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Libervurto

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Posted: 19th Jul 2013 17:10 Edited at: 21st Jul 2013 01:09

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I thought it might be fun to make some quick challenges for newbies to see how they are growing as problem-solvers. If you want to post your own solutions please put them in code tags.

Challenge 0 — Produce random numbers between 12 and 29, and demonstrate that your program works as required.
Solution:

+ Code Snippet

// x = rnd(17)+12
for n = 0 to 17
  print n+12
next n

[OUT OF ORDER: DO NOT ATTEMPT] Challenge 1 — This time, pretend you are working with a language that only produces random numbers from 0 to 0.99999, how are you going to create the same range from 12 to 29? Use this code for getting random numbers: n# = rnd(99999)/100000.0
Solution:

+ Code Snippet

n# = rnd(99999)/100000.0
x = int(n# * 18.00017 + 12) // int() is optional since the value will be cast to an integer anyway.

// Why do we not multiply by 17 or 18?
// Note that: 0.99999 × (n + n × 10⁻⁵) = n
// So to ensure all numbers in our range have the same probability we must give each value of n the range of n.00000 to n.99999
// To achieve this we increase the maximum n by 1 and subtract the smallest unit we are working with, in this case the smallest unit is 10⁻⁵
// So when you see: n# * 18.00017 + 12
// You're actually looking at: n# * (n + 1 + n * 10⁻⁵) + 12

I will add more when I think of them. Please post your own challenges and corrections to mine, if there are any mistakes. But please, only post original challenges, not things like FizzBuzz that are widely known.

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WickedX

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Posted: 19th Jul 2013 19:23

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Challenge 2 - Rewrite the following code using no, if conditional testing.

+ Code Snippet

   if rightkey()
      inc turn#, 0.25
      if turn# > 20
         turn# = 20
      endif
   endif

   if leftkey()
      dec turn#, 0.25
      if turn# < -20
         turn# = -20
      endif
   endif

Solution:

+ Code Snippet

   turn# = turn#+(rightkey()*(turn#<20)-leftkey()*(turn#>-20))*0.25

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thenerd

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Posted: 20th Jul 2013 05:50 Edited at: 20th Jul 2013 05:52

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Challenge 3 - Through code, list all prime numbers between 1 and 100.

Solution:

+ Code Snippet

for i = 1 to 100:
    prime = 1
    for n = 2 to (i-1):
        if i mod n = 0 then prime = 0
    next n
    if prime = 1 then print i
next i

If anyone has a better solution, that would be interesting. I actually made myself think trying to answer my own challenge!

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Libervurto

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Posted: 20th Jul 2013 12:13 Edited at: 20th Jul 2013 12:21

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@WickedX - Your solution doesn't function the same as your sample. I adjusted your sample to function the same as the solution:

+ Code Snippet

x = 0			//main variable
a = rightkey()		//input a
b = leftkey()		//input b
s = 2			//speed
max = 20		//maximum
min = -20		//minimum

//using if
if a and x < max then inc x, s
if b and x > min then dec x, s

//without if
inc x, (a*(x<max) - b*(x>min)) * s

Mathematical logic is certainly interesting and can be very useful, but I don't particularly like this example as it doesn't improve the code.

@thenerd - Here's my solution:

+ Code Snippet

lim = 100
dim p(lim)
for n = 2 to lim : p(n)=1 : next n //start with all numbers >1 on the primes list

rem -- main ----------------------------------
for i = 2 to lim
	if p(i)
		for j = 2 to i
			if p(j)
				k = 1
				while i * j^k <= lim
					p(i * j^k) = 0
					inc k
				endwhile
			endif
		next j
	endif
next i

I don't have DB installed so I had to dry run the code to check it. I sort of had to develop my own notation in the process so I hope it is easy to follow, maybe useful:

lim = 100
dim p(lim)
for n = 2 to lim : p(n)=1 : next n //start with all numbers >1 on the primes list

