Forum

CS2D

Scripts

The Ultimate Lua Challenge

Forums overview CS2D overview Scripts overview Log in to reply

The Ultimate Lua Challenge

89 replies

1 2 3 4 5

Re: The Ultimate Lua Challenge

Snurq
BANNED Off

Offline

05.07.12 03:24:26 pm

31:
Would take some time to do this in lua, so i did it by hand.
basically it came down to this in matrix notation:

f(n+2) = (4 12 1) * f(n+1)
f(n+1) = (1 0 0 ) * f(n)
f(n) = (0 1 0 ) * 5

f(n+2) = 4 * f(n+1) + 12* f(n) + 5

diagonalize:
A = P * D * P^-1
Where D is the diagonal matrix with the eigen values,
P is the matrix with the eigenvectors as columns.
D^100 is easy to compute because diagonal.

A ^100 = P * D ^100 * P = Q

now Q * [1 0 5]' and u have the answer.

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

05.07.12 10:56:17 pm

That's pretty much correct, however there's a small problem in the transformation matrix

http://mathbin.net/101651

34.

Find the last three digits of (25!)^2000 such that they are not all 0

34.b (Challenge)

As a corollary, show that for any n, the last three non-zero digits of (n!)^2000 can be expressed as 2^(2000*x) mod 1000 for some integer x < log_2(n!).

For example, for 25!, x = 16, and 2^(32000) mod 1000 = 376.

Re: The Ultimate Lua Challenge

SD
User

Offline

05.07.12 11:14:13 pm

I didn't really understand the first task. Could this be a correct answer?

Code

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

06.07.12 01:49:27 am

Not quite, it's actually simpler than what you are thinking. Essentially, if you have a table {"a", "b", "c", "d", ...}, the first element of that table is "a". How might you get the first element out of a table?

Re: The Ultimate Lua Challenge

SD
User

Offline

06.07.12 03:55:10 am

Like this.

1
2

arrayx = {"a", "b", "c"}
print(arrayx[1])

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

14.01.13 08:17:29 am

35. Suppose you are stacking a tower of boxes and each box has a weight that's equal to the capacity of weight it can support. For example, a box of weight 3 can support only up to 3 units of weight. What is the least weight of a 1000 box high tower?

36. Suppose that you are inside a grid and you can only walk either to the square below or to the right of you. How many paths are there between the origin and (x,y)?

37. Suppose that in 2020, UnrealSoftware users reorganized themselves into provinces and there are yearly elections for a moderator (out of two candidates). Suppose that there are 1000 provinces consisting of k_i users for each province i, and each province will put all of its votes into only one candidate. Given a randomly generated set of k_i, determine if it's possible for a tie to occur.

38. Suppose that China has recently decided to switch its language completely into pinyin (made up of the english alphabet), but are adamant about not putting spaces around words. For example, in english, the phrase "how are you doing" would become "howareyoudoing" if we're using the reformed chinese analogy. Suppose I give you an electronic dictionary of pinyin to chinese characters, write a program that can parse a stream of pinyin and output a character translation if one exists.

(Note: the obvious method for 38 may not work here; why?)

39. (Slightly challenging) Suppose that you live in a world where once a person makes a call, he cannot make another call for the rest of that minute. General Sherman was a general and had 10 officers that he can directly call, and those 10 officers each have some number of direct subordinates to whom they can call (think of this like a tree/hierarchy of command). Suppose the general wanted to everyone hi (i.e., each person is to utter the phrase "the general says hi, send it along" and hang up) which takes 0 seconds to do (assuming), then what is the least number of minutes it takes for everyone under Sherman's command to get the message if there are over 100 levels of command in this hypothetical world (generate a random test case)

For example, suppose that Sherman is in charge of Bob and Frank, Bob is not in charge of anyone else but Frank is in charge of Emily and Matt. If Sherman calls Bob first, then Frank, and Frank tells Emily and Matt, then it'll take a total of 4 minutes to finish the call. On the other hand, if Sherman calls Frank first, and while calling Bob next, Frank calls Emily, and finally Matt, then the whole process only takes 3 minutes. How do you compute the minimum time it takes? The brute force method will take too long when you have 100 levels of command.

