a void function to fill the array by converting integer to binary - c

int main(void)
{
/* Stop WDT */
MAP_WDT_A_holdTimer();
/* Selecting P1.2 and P1.3 in UART mode */
MAP_GPIO_setAsPeripheralModuleFunctionInputPin(GPIO_PORT_P1,
GPIO_PIN1 | GPIO_PIN2 | GPIO_PIN3, GPIO_PRIMARY_MODULE_FUNCTION);
/* Setting DCO to 12MHz */
CS_setDCOCenteredFrequency(CS_DCO_FREQUENCY_12);
/* Configuring UART Module */
MAP_UART_initModule(EUSCI_A0_BASE, &uartConfig);
/* Enable UART module */
MAP_UART_enableModule(EUSCI_A0_BASE);
/* Configuring GPIO2.4 as peripheral output for PWM and P6.7 for button
* interrupt */
MAP_GPIO_setAsPeripheralModuleFunctionOutputPin(GPIO_PORT_P2, GPIO_PIN4,
GPIO_PRIMARY_MODULE_FUNCTION);
redirect();
/* Configuring P1.0 as output */
MAP_GPIO_setAsOutputPin(GPIO_PORT_P1, GPIO_PIN0);
MAP_GPIO_setOutputLowOnPin(GPIO_PORT_P1, GPIO_PIN0);
/* Configuring Timer_A to have a period of approximately 500ms and
* an initial duty cycle of 10% of that (3200 ticks) */
//MAP_Interrupt_enableSleepOnIsrExit();
MAP_Interrupt_enableInterrupt(INT_TA0_0);
MAP_Timer_A_generatePWM(TIMER_A0_BASE,&pwmConfig);
MAP_Timer_A_clearInterruptFlag(TIMER_A0_BASE);
MAP_Timer_A_enableInterrupt(TIMER_A0_BASE);
MAP_Timer_A_enableCaptureCompareInterrupt
(TIMER_A0_BASE,TIMER_A_CAPTURECOMPARE_REGISTER_0);
/* Enabling MASTER interrupts */
MAP_Interrupt_enableMaster();
/* Sleeping when not in use */
while (1)
{
//MAP_PCM_gotoLPM0();
}
}
const int bit_length = 33;
int period;
int times[33];
int values[32];
x=598;
number_bit=10;
// this function - period,times,values
// x is the 32 bit integer , number_bit is how many bits in that integer
void int_To_Arr(uint32_t x,int number_bit){
int i = 0;
period = BIT_LENGTH * 67 ; // 15fps -> 1/15=66.67m
for (i = 0; i < number_bit ; i++) {
if (((x >> i) & 1) == 0) /* shift right by i-bits, check on/off */
values[i] = 1000; /* assign to values[i] based on result */
else
values[i] = 11000;
times[i] = BIT_LENGTH * i; /* set times[i] */
}
times[i] = BIT_LENGTH * i;
}
void TA0_0_IRQHandler(void)
{
int i,value;
MAP_Timer_A_clearCaptureCompareInterrupt(TIMER_A0_BASE,
TIMER_A_CAPTURECOMPARE_REGISTER_0);
time=time+1;
if(time>=period){
time=0;
MAP_GPIO_toggleOutputOnPin(GPIO_PORT_P1, GPIO_PIN0);
}
for(i=0;times[i]!=-1;i++){
if(times[i]>time){
break;
}
value=values[i];
}
MAP_Timer_A_setCompareValue(TIMER_A0_BASE,TIMER_A_CAPTURECOMPARE_REGISTER_1, value);
}
So this function will take x (32-bit integer) and number_bit(how many bits in the integer) and will fill the array in the main. if the bit is 1 , values = 11000 which will turn on the led. if the bit is 0 ,values =1000 .Unfortunately, the LED doesnt blink or do anything. Before this I do it manually, yes the LED blinking.
int time=0;
const int BIT_LENGTH = 33;
int period;
int times[33]; //{x}
int values[32];//{y}
//new 1001010110
const int period=667; //15fps - bitlength*67
const int times[]={0,67,200,267,333,400,467,600,667,-1};
const int values[]={11000,1000,11000,1000,11000,1000,11000,1000};

If through the extended discussions we have had, you are seeking to fill the values array based on the number_bits in the uint32_t x value passed to your int_to_array function by shifting the value x to the right number_bits times and at each iteration determining whether that bit in x is 0 or 1 and setting values[i] = 1000 if the bit is 0 and to 11000 if the bit is 1, then you could do something like the following:
#define BIT_LENGTH 33 /* if you need a constant, define one (or more) */
#define FPS_MULT 33
...
/* fill values based on nbits bit-values in x,
* fill times based on BIT_LENGTH and index.
* note: CHAR_BIT defined in limits.h
* (defined as 8 for virtually all common systems)
*/
void int_to_array (uint32_t x, int nbits)
{
int i = 0; /* loop variable - can be declared in loop for C99+ */
period = BIT_LENGTH * FPS_MULT; // your 30fps-1/30=0.033
/* validate nbits <= 32 */
if (nbits > (int)(sizeof x * CHAR_BIT)) {
fprintf (stderr, "error: nbits out of range of uint32_t\n");
return;
}
for (i = 0; i < nbits; i++) {
if (((x >> i) & 1) == 0) /* shift right by i-bits, check on/off */
values[i] = 1000; /* assign to values[i] based on result */
else
values[i] = 11000;
times[i] = BIT_LENGTH * i; /* set times[i] */
}
times[i] = BIT_LENGTH * i; /* final times[BIT_LENGTH - 1] */
}
While I am still unclear where those values come from, based on our discussion, that should be what you are looking for. Otherwise, I'm still uncertain. note I have added a validation check to insure nbits cannot exceed 32 (the number of bits in x)
Validation Test
If you wanted to write a short bit of validation code for your function, you could just write a short program that passes the value provided on the command line as the first argument to your function as x (nbits won't change, it will always be sizeof x * CHAR_BIT). The following code passes the first argument to the function (passing 10 by default if no argument is given on the command line)
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <errno.h>
#include <limits.h>
#define BIT_LENGTH 33 /* if you need a constant, define one (or more) */
#define FPS_MULT 33
int period;
int times[BIT_LENGTH];
int values[BIT_LENGTH - 1];
/* fill values based on nbits bit-values in x,
* fill times based on BIT_LENGTH and index.
* note: CHAR_BIT defined in limits.h
* (defined as 8 for virtually all common systems)
*/
void int_to_array (uint32_t x, int nbits)
{
int i = 0; /* loop variable - can be declared in loop for C99+ */
period = BIT_LENGTH * FPS_MULT; // your 30fps-1/30=0.033
/* validate nbits <= 32 */
if (nbits > (int)(sizeof x * CHAR_BIT)) {
fprintf (stderr, "error: nbits out of range of uint32_t\n");
return;
}
for (i = 0; i < nbits; i++) {
if (((x >> i) & 1) == 0) /* shift right by i-bits, check on/off */
values[i] = 1000; /* assign to values[i] based on result */
else
values[i] = 11000;
times[i] = BIT_LENGTH * i; /* set times[i] */
}
times[i] = BIT_LENGTH * i; /* final times[BIT_LENGTH - 1] */
}
int main (int argc, char **argv) {
unsigned long tmp = argc > 1 ? strtoul (argv[1], NULL, 0) : 10;
uint32_t x;
int i, nbits = sizeof x * CHAR_BIT;
if (errno || tmp > UINT32_MAX) {
fprintf (stderr, "error: conversion error or value out of range.\n");
return 1;
}
x = (uint32_t)tmp;
int_to_array (x, nbits);
printf ("x: %u\n\nperiod: %d\n\n", x, period);
for (i = 0; i < nbits; i++)
printf ("values[%2d]: %5d times[%2d]: %5d\n",
i, values[i], i, times[i]);
printf (" times[%2d]: %5d\n", i, times[i]);
return 0;
}
(note: I tweaked the function by addition optional parenthesis to a check to get rid of a -pedantic signed/unsigned comparison warning and added a final times[i] = BIT_LENGTH * i; after the loop to handle times having 1 more element than values)
Example Test
Simple default value of 10 (1010)
$ ./bin/values_times
x: 10
period: 1089
values[ 0]: 1000 times[ 0]: 0
values[ 1]: 11000 times[ 1]: 33
values[ 2]: 1000 times[ 2]: 66
values[ 3]: 11000 times[ 3]: 99
values[ 4]: 1000 times[ 4]: 132
values[ 5]: 1000 times[ 5]: 165
values[ 6]: 1000 times[ 6]: 198
values[ 7]: 1000 times[ 7]: 231
values[ 8]: 1000 times[ 8]: 264
values[ 9]: 1000 times[ 9]: 297
values[10]: 1000 times[10]: 330
values[11]: 1000 times[11]: 363
values[12]: 1000 times[12]: 396
values[13]: 1000 times[13]: 429
values[14]: 1000 times[14]: 462
values[15]: 1000 times[15]: 495
values[16]: 1000 times[16]: 528
values[17]: 1000 times[17]: 561
values[18]: 1000 times[18]: 594
values[19]: 1000 times[19]: 627
values[20]: 1000 times[20]: 660
values[21]: 1000 times[21]: 693
values[22]: 1000 times[22]: 726
values[23]: 1000 times[23]: 759
values[24]: 1000 times[24]: 792
values[25]: 1000 times[25]: 825
values[26]: 1000 times[26]: 858
values[27]: 1000 times[27]: 891
values[28]: 1000 times[28]: 924
values[29]: 1000 times[29]: 957
values[30]: 1000 times[30]: 990
values[31]: 1000 times[31]: 1023
times[32]: 1056
or a larger value, 0xdeadbeef (11011110101011011011111011101111)
$ ./bin/values_times 0xdeadbeef
x: 3735928559
period: 1089
values[ 0]: 11000 times[ 0]: 0
values[ 1]: 11000 times[ 1]: 33
values[ 2]: 11000 times[ 2]: 66
values[ 3]: 11000 times[ 3]: 99
values[ 4]: 1000 times[ 4]: 132
values[ 5]: 11000 times[ 5]: 165
values[ 6]: 11000 times[ 6]: 198
values[ 7]: 11000 times[ 7]: 231
values[ 8]: 1000 times[ 8]: 264
values[ 9]: 11000 times[ 9]: 297
values[10]: 11000 times[10]: 330
values[11]: 11000 times[11]: 363
values[12]: 11000 times[12]: 396
values[13]: 11000 times[13]: 429
values[14]: 1000 times[14]: 462
values[15]: 11000 times[15]: 495
values[16]: 11000 times[16]: 528
values[17]: 1000 times[17]: 561
values[18]: 11000 times[18]: 594
values[19]: 11000 times[19]: 627
values[20]: 1000 times[20]: 660
values[21]: 11000 times[21]: 693
values[22]: 1000 times[22]: 726
values[23]: 11000 times[23]: 759
values[24]: 1000 times[24]: 792
values[25]: 11000 times[25]: 825
values[26]: 11000 times[26]: 858
values[27]: 11000 times[27]: 891
values[28]: 11000 times[28]: 924
values[29]: 1000 times[29]: 957
values[30]: 11000 times[30]: 990
values[31]: 11000 times[31]: 1023
times[32]: 1056
Look things over and let me know if you have further questions.

