Consider:
#include "share/atspre_staload.hats"
fun only_zero(n: int(0)): void =
println!("This is definitely zero: ", n)
fun less_than{n,m:int | n < m}(n: int(n), m: int(m)): void =
println!(n, " is less than ", m)
implement main0() = (
only_zero(zeroify(n));
only_zero(m);
less_than(b, a);
less_than(f, e) where { val (f, e) = make_less_than((d, c)) };
) where {
val n = 5
val m = ~5
val (a, b, c, d) = (1, 2, 3, 4)
extern castfn zeroify{n:int}(n: int(n)): int(0)
extern praxi lemma_this_is_zero{n:int}(n: int(n)): [n == 0] void
extern castfn make_less_than{n,m:int}(t: (int(n), int(m))): [o,p:int | o < p] (int(o), int(p))
extern praxi lemma_less_than{n,m:int}(n: int(n), m: int(m)): [n < m] void
prval _ = lemma_this_is_zero(m)
}
which has this output:
This is definitely zero: 5
This is definitely zero: -5
2 is less than 1
4 is less than 3
Are there cases that demand one of these over the other?
If you use 'castfn', you need to make sure that there is a corresponding implicit cast function in the target language. For instance, int2double is a castfn if C is the target language.
On the other hand, praxi/prfun is completely erased, having no trace in the generated code.
I would say that praxi/prfun is more general, but int2double is definitely not a praxi/prfun.
Given a string and the constraint of matching on >= 3 characters, how can you ensure that the result string will be as small as possible?
edit with gassa's explicitness:
E.G.
'AAAABBBAC'
If I remove the B's first,
AAAA[BBB]AC -- > AAAAAC, then I can remove all of the A's from the resultant string and be left with:
[AAAAA]C --> C
'C'
If I just remove what is available first (the sequence of A's), I get:
[AAAA]BBBAC -- > [BBB]AC --> AC
'AC'
A tree would definitely get you the shortest string(s).
The tree solution:
Define a State (node) for each current string Input and all its removable sub-strings' int[] Indexes.
Create the tree: For each int index create another State and add it to the parent state State[] Children.
A State with no possible removable sub-strings has no children Children = null.
Get all Descendants State[] of your root State. Order them by their shortest string Input. And that is/are your answer(s).
Test cases:
string result = FindShortest("AAAABBBAC"); // AC
string result2 = FindShortest("AABBAAAC"); // AABBC
string result3 = FindShortest("BAABCCCBBA"); // B
The Code:
Note: Of-course everyone is welcome to enhance the following code in terms of performance and/or fixing any bug.
class Program
{
static void Main(string[] args)
{
string result = FindShortest("AAAABBBAC"); // AC
string result2 = FindShortest("AABBAAAC"); // AABBC
string result3 = FindShortest("BAABCCCBBA"); // B
}
// finds the FIRST shortest string for a given input
private static string FindShortest(string input)
{
// all possible removable strings' indexes
// for this given input
int[] indexes = RemovableIndexes(input);
// each input string and its possible removables are a state
var state = new State { Input = input, Indexes = indexes };
// create the tree
GetChildren(state);
// get the FIRST shortest
// i.e. there would be more than one answer sometimes
// this could be easily changed to get all possible results
var result =
Descendants(state)
.Where(d => d.Children == null || d.Children.Length == 0)
.OrderBy(d => d.Input.Length)
.FirstOrDefault().Input;
return result;
}
// simple get all descendants of a node/state in a tree
private static IEnumerable<State> Descendants(State root)
{
var states = new Stack<State>(new[] { root });
while (states.Any())
{
State node = states.Pop();
yield return node;
if (node.Children != null)
foreach (var n in node.Children) states.Push(n);
}
}
// creates the tree
private static void GetChildren(State state)
{
// for each an index there is a child
state.Children = state.Indexes.Select(
i =>
{
var input = RemoveAllAt(state.Input, i);
return input.Length < state.Input.Length && input.Length > 0
? new State
{
Input = input,
Indexes = RemovableIndexes(input)
}
: null;
}).ToArray();
foreach (var c in state.Children)
GetChildren(c);
}
// find all possible removable strings' indexes
private static int[] RemovableIndexes(string input)
{
var indexes = new List<int>();
char d = input[0];
int count = 1;
for (int i = 1; i < input.Length; i++)
{
if (d == input[i])
count++;
else
{
if (count >= 3)
indexes.Add(i - count);
// reset
d = input[i];
count = 1;
}
}
if (count >= 3)
indexes.Add(input.Length - count);
return indexes.ToArray();
}
// remove all duplicate chars starting from an index
private static string RemoveAllAt(string input, int startIndex)
{
string part1, part2;
int endIndex = startIndex + 1;
int i = endIndex;
for (; i < input.Length; i++)
if (input[i] != input[startIndex])
{
endIndex = i;
break;
}
if (i == input.Length && input[i - 1] == input[startIndex])
endIndex = input.Length;
part1 = startIndex > 0 ? input.Substring(0, startIndex) : string.Empty;
part2 = endIndex <= (input.Length - 1) ? input.Substring(endIndex) : string.Empty;
return part1 + part2;
}
// our node, which is
// an input string &
// all possible removable strings' indexes
// & its children
public class State
{
public string Input;
public int[] Indexes;
public State[] Children;
}
}
I propose O(n^2) solution with dynamic programming.