Re: The Ultimate Lua Challenge

gotya2
GAME BANNED Off

Offline

14.01.13 12:40:17 pm

Remember that this is the ultimate Lua challenge, and not the ultimate math challenge

ANyway for 35. I assume that the mimimum weight of a box is 1.

[1],
[1],
[2],
[4], this is what the top would look like
so the series of weight would look like

1 1 2^1 2^2 ... 2^(1000-2)
the sum of this is 2^(1000 - 1) = a huge number.

print(math.pow(2,999))

36.
combinational problem
http://en.wikipedia.org/wiki/Combination

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21

fact = function(n)
	local sum = 1
	while( n > 0 ) do
		sum = sum * n
		n = n - 1
	end
	return sum

end

paths = function(o_x,o_y,x,y)
	if( x < o_x or y < o_y) then
		return 0
	else
		local dx,dy = x - o_x, y - o_y
		return fact(dx+dy) / (fact(dx) * fact(dy))
	end
end


print(paths(3,3,6,6))

38 This is a problem of lexical analysis
http://en.wikipedia.org/wiki/Lexical_analysis

The tokens would be {"how", "are", "you", "doing", ... }

the obvious problem is that in this sentence, howareyoudoing, also the words "war" and "do" appear, also part of the tokens. You would have to write a regular expression parser that only accepts the longest match. not "do" but "doing".

edited 1×, last 14.01.13 12:53:03 pm

Re: The Ultimate Lua Challenge

Flacko
User Off

Offline

14.01.13 03:17:40 pm

1
2
3
4
5
6
7
8
9

function stackWeight(x,minw)
	local minw = minw or 1 --2^(x-1)
	for i=1, x-1 do
		minw = minw + minw
	end
	return minw
end
--output
print(stackWeight(1000))

1
2
3
4
5
6
7
8
9

function calcPaths(x,y) --assuming start tile is (0,0)
	local total = 0
	if x > 0 then total = total + calcPaths(x-1, y) end
	if y > 0 then total = total + calcPaths(x, y-1) end
	if x==0 and y==0 then return 1 end
	return total
end
--output
print(calcPaths(2,2))

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

k = {voters = 0}
for i=1, 1000 do
	local x = math.random(1,10)
	k[i] = x
	k.voters = k.voters + x
end
function isTiePossible()
	--assuming k[1 ... 500] will all vote candidate 1 while [500...1000] will vote 2
	local count = 0
	for i=1,500 do
		count = count + k[i]
	end
	return k.voters/count == 2
end
--output
print(tostring(isTiePossible()))

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

--english to spanish
dictionary = {
	{"war","guerra"},
	{"do","hacer"},
	{"how","como"},
	{"are","estas"},
	{"you","tu"},
	{"doing","haciendo"},
	{"other", "otro"},
	{"right", "correcto"},
	{"the", "lo"},
	{"thing", "cosa"},
}
table.sort(dictionary, function(a,b) return a[1]:len() > b[1]:len() end)
function translate(x,start)
	if not start then
		start = 1
		x = string.lower(x)
	end
	local m = {math.huge, math.huge, 0}
	for i=1,#dictionary do
		local s,e = x:find(dictionary[i][1], start, true)
		if s and e then
			if s < m[1] or (s == m[1] and e > m[2])then
				m = {s,e, i}
			end
		end
	end
	local str = ""
	if m[2] <= string.len(x) then 
		str = dictionary[m[3]][2] .. " " .. translate(x, m[2]+1)
	end
	return str
end
--output
print(translate("howareyoudoing"))
print(translate("dotherightthing"))

edited 1×, last 14.01.13 06:17:43 pm

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

15.01.13 02:02:36 am

gotya2: You know me :P, it's hard pressed for me to differentiate between the two

@ user

Flacko:

36 is a bit intractable for large values, it's possible to bring the time complexity down to O(n^2) rather than EXP by memoization.

Another cool thing, as user

gotya2 pointed out, this has a combinatorial form of either [(x+y) choose x] or [(x+y) choose y] because this has a natural reduction to a problem of shooting x balls into y baskets (can you see how to make this analogy)?