It seems you want to examine each bit of x and see if it is 1 or 0, using >> is correct however doing
x>>i; //x<<i
if(x&1==0){
will not work because x>>i just returns what the value of x shifted i bits is, it does not update the value of x. Instead, I would do the following
x = x >> 1;
if(x&1==0){
It will shift the value of x by one each time through the loop and then check the rightmost bit.
Also, your loop counter is incorrect, instead of
for(i=1;i<x;i++){
You most likely want
for(i = 1; i < number_bit; i++) {
Actually looking at this again, you special case setting times[0] and values[0] outside the loop, and then start your loop at 1, so if you want to keep the code structured like that, your loop counter must be
for(i = 1; i < number_bit - 1; i++) {
Though looking at the code, I see no reason to special case the setting of times[0] and values[0], if you examine it closely, you'll see you can loop from
for(i = 0; i < number_bit; i++) {
and include setting times[0] and values[0] within the loop. You would just have to do the shift AFTER setting values or use an if to not shift x when i is 0.

Related

Project Euler #2 (Even Fibonacci numbers): O(1) solution

Project Euler #2 Even Fibonacci Numbers
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... By considering the terms in the
Fibonacci sequence whose values do not exceed four million, find the
sum of the even-valued terms.
Question: Is it possible to solve this in constant time?
Here is a hand-wavy O(1)-ish solution to PE-2, see comment by #harold :( above. My solution is psuedo-mathematical and not rigorous but hey, it works!
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... By considering the terms in the
Fibonacci sequence whose values do not exceed four million, find the
sum of the even-valued terms.
Some building blocks:
Binet's formula allows us to compute Fn instantly
There is another well-known formula for the sum of the first n fibonacci number i.e. F0 + F1 + ... + Fn = Fn+2 - 1
We need a formula for computing the sum of first k even numbers of the fibonacci sequence. Let En denote the n-th even number in the sequence so E1 = 2. It is well-known that En = 4En-1 - En-2. Many solutions to this problem utilise this very formula. Doing a bit of math (proof in appendix), you can see that
E1 + E2 + ... + Ek = (Ek+2 - 3Ek+1 - 2) ÷ 4 = (F3(k+2) - 3F3(k+1) - 2) ÷ 4 since En = F3n
We can compute the above using Binet's formula. The only question that remains is given a bound X what should k be? We invert Binet's formula to solve for n and we get
Rather the ceiling, we floor to get the index n of the fibonacci number below X. Then we only have to count how many even numbers there are in {F0, ..., Fn} and use the formula in Point 3.
C Implementation:
I tested it against brute force solutions. Seems to be working fine.
#include <stdio.h>
#include <math.h>
// Input N
// Output N'th term of fibonacci
unsigned long long binet(int N)
{
long double phi = (1.0 + sqrt(5)) / 2.0;
return floor((pow(phi, N) / sqrt(5)) + 0.5);
}
int find_fib_n(unsigned long long F_n)
{
return round(log(sqrt(5) * F_n) / log((1.0 + sqrt(5)) / 2.0));
}
// Input N
// Count of even numbers in the first N fib nums
int count_even_in_fib_seq(int N)
{
if (N <= 3)
return 0;
int tmp = N;
tmp -= 4;
return 1 + floor(tmp / 3);
}
#define bound 4000000
int main(void)
{
int fib_approx_index = find_fib_n(bound);
printf("%d %llu %d\n", fib_approx_index, binet(fib_approx_index), bound);
// Number of even numbers below bound
int K = count_even_in_fib_seq(fib_approx_index + 1);
printf("count %d\n", K);
unsigned long long result = (-3 * binet(3 * (K + 1)) + binet(3 * (K + 2)) - 2) / 4;
printf("%llu", result);
}
Appendix: Formula for sum of the first k even fibonacci numbers
After surfacing the web for a while I found some other people who came up with similar solution. Check out these links for more information:
https://www.xarg.org/puzzle/project-euler/problem-2/
https://brokensandals.net/technical/programming-challenges/projecteuler-2/
Thanks for reading
OEIS also gives a constant solution:
a(n) = (-10 + (5 - 3*sqrt(5))*(2 - sqrt(5))^n + (2 + sqrt(5))^n*(5 + 3*sqrt(5)))/20
-- Colin Barker, Nov 26 2016
Tested in Java:
public class Main {
// https://oeis.org/A099919
static long a(int n) {
double sqrt5 = Math.sqrt(5.0);
return (long)(-10 + (5 - 3 * sqrt5) * Math.pow(2 - sqrt5, n) + Math.pow(2 + sqrt5, n)*(5 + 3*sqrt5))/20;
}
public static void main(String[] args) {
for (int n = 0; n < 20; n++) {
System.out.println(a(n));
}
}
}
prints:
0
2
10
44
188
798
3382
14328
60696
257114
1089154
4613732
19544084
82790070
350704366
1485607536
6293134512
26658145586
112925716858
478361013020
Is it possible to solve this in constant time?
The goal implies using an O(1) equation to calculate the sum rather than a simple puts("4613732");
Given the limit "terms in the Fibonacci sequence whose values do not exceed four million", we only need to deal with Fibonacci(0) to Fibonacci(38) (3,524,578). Note that such a small table is easy to make at compile time with the compiler calculating the values - so no run time cost.
An O(1) equation is demonstrated by others using floating point math (FP).
A problem with a floating point solution is that this is an integer problem that deserves an integer solution. Floating point math has many rounding and inexact computations that render FP approaches questionable for an exact correct answer, especially as N grows large.
It is not enough to simply code some math formula in C using FP math for an integer problem. We need to assess it and determine when do the wheels fall off.
I do not have a better than O(n) integer solution, yet the following can serve as a test harness and reference for those seeking to validate higher value N solutions.
It would be interesting to see how far a FP solution can go before it generates incorrect answers.
#include <assert.h>
#include <math.h>
#include <limits.h>
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#if 0
typedef uint64_t fib_t;
#define FIB_MAX ULLONG_MAX
#define FIB_BASE10_LEN 20
#define FIBONACCI_INDEX_MAX 93
#else
typedef unsigned __int128 fib_t;
#define FIB_MAX ((fib_t) -1)
#define FIB_BASE10_LEN 39
#define FIBONACCI_INDEX_MAX 186
#endif
#define FIB_STR_N (sizeof(fib_t)*CHAR_BIT + 1)
// v--compound literal--v
#define FIB_STR(x) fibonacci_to_string((char [FIB_STR_N]){""}, (x), 10)
char* fibonacci_to_string(char buf[FIB_STR_N], fib_t i, int base) {
assert(base >= 2 && base <= 36);
unsigned ubase = (unsigned) base;
char *s = &buf[FIB_STR_N - 1];
*s = '\0';
do {
s--;
unsigned digit = (unsigned) (i % ubase);
*s = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[digit];
i /= ubase;
} while (i);
return s;
}
fib_t fibonacci(unsigned index) {
static fib_t f[FIBONACCI_INDEX_MAX + 1];
static unsigned index_size = 0;
if (index >= index_size) {
if (index_size < 2) {
f[0] = 0;
f[1] = 1;
index_size = 2;
} else if (index > FIBONACCI_INDEX_MAX) {
fprintf(stderr, "Fibonacci index too great %u\n", index);
exit(EXIT_FAILURE);
}
while (index >= index_size) {
f[index_size] = f[index_size - 1] + f[index_size - 2];
index_size++;
}
}
return f[index];
}
fib_t sum_of_even_fibonacci(fib_t f_max) {
fib_t sum = 0;
fib_t f;
for (unsigned index = 0; (f = fibonacci(index)) <= f_max; index += 3) {
if (sum > FIB_MAX - f) {
fprintf(stderr, "Sum too large f:%s sum:%s\n", FIB_STR(f), FIB_STR(sum));
exit(EXIT_FAILURE);
}
sum += f;
}
return sum;
}
int main(void) {
for (unsigned i = 0; 1 && i <= FIBONACCI_INDEX_MAX; i++) {
printf("%3u %*s\n", i, FIB_BASE10_LEN, FIB_STR(fibonacci(i)));
}
printf("Sum of even Fibonacci numbers that do not exceed\n");
printf("%*s %*s\n", FIB_BASE10_LEN, "limit", FIB_BASE10_LEN, "sum");
fib_t f_max = FIB_MAX; // 4000000;
for (fib_t i = 4; i <= f_max; i *= 10) {
printf("%*s %*s\n", FIB_BASE10_LEN, FIB_STR(i), FIB_BASE10_LEN,
FIB_STR(sum_of_even_fibonacci(i)));
if (i >= FIB_MAX / 10)
break;
}
return 0;
}
Output
0 0
1 1
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55
11 89
12 144
13 233
14 377
15 610
16 987
17 1597
18 2584
19 4181
20 6765
21 10946
22 17711
23 28657
24 46368
25 75025
26 121393
27 196418
28 317811
29 514229
30 832040
31 1346269
32 2178309
33 3524578
34 5702887
35 9227465
36 14930352
37 24157817
38 39088169
39 63245986
40 102334155
41 165580141
42 267914296
43 433494437
44 701408733
45 1134903170
46 1836311903
47 2971215073
48 4807526976
49 7778742049
50 12586269025
51 20365011074
52 32951280099
53 53316291173
54 86267571272
55 139583862445
56 225851433717
57 365435296162
58 591286729879
59 956722026041
60 1548008755920
61 2504730781961
62 4052739537881
63 6557470319842
64 10610209857723
65 17167680177565
66 27777890035288
67 44945570212853
68 72723460248141
69 117669030460994
70 190392490709135
71 308061521170129
72 498454011879264
73 806515533049393
74 1304969544928657
75 2111485077978050
76 3416454622906707
77 5527939700884757
78 8944394323791464
79 14472334024676221
80 23416728348467685
81 37889062373143906
82 61305790721611591
83 99194853094755497
84 160500643816367088
85 259695496911122585
86 420196140727489673
87 679891637638612258
88 1100087778366101931
89 1779979416004714189
90 2880067194370816120
91 4660046610375530309
92 7540113804746346429
93 12200160415121876738
94 19740274219868223167
95 31940434634990099905
96 51680708854858323072