Let's introduce notation. Prefix and suffix of length l of string A denoted by P[l] and S[l]. And we call our procedure Rcd.
Rcd(A) = Rcd(Rcd(P[n-1])+S[1])
Rcd(A) = Rcd(P[1]+Rcd(S[n-1]))
Note that outer Rcd in the RHS is trivial. So, that's our optimal substructure. Based on this i came up with the following implementation:
#include <iostream>
#include <string>
#include <vector>
#include <cassert>
using namespace std;
string remdupright(string s, bool allowEmpty) {
if (s.size() >= 3) {
auto pos = s.find_last_not_of(s.back());
if (pos == string::npos && allowEmpty) s = "";
else if (pos != string::npos && s.size() - pos > 3) s = s.substr(0, pos + 1);
}
return s;
}
string remdupleft(string s, bool allowEmpty) {
if (s.size() >= 3) {
auto pos = s.find_first_not_of(s.front());
if (pos == string::npos && allowEmpty) s = "";
else if (pos != string::npos && pos >= 3) s = s.substr(pos);
}
return s;
}
string remdup(string s, bool allowEmpty) {
return remdupleft(remdupright(s, allowEmpty), allowEmpty);
}
string run(const string in) {
vector<vector<string>> table(in.size());
for (int i = 0; i < (int)table.size(); ++i) {
table[i].resize(in.size() - i);
}
for (int i = 0; i < (int)table[0].size(); ++i) {
table[0][i] = in.substr(i,1);
}
for (int len = 2; len <= (int)table.size(); ++len) {
for (int pos = 0; pos < (int)in.size() - len + 1; ++pos) {
string base(table[len - 2][pos]);
const char suffix = in[pos + len - 1];
if (base.size() && suffix != base.back()) {
base = remdupright(base, false);
}
const string opt1 = base + suffix;
base = table[len - 2][pos+1];
const char prefix = in[pos];
if (base.size() && prefix != base.front()) {
base = remdupleft(base, false);
}
const string opt2 = prefix + base;
const string nodupopt1 = remdup(opt1, true);
const string nodupopt2 = remdup(opt2, true);
table[len - 1][pos] = nodupopt1.size() > nodupopt2.size() ? opt2 : opt1;
assert(nodupopt1.size() != nodupopt2.size() || nodupopt1 == nodupopt2);
}
}
string& res = table[in.size() - 1][0];
return remdup(res, true);
}
void testRcd(string s, string expected) {
cout << s << " : " << run(s) << ", expected: " << expected << endl;
}
int main()
{
testRcd("BAABCCCBBA", "B");
testRcd("AABBAAAC", "AABBC");
testRcd("AAAA", "");
testRcd("AAAABBBAC", "C");
}
You can check default and run your tests here.
Clearly we are not concerned about any block of repeated characters longer than 2 characters. And there is only one way two blocks of the same character where at least one of the blocks is less than 3 in length can be combined - namely, if the sequence between them can be removed.
So (1) look at pairs of blocks of the same character where at least one is less than 3 in length, and (2) determine if the sequence between them can be removed.
We want to decide which pairs to join so as to minimize the total length of blocks less than 3 characters long. (Note that the number of pairs is bound by the size (and distribution) of the alphabet.)
Let f(b) represent the minimal total length of same-character blocks remaining up to the block b that are less than 3 characters in length. Then:
f(b):
p1 <- previous block of the same character
if b and p1 can combine:
if b.length + p1.length > 2:
f(b) = min(
// don't combine
(0 if b.length > 2 else b.length) +
f(block before b),
// combine
f(block before p1)
)
// b.length + p1.length < 3
else:
p2 <- block previous to p1 of the same character
if p1 and p2 can combine:
f(b) = min(
// don't combine
b.length + f(block before b),
// combine
f(block before p2)
)
else:
f(b) = b.length + f(block before b)
// b and p1 cannot combine
else:
f(b) = b.length + f(block before b)
for all p1 before b
The question is how can we efficiently determine if a block can be combined with the previous block of the same character (aside from the obvious recursion into the sub-block-list between the two blocks).
