambda function and sets - function

I need to remove all elements from set that satisfy the lambda function, is there a way to make it continue when the fisrt part of the statement returns error?
f = lambda x : x < "t" or x % 2 == 0
s = {"a", 2, "u", 4, 3}
def sort(fun, s):
l = []
for element in s:
try:
if (fun(element)) == True:
l.append(element)
except:
pass
for element in l:
s.remove(element)
my code returns:
s = {2, 3, 4, 'u'}
but I need it to return:
s = { 3, 'u'}


Maybe you can adjust a lambda function to check the type of the element.
Also, to remove values from set use set-comprehension:
f = lambda x: (isinstance(x, str) and x < "t") or (
isinstance(x, int) and x % 2 == 0
)
s = {"a", 2, "u", 4, 3}
s = {v for v in s if not f(v)}
print(s)
Prints:
{'u', 3}

Related

argument after * must be an iterable, not int

I have a small script that takes in a list and a value that represents what size sublists to break the list into:
def chunk(alist, n):
i = 0
j = n
while j < (len(alist) + 2):
sub = alist[i:j]
i += n
j += n
print(sub)
chunk([1, 2, 3, 4, 5], 2)
This works. I get [1, 2] [3, 4] [5]
But if I try to return sub from the function and print it in a loop it fails with: argument after * must be an iterable, not int
def chunks(alist, n):
i = 0
j = n
while j < (len(alist) + 2):
sub = alist[i:j]
i += n
j += n
return sub
for chunk in chunks([1, 2, 3, 4, 5], 2):
print(*chunk)
sub is an iterable (a list). Not sure what I am doing wrong.
NOTE cannot use itertools or any other add ons.

Your sub variable holds only a slice of the list. It should instead be appended to a list of lists as the returning value of the chunks function:
def chunks(alist, n):
i = 0
j = n
output = []
while j < (len(alist) + 2):
output.append(alist[i:j])
i += n
j += n
return output
for chunk in chunks([1, 2, 3, 4, 5], 2):
print(*chunk)
This outputs:
1 2
3 4
5

Modifying global variable from within function that have the same parameter name in python

I have this function:
def applychange(L):
result=3+2
L=[4,5,6]
return result
L = [1,2,3]
print(applychange(L))
print(L)
and the result is :
5
[1, 2, 3]
how to change the L variable to new value to have this result:
5
[4, 5, 6]

If your goal is to modify L as global variable you need to declare it as such.
For your code to run as intended, insert global L after the result assignment in your function.
Additionally, there is no need to pass L to the function in order to modify L.
def applychange():
result = 3 + 2
global L
L = [4, 5, 6]
return result
L = [1, 2, 3]
print(applychange())
print(L)
Your code written this way will result in the intended output:
5
[4, 5, 6]

The problem is that you overwrite the variable L, instead of modifying it. You can use
L[:] = [4,5,6]
instead of
L = [4,5,6]

z3py: how to make constraints for "else" value when inferring a function

I am inferring a function using z3py as follows
f = Function('f',IntSort(),IntSort(),IntSort())
After asserting a set of constraints like:
s.add(f(a1,b1)==c1)
s.add(f(a2,b2)==c2)
s.add(f(a3,b3)==c3)...
The function is inferred as
[(a1,b1) -> c1,
(a2,b2) -> c2,
(a3,b3) -> c3,
...,
else -> 2]
How could I constraint the "else" value to a fixed number? So that the output of inferred f will be
[(a1,b1) -> c1,
(a2,b2) -> c2,
(a3,b3) -> c3,
...,
else -> some number that I assert]
Edit:
from z3 import *
s = Solver()
k = Int('k')
f = Function('f',IntSort(),IntSort())
s.add(And(f(1)==1,f(2)==2))
list1 = []
list1.append(k!=1)
list1.append(k!=2)
s.add(ForAll(k,Implies(And(list1),f(k)==5)))
print s.check()
print s.model()
The output is
sat
[f = [1 -> 1, 2 -> 2, else -> 5]]
This seems to work fine for this simple case.
However, when the input for function 'f' in the constraints is undecided. The output can be weird. For example
from z3 import *
s = Solver()
f = Function('f',IntSort(),IntSort(),IntSort())
i = Int('i')
j = Int('j')
k = Int('k')
l = Int('l')
s.add(i>=0,i<5)
s.add(j>=0,j<5)
s.add(And(f(j,1)==i,f(i,2)==j))
list1 = []
list1.append(And(k==1,l==j))
list1.append(And(k==2,l==i))
s.add(ForAll([k,l],Implies(Not(Or(list1)),f(l,k)==5)))
print s.check()
print s.model()
The output is
[i = 0,
k!6 = 0,
j = 3,
k!12 = 6,
f = [else -> f!15(k!13(Var(0)), k!14(Var(1)))],
f!15 = [(3, 1) -> 0, (0, 2) -> 3, else -> 5],
k!13 = [0 -> 0, 3 -> 3, else -> 6],
k!14 = [2 -> 2, 1 -> 1, else -> 0]]
Then it is hard to interpret the inferred f.

It is a great question and very informed analysis. Yes, you can control the default values by using quantifiers. Z3 will have no choice but agree.
However, the encoding of models is based on how the quantifier instantiation engine (see Complete quantifier instantiation by Yeting Ge and Leonardo de Moura).
Z3 does not beta-reduce the expressions in the models and has left it to applications to perform the beta-reduction, if desired. You can have Z3 beta reduce else branches by plugging in your arguments to the parameters of the functions (use the substitution routines exposed by the API), and then call the model evaluator to reduce the resulting expression with respect to the model.

How can I chain a map generating a tuple with a filter, in Scala?

So, I have a simple collection, say:
val a = List(1, 2, 3, 4)
I want to map it to a list of tuples, and then filter over that, and then later map over the result of it, so it would be something like:
a.map(x => (x, x * x)).filter(tup => tup._2 < 10).map(tup => tup._1 + tup._2)
Except instead of using tup and tup._1, I want to use variable names like number and square, preferably in the arguments section.
Is it possible? How can I achieve that?

