If x = y And Not (z(0) = w(0) And z(1) = w(1) And z(2) = w(2)) then .....
I want to test if x = y and z(0) is not equal to w(0) and z(1) is not equal to w(0) and z(2) is not equal to w(2). Basicly if x=y and any of the rest is not equal to the other it should do the code in the if-statement
Will one And Not (....) work ?
Let me know if you have alternative solutions or if this will work thanks for the help
Your code is not doing what you say you want. It is checking if x = y and any of the other terms are unequal. Think about it, your if statement is asking if two things are true, the first is that x = y and the second is that the expression in parentheses is not true (and the expression in parentheses will not be true if and of the z and w terms are unequal).
If you want to check that all the other terms are unequal, you need
If x = y And z(0) <> w(0) And z(1) <> w(1) And z(2) <> w(2) then ...
Related
How the game works is that there is a 3-digit number, and you have to guess it. If you guess a digit in the right spot, you get a strike, and if you guess a digit but in the wrong spot you get a ball. I've coded it like this.
x = random.randint(1, 9)
y = random.randint(1, 9)
z = random.randint(1, 9)
userguessunlisted = input('What number do you want to guess?')
numbertoguess = list[x, y, z]
userguess = list(userguessunlisted)
b = 0
s = 0
while 0 == 0:
if userguess[0] == numbertoguess[0]:
s = s + 1
if userguess[0] == numbertoguess[1]:
b = b + 1
if userguess[0] == numbertoguess[2]:
b = b + 1
if userguess[1] == numbertoguess[0]:
b = b + 1
if userguess[1] == numbertoguess[1]:
s = s + 1
if userguess[1] == numbertoguess[2]:
b = b + 1
if userguess[2] == numbertoguess[0]:
b = b + 1
if userguess[2] == numbertoguess[1]:
b = b + 1
if userguess[2] == numbertoguess[2]:
s = s + 1
print(s + "S", b + "B")
if s != 3:
b = 0
s = 0
else:
print('you win!')
break
When you said list[x, y, z] on line 5, you used square brackets, which python interprets to be a type annotation. For example, if I wanted to specify that a variable is a list of ints, I could say
my_list_of_ints: list[int] = [1, 2, 3]
I think what you meant to do is create a new list from x, y, and z. One way to do this is
numbertoguess = list([x, y, z])
which is probably what you meant to write. This is valid because the list function takes an iterable as its one and only argument.
However, the list portion is redundant; square brackets on the right-hand side of an assignment statement already means "create a list with this content," so instead you should simply say
numbertoguess = [x, y, z]
A few other notes:
input will return a string, but you are comparing that string to integers further down, so none of the comparisons will ever be true. What you want to say is something like the following:
while True:
try:
userguessunlisted = int(input('What number do you want to guess?'))
except:
continue
break
What this code does is attempts to parse the string returned from input into an int. If it fails to do so, which would happen if the user inputted something other than a valid integer, an exception would be thrown, and the except block would be entered. continue means go to the top of the loop, so the input line runs repeatedly until a valid int is entered. When that happens, the except block is skipped, so break runs, which means "exit the loop."
userguessunlisted is only ever going to contain 1 number as written, so userguess will be a list of length 1, and all of the comparisons using userguess[1] and userguess[2] will throw an IndexError. Try to figure out how to wrap the code from (1) in another loop to gather multiple guesses from the user. Hint: use a for loop with range.
It might also be that you meant for the user to input a 3-digit number all at once. In that case, you can use a list comprehension to grab each character from the input and parse it into a separate int. This is probably a bit complicated for a beginner, so I'll help you out:
[int(char) for char in input('What number do you want to guess?')]
print(s + "S", b + "B") will throw TypeError: unsupported operand type(s) for +: 'int' and 'str'. There are lots of ways to combine non-string types with strings, but the most modern way is using f-strings. For example, to combine s with "S", you can say f"{s}S".
When adding some amount to a variable, instead of saying e.g. b = b + 1, you can use the += operator to more concisely say b += 1.
It's idiomatic in python to use snake_case for variables and Pascal case for classes. So instead of writing e.g. numbertoguess, you should use number_to_guess. This makes your code more readable and familiar to other python programmers.
Happy coding!
Here's my Fibonacci code using python 3.5
z = 0
x = 0
y = 1
while z != 317811:
x = x + y
z = x
print (z)
y = x + y
z = y
print (z)
I am wondering why this prints to infinity when setting the condition to
z != 317811
but works when it is below this number like
z != 196418
or a number greater than this like
z!= 514229
I tried a different approach (z <= 317811) but it prints up to 514229.
Thank you for your time.
KD
You're only testing alternate Fibonnaci numbers as the stopping condition: 317811 is getting missed.
One fix would be to test both x and y.
this is just logical problem
you are printing two
z != 317811
for this condition
"z"
is updated twice once in first z assignment i.e z = x
but "z" again get updated at second assignment z = y and then "z" is not sutisfying the condition(z != 317811) and not equal to 317811 but it is now 514229
Note: it will always work for number being printed at the second steps as this value of Z will be compared in while condition in loop
You are increasing z value twice a loop, but only checking once.