Python code:
import random
import time
def parse(length):
return length if length < 3 else 0
def f(string):
chars = {}
blocks = [[string[0], 1, 0]]
chars[string[0]] = {'indexes': [0]}
chars[string[0]][0] = {'prev': -1}
p = 0 # pointer to current block
for i in xrange(1, len(string)):
if blocks[len(blocks) - 1][0] == string[i]:
blocks[len(blocks) - 1][1] += 1
else:
p += 1
# [char, length, index, f(i), temp]
blocks.append([string[i], 1, p])
if string[i] in chars:
chars[string[i]][p] = {'prev': chars[string[i]]['indexes'][ len(chars[string[i]]['indexes']) - 1 ]}
chars[string[i]]['indexes'].append(p)
else:
chars[string[i]] = {'indexes': [p]}
chars[string[i]][p] = {'prev': -1}
#print blocks
#print
#print chars
#print
memo = [[None for j in xrange(len(blocks))] for i in xrange(len(blocks))]
def g(l, r, top_level=False):
####
####
#print "(l, r): (%s, %s)" % (l,r)
if l == r:
return parse(blocks[l][1])
if memo[l][r]:
return memo[l][r]
result = [parse(blocks[l][1])] + [None for k in xrange(r - l)]
if l < r:
for i in xrange(l + 1, r + 1):
result[i - l] = parse(blocks[i][1]) + result[i - l - 1]
for i in xrange(l, r + 1):
####
####
#print "\ni: %s" % i
[char, length, index] = blocks[i]
#p1 <- previous block of the same character
p1_idx = chars[char][index]['prev']
####
####
#print "(p1_idx, l, p1_idx >= l): (%s, %s, %s)" % (p1_idx, l, p1_idx >= l)
if p1_idx < l and index > l:
result[index - l] = parse(length) + result[index - l - 1]
while p1_idx >= l:
p1 = blocks[p1_idx]
####
####
#print "(b, p1, p1_idx, l): (%s, %s, %s, %s)\n" % (blocks[i], p1, p1_idx, l)
between = g(p1[2] + 1, index - 1)
####
####
#print "between: %s" % between
#if b and p1 can combine:
if between == 0:
if length + p1[1] > 2:
result[index - l] = min(
result[index - l],
# don't combine
parse(length) + (result[index - l - 1] if index - l > 0 else 0),
# combine: f(block before p1)
result[p1[2] - l - 1] if p1[2] > l else 0
)
# b.length + p1.length < 3
else:
#p2 <- block previous to p1 of the same character
p2_idx = chars[char][p1[2]]['prev']
if p2_idx < l:
p1_idx = chars[char][p1_idx]['prev']
continue
between2 = g(p2_idx + 1, p1[2] - 1)
#if p1 and p2 can combine:
if between2 == 0:
result[index - l] = min(
result[index - l],
# don't combine
parse(length) + (result[index - l - 1] if index - l > 0 else 0),
# combine the block, p1 and p2
result[p2_idx - l - 1] if p2_idx - l > 0 else 0
)
else:
#f(b) = b.length + f(block before b)
result[index - l] = min(
result[index - l],
parse(length) + (result[index - l - 1] if index - l > 0 else 0)
)
# b and p1 cannot combine
else:
#f(b) = b.length + f(block before b)
result[index - l] = min(
result[index - l],
parse(length) + (result[index - l - 1] if index - l > 0 else 0)
)
p1_idx = chars[char][p1_idx]['prev']
#print l,r,result
memo[l][r] = result[r - l]
"""if top_level:
return (result, blocks)
else:"""
return result[r - l]
if len(blocks) == 1:
return ([parse(blocks[0][1])], blocks)
else:
return g(0, len(blocks) - 1, True)
"""s = ""
for i in xrange(300):
s = s + ['A','B','C'][random.randint(0,2)]"""
print f("abcccbcccbacccab") # b
print
print f("AAAABBBAC"); # C
print
print f("CAAAABBBA"); # C
print
print f("AABBAAAC"); # AABBC
print
print f("BAABCCCBBA"); # B
print
print f("aaaa")
print
The string answers for these longer examples were computed using jdehesa's answer:
t0 = time.time()
print f("BCBCCBCCBCABBACCBABAABBBABBBACCBBBAABBACBCCCACABBCAABACBBBBCCCBBAACBAABACCBBCBBAABCCCCCAABBBBACBBAAACACCBCCBBBCCCCCCCACBABACCABBCBBBBBCBABABBACCAACBCBBAACBBBBBCCBABACBBABABAAABCCBBBAACBCACBAABAAAABABB")
# BCBCCBCCBCABBACCBABCCAABBACBACABBCAABACAACBAABACCBBCBBCACCBACBABACCABBCCBABABBACCAACBCBBAABABACBBABABBCCAACBCACBAABBABB
t1 = time.time()
total = t1-t0
print total
t0 = time.time()
print f("CBBACAAAAABBBBCAABBCBAABBBCBCBCACACBAABCBACBBABCABACCCCBACBCBBCBACBBACCCBAAAACACCABAACCACCBCBCABAACAABACBABACBCBAACACCBCBCCCABACABBCABBAAAAABBBBAABAABBCACACABBCBCBCACCCBABCAACBCAAAABCBCABACBABCABCBBBBABCBACABABABCCCBBCCBBCCBAAABCABBAAABBCAAABCCBAABAABCAACCCABBCAABCBCBCBBAACCBBBACBBBCABAABCABABABABCA")
# CBBACCAABBCBAACBCBCACACBAABCBACBBABCABABACBCBBCBACBBABCACCABAACCACCBCBCABAACAABACBABACBCBAACACCBCBABACABBCBBCACACABBCBCBCABABCAACBCBCBCABACBABCABCABCBACABABACCBBCCBBCACBCCBAABAABCBBCAABCBCBCBBAACCACCABAABCABABABABCA
t1 = time.time()
total = t1-t0
print total
t0 = time.time()
print f("AADBDBEBBBBCABCEBCDBBBBABABDCCBCEBABADDCABEEECCECCCADDACCEEAAACCABBECBAEDCEEBDDDBAAAECCBBCEECBAEBEEEECBEEBDACDDABEEABEEEECBABEDDABCDECDAABDAEADEECECEBCBDDAEEECCEEACCBBEACDDDDBDBCCAAECBEDAAAADBEADBAAECBDEACDEABABEBCABDCEEAABABABECDECADCEDAEEEBBBCEDECBCABDEDEBBBABABEEBDAEADBEDABCAEABCCBCCEDCBBEBCECCCA")
# AADBDBECABCEBCDABABDCCBCEBABADDCABCCEADDACCEECCABBECBAEDCEEBBECCBBCEECBAEBCBEEBDACDDABEEABCBABEDDABCDECDAABDAEADEECECEBCBDDACCEEACCBBEACBDBCCAAECBEDDBEADBAAECBDEACDEABABEBCABDCEEAABABABECDECADCEDACEDECBCABDEDEABABEEBDAEADBEDABCAEABCCBCCEDCBBEBCEA
t1 = time.time()
total = t1-t0
print total
Another scala answer, using memoization and tailcall optimization (partly) (updated).