97 83621143489848422977
98 135301852344706746049
99 218922995834555169026
100 354224848179261915075
101 573147844013817084101
102 927372692193078999176
103 1500520536206896083277
104 2427893228399975082453
105 3928413764606871165730
106 6356306993006846248183
107 10284720757613717413913
108 16641027750620563662096
109 26925748508234281076009
110 43566776258854844738105
111 70492524767089125814114
112 114059301025943970552219
113 184551825793033096366333
114 298611126818977066918552
115 483162952612010163284885
116 781774079430987230203437
117 1264937032042997393488322
118 2046711111473984623691759
119 3311648143516982017180081
120 5358359254990966640871840
121 8670007398507948658051921
122 14028366653498915298923761
123 22698374052006863956975682
124 36726740705505779255899443
125 59425114757512643212875125
126 96151855463018422468774568
127 155576970220531065681649693
128 251728825683549488150424261
129 407305795904080553832073954
130 659034621587630041982498215
131 1066340417491710595814572169
132 1725375039079340637797070384
133 2791715456571051233611642553
134 4517090495650391871408712937
135 7308805952221443105020355490
136 11825896447871834976429068427
137 19134702400093278081449423917
138 30960598847965113057878492344
139 50095301248058391139327916261
140 81055900096023504197206408605
141 131151201344081895336534324866
142 212207101440105399533740733471
143 343358302784187294870275058337
144 555565404224292694404015791808
145 898923707008479989274290850145
146 1454489111232772683678306641953
147 2353412818241252672952597492098
148 3807901929474025356630904134051
149 6161314747715278029583501626149
150 9969216677189303386214405760200
151 16130531424904581415797907386349
152 26099748102093884802012313146549
153 42230279526998466217810220532898
154 68330027629092351019822533679447
155 110560307156090817237632754212345
156 178890334785183168257455287891792
157 289450641941273985495088042104137
158 468340976726457153752543329995929
159 757791618667731139247631372100066
160 1226132595394188293000174702095995
161 1983924214061919432247806074196061
162 3210056809456107725247980776292056
163 5193981023518027157495786850488117
164 8404037832974134882743767626780173
165 13598018856492162040239554477268290
166 22002056689466296922983322104048463
167 35600075545958458963222876581316753
168 57602132235424755886206198685365216
169 93202207781383214849429075266681969
170 150804340016807970735635273952047185
171 244006547798191185585064349218729154
172 394810887814999156320699623170776339
173 638817435613190341905763972389505493
174 1033628323428189498226463595560281832
175 1672445759041379840132227567949787325
176 2706074082469569338358691163510069157
177 4378519841510949178490918731459856482
178 7084593923980518516849609894969925639
179 11463113765491467695340528626429782121
180 18547707689471986212190138521399707760
181 30010821454963453907530667147829489881
182 48558529144435440119720805669229197641
183 78569350599398894027251472817058687522
184 127127879743834334146972278486287885163
185 205697230343233228174223751303346572685
186 332825110087067562321196029789634457848
More output
Sum of even Fibonacci numbers that do not exceed
limit sum
4 2
40 44
400 188
4000 3382
40000 14328
400000 257114
4000000 4613732
40000000 19544084
400000000 350704366
4000000000 1485607536
40000000000 26658145586
400000000000 478361013020
4000000000000 2026369768940
40000000000000 36361730124070
400000000000000 154030760585064
4000000000000000 2763969850442378
40000000000000000 49597426547377748
400000000000000000 210098070363744836
4000000000000000000 3770056902373173214
40000000000000000000 15970217317495049952
400000000000000000000 286573922006908542050
4000000000000000000000 5142360378806858706956
40000000000000000000000 21783388129427422369052
400000000000000000000000 390887039715493615101718
4000000000000000000000000 1655824071758491008590040
40000000000000000000000000 29712557378756321606437562
400000000000000000000000000 125864412841774744075212130
4000000000000000000000000000 2258545247825195935704356468
40000000000000000000000000000 40527950048011752098603204302
400000000000000000000000000000 171679151392093647435137529168
4000000000000000000000000000000 3080657373857639014791750813074
40000000000000000000000000000000 13049874051046942401006156573274
400000000000000000000000000000000 234170488363228576876271664997964
4000000000000000000000000000000000 4202018916487067441371883813390086
40000000000000000000000000000000000 17800037772979229481611438290658376
400000000000000000000000000000000000 319408717806595170952881986194752746
4000000000000000000000000000000000000 1353037041234784669179345581755034578
40000000000000000000000000000000000000 24279264572217720059860402834614598820
By Binet's formula,
√5 F3m = p^m + q^m
where p is the cubed Golden ratio and q the cubed inverse.
Summing from 0 to m = n/3 one gets
√5 Σ F3m = ((p³)^(m+1) - 1) / (p³-1) - ((q³)^(m+1) - 1) / (q³-1)
If floating-point is allowed, this is computed in constant time.
Question: Is it possible to solve this in constant time?
I think OP means O(1) and not constant time as functions like pow(), log() are rarely ever constant time.
Floating point (FP) approach has troubles.
A common problem with FP occurs when one wants fast code (O(1)), yet does not realized how often the answer is incorrect or excessively far from the best answer.
Testing is needed.
Sometime rigorous testing is needed.
OP's answer was not apparently tested with various bounds.
correct bound sum
ref 0 0
ref 2 2
ref 8 10
ref 34 44
ref 144 188
ref 610 798
ref 2584 3382
ref 10946 14328
ref 46368 60696
ref 196418 257114
ref 832040 1089154
ref 3524578 4613732
Haziq 0 0
Haziq 2 2
Haziq 7 10 error
Haziq 27 44 error
Haziq 114 188 error
Haziq 480 798 error
Haziq 2032 3382 error
Haziq 8606 14328 error
Haziq 36453 60696 error
Haziq 154415 257114 error
Haziq 654110 1089154 error
Haziq 2770852 4613732 error
Incorrect results
find_fib_n() is a major culprit.
Using floor() instead of round() helps reduce the various bound errors:
// return round(log(sqrt(5) * F_n) / log((1.0 + sqrt(5)) / 2.0));
return floor(log(sqrt(5) * F_n) / log((1.0 + sqrt(5)) / 2.0));
Yet is still not fully correct.
Haziq 0 0
Haziq 2 2
Haziq 9 10 error
Haziq 34 44
Haziq 145 188 error
Haziq 610 798
Haziq 2585 3382 error
Haziq 10946 14328
Haziq 46369 60696 error
Haziq 196418 257114
Haziq 832041 1089154 ...
Haziq 3524578 4613732
To improve further is challenged by the slight errors in the argument to log() and that the formula used is not fully correct, but only mathematicaly approximate.
We could try:
F_n += (F_n & 1) == 0;
return floor(log2(sqrt(5) * F_n) / log2((1.0 + sqrt(5)) / 2.0));
Which returns the correct result over OP's target range (your results may differ). Yet that is not a vetted solution in general.
As it turns out, F_n += (F_n & 1) == 0; is better than first thought.
Deeper
OP's code hinges on returning a correct value from find_fib_n(bound). Although that function takes a integer, it could take a double. When bound = some_even_fib_number, it returns i, yet with some_even_fib_number - 0.00000001, it needs to return i - 1. Thus find_fib_n(bound) is very sensitive to calculation errors when bound = some_even_fib_number.
Notice that find_fib_n(some_even_bound) is always equal to find_fib_n(some_even_bound + 1) and the return value only change as bound increases from some_odd_bound.9999999 to the _next_even_bound.
Incrementing any even_bound, we push the floating point calculation away from that sensitive some_even_fib_number.00000 to some_even_fib_number + 1.
This nicely provides more error resilience to OP's calculation.
Combined with other silent improves consider:
unsigned long long binet(int N) {
// (sqrt(5) + 1)/2
#define PHI 1.61803398874989484820458683436564
#define root5i 0.44721359549995793928183473374626
double b = pow(PHI, N) * root5i;
assert(b < (double) LLONG_MAX);
return (unsigned long long) llround(b);
}
int find_fib_n(unsigned long long F_n) {
// 1/log2(PHI)
#define LN2PHIi (1.4404200904125564790175514995879)
F_n += (F_n & 1) == 0;
return (int) floor(log2(sqrt(5) * (double)F_n) * LN2PHIi);
}
// Input N
// Count of even numbers in the first N fib nums
int count_even_in_fib_seq(int N) {
if (N <= 3) {
return 0;
}
int tmp = N;
tmp -= 4;
return 1 + tmp / 3;
}
Moral of the story: Test code for correct functionality before worrying about speed.
For reference, follows is a correct and up to a 64-bit sum formed by using integer only math.
bound sum
ref 0 0
ref 2 2
ref 8 10
ref 34 44
ref 144 188
ref 610 798
ref 2584 3382
ref 10946 14328
ref 46368 60696
ref 196418 257114
ref 832040 1089154
ref 3524578 4613732
ref 14930352 19544084
ref 63245986 82790070
ref 267914296 350704366
ref 1134903170 1485607536
ref 4807526976 6293134512
ref 20365011074 26658145586
ref 86267571272 112925716858
ref 365435296162 478361013020
ref 1548008755920 2026369768940
ref 6557470319842 8583840088782
ref 27777890035288 36361730124070
ref 117669030460994 154030760585064
ref 498454011879264 652484772464328
ref 2111485077978050 2763969850442378
ref 8944394323791464 11708364174233842
ref 37889062373143906 49597426547377748
ref 160500643816367088 210098070363744836
ref 679891637638612258 889989708002357094
ref 2880067194370816120 3770056902373173214

how to display the table in c? [closed]

