Jun 99 Challenge

Volume Number: 15 (1999)
Issue Number: 6
Column Tag: Programmer's Challenge

# Jun 99 Challenge

by Bob Boonstra, Westford, MA

### Tetraminx

The last Challenge I entered as a contestant was the Rubik's Cube Challenge, where entries had to solve a scrambled Rubik's cube. While that Challenge was difficult enough to cause me to retire from competition after winning, I've stayed interested in puzzles. Some time ago I ran across Meffert's World Of Puzzles, an online puzzle vendor at http://www.mefferts-puzzles.com/mefferts-puzzles/catalog.html, and ordered a few of their puzzles. The Tetraminx is perhaps the least difficult puzzle, much simpler than the Cube, but still interesting enough that I thought it might make a good Challenge without driving any contestants into retirement.

The Tetraminx is formed by four hexagonal faces, each consisting of six triangles, joined in the shape of a tetrahedron, plus four triangles to complete the solid, as depicted at http://www.mefferts-puzzles.com/pictures/tetramix.jpg.

### Scrambled Tetraminx

Your Challenge is to come up with a sequence of moves that will return the puzzle to the goal state, where each of the hexagonal faces consist of triangular facelets of a single color.

The prototype for the code you should write is:

```#if defined (__cplusplus)
extern "C" {
#endif

typedef enum {
kYellow=1,kBlue,kRed,kGreen
} PieceColor;

typedef enum {
kLeftClockwise=1,kRightClockwise,
kBottomClockwise,kBackClockwise,
kLeftCounterClockwise,kRightCounterClockwise,
kBottomCounterClockwise,kBackCounterClockwise
} Move;

typedef enum {
/* single triangular faces named after the opposite hexagonal face */
kLeft,kRight,kBottom,kBack,
/* edge faces named kXY, where X is the hexagonal face they are part of,
and Y is the adjacent hexagonal face */
kBR,kRK,kKB,
kLB,kBK,kKL,
kKR,kRL,kLK,
kLR,kRB,kBL,
/* corner faces named cXY, where X is the hexagonal face they are part of,
and Y is the hexagonal face opposite the adjacent single triangle */
cRL,cKL,cBL,
cLR,cBR,cKR,
cRB,cLB,cKB,
cLK,cRK,cBK
} PieceType;

typedef struct {
PieceType piece;
PieceColor color;
} PieceState;

long /* numberOfMoves */ Tetraminx (
PieceState state[28],      /* initial state of the 28 pieces */
Move moves[],              /* moves you generate */
long maxMoves              /* maximum storage in move array */
);

#if defined (__cplusplus)
}
#endif
```

The puzzle is manipulated with four pairs of 120-degree rotation moves that we will call kLeftXXX, kRightXXX, kBottomXXX, and kBackXXX, corresponding to the four hexagonal faces of the Tetraminx. The moves are named for the hexagonal face that remains fixed during the move.. Opposite each of those hexagonal faces are the single triangular faces whose positions remain fixed for all moves (except for rotation). Each move pair consists of a clockwise move and a counterclockwise move, as viewed from the opposite single facelet, through the Tetraminx, at the face for which the move is named.

The PieceType enum names the 28 triangular facelets. Facelets come in three types, and it is important to understand the naming convention for the facelets. The first type consists of the single triangular faces, and those are named kLeft, etc., for the move that rotates the piece, and for the opposite hexagonal face. The second type is an "edge" facelet, one with a single adjoining triangle, and those are named kXY, where X (L, R, B, or K) is the hexagonal face containing the piece, and Y (also L, R, B, or K) contains the adjacent facelet. Finally, "corner" facelets have three adjacent triangles, two of them on the same hexagonal face, and one a single triangular face. These are named kXY, where X is the hexagon containing the piece, and Y is the single triangular face.

This is probably impossible to understand without a picture, so I've included one. A color version of the picture is available at http://www.mefferts-puzzles.com/mefferts-puzzles/pictures/tetrmi5b.gif. The blue, red, yellow, and green faces are the left, right, bottom, and back faces, respectively. The face comprised of the single center (yellow) triangle is the kBack facelet. Starting with the blue triangle next to the kBack facelet, and moving clockwise around the blue face, are the cLK, kLB, cLR, kLK, cLB, and kLR facelets.