import scala.collection.mutable.HashSet
import scala.annotation._
object StringCondense extends App {
#tailrec
def groupConsecutive (s: String, sofar: List[String]): List[String] = s.toList match {
// def groupConsecutive (s: String): List[String] = s.toList match {
case Nil => sofar
// case Nil => Nil
case c :: str => {
val (prefix, rest) = (c :: str).span (_ == c)
// Strings of equal characters, longer than 3, don't make a difference to just 3
groupConsecutive (rest.mkString(""), (prefix.take (3)).mkString ("") :: sofar)
// (prefix.take (3)).mkString ("") :: groupConsecutive (rest.mkString(""))
}
}
// to count the effect of memoization
var count = 0
// recursively try to eliminate every group of 3 or more, brute forcing
// but for "aabbaabbaaabbbaabb", many reductions will lead sooner or
// later to the same result, so we try to detect these and avoid duplicate
// work
def moreThan2consecutive (s: String, seenbefore: HashSet [String]): String = {
if (seenbefore.contains (s)) s
else
{
count += 1
seenbefore += s
val sublists = groupConsecutive (s, Nil)
// val sublists = groupConsecutive (s)
val atLeast3 = sublists.filter (_.size > 2)
atLeast3.length match {
case 0 => s
case 1 => {
val res = sublists.filter (_.size < 3)
moreThan2consecutive (res.mkString (""), seenbefore)
}
case _ => {
val shrinked = (
for {idx <- (0 until sublists.size)
if (sublists (idx).length >= 3)
pre = (sublists.take (idx)).mkString ("")
post= (sublists.drop (idx+1)).mkString ("")
} yield {
moreThan2consecutive (pre + post, seenbefore)
}
)
(shrinked.head /: shrinked.tail) ((a, b) => if (a.length <= b.length) a else b)
}
}
}
}
// don't know what Rcd means, adopted from other solution but modified
// kind of a unit test **update**: forgot to reset count
testRcd (s: String, expected: String) : Boolean = {
count = 0
val seenbefore = HashSet [String] ()
val result = moreThan2consecutive (s, seenbefore)
val hit = result.equals (expected)
println (s"Input: $s\t result: ${result}\t expected ${expected}\t $hit\t count: $count");
hit
}
// some test values from other users with expected result
// **upd:** more testcases
def testgroup () : Unit = {
testRcd ("baabcccbba", "b")
testRcd ("aabbaaac", "aabbc")
testRcd ("aaaa", "")
testRcd ("aaaabbbac", "c")
testRcd ("abcccbcccbacccab", "b")
testRcd ("AAAABBBAC", "C")
testRcd ("CAAAABBBA", "C")
testRcd ("AABBAAAC", "AABBC")
testRcd ("BAABCCCBBA", "B")
testRcd ("AAABBBAAABBBAAABBBC", "C") // 377 subcalls reported by Yola,
testRcd ("AAABBBAAABBBAAABBBAAABBBC", "C") // 4913 when preceeded with AAABBB
}
testgroup
def testBigs () : Unit = {
/*
testRcd ("BCBCCBCCBCABBACCBABAABBBABBBACCBBBAABBACBCCCACABBCAABACBBBBCCCBBAACBAABACCBBCBBAABCCCCCAABBBBACBBAAACACCBCCBBBCCCCCCCACBABACCABBCBBBBBCBABABBACCAACBCBBAACBBBBBCCBABACBBABABAAABCCBBBAACBCACBAABAAAABABB",
"BCBCCBCCBCABBACCBABCCAABBACBACABBCAABACAACBAABACCBBCBBCACCBACBABACCABBCCBABABBACCAACBCBBAABABACBBABABBCCAACBCACBAABBABB")
*/
testRcd ("CBBACAAAAABBBBCAABBCBAABBBCBCBCACACBAABCBACBBABCABACCCCBACBCBBCBACBBACCCBAAAACACCABAACCACCBCBCABAACAABACBABACBCBAACACCBCBCCCABACABBCABBAAAAABBBBAABAABBCACACABBCBCBCACCCBABCAACBCAAAABCBCABACBABCABCBBBBABCBACABABABCCCBBCCBBCCBAAABCABBAAABBCAAABCCBAABAABCAACCCABBCAABCBCBCBBAACCBBBACBBBCABAABCABABABABCA",
"CBBACCAABBCBAACBCBCACACBAABCBACBBABCABABACBCBBCBACBBABCACCABAACCACCBCBCABAACAABACBABACBCBAACACCBCBABACABBCBBCACACABBCBCBCABABCAACBCBCBCABACBABCABCABCBACABABACCBBCCBBCACBCCBAABAABCBBCAABCBCBCBBAACCACCABAABCABABABABCA")
/*testRcd ("AADBDBEBBBBCABCEBCDBBBBABABDCCBCEBABADDCABEEECCECCCADDACCEEAAACCABBECBAEDCEEBDDDBAAAECCBBCEECBAEBEEEECBEEBDACDDABEEABEEEECBABEDDABCDECDAABDAEADEECECEBCBDDAEEECCEEACCBBEACDDDDBDBCCAAECBEDAAAADBEADBAAECBDEACDEABABEBCABDCEEAABABABECDECADCEDAEEEBBBCEDECBCABDEDEBBBABABEEBDAEADBEDABCAEABCCBCCEDCBBEBCECCCA",
"AADBDBECABCEBCDABABDCCBCEBABADDCABCCEADDACCEECCABBECBAEDCEEBBECCBBCEECBAEBCBEEBDACDDABEEABCBABEDDABCDECDAABDAEADEECECEBCBDDACCEEACCBBEACBDBCCAAECBEDDBEADBAAECBDEACDEABABEBCABDCEEAABABABECDECADCEDACEDECBCABDEDEABABEEBDAEADBEDABCAEABCCBCCEDCBBEBCEA")