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Want to improve this question? Add details and clarify the problem by editing this post.
Closed 5 months ago.
Improve this question
#include <stdio.h>
#include <math.h>
int main(void)
{
// input value
int num, lower, upper;
double squareroot;
int square;
int cube;
printf("enter your number:\n");
scanf_s("%d", &num);
do
{
printf("the lower value limit is ");
scanf_s("%d", &lower);
} while (lower < 0 || lower > 50);
do
{
printf("the upper value limit is ");
scanf_s("%d", &upper);
} while (upper < 0 || upper > 50);
// the formular to find the squareroot, square, cube
squareroot = sqrt(num);
square = num * num;
cube = num * num * num;
//a for loop
for (num = 0; num <= upper; num++) {
printf("*base number* || *square root* || *square* || *cube*\n");
printf("*%d* || *%f* || *%ld* || *%ld*\n",
lower, squareroot, square, cube);
}
return 0;
}
i try to make a table to display the base number, square root, square, and cube
and set a limit for the table.
for example, if I input the lower number is 1 and the upper number is 5 then the table will stop at 5 then display the square root, square, and cube
At least this problem:
Mismatched specifiers/type
int square;
int cube;
...
printf("*%d* || *%f* || *%ld* || *%ld*\n",
lower, squareroot, square, cube);
Use "%d" with int, not "%ld".
Move assignments
Following assignments need to be inside the loop.
for (num = 0; num <= upper; num++) {
square = num * num;
cube = num * num * num;
Save time. Enable all compiler warnings
Try the Below Code Make all Changes I have added Comments to Clarify why I made The
#include <stdio.h>
#include <math.h>
int main(void)
{
// input value
int lower, upper;
double squareroot;
int square;
int cube;
//Read the Limits First
printf("Enter the Lower Limit: ");
scanf("%d", &lower);
printf("Enter the Upper Limit: ");
scanf("%d", &upper);
//Instead of Declaring an Entire Loop You Can Just Use an If Statement this Reduces Code Complexity
//You can also Set Your Limits
if(upper > 0 && upper < 50 && lower > 0 && lower < 50)
{
//THen Enter Actual Code
//Also Dont Set Lower to 0 It will Change Your Actual Value Instead Take another Loop Var
for (int i = lower; i <= upper; i++)
{
//Then Perform all Functions on i
squareroot = sqrt(i);
square = i * i;
cube = i * i * i;
printf("*base number* || *square root* || *square* || *cube*\n");
printf("*%d* || *%f* || *%d* || *%d*\n", i, squareroot, square, cube);
}
}
return 0;
}
Several issues:
You go to the trouble of asking for lower, but you don't use it to control your loop - your loop should befor( int num = lower; num <= upper; num++)
{
...
}
Put another way, if you always intend for your loop to start from zero, then you don't need lower at all.
You need compute the square root, square, and cube of num for each iteration of the loop. Instead, you're doing it once before the loop and just printing those same values over and over again;
You only need to print your table header once, outside the body of the loop;
You can compute your square root, square, and cube all as part of the printf statement:
printf( "%d %f %d %d\n", num, sqrt((double) num), num * num, num * num * num );
Field width specifiers are your friends - you can tell printf exactly how wide you want each column to be. Example:
printf( "%6d%12.2f%12d%12d", num, sqrt((double) num), num * num, num * num * num );
This means the column for num is 6 characters wide, the column for square root is 12 characters wide, with 3 characters reserved for the decimal point and two following digits, and the columns for square and cube are 12 characters wide. Example:
printf( "%6s%12s%12s%12s\n", "base", "root", "square", "cube" );
printf( "%6s%12s%12s%12s\n", "----", "----", "------", "----" );
for (int i = lower; i <= upper; i++ )
printf( "%6d%12.2f%12d%12d\n", i, sqrt( (double) i ), i*i, i*i*i );
Which gives output like this (lower == 1, upper == 10):
base root square cube
---- ---- ------ ----
1 1.00 1 1
2 1.41 4 8
3 1.73 9 27
4 2.00 16 64
5 2.24 25 125
6 2.45 36 216
7 2.65 49 343
8 2.83 64 512
9 3.00 81 729
10 3.16 100 1000
Full example:
#include <stdio.h>
#include <math.h>
int main( void )
{
int lower = 0, upper = 0;
printf( "Gimme a lower value: " );
while ( scanf( "%d", &lower ) != 1 || (lower < 0 || lower > 50 ))
{
/**
* Clear any non-numeric characters from the input stream
*/
while ( getchar() != '\n' )
; // empty loop
printf( "Nope, try again: " );
}
printf( "Gimme an upper value: " );
while ( scanf( "%d", &upper ) != 1 || (upper < lower || upper > 50 ))
{
while( getchar() != '\n' )
; // empty loop
printf( "Nope, try again: " );
}
printf( "%6s%12s%12s%12s\n", "base", "root", "square", "cube" );
printf( "%6s%12s%12s%12s\n", "----", "----", "------", "----" );
for (int i = lower; i <= upper; i++ )
printf( "%6d%12.2f%12d%12d\n", i, sqrt( (double) i ), i*i, i*i*i );
return 0;
}
I've written the input section such that it will reject inputs like foo or a123. It will not properly handle inputs like 12w4, but that would make the example more complicated than it needs to be (you're not asking about input validation, you're asking about computation and formatting). Example run:
$ ./table
Gimme a lower value: foo
Nope, try again: a123
Nope, try again: 123
Nope, try again: 1
Gimme an upper value: 100
Nope, try again: 50
base root square cube
---- ---- ------ ----
1 1.00 1 1
2 1.41 4 8
3 1.73 9 27
4 2.00 16 64
5 2.24 25 125
6 2.45 36 216
7 2.65 49 343
8 2.83 64 512
9 3.00 81 729
10 3.16 100 1000
11 3.32 121 1331
12 3.46 144 1728
13 3.61 169 2197
14 3.74 196 2744
15 3.87 225 3375
16 4.00 256 4096
17 4.12 289 4913
18 4.24 324 5832
19 4.36 361 6859
20 4.47 400 8000
21 4.58 441 9261
22 4.69 484 10648
23 4.80 529 12167
24 4.90 576 13824
25 5.00 625 15625
26 5.10 676 17576
27 5.20 729 19683
28 5.29 784 21952
29 5.39 841 24389
30 5.48 900 27000
31 5.57 961 29791
32 5.66 1024 32768
33 5.74 1089 35937
34 5.83 1156 39304
35 5.92 1225 42875
36 6.00 1296 46656
37 6.08 1369 50653
38 6.16 1444 54872
39 6.24 1521 59319
40 6.32 1600 64000
41 6.40 1681 68921
42 6.48 1764 74088
43 6.56 1849 79507
44 6.63 1936 85184
45 6.71 2025 91125
46 6.78 2116 97336
47 6.86 2209 103823
48 6.93 2304 110592
49 7.00 2401 117649
50 7.07 2500 125000