40 A (significantly) harder version of this problem with no known closed combinatorial form asks how many paths there are to a point (n,n) such that no points (x,y) whose x is perfectly square, y is perfectly square, and the sum x+y is also perfectly square (i.e. eliminating all pythagorean pairs).

Hint: inclusion/exclusion

For

37, I'm not sure if this algorithm is correct for the general case. Supposing that all of the candidates are sorted in increasing order of their number of votes, then the following scenario [1,1,1,1,1,5] might constitute as a counterexample; in theory it could result in a tie if the first 5 candidates votes for 1 and the last candidate votes for two, but in which the sorted algorithm will return as being impossible to achieve.

For

38, both of you are slightly off (you are both seemingly trying to match the earliest and the longest matchable sequence first). Consider the following dictionary which consists of only three words, do, doing, and ingrid

1
2
3

dict.doing = "..."
dict.do = "..."
dict.ingrid = "..."

Suppose we have as an input the string "doingrid". The maximal munch technique will first match "doing" because that is the longest sequence, then it will have to match "rid", which has no known translation, and will then report that translation is not possible when it could be tokenized into "do" and "ingrid". Of course, as your dictionary becomes larger, such problems aren't as common and maximal much proves a good approximation. (Note that in general, this problem is NP-Hard, but it does not take an exponential number of operations to solve. The version which decides and translates the maximal subset of the string is both NP-Hard and take an exponential number of operations to solve; you can prove both by reduction from set cover and maximal set cover respectively)

Hint for

39: you can solve this recursively and use a simple heuristic (just sort something in decreasing order) which turns out to be optimal. Ponder on the Sherman-Bob-Frank example and draw out the tree, you'll probably see the solution.

Re: The Ultimate Lua Challenge

Flacko
User Off

Offline

15.01.13 04:07:41 am

Ok then, from what I ripped off from wikipedia:
36.

1
2
3
4
5
6
7
8
9
10
11
12
13

function calcPaths(x,y) --assuming start tile is (0,0)
	local fz = 1
	for i=y+1, x+y do
		fz = fz*i
	end
	local fx = 1
	for i=1, x do
		fx = fx*i
	end
	return fz/fx
end
--output
print(calcPaths(3,3))

37. (ugly and inefficient, as I didn't understand wikipedia's subset sum article)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40

k = {voters = 0}
for i=1, 1000 do
	local x = math.random(1,5)
	k[#k+1] = x
	k.voters = k.voters + x
end
function isTiePossible()
	if k.voters % 2 ~= 0 then
		return false
	end
	local selections = {total=0}
	while selections.total ~= k.voters / 2 do
		if #k == 0 then return false end
		local lasti = 0
		for i=1,#k do --look for the biggest possible item to add
			if k[i]+selections.total <= k.voters/2 and k[i] > (k[lasti] or 0) then
				lasti = i
			end
		end
		if lasti == 0 then --couldn't fit a new item? discard the biggest element in our list
			local i = table.biggest(selections)
			selections.total = selections.total - selections[i]
			table.remove(selections,i)
		else --add new item to selections
			selections[#selections+1] = k[lasti]
			selections.total = selections.total + selections[#selections]
			table.remove(k,lasti)
		end
	end
	return selections
end
function table.biggest(t)
	local p = 1
	for i=2, #t do
		if t[i] > t[p] then p=i	end
	end
	return p
end
--output
print(isTiePossible())

Re: The Ultimate Lua Challenge

FlooD
GAME BANNED Off

Offline

15.01.13 05:37:44 am

wow
it's hard to remember the time when i wasn't so lazy and would actually try these

kkkk fine
didn't google or look at anyone's answer yet
35. trivial. 2^(1000-1)
36. trivial. x+y choose x
37. a bit nontrivial. coding on iphone sucks. one space indent, umad??