Consider using partial function and collect function:
scala> val a = List(1, 2, 3, 4)
a: List[Int] = List(1, 2, 3, 4)
scala> a.map(x => (x, x * x)) collect {
| case (number, square) if square < 10 =>
| number + square
| }
res0: List[Int] = List(2, 6, 12)
Collect in this case works like a combination of map and filter

Code Golf: New Year's Fireworks [closed]

Closed. This question is off-topic. It is not currently accepting answers.
Closed 11 years ago.
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 year 2009 is coming to an end, and with the economy and all, we'll save our money and instead of buying expensive fireworks, we'll celebrate in ASCII art this year.
The challenge
Given a set of fireworks and a time, take a picture of the firework at that very time and draw it to the console.
The best solution entered before midnight on New Year's Eve (UTC) will receive a bounty of 500 rep. This is code golf, so the number of characters counts heavily; however so do community votes, and I reserve the ultimate decision as to what is best/coolest/most creative/etc.
Input Data
Note that our coordinate system is left-to-right, bottom-to-top, so all fireworks are launched at a y-coordinate of 0 (zero).
The input data consists of fireworks of the form
(x, speed_x, speed_y, launch_time, detonation_time)
where
x is the position (column) where the firework is launched,
speed_x and speed_y are the horizontal and vertical velocity of the firework at launch time,
launch_time is the point in time that this firework is launched,
detonation_time is the point in time that this firework will detonate.
The firework data may be hardcoded in your program as a list of 5-tuples (or the equivalent in your language), not counting towards your character count. It must, however, be easy to change this data.
You may make the following assumptions:
there is a reasonable amount of fireworks (say, fewer then a hundred)
for each firework, all five numbers are integers within a reasonable range (say, 16 bits would suffice for each),
-20 <= x <= 820
-20 <= speed_x <= 20
0 < speed_y <= 20
launch_time >= 0
launch_time < detonation_time < launch_time + 50
The single additional piece of input data is the point of time which is supposed to be rendered. This is a non-negative integer that is given to you via standard input or command line argument (whichever you choose).
The idea is that (assuming your program is a python script called firework.py) this bash script gives you a nice firework animation:
#!/bin/bash
I=0
while (( 1 )) ; do
python firework.py $I
I=$(( $I + 1 ))
done
(feel free to put the equivalent .BAT file here).
Life of a firework
The life of a firework is as follows:
Before the launch time, it can be ignored.
At launch time, the rocket has the position (x, 0) and the speed vector (speed_x, speed_y).
For each time step, the speed vector is added to the position. With a little stretch applied to Newton's laws, we assume that the speed stays constant.
At detonation time, the rocket explodes into nine sparks. All nine sparks have the same position at this point in time (which is the position that the rocket would have, hadn't it exploded),
but their speeds differ. Each speed is based on the rocket's speed, with -20, 0, or 20 added to speed_x and -10, 0, or 10 added to speed_y. That's nine possible combinations.
After detonation time, gravity starts to pull: With each time step, the gravitational constant, which happens to be 2 (two), is subtracted from every spark's speed_y.
The horizontal speed_x stays constant.
For each time step after the detonation time, you first add the speed vector to the position, then subtract 2 from speed_y.
When a spark's y position drops below zero, you may forget about it.
Output
What we want is a picture of the firework the way it looks at the given point in time. We only look at the frame 0 <= x <= 789 and 0 <= y <= 239, mapping it to a 79x24 character output.
So if a rocket or spark has the position (247, 130), we draw a character in column 24 (zero-based, so it's the 25th column), row 13 (zero-based and counting from the bottom, so it's line 23 - 13 = 10, the 11th line
of the output).
Which character gets drawn depends on the current speed of the rocket / spark:
If the movement is horizontal*, i.e. speed_y == 0 or abs(speed_x) / abs(speed_y) > 2, the character is "-".
If the movement is vertical*, i.e. speed_x == 0 or abs(speed_y) / abs(speed_x) > 2, the character is "|".
Otherwise the movement is diagonal, and the character is "\" or "/" (you'll guess the right one).
If the same position gets drawn to more than once (even if it's the same character), we put "X" instead. So assuming you have a spark at (536, 119) and one at (531, 115), you draw an "X", regardless of their speeds.
* update: these are integer divisions, so the slope has to be at least 3, or at most 1/3, respectively
The output (written to standard output) is 24 lines, each terminated by a newline character. Trailing spaces are ignored, so you may, but don't need to, pad to a width of 79. The lines may not be longer than 79 characters (excluding the newline). All interior spacing must be space characters (ASCII 32).
Sample Data
Fireworks:
fireworks = [(628, 6, 6, 3, 33),
(586, 7, 11, 11, 23),
(185, -1, 17, 24, 28),
(189, 14, 10, 50, 83),
(180, 7, 5, 70, 77),
(538, -7, 7, 70, 105),
(510, -11, 19, 71, 106),
(220, -9, 7, 77, 100),
(136, 4, 14, 80, 91),
(337, -13, 20, 106, 128)]
Output at time 33:
\ | /
/ \
- | /
- | -
/ \
Output at time 77:
\
\
X
\
Output at time 93:
\ | /
\ / /
- - - \
\
/ \ \
Update: I have uploaded the expected output at the times 0 thru 99 to firework.ü-wie-geek.de/NUMBER.html, where NUMBER is the time. It includes debug information; click on a particle to see its current position, speed, etc. And yes, it's an umlaut domain. If your browser can't handle that (as obviously neither can Stack Overflow), try firework.xn---wie-geek-p9a.de.
Another update: As hinted at in the comments below, a longer firework is now available on YouTube. It was created with a modified version of MizardX' entry, with a total fireworks count of 170 (yes, that's more than the spec asked for, but the program handled it gracefully). Except for the color, the music, and the end screen, the animation can be recreated by any entry to this code golf. So, if you're geeky enough to enjoy an ASCII art firework (you know you are): Have fun, and a happy new year to all!

Heres my solution in Python:
c = [(628, 6, 6, 3, 33),
(586, 7, 11, 11, 23),
(185, -1, 17, 24, 28),
(189, 14, 10, 50, 83),
(180, 7, 5, 70, 77),
(538, -7, 7, 70, 105),
(510, -11, 19, 71, 106),
(220, -9, 7, 77, 100),
(136, 4, 14, 80, 91),
(337, -13, 20, 106, 128)]
t=input()
z=' '
s=([z]*79+['\n'])*23+[z]*79
def p(x,y,i,j):
if 0<=x<790 and 0<=y<240:p=x/10+(23-y/10)*80;I=abs(i);J=abs(j);s[p]='X-|\\/'[[s[p]!=z,I>=3*J,J>=3*I,i*j<0,1].index(1)]
for x,i,j,l,d in c:
T=t-l;x+=i*T
if t>=d:e=t-d;[p(x+X*e,j*T+e*(Y-e+1),i+X,j+Y-2*e)for X in -20,0,20 for Y in -10,0,10]
elif t>=l:p(x,j*T,i,j)
print ''.join(s)
Takes the time from the stdin and it has the nice number of 342 characters. I'm still trying to imagine how the OP got 320 :P
Edit:
This is the best I could get, 322 chars acording to wc
t=input()
s=([' ']*79+['\n'])*24
def p(x,y,i,j):
if 790>x>-1<y<240:p=x/10+(23-y/10)*80;I=abs(i);J=abs(j);s[p]='X-|\\/'[[s[p]>' ',I>=3*J,J>=3*I,i*j<0,1].index(1)]
for x,i,j,l,d in c:
T=t-l;x+=i*T;e=t-d
if t>=d:[p(x+X*e,j*T+e*(Y-e+1),i+X,j+Y-2*e)for X in-20,0,20for Y in-10,0,10]
elif t>=l:p(x,j*T,i,j)
print''.join(s),