Generate Unique Values

I want to create a C program to generate numbers from 0 to 999999, keeping in mind that the number generated should not have any digits that are repetitive within it. For example, "123" is an acceptable value but not "121" as the '1' is repeated. I have sourced other program codes that check if an integer has repeated digits:
Check if integer has repeating digits. No string methods or arrays
What is the fastest way to check for duplicate digits of a number?
However these do not really solve my problem and they are very inefficient solutions if I were to perform the check for 1,000,000 different values. Moreover, the solution provided is for int and not char[] and char *, which I use in my program. Below is my code thus far. As you can see I have no problems handling values of up to "012", however the possibilities for values with 3 digits and above are too many to list and too inefficient to code. Would appreciate some help.
int i, j;
char genNext[7] = "0";
printf("%s\n", genNext);
// loop through to return next pass in sequence
while (1) {
for (i = 0; i < sizeof(genNext) / sizeof(char); i++) {
if (genNext[i] == '9') {
char * thisPass = strndup(genNext, sizeof(genNext));
int countDigit = (int) strlen(thisPass);
switch (countDigit) {
case 1:
genNext = "01";
break;
case 2:
if (strcmp(genNext, "98")) {
if (i == 0) {
genNext[1] += 1;
} else {
genNext[0] += 1;
genNext[1] == '0';
}
} else {
genNext = "012";
}
break;
case 3:
if (strcmp(genNext, "987")) {
// code to handle all cases
} else {
genNext = "0123";
}
break;
case 4:
case 5:
case 6:
// insert code here
}
break;
} else if (genNext[i] == '\0') {
break;
} else if (genNext[i+1] == '\0') {
genNext[i] += 1;
for (j = 0; j < i; j++) {
if (genNext[i] == genNext[j]) {
genNext[i] += 1;
}
}
} else {
continue;
}
}
printf("%s\n", genNext);
if (strcmp(genNext, "987654") == 0) {
break;
}
}
The main problem that I am facing is the cases when '9' is part of the value that is being tested. For example, the next value in the sequence after "897" is "901" and after "067895" comes "067912" based on the rules of non-repetitiveness as well as sequential returning of the result.
A desired output would be as follows:
0
1
2
3
...
8
9
01
02
03
...
09
10
12
13
...
97
98
012
013
014
...
098
102
103
...
985
986
987
0123
0124
...
etc etc.
Any assistance is appreciated, and if any part of my question was unclear, feel free to clarify. Thanks!
EDIT: How do I generate all permutations of a list of numbers? does not solve my question as the increment from "120398" to "120435" as the next "legal" value in the sequence.
EDIT 2: Updated question to include desired output
There are three variant algorithms below. Adapt variant 3 to suit your requirements.
Variant 1
This is one way to do it. It implements a minor variant of the initialize a table of 10 digit counts to 0; scan the digits, increment the count for each digit encountered, then check whether any of the digit counts is more than 1 algorithm I suggested in a comment. The test function returns as soon as a duplicate digit is spotted.
#include <stdio.h>
#include <stdbool.h>
enum { MAX_ITERATION = 1000000 };
static bool duplicate_digits_1(int value)
{
char buffer[12];
snprintf(buffer, sizeof(buffer), "%d", value);
char digits[10] = { 0 };
char *ptr = buffer;
char c;
while ((c = *ptr++) != '\0')
{
if (++digits[c - '0'] > 1)
return true;
}
return false;
}
int main(void)
{
int count = 0;
for (int i = 0; i < MAX_ITERATION; i++)
{
if (!duplicate_digits_1(i))
{
count += printf(" %d", i);
if (count > 72)
{
putchar('\n');
count = 0;
}
}
}
putchar('\n');
return 0;
}
When run, it produces 168,571 values between 0 and 1,000,000, starting:
0 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 23 24 25 26 27 28 29
30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47 48 49 50 51 52 53 54 56 57
58 59 60 61 62 63 64 65 67 68 69 70 71 72 73 74 75 76 78 79 80 81 82 83 84
85 86 87 89 90 91 92 93 94 95 96 97 98 102 103 104 105 106 107 108 109 120
123 124 125 126 127 128 129 130 132 134 135 136 137 138 139 140 142 143 145
146 147 148 149 150 152 153 154 156 157 158 159 160 162 163 164 165 167 168
169 170 172 173 174 175 176 178 179 180 182 183 184 185 186 187 189 190 192
193 194 195 196 197 198 201 203 204 205 206 207 208 209 210 213 214 215 216
217 218 219 230 231 234 235 236 237 238 239 240 241 243 245 246 247 248 249
250 251 253 254 256 257 258 259 260 261 263 264 265 267 268 269 270 271 273
…
987340 987341 987342 987345 987346 987350 987351 987352 987354 987356 987360
987361 987362 987364 987365 987401 987402 987403 987405 987406 987410 987412
987413 987415 987416 987420 987421 987423 987425 987426 987430 987431 987432
987435 987436 987450 987451 987452 987453 987456 987460 987461 987462 987463
987465 987501 987502 987503 987504 987506 987510 987512 987513 987514 987516
987520 987521 987523 987524 987526 987530 987531 987532 987534 987536 987540
987541 987542 987543 987546 987560 987561 987562 987563 987564 987601 987602
987603 987604 987605 987610 987612 987613 987614 987615 987620 987621 987623
987624 987625 987630 987631 987632 987634 987635 987640 987641 987642 987643
987645 987650 987651 987652 987653 987654
Before you decide this is 'not efficient', measure it. Are you really exercising it often enough that the performance is a real problem?
Variant 2
Creating the alternative version I suggested in the comments: use strchr() iteratively, checking whether the first digit appears in the tail, and if not whether the second digit appears in the tail, and so on is very easy to implement given the framework of the first answer:
static bool duplicate_digits_2(int value)
{
char buffer[12];
snprintf(buffer, sizeof(buffer), "%d", value);
char *ptr = buffer;
char c;
while ((c = *ptr++) != '\0')
{
if (strchr(ptr, c) != NULL)
return true;
}
return false;
}
When the times are compared I got these results (ng41 uses duplicate_digits_1() and ng43 uses duplicate_digits_2().
$ time ng41 > /dev/null
real 0m0.175s
user 0m0.169s
sys 0m0.002s
$ time ng43 > /dev/null
real 0m0.201s
user 0m0.193s
sys 0m0.003s
$
Repeated timings generally showed similar results, but sometimes I got ng43 running faster than ng41 — the timing on just one set of one million numbers isn't clear cut (so YMMV — your mileage may vary!).
Variant 3
You could also use this technique, which is analogous to 'count digits' but without the conversion to string first (so it should be quicker).
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
enum { MAX_ITERATION = 1000000 };
static bool duplicate_digits_3(int value)
{
char digits[10] = { 0 };
while (value > 0)
{
if (++digits[value % 10] > 1)
return true;
value /= 10;
}
return false;
}
int main(void)
{
int count = 0;
const char *pad = "";
for (int i = 0; i < MAX_ITERATION; i++)
{
if (!duplicate_digits_3(i))
{
count += printf("%s%d", pad, i);
pad = " ";
if (count > 72)
{
putchar('\n');
count = 0;
pad = "";
}
}
}
putchar('\n');
return 0;
}
Because it avoids conversions to strings, it is much faster. The slowest timing I got from a series of 3 runs was:
real 0m0.063s
user 0m0.060s
sys 0m0.001s
which is roughly three times as fast as either of the other two.
Extra timing
I also changed the value of MAX_ITERATION to 10,000,000 and ran timing. There are many more rejected outputs, of course.
$ time ng41 >/dev/null
real 0m1.721s
user 0m1.707s
sys 0m0.006s
$ time ng43 >/dev/null
real 0m1.958s
user 0m1.942s
sys 0m0.008s
$ time ng47 >/dev/null
real 0m0.463s
user 0m0.454s
sys 0m0.004s
$ ng41 | wc
69237 712891 5495951
$ ng43 | wc
69237 712891 5495951
$ ng47 | wc
69237 712891 5495951
$ cmp <(ng41) <(ng43)
$ cmp <(ng41) <(ng47)
$ cmp <(ng43) <(ng47)
$
These timings were more stable; variant 1 (ng41) was always quicker than variant 2 (ng43), but variant 3 (ng47) beats both by a significant margin.
JFTR: testing was done on macOS Sierra 10.12.1 with GCC 6.2.0 on an old 17" MacBook Pro — Early 2011, 2.3GHz Intel Core i7 with 16 GB 1333 MHz DDR3 RAM — not that memory is an issue with this code. The program numbers are consecutive 2-digit primes, in case you're wondering.
Leading zeros too
This code generates the sequence of numbers you want (though it is only configured to run up to 100,000 — the change for 1,000,000 is trivial). It's fun in a masochistic sort of way.
#include <assert.h>
#include <stdbool.h>
#include <stdio.h>
#include <string.h>
enum { MAX_ITERATIONS = 100000 };
/* lz = 1 or 0 - consider that the number has a leading zero or not */
static bool has_duplicate_digits(int value, int lz)
{
assert(value >= 0 && value < MAX_ITERATIONS + 1);
assert(lz == 0 || lz == 1);
char digits[10] = { [0] = lz };
