Mailman 3 November 2013 - Edu-sig

a short essay about programming
by kirby urner Dec. 8, 2022

Dec. 8, 2022

A common "error" (not too serious) that I see in beginning Python (and no doubt other languages, but Python is the one I'm teaching), is having a while condition that appears to put a lid on things, but then the flow all leaks away through break statements, such that the "front door" condition is never revisited. while guess != secret: guess = int(input("Guess?: ") if guess == secret: print("You guessed it!") break else: print("Nope, try again...") What's messed up about the above code is you never really go back to the top in the case where you'd get to leave. Rather, you exit by the back door, through break. So in that case, wouldn't have been simpler and more elegant, even more "correct" (dare I say it) to have gone: while True: # no ifs ands or buts guess = int(input("Guess?: ") if guess == secret: print("You guessed it!") break else: print("Nope, try again...") I see lots of heads nodding, and that could be the end of the matter, but then a next question arises: wouldn't this also be a more correct solution?: while guess != secret: guess = int(input("Guess?: ") if guess == secret: print("You guessed it!") continue # instead of break else: print("Nope, try again...") We're back to having a variable while condition, not a boolean constant, but this time we actually exit by means of it, thanks to continue or... while guess != secret: guess = int(input("Guess?: ") if guess == secret: print("You guessed it!") else: print("Nope, try again...") ... thanks to no continue. This last one is getting a thumbs up, but then I'd pause here and say "continue" can be easier on the eyes. It's unambiguous where it takes you, in contrast to having to scan on down the gauntlet, looking for possibly other open doors. "What happens next" should not require scanning ahead too far. Help us not to get lost. Be a civic-minded coder. I'm thinking of a programming style that advises two things: (a) if you use a while condition that's variable, that's expected to change, then your goal should be to always exit because of that, i.e. that should be your only exit point. Even if some other criterion suggests exiting, you have the option to flip that "lid" at the top, to crack that front door, and bounce the ball out. (b) which is why 'continue' is your friend. You are showing the user where your 'break' statements are, except you don't use "break" statements, as you've given a non-constant condition, and your aim is to make that your ONLY exit point. In short: never use break to exit a while loop unless your condition is while True. Instead, always flip the condition and exit through the font door. However, as soon as I make that rule I can think of good reasons to break it. The other programmers in the room are shaking their heads. Won't we lose information, needed elsewhere in the program, if we "artificially" force a True condition to False. Aren't we, in effect, lying? That concern could be addressed. Keep all the info true, just treat the "lid condition" (it "keeps a lid on it") as a flag. Go ahead and remember the user's guess. allowed = 5 tries = 1 exit = False while not exit: # no other way out guess = int(input("Guess?: ") if guess == secret: print("You guessed it!") exit = True # instead of break continue print("Nope, try again...") tries += 1 if tries == allowed: print("You've maxed out") exit = True continue I think we all understand the main issue: writing reader-friendly code, and rules for doing so. There's something comforting about approaching a while loop and knowing its not a leaky sieve, riddled with back doors, maybe even an exit( ) time bomb. But in the recursion world we want a minimum of two exits usually: when another round is called for versus when we've "hit bottom". Can we have it both ways? Conclusions: Lets not be too hasty with rules of thumb and: Lets keep the reader in mind when writing code. Just because the interpreter knows to compute the flow unambiguously, doesn't mean all ways of writing it are equally reader-friendly. What may seem a gratuitous gesture, an unnecessary flourish, may actually promote reader comprehension of your code, and that should be a goal as much as satisfying the interpreter. Kirby

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Re: [Edu-sig] RurPLE-NG is nice
by Jurgis Pralgauskis April 5, 2014

April 5, 2014

Some updates: - separated Variables Watch and Output panes, to be seen simultaneusly - implemented Breakpoints :) - beta style Todo: - let enter arbitrary watch expressions http://ftp.akl.lt/users/jurgis/python/rurple-NG053.png http://ftp.akl.lt/users/jurgis/python/rurple-ng-dz0-SplitLogVars_and_Breakp… for windows one doesn't need to install python separately, just unzip and run :) http://ftp.akl.lt/users/jurgis/python/RURPLE-NG-0.53%20%5bindependent%20exe… ps.: RurPLE-NG uses py3 unicode_literals, with_statement, division, print_function so code is generally py3 compliant. when wxpy will work well with py3 it shouldn't be too hard to port fully... >> >> just wanted to mention, >> that http://dev.lshift.net/paul/rurple/ >> is a very nice and lightweit PLE. >> >> what I like most (compared to RurPLE) are: >> 1) variables watch list (if it had breakpoints, it would rock :)

