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javascripttimsort

How to use Timsort in javascript?


How can I use Timsort in Javascript format? there are a lot of documentations in Java, Python and C++, is it doable in JS also?


Solution

  • Timsort Javascript


    Array.prototype.timsort = function(comp){
    
        var global_a=this
        var MIN_MERGE = 32;
        var MIN_GALLOP = 7
        var runBase=[];
        var runLen=[];
        var stackSize = 0;
        var compare = comp;
    
        sort(this,0,this.length,compare);
    
        /*
             * The next two methods (which are package private and static) constitute the entire API of this class. Each of these methods
             * obeys the contract of the public method with the same signature in java.util.Arrays.
             */
    
        function sort (a, lo, hi, compare) {
    
            if (typeof compare != "function") {
                throw new Error("Compare is not a function.");
                return;
            }
    
            stackSize = 0;
            runBase=[];
            runLen=[];
    
            rangeCheck(a.length, lo, hi);
            var nRemaining = hi - lo;
            if (nRemaining < 2) return; // Arrays of size 0 and 1 are always sorted
    
            // If array is small, do a "mini-TimSort" with no merges
            if (nRemaining < MIN_MERGE) {
                var initRunLen = countRunAndMakeAscending(a, lo, hi, compare);
                binarySort(a, lo, hi, lo + initRunLen, compare);
                return;
            }
    
            /**
                     * March over the array once, left to right, finding natural runs, extending short natural runs to minRun elements, and
                     * merging runs to maintain stack invariant.
                     */
            var ts = [];
            var minRun = minRunLength(nRemaining);
            do {
                // Identify next run
                var runLenVar = countRunAndMakeAscending(a, lo, hi, compare);
    
                // If run is short, extend to min(minRun, nRemaining)
                if (runLenVar < minRun) {
                    var force = nRemaining <= minRun ? nRemaining : minRun;
                    binarySort(a, lo, lo + force, lo + runLenVar, compare);
                    runLenVar = force;
                }
    
                // Push run onto pending-run stack, and maybe merge
                pushRun(lo, runLenVar);
                mergeCollapse();
    
                // Advance to find next run
                lo += runLenVar;
                nRemaining -= runLenVar;
            } while (nRemaining != 0);
    
            // Merge all remaining runs to complete sort
            mergeForceCollapse();
        }
    
    
        /**
             * Sorts the specified portion of the specified array using a binary insertion sort. This is the best method for sorting small
             * numbers of elements. It requires O(n log n) compares, but O(n^2) data movement (worst case).
             *
             * If the initial part of the specified range is already sorted, this method can take advantage of it: the method assumes that
             * the elements from index {@code lo}, inclusive, to {@code start}, exclusive are already sorted.
             *
             * @param a the array in which a range is to be sorted
             * @param lo the index of the first element in the range to be sorted
             * @param hi the index after the last element in the range to be sorted
             * @param start the index of the first element in the range that is not already known to be sorted (@code lo <= start <= hi}
             * @param c comparator to used for the sort
             */
        function binarySort (a, lo, hi, start, compare) {
            if (start == lo) start++;
            for (; start < hi; start++) {
                var pivot = a[start];
    
                // Set left (and right) to the index where a[start] (pivot) belongs
                var left = lo;
                var right = start;
                /*
                * Invariants: pivot >= all in [lo, left). pivot < all in [right, start).
                */
                while (left < right) {
                    var mid = (left + right) >>> 1;
                    if (compare(pivot, a[mid]) < 0)
                        right = mid;
                    else
                        left = mid + 1;
                }
                /*
                * The invariants still hold: pivot >= all in [lo, left) and pivot < all in [left, start), so pivot belongs at left. Note
                * that if there are elements equal to pivot, left points to the first slot after them -- that's why this sort is stable.
                * Slide elements over to make room to make room for pivot.
                */
                var n = start - left; // The number of elements to move
                // Switch is just an optimization for arraycopy in default case
                switch (n) {
                case 2:
                    a[left + 2] = a[left + 1];
                case 1:
                    a[left + 1] = a[left];
                    break;
                default:
                arraycopy(a, left, a, left + 1, n);
                }
                a[left] = pivot;
            }
        }
    
