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    <p>For the function</p> <pre><code> _________ 77 if(x&lt;=5) / / foo(x)- \ \_________ foo(x-1) - foo(x-1) if(x&gt;5) let f(x) be time function for foo(x) f(x) = f(x-1) - f(x-1) // 2 (2^1)calls of f(x-2) happened for 1 level depth f(x) = [f(x-2) - f(x-2)] - [ f(x-2) - f(x-2)] (expanding f(x-1)) // 4(2^2) calls of f(x-2) happened for 2nd level depth f(x)={[ f(x-3) - f(x-3)]-[ f(x-3) - f(x-3)]} - {[ f(x-3) - f(x-3)]-[ f(x-3) - f(x-3)]} // 8(2^3) calls of f(x-2) happened for 3rd level depth </code></pre> <p>let i calls has happened to complete program.... </p> <pre><code>but program terminates when x&lt;=5, so program terminates in call f(x-i) when x-i&lt;=5 that means i&gt;=x-5 , at level i there are 2power(i) calls ==&gt; 2^i calls of f(x-i). since f(&lt;=5)=1(1 is unit of measure) for i&gt;=x-5 </code></pre> <p>so f(n) = 2^(x-5)*(1) =>O(2^x) where x is input size . if we replace x with n complexity is O(2^n) .</p> <p>for second question </p> <pre><code> _________ 0 if(b&lt;=0) / / bar(a,b) \ \_________ foo( bar(a,b+1) ,b-1 ) if(b&gt;0) </code></pre> <p>let t(n) be time function for bar(a,b) where n is proportional to b as b is deciding factor for termination .</p> <p>expanding reccurence</p> <pre><code>t(a,b) = t( t(a,b+1), b-1) . first t(a,b+1) is executed it inturn calls t(a,b+2) and so on.... it will be infinite recursion ..... for b &gt; 0 . </code></pre> <p>To my knowledge ,since we don't have limit for infinity(neither lower limit nor upper limit, so no theta notation and no big-oh notation so as omega notation) we cant measure complexity function as well .(Please correct me if i'm wrong) </p> <p>But if b&lt;0 then it will be done in O(1) time...</p>
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