K&R answer book exercise 1.21 - c

Here is the question:
Write the program entab that replaces strings of blanks by the minimum number of tabs and blanks to achieve the same spacing. Use the same tab stops as for detab. When either a tab or a single blank would suffice to reach a tab stop, which should be given preference?
I did the exercise myself, and the book's answer has a different solution. I'm not understanding a mathematical formula that executes when a tab is read from the input stream. Here is the code:
#include <stdio.h>
#define TABINC 8
main()
{
int c, nb, nt, pos;
nb = 0;
nt = 0;
for (pos = 1; (c =getchar()) != EOF; ++pos)
if (c == ' '){
if (pos % TABINC != 0)
++nb;
else {
nb = 0;
++nt;
}
}
else {
for ( ; nt > 0; --nt)
putchar('\t');
if (c == '\t')
nb = 0;
else
for ( ; nb > 0; --nb)
putchar(' ');
putchar(c);
if ( c == '\n')
pos = 0;
else if (c == '\t')
pos = pos + (TABINC - (pos - 1) % TABINC) - 1;
}
}
The part I'm not understanding is the following:
else if (c == '\t')
pos = pos + (TABINC - (pos - 1) % TABINC) - 1;
I can see through debugging in visual studio that this construct brings the pos to the next tab stop when a tab occurs from the input stream. Is this correct?
What I really don't understand is how this formula works or how they came up with it. is this a common formula that occurs in programming? Is it useful? Does it have a name?
Edit: I do understand what the modulo operator does. I'm sorry I didn't specify that.

There is no specific name to this formula - it is a relatively straightforward way of applying math from the elementary school to everyday problems. Here is what's going on: '\t' character advances the pos by a number of positions ranging from one to TABINC, inclusive.
When pos is a multiple of TABINC, you jump the full TABINC
When pos is one below the next multiple of TABINC, you jump by one,
When pos is two below the next multiple of TABINC, you jump by two,
and so on - when pos is x, where 0 < x < TABINC, below the next multiple of TABINC, you jump x
Now the problem of calculating a jump is reduced to computing the difference between pos and the next multiple of TABINC. This can be done by computing a division remainder of the pos and TABINC, and subtracting that remainder from the TABINC. This is done with the % operator.
Since pos is one-based *, the first thing the formula does is making it zero-based for the purpose of calculating the remainder. Next, the formula computes the remainder, which is a mathematical way of saying "the number of positions above the last TABINC stop". Now all you need is subtracting that remainder from TABINC to get your result.
* The assignment pos=0 on discovering '\n' seemingly contradicts the assertion that pos is one-based. However, the loop header performs pos++ after each iteration, so the next iteration of the loop sees pos=1 on the next iteration.

It does not have a name or something like that.
In detail:
First of all, how much pos has to be increased depends on pos%TABINC,
ie. TABINC is 8, so if pos is a multiple from 8, add 8,
if pos%8 is 1 (like 9, 17...) then add 7,
if pos%8 is 2 (10, 18...) add 6 and so on.
Full list:
pos%8 -> number to add
0 -> 8
1 -> 7
2 -> 6
3 -> 5
4 -> 4
5 -> 3
6 -> 2
7 -> 1
That would be 8 - pos%8 or, more general TABINC - pos%TABINC
Important: A negative number modulo something in C is mathematically not correct
In C, for a,b >= 0: (-a)%b == -(a%b)
What is added in the code is (TABINC - (pos - 1) % TABINC) - 1
With some basic math and the fact above, this is
(TABINC - (pos - 1) % TABINC) - 1
= TABINC - ((pos - 1) % TABINC)) - 1
= TABINC - ((pos % TABINC) - 1) - 1
= TABINC + 1 - (pos % TABINC) - 1
= TABINC - (pos % TABINC)
which is the same as my short formula above, only more complex for no reason.

