We can use the following code to solve the Dog,Cat,Mouse puzzle in the tutorial.
dog, cat, mouse = Ints('dog cat mouse')
s = Solver();
s.add(dog>=1)
s.add(cat>=1)
s.add(mouse>=1)
s.add(dog+cat+mouse==100)
s.add(1500 * dog + 100 * cat + 25 * mouse == 10000)
print s.check()
print s.model()
Well, I know I can use
m=s.model
for d in m.decls():
print "%s = %s" % (d.name(), m[d])
to get the names and values of the variables. For example, cat = 41. I wonder can I create a new assertion from the names and values such as cat != 41. I used
s.add(d.name != m[d])
s.add("%s != %s" % (d.name(), m[d]))
But, none of them can work. Any one knows how to add a new assertion from reading the names and values of the model? Many thanks.
In for d in m.decls():, d is a func_decl, i.e., only a declaration, not a variable (constant function) yet, so we need to apply it to it's arguments, which here are empty. Thus we can do:
m = s.model()
for d in m.decls():
v = d() # <-- Note parenthesis ()
print("%s != %s" % (v, m[d]))
s.add(v != m[d])
print(s)
print(s.check())
to get
...
cat != 41
mouse != 56
dog != 3
[dog >= 1,
cat >= 1,
mouse >= 1,
dog + cat + mouse == 100,
1500*dog + 100*cat + 25*mouse == 10000,
41 != cat,
56 != mouse,
3 != dog]
unsat
to get
Related
I have a Lua script that turns a table into segments:
function tablecut(t, n)
local result = {}
local j = 0
for i = 1, #t do
if (i-1) % n == 0 then
j = j + 1
result[j] = {}
end
result[j][#result[j]+1] = t[i]
end
return result
end
output = tablecut({'15', '62', '14', '91', '33', '55', '29', '4'}, 4)
for i = 1, #output do
for j = 1, #output[i] do
io.write(tostring(output[i][j])..' ')
end
print()
end
output:
15 62 14 91
33 55 29 4
And I am trying to find the minima from the cut lists so the output would look like this:
15 62 14 91
min = 14
33 55 29 4
min = 4
Edit: If its of any importance this is how I got it to work on Lua 5.3 but there is no table.move function on Lua 5.1. I can't remember how my thought function worked when I wrote this code.
function indexOf(array, value)
for i, v in ipairs(array) do
if v == value then
return i
end
end
return nil
end
Indicies = {}
Answers = {}
function chunks(lst, size)
local i = 1
local count = 0
return function()
if i > #lst then return end
local chunk = table.move(lst, i, i + size -1, 1, {})
i = i + size
count = count + 1
return count, chunk
end
end
local a = {91,52,19,59,38,29,58,11,717,91,456,49,30,62,43,8,17,15,26,22,13,10,2,23} --Test list
for i, chunk in chunks(a, 4) do
x=math.min(a)
print(string.format("#%d: %s", i, table.concat(chunk, ",")))
table.sort(chunk)
print(math.min(chunk[1]))
table.insert(Answers, chunk[1])
table.insert(Indicies, (indexOf(a, chunk[1])))
Output:
#1: 91,52,19,59
19
#2: 38,29,58,11
11
#3: 717,91,456,49
49
your table cut function could be simplified, and your output for loop needs you use an iterator if you want to get an output simply like you do in your 5.3 script.
function cuttable(t,n)
local binned = {}
for i=1,#t,n do
local bin = {}
for j=1,n do
table.insert(bin, t[i + ((j - 1) % n)])
end
table.insert(binned, bin)
end
return binned
end
For the for loop, we can use ipairs on the output of cuttable keeping things pretty simple, then we just do the same steps of concat then sort and print out our results.
for k, bin in ipairs(cuttable(a,4)) do
local output = "#" .. k .. ":" .. table.concat(bin, ",")
table.sort(bin)
print(output)
print(bin[1])
end
Output
#1:91,52,19,59
19
#2:38,29,58,11
11
#3:717,91,456,49
49
#4:30,62,43,8
8
#5:17,15,26,22
15
#6:13,10,2,23
2
One way to implement the cutting would be using a for loop & unpack. I have handled the case of the length not being divisible by 4 after the for loop to (1) maximize performance (check doesn't need to be done every iteration) and (2) be able to directly pass the values to math.min, which doesn't accept nils.
for i = 1, math.floor(#t / 4), 4 do
print(unpack(t, i, i+4))
print("min = " .. math.min(unpack(t, i, i+4)))
end
-- If #t is not divisible by 4, deal with the remaining elements
local remaining = #t % 4
if remaining > 0 then
print(unpack(t, #t - remaining, remaining))
print("min = " .. math.min(unpack(t, #t - remaining, remaining)))
end
I'm trying to reverse a decode function. This function takes a string and a key and encodes the string with that key. This is the code:
function decode(key, code)
return (code:gsub("..", function(h)
return string.char((tonumber(h, 16) + 256 - 13 - key + 255999744) % 256)
end))
end
If I input 7A as code and 9990 as key, it returns g
I tried reversing the operators and fed back the output of the decode function but I get an error becauase tonumber() returns nil. How can I reverse this function?
By using the answer to this Lua base coverter and flipping the operators of the decode function, I was able to convert back the input.
