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TI-Basic Brain Teasers => TI-BASIC
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ztrumpet

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Joined: 06 May 2009
Posts: 555

 Posted: 18 Oct 2009 09:57:40 am    Post subject: Simplify a radical with an input in Ans and an output in Ans. For example: {1,8} prgmR {2,2} Read: One square root of eight is simplified to two square roots of two. I've got 151 bytes in the Memory Managment Menu with a one letter program name. Who can beat it?Last edited by Guest on 05 Jul 2010 07:49:16 am; edited 1 time in total
thornahawk
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Joined: 27 Mar 2005
Posts: 569

 Posted: 18 Oct 2009 11:05:00 am    Post subject: Quote: Ans→A ∟A(1→M:∟A(2→N int(√(N→K While fPart(N/KČ K-1→K End {MK,N/KČ Maybe somebody has a better idea? :) A stiffer challenge might be to write a program that processes general radicals, and not just square roots... thornahawk
DarkerLine
ceci n'est pas une |

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Joined: 04 Nov 2003
Posts: 8328

 Posted: 18 Oct 2009 11:47:17 am    Post subject: Here's a cheap way to simplify square roots that can be generalized, though you wouldn't want to: Quote: For(A,1,max(Ans For(B,1,max(Ans If A2B=prod(Ans^{2,1 {A,B End End AnsLast edited by Guest on 05 Jul 2010 07:49:30 am; edited 1 time in total
ztrumpet

Active Member

Joined: 06 May 2009
Posts: 555

 Posted: 18 Oct 2009 12:00:04 pm    Post subject: For some reason I can't get either of yours to work right. It's probably my fault though. :/
bfr

Member

Joined: 13 Feb 2006
Posts: 108

 Posted: 18 Oct 2009 12:05:06 pm    Post subject: Quote: For(A,2,Ans(2 If not(fPart(A-12Ans(2 {AAns(1),A-12Ans(2 End 44 bytes with 1 character name, a relatively straightforward approach that I guess works if you're going for reasonably small size with decent efficiency for smaller numbers. Otherwise, thornahawk's is probably nicer for larger numbers, especially if the number can't be simplified much.Last edited by Guest on 18 Oct 2009 12:13:09 pm; edited 1 time in total
Weregoose
Authentic INTJ

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Joined: 25 Nov 2004
Posts: 3976

 Posted: 18 Oct 2009 06:48:23 pm    Post subject: Here's a fun way: type √(845→X on the home screen, press enter, and then write XAns/R►Pr(X,Ans. Now, hit enter repeatedly. The first answer to not have a fractional part (be careful with rounding issues here) is the new coefficient to your radical.Last edited by Guest on 05 Jul 2010 07:49:49 am; edited 1 time in total
ztrumpet

Active Member

Joined: 06 May 2009
Posts: 555

 Posted: 18 Oct 2009 06:57:23 pm    Post subject: That's pretty cool. How did you figure this out?
thornahawk
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Active Member

Joined: 27 Mar 2005
Posts: 569

 Posted: 18 Oct 2009 07:07:21 pm    Post subject: For Goose's approach, one plus the number of times to repeat that line is the number inside the square root. :) thornahawk
GloryMXE7
Puzzleman 3000

Active Member

Joined: 02 Nov 2008
Posts: 604

 Posted: 18 Oct 2009 07:51:05 pm    Post subject: if you put it in a while loop such as while fpart(ans =/= 0 would it work the same with out having to repeatedly hit the enter button although its not really a burden.
Weregoose
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Joined: 25 Nov 2004
Posts: 3976

 Posted: 18 Oct 2009 08:24:40 pm    Post subject: ztrumpet wrote:That's pretty cool. How did you figure this out? Normally, one might do the following (note that X is already the result of the radical): For(N,1,X Pause √(X[font="verdana"]²/N) End The idea is getting to the next N not from the loop, but from the expression. That is, we're going to take the last answer that the expression gave us, and then compute "N" using that. A little elementary algebra: Ans=√(X[font="verdana"]²/N) Ans[font="verdana"]²=X[font="verdana"]²/N X[font="verdana"]²/Ans[font="verdana"]²=N We'll now add 1 to "step" the variable... X[font="verdana"]²/Ans[font="verdana"]²+1 ...and then bundle it up to form a recursion: √(X[font="verdana"]²/(X[font="verdana"]²/Ans[font="verdana"]²+1)) This is what the program looks like so far: √(X[font="verdana"]²/1) While 1 Pause √(X[font="verdana"]²/(X[font="verdana"]²/Ans[font="verdana"]²+1)) End Basically, the expression breaks itself back down to the "N", increments it, and then uses the original expression on itself to compute the next answer. It'd be a shame to leave something like that unsimplified. :) X[font="verdana"]²/Ans[font="verdana"]²+1 can be factored like so: X[font="verdana"]²/Ans[font="verdana"]² + 1 X[font="verdana"]²/Ans[font="verdana"]² + (Ans[font="verdana"]²/Ans[font="verdana"]²) (X[font="verdana"]² + Ans[font="verdana"]²)/Ans[font="verdana"]² X[font="verdana"]² over that gives us X[font="verdana"]²/((X[font="verdana"]²+Ans[font="verdana"]²)/Ans[font="verdana"]²) = X[font="verdana"]²Ans[font="verdana"]²/(X[font="verdana"]²+Ans[font="verdana"]²). Let's extract the perfect squares from inside the square root, shall we? √(X[font="verdana"]²Ans[font="verdana"]²/(X[font="verdana"]²+Ans[font="verdana"]²)) = XAns√(1/(X[font="verdana"]²+Ans[font="verdana"]²)) ...which of course translates to XAns√(1)/√(X[font="verdana"]²+Ans[font="verdana"]²) = XAns/√(X[font="verdana"]²+Ans[font="verdana"]²). Now for the fun part. See √(X[font="verdana"]²+Ans[font="verdana"]²)? That's how you find the hypotenuse using the Pythagorean theorem! But, if you put the triangle in a circle like so: ...you can use the relevant angle function on your calculator to find the length in radians given X and Ans, bringing our program down to: √(X[font="verdana"]²/1) While 1 Pause XAns/R►Pr(X,Ans End Anyway, my submission is 41 bytes: Spoiler wrote:Ans(1)prod(Ans→A For(B,1,Ans √(A/B If fPart(Ans End {Ans,B Last edited by Guest on 05 Jul 2010 07:48:46 am; edited 1 time in total
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