Now that the winner is chosen – congratulations to Juan – here is my own solution, 304 characters in Python:
t=input()
Q=range
for y in Q(24):print"".join((["\\/|-"[3*(h*h>=9*v*v)or(v*v>=9*h*h)*2or h*v>0]for X,H,V,L,D in F for s,Z,K,P,u in[(t-D,t>D,t-L,G%3*20-20,G/3*10-10)for G in[Q(9),[4]][t<D]]for h,v in[[P+H,u+V-2*s*Z]]if((X+H*K+P*s)/10,23-(V*K-s*(Z*s-Z-u))/10)==(x,y)][:2]+[" ","X"])[::3][-1]for x in Q(79))
This is not really fast, because for each point in the 79x24 display, it loops through all fireworks to see if any of them is visible at this point.
Here is a version that tries to explain what's going on:
t=input()
Q=range
for y in Q(24):
line = ""
for x in Q(79):
chars = [] # will hold all characters that should be drawn at (x, y)
for X,H,V,L,D in F: # loop through the fireworks
s = t - D
Z = t > D
K = t - L
# if t < D, i.e. the rocket hasn't exploded yet, this is just [(0, 0)];
# otherwise it's all combinations of (-20, 0, 20) for x and (-10, 0, 10)
speed_deltas = [(G % 3 * 20 - 20, G / 3 * 10 -10) for G in [Q(9), [4]][t < D]]
for P, u in speed_deltas:
if x == (X + H*K + P*s)/10 and y == 23 - (V*K - s*(Z*s - Z - u))/10:
# the current horizontal and vertical speed of the particle
h = P + H
v = u + V - 2*s*Z
# this is identical to (but shorter than) abs(h) >= 3 * abs(v)
is_horizontal = h*h >= 9*v*v
is_vertical = v*v >= 9*h*h
is_northeast_southwest = h*v > 0
# a shorter way of saying
# char_index = (3 if is_horizontal else 2 if is_vertical else 1
# if is_northeast_southwest else 0)
char_index = 3 * is_horizontal or 2 * is_vertical or is_northeast_southwest
chars.append("\\/|-"[char_index])
# chars now contains all characters to be drawn to this point. So we have
# three possibilities: If chars is empty, we draw a space. If chars has
# one element, that's what we draw. And if chars has more than one element,
# we draw an "X".
actual_char = (chars[:2] + [" ", "X"])[::3][-1] # Yes, this does the trick.
line += actual_char
print line

Python:
fireworks = [(628, 6, 6, 3, 33),
(586, 7, 11, 11, 23),
(185, -1, 17, 24, 28),
(189, 14, 10, 50, 83),
(180, 7, 5, 70, 77),
(538, -7, 7, 70, 105),
(510, -11, 19, 71, 106),
(220, -9, 7, 77, 100),
(136, 4, 14, 80, 91),
(337, -13, 20, 106, 128)]
import sys
t = int(sys.argv[1])
particles = []
for x, speed_x, speed_y, launch_time, detonation_time in fireworks:
if t < launch_time:
pass
elif t < detonation_time:
x += speed_x * (t - launch_time)
y = speed_y * (t - launch_time)
particles.append((x, y, speed_x, speed_y))
else:
travel_time = t - detonation_time
x += (t - launch_time) * speed_x
y = (t - launch_time) * speed_y - travel_time * (travel_time - 1)
for dx in (-20, 0, 20):
for dy in (-10, 0, 10):
x1 = x + dx * travel_time
y1 = y + dy * travel_time
speed_x_1 = speed_x + dx
speed_y_1 = speed_y + dy - 2 * travel_time
particles.append((x1, y1, speed_x_1, speed_y_1))
rows = [[' '] * 79 for y in xrange(24)]
for x, y, speed_x, speed_y in particles:
x, y = x // 10, y // 10
if 0 <= x < 79 and 0 <= y < 24:
row = rows[23 - y]
if row[x] != ' ': row[x] = 'X'
elif speed_y == 0 or abs(speed_x) // abs(speed_y) > 2: row[x] = '-'
elif speed_x == 0 or abs(speed_y) // abs(speed_x) > 2: row[x] = '|'
elif speed_x * speed_y < 0: row[x] = '\\'
else: row[x] = '/'
print '\n'.join(''.join(row) for row in rows)
If you remove the initial fireworks declaration, compress variable-names to single characters, and whitespace to a minimum, you can get 590 characters.