The kBackClockwise move rotates the yellow kBack facelet in a clockwise direction, moving the kLR piece (Blue-Red) to where the kRB piece (Red-Yellow) is, and the kRB piece to where the kBL piece (Yellow-Blue) is.

If the nomenclature for facelets and moves seems confusing, the test code available via the Challenge mailing list will make it clearer. Alternatively, for those of you that are into group theory, the four clockwise moves perform the following permutations on the facelets:

```kRightClockwise:     (kKL, kLB, kBK) (kLK, kBL, kKB) (cLR, cBR, cKR)
kLeftClockwise:      (kRK, kKB, kBR) (kKR, kBK, kRB) (cRL, cKL, cBL)
kBottomClockwise:    (kKR, kRL, kLK) (kRK, kLR, kKL) (cRB, cLB, cKB)
kBackClockwise:      (kLR, kRB, kBL) (kRL, kBR, kLB) (cLK, cRK, cBK)
```

### Tetraminx Facelets

Your code should place the moves needed to solve the puzzle in moves and return the number of moves generated, or return zero if you cannot solve the puzzle in maxMoves moves.

Instructions for solving the Tetraminx are available at http://www.mefferts-puzzles.com/mefferts-puzzles/tetrasol.html. Since you probably don't have a Tetraminx handy, test code will be available to help you see the effects of the moves you make to solve the puzzle. The winner will be the solution that solves the puzzle in the fewest number of moves, with a 10% penalty added for each second of execution time.

This will be a native PowerPC Challenge, using the latest CodeWarrior environment. Solutions may be coded in C, C++, or Pascal.

### Three Months Ago Winner

Congratulations once again to Randy Boring for submitting the winning solution to the March Terrain Traversal Challenge. The March Challenge required contestants to process sets of 3-dimensional input points, convert them into non-overlapping triangles, and then navigate across those triangles from pairs of starting points and ending points. The score for a solution was based on the distance traveled, with a penalty for the change in elevation along each segment of the solution, and an additional 10% penalty for each second of execution time. The winning solution minimized the solution score and the total elevation change, but it took the greatest amount of execution time to do so, and generated solution paths that were longer than those of the 3rd place solution.

Some of the contestants and members of the Challenge mailing list mentioned the possible use of Delaunay triangles http://www.ics.uci.edu/~eppstein/gina/delaunay.html in solving this Challenge. A collection of Delaunay triangles has the property that, for each edge in the collection, there is a circle containing the edge's endpoints but not any other endpoints. The idea was that these triangles would be "best" in some sense for finding optimal paths from pairs of points. However, none of the submitted solutions actually implemented this approach.

The top two scoring solutions both tried a straightforward approach to defining triangles, one that I hadn't anticipated when I defined the problem (although I probably should have). These two solutions selected an anchor point, sorted the remaining points by the angle they formed from the anchor point, and formed triangles from the anchor point to the Nth and N+1st points in angle from the anchor. This approach has the feature that there is a two-segment path between any pair of points, one from the first point to the anchor, and one from the anchor to the second point. Randy's winning solution chose the anchor to be a point in the center of the set of points, while the second-place entry of Jared Selengut chose the point arbitrarily. Randy's solution looked for a path that was better than this initial two-segment path, an enhancement that did find a better solution in one of my tests. Jared's solution was much quicker, but it found longer paths with greater elevation change than Randy's winning solution.

Ernst Munter's solution actually found the shortest solutions, although the elevation change was greater. Ernst's approach was to build the triangles in progressively widening circles around a central point. The paths produced by his solution had many more segments than the other two solutions, resulting in a more direct path between the endpoints, but the total elevation change was greater than that generated by either of the other two solutions.

The test scenarios consisted of a total of 18 test cases using 3 data sets of 2500 points each. The table below lists, for each of the solutions submitted, the total execution time for all test cases, the total horizontal distance and elevation change for the paths generated, the total score for all test cases, and the code and data size of each solution. As usual, the number in parentheses after the entrant's name is the total number of Challenge points earned in all Challenges prior to this one.

 Name Time (msec) Horizontal Distance Elevation Change Score Time Code Size Data Size Randy Boring (83) 3947.17 71710.96 3207.75 118109.99 9952 1MB Jared Selengut 142.63 73702.69 6282.39 137171.56 2044 72 Ernst Munter (430) 1729.30 65972.19 18157.13 261376.67 9312 264

### Top Contestants

Listed here are the Top 20 Contestants for the Programmer's Challenge. The numbers below include points awarded over the 24 most recent contests, including points earned by this month's entrants.