*/
}
// for generated input, but with fixed seed, to compare the count with
// and without memoization
import util.Random
val r = new Random (31415)
// generate Strings but with high chances to produce some triples and
// longer sequences of char clones
def genRandomString () : String = {
(1 to 20).map (_ => r.nextInt (6) match {
case 0 => "t"
case 1 => "r"
case 2 => "-"
case 3 => "tt"
case 4 => "rr"
case 5 => "--"
}).mkString ("")
}
def testRandom () : Unit = {
(1 to 10).map (i=> testRcd (genRandomString, "random mode - false might be true"))
}
testRandom
testgroup
testRandom
// testBigs
}
Comparing the effect of memoization lead to interesting results:
Updated measurements. In the old values, I forgot to reset the counter, which leaded to much higher results. Now the spreading of results
is much more impressive and in total, the values are smaller.
No seenbefore:
Input: baabcccbba result: b expected b true count: 4
Input: aabbaaac result: aabbc expected aabbc true count: 2
Input: aaaa result: expected true count: 2
Input: aaaabbbac result: c expected c true count: 5
Input: abcccbcccbacccab result: b expected b true count: 34
Input: AAAABBBAC result: C expected C true count: 5
Input: CAAAABBBA result: C expected C true count: 5
Input: AABBAAAC result: AABBC expected AABBC true count: 2
Input: BAABCCCBBA result: B expected B true count: 4
Input: AAABBBAAABBBAAABBBC res: C expected C true count: 377
Input: AAABBBAAABBBAAABBBAAABBBC r: C expected C true count: 4913
Input: r--t----ttrrrrrr--tttrtttt--rr----result: rr--rr expected ? unknown ? false count: 1959
Input: ttrtt----tr---rrrtttttttrtr--rr result: r--rr expected ? unknown ? false count: 213
Input: tt----r-----ttrr----ttrr-rr--rr-- result: ttrttrrttrr-rr--rr-- ex ? unknown ? false count: 16
Input: --rr---rrrrrrr-r--rr-r--tt--rrrrr result: rr-r--tt-- expected ? unknown ? false count: 32
Input: tt-rrrrr--r--tt--rrtrrr------- result: ttr--tt--rrt expected ? unknown ? false count: 35
Input: --t-ttt-ttt--rrrrrt-rrtrttrr result: --tt-rrtrttrr expected ? unknown ? false count: 35
Input: rrt--rrrr----trrr-rttttrrtttrr result: rrtt- expected ? unknown ? false count: 1310
Input: ---tttrrrrrttrrttrr---tt-----tt result: rrttrr expected ? unknown ? false count: 1011
Input: -rrtt--rrtt---t-r--r---rttr-- result: -rrtt--rr-r--rrttr-- ex ? unknown ? false count: 9
Input: rtttt--rrrrrrrt-rrttt--tt--t result: r--t-rr--tt--t expectd ? unknown ? false count: 16
real 0m0.607s (without testBigs)
user 0m1.276s
sys 0m0.056s
With seenbefore:
Input: baabcccbba result: b expected b true count: 4
Input: aabbaaac result: aabbc expected aabbc true count: 2
Input: aaaa result: expected true count: 2
Input: aaaabbbac result: c expected c true count: 5
Input: abcccbcccbacccab result: b expected b true count: 11
Input: AAAABBBAC result: C expected C true count: 5
Input: CAAAABBBA result: C expected C true count: 5
Input: AABBAAAC result: AABBC expected AABBC true count: 2
Input: BAABCCCBBA result: B expected B true count: 4
Input: AAABBBAAABBBAAABBBC rest: C expected C true count: 28
Input: AAABBBAAABBBAAABBBAAABBBC C expected C true count: 52
Input: r--t----ttrrrrrr--tttrtttt--rr----result: rr--rr expected ? unknown ? false count: 63
Input: ttrtt----tr---rrrtttttttrtr--rr result: r--rr expected ? unknown ? false count: 48
Input: tt----r-----ttrr----ttrr-rr--rr-- result: ttrttrrttrr-rr--rr-- xpe? unknown ? false count: 8
Input: --rr---rrrrrrr-r--rr-r--tt--rrrrr result: rr-r--tt-- expected ? unknown ? false count: 19
Input: tt-rrrrr--r--tt--rrtrrr------- result: ttr--tt--rrt expected ? unknown ? false count: 12
Input: --t-ttt-ttt--rrrrrt-rrtrttrr result: --tt-rrtrttrr expected ? unknown ? false count: 16
Input: rrt--rrrr----trrr-rttttrrtttrr result: rrtt- expected ? unknown ? false count: 133
Input: ---tttrrrrrttrrttrr---tt-----tt result: rrttrr expected ? unknown ? false count: 89