This is the whole code:
function encodes(key, code)
return (code:gsub("..", function(h)
return string.char((tonumber(h, 16) + 256 - 13 - key + 255999744) % 256)
end))
end
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
function decodes(key, code)
return (code:gsub(".", function(h)
out = (string.byte(h) - 256 + 13 + key - 255999744) % 256
return basen(out,16)
end))
end
a = encodes(9999, "7c7A")
print(a) --prints: `^
print("----------")
b = decodes(9999, a)
print(b) --prints: 7C7A
I have a function that determines whether a value is divisible by 2 or 3, but **NOT** 5:
let ttnf x =
if (x % 2 = 0) || (x % 3 = 0) && not(x % 5 = 0) then true
else
false
I'm getting a weird response from Visual Studio 2015 in the interactive panel.
I execute the above code in the F# interactive panel then enter say...
ttnf 15
Hit enter, nothing...
hit alt + enter then it returns it on the second time.
Any idea why it isn't returning true/false from entering:
ttnf 15
The first time?
Thanks.
#ildjarn commented about the error in your code, but about F# interactive's behavior: when you type code directly into fsi, you need to terminate each declaration with ;; to tell fsi to interpret it, otherwise it will just wait for you to continue your input (as you experienced). So:
> let ttnf x =
if (x % 2 = 0 || x % 3 = 0) && not(x % 5 = 0) then true
else
false;;
val ttnf : x:int -> bool
> ttnf 15;;
val it : bool = false
>
I have the following problem:
Solve the following recurrence relation, simplifying your final answer
using 'O' notation.
f(0)=3
f(1)=12
f(n)=6f(n-1)-9f(n-2)
We know this is a homogeneous 2nd order relation so we write the characteristic equation: a^2-6a+9=0 and the solutions are a1,2=3.
The problem is when I replace these values I get:
f(n)=c1*3^n+c2*3^n
and using the 2 initial relations I have:
f(0)=c1+c2=3
f(1)=3(c1+c2)=12
which gives me that there no values such that c1 and c2 such that these 2 relation are true.
Am I doing something wrong? Is the way it should be solved different when it comes to identical roots for the characteristic equation?
You can't solve it this way, because your matrix A is not diagonalizable.
However, here is what you get if you use Jordan's normal form instead:
f(n) = 3^{n-1}(3n + 9)
The Jordan matrix and the basis (with notation from wikipedia + Octave) is:
J := [3,1;0,3]
P := [3,4;1,1]
such that PJP^{-1} = A, where
A := [6,-9;1,0]
is your recurrence matrix. Furthermore, the Jordan matrix is almost as good as a diagonal matrix for computing powers:
J^n = 3^(n-1) * [3,n;0,3].
The recurrence is then:
[f(n+1); f(n)] = A^n [12,3] = PJ^nP^-1[12,3] = (<whatever>, 3^(n-1)*(3n+9)).
Here a quick numerical check (Scala, but you can take whatever you want, Octave or I whatever you like):
scala> def f(n: Int): Int = { if (n == 0) 3 else if (n == 1) 12 else (6 * f(n-1) - 9 * f(n-2)) }
f: (n: Int)Int
scala> for (i <- 0 until 20) println(f(i))
3
12
45
162
567
1944
6561
21870
72171
236196
767637
2480058
7971615
25509168
81310473
258280326
817887699
^
scala> def explicit(n: Int): Int = (Math.pow(3, n -1) * (3 * n + 9)).toInt
explicit: (n: Int)Int
scala> for (i <- 0 until 20) println(explicit(i))
3
12
45
162
567
1944
6561
21870
72171
236196
767637
2480058
7971615
25509168
81310473
258280326
817887699
Certain problem in transport Phenomena is solved using the following code:
T_max, T_0, S, R, k, I, k_e, L, R, E, a = Reals('T_max T_0 S R k I k_e L R E a')
k = a*k_e*T_0
I = k_e*E/L
S = (I**2)/k_e
eq = T_0 + S* R**2/(4*k)
print eq
equations = [
T_max == eq,
]
print "Temperature equations:"
print equations
problem = [
R == 2, L == 5000,
T_0 == 20 + 273,
T_max == 30 + 273, k_e == 1,
a == 2.23*10**(-8), E > 0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
using this code online we obtain
This output gives the correct answer but there are two issues in the code: a) the expresion named "eq" is not fully simplified and then it is necessary to give an arbitrary value for k_e . My question is: How to simplify the expression "eq" in such way that k_e be eliminated from "eq"?
Other example: To determine the radius of a tube
Code:
def inte(n,a,b):
return (b**(n+1))/(n+1)-(a**(n+1))/(n+1)
P_0, P_1, L, R, mu, q, C = Reals('P_0 P_1 L R mu q C')
k = (P_0 - P_1)/(2*mu*L)
equations = [0 == -k*inte(1,0,R) +C,
q == 2*3.1416*(-(k/2)*inte(3,0,R) + C*inte(1,0,R))]
print "Fluid equations:"
print equations
problem = [
L == 50.02/100, mu == (4.03*10**(-5)),
P_0 == 4.829*10**5, P_1==0,
q == 2.997*10**(-3), R >0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
Output:
Fluid equations:
[-((P_0 - P_1)/(2·mu·L))·(R2/2 - 0) + C = 0, q =
3927/625·
(-(((P_0 - P_1)/(2·mu·L))/2)·(R4/4 - 0) + C·(R2/2 - 0))]
Problem:
[L = 2501/5000, mu = 403/10000000, P_0 = 482900, P_1 = 0, q = 2997/1000000, R > 0]
Solution:
[R = 0.0007512843?,
q = 2997/1000000,
P_1 = 0,
P_0 = 482900,
mu = 403/10000000,
L = 2501/5000,
C = 3380.3149444289?]