C:
With all unnecessary whitespace removed (632 bytes excluding the fireworks declaration):
#define N 10
int F[][5]={628,6,6,3,33,586,7,11,11,23,185,-1,17,24,28,189,14,10,50,83,180,7,5,70,77,538,-7,7,70,105,510,-11,19,71,106,220,-9,7,77,100,136,4,14,80,91,337,-13,20,106,128};
#define G F[i]
#define R P[p]
g(x,y){if(y==0||abs(x)/abs(y)>2)return 45;if(x==0||abs(y)/abs(x)>2)return'|';if(x*y<0)return 92;return 47;}main(int A,char**B){int a,b,c,C[24][79]={},d,i,j,p=0,P[N*9][3],Q,t=atoi(B[1]),x,y;for(i=0;i<N;i++){if(t>=G[3]){a=t-G[3];x=G[0]+G[1]*a;y=G[2]*a;if(t<G[4]){R[0]=x;R[1]=y;R[2]=g(G[1],G[2]);p++;}else{b=t-G[4];y-=b*(b-1);for(c=-20;c<=20;c+=20){for(d=-10;d<=10;d+=10){R[0]=x+c*b;R[1]=y+d*b;R[2]=g(G[1]+c,G[2]+d-2*b);p++;}}}}}Q=p;for(p=0;p<Q;p++){x=R[0]/10;y=R[1]/10;if(R[0]>=0&&x<79&&R[1]>=0&&y<24)C[y][x]=C[y][x]?88:R[2];}for(i=23;i>=0;i--){for(j=0;j<79;j++)putchar(C[i][j]?C[i][j]:32);putchar(10);}}
And here's the exact same code with whitespace added for readability:
#define N 10
int F[][5] = {
628, 6, 6, 3, 33,
586, 7, 11, 11, 23,
185, -1, 17, 24, 28,
189, 14, 10, 50, 83,
180, 7, 5, 70, 77,
538, -7, 7, 70, 105,
510, -11, 19, 71, 106,
220, -9, 7, 77, 100,
136, 4, 14, 80, 91,
337, -13, 20, 106, 128
};
#define G F[i]
#define R P[p]
g(x, y) {
if(y == 0 || abs(x)/abs(y) > 2)
return 45;
if(x == 0 || abs(y)/abs(x) > 2)
return '|';
if(x*y < 0)
return 92;
return 47;
}
main(int A, char**B){
int a, b, c, C[24][79] = {}, d, i, j, p = 0, P[N*9][3], Q, t = atoi(B[1]), x, y;
for(i = 0; i < N; i++) {
if(t >= G[3]) {
a = t - G[3];
x = G[0] + G[1]*a;
y = G[2]*a;
if(t < G[4]) {
R[0] = x;
R[1] = y;
R[2] = g(G[1], G[2]);
p++;
} else {
b = t - G[4];
y -= b*(b-1);
for(c = -20; c <= 20; c += 20) {
for(d =- 10; d <= 10; d += 10) {
R[0] = x + c*b;
R[1] = y + d*b;
R[2] = g(G[1] + c, G[2] + d - 2*b);
p++;
}
}
}
}
}
Q = p;
for(p = 0; p < Q; p++) {
x = R[0]/10;
y = R[1]/10;
if(R[0] >= 0 && x < 79 && R[1] >= 0 && y < 24)
C[y][x] = C[y][x] ? 88 : R[2];
}
for(i = 23; i >= 0; i--) {
for(j = 0; j < 79; j++)
putchar(C[i][j] ? C[i][j] : 32);
putchar(10);
}
}

For Python, #MizardX's solution is nice, but clearly not codegolf-optimized -- besides the "don't really count" 333 characters of the prefix, namely:
fireworks = [(628, 6, 6, 3, 33),
(586, 7, 11, 11, 23),
(185, -1, 17, 24, 28),
(189, 14, 10, 50, 83),
(180, 7, 5, 70, 77),
(538, -7, 7, 70, 105),
(510, -11, 19, 71, 106),
(220, -9, 7, 77, 100),
(136, 4, 14, 80, 91),
(337, -13, 20, 106, 128)]
f = fireworks
### int sys argv append abs join f xrange
(the last comment is a helper for a little codegolf-aux script of mine that makes all feasible names 1-char mechanically -- it needs to be told what names NOT to minify;-), the shortest I can make that solution by squeezing whitespace is 592 characters (close enough to the 590 #MizardX claims).
Pulling out all the stops ("refactoring" the code in a codegolf mood), I get, after the prefix (I've used lowercase for single-character names I'm manually introducing or substituting, uppercase for those my codegolf-aux script substituted automatically):
import sys
Z=int(sys.argv[1])
Y=[]
e=Y.extend
for X,W,V,U,T in f:
if Z>=U:
z=Z-U;X+=W*z
if Z<T:e(((X,V*z,W,V),))
else:R=Z-T;e((X+Q*R,z*V-R*(R-1)+P*R,W+Q,V+P-2*R)for Q in(-20,0,20)for P in(-10,0,10))
K=[79*[' ']for S in range(24)]
for X,S,W,V in Y:
X,S=X/10,S/10
if(0<=X<79)&(0<=S<24):
J=K[23-S];v=abs(V);w=abs(W)
J[X]='X'if J[X]!=' 'else'-'if V==0 or w/v>2 else'|'if W==0 or v/w>2 else '\\'if W*V<0 else'/'
print '\n'.join(''.join(J)for J in K)
which measures in at 460 characters -- that's a reduction of 130, i.e. 130/590 = 22%.
Beyond 1-character names and obvious ways to minimize spacing, the key ideas include: single / for division (same as the nicer // for ints in Python 2.*), an if/else expression in lieu of an if/elif/else statement, extend with a genexp rather than a nested loop with append (allows the removal of some spaces and punctuation), not binding to a name subexpressions that occur just once, binding to a name subexpressions that would otherwise get repeated (including the .extend attribute lookup), semicolons rather than newlines where feasible (only if the separate lines would have to be indented, otherwise, counting a newline as 1 character, there is no saving).
Yep, readability suffers a bit, but that's hardly surprising in code golf;-).
Edit: after a lot more tightening, I now have a smaller program (same prefix):
Z=input()
K=[79*[' ']for S in range(24)];a=-10,0,10
def g(X,S,W,V):
X/=10;S/=10
if(0<=X<79)&(0<=S<24):J=K[23-S];v=abs(V);w=abs(W);J[X]=[[['/\\'[W*V<0],'|'][v>2.9*w],'-'][w>2.9*v],'X'][J[X]!=' ']
for X,W,V,U,T in f:
if Z>=U:
z=Z-U;X+=W*z
if Z<T:g(X,V*z,W,V)
else:R=Z-T;[g(X+Q*2*R,z*V-R*(R-1)+P*R,W+Q*2,V+P-2*R)for Q in a for P in a]
print'\n'.join(''.join(J)for J in K)
Still the same output, but now 360 characters -- exactly 100 fewer than my previous solution, which i've left as the first part of this answer (still well above the 320 the OP says he has, though!-).
I've taken advantage of the degree of freedom allowing the input-time value to come from stdin (input is much tighter than importing sys and using sys.argv[1]!-), eliminated the intermediate list (w/the extend calls and a final loop of it) in favor of the new function g which gets called directly and updates K as we go, found and removed some commonality, refactored the nested if/else expression into a complicated (but more concise;-) building and indexing of nested lists, used the fact that v>2.9*w is more concise than w==0 or v/w>2 (and always gives the same result in the range of values that are to be considered).
Edit: making K (the "screen image") into a 1-D list saves a further 26 characters, shrinking the following solution to 334 (still 14 above the OP's, but closing up...!-):
Z=input()
K=list(24*(' '*79+'\n'))
a=-10,0,10
def g(X,S,W,V):
if(0<=X<790)&(0<=S<240):j=80*(23-S/10)+X/10;v=abs(V);w=abs(W);K[j]=[[['/\\'[W*V<0],'|'][v>2.9*w],'-'][w>2.9*v],'X'][K[j]!=' ']
for X,W,V,U,T in f:
if Z>=U:
z=Z-U;X+=W*z
if Z<T:g(X,V*z,W,V)
else:R=Z-T;[g(X+Q*2*R,z*V-R*(R-1)+P*R,W+Q*2,V+P-2*R)for Q in a for P in a]
print ''.join(K),