Input: -rrtt--rrtt---t-r--r---rttr-- result: -rrtt--rr-r--rrttr-- ex ? unknown ? false count: 6
Input: rtttt--rrrrrrrt-rrttt--tt--t result: r--t-rr--tt--t expected ? unknown ? false count: 8
real 0m0.474s (without testBigs)
user 0m0.852s
sys 0m0.060s
With tailcall:
real 0m0.478s (without testBigs)
user 0m0.860s
sys 0m0.060s
For some random strings, the difference is bigger than a 10fold.
For long Strings with many groups one could, as an improvement, eliminate all groups which are the only group of that character, for instance:
aa bbb aa ccc xx ddd aa eee aa fff xx
The groups bbb, ccc, ddd, eee and fff are unique in the string, so they can't fit to something else and could all be eliminated, and the order of removal is will not matter. This would lead to the intermediate result
aaaa xx aaaa xx
and a fast solution. Maybe I try to implement it too. However, I guess, it will be possible to produce random Strings, where this will have a big impact and by a different form of random generated strings, to distributions, where the impact is low.
Here is a Python solution (function reduce_min), not particularly smart but I think fairly easy to understand (excessive amount of comments added for answer clarity):
def reductions(s, min_len):
"""
Yields every possible reduction of s by eliminating contiguous blocks
of l or more repeated characters.
For example, reductions('AAABBCCCCBAAC', 3) yields
'BBCCCCBAAC' and 'AAABBBAAC'.
"""
# Current character
curr = ''
# Length of current block
n = 0
# Start position of current block
idx = 0
# For each character
for i, c in enumerate(s):
if c != curr:
# New block begins
if n >= min_len:
# If previous block was long enough
# yield reduced string without it
yield s[:idx] + s[i:]
# Start new block
curr = c
n = 1
idx = i
else:
# Still in the same block
n += 1
# Yield reduction without last block if it was long enough
if n >= min_len:
yield s[:idx]
def reduce_min(s, min_len):
"""
Finds the smallest possible reduction of s by successive
elimination of contiguous blocks of min_len or more repeated
characters.
"""
# Current set of possible reductions
rs = set([s])
# Current best solution
result = s
# While there are strings to reduce
while rs:
# Get one element
r = rs.pop()
# Find reductions
r_red = list(reductions(r, min_len))
# If no reductions are found it is irreducible
if len(r_red) == 0 and len(r) < len(result):
# Replace if shorter than current best
result = r
else:
# Save reductions for next iterations
rs.update(r_red)
return result
assert reduce_min("BAABCCCBBA", 3) == "B"
assert reduce_min("AABBAAAC", 3) == "AABBC"
assert reduce_min("AAAA", 3) == ""
assert reduce_min("AAAABBBAC", 3) == "C"
EDIT: Since people seem to be posting C++ solutions, here is mine in C++ (again, function reduce_min):
#include <string>
#include <vector>
#include <unordered_set>
#include <iterator>
#include <utility>
#include <cassert>
using namespace std;
void reductions(const string &s, unsigned int min_len, vector<string> &rs)
{
char curr = '\0';
unsigned int n = 0;
unsigned int idx = 0;
for (auto it = s.begin(); it != s.end(); ++it)
{
if (curr != *it)
{
auto i = distance(s.begin(), it);
if (n >= min_len)
{
rs.push_back(s.substr(0, idx) + s.substr(i));
}
curr = *it;
n = 1;
idx = i;
}
else
{
n += 1;
}
}
if (n >= min_len)
{
rs.push_back(s.substr(0, idx));
}
}
string reduce_min(const string &s, unsigned int min_len)
{
unordered_set<string> rs { s };
string result = s;
vector<string> rs_new;
while (!rs.empty())
{
auto it = rs.begin();
auto r = *it;
rs.erase(it);
rs_new.clear();
reductions(r, min_len, rs_new);
if (rs_new.empty() && r.size() < result.size())
{
result = move(r);
}
else
{
rs.insert(rs_new.begin(), rs_new.end());
}
}
return result;
}
int main(int argc, char **argv)
{
assert(reduce_min("BAABCCCBBA", 3) == "B");
assert(reduce_min("AABBAAAC", 3) == "AABBC");
assert(reduce_min("AAAA", 3) == "");
assert(reduce_min("AAAABBBAC", 3) == "C");
return 0;
}
If you can use C++17 you can save memory by using string views.