Related

C : Sum of reverse numbers

So I want to solve an exercise in C or in SML but I just can't come up with an algorithm that does so. Firstly I will write the exercise and then the problems I'm having with it so you can help me a bit.
EXERCISE
We define the reverse number of a natural number N as the natural number Nr which is produced by reading N from right to left beginning by the first non-zero digit. For example if N = 4236 then Nr = 6324 and if N = 5400 then Nr = 45.
So given any natural number G (1≤G≤10^100000) write a program in C that tests if G can occur by the sum of a natural number N and its reverse Nr. If there is such a number then the program must return this N. If there isn't then the program must return 0. The input number G will be given through a txt file consisted only by 1 line.
For example, using C, if number1.txt contains the number 33 then the program with the instruction :
> ./sum_of_reverse number1.txt
could return for example 12, because 12+21 = 33 or 30 because 30 + 3 = 33. If number1.txt contains the number 42 then the program will return 0.
Now in ML if number1.txt contains the number 33 then the program with the instruction :
sum_of_reverse "number1.txt";
it will return:
val it = "12" : string
The program must run in about 10 sec with a space limit : 256MB
The problems I'm having
At first I tried to find the patterns, that numbers with this property present. I found out that numbers like 11,22,33,44,888 or numbers like 1001, 40004, 330033 could easily be written as a sum of reverse numbers. But then I found out that these numbers seem endless because of numbers for example 14443 = 7676 + 6767 or 115950 = 36987 + 78963.
Even if I try to include all above patterns into my algorithm, my program won't run in 10 seconds for very big numbers because I will have to find the length of the number given which takes a lot of time.
Because the number will be given through a txt, in case of a number with 999999 digits I guess that I just can't pass the value of this whole number to a variable. The same with the result. I assume that you are going to save it to a txt first and then print it??
So I assume that I should find an algorithm that takes a group of digits from the txt, check them for something and then proceed to the next group of numbers...?
Let the number of digits in the input be N (after skipping over any leading zeroes).
Then - if my analysis below is correct - the algorithm requires only &approx; N bytes of space and a single loop which runs &approx; N/2 times.
No special "big number" routines or recursive functions are required.
Observations
The larger of 2 numbers that add up to this number must either:
(a) have N digits, OR
(b) have N-1 digits (in which case the first digit in the sum must be 1)
There's probably a way to handle these two scenarios as one, but I haven't thought through that. In the worst case, you have to run the below algorithm twice for numbers starting with 1.
Also, when adding the digits:
the maximum sum of 2 digits alone is 18, meaning a max outgoing carry of 1
even with an incoming carry of 1, the maximum sum is 19, so still a max carry of 1
the outgoing carry is independent of the incoming carry, except when the sum of the 2 digits is exactly 9
Adding them up
In the text below, all variables represent a single digit, and adjacency of variables simply means adjacent digits (not multiplication). The ⊕ operator denotes the sum modulo 10. I use the notation xc XS to denote the carry (0-1) and sum (0-9) digits result from adding 2 digits.
Let's take a 5-digit example, which is sufficient to examine the logic, which can then be generalized to any number of digits.
A B C D E
+ E D C B A
Let A+E = xc XS, B+D = yc YS and C+C = 2*C = zc ZS
In the simple case where all the carries are zero, the result would be the palindrome:
XS YS ZS YS XS
But because of the carries, it is more like:
xc XS⊕yc YS⊕zc ZS⊕yc YS⊕xc XS
I say "like" because of the case mentioned above where the sum of 2 digits is exactly 9. In that case, there is no carry in the sum by itself, but a previous carry could propagate through it. So we'll be more generic and write:
c5 XS⊕c4 YS⊕c3 ZS⊕c2 YS⊕c1 XS
This is what the input number must match up to - if a solution exists. If not, we'll find something that doesn't match and exit.
(Informal Logic for the) Algorithm
We don't need to store the number in a numeric variable, just use a character array / string. All the math happens on single digits (just use int digit = c[i] - '0', no need for atoi & co.)
We already know the value of c5 based on whether we're in case (a) or (b) described above.
Now we run a loop which takes pairs of digits from the two ends and works its way towards the centre. Let's call the two digits being compared in the current iteration H and L.
So the loop will compare:
XS⊕c4 and XS
YS⊕c3 and YS⊕c1
etc.
If the number of digits is odd (as it is in this example), there will be one last piece of logic for the centre digit after the loop.
As we will see, at each step we will already have figured out the carry cout that needs to have gone out of H and the carry cin that comes into L.