What actually is happening is that z is increasing with the fibonacci serie. Last values of z are:
196418
317811
514229
But you are only checking the stop condition once every two assignment. In this case you are checking that 196418 != 317811 and 514229 != 317811, thus never matching it.
One possible fix could be to test if z != 317811 after the first print. Even if in this case I would prefer testing "<" instead of "!="
I'm doing a query on a complex db:
SELECT *
FROM
table1,
table2,
table3,
table4,
table5,
table6,
table7,
table8
WHERE
a = b and
c = d and
e = d and
(
(strfldvar = 'BROKEN_ARROW' AND x = g)
OR (strfldvar = 'BROKEN_BOX' AND y = g)
) and
f = h and
i = j
Only works when strfldvar = 'BROKEN_BOX' and not when strfldvar = 'BROKEN_ARROW'. When I replace
(
(strfldvar = 'BROKEN_ARROW' AND x = g)
OR (strfldvar = 'BROKEN_BOX' AND y = g)
) and
with either x = g and or y = g and it works fine in two seperate queries runs like that. The error message for the case strfldvar = 'BROKEN_ARROW' is:
ORA-01013: user requested cancel of current operation
Before this error message comes the computer goes into deep thought for I guess 2 minutes.
What am I doing wrong here?
f.y.i. I looked at the names of the fields of the of the two seperate runs and they appear idendical. I mean the scema of the output looks the same for both. But I'm not 100% sure they are the same, if that matters i.e.
Thanks for your help
When strfldvar = 'BROKEN_ARROW' AND x = g (or if strfldvar is not BROKEN_ARROW or BROKEN_BOX), the y = g part is not evaluated, which seems to be causing the query to run for longer than you expect - until it's eventually killed by you, your client or resource limits. I suspect that's the only join condition for whichever table y is from, so you end up with a cartesian product.
When strfldvar = 'BROKEN_BOX' then both x = g and y = g will be evaluated, so you wouldn't get the same cartesian product, against either of the tables providing x and y.
If you are essentially deciding which table to include in the query based on that flag then you'll need to redesign this; possibly with a union of two queries, one which joins to x and the other on y; or with separate queries and you decide which to run; or maybe even with outer joins. But it depends on what you're really trying to do and what the data looks like. The code you have shown is a too generic to guess what will be appropriate.
I'm sorry the title is so confusingly worded, but it's hard to condense this problem down to a few words.
I'm trying to find the minimum value of a specific equation. At first I'm looping through the equation, which for our purposes here can be something like y = .245x^3-.67x^2+5x+12. I want to design a loop where the "steps" through the loop get smaller and smaller.
For example, the first time it loops through, it uses a step of 1. I will get about 30 values. What I need help on is how do I Use the three smallest values I receive from this first loop?
Here's an example of the values I might get from the first loop: (I should note this isn't supposed to be actual code at all. It's just a brief description of what's happening)
loop from x = 1 to 8 with step 1
results:
x = 1 -> y = 30
x = 2 -> y = 28
x = 3 -> y = 25
x = 4 -> y = 21
x = 5 -> y = 18
x = 6 -> y = 22
x = 7 -> y = 27
x = 8 -> y = 33
I want something that can detect the lowest three values and create a loop. From theses results, the values of x that get the smallest three results for y are x = 4, 5, and 6.
So my "guess" at this point would be x = 5. To get a better "guess" I'd like a loop that now does:
loop from x = 4 to x = 6 with step .5
I could keep this pattern going until I get an absurdly accurate guess for the minimum value of x.
Does anybody know of a way I can do this? I know the values I'm going to get are going to be able to be modeled by a parabola opening up, so this format will definitely work. I was thinking that the values could be put into a column. It wouldn't be hard to make something that returns the smallest value for y in that column, and the corresponding x-value.
If I'm being too vague, just let me know, and I can answer any questions you might have.
nice question. Here's at least a start for what I think you should do for this:
Sub findMin()
Dim lowest As Integer
Dim middle As Integer
Dim highest As Integer
lowest = 999
middle = 999
hightest = 999
Dim i As Integer
i = 1
Do While i < 9
If (retVal(i) < retVal(lowest)) Then
highest = middle
middle = lowest
lowest = i
Else
If (retVal(i) < retVal(middle)) Then
highest = middle
middle = i
Else
If (retVal(i) < retVal(highest)) Then
highest = i
End If
End If
End If
i = i + 1
Loop
End Sub
Function retVal(num As Integer) As Double
retVal = 0.245 * Math.Sqr(num) * num - 0.67 * Math.Sqr(num) + 5 * num + 12
End Function
What I've done here is set three Integers as your three Min values: lowest, middle, and highest. You loop through the values you're plugging into the formula (here, the retVal function) and comparing the return value of retVal (hence the name) to the values of retVal(lowest), retVal(middle), and retVal(highest), replacing them as necessary. I'm just beginning with VBA so what I've done likely isn't very elegant, but it does at least identify the Integers that result in the lowest values of the function. You may have to play around with the values of lowest, middle, and highest a bit to make it work. I know this isn't EXACTLY what you're looking for, but it's something along the lines of what I think you should do.