while (value > 0)
{
if (++digits[value % 10] > 1)
return true;
value /= 10;
}
return false;
}
int main(void)
{
int lz = 0;
int p10 = 1;
int log_p10 = 0; /* log10(0) is -infinity - but 0 works better */
int linelen = 0;
const char *pad = "";
/* The + 1 allows the cycle to reset for the leading zero pass */
for (int i = 0; i < MAX_ITERATIONS + 1; i++)
{
if (i >= 10 * p10 && lz == 0)
{
/* Passed through range p10 .. (10*p10-1) once without leading zeros */
/* Repeat, adding leading zeros this time */
lz = 1;
i = p10;
}
else if (i >= 10 * p10)
{
/* Passed through range p10 .. (10*p10-1) without and with leading zeros */
/* Continue through next range, without leading zeros to start with */
p10 *= 10;
log_p10++;
lz = 0;
}
if (!has_duplicate_digits(i, lz))
{
/* Adds a leading zero if lz == 1; otherwise, it doesn't */
linelen += printf("%s%.*d", pad, log_p10 + lz + 1, i);
pad = " ";
if (linelen > 72)
{
putchar('\n');
pad = "";
linelen = 0;
}
}
}
putchar('\n');
return 0;
}
Sample output (to 100,000):
0 1 2 3 4 5 6 7 8 9 01 02 03 04 05 06 07 08 09 10 12 13 14 15 16 17 18 19
20 21 23 24 25 26 27 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47
48 49 50 51 52 53 54 56 57 58 59 60 61 62 63 64 65 67 68 69 70 71 72 73 74
75 76 78 79 80 81 82 83 84 85 86 87 89 90 91 92 93 94 95 96 97 98 012 013
014 015 016 017 018 019 021 023 024 025 026 027 028 029 031 032 034 035 036
037 038 039 041 042 043 045 046 047 048 049 051 052 053 054 056 057 058 059
061 062 063 064 065 067 068 069 071 072 073 074 075 076 078 079 081 082 083
084 085 086 087 089 091 092 093 094 095 096 097 098 102 103 104 105 106 107
108 109 120 123 124 125 126 127 128 129 130 132 134 135 136 137 138 139 140
…
901 902 903 904 905 906 907 908 910 912 913 914 915 916 917 918 920 921 923
924 925 926 927 928 930 931 932 934 935 936 937 938 940 941 942 943 945 946
947 948 950 951 952 953 954 956 957 958 960 961 962 963 964 965 967 968 970
971 972 973 974 975 976 978 980 981 982 983 984 985 986 987 0123 0124 0125
0126 0127 0128 0129 0132 0134 0135 0136 0137 0138 0139 0142 0143 0145 0146
0147 0148 0149 0152 0153 0154 0156 0157 0158 0159 0162 0163 0164 0165 0167
…
0917 0918 0921 0923 0924 0925 0926 0927 0928 0931 0932 0934 0935 0936 0937
0938 0941 0942 0943 0945 0946 0947 0948 0951 0952 0953 0954 0956 0957 0958
0961 0962 0963 0964 0965 0967 0968 0971 0972 0973 0974 0975 0976 0978 0981
0982 0983 0984 0985 0986 0987 1023 1024 1025 1026 1027 1028 1029 1032 1034
1035 1036 1037 1038 1039 1042 1043 1045 1046 1047 1048 1049 1052 1053 1054
1056 1057 1058 1059 1062 1063 1064 1065 1067 1068 1069 1072 1073 1074 1075
…
9820 9821 9823 9824 9825 9826 9827 9830 9831 9832 9834 9835 9836 9837 9840
9841 9842 9843 9845 9846 9847 9850 9851 9852 9853 9854 9856 9857 9860 9861
9862 9863 9864 9865 9867 9870 9871 9872 9873 9874 9875 9876 01234 01235 01236
01237 01238 01239 01243 01245 01246 01247 01248 01249 01253 01254 01256 01257
01258 01259 01263 01264 01265 01267 01268 01269 01273 01274 01275 01276 01278
01279 01283 01284 01285 01286 01287 01289 01293 01294 01295 01296 01297 01298
…
09827 09831 09832 09834 09835 09836 09837 09841 09842 09843 09845 09846 09847
09851 09852 09853 09854 09856 09857 09861 09862 09863 09864 09865 09867 09871
09872 09873 09874 09875 09876 10234 10235 10236 10237 10238 10239 10243 10245
10246 10247 10248 10249 10253 10254 10256 10257 10258 10259 10263 10264 10265
…
98705 98706 98710 98712 98713 98714 98715 98716 98720 98721 98723 98724 98725
98726 98730 98731 98732 98734 98735 98736 98740 98741 98742 98743 98745 98746
98750 98751 98752 98753 98754 98756 98760 98761 98762 98763 98764 98765 012345
012346 012347 012348 012349 012354 012356 012357 012358 012359 012364 012365
012367 012368 012369 012374 012375 012376 012378 012379 012384 012385 012386
…
098653 098654 098657 098671 098672 098673 098674 098675 098712 098713 098714
098715 098716 098721 098723 098724 098725 098726 098731 098732 098734 098735
098736 098741 098742 098743 098745 098746 098751 098752 098753 098754 098756
098761 098762 098763 098764 098765
Using a loop (from 0 to 999,999, inclusive), and rejecting values with repeating digits sounds like the most straightforward implementation to me.
The reject-if-duplicate-digits function can be made to be pretty fast. Consider, for example,
int has_duplicate_digits(unsigned int value)
{
unsigned int digit_mask = 0U;
do {
/* (value % 10U) is the value of the rightmost
decimal digit of (value).
1U << (value % 10U) refers to the value of
the corresponding bit -- bit 0 to bit 9. */
const unsigned int mask = 1U << (value % 10U);
/* If the bit is already set in digit_mask,
we have a duplicate digit in value. */
if (mask & digit_mask)
return 1;
/* Mark this digit as seen in the digit_mask. */
digit_mask |= mask;
/* Drop the rightmost digit off value.
Note that this is integer division. */
value /= 10U;
/* If we have additional digits, repeat loop. */
} while (value);
/* No duplicate digits found. */
return 0;
}
This is actually a classical combinatorial problem. Below is a proof of concept implementation using Algorithm L in TAOCP 7.2.1.2 and Algorithm T in TAOCP 7.2.1.3. There might be some errors. Refer to the algorithms for details.
Here is a bit of explanation. Let t be the number of digits. For t == 10, the problem is to generate all t! permutations of the set {0,1,2,...,9} in lexicographic order (Algorithm L).
For t > 0 and t < 10, this breaks down to 1) Generate all combinations of t digits from the 10 possible digits. 2). For each combination, generate all t! permutations.
Last, you can sort all 10! + 10! / 2 + 10! / 3! + .. + 10 results. The sorting might look expensive at first. But first, the combination generating is already in lexical order. Second, the permutation generating is also in lexical order. So the sequence is actually highly regular. A QSort is not really too bad here.
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
static inline int compare_str(const void *a, const void *b)
{
return strcmp(a, b);
}
static inline int compare_char(const void *a, const void *b)
{
char ca = *((char *) a);
char cb = *((char *) b);
if (ca < cb)
return -1;
if (ca > cb)
return 1;
return 0;
}
// Algorithm L in TAOCP 7.2.1.2
static inline char *algorithm_l(int n, const char *c, char *r)
{
char a[n + 1];
memcpy(a, c, n);
a[n] = '\0';
qsort(a, n, 1, compare_char);
while (1) {
// L1. [Visit]
memcpy(r, a, n + 1);
r += n + 1;
// L2. [Find j]
int j = n - 1;
while (j > 0 && a[j - 1] >= a[j])
--j;
if (j == 0)
break;
// L3. [Increase a[j - 1]]
int l = n;
while (l >= 0 && a[j - 1] >= a[l - 1])
--l;
char tmp = a[j - 1];
a[j - 1] = a[l - 1];
a[l - 1] = tmp;
// L4. [Reverse a[j]...a[n-1]]
int k = j + 1;
l = n;
while (k < l) {
char tmp = a[k - 1];
a[k - 1] = a[l - 1];
a[l - 1] = tmp;
++k;
--l;
}
}
return r;
}
// Algorithm T in TAOCP 7.2.1.2
static inline void algorithm_t(int t, char *r)
{
assert(t > 0);
assert(t < 10);
// Algorithm T in TAOCP 7.2.1.3
// T1. [Initialize]
char c[12]; // the digits
for (int i = 0; i < t; ++i)
c[i] = '0' + i;
c[t] = '9' + 1;
c[t + 1] = '0';
char j = t;
char x = '0';
while (1) {
// T2. [Visit]
r = algorithm_l(t, c, r);
if (j > 0) {
x = '0' + j;
} else {
// T3. [Easy case?]
if (c[0] + 1 < c[1]) {
++c[0];
continue;
}
j = 2;
// T4. [Find j]
while (1) {
c[j - 2] = '0' + j - 2;
x = c[j - 1] + 1;
if (x != c[j])
break;
++j;
}
// T5. [Done?]
if (j > t)
break;
}
// T6. [Increase c[j - 1]]
c[j - 1] = x;
--j;
}
}
static inline void generate(int t)
{
assert(t >= 0 && t <= 10);
if (t == 0)
return;
int n = 1;
int k = 10;
for (int i = 1; i <= t; ++i, --k)
n *= k;
char *r = (char *) malloc((t + 1) * n);
if (t == 10) {
algorithm_l(10, "0123456789", r);
} else {
algorithm_t(t, r);
}
qsort(r, n, t + 1, strcmpv);
for (int i = 0; i < n; ++i, r += t + 1)
printf("%s\n", r);
}
int main()
{
for (int t = 1; t <= 10; ++t)
generate(t);
}
Efficiency: This is implementation is not very efficient. It is a direct translation from the algorithm, for easier understanding. However it is still a lot more efficient than iterating over 10^10 numbers. It takes about 2.5 seconds to generate all numbers from 0 to 9876543210. This includes the time of writing them to a file, a 94MB output file, with one number a line. For up to 6 digits, it takes about 0.05 seconds.
If you want these numbers come in the order you want in program, it might be better to generate the numbers as above to prepare a table and use the table later. Even for the table from 0 to 9876543210, there are less than ten million numbers, which is not a really big number in today's computers. In your case, up to six digits, there are only less than one million numbers.