rem -- dry run ----------------------------------
for i = 2 to lim → i=2 → TRUE
	if p(i) → if p(2)=1 → TRUE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 2·2¹… → 4 <= 100 → TRUE
					p(i * j^k) = 0 → p(4)=0
					inc k → k=2
				→ while i * j^k <= lim → 2·2²… → 8 <= 100 → TRUE
					p(i * j^k) = 0 → p(8)=0
					inc k → k=3
				… → k=5
				→ while i * j^k <= lim → 2·2⁵… → 64 <= 100 → TRUE
					p(i * j^k) = 0 → p(64)=0
					inc k → k=6
				→ while i * j^k <= lim → 2·2⁶… → 128 <= 100 → FALSE
			→ endif
		→ j=3 → FALSE
	→ endif
→ i=3 → TRUE
	if p(i) → if p(3)=1 → TRUE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 3·2¹… → 6 <= 100 → TRUE
					p(i * j^k) = 0 → p(6)=0
					inc k → k=2
				→ while i * j^k <= lim → 3·2²… → 12 <= 100 → TRUE
					p(i * j^k) = 0 → p(12)=0
					inc k → k=3
				… → k=5
				→ while i * j^k <= lim → 3·2⁵… → 96 <= 100 → TRUE
					p(i * j^k) = 0 → p(96)=0
					inc k → k=6
				→ while i * j^k <= lim → 3·2⁶… → 192 <= 100 → FALSE
			→ endif
		→ j=3 → TRUE
			if p(j) → if p(3)=1 → TRUE
				k = 1
				while i * j^k <= lim → 3·3¹… → 9 <= 100 → TRUE
					p(i * j^k) = 0 → p(9)=0
					inc k → k=2
				… → k=3
				→ while i * j^k <= lim → 3·3³… → 81 <= 100 → TRUE
					p(i * j^k) = 0 → p(81)=0
					inc k → k=4
				→ while i * j^k <= lim → 3·3⁴… → 243 <= 100 → FALSE
			→ endif
		→ j=4 → FALSE
	→ endif
→ i=4 → TRUE
	if p(i) → if p(4)=1 → FALSE
→ i=5 → TRUE
	if p(i) → if p(5)=1 → FALSE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·2¹… → 10 <= 100 → TRUE
					p(i * j^k) = 0 → p(10)=0
					inc k → k=2
				… → k=4
				→ while i * j^k <= lim → 5·2⁴… → 80 <= 100 → TRUE
					p(i * j^k) = 0 → p(80)=0
					inc k → k=5
				→ while i * j^k <= lim → 5·2⁵… → 160 <= 100 → FALSE
			→ endif
		→ j=3 → TRUE
			if p(j) → if p(3)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·3¹… → 15 <= 100 → TRUE
					p(i * j^k) = 0 → p(15)=0
					inc k → k=2
				→ while i * j^k <= lim → 5·3²… → 45 <= 100 → TRUE
					p(i * j^k) = 0 → p(45)=0
					inc k → k=3
				→ while i * j^k <= lim → 5·3³… → 135 <= 100 → FALSE
			→ endif
		→ j=4 → TRUE
			if p(j) → if p(4)=1 → FALSE
		→ j=5 → TRUE
			if p(j) → if p(5)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·5¹… → 25 <= 100 → TRUE
					p(i * j^k) = 0 → p(25)=0
					inc k → k=2
				→ while i * j^k <= lim → 5·5²… → 125 <= 100 → FALSE
			→ endif
		→ j=6 → FALSE
	→ endif
… → i=101 → FALSE

+ Code Snippet

lim = 100
dim p(lim)
for n = 2 to lim : p(n)=1 : next n //start with all numbers >1 on the primes list

