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Due NovOct.25 (Fri) in class
Hardcopy, and on D2L.
Note: Some point-values will be increased from what is currently shown, by about 25-50%.
Reading: Scott §3.6.4 (Lambda expressions); §11, but skip 11.4 and 11.5 (OCaml and Evaluation Order, respectively).
Your name and the assignment-number must be in a comment at the start of the file, and your hardcopy must be stapled. All functions/data must include the appropriate steps1 of the design recipe. In particular, test cases alone might be worth half the credit for a function. Unless otherwise indicated, two test cases will suffice for most problems, if they are testing different situations.
For this (and future) assignments, bump up your DrRacket language-level to
Intermediate Student with
(4pts)
Write
data def: A natural-number is:
The predicates
- 0, OR
- (+ 1 [natural-number])
zero? andpositive? are often used to distinguish the two branches, though of course= and> could be used equally-well.
And/or, see the lecture notes on natnums as recursive data, about viewingsub1 as thegetterto pull out thenatural-number field, which a positive is built out of.
Note: Note that being tail-recursive is a property of a function’s source-code. The fact that a tail-recursive function can be optimized to not unnecessarily-allocate stack space is a compiler-implementation issue — albeit it’s what makes the concept of tail-recursion important.
Reading: Scott also discusses recursion and tail-recursion, in §6.6.1 (both 3rd and 4th eds).
#| /** Return the smallest number in a list. * @pre <code>!nums.isEmpty()</code> * @return the smallest number in `nums`. */ static Double myMin( List<Double> nums ) { // initialize our loop-variables: double minSoFar = nums.get( ); List<Double> numsRemaining = nums.subList( ,nums.size()); while ( ) { double a = numsRemaining.get(0); // corresponding to Scott’s variable `a` minSoFar = ( ? : ); numsRemaining = ; } return minSoFar; } |# |
Btw: The book’s starting-code calls(empty? (rest l)) — something not in the template. It’s a bit of a hack to stray from the template: Scott wants to avoid making a helper-function, but still return a sentinel-value answer for empty-lists.
However, when converting to tail-recursion, this difference ends up being moot: as you make a helper/wrapper function for the tail-recursion, that extra check disappears.
scheme vs racket: The book’s scheme code uses:car ,cdr ,null? , and#t . In racket, these names are (respectively):first ,rest ,empty? , and#true .
(check-expect (natnum->string/binary 0) "") (check-expect (natnum->string/binary 1) "1") (check-expect (natnum->string/binary 2) "10") (check-expect (natnum->string/binary 3) "11") (check-expect (natnum->string/binary 4) "100") (check-expect (natnum->string/binary 5) "101") (check-expect (natnum->string/binary 15) "1111") (check-expect (natnum->string/binary 16) "10000") ; natnum->string/binary : natnum -> string ; Return the binary-numeral representing n (without any leading zeroes). ; Note that the numeral for zero, without leading zeros, is the empty-string! ; (define (natnum->string/binary n) (cond [(zero? n) ""] [(positive? n) (string-append (natnum->string/binary (quotient n 2)) (if (even? n) "0" "1"))])) |
Btw: This code doesn’t quite follow the design-recipe for natural-numbers, because it recurs on(quotient n 2) rather than(sub1 n) . But it still works fine because it “reduces” the natnum to a smaller one. To reason about this code, you wouldn’t use straight-up mathematical induction; instead you'd call it “strong induction”.
Recall:
the scope of identifiers introduced with
Recall: a variable-use's binding-occurrence is the place where that variable is defined.
Hint: The binding-occurrence itself is not considered a use of the variable. There are a total of seven uses ofa , amongst the four parts.
line 01 (define a 5) line 02 (define b 10) |
line A1 (let {[z 50] line A2 [a 51] line A3 } line A4 (+ a b z)) ; evaluates to: |
line B1 (let {[z 50] line B2 [a 51] line B3 [b (* a 3)] line B4 } line B5 (+ a b z)) ; evaluates to: |
line C1 (define (foo a) line C2 (let {[z 50] line C3 [a 51] line C4 [b (* a 3)] line C5 } line C6 (+ a b z))) line C7 (foo 1001) ; evaluates to: |
line D1 (let* {[z 50] line D2 [a 51] line D3 [b (* a 3)] line D4 } line D5 (+ a b z)) ; evaluates to: |
Notice that
several of the image-creating functions imported via
(3pts)
Let’s write a function
(check-expect (shapes-in-a-row |
(2pts) Pass your function a list containing
Hint: Using the name “keep? ” for one of your parameters is a good, descriptive name.
(2pts): using
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