    
        /**
             * Returns the length of the run beginning at the specified position in the specified array and reverses the run if it is
             * descending (ensuring that the run will always be ascending when the method returns).
             *
             * A run is the longest ascending sequence with:
             *
             * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
             *
             * or the longest descending sequence with:
             *
             * a[lo] > a[lo + 1] > a[lo + 2] > ...
             *
             * For its intended use in a stable mergesort, the strictness of the definition of "descending" is needed so that the call can
             * safely reverse a descending sequence without violating stability.
             *
             * @param a the array in which a run is to be counted and possibly reversed
             * @param lo index of the first element in the run
             * @param hi index after the last element that may be contained in the run. It is required that @code{lo < hi}.
             * @param c the comparator to used for the sort
             * @return the length of the run beginning at the specified position in the specified array
             */
        function countRunAndMakeAscending (a, lo, hi, compare) {
            var runHi = lo + 1;
    
            // Find end of run, and reverse range if descending
            if (compare(a[runHi++], a[lo]) < 0) { // Descending
                while (runHi < hi && compare(a[runHi], a[runHi - 1]) < 0){
                    runHi++;
                }
                reverseRange(a, lo, runHi);
            } else { // Ascending
                while (runHi < hi && compare(a[runHi], a[runHi - 1]) >= 0){
                    runHi++;
                }
            }
    
            return runHi - lo;
        }
    
        /**
             * Reverse the specified range of the specified array.
             *
             * @param a the array in which a range is to be reversed
             * @param lo the index of the first element in the range to be reversed
             * @param hi the index after the last element in the range to be reversed
             */
        function /*private static void*/ reverseRange (/*Object[]*/ a, /*int*/ lo, /*int*/ hi) {
            hi--;
            while (lo < hi) {
                var t = a[lo];
                a[lo++] = a[hi];
                a[hi--] = t;
            }
        }
    
    
        /**
             * Returns the minimum acceptable run length for an array of the specified length. Natural runs shorter than this will be
             * extended with {@link #binarySort}.
             *
             * Roughly speaking, the computation is:
             *
             * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff). Else if n is an exact power of 2, return
             * MIN_MERGE/2. Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k is close to, but strictly less than, an
             * exact power of 2.
             *
             * For the rationale, see listsort.txt.
             *
             * @param n the length of the array to be sorted
             * @return the length of the minimum run to be merged
             */
        function /*private static int*/ minRunLength (/*int*/ n) {
            //var v=0;
            var r = 0; // Becomes 1 if any 1 bits are shifted off
            /*while (n >= MIN_MERGE) { v++;
                r |= (n & 1);
                n >>= 1;
            }*/
            //console.log("minRunLength("+n+") "+v+" vueltas, result="+(n+r));
            //return n + r;
            return n + 1;
        }
    
        /**
             * Pushes the specified run onto the pending-run stack.
             *
             * @param runBase index of the first element in the run
             * @param runLen the number of elements in the run
             */
        function pushRun (runBaseArg, runLenArg) {
            //console.log("pushRun("+runBaseArg+","+runLenArg+")");
            //this.runBase[stackSize] = runBase;
            //runBase.push(runBaseArg);
            runBase[stackSize] = runBaseArg;
    
            //this.runLen[stackSize] = runLen;
            //runLen.push(runLenArg);
            runLen[stackSize] = runLenArg;
            stackSize++;
        }
    
        /**
             * Examines the stack of runs waiting to be merged and merges adjacent runs until the stack invariants are reestablished:
             *
             * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 2. runLen[i - 2] > runLen[i - 1]
             *
             * This method is called each time a new run is pushed onto the stack, so the invariants are guaranteed to hold for i <
             * stackSize upon entry to the method.
             */
        function mergeCollapse () {
            while (stackSize > 1) {
                var n = stackSize - 2;
                if (n > 0 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) {
                    if (runLen[n - 1] < runLen[n + 1]) n--;
                    mergeAt(n);
                } else if (runLen[n] <= runLen[n + 1]) {
                    mergeAt(n);
                } else {
                    break; // Invariant is established
                }
            }
        }
    
        /**
             * Merges all runs on the stack until only one remains. This method is called once, to complete the sort.
             */
        function mergeForceCollapse () {
            while (stackSize > 1) {
                var n = stackSize - 2;
                if (n > 0 && runLen[n - 1] < runLen[n + 1]) n--;
                mergeAt(n);
            }
        }
    