Division of very big numbers using arrays in C

I'm trying to make a calculator for very big numbers (even bigger than long long) and I'm using arrays to make it work.
So far I have done addition, subtraction and multiplication. But I'm really stuck in division part.
EDIT:
new progress. as a friend mentioned i need to compare result array with divisor each time so i can stop the progress any time divisor is larger than dividend. I managed to make a nice function to compare it every time. this function is tested separately and it's working fine. OK. now i'm starting to make REAL progress. i got the quotient. now i will try to put quotient in array so that we can work with LARGER numbers!
#define MAX_SIZE 50
#define SIZE_USE (MAX_SIZE-1)
int div(int inum_first[], int inum_second[], int div_result[], int firstlen, int secondlen)
{
int i;
int check1 = 0, check2 = 0;
int zeroC = 0;
int tmp[MAX_SIZE];
for (i = 0; i <= SIZE_USE; i++)
{
tmp[i] = 0;
}
int inum_firstCP[MAX_SIZE] = { 0 };
for (i = 0; i <= 1; i++)
{
inum_firstCP[i] = inum_first[i]; // create a copy of inum_first
}
for (i = 0; i <= SIZE_USE; i++)
{
if (inum_first[i] != 0)
check1++;
if (inum_second[i] != 0)
check2++;
}
if (secondlen > firstlen)
{
zeroC++;
goto EOI;
}
if (check2 == 0)
{
puts("\nExpected error\n");
return -1;
}
int j = 0, p = 0;
int s = 0;
int o = 1; // o is Quotient!
do
{
for (i = SIZE_USE; i >= 0; i--)
{
if (tmp[i] = inum_firstCP[i] - inum_second[i] >= 0)
{
tmp[i] = inum_firstCP[i] - inum_second[i];
}
else
{
inum_firstCP[i - 1] = inum_firstCP[i - 1] - 1;
tmp[i] = (inum_firstCP[i] + 10) - inum_second[i];
}
inum_firstCP[i] = tmp[i];
}
if (compare(inum_firstCP, inum_second, firstlen, secondlen) < 0) break;
j++;
o++;
} while (j<MAX_SIZE); // anything else will also work
EOI:
return 0;
}
int compare(int inum_firstCP[], int inum_second[], int firstlen, int secondlen)
{
int c = 0, d = 0;
int i;
firstlen = MAX_SIZE, secondlen = MAX_SIZE; // temporary. will provide a better solution ASAP
if (firstlen > secondlen)
{
return 1;
}
else if (secondlen > firstlen)
{
return -1;
}
else
{
for (i = 0; i < firstlen; i++)
{
if (inum_firstCP[i] > inum_second[i]) c++;
else if (inum_second[i] > inum_firstCP[i]) d++;
}
if (c>d) return 1;
else if (d>c) return -1;
}
return 0; // else
}
If you have the subtraction of those big numbers the easiest solution is to take the two numbers and substract one from the other until you are left with something less then zero. It is the basic solution, it works but is a bit slow.
To make it faster you can do the following, take the divisor, multiply it by 2, if it is less then the dividend, keep on multiplying. When you will reach the first number bigger then a dividend set the corresponding bit to 1, subtract the multiplied dividend then do the same for the result.
There is the same thing nicely described on wiki.
In order to make it work you need to implement your own comparing function.
Assuming you will store the size of the malloc allocation in your structure in filed len you can do something like this:
int compare( mynum &a, mynum &b){
if (a.len() > b.len()){
return 1;
} else (if b.len() > a.len()){
return -1;
} else(){
for(int i = b.len(); i > 0; i--){
if (a[i] > b[i]){
return 1;
} else if(b[i] > a[i]){
return -1;
}
}
#if we get there the numbers are the same
return 0;
}
}
I've done this before and was very happy to implement it the same way as you'd do it by hand, with a small modification of multiple subtraction at each step. The algorithm is like that:
Multiply divisor by ten as often as you can without divisor becoming bigger than dividend.
Subtract divisor from dividend as often as you can and remember how many times.
The rest of all the subtractions is the new dividend.
Repeat at step (1) until dividend is smaller than divisor.
The current dividend is the "rest".
All the numbers remembered at step (3) are the "result" when ordered left to right (left calculated first).
Okay, let's try it by example:
E.g. you have 25391 and want to divide it by 71.
(1) 25391 and 71 * 10 = 710
25391 and 710 * 10 = 7100
25391 and 7100 * 10 = 71000 <-- TOO BIG
(2) 25391 - 7100 => X
18291 - 7100 => X
11191 - 7100 => X
4091 - 7100 <--- NOT POSSIBLE
(3) Number of X: 3
(4) 4091 > 71, okay, back to step 1.
(1) 4091 and 71 * 10 = 710
4091 and 710 * 10 = 7100 <--- TOO BIG
(2) 4091 - 710 => X
3381 - 710 => X
2671 - 710 => X
1961 - 710 => X
1251 - 710 => X
541 - 710 <--- NOT POSSIBLE
(3) Number of X: 5
(4) 541 > 71, okay, back to step 1
(1) 541 and 71 * 10 = 710 <--- TOO BIG
(2) 541 - 71 => X
470 - 71 => X
399 - 71 => X
328 - 71 => X
257 - 71 => X
186 - 71 => X
115 - 71 => X
44 - 71 <--- NOT POSSIBLE
(3) Number of X: 7
(4) 44 > 71, WRONG, continue with step 5
(5) Rest is 44
(6) Result is 357
If you had just tested how often you can subtract 71 from 25391, this loop would have had 357 iterations! Of course, my solution uses multiplication, but honestly, multiplying by 10 is no real multiplication, just shift all digits one position to the left and put a zero at the top right one.
The algorithm will need as many iterations as the result has digits and it will need at most 9 iterations (with subtraction) per digit.
#Mecki Try with 54 664 455 645 655 divided by 5 465 126 544, it fails. At step 3 you must add a number of '0' corresponding to the difference of length between the divisor (x n x 10) and the "rest". ie if the rest is 13 190 205 655 (11 digits length) and divisor is 54 651 265 440 000 (14 digits length) then three '0' must be added to the result before performing the next loop.

