I got this question in a programming test. Do you think this question is even correct? Look at the answer choices. (2^x means 2 raised to x)
Consider the following pseudocode.
x := 1;
i := 1;
while (x >= 1000)
begin
x := 2^x;
i := i + 1;
end;
What is the value of i at the end of the pseudocode?
a) 4
b) 5
c) 6
d) 7
e) 8
I am sure that the value of i will 1. I told the examiner of the discrepancy and he advised me the leave the question unanswered if I felt it was incorrect. What else could I have done?
1
X < 1000, so it doesn't enter the while.
Or there is an error in the Question (and X should be <= 1000 and not >=1000)
If it's <= 1000 it should be 5:
2 - 4 - 16 - 65K
2 - 3 - 4 - 5
This question tests two things:
can you read code
can you communicate / interact
Since you asked about the discrepancy, you showed 1. to be true. I'm not so sure if you passed 2, it depends too much on the situation / expectations.
I believe I would have left a note on the answer sheet stating 'none of the given'.
Not an easy situation!
As written, the answer would be 1.
Had the test on the while been reversed (i.e. x < 1000), then the series is:
At the end of each loop iteration
i = 2, x = 2
i = 3, x = 2^2 = 4
i = 4, x = 2^4 = 16
i = 5, x = 2^16 = 65,536
So i would be 5
If I where you, I would say that it is none of the above and basically say that since x is less than 1000 when the while loop starts, the value of i is never modified. I think that it is a bad thing to leave blank answers in a quiz, it is always better to write something which you think is relevant. If you think that there is a mistake, you can always state an assumption before stating your answer, so in this case you can either say that the while loop never works, or else, you explicitly state an assumption, in this case, it would be something like "Assuming that there is a mistake in the question, and that "while (x >= 1000)" should in fact be "while (x <= 1000)"...
Then, you proceed with your working.
Related
I have to use borland TurboC++ for C programming in my college.
They say our examination board recommends it. I have to use it..
The problem is that they gave this operator precedence related question:
int a=10,b=20,result;
result1 = ++a + b-- - a++ * b++ + a * ++b;
printf("result=%d",);
printf("\n a=%d",a);
printf("\n b=%d",b);
Other compilers like gcc can't perform this operation. But turbo C can and gives us:
result=32
a=12
b=21
I made mistake in my test. My teacher tried to explain what's going on. But I am not convinced. Is it some kind of weird behavior of turbo C or in older days it used to be totally fine with all compilers. If so, what are the steps to understand what is going on and how to understand.
To solve these kind of problem, turbo-c do it in manner as follows :
1) Consider the initial value of variables used.
a=10
b=20
2) Count all the pre-increment and decrements for each variable and store all post on stack separate for each variable.
for variable a
pre increment = 1 therefore change the value of a to 11
post = 1 stored to stack
for variable b
pre increment = 1 therefore change the value of b to 21
post = 2 stored to stack
3) Now replace all the pre and post with the current value of a and b
result = 11 + 21 - 11 * 21 + 11 * 21 ;
result = 11 + 21;
result = 32;
4) lastly pop the stack and perform the operation on the variable.
a = 12
b = 21
This the only way to solve this problem. You can check the procedure with any question of same kind. The result will came out same. g++ fails to solve because it probably cannot resolve the variable in the same way thus the precedence error came in picture. It might probably fail with ++ + and -- - because it cannot understand the increment or decrements operator and forms ambiguous trees.
I have a bit of math question for my sql code. I want to be able to work out the following in my code using a mathematical formula, if anyone knows how I would love to know.
Any number 1->99 : 10
Any number 100->999 : 100
Any number 1000->9999 : 1000
...
Is there anyway to work out the 10 multiplier just from the value? I feel like this should be an easy formula but I cant seem to get it.
Thanks
How about this?
SELECT POWER(10, CONVERT(INT, LOG10(#Input)))
It takes the log base 10 of the input value (which returns the value of the exponent to which you would have to raise 10 to in order to get the input value), then it lops off the decimal portion leaving only the whole number, and then raises 10 to that power.
You just need logs and their opposite (power)...
power(10, floor(log10(x)))
As follows...
log10(99) = 1.9956351946
floor(1.9956351946) = 1
power(10, 1) = 10
This does, however, assume that your example is wrong and that 1 -> 9 should "round" to 1...
log10(9) = 0.95424250943
floor(0.95424250943) = 0
power(10, 0) = 1
Suppose we have N (in this example N = 3) events that can happen depending on some variables. Each of them can generate certain profit or loses (event1 = 300, event2 = -100, event3 = 200), they are constrained by rules when they happen.
event 1 happens only when x > 5,
event 2 happens only when x = 2 and y = 3
event 3 happens only when x is odd.