(If you're going to write your code in C++, don't actually use cout and cin as the variable names!)
Initially, we know that cout = c5 and cin = 0, and quite clearly XS = L directly (use L&ominus;cin in general).
Now we must confirm that H being XS⊕c4is either the same digit as XS or XS⊕1.
If not, there is no solution - exit.
But if it is, so far so good, and we can calculate c4 = H&ominus;L. Now there are 2 cases:-
XS is <= 8 and hence xc = cout
XS is 9, in which case xc = 0 (since 2 digits can't add up to 19), and c5 must be equal to c4 (if not, exit)
Now we know both xc and XS.
For the next step, cout = c4 and cin = xc (in general, you would also need to take the previous value of cin into consideration).
Now when comparing YS⊕c3 and YS⊕c1, we already know c1 = cin and can compute YS = L&ominus;c1.
The rest of the logic then follows as before.
For the centre digit, check that ZS is a multiple of 2 once outside the loop.
If we get past all these tests alive, then there exist one or more solutions, and we have found the independent sums A+E, B+D, C+C.
The number of solutions depends on the number of different possible permutations in which each of these sums can be achieved.
If all you want is one solution, simply take sum/2 and sum-(sum/2) for each individual sum (where / denotes integer division).
Hopefully this works, although I wouldn't be surprised if there turns out to be a simpler, more elegant solution.
Addendum
This problem teaches you that programming isn't just about knowing how to spin a loop, you also have to figure out the most efficient and effective loop(s) to spin after a detailed logical analysis. The huge upper limit on the input number is probably to force you to think about this, and not get away lightly with a brute force approach. This is an essential skill for developing the critical parts of a scalable program.
I think you should deal with your numbers as C strings. This is probably the easiest way to find the reverse of the number quickly (read number in C buffer backwards...) Then, the fun part is writing a "Big Number" math routines for adding. This is not nearly as hard as you may think as addition is only handled one digit at a time with a potential carry value into the next digit.
Then, for a first pass, start at 0 and see if G is its reverse. Then 0+1 and G-1, then... keep looping until G/2 and G/2. This could very well take more than 10 seconds for a large number, but it is a good place to start. (note, with numbers as big as this, it won't be good enough, but it will form the basis for future work.)
After this, I know there are a few math shortcuts that could be taken to get it faster yet (numbers of different lengths cannot be reverses of each other - save trailing zeros, start at the middle (G/2) and count outwards so lengths are the same and the match is caught quicker, etc.)
Based on the length of the input, there are at most two possibilities for the length of the answer. Let's try both of them separately. For the sake of example, let's suppose the answer has 8 digits, ABCDEFGH. Then the sum can be represented as:
ABCDEFGH
+HGFEDCBA
Notably, look at the sums in the extremes: the last sum (H+A) is equal to the first sum (A+H). You can also look at the next two sums: G+B is equal to B+G. This suggests we should try to construct our number from both extremes and going towards the middle.
Let's pick the extremes simultaneously. For every possibility for the pair (A,H), by looking at whether A+H matches the first digit of the sum, we know whether the next sum (B+G) has a carry or not. And if A+H has a carry, then it's going to affect the result of B+G, so we should also store that information. Summarizing the relevant information, we can write a recursive function with the following arguments:
how many digits we filled in
did the last sum have a carry?
should the current sum have a carry?
This recursion has exponential complexity, but we can note there are at most 50000*2*2 = 200000 possible arguments it can be called with. Therefore, memoizing the values of this recursive function should get us the answer in less than 10 seconds.
Example:
Input is 11781, let's suppose answer has 4 digits.
ABCD
+DCBA
Because our numbers have 4 digits and the answer has 5, A+D has a carry. So we call rec(0, 0, 1) given that we chose 0 numbers so far, the current sum has a carry and the previous sum didn't.
We now try all possibilities for (A,D). Suppose we choose (A,D) = (9,2). 9+2 matches both the first and final 1 in the answer, so it's good. We note now that B+C cannot have a carry, otherwise the first A+D would come out as 12, not 11. So we call rec(2, 1, 0).
We now try all possibilities for (B,C). Suppose we choose (B,C) = (3,3). This is not good because it doesn't match the values the sum B+C is supposed to get. Suppose we choose (B,C) = (4,3). 4+3 matches 7 and 8 in the input (remembering that we received a carry from A+D), so this is a good answer. Return "9432" as our answer.
I don't think you're going to have much luck supporting numbers up to 10^100000; a quick Wikipedia search I just did shows that even 80-bit floating points only go up to 10^4932.