    
        /**
             * Merges the two runs at stack indices i and i+1. Run i must be the penultimate or antepenultimate run on the stack. In other
             * words, i must be equal to stackSize-2 or stackSize-3.
             *
             * @param i stack index of the first of the two runs to merge
             */
        function mergeAt (i) {
    
            var base1 = runBase[i];
            var len1 = runLen[i];
            var base2 = runBase[i + 1];
            var len2 = runLen[i + 1];
    
            /*
            * Record the length of the combined runs; if i is the 3rd-last run now, also slide over the last run (which isn't involved
            * in this merge). The current run (i+1) goes away in any case.
            */
            //var stackSize = runLen.length;
            runLen[i] = len1 + len2;
            if (i == stackSize  - 3) {
                runBase[i + 1] = runBase[i + 2];
                runLen[i + 1] = runLen[i + 2];
            }
            stackSize--;
    
            /*
            * Find where the first element of run2 goes in run1. Prior elements in run1 can be ignored (because they're already in
            * place).
            */
    
            var k = gallopRight(global_a[base2], global_a, base1, len1, 0, compare);
            base1 += k;
            len1 -= k;
            if (len1 == 0) return;
    
            /*
            * Find where the last element of run1 goes in run2. Subsequent elements in run2 can be ignored (because they're already in
            * place).
            */
            len2 = gallopLeft(global_a[base1 + len1 - 1], global_a, base2, len2, len2 - 1, compare);
    
            if (len2 == 0) return;
    
            // Merge remaining runs, using tmp array with min(len1, len2) elements
            if (len1 <= len2)
                mergeLo(base1, len1, base2, len2);
            else
                mergeHi(base1, len1, base2, len2);
        }
    
    
        /**
             * Locates the position at which to insert the specified key into the specified sorted range; if the range contains an element
             * equal to key, returns the index of the leftmost equal element.
             *
             * @param key the key whose insertion point to search for
             * @param a the array in which to search
             * @param base the index of the first element in the range
             * @param len the length of the range; must be > 0
             * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the result, the faster this method
             *           will run.
             * @param c the comparator used to order the range, and to search
             * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k], pretending that a[b - 1] is minus infinity and a[b
             *         + n] is infinity. In other words, key belongs at index b + k; or in other words, the first k elements of a should
             *         precede key, and the last n - k should follow it.
             */
        function gallopLeft (key, a, base, len, hint, compare) {
            var lastOfs = 0;
            var ofs = 1;
            if (compare(key, a[base + hint]) > 0) {
                // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
                var maxOfs = len - hint;
                while (ofs < maxOfs && compare(key, a[base + hint + ofs]) > 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0) // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs) ofs = maxOfs;
    
                // Make offsets relative to base
                lastOfs += hint;
                ofs += hint;
            } else { // key <= a[base + hint]
                // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
                var maxOfs = hint + 1;
                while (ofs < maxOfs && compare(key, a[base + hint - ofs]) <= 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0) // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs) ofs = maxOfs;
    
                // Make offsets relative to base
                var tmp = lastOfs;
                lastOfs = hint - ofs;
                ofs = hint - tmp;
            }
    
            /*
            * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
            * Do a binary search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
            */
            lastOfs++;
            while (lastOfs < ofs) {
                var m = lastOfs + ((ofs - lastOfs) >>> 1);
    
                if (compare(key, a[base + m]) > 0)
                    lastOfs = m + 1; // a[base + m] < key
                else
                    ofs = m; // key <= a[base + m]
            }
            return ofs;
        }
    
        /**
             * Like gallopLeft, except that if the range contains an element equal to key, gallopRight returns the index after the
             * rightmost equal element.
             *
             * @param key the key whose insertion point to search for
             * @param a the array [] in which to search
             * @param base the index of the first element in the range
             * @param len the length of the range; must be > 0
             * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the result, the faster this method
             *           will run.
             * @param c the comparator used to order the range, and to search
             * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
             */
        function gallopRight (key, a, base, len, hint,  compare) {
    
            var ofs = 1;
            var lastOfs = 0;
            if (compare(key, a[base + hint]) < 0) {
                // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
                var maxOfs = hint + 1;
                while (ofs < maxOfs && compare(key, a[base + hint - ofs]) < 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0) // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs) ofs = maxOfs;
    