Done in F# in 957* characters, and it's ugly as sin:
Array of fireworks:
let F = [(628,6,6,3,33);(586,7,11,11,23);(185,-1,17,24,28);(189,14,10,50,83);(180,7,5,70,77);(538,-7,7,70,105);(510,-11,19,71,106);(220,-9,7,77,100);(136,4,14,80,91);(337,-13,20,106,128)]
Remaining code
let M=List.map
let C=List.concat
let P=List.partition
let L t f r=(let s=P(fun(_,_,_,u,_)->not(t=u))f
(fst s, r#(M(fun(x,v,w,_,t)->x,0,v,w,t)(snd s))))
let X d e (x,y,v,w)=C(M(fun(x,y,v,w)->[x,y,v-d,w;x,y,v,w;x,y,v+d,w])[x,y,v,w-e;x,y,v,w;x,y,v,w+e])
let D t r s=(let P=P(fun(_,_,_,_,u)->not(t=u))r
(fst P,s#C(M(fun(x,y,v,w,_)->(X 20 10(x,y,v,w)))(snd P))))
let rec E t l f r s=(
let(a,m)=L t f (M(fun(x,y,v,w,t)->x+v,y+w,v,w,t)r)
let(b,c)=D t m (M(fun(x,y,v,w)->x+v,y+w,v,w-2)s)
if(t=l)then(a,b,c)else E(t+1)l a b c)
let N=printf
let G t=(
let(f,r,s)=E 0 t F [] []
let os=s#(M(fun(x,y,v,w,_)->(x,y,v,w))r)
for y=23 downto 0 do (
for x=0 to 79 do (
let o=List.filter(fun(v,w,_,_)->((v/10)=x)&&((w/10)=y))os
let l=o.Length
if l=0 then N" "
elif l=1 then
let(_,_,x,y)=o.Head
N(
if y=0||abs(x)/abs(y)>2 then"-"
elif x=0||abs(y)/abs(x)>2 then"|"
elif y*x>0 then"/"
else"\\")
elif o.Length>1 then N"X")
N"\n"))
[<EntryPointAttribute>]
let Z a=
G (int(a.[0]))
0
"Pretty" code:
let fxs = [(628,6,6,3,33);(586,7,11,11,23);(185,-1,17,24,28);(189,14,10,50,83);(180,7,5,70,77);(538,-7,7,70,105);(510,-11,19,71,106);(220,-9,7,77,100);(136,4,14,80,91);(337,-13,20,106,128)]
let movs xs =
List.map (fun (x, y, vx, vy) -> (x + vx, y + vy, vx, vy-2)) xs
let movr xs =
List.map (fun (x, y, vx, vy, dt) -> (x + vx, y + vy, vx, vy, dt)) xs
let launch t fs rs =
let split = List.partition(fun (lx, sx, sy, lt, dt) -> not (t = lt)) fs
(fst split, rs # (List.map(fun (lx, sx, sy, lt, dt) -> (lx, 0, sx, sy, dt)) (snd split)))
let split dx dy (x,y,sx,sy) =
List.concat (List.map (fun (x,y,sx,sy)->[(x,y,sx-dx,sy);(x,y,sx,sy);(x,y,sx+dx,sy)]) [(x,y,sx,sy-dy);(x,y,sx,sy);(x,y,sx,sy+dy)])
let detonate t rs ss =
let tmp = List.partition (fun (x, y, sx, sy, dt) -> not (t = dt)) rs
(fst tmp, ss # List.concat (List.map(fun (x, y, sx, sy, dt) -> (split 20 10 (x, y, sx, sy))) (snd tmp)))
let rec simulate t l fs rs ss =
let (nfs, trs) = launch t fs (movr rs)
let (nrs, nss) = detonate t trs (movs ss)
if (t = l) then (nfs,nrs,nss)
else
simulate (t+1) l nfs nrs nss
let screen t =
let (fs, rs, ss) = simulate 0 t fxs [] []
let os = ss # (List.map(fun (x, y, sx, sy,_) -> (x, y, sx, sy)) rs)
for y = 23 downto 0 do
for x = 0 to 79 do
let o = List.filter(fun (px,py,_,_)->((px/10)=x) && ((py/10)=y)) os
if o.Length = 0 then printf " "
elif o.Length = 1 then
let (_,_,sx,sy) = o.Head
printf (
if sy = 0 || abs(sx) / abs(sy) > 2 then "-"
elif sx = 0 || abs(sy) / abs(sx) > 2 then "|"
elif sy * sx > 0 then "/"
else"\\"
)
elif o.Length > 1 then printf "X"
printfn ""
[<EntryPointAttribute>]
let main args =
screen (int(args.[0]))
0
Completely stolenrewritten with new and improved logic. This is as close as I could get to Python. You can see the weakness of F# not being geared toward ad hoc scripting here, where I have to explicitly convert V and W to a float, declare a main function with an ugly attribute to get the command line args, and I have to reference the .NET System.Console.Write to get a pretty output.
Oh well, good exercise to learn a language with.
Here's the new code, at 544 bytes:
let Q p t f=if p then t else f
let K=[|for i in 1..1920->Q(i%80>0)' ''\n'|]
let g(X,S,W,V)=
if(X>=0&&X<790&&S>=0&&S<240)then(
let (j,v,w)=(80*(23-S/10)+X/10,abs(float V),abs(float W))
Array.set K j (Q(K.[j]=' ')(Q(w>2.9*v)'-'(Q(v>2.9*w)'|'(Q(W*V>0)'/''\\')))'X'))
let a=[-10;0;10]
[<EntryPointAttribute>]
let m s=
let Z=int s.[0]
for (X,W,V,U,T) in F do(
if Z>=U then
let z,R=Z-U,Z-T
let x=X+W*z
if(Z<T)then(g(x,V*z,W,V))else(for e in[|for i in a do for j in a->(x+j*2*R,z*V-R*(R-1)+i*R,W+j*2,V+i-2*R)|]do g e))
System.Console.Write K
0