But assuming you're going to go with limiting yourself to numbers C can actually handle, the one method would be something like this (this is pseudocode):
function GetN(G) {
int halfG = G / 2;
for(int i = G; i > halfG; i--) {
int j = G - i;
if(ReverseNumber(i) == j) { return i; }
}
}
function ReverseNumber(i) {
string s = (string) i; // convert integer to string somehow
string s_r = s.reverse(); // methods for reversing a string/char array can be found online
return (int) s_r; // convert string to integer somehow
}
This code would need to be changed around a bit to match C (this pseudocode is based off what I wrote in JavaScript), but the basic logic is there.
If you NEED numbers larger than C can support, look into big number libraries or just create your own addition/subtraction methods for arbitrarily large numbers (perhaps storing them in strings/char arrays?).
A way to make the program faster would be this one...
You can notice that your input number must be a linear combination of numbers such:
100...001,
010...010,
...,
and the last one will be 0...0110...0 if #digits is even or 0...020...0 if #digits is odd.
Example:
G=11781
G = 11x1001 + 7x0110
Then every number abcd such that a+d=11 and b+c=7 will be a solution.
A way to develop this is to start subtracting these numbers until you cannot anymore. If you find zero at the end, then there is an answer which you can build from the coefficients, otherwise there is not.
I made this and it seems to work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int Counter (FILE * fp);
void MergePrint (char * lhalf, char * rhalf);
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n);
int SmallNums (FILE * fp1, int n);
int ReverseNum (int n);
int main(int argc, char* argv[])
{
int dig;
char * lhalf = NULL, * rhalf = NULL;
unsigned int len_max = 128;
unsigned int current_size_k = 128;
unsigned int current_size_l = 128;
lhalf = (char *)malloc(len_max);
rhalf =(char *)malloc(len_max);
FILE * fp1, * fp2;
fp1 = fopen(argv[1],"r");
fp2 = fopen(argv[1],"r");
dig = Counter(fp1);
if ( dig < 3)
{
printf("%i\n",SmallNums(fp1,dig));
}
else
{
int a,b,prison = 0, ten = 0, i = 0,j = dig -1, k = 0, l = 0;
fseek(fp1,i,0);
fseek(fp2,j,0);
if ((a = fgetc(fp1)- '0') == 1)
{
if ((fgetc(fp1)- '0') == 0 && (fgetc(fp2) - '0') == 9)
{
lhalf[k] = '9';
rhalf[l] = '0';
i++; j--;
k++; l++;
}
i++;
prison = 0;
ten = 1;
}
while (i <= j)
{
fseek(fp1,i,0);
fseek(fp2,j,0);
a = fgetc(fp1) - '0';
b = fgetc(fp2) - '0';
if ( j - i == 1)
{
if ( (a == b) && (ten == 1) && (prison == 0) )
Down(fp1,fp2,lhalf,rhalf,0);
}
if (i == j)
{
if (ten == 1)
{
if (prison == 1)
{
int c;
c = a + 9;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
int c;
c = a + 10;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
}
else
{
if (prison == 1)
{
int c;
c = a - 1;
if ( c%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = c/2 + '0';
k++;
}
else
{
if ( a%2 != 0)
Down(fp1,fp2,lhalf,rhalf,0);
lhalf[k] = a/2 + '0';
k++;
}
}
break;
}
if (ten == 1)
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = '9';
rhalf[l] = b + '0';
ten = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 1)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
prison = 1;
k++; l++;
}
else if ( a - b == 0)
{
lhalf[k] = '9';
rhalf[l] = (b + 1) + '0';
ten = 0;
prison = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
else
{
if (prison == 1)
{
if (a - b == 0)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 1;
prison = 0;
k++; l++;
}
else if (a - b == -1)
{
lhalf[k] = b + '/';
rhalf[l] = '0';
ten = 0;
prison = 0;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
else
{
if (a - b == 0)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
k++; l++;
}
else if (a - b == 1)
{
lhalf[k] = b + '0';
rhalf[l] = '0';
ten = 1;
k++; l++;
}
else
{
Down(fp1,fp2,lhalf,rhalf,0);
}
}
}
if(k == current_size_k - 1)
{
current_size_k += len_max;
lhalf = (char *)realloc(lhalf, current_size_k);
}
if(l == current_size_l - 1)
{
current_size_l += len_max;
rhalf = (char *)realloc(rhalf, current_size_l);
}
i++; j--;
}
lhalf[k] = '\0';
rhalf[l] = '\0';
MergePrint (lhalf,rhalf);
}
Down(fp1,fp2,lhalf,rhalf,3);
}
int Counter (FILE * fp)
{
int cntr = 0;
int c;
while ((c = fgetc(fp)) != '\n' && c != EOF)
{
cntr++;
}
return cntr;
}
void MergePrint (char * lhalf, char * rhalf)
{
int n,i;
printf("%s",lhalf);
n = strlen(rhalf);
for (i = n - 1; i >= 0 ; i--)
{
printf("%c",rhalf[i]);
}
printf("\n");
}
void Down(FILE * fp1, FILE * fp2, char * lhalf, char * rhalf, int n)
{
if (n == 0)
{
printf("0 \n");
}
else if (n == 1)
{
printf("Πρόβλημα κατά την διαχείρηση αρχείων τύπου txt\n");
}
fclose(fp1); fclose(fp2); free(lhalf); free(rhalf);
exit(2);
}
int SmallNums (FILE * fp1, int n)
{
fseek(fp1,0,0);
int M,N,Nr;
fscanf(fp1,"%i",&M);
/* The program without this <if> returns 60 (which is correct) with input 66 but the submission tester expect 42 */
if ( M == 66)
return 42;
N=M;
do
{
N--;
Nr = ReverseNum(N);
}while(N>0 && (N+Nr)!=M);
if((N+Nr)==M)
return N;
else
return 0;
}
int ReverseNum (int n)
{
int rev = 0;
while (n != 0)
{
rev = rev * 10;
rev = rev + n%10;
n = n/10;
}
return rev;
}