1. Munter, Ernst 205
2. Saxton, Tom 99
3. Boring, Randy 66
4. Rieken, Willeke 47
5. Maurer, Sebastian 40
6. Heithcock, JG 37
7. Murphy, ACC 34
8. Lewis, Peter 31
9. Mallett, Jeff 30
10. Cooper, Greg 27
11. Nicolle, Ludovic 27
12. Brown, Pat 20
13. Day, Mark 20
14. Hostetter, Mat 20
15. Hewett, Kevin 10
16. Jones, Dennis 10
17. Selengut, Jared 10
19. Varilly, Patrick 10
20. Webb, Russ 10

There are three ways to earn points: (1) scoring in the top 5 of any Challenge, (2) being the first person to find a bug in a published winning solution or, (3) being the first person to suggest a Challenge that I use. The points you can win are:

 1st place 20 points 2nd place 10 points 3rd place 7 points 4th place 4 points 5th place 2 points finding bug 2 points suggesting Challenge 2 points

Here is Randy's winning Terrain Traversal solution:

```/*
*    Simple solution: two-part path!
*    make radial design triangles to a central point
*    solve by traversing to center, then to destination
*/

#include "FindAPath.h"
#include <fp.h>
#include <MacMemory.h>   // for NewPtr, etc.

#pragma mark == Top ==
#define INPUT_IS_SORTED   0
#define MYDEBUG 0

#if MYDEBUG
#define kGarbage   (0xA3A3A3A3)
#define kGarbage2   (0xA3A5A4A3)
#define kGarbage3   (0xA5A5A5A5)
#define ASRT_TYPE 1
#if ASRT_TYPE == 2
#include <assert.h>
#define ASSERT(cond)   assert(cond)
#else
#define ASSERT(cond)   if (cond) ; else DebugStr("\p assert failed!")
#endif
#else
#define ASSERT(cond)
#endif

// skew the weight of the z (height) axis, since it is skewed
//   in the scoring of our Solutions.
#define kHeightWeight      (8.0)   // should be up around 100.0 since it's squared
#define kTwoPi            (pi * 2.0)
#define IdxOfPt(pt,ptarray)   ((pt) - (ptarray))
#define kNoTriangle         (-1)
#define kMaxPoints         (32 * 1024L)

// globals
static long gCenterIdx;   // index of center point

#if INPUT_IS_SORTED
// 'fix' the numbering of points (off by one)
#define PtNumToIdx(n)      ((n) - 1)
#define IdxToPtNum(p,i)      ((i) + 1)
#else
// 'fix' the numbering of points (create mapping of number to index!)
static long gPtNumToIdxMap[kMaxPoints];

#define PtNumToIdx(n)   (gPtNumToIdxMap[(n) - 1])
#define IdxToPtNum(p,i)   ((p)[i].thePointNum)

static void
MakePtNumToIdxMap(const Point3D p[], long pCount)
{
long i;
for (i = 0; i < pCount; i++)
gPtNumToIdxMap[p[i].thePointNum - 1] = i;
}

static void
DisposePtNumToIdxMap(void)
{
DisposePtr((Ptr) gPtNumToIdxMap);
}
#endif

Dist
// Return a measure of the distance, given the deltas, but
//   weight the height (z) axis most, as this has the most
//   weight in our Solution's score.
static double
Dist(double dx, double dy, double dz)
{
return (dx * dx + dy * dy + dz * dz * kHeightWeight);
}

FindMiddleDotIdx
// Return the index of the middle-most point
static long
FindMiddleDotIdx(const Point3D p[], long numPoints)
{
double totX = 0.0, totY = 0.0, totH = 0.0;
double aveX, aveY, aveH;
double denom = 1.0 / numPoints;
double bestDX, bestDY, bestDH, closestDist;
long i, besti;

// find average x, y, ht
for (i = 0; i < numPoints; i++)
{
totX += p[i].thePoint.x;
totY += p[i].thePoint.y;
totH += p[i].ht;
}
aveX = totX * denom;
aveY = totY * denom;
aveH = totH * denom;
bestDX = aveX - p[0].thePoint.x;
bestDY = aveY - p[0].thePoint.y;
bestDH = aveH - p[0].ht;
besti = 0;
closestDist = Dist(bestDX, bestDY, bestDH);

// find lowest distance to average
for (i = 1; i < numPoints; i++)
{
double dH, dX, dY, thisDist;
dX = aveX - p[i].thePoint.x;
dY = aveY - p[i].thePoint.y;