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28

test={1,3,5,7,9,11}
function tie(t)
 local s=0
 for _,v in pairs(t) do s=s+v end
 if s%2==1 then return false end
 s=s/2 --problem?
 for _,v in pairs(t) do if v>s then return false end end
 return a(copy(t),s)
end
function copy(t)
 local tt={}
 for i,v in pairs(t) do tt[i]=v end
 return tt
end
function a(t,c)
 if c==0 then return true end
 local imax
 local max=0
 for i,v in pairs(t) do
  if v==c then return true
  elseif v>c then t[i]=0
  elseif v>max then imax,max=i,v
  end
 end
 if max==0 then return false end
 t[imax]=0
 return a(copy(t),c-max) or a(t,c)
end

inb4 stack overflow

38. just hire someone who reads pinyin...
39. tl;dr sorry
40. sentence fragment lol. sounds like boring problem so i'll pass

edit2: is flacko's 37 the same as mine? (while loop in place of crazy recursion)

edited 2×, last 15.01.13 10:00:08 am

Re: The Ultimate Lua Challenge

gotya2
GAME BANNED Off

Offline

15.01.13 12:49:02 pm

Lee has written

The outcome you gave in the example is not correct..
1. Sherman calls bob at minute 0, frank at minute 1. Frank calls emily at minute 1, and matt at minute 2. Process takes 2 minutes.
2. Sherman calls frank at minute 0, frank calls emily at minute 0 and matt at minute 1, sherman calls bob at minute 1. Whole process takes 1 minute.
The rule was that you could only "make" a phonecall once a minute, but can you receive and then make a phonecall ?
i.e. the problem is not very clear to me.

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

17.01.13 11:17:26 pm

I originally intended that a person can only make or receive one call per minute, but without loss of generality, if you are allowed to receive and then immediately make a call once a minute, then your optimal would be the optimal of the former minus the height of the chain or command hierarchy tree.

Re: The Ultimate Lua Challenge

FlooD
GAME BANNED Off

Offline

18.01.13 08:40:31 am

wat happened to cs2d.pastebin.com

btw question:

which is better?

1
2
3
4
5

local a
for _,b in pairs(mytable) do
	a=myfunction(b)
	--do something with a
end

1
2
3
4

for _,b in pairs(mytable) do
	local a=myfunction(b)
	--do something with a
end

i supposed the latter is better but idk

edited 1×, last 18.01.13 09:51:10 am

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

20.01.13 08:15:57 am

the latter is slightly faster, the former is an implicit table look up and the latter is stored directly onto one of the 255 registers allocated for you in Lua.

41. (Intermediate) A zig-zag permutation of order n is a permutation of the collection of numbers {1,2,...,n} such that within the zig-zag permutation, the first element is smaller than the second element, the second element is larger than the third element, the third element is smaller than the fourth, the fourth larger than the fifth, and so on, for n odd.

For example, one such permutation for n=9 is the following set: {4,8,6,7,5,9,1,3,2}
IMG:https://quicklatex.com/cache3/ql_8080fdda4f1b34cbcf423d093643f9a7_l3.png

IMG:https://quicklatex.com/cache3/ql_8080fdda4f1b34cbcf423d093643f9a7_l3.png

Count the number of zig-zag permutations for all odd n less than 11. (Answer: {1,2,16,272,7936,353792})

42. (Trivial) If I give you an arbitrary n, construct one such valid zig-zag permutation. Hint: A particular path through a binary tree with the property that the root node is always larger than the two sub-child can generate this; can you find it?

43. (Extremely challenging) Desiree had a calculus problem: generate the taylor expansion of tan(x). He found a startling discovery

tan(x) = 1 (z/1!) + 2 (z^3/3!) + 16 (z^5/5!) + 272 (z^7/7!) + 7936 (z^9/9!) + 353792 (z^11/11!)

