/* ===========================================================================
 * Initialize a new block.
 */
local void init_block(s)
    deflate_state *s;
{
    int n; /* iterates over tree elements */

    /* Initialize the trees. */
    for (n = 0; n < L_CODES;  n++) s->dyn_ltree[n].Freq = 0;
    for (n = 0; n < D_CODES;  n++) s->dyn_dtree[n].Freq = 0;
    for (n = 0; n < BL_CODES; n++) s->bl_tree[n].Freq = 0;

    s->dyn_ltree[END_BLOCK].Freq = 1;
    s->opt_len = s->static_len = 0L;
    s->last_lit = s->matches = 0;
}

#define SMALLEST 1
/* Index within the heap array of least frequent node in the Huffman tree */


/* ===========================================================================
 * Remove the smallest element from the heap and recreate the heap with
 * one less element. Updates heap and heap_len.
 */
#define pqremove(s, tree, top) \
{\
    top = s->heap[SMALLEST]; \
    s->heap[SMALLEST] = s->heap[s->heap_len--]; \
    pqdownheap(s, tree, SMALLEST); \
}

/* ===========================================================================
 * Compares to subtrees, using the tree depth as tie breaker when
 * the subtrees have equal frequency. This minimizes the worst case length.
 */
#define smaller(tree, n, m, depth) \
   (tree[n].Freq < tree[m].Freq || \
   (tree[n].Freq == tree[m].Freq && depth[n] <= depth[m]))

/* ===========================================================================
 * Restore the heap property by moving down the tree starting at node k,
 * exchanging a node with the smallest of its two sons if necessary, stopping
 * when the heap property is re-established (each father smaller than its
 * two sons).
 */
local void pqdownheap(s, tree, k)
    deflate_state *s;
    ct_data *tree;  /* the tree to restore */
    int k;               /* node to move down */
{
    int v = s->heap[k];
    int j = k << 1;  /* left son of k */
    while (j <= s->heap_len) {
        /* Set j to the smallest of the two sons: */
        if (j < s->heap_len &&
            smaller(tree, s->heap[j+1], s->heap[j], s->depth)) {
            j++;
        }
        /* Exit if v is smaller than both sons */
        if (smaller(tree, v, s->heap[j], s->depth)) break;

        /* Exchange v with the smallest son */
        s->heap[k] = s->heap[j];  k = j;

        /* And continue down the tree, setting j to the left son of k */
        j <<= 1;
    }
    s->heap[k] = v;
}

/* ===========================================================================
 * Compute the optimal bit lengths for a tree and update the total bit length
 * for the current block.
 * IN assertion: the fields freq and dad are set, heap[heap_max] and
 *    above are the tree nodes sorted by increasing frequency.
 * OUT assertions: the field len is set to the optimal bit length, the
 *     array bl_count contains the frequencies for each bit length.
 *     The length opt_len is updated; static_len is also updated if stree is
 *     not null.
 */
local void gen_bitlen(s, desc)
    deflate_state *s;
    tree_desc *desc;    /* the tree descriptor */
{
    ct_data *tree        = desc->dyn_tree;
    int max_code         = desc->max_code;
    const ct_data *stree = desc->stat_desc->static_tree;
    const intf *extra    = desc->stat_desc->extra_bits;
    int base             = desc->stat_desc->extra_base;
    int max_length       = desc->stat_desc->max_length;
    int h;              /* heap index */
    int n, m;           /* iterate over the tree elements */
    int bits;           /* bit length */
    int xbits;          /* extra bits */
    ush f;              /* frequency */
    int overflow = 0;   /* number of elements with bit length too large */

    for (bits = 0; bits <= MAX_BITS; bits++) s->bl_count[bits] = 0;