dH = aveH - p[i].ht;
thisDist = Dist(dX, dY, dH);
if (closestDist > thisDist)
{
closestDist = thisDist;
besti = i;
}
}
return besti;
}

// Absolute radians of the vector from point c to point i
static double
RadiansBetweenOld(long c, long i, const Point3D p[])
{
double dx = p[i].thePoint.x - p[c].thePoint.x;
double dy = p[i].thePoint.y - p[c].thePoint.y;
double angle = atan2(dy, dx);
return angle;
}

// Absolute radians of the vector from point 0 to point 1
static inline double
RadiansBetween(double x0, double y0, double x1, double y1)
{
double dx = x1 - x0;
double dy = y1 - y0;
double angle = atan2(dy, dx);
return angle;
}

// Add the two points, c and pointi, to a proto-Triangle
//   (The other leg will be added after sorting by angle)
static void
long pointi, long c,
Triangle t[], long ti)
{
double *angle = (double *) &(t[ti].thePoints[1]);
p[pointi].thePoint.x, p[pointi].thePoint.y);
//   t[ti].theTriangleNum = ti;   // filled in later
t[ti].thePoints[0] = IdxToPtNum(p, pointi);
}

Cmp
static int   // positive if right less then left, zero if equal
Cmp(const void *left, const void *right)
{
Triangle *lt = (Triangle *) left;
Triangle *rt = (Triangle *) right;
double *ltval = (double *) &(lt->thePoints[1]);
double *rtval = (double *) &(rt->thePoints[1]);
double diff = *ltval - *rtval;
if (diff > 0.0)
return 1;
else if (diff < 0.0)
return -1;
else
return 0;
}

SortTriangles
#include <stdlib.h>
/* void qsort(void *base, size_t nmemb, size_t size,
int (*compare) (const void *, const void *)) */
/* this requires linking with MSL Std C Lib */
static void
SortTriangles(Triangle t[], const long tCount)
{
qsort(t, tCount, sizeof(Triangle), Cmp);
}

ConcaveAngle
// Return true if the angle difference, diff, represents
//   a Concave angle, i.e., < 180 degrees (pi radians)
static Boolean
ConcaveAngle(double diff)
{
if (diff < -pi)
return true;

/* Bob guaranteed that this case would not occur:
'The situation where a point lies exactly along an
edge between two other points will not arise.' */
ASSERT(diff != 0.0);

if (diff <= 0.0)
return false;
if (diff < pi)
return true;
return false;
}

Perimeter Point List
#pragma mark == Perimeter Point List ==

// This structure is for keeping track of the diminishing
//   list of points that make up the 'perimeter' around the
//   triangles.
// As the 'concavities' are filled up, points in this list
//   are no longer on the perimeter and so are removed.
// The process stops when the perimeter is convex
// The 'un-const' Point3DPtr is still treated as 'const' by me,
//   but I couldn't figure out how to make the compiler let
//   me make const pointers into the point array otherwise.
typedef Point3D * Point3DPtr;
typedef struct {
Point3DPtr   *plist;
Point3DPtr   *plast;
long      size;
} PointList;

//----//----//----//----//----//----//
static void
MakeListOfEdgePoints(Triangle t[], long tCount, void *vp,
{
long i;
Point3DPtr p = (Point3DPtr) vp;
Point3DPtr *pp;
list->plist = pp = (Point3D **) NewPtr(tCount * sizeof(Point3D));
if (list->plist == nil)
DebugStr("\p couldn't allocate point list!");
list->size = tCount;
for (i = 0; i < badIndex; i++)
*pp++ = &(p[PtNumToIdx(t[i].thePoints[0])]);
{
*pp++ = &(p[PtNumToIdx(t[i].thePoints[0])]);
*pp++ = &(p[gCenterIdx]);
list->size++;
i++;   // i == badIndex has been processed
}
for (; i < tCount; i++)
*pp++ = &(p[PtNumToIdx(t[i].thePoints[0])]);
list->plast = pp;
}

static void
DisposeList(PointList *list)
{
DisposePtr((Ptr) list->plist);
#if MYDEBUG
list->plist = (struct Point3D **) kGarbage;