Suppose that we use the max-rooted binary tree encoding above, we can reduce the problem of counting all such permutations to finding all possible max-rooted binary trees with n nodes. One such combinatorial generator is specified by

Tree = Atom + (Tree, max, Tree)

This translates into a "generating function" y(x) = x + integral_0^z T(w)^2 dw. Use this to argue that tan(x) generates the zig-zag permutations and write a general function that approximates the number of zig-zag permutations for any arbitrary n.

edited 3×, last 31.01.13 10:10:40 pm

Re: The Ultimate Lua Challenge

gotya2
GAME BANNED Off

Offline

27.01.13 03:02:06 pm

For 5 numbers, i count more than 6..
{3,5,2,4,1} {3,5,1,4,2} {3,4,1,5,2} {3,4,2,5,1} {2,5,1,4,3}
{2,5,3,4,1} {1,5,3,4,2} { 1,5,2,4,3} etc... what's up with that?

Re: The Ultimate Lua Challenge

Lee
Moderator Off

Offline

31.01.13 10:10:09 pm

Oops, I mean to write 16 instead of 6 there.

Re: The Ultimate Lua Challenge

Infinite Rain
Reviewer Off

Offline

31.01.13 10:20:11 pm

It's ultimate math challenge.
Wouldn't be that enough?

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343

turretmod = {}
turretmod.spawnedturrets = {}
turretmod.spawnedprojectiles = {}
turretmod.buildingsites = {}
turretmod.player = {}
turretmod.game = {
	helptime = 0;
	helptimec = 12;
}

dofile('sys/lua/turretmod/turrets.lua')
dofile('sys/lua/turretmod/projectiles.lua')
dofile('sys/lua/turretmod/effects.lua')

addhook('always', 'turretmod.hook.always')
addhook('objectkill', 'turretmod.hook.objectkill')
addhook('ms100', 'turretmod.hook.ms100')
addhook('join', 'turretmod.hook.join')
addhook('team', 'turretmod.hook.team')
addhook('kill', 'turretmod.hook.kill')
addhook('menu', 'turretmod.hook.menu')
addhook('buildattempt', 'turretmod.hook.buildattempt')

--Functions
turretmod.func = {
	spawnturret = function(turret, x, y, objecttype, team, pl)
		if not tile(x, y, 'walkable') then return end
		parse('spawnobject '.. objecttype ..' '.. x ..' '.. y ..' 0 0 '.. team ..' '.. pl)
		local objectid
		for k, v in pairs(object(0, 'table')) do
			if object(v, 'tilex') == x and object(v, 'tiley') == y then
				objectid = v
			end
		end
		local tbl = {
			baseimage = image(turretmod.turrets[turret].baseimage, x * 32 + 16, y * 32 + 16, 1);
			lightimage = image('gfx/sprites/flare1.bmp', x * 32 + 16, y * 32 + 16, 1);
			gunimage = image(turretmod.turrets[turret].gunimage, x * 32 + 16, y * 32 + 16, 1);
			rootrotation = player(pl, 'rot');
			rotationvariable = 0;
			turret = turret;
			x = x;
			y = y;
			team = team;
			player = pl;
			objectid = objectid;
			time = 0;
			timec = 0;
			locked = false;
			length = 400;
		}
		tween_rotateconstantly(tbl.lightimage, 2)
		imageblend(tbl.lightimage, 1)
		local color = (team == 1 and {255, 0, 0} or {0, 0, 255})
		imagecolor(tbl.lightimage, color[1], color[2], color[3])
		table.insert(turretmod.spawnedturrets, tbl)
		if turretmod.turrets[turret].onspawn then turretmod.turrets[turret].onspawn(#turretmod.spawnedturrets) end
	end;
	
	spawnturretbs = function(turret, x, y, objecttype, team, pl, time)
		if not tile(x, y, 'walkable') then return end
		parse('spawnobject 10 '.. x ..' '.. y ..' 0 0 '.. team ..' '.. pl)
		local objectid
		for k, v in pairs(object(0, 'table')) do
			if object(v, 'tilex') == x and object(v, 'tiley') == y then
				objectid = v
			end
		end
		local tbl = {
			turret = turret;
			x = x; 
			y = y;
			ot = objecttype;
			id = objectid;
			team = team;
			pl = pl;
			time = os.clock();
			timec = time;
		}
		table.insert(turretmod.buildingsites, tbl)
	end;
	
