<div dir="ltr">Hi Patrick, <br><div class="gmail_extra"><br><div class="gmail_quote">On Tue, Sep 12, 2017 at 7:21 PM, PATRICK BROWNE <span dir="ltr"><<a href="mailto:patrick.browne@dit.ie" target="_blank">patrick.browne@dit.ie</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Mukesh,<div>Thanks for your reply.</div><div>Does the mean that my original sqrt0 makes (2^25=33554432) function calls?</div></div></blockquote><div> </div><div> Technically Yes, and more accurately for any given number n, you have 2 ^ (n - 1) call because your base case is 1. <br></div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div>How many function calls does the let/where versions make?</div></div></blockquote><div><br></div><div>With let/where version you store the value of function call in in variable, so no more second call and total functional in this scenario is linear in number n. For you case with n = 25 the number of function calls is just 24. <br></div><div><br></div><div>Best, <br></div><div>Mukesh Tiwari<br></div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div>Thanks,</div><div>Pat</div></div><div class="gmail_extra"><br><div class="gmail_quote">On 12 September 2017 at 01:36, mukesh tiwari <span dir="ltr"><<a href="mailto:mukeshtiwari.iiitm@gmail.com" target="_blank">mukeshtiwari.iiitm@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Hi Patrick, <br><div><div class="gmail_extra"><br><div class="gmail_quote"><span>On Mon, Sep 11, 2017 at 10:44 PM, PATRICK BROWNE <span dir="ltr"><<a href="mailto:patrick.browne@dit.ie" target="_blank">patrick.browne@dit.ie</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><div>Why is it the sqrt0 function is so much slower than sqrt1. Does the where clause allow intermediate values to be stored?</div><div>Regards,</div><div>Pat</div><div>sqrt0 :: Int -> Int</div><div>sqrt0 0 = 0 </div><div>sqrt0 1 = 1 </div><div>sqrt0 n = ((sqrt0 (n - 1)) + (n `quot` sqrt0 (n-1))) `quot` 2 </div><div>-- sqrt0 25 several minutes<br></div></div></blockquote><div> </div></span><div>In sqrt0, each function call with n > 1 creates two more function call, and this creates exponential blow up (factor of 2). You can make your code it faster by storing the intermediate result<br></div><div><br></div><div><span><div>sqrt0 :: Int -> Int</div><div>sqrt0 0 = 0 </div><div>sqrt0 1 = 1 </div></span><div>sqrt0 n = let k = sqrt0 (n - 1) in (k + (n `quot` k)) `quot` 2 <br></div></div><div> </div><div>This code is not blowing exponentially because of you storing intermediate result leading to faster computation.</div><div><br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div><div class="m_-1694153706539080484h5"><div dir="ltr"><div></div><div>sqrt1 :: Int -> Int</div><div>sqrt1 n </div><div> | n == 0 = 0</div><div> | n == 1 = 1 </div><div> | otherwise = div (k + ( div n k)) 2</div><div> where k = sqrt1(n-1)</div><div>-- sqrt1 25 instant<br></div><div><br></div></div><div class="gmail_extra"><br><div class="gmail_quote">On 9 September 2017 at 05:49, KC <span dir="ltr"><<a href="mailto:kc1956@gmail.com" target="_blank">kc1956@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="auto">One approach <div dir="auto"><br></div><div dir="auto">One function to compute the next iterate</div><div dir="auto"><br></div><div dir="auto">Another function to call the computation function until results are within some tolerance </div><div dir="auto"><br></div><div dir="auto">It's usually presented as separation of control and computation 😎<br><br><div dir="auto">--<br>Sent from an expensive device which will be obsolete in a few months<br>Casey</div></div></div><div class="m_-1694153706539080484m_-8162227982148736475gmail-m_-2211523315161561755HOEnZb"><div class="m_-1694153706539080484m_-8162227982148736475gmail-m_-2211523315161561755h5"><div class="gmail_extra"><br><div class="gmail_quote">On Sep 3, 2017 1:23 AM, "mike h" <<a href="mailto:mike_k_houghton@yahoo.co.uk" target="_blank">mike_k_houghton@yahoo.co.uk</a>> wrote:<br type="attribution"><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div>Hi,<div><br></div><div>To help me in learning Haskell I started blogging about some of the things I’ve looked at. </div><div>One such topic was calculating square roots ‘by hand’ and then deriving a Haskell algorithm. </div><div>I wrote about the well known technique here</div><div><a href="http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/" target="_blank">http://gitcommit.co.uk/2017/08<wbr>/25/the-root-of-the-problem-pa<wbr>rt-1/</a></div><div><br></div><div>and it it is really quite a simple method. </div><div><br></div><div>The second part of the post will be an implementation in Haskell. </div><div><br></div><div>I then tried implementing it and got something that works but really its not very pleasant to look at! And its something I don’t want to post! Some parts are fine but I think I locked myself into the notion that it had to be using State and really the end result is pretty poor. </div><div><br></div><div>I know this i perhaps a ‘big ask’ but I’d really appreciate any suggestions, solutions, hints etc. I will of course give full attribution. </div><div><br></div><div>I’ve created a gist of the code here</div><div><a href="https://gist.github.com/banditpig" target="_blank">https://gist.github.com/bandit<wbr>pig</a></div><div><br></div><div>Many Thanks</div><div><br></div><div>Mike</div></div><br>______________________________<wbr>_________________<br>
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