                // Make offsets relative to b
                var tmp = lastOfs;
                lastOfs = hint - ofs;
                ofs = hint - tmp;
            } else { // a[b + hint] <= key
                // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
                var maxOfs = len - hint;
                while (ofs < maxOfs && compare(key, a[base + hint + ofs]) >= 0) {
                    lastOfs = ofs;
                    ofs = (ofs << 1) + 1;
                    if (ofs <= 0) // int overflow
                        ofs = maxOfs;
                }
                if (ofs > maxOfs) ofs = maxOfs;
    
                // Make offsets relative to b
                lastOfs += hint;
                ofs += hint;
            }
    
            /*
            * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to the right of lastOfs but no farther right than ofs.
            * Do a binary search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
            */
            lastOfs++;
            while (lastOfs < ofs) {
                var m = lastOfs + ((ofs - lastOfs) >>> 1);
    
                if (compare(key, a[base + m]) < 0)
                    ofs = m; // key < a[b + m]
                else
                    lastOfs = m + 1; // a[b + m] <= key
            }
            return ofs;
        }
    
        /**
        * Merges two adjacent runs in place, in a stable fashion. The first element of the first run must be greater than the first
        * element of the second run (a[base1] > a[base2]), and the last element of the first run (a[base1 + len1-1]) must be greater
        * than all elements of the second run.
        *
        * For performance, this method should be called only when len1 <= len2; its twin, mergeHi should be called if len1 >= len2.
        * (Either method may be called if len1 == len2.)
        *
        * @param base1 index of first element in first run to be merged
        * @param len1 length of first run to be merged (must be > 0)
        * @param base2 index of first element in second run to be merged (must be aBase + aLen)
        * @param len2 length of second run to be merged (must be > 0)
        */
        function mergeLo (base1, len1, base2, len2) {
    
            // Copy first run into temp array
            var a = global_a;// For performance
            var tmp=a.slice(base1,base1+len1);
    
            var cursor1 = 0; // Indexes into tmp array
            var cursor2 = base2; // Indexes int a
            var dest = base1; // Indexes int a
    
            // Move first element of second run and deal with degenerate cases
            a[dest++] = a[cursor2++];
            if (--len2 == 0) {
                arraycopy(tmp, cursor1, a, dest, len1);
                return;
            }
            if (len1 == 1) {
                arraycopy(a, cursor2, a, dest, len2);
                a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
                return;
            }
    
            var c = compare;// Use local variable for performance
    
            var minGallop = MIN_GALLOP; // "    " "     " "
            outer:
            while (true) {
                var count1 = 0; // Number of times in a row that first run won
                var count2 = 0; // Number of times in a row that second run won
    
                /*
                * Do the straightforward thing until (if ever) one run starts winning consistently.
                */
                do {
                    if (compare(a[cursor2], tmp[cursor1]) < 0) {
                        a[dest++] = a[cursor2++];
                        count2++;
                        count1 = 0;
                        if (--len2 == 0) break outer;
                    } else {
                        a[dest++] = tmp[cursor1++];
                        count1++;
                        count2 = 0;
                        if (--len1 == 1) break outer;
                    }
                } while ((count1 | count2) < minGallop);
    
                /*
                * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
                * ever) neither run appears to be winning consistently anymore.
                */
                do {
                    count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                    if (count1 != 0) {
                        arraycopy(tmp, cursor1, a, dest, count1);
                        dest += count1;
                        cursor1 += count1;
                        len1 -= count1;
                        if (len1 <= 1) // len1 == 1 || len1 == 0
                            break outer;
                    }
                    a[dest++] = a[cursor2++];
                    if (--len2 == 0) break outer;
    
                    count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                    if (count2 != 0) {
                        arraycopy(a, cursor2, a, dest, count2);
                        dest += count2;
                        cursor2 += count2;
                        len2 -= count2;
                        if (len2 == 0) break outer;
                    }
                    a[dest++] = tmp[cursor1++];
                    if (--len1 == 1) break outer;
                    minGallop--;
                } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
                if (minGallop < 0) minGallop = 0;
                minGallop += 2; // Penalize for leaving gallop mode
            } // End of "outer" loop
            this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
    
            if (len1 == 1) {
                arraycopy(a, cursor2, a, dest, len2);
                a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            } else if (len1 == 0) {
                throw new Error("IllegalArgumentException. Comparison method violates its general contract!");
            } else {
                arraycopy(tmp, cursor1, a, dest, len1);
            }
        }
    
