Simplify radicands
$begingroup$
I am attempting to create a program that can, given an integer as input, simplify the radicand by removing all square factors. For use on my TI 84+ CE
Code: for clarity reasons I've indented conditional blocks and added a new line after Delvar.
If not(A:Input “√(“,A
1→Y
While not(remainder(A,4
A/4→A
2Y→Y
End
3→X
While X²<=A
X²→θ
While not(remainder(A,θ
A/θ→A
XY→Y
End
X+2→X
End
{Y,A
DelVar A
If 1=Ans(2:Then
Disp “√(“+toString(Ans(2))+”)
Else:If 1=Ans(1:Then
Disp Ans(1
Else
Disp toString(Ans(1))+“√(“+toString(Ans(2))+”)
The program asks for input if one is not already provided in the A variable. It also initializes a Y variable to store the removed factors (part outside the radical sign). It then proceeds, for X from 2 counting up until X is greater than the square root of A to successively divide out squares of X and multiply Y by X for each iteration. After doing this for 2, it only tests odd numbers. At the end, A stores the value left inside the radicand. If A=1, then it simply displays Y as the answer. If Y=1, then it displays √Y as the answer. Otherwise, it shows A√Y as the result.
The worst case input for the algorithm is any number without a square factor, because it then has to run all the iterations to complete.
Are there any optimizations I can make? As space is not a problem (yet) on my calculator, I would prefer a speed optimization over a space optimization, but both are welcome.
symbolic-math ti-basic
edited 3 hours ago
200_success
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
$endgroup$
add a comment |
$begingroup$
I am attempting to create a program that can, given an integer as input, simplify the radicand by removing all square factors. For use on my TI 84+ CE
Code: for clarity reasons I've indented conditional blocks and added a new line after Delvar.
If not(A:Input “√(“,A
1→Y
While not(remainder(A,4
A/4→A
2Y→Y
End
3→X
While X²<=A
X²→θ
While not(remainder(A,θ
A/θ→A
XY→Y
End
X+2→X
End
{Y,A
DelVar A
If 1=Ans(2:Then
Disp “√(“+toString(Ans(2))+”)
Else:If 1=Ans(1:Then
Disp Ans(1
Else
Disp toString(Ans(1))+“√(“+toString(Ans(2))+”)
The program asks for input if one is not already provided in the A variable. It also initializes a Y variable to store the removed factors (part outside the radical sign). It then proceeds, for X from 2 counting up until X is greater than the square root of A to successively divide out squares of X and multiply Y by X for each iteration. After doing this for 2, it only tests odd numbers. At the end, A stores the value left inside the radicand. If A=1, then it simply displays Y as the answer. If Y=1, then it displays √Y as the answer. Otherwise, it shows A√Y as the result.
The worst case input for the algorithm is any number without a square factor, because it then has to run all the iterations to complete.
Are there any optimizations I can make? As space is not a problem (yet) on my calculator, I would prefer a speed optimization over a space optimization, but both are welcome.
symbolic-math ti-basic
edited 3 hours ago
200_success
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
$endgroup$
1
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
1
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago
add a comment |
$begingroup$
I am attempting to create a program that can, given an integer as input, simplify the radicand by removing all square factors. For use on my TI 84+ CE
Code: for clarity reasons I've indented conditional blocks and added a new line after Delvar.
If not(A:Input “√(“,A
1→Y
While not(remainder(A,4
A/4→A
2Y→Y
End
3→X
While X²<=A
X²→θ
While not(remainder(A,θ
A/θ→A
XY→Y
End
X+2→X
End
{Y,A
DelVar A
If 1=Ans(2:Then
Disp “√(“+toString(Ans(2))+”)
Else:If 1=Ans(1:Then
Disp Ans(1
Else
Disp toString(Ans(1))+“√(“+toString(Ans(2))+”)
The program asks for input if one is not already provided in the A variable. It also initializes a Y variable to store the removed factors (part outside the radical sign). It then proceeds, for X from 2 counting up until X is greater than the square root of A to successively divide out squares of X and multiply Y by X for each iteration. After doing this for 2, it only tests odd numbers. At the end, A stores the value left inside the radicand. If A=1, then it simply displays Y as the answer. If Y=1, then it displays √Y as the answer. Otherwise, it shows A√Y as the result.
The worst case input for the algorithm is any number without a square factor, because it then has to run all the iterations to complete.
Are there any optimizations I can make? As space is not a problem (yet) on my calculator, I would prefer a speed optimization over a space optimization, but both are welcome.
symbolic-math ti-basic
edited 3 hours ago
200_success
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
$endgroup$
I am attempting to create a program that can, given an integer as input, simplify the radicand by removing all square factors. For use on my TI 84+ CE
Code: for clarity reasons I've indented conditional blocks and added a new line after Delvar.
If not(A:Input “√(“,A
1→Y
While not(remainder(A,4
A/4→A
2Y→Y
End
3→X
While X²<=A
X²→θ
While not(remainder(A,θ
A/θ→A
XY→Y
End
X+2→X
End
{Y,A
DelVar A
If 1=Ans(2:Then
Disp “√(“+toString(Ans(2))+”)
Else:If 1=Ans(1:Then
Disp Ans(1
Else
Disp toString(Ans(1))+“√(“+toString(Ans(2))+”)
The program asks for input if one is not already provided in the A variable. It also initializes a Y variable to store the removed factors (part outside the radical sign). It then proceeds, for X from 2 counting up until X is greater than the square root of A to successively divide out squares of X and multiply Y by X for each iteration. After doing this for 2, it only tests odd numbers. At the end, A stores the value left inside the radicand. If A=1, then it simply displays Y as the answer. If Y=1, then it displays √Y as the answer. Otherwise, it shows A√Y as the result.
The worst case input for the algorithm is any number without a square factor, because it then has to run all the iterations to complete.
Are there any optimizations I can make? As space is not a problem (yet) on my calculator, I would prefer a speed optimization over a space optimization, but both are welcome.
symbolic-math ti-basic
symbolic-math ti-basic
edited 3 hours ago
200_success
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
edited 3 hours ago
200_success
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
edited 3 hours ago
200_success
129k15153415
edited 3 hours ago
200_success
129k15153415
edited 3 hours ago
200_success
129k15153415
129k15153415
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
asked Jan 11 '18 at 6:41
Neil A.Neil A.
1785
1785
1
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
1
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago
add a comment |
1
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
1
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago
1
1
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
1
1
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago
add a comment |
0
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1
$begingroup$
It would be helpful if you gave a brief algorithm overview. given the formatting, and the fact that it is a kind of obscure language, it will let people give better advice.
$endgroup$
– Oscar Smith
Jan 11 '18 at 6:47
$begingroup$
Also indentation and adding closing brackets would help - at least here. The 84s display is probably not suited for the latter one.
$endgroup$
– Markus Deibel
Jan 11 '18 at 8:00
1
$begingroup$
If you had a list of prime numbers on my calculator (at some point I generated the first 999, which is the most that can fit in a list), you could iterate those instead of all the numbers (then iterate all the odds after the primes run out).
$endgroup$
– Solomon Ucko
Feb 28 '18 at 20:10
$begingroup$
How large is the integer? It seems like you need an integer factorisation algorithm, depending on the size of the number, different algorithms would be appropriate.
$endgroup$
– George Barwood
4 hours ago