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Easy Web Environment to post pupils Python Creations
by Philip Bagge April 1, 2014

April 1, 2014

Dear Colleagues, My Y6 pupils (10-11 year olds) have been creating some simple text based linear adventures using Python. Does anyone know of a way we can play their adventures online? They are not very complicated. You can see the planning here. https://sites.google.com/site/pythoncodecouk/python-adventure Also as a new person to your forum does anyone know of an easy way to search you archives just in case someone has already answered this question. Thanks in advance for any help Phil Bagge -- Primary ICT Advanced Skills Teacher HCC Teacher & IT Manager http://code-it.co.uk

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Re: [Edu-sig] Fibonacci Numbers and Phi (again)
by Litvin Nov. 23, 2013

Nov. 23, 2013

Kirby, You are right, a little program does help to clarify the statement and the formulas involved. But then, keep it really simple (no classes, no generators), so that understanding of the code doesn't get in the way. For example: from math import sqrt rt5 = sqrt(5) phi = (1 + rt5)/2 def neat_formula(f0, f1, f2): return (f0 + f2 + rt5*f1)/2 power_of_phi = phi**(-6) f0, f1, f2 = 13, -8, 5 for k in range(20): print("{0:8.5f} {1:8.5f}".format(power_of_phi, neat_formula(f0, f1, f2))) f0, f1, f2 = f1, f2, f1+f2 power_of_phi *= phi There are many other occasions to bring in more advanced tools. Gary Litvin www.skylit.com At 11:08 AM 11/23/2013, kirby urner wrote: > > On Fri, Nov 22, 2013 at 9:49 PM, Litvin > <<mailto:litvin@skylit.com>litvin(a)skylit.com> wrote: > > > Kirby, > > > > I am sorry to spoil all the fun, but this is a not a gem but a > > mathematical trick, and I think the way to deal with > > mathematical tricks is to explain them, not to verify with code. > > >Having an algebraic verification is more like the proof I did not >provide. Thanks for the reasoning Gary. > >I do think there's a two phase understanding for theorems where >phase one is too oft neglected, which is understanding what the >theorem is claiming or asserting, before attempting an analysis of "why?". > >Euler's Theorem by way of example, the one with the totient, a >generalization of Fermat's: it's too easy to just stare at the >assertion and not think of examples, not know what it says. > >This is where I think a Python program may come in handy, not as a >proof of a theorem but as a illustration or demonstration of its >consequences. See the theorem in action somehow. > >Seeing something in Python that's runnable may close some critical >circuits in the brain of the theorem's reader. "Now I better know >what it means" may be the aha experience. > >Most people who have read up on Fibonaccis know they can be extended >in the negative direction. Then the idea of starting with any two >numbers, not necessarily integers may be introduced. > >The limit of F[N+1]/F[N] and its relationship to PHI (equal as N -> >infinity) may then be discussed. > >I've taken exactly this same approach with one of Ramanujan's for >1/pi (just flip it for pi). A little Python generator will do the >trick. In this case the gem being verified has not to my knowledge >been fully explained or reasoned about (the proof has yet to be supplied): > ><http://worldgame.blogspot.com/2012/01/testing-math-ml.html>http://worldgame.blogspot.com/2012/01/testing-math-ml.html > >(this blogpost uses MathJax, a good resource for the JavaScript-enabled) > >I'm still without an explanation for this pi digits generator from >one of Michel Paul's students, but I share it around in hopes of >finding one -- and then I hope its one I can follow. > ><https://groups.google.com/forum/#!msg/mathfuture/LA0pMPC6-HE/MBGWxn4ENsUJ>https://groups.google.com/forum/#!msg/mathfuture/LA0pMPC6-HE/MBGWxn4ENsUJ > >Thanks again for clearing up any mysteries that may have attached to >the PHI ** N formula (I originally called it a formula, not a gem... >some gems are only semi-precious). > >Kirby >

