{"id":33380,"date":"2023-02-07T11:35:21","date_gmt":"2023-02-07T06:05:21","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-recursive-to-iterative-using-a-systematic-method-closed\/"},"modified":"2023-02-07T11:35:21","modified_gmt":"2023-02-07T06:05:21","slug":"solved-recursive-to-iterative-using-a-systematic-method-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-recursive-to-iterative-using-a-systematic-method-closed\/","title":{"rendered":"[Solved] Recursive to iterative using a systematic method [closed]"},"content":{"rendered":"

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So, to restate the question. We have a function f<\/em>, in our case fac<\/code>. <\/p>\n

def fac(n):\n    if n==0:\n        return 1\n    else:\n        return n*fac(n-1)\n<\/code><\/pre>\n
It is implemented recursively. We want to implement a function facOpt<\/code> that does the same thing but iteratively. fac<\/code> is written almost in the form we need. Let us rewrite it just a bit:<\/p>\n
def fac_(n, r):\n    return (n+1)*r\n\ndef fac(n):\n    if n==0:\n        return 1\n    else:\n        r = fac(n-1)\n        return fac_(n-1, r)\n<\/code><\/pre>\n
This is exactly the recursive definition from section 4.2. Now we need to rewrite it iteratively:<\/p>\n
def facOpt(n):\n    if n==0:\n        return 1\n    x = 1\n    r = 1\n    while x != n:\n        x = x + 1\n        r = fac_(x-1, r)\n    return r\n<\/code><\/pre>\n
This is exactly the iterative definition from section 4.2. Note that facOpt<\/code> does not call itself anywhere. Now, this is neither the most clear nor the most pythonic way of writing down this algorithm — this is just a way to transform one algorithm to another. We can implement the same algorithm differently, e.g. like that:<\/p>\n
def facOpt(n):\n    r = 1\n    for x in range(1, n+1):\n        r *= x \n    return r\n<\/code><\/pre>\n
Things get more interesting when we consider more complicated functions. Let us write fibObt<\/code> where fib<\/code> is :<\/p>\n
def fib(n):\n    if n==0:\n        return 0\n    elif n==1:\n        return 1\n    else:\n        return fib(n-1) + fib(n-2)\n<\/code><\/pre>\n
fib<\/code> calls itself two times, but the recursive pattern from the book allows only a single call. That is why we need to extend the function to returning not one, but two values. Fully reformated, fib<\/code> looks like this:<\/p>\n
def fibExt_(n, rExt):\n    return rExt[0] + rExt[1], rExt[0]\n\ndef fibExt(n):\n    if n == 0:\n        return 0, 0\n    elif n == 1:\n        return 1, 0\n    else:\n        rExt = fibExt(n-1)\n        return fibExt_(n-1, rExt)\n\ndef fib(n):\n    return fibExt(n)[0]\n<\/code><\/pre>\n
You may notice that the first argument to fibExt_<\/code> is never used. I just added it to follow the proposed structure exactly.
\nNow, it is again easy to turn fib<\/code> into an iterative version:<\/p>\n
def fibExtOpt(n):\n    if n == 0:\n        return 0, 0\n    if n == 1:\n        return 1, 0\n    x = 2\n    rExt = 1, 1\n    while x != n:\n        x = x + 1\n        rExt = fibExt_(x-1, rExt)\n    return rExt\n\ndef fibOpt(n):\n    return fibExtOpt(n)[0]\n<\/code><\/pre>\n
Again, the new version does not call itself. And again one can streamline it to this, for example:<\/p>\n
def fibOpt(n):\n    if n < 2:\n        return n\n    a, b = 1, 1\n    for i in range(n-2):\n        a, b = b, a+b\n    return b\n<\/code><\/pre>\n
The next function to translate to iterative version is bin<\/code>:<\/p>\n
def bin(n,k):\n    if k == 0 or k == n:\n        return 1\n    else:\n        return bin(n-1,k-1) + bin(n-1,k)\n<\/code><\/pre>\n
Now neither x<\/code> nor r<\/code> can be just numbers. The index (x<\/code>) has two components, and the cache (r<\/code>) has to be even larger. One (not quite so optimal) way would be to return the whole previous row of the Pascal triangle:<\/p>\n
def binExt_(r):\n    return [a + b for a,b in zip([0] + r, r + [0])]\n\ndef binExt(n):\n    if n == 0:\n        return [1]\n    else:\n        r = binExt(n-1)\n        return binExt_(r)\n\ndef bin(n, k):\n    return binExt(n)[k]\n<\/code><\/pre>\n
I have’t followed the pattern so strictly here and removed several useless variables. It is still possible to translate to an iterative version directly:<\/p>\n
def binExtOpt(n):\n    if n == 0:\n        return [1]\n    x = 1\n    r = [1, 1]\n    while x != n:\n        r = binExt_(r)\n        x += 1\n    return r\n\ndef binOpt(n, k):\n    return binExtOpt(n)[k]\n<\/code><\/pre>\n
For completeness, here is an optimized solution that caches only part of the row:<\/p>\n
def binExt_(n, k_from, k_to, r):\n    if k_from == 0 and k_to == n:\n        return [a + b for a, b in zip([0] + r, r + [0])]\n    elif k_from == 0:\n        return [a + b for a, b in zip([0] + r[:-1], r)]\n    elif k_to == n:\n        return [a + b for a, b in zip(r, r[1:] + [0])]\n    else:\n        return [a + b for a, b in zip(r[:-1], r[1:])]\n\ndef binExt(n, k_from, k_to):\n    if n == 0:\n        return [1]\n    else:\n        r = binExt(n-1, max(0, k_from-1), min(n-1, k_to+1) )\n        return binExt_(n, k_from, k_to, r)\n\ndef bin(n, k):\n    return binExt(n, k, k)[0]\n\ndef binExtOpt(n, k_from, k_to):\n    if n == 0:\n        return [1]\n    ks = [(k_from, k_to)]\n    for i in range(1,n):\n        ks.append((max(0, ks[-1][0]-1), min(n-i, ks[-1][1]+1)))\n    x = 0\n    r = [1]\n    while x != n:\n        x += 1\n        r = binExt_(x, *ks[n-x], r)\n    return r\n\ndef binOpt(n, k):\n    return binExtOpt(n, k, k)[0]\n<\/code><\/pre>\n
In the end, the most difficult task is not switching from recursive to iterative implementation, but to have a recursive implementation that follows the required pattern. So the real question is how to create fibExt'<\/code>, not fibExtOpt<\/code>.<\/p>\n<\/p><\/div>\n
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            4<\/span> <\/p><\/div>\n<\/div>\n
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solved Recursive to iterative using a systematic method [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] So, to restate the question. We have a function f, in our case fac. def fac(n): if n==0: return 1 else: return n*fac(n-1) It is implemented recursively. We want to implement a function facOpt that does the same thing but iteratively. fac is written almost in the form we need. Let us rewrite it … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[349,494],"class_list":["post-33380","post","type-post","status-publish","format-standard","hentry","category-solved","tag-python","tag-recursion"],"yoast_head":"\n[Solved] Recursive to iterative using a systematic method [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-recursive-to-iterative-using-a-systematic-method-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Recursive to iterative using a systematic method [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] So, to restate the question. We have a function f, in our case fac. def fac(n): if n==0: return 1 else: return n*fac(n-1) It is implemented recursively. We want to implement a function facOpt that does the same thing but iteratively. fac is written almost in the form we need. Let us rewrite it ... 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