Cody

Problem 24. Function Iterator

Solution 179140

Submitted on 18 Dec 2012 by Des Mc Manus
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Test Suite

Test Status Code Input and Output
1   Pass
%% noOp = @(x)x; fh2 = iterate_fcn(noOp, 50); assert(isequal(fh2(pi),pi));

h = @(x)x h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x))

2   Pass
%% addOne = @(x)x+1; addTen = iterate_fcn(addOne, 10); assert(isequal(addTen(3),13));

h = @(x)x+1 h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x))

3   Pass
%% addOne = @(x)x+1; addOne2 = iterate_fcn(addOne, 1); assert(isequal(addOne2(3),4));

h = @(x)x+1

4   Pass
%% squarer = @(a) a^2; fh2 = iterate_fcn(squarer, 3); assert(isequal(fh2(3),6561));

h = @(a)a^2 h = @(x)f(h(x)) h = @(x)f(h(x))

5   Pass
%% fh = @(y)sqrt(y+1); fh2 = iterate_fcn(fh,30); assert(abs(fh2(1) - (1+sqrt(5))/2) < 100*eps);

h = @(y)sqrt(y+1) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x)) h = @(x)f(h(x))

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