	spawnprojectile = function(projectile, x, y, dir, team, pl, speed, livelength) 
		local tbl = {
			image = image(turretmod.projectiles[projectile].image, x, y, 1);
			speed = speed;
			x = x;
			y = y;
			team = team;
			player = pl;
			dir = dir;
			projectile = projectile;
			livelength = livelength or "0";
		}
		table.insert(turretmod.spawnedprojectiles, tbl)
		if turretmod.projectiles[projectile].onspawn then turretmod.projectiles[projectile].onspawn(#turretmod.spawnedprojectiles) end
	end;
	
	spawnlaser = function(laser, x1, y1, x2, y2, team, pl)
		local dir = -math.deg(math.atan2(x1 - x2, y1 - y2))
		local dist = math.sqrt((x1 - x2)^2 + (y1 - y2)^2)
		local x, y = x1 + math.sin(math.rad(dir)) * dist/2, y1 - math.cos(math.rad(dir)) * dist/2
		local img = image(turretmod.lasers[laser].image, x, y, 1)
		imageblend(img, 0)
		imagepos(img, x, y, dir)
		imagescale(img, 1, dist)
		imagecolor(img, turretmod.lasers[laser].r, turretmod.lasers[laser].g, turretmod.lasers[laser].b)
		tween_alpha(img, 1000, 0)
		timer(1000, "freeimage", img)
		local ex, ey = math.sin(math.rad(dir)), -math.cos(math.rad(dir))
		local exc = {v}
		for k, v in pairs(player(0, 'team'.. team)) do
			local xx, yy = x1, y1
			for i = 1, math.floor(dist) do
				xx = xx + ex
				yy = yy + ey
				if math.floor(xx / 32) == player(v, 'tilex') and math.floor(yy / 32) == player(v, 'tiley') then
					turretmod.lasers[laser].onplayercollision(v, pl)
					table.insert(exc, v)
					break
				end
			end
		end
	end;
	
	affect = function(player, effect, source)
		for k, v in pairs(turretmod.player[player].effects) do
			if v.effect == effect then return end
		end
		local tbl = {
			effect = effect;
			expiretime = os.clock() + turretmod.effects[effect].time;
			source = source
		}
		table.insert(turretmod.player[player].effects, tbl)
		if turretmod.effects[effect].onspawn then turretmod.effects[effect].onspawn(player, #turretmod.player[player].effects) end
	end;
	
	killprojectile = function(pid)
		if not turretmod.spawnedprojectiles[pid] then return end
		if turretmod.projectiles[turretmod.spawnedprojectiles[pid].projectile].ondeath then turretmod.projectiles[turretmod.spawnedprojectiles[pid].projectile].ondeath(pid) end
		freeimage(turretmod.spawnedprojectiles[pid].image)
		table.remove(turretmod.spawnedprojectiles, pid)
	end;
	
	freeline = function(x1, y1, x2, y2)
		local dir = -math.deg(math.atan2(x1 - x2, y1 - y2))
		local ex, ey = math.sin(math.rad(dir)), -math.cos(math.rad(dir))
		x1 = x1 + ex * 23
		y1 = y1 + ey * 23
		repeat 
			x1 = x1 + ex
			y1 = y1 + ey
			if tile(math.floor(x1 / 32), math.floor(y1 / 32), "wall") then
				return false
			end
		until math.floor(x1 / 32) == math.floor(x2 / 32) and math.floor(y1 / 32) == math.floor(y2 / 32)
		return true
	end;
	
	openturretmenu = function(id, page, x, y)
		local page = page or 1
		local pages = math.ceil(#turretmod.turrets / 5)
		local str = "Build Turret P".. page
		for i = page * 5 - 4, page * 5 do
			str = str .. (turretmod.turrets[i] and ','.. turretmod.turrets[i].name ..'|Price: '.. turretmod.turrets[i].price or ',')
		end
		str = str ..','.. (page == pages and ',(Next Page -->)' or ',Next Page -->')
		str = str .. (page == 1 and ',(<-- Previous Page)' or ',<-- Previous Page')
		menu(id, str)
	end;
}