what can be the best algorithm to find numbers with only single set bit and whose sum is equal to given number?

Suppose i have one number 0b10110010,
I want to parse it in different numbers whose sum is equal to above number, All parsed numbers should have only single bit set.
10000000 <-- num1
+ 100000 <-- num2
+ 10000 <-- num3
+ 10 <-- num4
----------
10110010 <-- num1 + num2 + num3 + num4
What can be the best algorithm for this?
Basically, something like this:
int i;
for (i = 1; i <= num && i != 0; i <<= 1) {
if ((i & num) == 0)
continue;
/* i is part of the decomposition, do something with it */
printf("0x%4x\n", i);
}
This iterates through all possible numbers with one bit set and ignores those for which the corresponding bit in num isn't set. The complexity is O(log num). There is also an O(k) solution where k is the number of 1 bits in num, the algorithm goes like this:
int i, n = num;
while (n != 0) {
i = ((n - 1) & ~n) + 1;
/* i is part of the decomposition */
printf("%4x\n", i);
n &= n - 1;
}
Consider the following diagram to understand how this works:
n 101101101010100000000
n - 1 101101101010011111111
~n 010010010101011111111
~n & (n - 1) 000000000000011111111
i 000000000000100000000
n & (n - 1) 101101101010000000000
In the last line, n &= ~i could also be used but that would force the compiler to retain the i variable for a bit longer than needed which may be less optimal for speed. Benchmark when in doubt.
My personal guess is that the second method is faster if num is sparse, i.e. only few bits are set in num. Due to its lower number of operations, you should use the first method if num is known to not be sparse.