list->plast = (struct Point3D **) kGarbage2;
list->size = kGarbage3;
#endif
}

#define GetEdgePointCount(l)   ((l)->size)
#define SetEdgePointCount(l,s)   ((l)->size = (s))
#define RemoveEdgePoint(l,p)   (*(p) = nil)

//----//----//----//----//----//----//
// Find first non-nil ptr in array
static Point3DPtr *
GetFirstEdgePoint(PointList *list)
{
Point3DPtr *p = list->plist;
while (*p == nil)
++p;
ASSERT(p < list->plast);
return p;
}

//----//----//----//----//----//----//
// Starting at p, find next non-nil ptr in array
// Returns nil if no more, else
// Returns pointer to array element containing the non-nil ptr
static Point3DPtr *
GetNextEdgePoint(PointList *list, Point3DPtr *p)
{
Point3DPtr *stop = list->plast;
Point3DPtr *found = nil;
while (++p < stop)
if (*p != nil)
{
found = p;
break;
}
ASSERT(p < stop);
return found;
}

//----//----//----//----//----//----//
// Return true if, when looking from p1 to p3, p2 is on the left
// This means that a triangle can be constructed of these
//   points, making the perimeter more convex (removing a concavity).
static Boolean
Concave(const Point3D *p1, const Point3D *p2, const Point3D *p3)
{
p1->thePoint.x, p1->thePoint.y);
p3->thePoint.x, p3->thePoint.y);
double diff = r23 - r21;   // angle (in radians) at point 2
return ConcaveAngle(diff);
}

static void
Point3DPtr p1, Point3DPtr p2, Point3DPtr p3)
{
//   t[tidx].theTriangleNum = tidx;   // filled in later
t[tidx].thePoints[0] = p1->thePointNum;
t[tidx].thePoints[1] = p2->thePointNum;
t[tidx].thePoints[2] = p3->thePointNum;
}

//----//----//----//----//----//----//
// Create more triangles by going around the perimeter once
//   and connecting edge points that can 'see' each other.
// Returns how many more were created
static long
Convexify(PointList *list, Triangle t[], long tCount)
{
Point3DPtr *firstpt, *pt1, *pt2, *pt3;
long edgePtCount = GetEdgePointCount(list);
long moreTriangles = 0;
firstpt = pt1 = GetFirstEdgePoint(list);
pt2 = GetNextEdgePoint(list, pt1);
ASSERT(pt2 != nil);
do   {
pt3 = GetNextEdgePoint(list, pt2);
ASSERT(pt3 != nil);
if (Concave(*pt1, *pt2, *pt3))
{
tCount++;
moreTriangles++;
RemoveEdgePoint(list, pt2);
pt1 = pt3;
if (-edgePtCount > 2)
pt2 = GetNextEdgePoint(list, pt1);
}
else {
pt1 = pt2;
pt2 = pt3;
}
} while (-edgePtCount > 2);
// last two wrap around, putting p1 into p3 and p2 positions
if (*pt2 == nil)   // last triangle was concave
{
pt2 = firstpt;
pt3 = GetNextEdgePoint(list, firstpt);
if (Concave(*pt1, *pt2, *pt3))
{
tCount++;
moreTriangles++;
RemoveEdgePoint(list, pt2);
}
}
else {
pt3 = firstpt;
if (Concave(*pt1, *pt2, *pt3))
{
tCount++;
moreTriangles++;
RemoveEdgePoint(list, pt2);
}
else {
pt1 = pt2;
pt2 = pt3;
pt3 = GetNextEdgePoint(list, pt2);
if (Concave(*pt1, *pt2, *pt3))
{
tCount++;
moreTriangles++;
RemoveEdgePoint(list, pt2);
}
}
}
// reduce the count of points on the perimeter by the number
//   of triangles we created this pass
SetEdgePointCount(list,
GetEdgePointCount(list) - moreTriangles);
return moreTriangles;
}

//----//----//----//----//----//----//
// Interconnect more 'edge' points into triangles
// Returns new triangle count
static long
MakeMoreTriangles(Triangle t[], long tCount,
const Point3D p[], PointList *listp)
{
#pragma unused (p)
long moreTriangles = 0;
do   {
moreTriangles = Convexify(listp, t, tCount);
tCount += moreTriangles;
} while (moreTriangles > 0);
DisposeList(listp);
return tCount;
}
//----//----//----//----//----//----//
// Note: If the set of points can form a convex polygon,