Haskell
import Data.List
f=[(628,6,6,3,33),(586,7,11,11,23),(185,-1,17,24,28),(189,14,10,50,83),(180,7,5,70,77),(538,-7,7,70,105),(510,-11,19,71,106),(220,-9,7,77,100),(136,4,14,80,91),(337,-13,20,106,128)]
c=filter
d=True
e=map
a(_,_,_,t,_)=t
b(_,_,_,_,t)=t
aa(_,y,_,_)=y
ab(x,t,y,_,u)=(x,0,t,y,u)
ac(x,y,t,u,_)=[(x,y,t+20,u+10),(x,y,t,u+10),(x,y,t-20,u+10),(x,y,t+20,u),(x,y,t,u),(x,y,t-20,u),(x,y,t+20,u-10),(x,y,t,u-10),(x,y,t-20,u-10)]
g(x,y,t,u,v)=(x+t,y+u,t,u,v)
h(x,y,t,u)=(x+t,y+u,t,u-2)
i=(1,f,[],[])
j s 0=s
j(t,u,v,w)i=j(t+1,c((/=t).a)u,c((> t).b)x++(e ab.c((==t).a))u,c((>0).aa)(e h w)++(concat.e ac.c((==t).b))x)(i-1)
where x=e g v
k x y
|x==0='|'
|3*abs y<=abs x='-'
|3*abs x<=abs y='|'
|(y<0&&x>0)||(y>0&&x<0)='\\'
|d='/'
l(x,y,t,u,_)=m(x,y,t,u)
m(x,y,t,u)=(div x 10,23-div y 10,k t u)
n(x,y,_)(u,v,_)
|z==EQ=compare x u
|d=z
where z=compare y v
o((x,y,t):(u,v,w):z)
|x==u&&y==v=o((x,y,'X'):z)
|d=(x,y,t):(o((u,v,w):z))
o x=x
q _ y []
|y==23=""
|d='\n':(q 0(y+1)[])
q v u((x,y,z):t)
|u>22=""
|v>78='\n':(q 0(u+1)((x,y,z):t))
|u/=y='\n':(q 0(u+1)((x,y,z):t))
|v/=x=' ':(q(v+1)u((x,y,z):t))
|d = z:(q(v+1)u t)
p(_,_,v,w)=q 0 0((c z.o.sortBy n)((e l v)++(e m w)))
where z(x,y,_)=x>=0&&x<79&&y>=0
r x=do{z <- getChar;(putStr.p)x}
s=e(r.j i)[1..]
main=foldr(>>)(return())s
Not nearly as impressive as MizardX's, coming in at 1068 characters if you remove the f=… declaration, but hell, it was fun. It's been a while since I've had a chance to play with Haskell.
The (slightly) prettier version is also available.
Edit: Ack. Rereading, I don't quite meet the spec.: this version prints a new screen of firework display every time you press a key, and requires ^C to quit; it doesn't take a command line argument and print out the relevant screen.

Perl
Assuming firework data is defined as:
#f = (
[628, 6, 6, 3, 33],
[586, 7, 11, 11, 23],
[185, -1, 17, 24, 28],
[189, 14, 10, 50, 83],
[180, 7, 5, 70, 77],
[538, -7, 7, 70, 105],
[510, -11, 19, 71, 106],
[220, -9, 7, 77, 100],
[136, 4, 14, 80, 91],
[337, -13, 20, 106, 128]
);
$t=shift;
for(#f){
($x,$c,$d,$l,$e)=#$_;
$u=$t-$l;
next if$u<0;
$x+=$c*$u;
$h=$t-$e;
push#p,$t<$e?[$x,$d*$u,$c,$d]:map{$f=$_;map{[$x+$f*$h,($u*$d-$h*($h-1))+$_*$h,$c+$f,$d+$_-2*$h]}(-10,0,10)}(-20,0,20)
}
push#r,[($")x79]for(1..24);
for(#p){
($x,$y,$c,$d)=#$_;
if (0 <= $x && ($x=int$x/10) < 79 && 0 <= $y && ($y=int$y/10) < 24) {
#$_[$x]=#$_[$x]ne$"?'X':!$d||abs int$c/$d>2?'-':!$c||abs int$d/$c>2?'|':$c*$d<0?'\\':'/'for$r[23 - $y]
}
}
$"='';
print$.,map{"#$_\n"}#r
Compressed, it comes in at 433 characters. (see edits for history)
This is based off of pieces of multiple previous answers (mostly MizardX's) and can definitely be improved upon. The guilt of procrastinating other, job-related tasks means i have to give up for now.
Forgive the edit -- pulling out all of the tricks I know, this can be compressed to 356 char:
sub p{
($X,$=,$C,$D)=#_;
if(0<=$X&($X/=10)<79&0<=$=&($=/=10)<24){
#$_[$X]=#$_[$X]ne$"?X:$D&&abs$C/$D<3?$C&&abs$D/$C<3?
$C*$D<0?'\\':'/':'|':'-'for$r[23-$=]
}
}
#r=map[($")x79],1..24;
$t=pop;
for(#f){
($x,$c,$d,$u,$e)=#$_;
$x-=$c*($u-=$t);
$u>0?1:($h=$t-$e)<0
?p$x,-$d*$u,$c,$d
:map{for$g(-10,0,10){p$x+$_*$h,$h*(1-$h+$g)-$u*$d,$c+$_,$d+$g-2*$h}}-20,0,20
}
print#$_,$/for#r
$= is a special Perl variable (along with $%, $-, and $?) that can only take on integer values. Using it eliminates the need to use the int function.