    
        /**
             * Like mergeLo, except that this method should be called only if len1 >= len2; mergeLo should be called if len1 <= len2.
             * (Either method may be called if len1 == len2.)
             *
             * @param base1 index of first element in first run to be merged
             * @param len1 length of first run to be merged (must be > 0)
             * @param base2 index of first element in second run to be merged (must be aBase + aLen)
             * @param len2 length of second run to be merged (must be > 0)
             */
        function mergeHi ( base1, len1, base2, len2) {
    
            // Copy second run into temp array
            var a = global_a;// For performance
            var tmp=a.slice(base2, base2+len2);
    
            var cursor1 = base1 + len1 - 1; // Indexes into a
            var cursor2 = len2 - 1; // Indexes into tmp array
            var dest = base2 + len2 - 1; // Indexes into a
    
            // Move last element of first run and deal with degenerate cases
            a[dest--] = a[cursor1--];
            if (--len1 == 0) {
                arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
                return;
            }
            if (len2 == 1) {
                dest -= len1;
                cursor1 -= len1;
                arraycopy(a, cursor1 + 1, a, dest + 1, len1);
                a[dest] = tmp[cursor2];
                return;
            }
    
            var c = compare;// Use local variable for performance
    
            var minGallop = MIN_GALLOP; // "    " "     " "
            outer:
            while (true) {
                var count1 = 0; // Number of times in a row that first run won
                var count2 = 0; // Number of times in a row that second run won
    
                /*
                * Do the straightforward thing until (if ever) one run appears to win consistently.
                */
                do {
                    if (compare(tmp[cursor2], a[cursor1]) < 0) {
                        a[dest--] = a[cursor1--];
                        count1++;
                        count2 = 0;
                        if (--len1 == 0) break outer;
                        } else {
                            a[dest--] = tmp[cursor2--];
                            count2++;
                            count1 = 0;
                            if (--len2 == 1) break outer;
                        }
                } while ((count1 | count2) < minGallop);
    
                /*
                * One run is winning so consistently that galloping may be a huge win. So try that, and continue galloping until (if
                * ever) neither run appears to be winning consistently anymore.
                */
                do {
                    count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                    if (count1 != 0) {
                        dest -= count1;
                        cursor1 -= count1;
                        len1 -= count1;
                        arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                        if (len1 == 0) break outer;
                    }
                    a[dest--] = tmp[cursor2--];
                    if (--len2 == 1) break outer;
    
                    count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
                    if (count2 != 0) {
                        dest -= count2;
                        cursor2 -= count2;
                        len2 -= count2;
                        arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                        if (len2 <= 1) // len2 == 1 || len2 == 0
                            break outer;
                    }
                    a[dest--] = a[cursor1--];
                    if (--len1 == 0) break outer;
                        minGallop--;
                } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
                if (minGallop < 0) minGallop = 0;
                minGallop += 2; // Penalize for leaving gallop mode
            } // End of "outer" loop
            this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field
    
            if (len2 == 1) {
                dest -= len1;
                cursor1 -= len1;
                arraycopy(a, cursor1 + 1, a, dest + 1, len1);
                a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
            } else if (len2 == 0) {
                throw new Error("IllegalArgumentException. Comparison method violates its general contract!");
            } else {
                arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
            }
        }
    
    
        /**
        * Checks that fromIndex and toIndex are in range, and throws an appropriate exception if they aren't.
        *
        * @param arrayLen the length of the array
        * @param fromIndex the index of the first element of the range
        * @param toIndex the index after the last element of the range
        * @throws IllegalArgumentException if fromIndex > toIndex
        * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex > arrayLen
        */
        function rangeCheck (arrayLen, fromIndex, toIndex) {
            if (fromIndex > toIndex) throw new Error( "IllegalArgument fromIndex(" + fromIndex + ") > toIndex(" + toIndex + ")");
            if (fromIndex < 0) throw new Error( "ArrayIndexOutOfBounds "+fromIndex);
            if (toIndex > arrayLen) throw new Error( "ArrayIndexOutOfBounds "+toIndex);
        }
    }
    
    // java System.arraycopy(Object src, int srcPos, Object dest, int destPos, int length)
    function arraycopy(s,spos,d,dpos,len){
        var a=s.slice(spos,spos+len);
        while(len--){
            d[dpos+len]=a[len];
        }
    }
    

    Active Referencee

    https://github.com/bellbind/stepbystep-timsort
    https://github.com/Scipion/interesting-javascript-codes