rem -- dry run ----------------------------------
for i = 2 to lim → i=2 → TRUE
	if p(i) → if p(2)=1 → TRUE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 2·2¹… → 4 <= 100 → TRUE
					p(i * j^k) = 0 → p(4)=0
					inc k → k=2
				→ while i * j^k <= lim → 2·2²… → 8 <= 100 → TRUE
					p(i * j^k) = 0 → p(8)=0
					inc k → k=3
				… → k=5
				→ while i * j^k <= lim → 2·2⁵… → 64 <= 100 → TRUE
					p(i * j^k) = 0 → p(64)=0
					inc k → k=6
				→ while i * j^k <= lim → 2·2⁶… → 128 <= 100 → FALSE
			→ endif
		→ j=3 → FALSE
	→ endif
→ i=3 → TRUE
	if p(i) → if p(3)=1 → TRUE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 3·2¹… → 6 <= 100 → TRUE
					p(i * j^k) = 0 → p(6)=0
					inc k → k=2
				→ while i * j^k <= lim → 3·2²… → 12 <= 100 → TRUE
					p(i * j^k) = 0 → p(12)=0
					inc k → k=3
				… → k=5
				→ while i * j^k <= lim → 3·2⁵… → 96 <= 100 → TRUE
					p(i * j^k) = 0 → p(96)=0
					inc k → k=6
				→ while i * j^k <= lim → 3·2⁶… → 192 <= 100 → FALSE
			→ endif
		→ j=3 → TRUE
			if p(j) → if p(3)=1 → TRUE
				k = 1
				while i * j^k <= lim → 3·3¹… → 9 <= 100 → TRUE
					p(i * j^k) = 0 → p(9)=0
					inc k → k=2
				… → k=3
				→ while i * j^k <= lim → 3·3³… → 81 <= 100 → TRUE
					p(i * j^k) = 0 → p(81)=0
					inc k → k=4
				→ while i * j^k <= lim → 3·3⁴… → 243 <= 100 → FALSE
			→ endif
		→ j=4 → FALSE
	→ endif
→ i=4 → TRUE
	if p(i) → if p(4)=1 → FALSE
→ i=5 → TRUE
	if p(i) → if p(5)=1 → FALSE
		for j = 2 to i → j=2 → TRUE
			if p(j) → if p(2)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·2¹… → 10 <= 100 → TRUE
					p(i * j^k) = 0 → p(10)=0
					inc k → k=2
				… → k=4
				→ while i * j^k <= lim → 5·2⁴… → 80 <= 100 → TRUE
					p(i * j^k) = 0 → p(80)=0
					inc k → k=5
				→ while i * j^k <= lim → 5·2⁵… → 160 <= 100 → FALSE
			→ endif
		→ j=3 → TRUE
			if p(j) → if p(3)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·3¹… → 15 <= 100 → TRUE
					p(i * j^k) = 0 → p(15)=0
					inc k → k=2
				→ while i * j^k <= lim → 5·3²… → 45 <= 100 → TRUE
					p(i * j^k) = 0 → p(45)=0
					inc k → k=3
				→ while i * j^k <= lim → 5·3³… → 135 <= 100 → FALSE
			→ endif
		→ j=4 → TRUE
			if p(j) → if p(4)=1 → FALSE
		→ j=5 → TRUE
			if p(j) → if p(5)=1 → TRUE
				k = 1
				while i * j^k <= lim → 5·5¹… → 25 <= 100 → TRUE
					p(i * j^k) = 0 → p(25)=0
					inc k → k=2
				→ while i * j^k <= lim → 5·5²… → 125 <= 100 → FALSE
			→ endif
		→ j=6 → FALSE
	→ endif
… → i=101 → FALSE

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WickedX

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Joined: 8th Feb 2009

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Posted: 20th Jul 2013 18:16 Edited at: 20th Jul 2013 19:38

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@OBese87 - The sample and solution function exactly the same.
Test:

+ Code Snippet

sync on
sync rate 5

backdrop on

do
   set cursor 0, 0
   print turn#

   ` using if
   `remstart
   if rightkey()
      inc turn#, 0.25
      if turn# > 20
         turn# = 20
      endif
   endif

   if leftkey()
      dec turn#, 0.25
      if turn# < -20
         turn# = -20
      endif
   endif
   `remend

   ` not using if
   `turn# = turn#+(rightkey()*(turn#<20)-leftkey()*(turn#>-20))*0.25

   sync

loop

Please explain to me how this is not an improvement?

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Libervurto

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Posted: 20th Jul 2013 23:16 Edited at: 20th Jul 2013 23:17

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@Wicked
It's not quite the same, you'll get the same output but logically they are different. The mathematical method checks bounds before incrementing; the conditional method checks bounds after incrementing. This probably would have been clearer if I hadn't totally changed your code. This functions exactly the same as the mathematical method:

+ Code Snippet

  if rightkey()
    if turn# < 20 then inc turn#, 0.25
  endif
  if leftkey()
    if turn# > -20 then dec turn#, 0.25
  endif

Quote: "Please explain to me how this is not an improvement?"

There seems to always be unnecessary calculations, it is always calculating for both keys whether one, both or neither are pressed. Have you run a speed test? I might be wrong, math logic can be surprisingly quick sometimes.

Here's another version:

+ Code Snippet

x = 0			//main variable
a = rightkey()		//input a
b = leftkey()		//input b
s = 2			//speed
max = 20		//maximum
min = -20		//minimum
//c,y are locally created variables

c = a-b
if c
  y = x+c*s
  if y < max and y > min then x=y
endif

Pretty amazing mow many different ways you can code the same thing.

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WickedX

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Posted: 21st Jul 2013 00:07 Edited at: 21st Jul 2013 00:11

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After running a speed test, you are correct it doesn’t perform as well. Your first snippet above is probably the best choice.

Quote: "Pretty amazing how many different ways you can code the same thing. "

Thats the reason I enjoy programming, there are always alternitive ways to code.

I have a question concerning your first challenge. The line you have commented is correct. The code that follows just prints a range of number from 12 to 29. The second challenge, 18 would have worked the same as 18.00017.

Thank you.

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Libervurto

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Posted: 21st Jul 2013 01:06

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Quote: "The second challenge, 18 would have worked the same as 18.00017."

I seem to have made a mistake there. It was supposed to be an example of truncation causing bias. I'll have to look at that and change it. Thanks.

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WickedX

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Posted: 21st Jul 2013 01:13

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Quote: "I seem to have made a mistake there."

We all make them

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