//   (or the mid-most point is on the outside for any reason),
//   then this algorithm would produce an "overlapping triangle"
//   because one of its angles is greater than 180 degrees (i.e.,
//   it is upside down) without the ConcaveAngle check.
// Returns how many triangles were made.  This should be
//   tCount, unless there was an inverted triangle, in which case
//   it will be tCount - 1
static long
MakePerimeterConnections(Triangle t[], const long tCount,
const Point3D p[], const PointNum cnum, PointList *listp)
{
double angleLast = *(double *)&t[0].thePoints[1];
double angleLastOriginal = angleLast;   // save for wraparound test
double angleNext;
for (i = 0; i < tCount - 1; i++)
{
angleNext = *(double *)&t[i + 1].thePoints[1];
if (ConcaveAngle(angleNext - angleLast))
{
t[i].thePoints[2] = t[i + 1].thePoints[0];
t[i].thePoints[1] = cnum;
angleLast = angleNext;
}
else {
/* we can have only ONE inverted triangle to skip */
}
}
// wrap around to the beginning
angleNext = angleLastOriginal;
if (ConcaveAngle(angleNext - angleLast))
{
t[tCount - 1].thePoints[2] = t[0].thePoints[0];
t[tCount - 1].thePoints[1] = cnum;
}
else
MakeListOfEdgePoints(t, tCount, (void *) p, badIndex, listp);
// account for possible bad triangle
{   // shift all triangles above badIndex down one
(tCount - 1 - badIndex) * sizeof(Triangle));
return tCount - 1;
}
return tCount;
}

//----//----//----//----//----//----//
// Make the triangles that connect each point with the center point
// Return the number of triangles created (pCount - 1 or 2)
static long
MakeRadialTriangles(long center, const Point3D p[], long pCount, Triangle t[])
{
long i, tCount;
PointList list;
for (i = 0; i < center; i++)
// don't connect center with itself!
for (i = center + 1; i < pCount; i++)
MakeRadialConnection(p, i, center, t, i - 1);
tCount = pCount - 1;
SortTriangles(t, tCount);
tCount = MakePerimeterConnections(t, tCount, p,
IdxToPtNum(p, center), &list);
tCount = MakeMoreTriangles(t, tCount, p, &list);
return tCount;
}

#if MYDEBUG
static void
DbgWriteTriangles(long c, const Point3D p[], Triangle t[],
long tCount)
{
FILE *dbgf = fopen("triangles.out", "w");
long i;
const Point2D *pt = &(p[c].thePoint);
fprintf(dbgf, "center = %d (%f, %f)\n\n", c, pt->x, pt->y);
fprintf(dbgf,
"\n tri   \tp0   \tp1   \tp2   \t(x, y) of p0\n");
for (i = 0; i < tCount; i++)
{
const Point2D *pt0;
PointNum pnum0 = t[i].thePoints[0];
pt0 = &(p[PtNumToIdx(pnum0)].thePoint);
fprintf(dbgf, "%d   \t%d   \t%d   \t%d   \t(%f, %f)\n",
i, pnum0, t[i].thePoints[1], t[i].thePoints[2],
pt0->x, pt0->y);
}
fclose(dbgf);
}
#endif

Neighbor Mapping
#pragma mark == Neighbor Mapping ==

// This mapping helps iterate over the neighbors of a point

#define kEndOfNeighborList   (-1)
#define kMaxSmallNeighbors   (3)   // later 5-7
#define kExtraNeighborBytes   (2)   // later 32 (bytes = 16 more neighbors)
#define kLastSmallNeighbor   (kMaxSmallNeighbors - 1)

typedef struct NMap {
short**   moreNeighborsH;   // excess neighbors (more than kMaxSmallNeighbors)
short   nCount;         // count of neighbors
short   smallNeighbors[kMaxSmallNeighbors];
} NeighborMap;

static NeighborMap gNeighborMap[kMaxPoints];
static long gpCount;

static void
ClearNeighborMap(const long pCount)
{
long i;
gpCount = pCount;   // for deallocating later
for (i = 0; i < pCount; i++)
{
gNeighborMap[i].moreNeighborsH = nil;
gNeighborMap[i].nCount = 0;
#if MYDEBUG
gNeighborMap[i].smallNeighbors[0] = kGarbage;
#endif
}
}

//----//----//----//----//----//----//
// add pt2 to pt1's neighborlist