    /* In a first pass, compute the optimal bit lengths (which may
     * overflow in the case of the bit length tree).
     */
    tree[s->heap[s->heap_max]].Len = 0; /* root of the heap */

    for (h = s->heap_max+1; h < HEAP_SIZE; h++) {
        n = s->heap[h];
        bits = tree[tree[n].Dad].Len + 1;
        if (bits > max_length) bits = max_length, overflow++;
        tree[n].Len = (ush)bits;
        /* We overwrite tree[n].Dad which is no longer needed */

        if (n > max_code) continue; /* not a leaf node */

        s->bl_count[bits]++;
        xbits = 0;
        if (n >= base) xbits = extra[n-base];
        f = tree[n].Freq;
        s->opt_len += (ulg)f * (bits + xbits);
        if (stree) s->static_len += (ulg)f * (stree[n].Len + xbits);
    }
    if (overflow == 0) return;

    Trace((stderr,"\nbit length overflow\n"));
    /* This happens for example on obj2 and pic of the Calgary corpus */

    /* Find the first bit length which could increase: */
    do {
        bits = max_length-1;
        while (s->bl_count[bits] == 0) bits--;
        s->bl_count[bits]--;      /* move one leaf down the tree */
        s->bl_count[bits+1] += 2; /* move one overflow item as its brother */
        s->bl_count[max_length]--;
        /* The brother of the overflow item also moves one step up,
         * but this does not affect bl_count[max_length]
         */
        overflow -= 2;
    } while (overflow > 0);

    /* Now recompute all bit lengths, scanning in increasing frequency.
     * h is still equal to HEAP_SIZE. (It is simpler to reconstruct all
     * lengths instead of fixing only the wrong ones. This idea is taken
     * from 'ar' written by Haruhiko Okumura.)
     */
    for (bits = max_length; bits != 0; bits--) {
        n = s->bl_count[bits];
        while (n != 0) {
            m = s->heap[--h];
            if (m > max_code) continue;
            if ((unsigned) tree[m].Len != (unsigned) bits) {
                Trace((stderr,"code %d bits %d->%d\n", m, tree[m].Len, bits));
                s->opt_len += ((long)bits - (long)tree[m].Len)
                              *(long)tree[m].Freq;
                tree[m].Len = (ush)bits;
            }
            n--;
        }
    }
}

/* ===========================================================================
 * Generate the codes for a given tree and bit counts (which need not be
 * optimal).
 * IN assertion: the array bl_count contains the bit length statistics for
 * the given tree and the field len is set for all tree elements.
 * OUT assertion: the field code is set for all tree elements of non
 *     zero code length.
 */
local void gen_codes (tree, max_code, bl_count)
    ct_data *tree;             /* the tree to decorate */
    int max_code;              /* largest code with non zero frequency */
    ushf *bl_count;            /* number of codes at each bit length */
{
    ush next_code[MAX_BITS+1]; /* next code value for each bit length */
    ush code = 0;              /* running code value */
    int bits;                  /* bit index */
    int n;                     /* code index */

    /* The distribution counts are first used to generate the code values
     * without bit reversal.
     */
    for (bits = 1; bits <= MAX_BITS; bits++) {
        next_code[bits] = code = (code + bl_count[bits-1]) << 1;
    }
    /* Check that the bit counts in bl_count are consistent. The last code
     * must be all ones.
     */
    Assert (code + bl_count[MAX_BITS]-1 == (1<<MAX_BITS)-1,
            "inconsistent bit counts");
    Tracev((stderr,"\ngen_codes: max_code %d ", max_code));

    for (n = 0;  n <= max_code; n++) {
        int len = tree[n].Len;
        if (len == 0) continue;
        /* Now reverse the bits */
        tree[n].Code = bi_reverse(next_code[len]++, len);

        Tracecv(tree != static_ltree, (stderr,"\nn %3d %c l %2d c %4x (%x) ",
             n, (isgraph(n) ? n : ' '), len, tree[n].Code, next_code[len]-1));
    }
}