The problem is to know the maximum profit.
Assume x, y are integer numbers >= 0
In the real problem there are many events and many dimensions.
(the solution should not be specific)
My question is:
Is this linear programming problem? If yes please provide solution to the example problem using this approach. If no please suggest some algorithms to optimize such problem.
This can be formulated as a mixed integer linear program. This is a linear program where some of the variables are constrained to be integer. Contrary to linear programs, solving the general integer program is NP-hard. However, there are many commercial or open source solvers that can solve efficiently large-scale problems. For up to 300 variables and constraints, you can use excel's solver.
Here is a way to formulate the above constraints:
If you go down this route, you might find this document useful.
the last constraint in an interesting one. I am assuming that x has to be integer, but if x can be either integer or continuous I will edit the answer accordingly.
I hope this helps!
Edit: L and U above should be interpreted as L1 and U1.
Edit 2: z2 needs to changed to (1-z2) on the 3rd and 4th constraint.
A specific answer:
seems more like a mathematical calculation than a programming problem, can't you just run a loop for x= 1->1000 to see what results occur?
for the example:
as x = 2 or 3 = -200 then x > 2 or 3, and if x < 5 doesn't get the 300, so all that is really happening is x > 5 and x = odd = maximum results.
x = 7 = 300 + 200 . = maximum profit for x
A general answer:
I don't see how to answer the question without seeing what the events are and how the events effect X ? Weather it's a linear or functional (mathematical) answer seems rather beside the point of finding the desired solution.
I have a set of decimal numbers. I need to check if a specific bit is set in each of them. If the bit is set, I need to return 1, otherwise return 0.
I am looking for a simple and fast way to do that.
Say, for example, I am checking if the third bit is set. I can do (number AND (2^2)), it will return 4 if the bit is set, otherwise it will return 0. How do I make it to return 1 instead of 4?
Thank you!
if ((number AND (2^bitnumber) <> 0) then return 1 else return 0 end if
If you can change your return type to boolean then this is more elegant
return ((number AND (2^bitnumber)) <> 0)
While the division solution is a simple one, I would think a bit-shift operation would be more efficient. You'd have to test it to be sure, though. For instance, if you are using 1 based bit indexes, you could do this:
Dim oneOrZero As Integer = (k And 2 ^ (n - 1)) >> (n - 1)
(Where k is the number and n is the bit index). Of, if you are using 0 based bit indexes, you could just do this:
Dim oneOrZero As Integer = (k And 2 ^ n) >> n
Sorry, guys, I am too slow today.
To test a bit number "n" in a decimal number "k":
(k AND 2^(n-1))/(2^(n-1))
will return 1 if the bit is set, otherwise will return 0.
=====================================================
Hi again, guys!
I compared the performance of the three proposed solutions with zero-based indexes, and here are the results:
"bit-shift solution" - 8.31 seconds
"if...then solution" - 8.44 seconds
"division solution" - 9.41 seconds
The times are average of the four consecutive runs.
Surprisingly for me, the second solution outperformed the third one.
However, after I modified the "division solution" this way:
p = 2 ^ n : oneOrZero = (k And p) / p
it started to run in 7.48 seconds.
So, this is the fastest of the proposed solutions (despite of what Keith says :-).
Thanks everybody for the help!
I really don't know if it can help anyone more than the above, but, here we go.
When I need to fast check a bit in number I compare the decimal-value of this bit directly.
I mean, if I would need to see of the 6th bit is on (32), I check its decimal value, like this:
if x and 32 = 32 then "the bit is ON"
Try for instance check 38 with 32, 4 and 2... and the other bits.
You will see only the actual bits turned on.
I hope it can help.
Yes! Simply use a bit mask. I enumerate the bits, then AND the number with the bit value. Very little math on the PC side as it uses lookup tables instead. The AND basically shuts off all the other bits except the one you are interested in. Then you check it against itself to see if it's on/off.
Enum validate
bit1 = 1
bit2 = 2
bit3 = 4
bit4 = 8
bit5 = 16
bit6 = 32
bit7 = 64
bit8 = 128
End Enum
If num And validate.bit3 = validate.bit3 Then true
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.