EDIT 2: About the complexity of the algorithm. It is not straightforward to figure out, and as I said the algorithm is meant to be simple more than anything, but let's see. In the end, it is more or less the same as a breadth-first search. Let's say the string length is n, and, for generality, let's say the minimum block length (value 3 in the question) is m. In the first level, we can generate up to n / m reductions in the worst case. For each of these, we can generate up to (n - m) / m reductions, and so on. So basically, at "level" i (loop iteration i) we create up to (n - i * m) / m reductions per string we had, and each of these will take O(n - i * m) time to process. The maximum number of levels we can have is, again, n / m. So the complexity of the algorithm (if I'm not making mistakes) should have the form:
O( sum {i = 0 .. n / m} ( O(n - i * m) * prod {j = 0 .. i} ((n - i * m) / m) ))
|-Outer iters--| |---Cost---| |-Prev lvl-| |---Branching---|
Whew. So this should be something like:
O( sum {i = 0 .. n / m} (n - i * m) * O(n^i / m^i) )
Which in turn would collapse to:
O((n / m)^(n / m))
So yeah, the algorithm is more or less simple, but it can run into exponential cost cases (the bad cases would be strings made entirely of exactly m-long blocks, like AAABBBCCCAAACCC... for m = 3).
For instance, how can I construct an array in ATS containing all of the letters in the upper case from A to Z? In C, this can be done as follows:
char *Letters()
{
int i;
char *cs = (char *)malloc(26);
assert(cs != 0);
for (i = 0; i < 26; i += 1) cs[i] = 'A' + i;
return cs;
}
You could use the tabulate function for creating linear arrays. For instance,
extern
fun
Letters(): arrayptr(char, 26)
implement
Letters() =
arrayptr_tabulate_cloref<char>
(i2sz(26), lam(i) => 'A' + sz2i(i))
If you don't want to use a higher-order function, you can try the following template-based solutioin:
implement
Letters() =
arrayptr_tabulate<char>(i2sz(26)) where
{
implement array_tabulate$fopr<char> (i) = 'A' + sz2i(i)
}
Well, here's one way, although it's extremely complicated, because it follows your outlined approach to the letter: it involves linear proofs for arrays (aka dataviews), memory allocation, and array initialization via a while loop.
extern
fun
Letters (): arrayptr (char, 26)
implement
Letters () = let
val (pf_arr, pf_gc | p_arr) = array_ptr_alloc<char> ((i2sz)26)
var i: int = 0
prval [larr:addr] EQADDR () = eqaddr_make_ptr (p_arr)
var p = p_arr
prvar pf0 = array_v_nil {char} ()
prvar pf1 = pf_arr
//
val () =
while* {i:nat | i <= 26} .<26-i>. (
i: int (i)
, p: ptr (larr + i*sizeof(char))
, pf0: array_v (char, larr, i)
, pf1: array_v (char?, larr+i*sizeof(char), 26-i)
) : (
pf0: array_v (char, larr, 26)
, pf1: array_v (char?, larr+i*sizeof(char), 0)
) => (
i < 26
) {
//
prval (pf_at, pf1_res) = array_v_uncons {char?} (pf1)
prval () = pf1 := pf1_res
//
val c = 'A' + (g0ofg1)i
val () = ptr_set<char> (pf_at | p, c)
val () = p := ptr1_succ<char> (p)
//
prval () = pf0 := array_v_extend {char} (pf0, pf_at)
val () = i := i + 1
//
} // end of [val]
//
prval () = pf_arr := pf0
prval () = array_v_unnil {char?} (pf1)
//
val res = arrayptr_encode (pf_arr, pf_gc | p_arr)
in
res
end // end of [Letters]
You can run the code at Glot.io
Sought is an efficient algorithm that finds the unique integer in an interval [a, b] which has the maximum number of trailing zeros in its binary representation (a and b are integers > 0):
def bruteForce(a: Int, b: Int): Int =
(a to b).maxBy(Integer.numberOfTrailingZeros(_))
def binSplit(a: Int, b: Int): Int = {
require(a > 0 && a <= b)
val res = ???
assert(res == bruteForce(a, b))
res
}
here are some examples
bruteForce( 5, 7) == 6 // binary 110 (1 trailing zero)
bruteForce( 1, 255) == 128 // binary 10000000
bruteForce(129, 255) == 192 // binary 11000000
etc.