/* ===========================================================================
 * Construct one Huffman tree and assigns the code bit strings and lengths.
 * Update the total bit length for the current block.
 * IN assertion: the field freq is set for all tree elements.
 * OUT assertions: the fields len and code are set to the optimal bit length
 *     and corresponding code. The length opt_len is updated; static_len is
 *     also updated if stree is not null. The field max_code is set.
 */
local void build_tree(s, desc)
    deflate_state *s;
    tree_desc *desc; /* the tree descriptor */
{
    ct_data *tree         = desc->dyn_tree;
    const ct_data *stree  = desc->stat_desc->static_tree;
    int elems             = desc->stat_desc->elems;
    int n, m;          /* iterate over heap elements */
    int max_code = -1; /* largest code with non zero frequency */
    int node;          /* new node being created */

    /* Construct the initial heap, with least frequent element in
     * heap[SMALLEST]. The sons of heap[n] are heap[2*n] and heap[2*n+1].
     * heap[0] is not used.
     */
    s->heap_len = 0, s->heap_max = HEAP_SIZE;

    for (n = 0; n < elems; n++) {
        if (tree[n].Freq != 0) {
            s->heap[++(s->heap_len)] = max_code = n;
            s->depth[n] = 0;
        } else {
            tree[n].Len = 0;
        }
    }

    /* The pkzip format requires that at least one distance code exists,
     * and that at least one bit should be sent even if there is only one
     * possible code. So to avoid special checks later on we force at least
     * two codes of non zero frequency.
     */
    while (s->heap_len < 2) {
        node = s->heap[++(s->heap_len)] = (max_code < 2 ? ++max_code : 0);
        tree[node].Freq = 1;
        s->depth[node] = 0;
        s->opt_len--; if (stree) s->static_len -= stree[node].Len;
        /* node is 0 or 1 so it does not have extra bits */
    }
    desc->max_code = max_code;

    /* The elements heap[heap_len/2+1 .. heap_len] are leaves of the tree,
     * establish sub-heaps of increasing lengths:
     */
    for (n = s->heap_len/2; n >= 1; n--) pqdownheap(s, tree, n);

    /* Construct the Huffman tree by repeatedly combining the least two
     * frequent nodes.
     */
    node = elems;              /* next internal node of the tree */
    do {
        pqremove(s, tree, n);  /* n = node of least frequency */
        m = s->heap[SMALLEST]; /* m = node of next least frequency */

        s->heap[--(s->heap_max)] = n; /* keep the nodes sorted by frequency */
        s->heap[--(s->heap_max)] = m;

        /* Create a new node father of n and m */
        tree[node].Freq = tree[n].Freq + tree[m].Freq;
        s->depth[node] = (uch)((s->depth[n] >= s->depth[m] ?
                                s->depth[n] : s->depth[m]) + 1);
        tree[n].Dad = tree[m].Dad = (ush)node;
#ifdef DUMP_BL_TREE
        if (tree == s->bl_tree) {
            fprintf(stderr,"\nnode %d(%d), sons %d(%d) %d(%d)",
                    node, tree[node].Freq, n, tree[n].Freq, m, tree[m].Freq);
        }
#endif
        /* and insert the new node in the heap */
        s->heap[SMALLEST] = node++;
        pqdownheap(s, tree, SMALLEST);

    } while (s->heap_len >= 2);

    s->heap[--(s->heap_max)] = s->heap[SMALLEST];

    /* At this point, the fields freq and dad are set. We can now
     * generate the bit lengths.
     */
    gen_bitlen(s, (tree_desc *)desc);

    /* The field len is now set, we can generate the bit codes */
    gen_codes ((ct_data *)tree, max_code, s->bl_count);
}