--Hooked functions
turretmod.hook = {
	join = function(id)
		turretmod.player[id] = {joined = false};
	end;
	
	team = function(id, team, look)
		if player(id, 'bot') then return end
		if not turretmod.player[id].joined then
			turretmod.player[id] = {
				joined = true;
				turretpoints = 0;
				bposx = 0;
				bposy = 0;
				effects = {};
			}
			parse('hudtxt2 '.. id ..' 1 "'.. string.char(169) ..'000150255Turret points: 0" 5 410 0')
		end
	end;
	
	kill = function(killer, victim, weapon, x, y)
		turretmod.player[killer].turretpoints = turretmod.player[killer].turretpoints + 1
		parse('hudtxt2 '.. killer ..' 1 "'.. string.char(169) ..'000150255Turret points: '.. turretmod.player[killer].turretpoints ..'" 5 410 0')
	end;
	
	always = function()
		for k, v in pairs(turretmod.spawnedturrets) do
			if v.timec + turretmod.turrets[v.turret].shootingcooldown < os.clock() and not v.locked then
				v.rotationvariable = v.rotationvariable + 1
				imagepos(v.gunimage, v.x * 32 + 16, v.y * 32 + 16, v.rootrotation + math.sin(math.rad(v.rotationvariable)) * 60)
			end
		end	
		
		for k, v in pairs(turretmod.spawnedprojectiles) do
			local nx, ny = v.x + math.sin(math.rad(v.dir)) * v.speed, v.y - math.cos(math.rad(v.dir)) * v.speed
			if tile(math.floor(nx / 32), math.floor(ny / 32), 'wall') then
				if turretmod.projectiles[v.projectile].onwallcollision then turretmod.projectiles[v.projectile].onwallcollision(k) end
			end
			for _, id in pairs(player(0, 'table')) do
				if player(id, 'tilex') == math.floor(nx / 32) and player(id, 'tiley') == math.floor(ny / 32) then
					if turretmod.projectiles[v.projectile].onplayercollision then turretmod.projectiles[v.projectile].onplayercollision(k, id) end
				end
			end
			v.x = nx
			v.y = ny
			imagepos(v.image, v.x, v.y, v.dir)	
			
			if v.livelength ~= "0" then
				v.livelength = v.livelength - v.speed
				if v.livelength <= 0 then
					turretmod.func.killprojectile(k)
				end
			end
		end
	end;
	
	objectkill = function(id, pl)
		for k, v in pairs(turretmod.spawnedturrets) do
			if v.objectid == id then
				freeimage(v.gunimage)
				freeimage(v.baseimage)
				freeimage(v.lightimage)
				turretmod.turrets[v.turret].ondeath(k)
				table.remove(turretmod.spawnedturrets, k)
				break
			end
		end
		
		for k, v in pairs(turretmod.buildingsites) do
			if v.id == id then
				table.remove(turretmod.buildingsites, k)
				break
			end
		end
	end;
	
	ms100 = function()
		for k, v in pairs(turretmod.spawnedturrets) do
			local team = v.team
			if not turretmod.turrets[v.turret].forallies then 
				if team == 1 then team = 2 else team = 1 end
			end
			local closest
			local length = v.length
			for _, id in pairs(player(0, 'team'.. team ..'living')) do
				if player(id, 'x') > (v.x*32+16)-320 and player(id, 'y') > (v.y*32+16)-240 and player(id, 'x') < (v.x*32+16)+320 and player(id, 'y') < (v.y*32+16)+240 then
					if math.sqrt(((v.x * 32 + 16) - player(id, 'x'))^2 + ((v.y * 32 + 16) - player(id, 'y'))^2) <= length then
						if turretmod.func.freeline(v.x*32+16, v.y*32+16, player(id, 'x'), player(id, 'y')) then
							closest = id
							length = math.sqrt(((v.x * 32 + 16) - player(id, 'x'))^2 + ((v.y * 32 + 16) - player(id, 'y'))^2)
						end
					end
				end
			end
			if closest then 
				v.rootrotation = -math.deg(math.atan2(v.x*32+16 - player(closest, 'x'), v.y*32+16 - player(closest, 'y')))
				v.locked = true
				v.rotationvariable = 0
				if v.time + turretmod.turrets[v.turret].shootingcooldown < os.clock() then
					turretmod.turrets[v.turret].whenclosest(k, closest)
					v.time = os.clock()
					v.timec = os.clock()
				end
			else
				v.locked = false
				if v.time + turretmod.turrets[v.turret].shootingcooldown < os.clock() then
					v.time = os.clock()
				end
			end
			imagepos(v.gunimage, v.x * 32 + 16, v.y * 32 + 16, v.rootrotation)
		end
		
