function a(m, n:integer):integer
begin
if m = 0 then a := n + 1 ;
if n = 0 then
a := a(m-1, 1)
else
a := a(m-1, a(m, n-1));
end;
I can't figure out why this function does not working. I made some test and everything crashes when it get's to a(0-1, a(0,1-1)) but it this means it comes from a(0,1) ... this a(0,1) should go with the first if if m = 0 then a := n +1; but it goes with the else case instead.
You are assuming that the second if is only executed when the first one fails, but nothing in your code indicates that is what should happen.
Related
I need to write a PL/SQL program to print out the first 10 numbers of Lazy Caterer's sequence. I'm having trouble. I don't understand how to do this
Lazy caterer's sequence has the formula F(1)=2, F(n)=F(n-1)+n.
E.g., F(2) = F(1) + 2 = 2+2=4; F(3) = F(2)+3 = 4+3=7
Please use a loop. The numbers you print out should look like
2
4
7
...
56
DECLARE
n NUMBER := 1;
BEGIN
FOR i IN 1..10 LOOP
n := n + i;
DBMS_OUTPUT.PUT_LINE(n);
END LOOP;
END;
/
I cheated and looked up Lazy Caterer's algorithm on Wikipedia. This is an implementation of the simplistic version of the formula:
create or replace function lc_algo
(n in number)
return number as
begin
return (power(n,2) + n + 2) / 2;
end;
/
I admit it doesn't use a loop, but it's hard to see why anybody would need to. However, if your assignment insists on loops (because performance is not your bag) here you go:
create or replace function lc_algo
(n in number)
return number
as
rv simple_integer := 1;
begin
for i in 1..n loop
rv := rv + i;
end loop;
return rv;
end;
/
To get the first ten numbers (with whatever version):
select lc_algo(level) -- assuming you don't want to start with f(0)
from dual
connect by level <= 10
/
I've beeng having some trouble with this code... I need to create an algorithm which makes the user input a number (X), and then the program calculates the sum of all the odd numbers below (x).
This what I've tried so far, but can't really wrap my head around the logic behind it:
Program odd_numbers;
Var
Num, Limite, Soma: integer;
Begin;
Soma := 0;
Writeln('Choose a limit:');
Readln(Limite);
While (Limite / 2 > 0) do
Begin;
Soma := ((Num < Limite) mod 2 > 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
if (Limite mod 2 = 0) then
Begin;
Soma := ((Num < Limite) mod 2 = 0);
Writeln('The sum of odd numbers from 0 to ', Limite, ' é ', Soma);
End;
End.
*PS: Been writing the code with variables in Portuguese, so don't mind the variables appearing weird to understand. *
I see that everyone is happily looping, but this is not necessary. This is a simple arithmetic sequence, and the sum can be calculated without a loop.
Just think of the following:
1 + 3 = 2 * (1 + 3) / 2 = 2 * 2 = 4 ; limits 3 and 4
1 + 3 + 5 = 3 * (1 + 5) / 2 = 3 * 3 = 9 ; limits 5 and 6
1 + 3 + 5 + 7 = 4 * (1 + 7) / 2 = 4 * 4 = 16 ; limits 7 and 8
1 + 3 + 5 + 7 + 9 = 5 * (1 + 9) / 2 = 5 * 5 = 25 ; limits 9 and 10
1 + 3 + 5 + 7 + 9 + 11 = 6 * (1 + 11) / 2 = 6 * 6 = 36 ; limits 11 and 12
But not only that, you'll see that it is in fact always a perfect square: Sqr((n+1) div 2).
So just calculate:
program odd_numbers;
var
Num, Limite, Soma: Integer;
begin
Write('Choose a limit: ');
Readln(Limite);
Num := (Limite + 1) div 2;
Soma := Num * Num;
Writeln('The sum of odd numbers from 0 to ', Limite, ' is ', Soma);
end.
Looks a little simpler than what the others propose.
The loop While (Limite / 2 > 0) do ... uses real arithmetic and not integer arithmetic. I guess you mean While (Limite div 2 > 0) do ... And you should change Limite in the loop otherwise you get stuck because the exit condition can never be reached.
After you have asked the user to enter a number, Limite, you need to keep that unchanged, because you need it in the final message. You also need a loop where you go through all numbers from Limite towards 0.