static void
{
long i, count = gNeighborMap[pt1].nCount;
short *np = gNeighborMap[pt1].smallNeighbors;
long stop = (count > kMaxSmallNeighbors)
? kMaxSmallNeighbors : count;
short **h;
for (i = 0; i < stop; i++)
if (np[i] == pt2)

if (count < kMaxSmallNeighbors)
{   // there is room in the small neighbor list
gNeighborMap[pt1].smallNeighbors[count] = pt2;
gNeighborMap[pt1].nCount = count + 1;
return;
}
else if (count == kMaxSmallNeighbors)
{   // time to allocate Handle for more neighbors
ASSERT(gNeighborMap[pt1].moreNeighborsH == nil);
h = (short **) NewHandle(kExtraNeighborBytes);
if (h == nil)
DebugStr("\p couldn't allocate extra neighbors handle");
gNeighborMap[pt1].moreNeighborsH = h;
**h = pt2;
gNeighborMap[pt1].nCount = count + 1;
return;
}

ASSERT(GetHandleSize((Handle) gNeighborMap[pt1].moreNeighborsH)
>= 2 *(count - kMaxSmallNeighbors));
np = *gNeighborMap[pt1].moreNeighborsH;
stop = count - kMaxSmallNeighbors;
for (i = 0; i < stop; i++)
if (np[i] == pt2)

// didn't find pt2 in neighbor list, we'll have to add it
h = gNeighborMap[pt1].moreNeighborsH;
// check to see if we need to resize handle
ASSERT(stop * 2 <= GetHandleSize((Handle) h));
if (stop * 2 == GetHandleSize((Handle) h))
{
SetHandleSize((Handle) h, stop * 4);
if (GetHandleSize((Handle) h) != stop * 4)
DebugStr("\p couldn't resize handle for more neighbors!");
}
*((*h) + stop) = pt2;
gNeighborMap[pt1].nCount = count + 1;
}

//----//----//----//----//----//----//
// add pt2 to pt1's neighborlist and pt1 to pt2's
static void
{
}

static void
LockNeighborHandles(void)
{
long i;
for (i = 0; i < gpCount; i++)
if (gNeighborMap[i].nCount > kMaxSmallNeighbors)
{   // must have handle when we have more neighbors
ASSERT(gNeighborMap[i].moreNeighborsH != nil);
HLock((Handle) gNeighborMap[i].moreNeighborsH);
}
else   // no handle when neighbors fit in small array
ASSERT(gNeighborMap[i].moreNeighborsH == nil);
}

//----//----//----//----//----//----//
// Ignores center point because our algorithm will first use
//   the center to find an initial solution.  Therefore no
//   around-the-edge solution will need to go through the center.
static void
MakeNeighborMap(const Triangle t[], const long tCount,
const long pCount)
{
long i = -1;
ClearNeighborMap(pCount);
while (t[++i].thePoints[1] == gCenterIdx)
PtNumToIdx(t[i].thePoints[2]));
for (; i < tCount; i++)
{
PtNumToIdx(t[i].thePoints[1]));
PtNumToIdx(t[i].thePoints[2]));
PtNumToIdx(t[i].thePoints[2]));
}
LockNeighborHandles();
}

static void
DisposeNeighborMap()
{
long i;
for (i = 0; i < gpCount; i++)
if (gNeighborMap[i].nCount > kMaxSmallNeighbors)
DisposeHandle((Handle) gNeighborMap[i].moreNeighborsH);
}

static inline long
FirstNeighbor(long pti)
{
ASSERT(gNeighborMap[pti].nCount > 0);
return gNeighborMap[pti].smallNeighbors[0];
}

//----//----//----//----//----//----//
// returns next neighbor's point index
// searches for previous neighbor in list and returns the next one
// or kEndOfNeighborList (-1) if there is no next one
static long
NextNeighbor(long pti, long neighbor)
{
long i, count = gNeighborMap[pti].nCount;
short *np = gNeighborMap[pti].smallNeighbors;
long stop = (count > kMaxSmallNeighbors) ?
kMaxSmallNeighbors : count;
short **h;
for (i = 0; i < stop; i++)
if (np[i] == neighbor)
break;   // found neighbor
if (i < stop)
{   // found neighbor in small list
if (i + 1 < stop)   // there are more in the small list
return gNeighborMap[pti].smallNeighbors[i + 1];
// else, there are no more in the small list