This one finds the number of zeros:
// Requires a>0
def mtz(a: Int, b: Int, mask: Int = 0xFFFFFFFE, n: Int = 0): Int = {
if (a > (b & mask)) n
else mtz(a, b, mask<<1, n+1)
}
This one returns the number with those zeros:
// Requires a > 0
def nmtz(a: Int, b: Int, mask: Int = 0xFFFFFFFE): Int = {
if (a > (b & mask)) b & (mask>>1)
else nmtz(a, b, mask<<1)
}
I doubt the log(log(n)) solution has a small enough constant term to beat this. (But you could do binary search on the number of zeros to get log(log(n)).)
I decided to take Rex's challenge and produce something faster. :-)
// requires a > 0
def mtz2(a: Int, b: Int, mask: Int = 0xffff0000, shift: Int = 8, n: Int = 16): Int = {
if (shift == 0) if (a > (b & mask)) n - 1 else n
else if (a > (b & mask)) mtz2(a, b, mask >> shift, shift / 2, n - shift)
else mtz2(a, b, mask << shift, shift / 2, n + shift)
}
Benchmarked with
import System.{currentTimeMillis => now}
def time[T](f: => T): T = {
val start = now
try { f } finally { println("Elapsed: " + (now - start)/1000.0 + " s") }
}
val range = 1 to 200
time(f((a, b) => mtz(a, b)))
time(f((a, b) => mtz2(a, b)))
First see if there is a power of two that lies within your interval. If there is at least one, the largest one wins.
Otherwise, choose the largest power of two that is less than your minimum bound.
Does 1100000...0 lie in your bound? If yes, you've won. If it's still less than your minimum bound, try 1110000...0; otherwise, if it's greater than your maximum bound, try 1010000...0.
And so forth, until you win.
as a conclusion, here is my variant of Rex' answer which gives both the center value and also an 'extent' which is the minimum power of two distance from the center which covers both a in the one direction and b in the other.
#tailrec def binSplit(a: Int, b: Int, mask: Int = 0xFFFFFFFF): (Int, Int) = {
val mask2 = mask << 1
if (a > (b & mask2)) (b & mask, -mask)
else binSplit(a, b, mask2)
}
def test(): Unit = {
val Seq(r1, r2) = Seq.fill(2)(util.Random.nextInt(0x3FFFFFFF) + 1)
val (a, b) = if (r1 <= r2) (r1, r2) else (r2, r1)
val (center, extent) = binSplit(a, b)
assert((center >= a) && (center <= b) && (center - extent) <= a &&
(center - extent) >= 0 && (center + extent) > b, (a, b, center, extent))
}
for (i <- 0 to 100000) { test() }
I am writing a string parser and the thought occurred to me that there might be some really interesting ways to convert an ASCII hexadecimal character [0-9A-Fa-f] to it's numeric value.
What are the quickest, shortest, most elegant or most obscure ways to convert [0-9A-Fa-f] to it's value between 0 and 15?
Assume, if you like, that the character is a valid hex character.
I have no chance so I'll have a go at the most boring.
( c <= '9' ) ? ( c - '0' ) : ( (c | '\x60') - 'a' + 10 )
(c&15)+(c>>6)*9
In response to "How does it work", it throws away enough bits so that the numbers map to [0:9] and the letters map to [1:6], then adds 9 for the letters. The c>>6 is a stand-in for if (c >= 64) ....
One simple way is to look for it in a string:
int n = "0123456789ABCDEF".IndexOf(Char.ToUpper(c));
Another way is to convert it as a digit, and then check if it's a character:
int n = Char.ToUpper(c) - '0';
if (n > 9) n -= 7;
private byte[] HexStringToBytes( string hexString )
{
byte[] bytes = ASCIIEncoding.ASCII.GetBytes(hexString);
byte[] ret = new byte[bytes.Length / 2];
for (int i = 0; i < bytes.Length; i += 2)
{
byte hi = (byte)(((bytes[i] & 0x40) == 0) ? bytes[i] & 0x0F : bytes[i] & 0x0F + 9);
byte lo = (byte)(((bytes[i+1] & 0x40) == 0) ? bytes[i+1] & 0x0F : bytes[i+1] & 0x0F + 9);
ret[i / 2] = (byte)((hi << 4) | lo);
}
return ret;
}
In C you can so something like:
if(isdigit(c))
num = c -'0';
else if(c>='a' && c<='f')
num = 10 + c - 'a';
else if(c>='A' && c<='F')
num = 10 + c - 'A';
In JavaScript:
num = parseInt(hex, 16);
Here's the boring C version (and my C is very rusty, so it's probably wrong as well).
char parseIntChar(const char p) {
char c[2];
c[0]=p;
c[1]=0;
return strtol(c,0,16);
}