You started with a while loop which is ok, you are just missing the loop control variable. That is a variable that eventually gets a terminating value which then stops the loop. Use for example the Num variable you already have declared. You can use the same variable to investigate the numbers between user input and 0, for being odd values.
num := limite-1; // give num a start value based on user input (-1 because of "... numbers below (x)")
while num > 0 do // stop the loop when 0 is reached
begin
// here you investigate if `num` is a odd number (e.g. using `mod` operator or
// possibly your pascal has a built in `function Odd(value: integer): boolean;`)
// and add it to `Soma` if it is
num := num - 1;// decrement num at every iteration
end;
Finally you need to consider changes to the above, to handle negative input from the user.
To test if an integer is an odd value, you could use following function:
function IsOdd( value : Integer) : Boolean;
begin
IsOdd := (value mod 2) <> 0;
end;
Many pascal compilers have a built-in function called Odd(), which you could use.
A while loop works well to solve this problem. If you start with lowest odd number above zero, i.e. one and continue upwards so long we do not exceed the limit value we have a simple start:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
if IsOdd(i) then begin
sum := sum + i; // add to sum
end;
i := i + 1; // Increment i
end;
GetOddSumBelowX := sum;
end;
Now, that was simple enough. Next step to simplify the loop is to increment the i variable by two instead, just to jump between all odd numbers:
function GetOddSumBelowX( X : Integer) : Integer;
var
i,sum: Integer;
begin
i := 1; // Start with first odd number
sum := 0;
while (i < X) do begin // as long as i less than X, loop
sum := sum + i; // add to sum
i := i + 2; // Increment to next odd number
end;
GetOddSumBelowX := sum;
end;
Part of the program I have checks if an input number is a perfect number. We're supposed to find a solution that runs in O(sqrt(n)). The rest of my program runs in constant time, but this function is holding me back.
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer=0;
begin
for i := 1 to x-1 do
if (x mod i = 0) then
sum := sum + i;
if sum = x then
exit(true)
else
exit(false);
end;
This runs in O(n) time, and I need to cut it down to O(sqrt(n)) time.
These are the options I've come up with:
(1) Find a way to make the for loop go from 1 to sqrt(x)...
(2) Find a way to check for a perfect number that doesn't use a for loop...
Any suggestions? I appreciate any hints, tips, instruction, etc. :)
You need to iterate the cycle not for i := 1 to x-1 but for i := 2 to trunc(sqrt(x)).
The highest integer divisor is x but we do not take it in into account when looking for perfect numbers. We increment sum by 1 instead (or initialize it with 1 - not 0).
The code if (x mod i = 0) then sum := sum + i; for this purpose can be converted to:
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i);
end;
And so we get the following code:
function Perfect(x: integer): boolean;
var
i: integer;
sum: integer = 1;
sqrtx: integer;
begin
sqrtx := trunc(sqrt(x));
i := 2;
while i <= sqrtx do
begin
if (x mod i = 0) then
begin
sum := sum + i;
sum := sum + (x div i) // you can also compare i and x div i
//to avoid adding the same number twice
//for example when x = 4 both 2 and 4 div 2 will be added
end;
inc(i);
end;
if sum = x then
exit(true)
else
exit(false);
end;
I have an array "line" which has a string contained in it of length "l" and an array "nl" which has a string contained in it of length "p".
Note: "l" and "p" don't necessarily have to be the length of each correspondent array.The parameter "at" will be position where the insertion will be made inside "line".
Resuming: An array of length "p" will be inserted into "line", moving all chars of "line" between position (at,i,at+p),'p' positions to the right in order to make the insertion.
My logic for the ensures is to check if the elements inserted in "line" have the same order and are the same that the chars contained in "nl".
Here is the code:
method insert(line:array<char>, l:int, nl:array<char>, p:int, at:int)
requires line != null && nl != null;
requires 0 <= l+p <= line.Length && 0 <= p <= nl.Length ;
requires 0 <= at <= l;
modifies line;
ensures forall i :: (0<=i<p) ==> line[at+i] == nl[i]; // error
{
var i:int := 0;
var positionAt:int := at;
while(i<l && positionAt < l)
invariant 0<=i<l+1;
invariant at<=positionAt<=l;
{
line[positionAt+p] := line[positionAt];
line[positionAt] := ' ';
positionAt := positionAt + 1;
i := i + 1;
}
positionAt := at;
i := 0;
while(i<p && positionAt < l)
invariant 0<=i<=p;
invariant at<=positionAt<=l;
{
line[positionAt] := nl[i];
positionAt := positionAt + 1;
i := i + 1;
}
}
Here are the errors that i am receiving.
