MULTI-DIMENSIONAL ARRAYS SEEN AS ONE-DIMENSIONAL ONE
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:confused: Hello. I know that a two-dimensional array can be handles as one-dimensional array by converting the indexes to a single one. Supose you have the following: int *table; table=new int table[3*3]; When referencig table[2][2] by a single index index=(2*3+2); then table[index] can be recalled. Now, I'm having a three-dimensional array and because is faster to handle dynamic arrays, I can define it as: int table1; table1=new int table[3*3*3]; but I got lost finding the index conversion formulae... :confused: Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
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:confused: Hello. I know that a two-dimensional array can be handles as one-dimensional array by converting the indexes to a single one. Supose you have the following: int *table; table=new int table[3*3]; When referencig table[2][2] by a single index index=(2*3+2); then table[index] can be recalled. Now, I'm having a three-dimensional array and because is faster to handle dynamic arrays, I can define it as: int table1; table1=new int table[3*3*3]; but I got lost finding the index conversion formulae... :confused: Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
Your old formulae was xPos + (yPos x Height). I believe your new formulae would therefore be: (xPos + (yPos x Width)) + (zPos x (Height x Width)). This could probably be further reduced. "When I left you I was but the learner, now I am the master" - Darth Vader
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:confused: Hello. I know that a two-dimensional array can be handles as one-dimensional array by converting the indexes to a single one. Supose you have the following: int *table; table=new int table[3*3]; When referencig table[2][2] by a single index index=(2*3+2); then table[index] can be recalled. Now, I'm having a three-dimensional array and because is faster to handle dynamic arrays, I can define it as: int table1; table1=new int table[3*3*3]; but I got lost finding the index conversion formulae... :confused: Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
Try this: int d1 = 3; int d2 = 4; int d3 = 5; BYTE x = 0; BYTE *b = new BYTE[d1*d2*d3]; for (int i = 0; i < d1; i++) { for (int j = 0; j < d2; j++) { for (int k = 0; k < d3; k++) { b[i*(d2*d3) + j*d3 + k] = x; x++; } } } I hope it does what you need.
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Your old formulae was xPos + (yPos x Height). I believe your new formulae would therefore be: (xPos + (yPos x Width)) + (zPos x (Height x Width)). This could probably be further reduced. "When I left you I was but the learner, now I am the master" - Darth Vader
Thanks for your help....I tried that but it did not work...however, you gave an idea and I put myself to think deeply and I just found this: Let be a multidimensional array MArray[max1stdim][max2nddim]...[maxnthdim], the indexes referencing certain location be i1,i2,...,in. If MArray is declared as int MArray[max1stdim*max2nddim*...*maxnthdim]; then the index referencing the location (i1,i2,...in-1,in) will be: index=(i1*max2nddim*max3rddim*...*maxnthdim)+(i2*max3rddim*...*maxnthdim)+...+(in-1)*maxnthdim+in Then for an array MArray[4][3][2] declared as MArray[3*3*3], the single index to reference (i,j,k)will be: index=(i*4*3)+(j*2)+k I have tried this and it works! Thanks again..... :) Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
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Try this: int d1 = 3; int d2 = 4; int d3 = 5; BYTE x = 0; BYTE *b = new BYTE[d1*d2*d3]; for (int i = 0; i < d1; i++) { for (int j = 0; j < d2; j++) { for (int k = 0; k < d3; k++) { b[i*(d2*d3) + j*d3 + k] = x; x++; } } } I hope it does what you need.
Thanks for your help....I have tried that and it did work...therefore, making a generalization for this, I just found this: Let be a multidimensional array MArray[max1stdim][max2nddim]...[maxnthdim], the indexes referencing certain location be i1,i2,...,in. If MArray is declared as int MArray[max1stdim*max2nddim*...*maxnthdim]; then the index referencing the location (i1,i2,...in-1,in) will be: index=(i1*max2nddim*max3rddim*...*maxnthdim)+(i2*max3rddim*...*maxnthdim)+...+(in-1)*maxnthdim+in Then for an array MArray[4][3][2] declared as MArray[3*3*3], the single index to reference (i,j,k)will be: index=(i*4*3)+(j*2)+k I have tried this and it works! Thanks again..... :) ;) Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
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Thanks for your help....I have tried that and it did work...therefore, making a generalization for this, I just found this: Let be a multidimensional array MArray[max1stdim][max2nddim]...[maxnthdim], the indexes referencing certain location be i1,i2,...,in. If MArray is declared as int MArray[max1stdim*max2nddim*...*maxnthdim]; then the index referencing the location (i1,i2,...in-1,in) will be: index=(i1*max2nddim*max3rddim*...*maxnthdim)+(i2*max3rddim*...*maxnthdim)+...+(in-1)*maxnthdim+in Then for an array MArray[4][3][2] declared as MArray[3*3*3], the single index to reference (i,j,k)will be: index=(i*4*3)+(j*2)+k I have tried this and it works! Thanks again..... :) ;) Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
Would be nice if you will write article about it and compare performance between different types of allocations and accesseing that memory.