/* ===========================================================================
 * Scan a literal or distance tree to determine the frequencies of the codes
 * in the bit length tree.
 */
local void scan_tree (s, tree, max_code)
    deflate_state *s;
    ct_data *tree;   /* the tree to be scanned */
    int max_code;    /* and its largest code of non zero frequency */
{
    int n;                     /* iterates over all tree elements */
    int prevlen = -1;          /* last emitted length */
    int curlen;                /* length of current code */
    int nextlen = tree[0].Len; /* length of next code */
    int count = 0;             /* repeat count of the current code */
    int max_count = 7;         /* max repeat count */
    int min_count = 4;         /* min repeat count */

    if (nextlen == 0) max_count = 138, min_count = 3;
    tree[max_code+1].Len = (ush)0xffff; /* guard */

    for (n = 0; n <= max_code; n++) {
        curlen = nextlen; nextlen = tree[n+1].Len;
        if (++count < max_count && curlen == nextlen) {
            continue;
        } else if (count < min_count) {
            s->bl_tree[curlen].Freq += count;
        } else if (curlen != 0) {
            if (curlen != prevlen) s->bl_tree[curlen].Freq++;
            s->bl_tree[REP_3_6].Freq++;
        } else if (count <= 10) {
            s->bl_tree[REPZ_3_10].Freq++;
        } else {
            s->bl_tree[REPZ_11_138].Freq++;
        }
        count = 0; prevlen = curlen;
        if (nextlen == 0) {
            max_count = 138, min_count = 3;
        } else if (curlen == nextlen) {
            max_count = 6, min_count = 3;
        } else {
            max_count = 7, min_count = 4;
        }
    }
}

/* ===========================================================================
 * Send a literal or distance tree in compressed form, using the codes in
 * bl_tree.
 */
local void send_tree (s, tree, max_code)
    deflate_state *s;
    ct_data *tree; /* the tree to be scanned */
    int max_code;       /* and its largest code of non zero frequency */
{
    int n;                     /* iterates over all tree elements */
    int prevlen = -1;          /* last emitted length */
    int curlen;                /* length of current code */
    int nextlen = tree[0].Len; /* length of next code */
    int count = 0;             /* repeat count of the current code */
    int max_count = 7;         /* max repeat count */
    int min_count = 4;         /* min repeat count */

    /* tree[max_code+1].Len = -1; */  /* guard already set */
    if (nextlen == 0) max_count = 138, min_count = 3;

    for (n = 0; n <= max_code; n++) {
        curlen = nextlen; nextlen = tree[n+1].Len;
        if (++count < max_count && curlen == nextlen) {
            continue;
        } else if (count < min_count) {
            do { send_code(s, curlen, s->bl_tree); } while (--count != 0);

        } else if (curlen != 0) {
            if (curlen != prevlen) {
                send_code(s, curlen, s->bl_tree); count--;
            }
            Assert(count >= 3 && count <= 6, " 3_6?");
            send_code(s, REP_3_6, s->bl_tree); send_bits(s, count-3, 2);

        } else if (count <= 10) {
            send_code(s, REPZ_3_10, s->bl_tree); send_bits(s, count-3, 3);

        } else {
            send_code(s, REPZ_11_138, s->bl_tree); send_bits(s, count-11, 7);
        }
        count = 0; prevlen = curlen;
        if (nextlen == 0) {
            max_count = 138, min_count = 3;
        } else if (curlen == nextlen) {
            max_count = 6, min_count = 3;
        } else {
            max_count = 7, min_count = 4;
        }
    }
}

/* ===========================================================================
 * Construct the Huffman tree for the bit lengths and return the index in
 * bl_order of the last bit length code to send.
 */
local int build_bl_tree(s)
    deflate_state *s;
{
    int max_blindex;  /* index of last bit length code of non zero freq */

    /* Determine the bit length frequencies for literal and distance trees */
    scan_tree(s, (ct_data *)s->dyn_ltree, s->l_desc.max_code);
    scan_tree(s, (ct_data *)s->dyn_dtree, s->d_desc.max_code);

    /* Build the bit length tree: */
    build_tree(s, (tree_desc *)(&(s->bl_desc)));
    /* opt_len now includes the length of the tree representations, except