- Write a procedure named 'sum-of-min-two' which accepts three
arguments and will figure out the sum of the minimum two arguments.
$ scheme48 > ,load ex1.scm > (sum-of-min-two 1 2 3) 3 > (sum-of-min-two 2 1 3) 3 > (sum-of-min-two 1 1 3) 2 > (sum-of-min-two 1 3 1) 2 > ,exit $ - Put your code in ~/scheme-1051/exer1
- Deadline: 2016/Sep/29 00:05 AM.
- Write a procedure, named (triangle? a b c), which determine if three input numbers can form a triangle.
- Write a procedure, named (right-tri? a b c), which determine if three input numbers can form a triangle and a^2 + b^2 = c^2.
- Write a procedure, named (area-of-right-tri a b c), which will
calculate the area of the triangle if a, b, and c can
form a right triagle. If a, b, and c can not form a right triangle,
this procedure returns zero.
$ scheme48 > ,load ex2.scm > (triangle? 1 2 3) #f > (triangle? 1.2 2 3) #t > (right-tri? 3 4 5) #t > (right-tri? 4 5 3) #f > (right-tri? -3 4 5) #f > (area-of-right-tri 3 4 5) 6.0 > (area-of-right-tri 13 5 12) 30.0 > (area-of-right-tri 6 6 6) 0 > ,exit $ - Put your code in ~/scheme-1051/exer2
- Deadline: 2016/Oct/06 00:05 AM.
- Write a procedure, named (cube-root x), which will return y such that y^3 = x.
- For improving a guess, check exercise 1.8.
- For estimating if a guess is good enough, please use the idea discussed in exercise 1.7.
- Put your code in ~/scheme-1051/exer3
- Deadline: 2016/Oct/013 00:05 AM.
- Write a procedure, named (sum-seq a b step), in two ways.
- One is written in linear recursive way, another in linear iterative way.
- Each version should be stored in a separate file.
- The "sum-seq" will calculate the sum: a+(a+step)+(a+2*step)....
- The summing will be terminated when a+k*step exceeds b.
That is, when step is positive, a+k*step > b,
when step is negative, a+k*step < b.
$ scheme48 > ,load ex4.scm > (sum-seq 1 11 3) 22 > (sum-seq 11 1 3) 0 > (sum-seq 11 1 -3) 26 > (sum-seq 11 13 -1) 0 > (sum-seq 11 11 3) 11 > (sum-seq 11 11 -3) 11 > ,exit $ - Put your code in ~/scheme-1051/exer4
- Deadline: 2016/Oct/20 00:05 AM.
- Refer to exercise 1.11 but the formula is changed to
f(n) = n if n < 3
f(n) = f(n-1) + 2*f(n-2) + 3*f(n-3) if n ≥ 3 and n is even
f(n) = f(n-1) - 2*f(n-2) + 3*f(n-3) if n ≥ 3 and n is odd
- Note that you have to write two procedures. One is written in recursive way and another in linear iterative way.
- Your program should be put under ~/scheme-1051/exer5
- Deadline: 2016/Oct/27 00:05 AM.
- Write a linear-iteraitve expmod.
- Check exercise 1.16 for how to convert 'recursive' to 'iterative'.
- Your program should be put under ~/scheme-1051/exer6
- Deadline: 2016/Nov/03 00:05 AM.
- Refer to exercise 1.37.
- You have to write two procedures. One is (cont-frac n d k) and another is (pi epsilon).
- The "cont-frac" can be implemented in either linear-recurse or linear iterative way. Only one version is required.
- To calculate pi, the sequence N_i is: 4, 1^2, 2^2, 3^2, ...
and the sequence D_i is: 1, 3, 5, 7, ... - To approximate pi, calculate (cont-frac n d 1), (cont-fract n d 2), ..
until two consecutive values are close enough, that is, the distance
between them is less than epsilon.
$ scheme48 > ,load ex7.scm > (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 100) 0.6180339887498948 > (pi 0.001) 3.141463414634146 > (pi 0.0001) 3.1415888250921244 > (pi 0.0000000000001) 3.141592653589791 > ,exit $ - Deadline: 2016 Nov 24 00:05 am
- Put your file(s) under ~/scheme-1051/exer7
- Refer to exercise 1.42, 1,43, and 1.45.
- First you have to write `compose' (exercise 1.42) and use it to write "repeated" (exercise 1.43).
- Then with compose and repeated, write "nth-root" (exercise 1.45).
You should also use average-damp and fixed-point. - You have to determine how many times average-damp is called such that the converted function when passed to fixed-point will converge. You have to do some experiments. Just using n is no good.
- Use the number of invocation of average-damp to write your nth-root.
- Summary: compose, repeated, number of invocation of average-damp,
nth-root
--prompt--> scheme48 > ,load ex7-nth-root.scm > (nth-root (fast-expt 2 16) 8) 4.0 > (nth-root (fast-expt 2 32) 16) 4.0 > (nth-root (fast-expt 2 100) 50) 3.9999999986136436 > (nth-root (fast-expt 5 10) 10) 4.9999999992832524 > ,exit --prompt--> - Put your file(s) under ~/scheme-1051/exer8
- Deadline 2016/Dec/01, 00:05 am
- Refer to exercise 2.5.
- Note the requirement of nonnegative is removed. Your method should accept two integers.
- Deadline: 2016 Dec 08 00:05
- Put your file(s) under ~/scheme-1051/exer9
- A tree with a single node is like: '(a)
- A tree with root a and two right b c is like: '(a (b) (c))
- For example:
$ scheme48 > ,load ex10-pre.scm > (preorder '(a)) (a) > (preorder '(a (b) (c))) (a b c) > (preorder '(a (b (c) (d)) (e))) (a b c d e) > ,exit $ - Deadline: 2016 Dec 15 00:05
- Put your file(s) under ~/scheme-1051/exer10
- Write a function, (sublist L start-index sublen), which
returns a sublist from L, starting from start-index and with
length=sublen.For example:
$ scheme48 > > (sublist '(0 1 2 3 4) 1 3) '(1 2 3) > (sublist '(0 1 2 3 4) 0 2) '(0 1) > (sublist '(0 1 2 3 4) 1 0) '() > (sublist '(0 1 2 3 4) 4 2) '(4) > (sublist '(0 1 2 3 4) 4 1) '(4) > (sublist '(0 1 2 3 4) 5 2) '() > (sublist '(0 1 2 3 4) 2 5) '(2 3 4) > ,exit $ - Deadline: 2016 Dec 22 00:05
- Put your file(s) under ~/scheme-1051/exer11
- Refer to exercise 3.2 but your `make-monitored' should be
able to handle variable number of arguments. For example,
$ scheme48 > ,load ex12-mk-monitor.scm > > (define plus (make-monitored +)) ; no values returned > (plus 1 2 3) 6 > (plus 1) 1 > (plus 2 3) 5 > (plus 'how-many-calls?) 3 > ,exit $ - Deadline: 2016 Dec 29 00:05
- Put your file(s) under ~/scheme-1051/exer12