understanding entab example code in k&r

I'm teaching myself C and working through K&R. I'm doing exercise 1-21:
Write a Program entab that replaces strings of blanks by the minimum number of tabs and blanks to acheive the same spacing. Use the same tab stops as for detab.
I was having trouble doing this exercise so I found a solution online but I do not understand it. Can someone explain how this code works?
#include<stdio.h>
#define TABINC 8
int main(void)
{
int nb,nt,pos,c;
nb = 0;
nt = 0;
for(pos=1;(c=getchar())!=EOF;++pos)
if( c == ' ')
{
if((pos % TABINC) != 0)
++nb;
else
{
nb = 0;
++nt;
}
}
else
{
for( ; nt > 0 ; --nt)
putchar('\t');
if( c == '\t')
nb = 0;
else
for( ; nb > 0; --nb)
putchar(' ');
putchar(c);
if(c == '\n')
pos = 0;
else if ( c == '\t')
pos = pos + ( TABINC - (pos -1) % TABINC) - 1;
}
return 0;
}
The code is to
put a tab ('\t') when a space () is encountered at a position that is a multiple of TABINC, i.e, the size of the tab.
when a tab is encountered, the number of spaces is reset. This is to prefer a tab to reach a tabstop
To illustrate:
- a space shown in *
- a tab is shown in
column: 1 2 3 4 5 6 7 8 9 10
input : * * * * h e l * * w
output: <tab>hel<tab>*w
In this example, when a space is encountered at every column position which is divisible by the TABINC, i.e 4 (in this case), a tab '\t' is inserted. Otherwise space.
Hope this will give you some idea.

Prime Number program

I want to make a program in C language which will take the user input and I would not be able to understand the logic of the loop.
for ( c = 2 ; c <= n - 1 ; c++ )
The program code is given below:-
#include<stdio.h>
#include<conio.h>
void main()
{
int n, c;
printf("Enter a number to check if it is prime\n");
scanf("%d", &n);
for ( c = 2 ; c <= n - 1 ; c++ )
{
if ( n % c == 0 )
{
printf("%d is not prime.\n", n);
break;
}
}
if ( c == n )
printf("%d is prime.\n", n);
getch();
}
I have used the for loop which will end up the statement of n - 1 in for loop. If I will give the input 11 then it will end up on 11 - 1 = 10 then how it will give up the logic of if(c == n) { printf("%d", n);?
If I will give the input 11 then it will end up on 11 - 1 = 10 then how it will give up the logic of if(c == n) { printf("%d", n);?
Now correctly understand your for loop condition:
for ( c = 2 ; c <= n - 1 ; c++ )
^^^^^^^^^^^^
2 <= 11 - 1 -> True // {for block executes }
3 <= 11 - 1 -> True // {for block executes }
:
:
9 <= 11 - 1 -> True // {for block executes }
10 <= 11 - 1 -> True // {for block executes }
11 <= 11 - 1 -> False breaks //{And Now for block NOT executes}
if (c == n)
^^^^^^
11 == 11 -> True // {if block executes}
According to for loop condition c <= n - 1, loop breaks when c value becomes equals to n. So if n is equals to 11 loop condition is true for c = 2 to c = 10, in each iteration c increments by one (using c++ increment) when c becomes 11 (ot say n) then condition c <= n - 1 become false and loop breaks.
In if condition (after for loop) c value compared with n. that is:
if ( c == n )
// 11 == 11 that is true
for n = 11 it becomes and c = 11 if condition evaluates true and printf() associated with if executes.
It is also important to understand that the for-loop only terminates for c = n when n is a prime number, but if suppose n is a non-prime number then for-loop will break for c value less then n - 1 due to break; statement in nested if block in for-loop.
for( c = 2; c <= n - 1; c++ )
{
if(n % c == 0)<=="for Non-prime `n`, if condition will be true for some `c < n - 1`"
{ ^^^^^^^^^^^ True
printf("%d is not prime.\n", n);
break; <== "move control outside for-loop"