FORTRAN 77
From the prehistoric languages department, here's my entry – in FORTRAN 77.
2570 chars including the initialization, a handful of spaces and some unnecessary whitespace, but I don't think it's likely to win for brevity. Especially since e.g. 6 leading spaces in each line are mandatory.
I called this file fireworks.ftn and compiled it with gfortran on a Linux system.
implicit integer(a-z)
parameter (n=10)
integer fw(5,n) /
+ 628, 6, 6, 3, 33,
+ 586, 7, 11, 11, 23,
+ 185, -1, 17, 24, 28,
+ 189, 14, 10, 50, 83,
+ 180, 7, 5, 70, 77,
+ 538, -7, 7, 70, 105,
+ 510, -11, 19, 71, 106,
+ 220, -9, 7, 77, 100,
+ 136, 4, 14, 80, 91,
+ 337, -13, 20, 106, 128
+ /
integer p(6, 1000) / 6000 * -1 /
character*79 s(0:23)
character z
c Transform input
do 10 r=1,n
p(1, r) = 0
do 10 c=1,5
10 p(c+1, r) = fw(c, r)
c Input end time
read *, t9
c Iterate from 1 to end time
do 62 t=1,t9
do 61 q=1,1000
if (p(1,q) .lt. 0 .or. t .lt. p(5,q)) goto 61
if (p(6,q).gt.0.and.t.gt.p(5,q) .or. t.gt.abs(p(6,q))) then
p(1,q) = p(1,q) + p(4,q)
p(2,q) = p(2,q) + p(3,q)
endif
if (t .lt. abs(p(6,q))) goto 61
if (t .gt. abs(p(6,q))) then
p(4,q) = p(4,q) - 2
elseif (t .eq. p(6,q)) then
c Detonation: Build 9 sparks
do 52 m=-1,1
do 51 k=-1,1
c Find a free entry in p and fill it with a spark
do 40 f=1,1000
if (p(1,f) .lt. 0) then
do 20 j=1,6
20 p(j,f) = p(j,q)
p(3,f) = p(3,q) + 20 * m
p(4,f) = p(4,q) + 10 * k
p(6,f) = -p(6,q)
goto 51
endif
40 continue
51 continue
52 continue
c Delete the original firework
p(1,q) = -1
endif
61 continue
62 continue
c Prepare output
do 70 r=0,23
70 s(r) = ' '
do 80 q=1,1000
if (p(1,q) .lt. 0) goto 80
if (p(5,q) .gt. t9) goto 80
y = p(1,q) / 10
if (y .lt. 0 .or. y .gt. 23) goto 80
x = p(2,q) / 10
if (x .lt. 0 .or. x .gt. 79) goto 80
if (s(y)(x+1:x+1) .ne. ' ') then
z = 'X'
elseif ((p(4,q) .eq. 0) .or. abs(p(3,q) / p(4,q)) .gt. 2) then
z = '-'
elseif ((p(3,q) .eq. 0) .or. abs(p(4,q) / p(3,q)) .gt. 2) then
z = '|'
elseif (sign(1, p(3,q)) .eq. sign(1, p(4,q))) then
z = '/'
else
z = '\'
endif
s(y)(x+1:x+1) = z
80 continue
c Output
do 90 r=23,0,-1
90 print *, s(r)
end

Here's a smaller Haskell implementation. It's 911 characters; minus the fireworks definition, it's 732 characters:
import System
z=789
w=239
r=replicate
i=foldl
main=do{a<-getArgs;p(f[(628,6,6,3,33),(586,7,11,11,23),(185,-1,17,24,28),(189,14,10,50,83),(180,7,5,70,77),(538,-7,7,70,105),(510,-11,19,71,106),(220,-9,7,77,100),(136,4,14,80,91),(337,-13,20,106,128)](read(a!!0)::Int));}
p[]=return()
p(f:g)=do{putStrLn f;p g}
f s t=i(a t)(r 24(r 79' '))s
a t f(x,s,y,l,d)=if t<l then f else if t<d then c f((x+s*u,y*u),(s,y))else i c f(map(v(t-d)(o(d-l)(x,0)(s,y)))[(g s,h y)|g<-[id,(subtract 20),(+20)],h<-[id,(subtract 10),(+10)]])where u=t-l
v 0(x,y)(vx,vy)=((x,y),(vx,vy))
v t(x,y)(vx,vy)=v(t-1)(x+vx,y+vy)(vx,vy-2)
o t(x,y)(vx,vy)=(x+(vx*t),y+(vy*t))
c f((x,y),(vx,vy))=if x<0||x>=z||y<0||y>=w then f else(take m f)++[(take n r)++[if d/=' 'then 'x'else if vy==0||abs(vx`div`vy)>2 then '-'else if vx==0||abs(vy`div`vx)>2 then '|'else if vx*vy>=0 then '/'else '\\']++(drop(n+1)r)]++(drop(m+1)f)where{s=w-y;n=x`div`10;m=s`div`10;r=f!!m;d=r!!n}
Here's the non-compressed version for the curious:
import System
sizeX = 789
sizeY = 239
main = do
args <- getArgs
printFrame (frame fireworks (read (args !! 0) :: Int))
where
fireworks = [
(628, 6, 6, 3, 33),
(586, 7, 11, 11, 23),
(185, -1, 17, 24, 28),
(189, 14, 10, 50, 83),
(180, 7, 5, 70, 77),
(538, -7, 7, 70, 105),
(510, -11, 19, 71, 106),
(220, -9, 7, 77, 100),
(136, 4, 14, 80, 91),
(337, -13, 20, 106, 128)]
printFrame :: [String] -> IO ()
printFrame [] = return ()
printFrame (f:fs) = do
putStrLn f
printFrame fs
frame :: [(Int,Int,Int,Int,Int)] -> Int -> [String]
frame specs time =
foldl (applyFirework time)
(replicate 24 (replicate 79 ' ')) specs
applyFirework :: Int -> [String] -> (Int,Int,Int,Int,Int) -> [String]
applyFirework time frame (x,sx,sy,lt,dt) =
if time < lt then frame
else if time < dt then
drawChar frame
((x + sx * timeSinceLaunch, sy * timeSinceLaunch), (sx,sy))
else
foldl drawChar frame
(
map
(
posVelOverTime (time - dt)
(posOverTime (dt - lt) (x,0) (sx, sy))
)
[
(fx sx, fy sy) |
fx <- [id,(subtract 20),(+20)],
fy <- [id,(subtract 10),(+10)]
]
)
where timeSinceLaunch = time - lt
posVelOverTime :: Int -> (Int,Int) -> (Int,Int) -> ((Int,Int),(Int,Int))
posVelOverTime 0 (x,y) (vx,vy) = ((x,y),(vx,vy))
posVelOverTime time (x,y) (vx,vy) =
posVelOverTime (time - 1) (x+vx, y+vy) (vx, vy - 2)
posOverTime :: Int -> (Int,Int) -> (Int,Int) -> (Int,Int)
posOverTime time (x,y) (vx, vy) = (x + (vx * time), y + (vy * time))
drawChar :: [String] -> ((Int,Int),(Int,Int)) -> [String]
drawChar frame ((x,y),(vx,vy)) =
if x < 0 || x >= sizeX || y < 0 || y >= sizeY then frame
else
(take mappedY frame)
++
[
(take mappedX row)
++
[
if char /= ' ' then 'x'
else if vy == 0 || abs (vx `div` vy) > 2 then '-'
else if vx == 0 || abs (vy `div` vx) > 2 then '|'
else if vx * vy >= 0 then '/'
else '\\'
]
++ (drop (mappedX + 1) row)
]
++ (drop (mappedY + 1) frame)
where
reversedY = sizeY - y
mappedX = x `div` 10
mappedY = reversedY `div` 10
row = frame !! mappedY
char = row !! mappedX