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Re: [Edu-sig] Fibonacci Numbers and Phi (again)
by Litvin Nov. 23, 2013

Nov. 23, 2013

Kirby, I am sorry to spoil all the fun, but this is a not a gem but a mathematical trick, and I think the way to deal with mathematical tricks is to explain them, not to verify with code. There are two Fibonacci sequences that are also geometric sequences 1, x, x^2, ..., x^n,...: when x = phi or x = -1/phi, because phi and -1/phi are the roots of the quadratic equation x^2 = x+1. Any Fibonacci sequence is a linear combination of these two: F[n] = a*phi^n + b*(-1/phi)^n. a and b can be determined from the first two terms of the sequence: a + b = F[0] a*phi - b/phi = F[1] If you choose F[0] and F[1] (not necessarily integers) in a tricky way, so that b = 0, then F[n] = a*phi^n. In addition, if a = phi^e, then F[n] = phi^(e+n). Working in reverse, you can find F[0] and F[1] that satisfy these conditions and get the trick going. Here F[0] = (13 + 5 - 8*sqrt(5))/2 = 9 - 4*sqrt(5) F[1] = (-8 - 3 + 5*sqrt(5))/2 = (-11+5sqrt(5))/2 Easy to verify that F0*phi = F1, so, indeed, b = 0 and F[n] is a geometric sequence. Also, phi^6 = 9 + 4*sqrt(5), so F[0]*phi^6 = 1 and a = F[0] = phi^(-6). You get F[n] = phi^(n-6). Gary Litvin www.skylit.com At 10:04 PM 11/22/2013, you wrote: >Here's something mathematical from the Poly list [1], adapting >something by David Koski. > >The code shows how one might use a program to express a >cool relationship, without offering a proof. > >The unit-test shows use raising PHI from the -6 to the 29th >power and finding an equivalent expression built solely from >three consecutive Fibonacci numbers and the sqrt(5). > >Something like: > >LOOP: > PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2 > E += 1 > N += 1 > >Start with: > >E = -6 > >FIB[N-2] = 13 >FIB[N-1] = -8 >FIB[N] = 5 > >in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . . > >=== > > >""" >Encapsulating a discovery -- not a proof. >By David Koski (Python by K. Urner) > >Failure at around 37th power is due to floating point limitations. >""" > >import unittest > >def fibonacci(a=0, b=1): > while True: > yield a > a, b = b, a + b > > >series1 = fibonacci(5, -3) # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 >series2 = fibonacci(-8, 5) # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 >series3 = fibonacci(13,-8) # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 > >RT5 = pow(5, .5) # "square root" of five >PHI = (1 + RT5)/2 # golden proportion > >def cool_formula(): > """ > Two fib numbers, two apart, plus the one in the middle * sqrt(5) > all over 2, gives PHI to a power. Advancing all three sequences > gives successive powers. > """ > while True: > yield (next(series1) + next(series3) + next(series2) * RT5)/2.0 > >class TestPhi(unittest.TestCase): > > def test_loop(self): > gem = cool_formula() > for e in range(-6, 30): # adjust range (fails about 37th power) > answer = next( gem ) > self.assertAlmostEqual( PHI ** e, answer) > >if __name__ == "__main__": > unittest.main() > >[1] Polyhedron mailing list >Contributions to ><mailto:polyhedron@lists.mathconsult.ch>mailto:polyhedron@lists.mathconsult.ch >Admin: ><http://lists.mathconsult.ch/mailman/listinfo/polyhedron>http://lists.mathconsult.ch/mailman/listinfo/polyhedron >Archive: ><http://lists.mathconsult.ch/mailman/private/polyhedron/>http://lists.mathconsult.ch/mailman/private/polyhedron/ > >_______________________________________________ >Edu-sig mailing list >Edu-sig(a)python.org >https://mail.python.org/mailman/listinfo/edu-sig At 10:04 PM 11/22/2013, kirby urner wrote: >Here's something mathematical from the Poly list [1], adapting >something by David Koski. > >The code shows how one might use a program to express a >cool relationship, without offering a proof. > >The unit-test shows use raising PHI from the -6 to the 29th >power and finding an equivalent expression built solely from >three consecutive Fibonacci numbers and the sqrt(5). > >Something like: > >LOOP: > PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2 > E += 1 > N += 1 > >Start with: > >E = -6 > >FIB[N-2] = 13 >FIB[N-1] = -8 >FIB[N] = 5 > >in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . . > >=== > > >""" >Encapsulating a discovery -- not a proof. >By David Koski (Python by K. Urner) > >Failure at around 37th power is due to floating point limitations. >""" > >import unittest > >def fibonacci(a=0, b=1): > while True: > yield a > a, b = b, a + b > > >series1 = fibonacci(5, -3) # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 >series2 = fibonacci(-8, 5) # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 >series3 = fibonacci(13,-8) # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 > >RT5 = pow(5, .5) # "square root" of five >PHI = (1 + RT5)/2 # golden proportion > >def cool_formula(): > """ > Two fib numbers, two apart, plus the one in the middle * sqrt(5) > all over 2, gives PHI to a power. Advancing all three sequences > gives successive powers. > """ > while True: > yield (next(series1) + next(series3) + next(series2) * RT5)/2.0 > >class TestPhi(unittest.TestCase): > > def test_loop(self): > gem = cool_formula() > for e in range(-6, 30): # adjust range (fails about 37th power) > answer = next( gem ) > self.assertAlmostEqual( PHI ** e, answer) > >if __name__ == "__main__": > unittest.main() > >[1] Polyhedron mailing list >Contributions to ><mailto:polyhedron@lists.mathconsult.ch>mailto:polyhedron@lists.mathconsult.ch >Admin: ><http://lists.mathconsult.ch/mailman/listinfo/polyhedron>http://lists.mathconsult.ch/mailman/listinfo/polyhedron >Archive: ><http://lists.mathconsult.ch/mailman/private/polyhedron/>http://lists.mathconsult.ch/mailman/private/polyhedron/ > >_______________________________________________ >Edu-sig mailing list >Edu-sig(a)python.org >https://mail.python.org/mailman/listinfo/edu-sig