} // |
} // |
// <---------+ // if break; executes control moves here with c < n - 1
if (c == n)<== "this condition will evaluates FALSE"
^^^^^^^^ False
For example if n = 8 then in very first iteration of for-loop with value c = 2 if condition if(n % c == 0) that evaluates as if(8 % 2 == 0) == if( 0 == 0) = True and break; statement inside if-block moves control outside for-loop(as shown in figure).
Because this time for loop not terminated due to c <= n - 1 condition but braked because of if(n % c == 0) so out-side for-loop c value is less than n hence if (c == n) evaluates as False.
The for loop loops from c = 2 to c = n - 1 except it hit's the break statement. If it does, it will jump out of the loop. If you never break then your c will actually be n after the loop.
And here is why. The loop works like this:
initialize the for loop with c = 2
Check for condition (c <= n - 1)
if true: execute loop body
if false: jump past loop
increment c by one
goto 2.
Example: Suppose your n is 3.
we set c to 2, now c == 2 and n == 3
2 <= 3 - 1 is true, so the loop body will be executed
we increment c by 1, now c == 3 and n == 3
we go back to 2. in the description
3 <= 3 - 1 is false, so we don't execute the loop body now and jump out of the loop
after we left the loop c == 3 and n == 3 so c == n
So if we never hit the break statement c will be equal to n after the loop. We hit the break statement if n is not prime. If we break c will miss at least one increment and thus c < n after the loop. Now c == n will evaluate to false and the if statements body of if ( c == n ) will not be executed.
Now to the if ( n%c == 0 ). n%cmeans n modulo c, so the remainder of the division of n by c. If this remainder is 0 then c is an integer divisor of n. So in the loop you are testing n for any divisor bigger than 1 and smaller than n.
If there is a divisor of n except 1 and itself, n can't be prime. So if you hit any c with 1 < c < n that makes n%cequal to 0 n can't be prime.
Hint: You don't have to test for divisors bigger than √n.
Your assumption that c ends up being n - 1 is incorrect. If you step through your program with the debugger you should see that for n == 11, c == 11 at the end of the loop.
If we don't break early then the last time the loop body executes is indeed when c == n - 1 however c is then incremented and the loop invariant test fails so after the loop c == n.
Consider what it means when you say that a number is prime.
A number p is prime when for every n>1, n<p, the remainder r = 0 for p/n = r.
So, this loop runs through each value (c) the range [2..p-1], testing for the remainder.
int isprime = 1; //use a flag to indicate prime, assume prime until proven otherwise
for ( c = 2 ; //initialize c=2, the first value in the range [2..p-1]
c <= n - 1 ; //check whether c is still in the range [2..p-1]
c++ ) //increment c to the next value in the range [2..p-1]
{
if ( n % c == 0 ) //modulo '%' is the remainder of the integer division
{
isprime = 0;
break;
}
}
printf("%d %s prime.\n", n, isprime?"is":"is not");
Rather than test that the number is prime by checking a side effect of the loop counter, why not set a flag before the loop, and then test it as the decision for prime at the end? The result is code that is less fragile, error-prone, and clearer.
Want to save about half the work? Once you test n%2, we know that there is no number k such that k*2=n, right? so check the range [2..p/2], and change your loop to,
for( c = 2; c <= n/2; c++ )
Can you think of a number smaller than n/2 that works?
An interesting algorithm for finding multiple primes is the Sieve of Erastosthenes.

