How would you create n nested loops for math?
1
So, I am trying to wrap my head around understanding how you can use a variable to denote how many times a loop is nested.
Here is an example I write up to simulate the output of dimensions = 4:
static void Main(string args)
{
int dimensions = 4; // e.g. for (1, 2, 3, 4), dimensions = 4
Console.WriteLine($"{addNumbers(dimensions)}");
Console.ReadKey();
}
static long addNumbers(int dimensions)
{
long number = 0;
// hard coded to be dimensions = 4
for (int h = 0; h <= dimensions; h++)
for (int i = 0; i <= dimensions; i++)
for (int j = 0; j <= dimensions; j++)
for (int k = 0; k <= dimensions; k++)
number += h + i + j + k; // just some random math
return number;
}
This will present the expected output of:
5000
So to readdress the problem, how can I code to allow this for n dimensions? Thanks for your help!
c# loops nested nested-loops
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
add a comment |
1
So, I am trying to wrap my head around understanding how you can use a variable to denote how many times a loop is nested.
Here is an example I write up to simulate the output of dimensions = 4:
static void Main(string args)
{
int dimensions = 4; // e.g. for (1, 2, 3, 4), dimensions = 4
Console.WriteLine($"{addNumbers(dimensions)}");
Console.ReadKey();
}
static long addNumbers(int dimensions)
{
long number = 0;
// hard coded to be dimensions = 4
for (int h = 0; h <= dimensions; h++)
for (int i = 0; i <= dimensions; i++)
for (int j = 0; j <= dimensions; j++)
for (int k = 0; k <= dimensions; k++)
number += h + i + j + k; // just some random math
return number;
}
This will present the expected output of:
5000
So to readdress the problem, how can I code to allow this for n dimensions? Thanks for your help!
c# loops nested nested-loops
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
1
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03
add a comment |
1
1
1
So, I am trying to wrap my head around understanding how you can use a variable to denote how many times a loop is nested.
Here is an example I write up to simulate the output of dimensions = 4:
static void Main(string args)
{
int dimensions = 4; // e.g. for (1, 2, 3, 4), dimensions = 4
Console.WriteLine($"{addNumbers(dimensions)}");
Console.ReadKey();
}
static long addNumbers(int dimensions)
{
long number = 0;
// hard coded to be dimensions = 4
for (int h = 0; h <= dimensions; h++)
for (int i = 0; i <= dimensions; i++)
for (int j = 0; j <= dimensions; j++)
for (int k = 0; k <= dimensions; k++)
number += h + i + j + k; // just some random math
return number;
}
This will present the expected output of:
5000
So to readdress the problem, how can I code to allow this for n dimensions? Thanks for your help!
c# loops nested nested-loops
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
So, I am trying to wrap my head around understanding how you can use a variable to denote how many times a loop is nested.
Here is an example I write up to simulate the output of dimensions = 4:
static void Main(string args)
{
int dimensions = 4; // e.g. for (1, 2, 3, 4), dimensions = 4
Console.WriteLine($"{addNumbers(dimensions)}");
Console.ReadKey();
}
static long addNumbers(int dimensions)
{
long number = 0;
// hard coded to be dimensions = 4
for (int h = 0; h <= dimensions; h++)
for (int i = 0; i <= dimensions; i++)
for (int j = 0; j <= dimensions; j++)
for (int k = 0; k <= dimensions; k++)
number += h + i + j + k; // just some random math
return number;
}
This will present the expected output of:
5000
So to readdress the problem, how can I code to allow this for n dimensions? Thanks for your help!
c# loops nested nested-loops
c# loops nested nested-loops
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
edited Nov 23 '18 at 9:22
Dmitry Bychenko
111k1099141
111k1099141
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
asked Nov 23 '18 at 8:46
TimeTravelPenguinTimeTravelPenguin
287
287
1
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03
add a comment |
1
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03
1
1
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03
add a comment |
1 Answer
1
active
oldest
votes
1
For arbitrary n dimensions you can loop with a help of array int address which represents n dimensions:
static long addNumbers(int dimensions) {
int address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: sincen = 1case solution (32) was very suspicious
– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
For arbitrary n dimensions you can loop with a help of array int address which represents n dimensions:
static long addNumbers(int dimensions) {
int address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: sincen = 1case solution (32) was very suspicious
– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
add a comment |
1
For arbitrary n dimensions you can loop with a help of array int address which represents n dimensions:
static long addNumbers(int dimensions) {
int address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: sincen = 1case solution (32) was very suspicious
– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
add a comment |
1
1
1
For arbitrary n dimensions you can loop with a help of array int address which represents n dimensions:
static long addNumbers(int dimensions) {
int address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
For arbitrary n dimensions you can loop with a help of array int address which represents n dimensions:
static long addNumbers(int dimensions) {
int address = new int[dimensions];
// size of each dimension; not necessary equals to dimensions
// + 1 : in your code, int the loops you have i <= dimensions, j <= dimensions etc.
int size = dimensions + 1;
long number = 0;
do {
//TODO: some math here
// i == address[0]; j = address[1]; ... etc.
number += address.Sum();
// next address: adding 1 to array
for (int i = 0; i < address.Length; ++i) {
if (address[i] >= size - 1)
address[i] = 0;
else {
address[i] += 1;
break;
}
}
}
while (!address.All(index => index == 0)); // all 0 address - items're exhausted
return number;
}
Finally, let's add some Linq to look at the results:
int upTo = 5;
string report = string.Join(Environment.NewLine, Enumerable
.Range(1, upTo)
.Select(i => $"{i} -> {addNumbers(i),6}"));
Console.Write(report);
Outcome:
1 -> 1
2 -> 18
3 -> 288
4 -> 5000 // <- We've got it: 5000 for 4 dimensions
5 -> 97200
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
edited Nov 23 '18 at 9:21
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
answered Nov 23 '18 at 8:56
Dmitry BychenkoDmitry Bychenko
111k1099141
111k1099141
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: sincen = 1case solution (32) was very suspicious
– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
add a comment |
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: sincen = 1case solution (32) was very suspicious
– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Thanks! This seems to be it!
– TimeTravelPenguin
Nov 23 '18 at 9:02
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Using array for addressing is a typical way out in case of arbitrary n dimension space, array, collection etc.: stackoverflow.com/questions/53431650/…
– Dmitry Bychenko
Nov 23 '18 at 9:04
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
Yeah, I just fixed the post! I didn't need the loop to print the console, my mistake!
– TimeTravelPenguin
Nov 23 '18 at 9:19
1
1
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: since
n = 1 case solution (32) was very suspicious– Dmitry Bychenko
Nov 23 '18 at 9:26
@TimeTravelPenguin: Thank you! I see; I've added Linq for debugging purpose: since
n = 1 case solution (32) was very suspicious– Dmitry Bychenko
Nov 23 '18 at 9:26
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
I've not learnt Linq yet, so it's something new to learn!
– TimeTravelPenguin
Nov 23 '18 at 9:29
add a comment |
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1
Let Recursion be your friend
– Michael Randall
Nov 23 '18 at 8:48
Yeah, I've thought about recursion, but I cannot figure out how I would do it.
– TimeTravelPenguin
Nov 23 '18 at 8:49
How important is the "number += h + i + j + k" line, because that's really the only thing that makes it slightly tricky (you'd need an array to hold all the counters instead)?
– Dylan Nicholson
Nov 23 '18 at 8:56
This is just filler code for now, I want to do more complex math at a later date
– TimeTravelPenguin
Nov 23 '18 at 9:03