Multithreaded Sorting Application

I am new to multithreaded programming and so I thought I would work on a project to help me learn it. Here are the details of the project:
Write a multithreaded sorting program in c that works as follows: A list of integers is divided into two smaller lists of equal size. Two separate threads (which we will term sorting threads) sort each sublist using a sorting algorithm of your choice. The two sublists are then merged by a third thread - a merging thread - which merges the two sublists into a single sorted list.
//Sort a list of numbers using two separate threads
//by sorting half of each list separately then
//recombining the lists
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#define SIZE 10
#define NUMBER_OF_THREADS 3
void *sorter(void *params); /* thread that performs basic sorting algorithm */
void *merger(void *params); /* thread that performs merging of results */
int list[SIZE] = {7,12,19,3,18,4,2,6,15,8};
int result[SIZE];
typedef struct
{
int from_index;
int to_index;
} parameters;
int main (int argc, const char * argv[])
{
int i;
pthread_t workers[NUMBER_OF_THREADS];
/* establish the first sorting thread */
parameters *data = (parameters *) malloc (sizeof(parameters));
data->from_index = 0;
data->to_index = (SIZE/2) - 1;
pthread_create(&workers[0], 0, sorter, data);
/* establish the second sorting thread */
data = (parameters *) malloc (sizeof(parameters));
data->from_index = (SIZE/2);
data->to_index = SIZE - 1;
pthread_create(&workers[1], 0, sorter, data);
/* now wait for the 2 sorting threads to finish */
for (i = 0; i < NUMBER_OF_THREADS - 1; i++)
pthread_join(workers[i], NULL);
/* establish the merge thread */
data = (parameters *) malloc(sizeof(parameters));
data->from_index = 0;
data->to_index = (SIZE/2);
pthread_create(&workers[2], 0, merger, data);
/* wait for the merge thread to finish */
pthread_join(workers[2], NULL);
/* output the sorted array */
return 0;
}
void *sorter(void *params)
{
parameters* p = (parameters *)params;
//SORT
int begin = p->from_index;
int end = p->to_index+1;
int z;
for(z = begin; z < end; z++){
printf("The array recieved is: %d\n", list[z]);
}
printf("\n");
int i,j,t,k;
for(i=begin; i< end; i++)
{
for(j=begin; j< end-i-1; j++)
{
if(list[j] > list[j+1])
{
t = list[j];
list[j] = list[j+1];
list[j+1] = t;
}
}
}
for(k = begin; k< end; k++){
printf("The sorted array: %d\n", list[k]);
}
int x;
for(x=begin; x<end; x++)
{
list[x] = result[x];
}
printf("\n");
pthread_exit(0);
}
void *merger(void *params)
{
parameters* p = (parameters *)params;
//MERGE
int begin = p->from_index;
int end = p->to_index+1;
int i,j,t;
printf("list[1]: %d",list[1]);
printf("result[1]: %d",result[1]);
for(i=begin; i< end; i++)
{
for(j=begin; j< end-i; j++)
{
if(result[j] > result[j+1])
{
t = result[j];
result[j] = result[j+1];
result[j+1] = t;
}
}
}
int d;
for(d=0; d<SIZE; d++)
{
printf("The final resulting array is: %d\n", result[d]);
}
pthread_exit(0);
}
I'm not sure what I'm doing wrong in my algorithms that its not working. It doesn't seem to catch the new sorted array. Any help on this problem would be appreciated VERY much! Thanks again in advance for all your help!
Your approach is incorrect. You should be splitting your partitions, then recursing or threading into them, joining the results, then merging. Its easy to screw this algorithm up, believe me.
Before anything else, make sure your merge algorithm is solid. If your merge has issues in a single-threaded arena, adding threads is only going to make it worse. In your case, you're making it worse because your merge thread appears to be running concurrently with your sorter threads.
That said, step back and consider this. Mergesort is about divide and conquer. To thread up a merge sort you should be doing the following:
Establish a maximum number of threads. Believe me, the last thing you want happening is spinning a thread for each partition. a sequence of 1024 values has 1023 partitions if you crunch the math hard enough. that many threads is not a solution. Establish some boundaries.
Establish a minimum partition size that you're willing to spin a thread for. This is as important as the first item above. Just like you don't want to be spinning 1023 threads to sort a 1024-slot sequence, you also don't want to be spinning a thread just to sort a sequence that has two items. There is zero benefit and much cost.
Have a solid merge algorithm. There are many efficient ways to do it, but do something simple and enhance it later. Right now we're just interested in getting the general threading down right. There is always time to enhance this with a fancy merge algorithm (like in-place, which believe me is harder than it sounds).
Having the above the idea is this:
The merge sort algorithm will have three parameters: a starting pointer, a length, and a thread-depth. For our purposes the thread depth will be N in a situation where we are using at-most 2N-1 threads. (more on that later, but trust me, it makes it easier to do the math this way).
If the thread depth has reached zero OR the sequence length is below a minimum threshold *we set), do not setup and run a new thread. Just recurse into our function again.
Otherwise, split the partition. Setup a structure that holds a partition definition (which for us will be a starting point and a length as well as the thread depth which will be N/2), launch a thread with that parameter block, then do NOT launch another thread. instead use the current thread to recurse into merge_sort_mt() for the "other" half.
Once the current thread returns from its recursion is must wait on the other thread via a join. once that is done both partitions will be done and they can be merged using your trivial merge algorithm.
Whew. Ok. so how does it look in practice:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <pthread.h>
struct Params
{
int *start;
size_t len;
int depth;
};
// only used for synchronizing stdout from overlap.
pthread_mutex_t mtx = PTHREAD_MUTEX_INITIALIZER;
// forward declare our thread proc
void *merge_sort_thread(void *pv);
// a simple merge algorithm. there are *several* more efficient ways
// of doing this, but the purpose of this exercise is to establish
// merge-threading, so we stick with simple for now.
void merge(int *start, int *mid, int *end)
{
int *res = malloc((end - start)*sizeof(*res));
int *lhs = start, *rhs = mid, *dst = res;
while (lhs != mid && rhs != end)
*dst++ = (*lhs < *rhs) ? *lhs++ : *rhs++;
while (lhs != mid)
*dst++ = *lhs++;
// copy results
memcpy(start, res, (rhs - start) * sizeof *res);
free(res);
}
// our multi-threaded entry point.
void merge_sort_mt(int *start, size_t len, int depth)
{
if (len < 2)
return;
if (depth <= 0 || len < 4)
{
merge_sort_mt(start, len/2, 0);
merge_sort_mt(start+len/2, len-len/2, 0);
}
else
{
struct Params params = { start, len/2, depth/2 };
pthread_t thrd;
pthread_mutex_lock(&mtx);
printf("Starting subthread...\n");
pthread_mutex_unlock(&mtx);
// create our thread
pthread_create(&thrd, NULL, merge_sort_thread, &params);
// recurse into our top-end parition
merge_sort_mt(start+len/2, len-len/2, depth/2);
// join on the launched thread
pthread_join(thrd, NULL);
pthread_mutex_lock(&mtx);
printf("Finished subthread.\n");
pthread_mutex_unlock(&mtx);
}
// merge the partitions.
merge(start, start+len/2, start+len);
}
// our thread-proc that invokes merge_sort. this just passes the
// given parameters off to our merge_sort algorithm
void *merge_sort_thread(void *pv)
{
struct Params *params = pv;
merge_sort_mt(params->start, params->len, params->depth);
return pv;
}
// public-facing api
void merge_sort(int *start, size_t len)
{
merge_sort_mt(start, len, 4); // 4 is a nice number, will use 7 threads.
}
int main()
{
static const unsigned int N = 2048;
int *data = malloc(N * sizeof(*data));
unsigned int i;
srand((unsigned)time(0));
for (i=0; i<N; ++i)
{
data[i] = rand() % 1024;
printf("%4d ", data[i]);
if ((i+1)%8 == 0)
printf("\n");
}
printf("\n");
// invoke our multi-threaded merge-sort
merge_sort(data, N);
for (i=0; i<N; ++i)
{
printf("%4d ", data[i]);
if ((i+1)%8 == 0)
printf("\n");
}
printf("\n");
free(data);
return 0;
}
The output for this looks something like this:
825 405 691 290 900 715 125 969
534 809 783 820 933 895 310 687
152 19 659 856 46 765 497 371
339 660 297 509 152 796 230 465
502 948 278 317 144 941 195 208
617 428 118 505 719 161 53 292
....
994 154 745 666 590 356 894 741
881 129 439 237 83 181 33 310
549 484 12 524 753 820 443 275
17 731 825 709 725 663 647 257
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
0 0 1 1 1 2 3 3
5 5 5 5 6 6 7 7
7 7 7 8 8 10 10 11
11 11 12 12 12 13 14 14
15 15 15 15 16 17 17 17
17 18 18 19 19 19 20 21
21 21 22 22 23 24 24 24
25 25 25 26 26 28 28 29
29 29 30 30 30 30 30 31
....
994 995 996 998 1000 1001 1001 1003
1003 1003 1003 1004 1004 1005 1007 1007
1010 1010 1010 1010 1011 1012 1012 1012
1012 1013 1013 1013 1015 1015 1016 1016
1016 1017 1018 1019 1019 1019 1020 1020
1020 1021 1021 1021 1021 1022 1023 1023
The most important part of this is the limiters that keep us from going thread-wild (which is easy to accidentally do with recursive threaded algorithms), and the join of the threads before merging their content with the other half of the partition (which we sorted on our thread, and may also have done the same thing).
It's a fun exercise, and I hope you got something out of it. Best of luck.
Update: Integrating qsort()
An interesting task would be performing this functionality using qsort() for sorting the smaller partitions or once the thread pool reaches exhaustion. qsort() is a pretty big hammer to bring to this party, and as such you're going to want to raise the minimum partition size to something respectful (in the example below, we use 256 elements).
So what would it take to integrate qsort() the the sub partitions rather than a hand-rolled merge-sort? Surprisingly, not much. Start with a qsort() compatible comparator:
// comparator for qsort
int cmp_proc(const void *arg1, const void* arg2)
{
const int *lhs = arg1;
const int *rhs = arg2;
return (*lhs < *rhs) ? -1 : (*rhs < *lhs ? 1 : 0);
}
Pretty brain-dead. Now, modify the mt-wrapper to look something like this:
// our multi-threaded entry point.
void merge_sort_mt(int *start, size_t len, int depth)
{
if (len < 2)
return;
// invoke qsort on the partition. no need for merge
if (depth <= 0 || len <= 256)
{
qsort(start, len, sizeof(*start), cmp_proc);
return;
}
struct Params params = { start, len/2, depth/2 };
pthread_t thrd;
pthread_mutex_lock(&mtx);
printf("Starting subthread...\n");
pthread_mutex_unlock(&mtx);
// create our thread
pthread_create(&thrd, NULL, merge_sort_thread, &params);
// recurse into our top-end parition
merge_sort_mt(start+len/2, len-len/2, depth/2);
// join on the launched thread
pthread_join(thrd, NULL);
pthread_mutex_lock(&mtx);
printf("Finished subthread.\n");
pthread_mutex_unlock(&mtx);
// merge the paritions.
merge(start, start+len/2, start+len);
}
That's it. Seriously. That is all it takes. Proving this works is a simple test run with the original program, shown below:
986 774 60 596 832 171 659 753
638 680 973 352 340 221 836 390
930 38 564 277 544 785 795 451
94 602 724 154 752 381 433 990
539 587 194 963 558 797 800 355
420 376 501 429 203 470 670 683
....
216 748 534 482 217 178 541 242
118 421 457 810 14 544 100 388
291 29 562 718 534 243 322 187
502 203 912 717 1018 749 742 430
172 831 341 331 914 866 931 368
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Starting subthread...
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
Finished subthread.
0 0 1 1 1 1 3 3
3 4 5 5 6 6 6 6
7 7 8 9 10 10 10 10
11 12 12 12 13 13 14 14
14 15 15 15 16 17 17 19
19 20 20 21 21 21 22 22
23 23 23 24 24 24 25 26
26 26 26 27 28 28 28 28
....
1000 1000 1000 1001 1001 1002 1003 1003
1004 1004 1004 1005 1005 1005 1006 1007
1008 1010 1010 1010 1010 1010 1011 1011
1011 1012 1012 1012 1012 1013 1013 1013
1015 1015 1015 1016 1016 1017 1017 1017
1018 1018 1018 1019 1019 1021 1021 1022
As you can see, the results are similar.
A couple of issues:
1 - What do you think this code is doing:
int x;
for(x=begin; x<end; x++)
{
list[x] = result[x];
}
2 - Your merger currently looks exactly like your sorter. It should instead be merging the sorted values from the first half of the list and the second half of the list into the result.
Your code is correct i have modified your code and tried to figure out the error,
the loop indexes are not correctly mapped and you are assigning the null result list into actual data in one loop, so the list is taking zeroes.
Find below the modified code and output.
//Sort a list of numbers using two separate threads
//by sorting half of each list separately then
//recombining the lists
void *sort(void *params)
{
parameters* p = (parameters *)params;
//SORT
int begin = p->fromVal;
int end = p->toVal+1;
for(int i = begin; i < end; i++){
printf("The array recieved is: %d\n", list[i]);
}
printf("\n");
int temp=0;
for(int i=begin; i< end; i++)
{
for(int j=begin; j< end-1; j++)
{
if(list[j] > list[j+1])
{
temp = list[j];
list[j] = list[j+1];
list[j+1] = temp;
}
}
}
for(int k = begin; k< end; k++){
printf("The sorted array: %d\n", list[k]);
}
for(int i=begin; i<end; i++)
{
result[i] = list[i];
}
printf("\n");
pthread_exit(NULL);
}
void *merging(void *params)
{
parameters* p = (parameters *)params;
//MERGE
int begin = p->fromVal;
int end = p->toVal+1;
int temp;
for(int i=begin; i< end; i++)
{
for(int j=begin; j< end-1; j++)
{
if(result[j] > result[j+1])
{
temp= result[j];
result[j] = result[j+1];
result[j+1] = temp;
}
}
}
printf("\n\nFINAL RESULT IS:\n");
for(int d=begin+1; d<end; d++)
{
printf("The final resulting array is: %d\n", result[d]);
}
pthread_exit(NULL);
}
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <pthread.h>
#include <time.h>
/*globle variables*/
/* structure for passing data to threads */
typedef struct
{
int *start;
int end;
int size;
} parameters;
int t = 1;
int *arr1, *arr2;
//for using quicksort
int comparator (const void * a, const void * b) {
return ( *(int*)a - *(int*)b );
}
void *merge(void *params){
//get data
int *len = params;
//SORT
int start = 0;
int end = *len/2;
int counter = end;
int size = *len;
int index = 0;
while (start < end && counter < size)
{
if (arr1[start] < arr1[counter])
{
arr2[index] = arr1[start];
start ++;
}
else
{
arr2[index] = arr1[counter];
counter ++;
}
index ++;
}
/* Copy the remaining elements , if there
are any */
while ( start < end)
{
arr2[index] = arr1[start];
start ++;
index ++;
}
/* Copy the remaining elements , if there
are any */
while ( counter < size)
{
arr2[index] = arr1[counter];
counter ++;
index ++;
}
}
void *sorting_thread(void *params){
printf("Thread %d ......\n", t);
t++;
//get data
parameters* data = (parameters *)params;
//SORT
int end = data->end;
int size = data->size;
//qsort
qsort(data->start, end, sizeof(*data->start), comparator);
printf("The array after sort : \n");
for(int i = size - end; i < size; i ++){
printf("arr1[%d]:%d, \n", i,arr1[i]);
}
printf("\n");
pthread_exit(0);
}
void *merge_sort_thread(void *params){
int *len = params;
//varaible allocation for two sorting threads.
parameters *data = (parameters *) malloc (sizeof(parameters));
parameters *data1 = (parameters *) malloc (sizeof(parameters));
if(data == NULL&& data1 == NULL){
printf("Memory not allocated. \n");
exit(0);
}
//value for data passing
data->start= arr1;
data->end = *len/2;
data->size = *len/2;
data1->start = arr1 + *len/2;
data1->end = *len-*len/2;
data1->size = *len;
pthread_t left, right;/* the thread identifier */
printf("Entering merge_Sorting..\n");
/* create the sorting thread */
pthread_create(&left, NULL, sorting_thread, data);
pthread_create(&right, NULL, sorting_thread, data1);
/* wait for the thread to exit */
pthread_join(left, NULL);
//free memory
free(data);
pthread_join(right, NULL);
printf("Merging Thread %d ......\n", t);
merge(len);
printf("Process is done.\n");
printf("The final output: \n");
for(int i = 0; i < *len; i ++){
if(i%10==0){
printf("\n");
}
printf("%d, ", arr2[i]);
}
printf("\n");
//free memory
free(data1);
pthread_exit(0);
}
int main( int argc, char *argv[] ) {
long len;
int temp, c, j, k;
char *ptr;
//
//check if the right amount of argument
if( argc == 2 ) {
printf("The input array size is %s\n", argv[1]);
//covert the user input to integer
len = strtol(argv[1], &ptr, 10);
//check if the input is valid.
if(len == 0) {//if not, leave the program.
printf("Please enter a proper number. Leaving the program...\n");
}else{
//dynamically allocate memory
arr1 = (int*)malloc(len * sizeof(int));
arr2 = (int*)malloc(len * sizeof(int));
//check Memory
if(arr1 == NULL && arr2 == NULL){
printf("Memory not allocated. \n");
exit(0);
}
printf("Memory allocated. \n");
//decide the value of data.
//generate random number to 100
srand(time(0));
printf("The array before sorting is: \n");
for(int i = 0; i < len; i ++){
arr1[i] = rand() % 100;
if(i%10==0){
printf("\n");
}
printf("%d, ", arr1[i]);
}
printf(" \n");
//merge sort handle all the threads
pthread_t tid;/* the thread identifier */
/* create the parent sorting thread */
pthread_create(&tid, NULL, merge_sort_thread, &len);
//wait for children thread
pthread_join(tid, NULL);
//printout array after merging threading
printf("\nThe program is finished. \n");
//free memory space
free(arr2);
free(arr1);
}
}
else if( argc > 2 ) {
printf("Too many arguments supplied.\n");
}
else {
printf("One argument expected.\n");
}
return 0;
}

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