		for k, v in pairs(turretmod.buildingsites) do
			if v.time + v.timec < os.clock() then
				parse('killobject '.. v.id)
				turretmod.func.spawnturret(v.turret, v.x, v.y, v.ot, v.team, v.pl)
				table.remove(turretmod.buildingsites, k)
			end
		end
		
		if turretmod.game.helptime + turretmod.game.helptimec < os.clock() then
			turretmod.game.helptime = os.clock()
			parse('hudtxt 0 "'.. string.char(169) ..'000255000Commands: !givepoints [id] [ammount] - Share your turretpoints with another player. | !info [id] - See the information about an another player. | !turretlist - See the list of the avaible turrets." 650 395 0')
			parse('hudtxtmove 0 0 80000 -10000 395')
		end
		
		for k, v in pairs(player(0, 'tableliving')) do
			for _, effect in pairs(turretmod.player[v].effects) do
				if turretmod.effects[effect.effect].update then turretmod.effects[effect.effect].update(v) end
				if os.clock() > effect.expiretime then
					table.remove(turretmod.player[v].effects, _)
					if turretmod.effects[effect.effect].onexpire then turretmod.effects[effect.effect].onexpire(v) end
				end
			end	
		end
	end;
	
	menu = function(id, title, button)
		if title:sub(1, 14) == 'Build Turret P' then
			local page = tonumber(title:sub(15))
			if button == 7 then 
				turretmod.func.openturretmenu(id, page + 1)
			elseif button == 8 then
				turretmod.func.openturretmenu(id, page - 1)
			else
				local i = (page * 5 - 5) + button
				if turretmod.player[id].turretpoints >= turretmod.turrets[i].price then
					local ptx, pty, btx, bty = player(id, 'tilex'), player(id, 'tiley'), turretmod.player[id].bposx, turretmod.player[id].bposy
					if ptx >= btx - 1 and pty >= bty - 1 and ptx <= btx + 1 and pty <= bty + 1 then  
						turretmod.player[id].turretpoints = turretmod.player[id].turretpoints - turretmod.turrets[i].price
						turretmod.func.spawnturretbs(i, turretmod.player[id].bposx, turretmod.player[id].bposy, 5, player(id, 'team'), id, 3) 
						parse('hudtxt2 '.. id ..' 1 "'.. string.char(169) ..'000150255Turret points: '.. turretmod.player[id].turretpoints ..'" 5 410 0')
					else
						msg2(id, string.char(169) .."255000000You can't build like that@C")
					end
				else
					msg2(id, string.char(169) .."255000000You have insufficient funds@C")
				end
			end
		end
	end;
	
	buildattempt = function(id, type, x, y, mode)
		if type == 8 then
			turretmod.player[id].bposx, turretmod.player[id].bposy = x, y
			turretmod.func.openturretmenu(id, 1)
			return 1
		end
	end;
}

Just asking.

Re: The Ultimate Lua Challenge

Alistaire
User Off

Offline

01.02.13 07:39:15 am

Infinite Rain has written

It's ultimate math challenge.
Wouldn't be that enough?

Brag brag olol see my script

Just asking.

We thank you for bragging and at the same time teasing your unreleased turretmod. Actually we don't.

Re: The Ultimate Lua Challenge

Yates
Reviewer Off

Offline

01.02.13 09:32:42 am

I second that. You just spammed a thread for the most ridiculous reason ever. To brag about your code.

You know, good code is shown by what it does, and not what it is. And stop using semi-colons (;), it's completely useless as comma does the exact same thing (Apart from in other coding languages which you probably don't know anyway).

Thanks.

1 2 3 4 5