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Would be nice if you will write article about it and compare performance between different types of allocations and accesseing that memory.
How will I do that? What allocation and accesing methods?. Where will I submit the article?...Would you like to help me? Thanks, Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
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How will I do that? What allocation and accesing methods?. Where will I submit the article?...Would you like to help me? Thanks, Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
I understood you think addressing multi-dimensional array with one index you are computing by yourself is a faster method than addressing multi-dimensional array with pointers. So what I think is a demonstration program using two arrays of the same size but different access methods and allocation and comparison between speeds. And obviously, you will submit article with that demo program here at CP.
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Thanks for your help....I tried that but it did not work...however, you gave an idea and I put myself to think deeply and I just found this: Let be a multidimensional array MArray[max1stdim][max2nddim]...[maxnthdim], the indexes referencing certain location be i1,i2,...,in. If MArray is declared as int MArray[max1stdim*max2nddim*...*maxnthdim]; then the index referencing the location (i1,i2,...in-1,in) will be: index=(i1*max2nddim*max3rddim*...*maxnthdim)+(i2*max3rddim*...*maxnthdim)+...+(in-1)*maxnthdim+in Then for an array MArray[4][3][2] declared as MArray[3*3*3], the single index to reference (i,j,k)will be: index=(i*4*3)+(j*2)+k I have tried this and it works! Thanks again..... :) Eric Manuel Rosales Pena Alfaro PhD student Unversity of Essex Wivenhoe Park Colchester, CO4 3SQ Essex, Uk email: emrosa@essex.ac.uk tel: +44-01206-87311
Can ask one question? how can you have "an array MArray[4][3][2] declared as MArray[3*3*3]"? If you are trying to access the 3*3*3 MArray at position [4][3][2] using a single index then that is impossible, because the rows and columns will be indexed from 0 through to 2 for each x,y and z column, so 4 and 3 are invalid index positions. If you were trying to access the last record in a 3*3*3 matrix, the single index value you require is 26 (0 to 8, 9 to 17, 18 to 26). Using my equation earlier, if you were trying to index the last position in the 3d array, so XPos = 2, YPos = 2 and ZPos = 2, you would get: (2 + (2 * 3)) + (2 * 3 * 3)) = 26. the index position you desire. Martin has implemented the coded version of the equation (but in the opposite direction to the way I wrote my equation) :) : b[i*(d2*d3) + j*d3 + k] = x; i = ZPos j = YPos k = XPos d2 = Height d3 = Width *Please note I made a small error before when I wrote (yPos + Height), this is probably why you couldn't get it to work, it should have read (yPos + Width), which I have now changed it to. Without this correction the formula would only *work* if the 3d array was a cube. It should now work regardless of dimensions. Alan. "When I left you I was but the learner, now I am the master" - Darth Vader
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Can ask one question? how can you have "an array MArray[4][3][2] declared as MArray[3*3*3]"? If you are trying to access the 3*3*3 MArray at position [4][3][2] using a single index then that is impossible, because the rows and columns will be indexed from 0 through to 2 for each x,y and z column, so 4 and 3 are invalid index positions. If you were trying to access the last record in a 3*3*3 matrix, the single index value you require is 26 (0 to 8, 9 to 17, 18 to 26). Using my equation earlier, if you were trying to index the last position in the 3d array, so XPos = 2, YPos = 2 and ZPos = 2, you would get: (2 + (2 * 3)) + (2 * 3 * 3)) = 26. the index position you desire. Martin has implemented the coded version of the equation (but in the opposite direction to the way I wrote my equation) :) : b[i*(d2*d3) + j*d3 + k] = x; i = ZPos j = YPos k = XPos d2 = Height d3 = Width *Please note I made a small error before when I wrote (yPos + Height), this is probably why you couldn't get it to work, it should have read (yPos + Width), which I have now changed it to. Without this correction the formula would only *work* if the 3d array was a cube. It should now work regardless of dimensions. Alan. "When I left you I was but the learner, now I am the master" - Darth Vader