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Fibonacci Numbers and Phi (again)
by kirby urner Nov. 23, 2013

Nov. 23, 2013

Here's something mathematical from the Poly list [1], adapting something by David Koski. The code shows how one might use a program to express a cool relationship, without offering a proof. The unit-test shows use raising PHI from the -6 to the 29th power and finding an equivalent expression built solely from three consecutive Fibonacci numbers and the sqrt(5). Something like: LOOP: PHI ** E = (FIB[N] + FIB[N-2] + FIB[N-1]*RT5) / 2 E += 1 N += 1 Start with: E = -6 FIB[N-2] = 13 FIB[N-1] = -8 FIB[N] = 5 in . . . 13, -8, 5, -3, 2, -1,1, 0, 1, 2, 3, 5, 8,13 . . . === """ Encapsulating a discovery -- not a proof. By David Koski (Python by K. Urner) Failure at around 37th power is due to floating point limitations. """ import unittest def fibonacci(a=0, b=1): while True: yield a a, b = b, a + b series1 = fibonacci(5, -3) # 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 series2 = fibonacci(-8, 5) # -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 series3 = fibonacci(13,-8) # 13, -8, 5, -3, 2, -1, 1, 0 , 1, 2, 3, 5, 8 RT5 = pow(5, .5) # "square root" of five PHI = (1 + RT5)/2 # golden proportion def cool_formula(): """ Two fib numbers, two apart, plus the one in the middle * sqrt(5) all over 2, gives PHI to a power. Advancing all three sequences gives successive powers. """ while True: yield (next(series1) + next(series3) + next(series2) * RT5)/2.0 class TestPhi(unittest.TestCase): def test_loop(self): gem = cool_formula() for e in range(-6, 30): # adjust range (fails about 37th power) answer = next( gem ) self.assertAlmostEqual( PHI ** e, answer) if __name__ == "__main__": unittest.main() [1] Polyhedron mailing list Contributions to mailto:polyhedron@lists.mathconsult.ch<polyhedron(a)lists.mathconsult.ch> Admin: http://lists.mathconsult.ch/mailman/listinfo/polyhedron Archive: http://lists.mathconsult.ch/mailman/private/polyhedron/

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Syllabus
by Jay Bloodworth Nov. 13, 2013

Nov. 13, 2013

Does anyone have a link to a syllabus that might be appropriate for a 9 week (give or take) introductory programming course? Thanks, Jay

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