Thanks.
I suspect that your algorithm is not correct, because it does not seem to take into account the fact that shifting the characters starting at position at by p places might write them over the end of the string in line.
My experience has been that in order to be successful with verification
Good standards of code development are crucial. Good variable naming, code formatting, and other code conventions are even more important than usual.
Writing code that is logically simple is really helpful. Try to avoid extraneous extra variables. Try to simplify arithmetic and logical expressions wherever practical.
Starting with a correct algorithm makes verification easier. Of course, this is easier said than done!
It is often helpful to write out the strongest loop invariants you can think of.
Working backwards from the postcondition is often helpful. In your case, take the postcondition and the negation of the final loop condition - and use these to work out what the invariant of the final loop must be in order to imply the postcondition. Then work backwards from that to the previous loop, etc
When manipulating arrays, using a ghost variable which contains the original value of the array as a sequence is very often an effective strategy. Ghost variables do not appear in the compiler output so will not effect the performance of your program.
It is often helpful to write down assertions for the exact state of the array, even if the postcondition only requires some weaker property.
Here is a verified implementation of your desired procedure:
// l is length of the string in line
// p is length of the string in nl
// at is the position to insert nl into line
method insert(line:array<char>, l:int, nl:array<char>, p:int, at:int)
requires line != null && nl != null
requires 0 <= l+p <= line.Length // line has enough space
requires 0 <= p <= nl.Length // string in nl is shorter than nl
requires 0 <= at <= l // insert position within line
modifies line
ensures forall i :: (0<=i<p) ==> line[at+i] == nl[i] // ok now
{
ghost var initialLine := line[..];
// first we need to move the characters to the right
var i:int := l;
while(i>at)
invariant line[0..i] == initialLine[0..i]
invariant line[i+p..l+p] == initialLine[i..l]
invariant at<=i<=l
{
i := i - 1;
line[i+p] := line[i];
}
assert line[0..at] == initialLine[0..at];
assert line[at+p..l+p] == initialLine[at..l];
i := 0;
while(i<p)
invariant 0<=i<=p
invariant line[0..at] == initialLine[0..at]
invariant line[at..at+i] == nl[0..i]
invariant line[at+p..l+p] == initialLine[at..l]
{
line[at + i] := nl[i];
i := i + 1;
}
assert line[0..at] == initialLine[0..at];
assert line[at..at+p] == nl[0..p];
assert line[at+p..l+p] == initialLine[at..l];
}
http://rise4fun.com/Dafny/ZoCv
I need to calculate odd/even sum, here is what I've got so far:
PROGRAM EvenOddSum;
USES
WinCrt;
VAR
odd, even, x: INTEGER;
BEGIN
WriteLn('Calculation of sum');
WriteLn;
odd := 0;
even := 0;
Write('Enter value(s)');
WHILE x > 0 DO BEGIN
IF x mod 2:= 0 THEN BEGIN
even := even + x;
ELSE
odd := odd + x;
ReadLn(x);
END;
WriteLn;
WriteLn('Even sum is = ', even);
WriteLn('Odd sum is =', odd);
END.
I use freepascal.org compiler and I get this error :
SYNTAX error THEN expected but := found
And I just can't see the problem with this code.
In Pascal, := is the assignment operator. Replace it with = on the line that reads IF x mod 2:= 0 THEN BEGIN.
Also, remove the BEGIN. The result should read:
IF x mod 2 = 0 THEN
It's in here:
IF x mod 2:= 0 THEN BEGIN
The := is used for assignment, use '=' or '==' for comparison.
(Off the top of my head, I don't know if Pascal uses '=', '==', or both for comparisons. One of them should do the trick).
If am not wrong, := is used for declaring & assigning the value. For condition, you should use = .
change x mod 2:= 0 to x mod 2 = 0
You could say:
x mod 2 > 0 then writeln(x);
This will print all the odd numbers.