There is no trivial way to approach this unless the problem domain is narrowed.
The example polynomial given in fact has no minimum, which is readily determined by observing y'>0 (hence, y is always increasing WRT x).
Given the wide interpretation of
[an] equation, which for our purposes here can be something like y =
.245x^3-.67x^2+5x+12
many conditions need to be checked, even assuming the domain is limited to polynomials.
The polynomial order is significant, and the order determines what conditions are necessary to check for how many solutions are possible, or whether any solution is possible at all.
Without taking this complexity into account, an iterative approach could yield an incorrect solution due to underflow error, or an unfortunate choice of iteration steps or bounds.
I'm not trying to be hard here, I think your idea is neat. In practice it is more complicated than you think.
I am trying to verify that a variable is NOT equal to either this or that. I tried using the following codes, but neither works:
if x ~=(0 or 1) then
print( "X must be equal to 1 or 0" )
return
end
if x ~= 0 or 1 then
print( "X must be equal to 1 or 0" )
return
end
Is there a way to do this?
Your problem stems from a misunderstanding of the or operator that is common to people learning programming languages like this. Yes, your immediate problem can be solved by writing x ~= 0 and x ~= 1, but I'll go into a little more detail about why your attempted solution doesn't work.
When you read x ~=(0 or 1) or x ~= 0 or 1 it's natural to parse this as you would the sentence "x is not equal to zero or one". In the ordinary understanding of that statement, "x" is the subject, "is not equal to" is the predicate or verb phrase, and "zero or one" is the object, a set of possibilities joined by a conjunction. You apply the subject with the verb to each item in the set.
However, Lua does not parse this based on the rules of English grammar, it parses it in binary comparisons of two elements based on its order of operations. Each operator has a precedence which determines the order in which it will be evaluated. or has a lower precedence than ~=, just as addition in mathematics has a lower precedence than multiplication. Everything has a lower precedence than parentheses.
As a result, when evaluating x ~=(0 or 1), the interpreter will first compute 0 or 1 (because of the parentheses) and then x ~= the result of the first computation, and in the second example, it will compute x ~= 0 and then apply the result of that computation to or 1.
The logical operator or "returns its first argument if this value is different from nil and false; otherwise, or returns its second argument". The relational operator ~= is the inverse of the equality operator ==; it returns true if its arguments are different types (x is a number, right?), and otherwise compares its arguments normally.
Using these rules, x ~=(0 or 1) will decompose to x ~= 0 (after applying the or operator) and this will return 'true' if x is anything other than 0, including 1, which is undesirable. The other form, x ~= 0 or 1 will first evaluate x ~= 0 (which may return true or false, depending on the value of x). Then, it will decompose to one of false or 1 or true or 1. In the first case, the statement will return 1, and in the second case, the statement will return true. Because control structures in Lua only consider nil and false to be false, and anything else to be true, this will always enter the if statement, which is not what you want either.
There is no way that you can use binary operators like those provided in programming languages to compare a single variable to a list of values. Instead, you need to compare the variable to each value one by one. There are a few ways to do this. The simplest way is to use De Morgan's laws to express the statement 'not one or zero' (which can't be evaluated with binary operators) as 'not one and not zero', which can trivially be written with binary operators:
if x ~= 1 and x ~= 0 then
print( "X must be equal to 1 or 0" )
return
end
Alternatively, you can use a loop to check these values:
local x_is_ok = false
for i = 0,1 do
if x == i then
x_is_ok = true
end
end
if not x_is_ok then
print( "X must be equal to 1 or 0" )
return
end
Finally, you could use relational operators to check a range and then test that x was an integer in the range (you don't want 0.5, right?)
if not (x >= 0 and x <= 1 and math.floor(x) == x) then
print( "X must be equal to 1 or 0" )
return
end
Note that I wrote x >= 0 and x <= 1. If you understood the above explanation, you should now be able to explain why I didn't write 0 <= x <= 1, and what this erroneous expression would return!
For testing only two values, I'd personally do this:
if x ~= 0 and x ~= 1 then
print( "X must be equal to 1 or 0" )
return
end
If you need to test against more than two values, I'd stuff your choices in a table acting like a set, like so:
choices = {[0]=true, [1]=true, [3]=true, [5]=true, [7]=true, [11]=true}
if not choices[x] then
print("x must be in the first six prime numbers")
return
end
x ~= 0 or 1 is the same as ((x ~= 0) or 1)
x ~=(0 or 1) is the same as (x ~= 0).
try something like this instead.
function isNot0Or1(x)
return (x ~= 0 and x ~= 1)
end
print( isNot0Or1(-1) == true )
print( isNot0Or1(0) == false )
print( isNot0Or1(1) == false )