First draft in Tcl8.5 913 bytes excluding fireworks definition:
set F {
628 6 6 3 33
586 7 11 11 23
185 -1 17 24 28
189 14 10 50 83
180 7 5 70 77
538 -7 7 70 105
510 -11 19 71 106
220 -9 7 77 100
136 4 14 80 91
337 -13 20 106 128
}
namespace import tcl::mathop::*
proc # {a args} {interp alias {} $a {} {*}$args}
# : proc
# = set
# D d p
# up upvar 1
# < append out
# _ foreach
# e info exists
# ? if
: P {s d t l} {+ $s [* $d [- $t $l]]}
: > x {= x}
: d {P x X y Y} {up $P p
= x [/ $x 10]
= y [/ $y 10]
= p($x,$y) [? [e p($x,$y)] {> X} elseif {
$Y==0||abs($X)/abs($Y)>2} {> -} elseif {
$X==0||abs($Y)/abs($X)>2} {> |} elseif {
$X*$Y<0} {> \\} {> /}]}
: r {P} {up $P p
= out ""
for {= y 23} {$y >= 0} {incr y -1} {
for {= x 0} {$x < 79} {incr x} {? {[e p($x,$y)]} {< $p($x,$y)} {< " "}}
< "\n"}
puts $out}
: s {F t} {array set p {}
_ {x X Y l d} $F {? {$t >= $l} {? {$t < $d} {= x [P $x $X $t $l]
= y [P 0 $Y $t $l]
D $x $X $y $Y} {= x [P $x $X $d $l]
= y [P 0 $Y $d $l]
= v [- $t $d]
_ dx {-20 0 20} {_ dy {-10 0 10} {= A [+ $X $dx]
= B [- [+ $Y $dy] [* 2 $v]]
= xx [P $x $A $v 0]
= yy [P $y $B $v 0]
D $xx $A $yy $B}}}}}
r p}
s $F [lindex $argv 0]
Optimized to the point of unreadability. Still looking for room to improve. Most of the compression basically uses command aliasing substituting single characters for command names. For example, function definitions are done using Forth-like : syntax.
Here's the uncompressed version:
namespace import tcl::mathop::*
set fireworks {
628 6 6 3 33
586 7 11 11 23
185 -1 17 24 28
189 14 10 50 83
180 7 5 70 77
538 -7 7 70 105
510 -11 19 71 106
220 -9 7 77 100
136 4 14 80 91
337 -13 20 106 128
}
proc position {start speed time launch} {
+ $start [* $speed [- $time $launch]]
}
proc give {x} {return $x}
proc draw {particles x speedX y speedY} {
upvar 1 $particles p
set x [/ $x 10]
set y [/ $y 10]
set p($x,$y) [if [info exists p($x,$y)] {
give X
} elseif {$speedY == 0 || abs(double($speedX))/abs($speedY) > 2} {
give -
} elseif {$speedX == 0 || abs(double($speedY))/abs($speedX) > 2} {
give |
} elseif {$speedX * $speedY < 0} {
give \\
} else {
give /
}
]
}
proc render {particles} {
upvar 1 $particles p
set out ""
for {set y 23} {$y >= 0} {incr y -1} {
for {set x 0} {$x < 79} {incr x} {
if {[info exists p($x,$y)]} {
append out $p($x,$y)
} else {
append out " "
}
}
append out "\n"
}
puts $out
}
proc show {fireworks time} {
array set particles {}
foreach {x speedX speedY launch detonate} $fireworks {
if {$time >= $launch} {
if {$time < $detonate} {
set x [position $x $speedX $time $launch]
set y [position 0 $speedY $time $launch]
draw particles $x $speedX $y $speedY
} else {
set x [position $x $speedX $detonate $launch]
set y [position 0 $speedY $detonate $launch]
set travel [- $time $detonate]
foreach dx {-20 0 20} {
foreach dy {-10 0 10} {
set speedXX [+ $speedX $dx]
set speedYY [- [+ $speedY $dy] [* 2 $travel]]
set xx [position $x $speedXX $travel 0]
set yy [position $y $speedYY $travel 0]
draw particles $xx $speedXX $yy $speedYY
}
}
}
}
}
render particles
}
show $fireworks [lindex $argv 0]

First Post hahaha
http://zipts.com/position.php?s=0
not my final submission but could not resist
Btw: Characters 937 not counting spaces (do we count spaces? )

My answer is at http://www.starenterprise.se/fireworks.html
all done in javascript.
and no I didn't bother to make it ashortap, I just wanted to see if I could.

Clojure
Unindented, without input output and unnecessary whitespace, it comes to 640 characters - exactly double the best value :( Thus, I'm not providing a "blank optimized" version in an attempt to win at brevity.
(def fw [
[628 6 6 3 33]
[586 7 11 11 23]
[185 -1 17 24 28]
[189 14 10 50 83]
[180 7 5 70 77]
[538 -7 7 70 105]
[510 -11 19 71 106]
[220 -9 7 77 100]
[136 4 14 80 91]
[337 -13 20 106 128]
])
(defn rr [x y u v dt g] (if (<= dt 0) [x y u v] (recur (+ x u) (+ y v) u (+ v g) (dec dt) g)))
(defn pp [t f]
(let [y 0 [x u v a d] f r1 (rr x y u v (- (min t d) a) 0)]
(if (< t a)
'()
(if (< t d)
(list r1)
(for [m '(-20 0 20) n '(-10 0 10)]
(let [[x y u v] r1]
(rr x y (+ u m) (+ v n) (- t d) -2)))))))
(defn at [x y t]
(filter #(and (= x (quot (first %) 10)) (= y (quot (second %) 10))) (apply concat (map #(pp t %) fw))))
(defn g [h]
(if (empty? h) \space
(if (next h) \X
(let [[x y u v] (first h)]
(cond
(or (zero? v) (> (* (/ u v) (/ u v)) 4)) \-
(or (zero? u) (> (* (/ v u) (/ v u)) 4)) \|
(= (neg? u) (neg? v)) \/
:else \\
)))))
(defn q [t]
(doseq [r (range 23 -1 -1)]
(doseq [c (range 0 80)]
(print (g (at c r t))))
(println)))
(q 93)