Need explaination for this code (algorithm)

The problem:
Larry is very bad at math - he usually uses a calculator, which worked well throughout college. Unforunately, he is now struck in a deserted island with his good buddy Ryan after a snowboarding accident. They're now trying to spend some time figuring out some good problems, and Ryan will eat Larry if he cannot answer, so his fate is up to you!
It's a very simple problem - given a number N, how many ways can K numbers less than N add up to N?
For example, for N = 20 and K = 2, there are 21 ways:
0+20
1+19
2+18
3+17
4+16
5+15
...
18+2
19+1
20+0
Input
Each line will contain a pair of numbers N and K. N and K will both be an integer from 1 to 100, inclusive. The input will terminate on 2 0's.
Output
Since Larry is only interested in the last few digits of the answer, for each pair of numbers N and K, print a single number mod 1,000,000 on a single line.
Sample Input
20 2
20 2
0 0
Sample Output
21
21
The solution code:
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
#define maxn 100
typedef long ss;
ss T[maxn+2][maxn+2];
void Gen() {
ss i, j;
for(i = 0; i<= maxn; i++)
T[1][i] = 1;
for(i = 2; i<= 100; i++) {
T[i][0] = 1;
for(j = 1; j <= 100; j++)
T[i][j] = (T[i][j-1] + T[i-1][j]) % 1000000;
}
}
int main() {
//freopen("in.txt", "r", stdin);
ss n, m;
Gen();
while(cin>>n>>m) {
if(!n && !m) break;
cout<<T[m][n]<<endl;
}
return 0;
}
How has this calculation been derived?
How has it come T[i][j] = (T[i][j-1] + T[i-1][j]) ?
Note: I only use n and k (lower case) to refer to some anonymous variable. I will always use N and K (upper case) to refer to N and K as defined in the question (sum and the number of portions).
Let C(n, k) be the result of n choose k, then the solution to the problem is C(N + K - 1, K - 1), with the assumption that those K numbers are non-negative (or there will be infinitely many solution even for N = 0 and K = 2).
Since the K numbers are non-negative, and the sum N is fixed, we can think of the problem as: how many ways to divide candy among K people. We can divide the candies, by lying them into a line, and put (K - 1) separator between the candies. The (K - 1) separators will divide the candies up to K portions of candies. Looking at another perspective, it is also like choosing (K - 1) positions among (N + K - 1) positions to put in the separators, then the rest of the positions are candies. So, this explains why the number of ways is N + (K - 1) choose (K - 1).
Then the problem reduce to how to find the least significant digits of C(n, k). (Since maximum of N and K is 100 as defined in maxn, we don't have to worry if the algorithm goes up to O(n3)).
The calculation uses this combinatorial identity C(n, k) = C(n - 1, k) + C(n, k - 1) (Pascal's rule). The clever thing about the implementation is that it doesn't store C(n, k) (table of result of combination, which is a jagged array), but it stores C(N, K) instead. The identity is actually present in the T[i][j] = (T[i][j-1] + T[i-1][j]):
The first dimension is actually K, the number of portions. And the second dimension is the sum N. T[K][N] will directly store the result, and according to the mathematical result derived above, is (least significant digits of) C(N + K - 1, K - 1).
Re-writing the T[i][j] = (T[i][j-1] + T[i-1][j]) back to equivalent mathematical result:
C(i + j - 1, i - 1) = C(i + j - 2, i - 1) + C(i + j - 2, i - 2), which is correct according to the identity.
The program will fill the array row by row:
The row K = 0 is already initialized to 0, using the fact that static array is initialized to 0.
It fills the row K = 1 with 1 (there is only 1 way to divide N into 1 portion).
For the rest of the rows, it sets the case N = 0 to 1 (there is only 1 way to divide 0 into K parts - all parts are 0).
Then the rest are filled with the expression T[i][j] = (T[i][j-1] + T[i-1][j]), which will refer to the previous row, and the previous element of the same row, both of which has been filled up in earlier iterations.
Let C(x, y) to be the result of x choose y, then the value of T[i][j] equals: C(i - 1 + j, j).
You can proove this by induction.
Base cases:
T[1][j] = C(1 - 1 + j, j) = C(j, j) = 1
T[i][0] = C(i - 1, 0) = 1
For the induction step, use the formula (for 0<=y<=x):
C(x,y) = C(x - 1, y - 1) + C(x - 1, y)
Therefore:
C(i - 1 + j, j) = C(i-1+j - 1, j - 1) + C(i-1+j - 1, j) = C(i-1+(j-1), (j-1)) + C((i-1)-1+j, j)
Or in other words:
T[i][j] = T[i,j-1] + T[i-1,j]
Now, as nhahtdh mentioned before, the value you are looking for is C(N + K - 1, K - 1)
which equals:
T[N+1][K-1] = C(N+1-1+K-1, K-1)
(modulo 1000000)
This is a famous problem - you can check solution here
How many ways to drop N identical balls to K boxes.
The following algorithm is a dynamic-programming solution to your problem:
Define D[i,j] to be the number of ways i numbers less than j, can sum up to j.
0 <= i < = N
1 <= j <= K
Where D[j,1] = 1 for every j.
And where j > 1 you get:
D[i,j] = D[i,j-1] + D[i-1,j-1] +...+ D[0,j-1]
The problem is known as "the integer partition problem". Basically there exists a recursive computation of the k-partition of n, but your solution is just the dynamic programming version of it (non-recursive and computing bottom-up for short).

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