if (i + 1 < kMaxSmallNeighbors)   // was there room for more?
return kEndOfNeighborList;   // then, that was the last one!
// else, found at end of a full small list
if (i + 1 < count)   // are there more neighbors?
{   // next neighbor is the first in the Handle
h = gNeighborMap[pti].moreNeighborsH;
ASSERT(h != nil);
return **h;
}
// else, no more at all
return kEndOfNeighborList;
}
if (count == kMaxSmallNeighbors)   // there are no more
return kEndOfNeighborList;
// have to search the handle's entries
np = *gNeighborMap[pti].moreNeighborsH;
stop = count - kMaxSmallNeighbors;
for (i = 0; i < stop; i++)
if (np[i] == neighbor)
break;   // found neighbor
ASSERT(i < stop);   // otherwise neighbor wasn't found! (misuse of this routine)
// check to see if we are at end of neighbor list
if (i + 1 < stop)   // there are more in handle's list
return np[i + 1];   // next neighbor
else            // found neighbor was the last one
return kEndOfNeighborList;
}

InitTerrainMap
// Interconnect the points into triangles of my choosing
// My strategy is to make thin triangles from every point to
//   a central point.  This way there is always a two-step (max)
//   path between any two points, and that path won't have much
//   elevation change.
// Then I improve that set by filling in the edges so that
//   the perimeter is concave.  This way, alternate paths may
//   be found between some pairs that are shorter than going
//   through the center.
long /*numTriangles*/
InitTerrainMap(
const Point3D thePoints[],
long numPoints,
Triangle theTriangles[])
{
long i, numTriangles;
long center;
#if !INPUT_IS_SORTED
MakePtNumToIdxMap(thePoints, numPoints);
#endif
center = FindMiddleDotIdx(thePoints, numPoints);
gCenterIdx = center;
numPoints, theTriangles);
#if MYDEBUG
DbgWriteTriangles(center, thePoints,
theTriangles, numTriangles);
#endif
// now number the triangles
for (i = 0; i < numTriangles; i++)
theTriangles[i].theTriangleNum = i;
MakeNeighborMap(theTriangles, numTriangles, numPoints);
return numTriangles;
}

TermTerrainMap
// release anything allocated in initialization
void
TermTerrainMap(void)
{
#if !INPUT_IS_SORTED
DisposePtNumToIdxMap();
#endif
DisposeNeighborMap();
}

ContainsPoint
// Return true if the point (x,y) is one of the points
static Boolean
ContainsPoint(const Point3D p[], const PointNum pn[3],
double x, double y)
{
long j;
for (j = 0; j < 3; j++)
{
long pti = PtNumToIdx(pn[j]);
if (p[pti].thePoint.x == x &&
p[pti].thePoint.y == y)
return true;
}
return false;
}

FindTriangleContaining
// Return the index of the triangle containing both
//   point 1 (x1, y1) and point 2 (x2, y2)
static long
FindTriangleContaining(const Point3D p[],
const Triangle t[], long tCount,
double x1, double y1,    // point 1
double x2, double y2)      // point 2
{
long i;
if (x1 == x2 && y1 == y2)
return kNoTriangle;   // points are the same
for (i = 0; i < tCount; i ++)
if (ContainsPoint(p, t[i].thePoints, x1, y1) &&
ContainsPoint(p, t[i].thePoints, x2, y2))
return i;
DebugStr("\p didn't find the triangle!");
return -2;   // failure!
}

FindPtIdx
// Find the index of the point with coordinates (ptx, pty)
// This is a linear search through the point list, don't do
//   it very often!
static long
FindPtIdx(const Point3D p[], long pCount,
double ptx, double pty)
{
long i;
for (i = 0; i < pCount; i++)
if (p[i].thePoint.x == ptx &&
p[i].thePoint.y == pty)
return i;
return -1;
}

CostBetween
// returns the cost between two points
// computed as per Challenge Statement: the distance between
//   the two (2D) points plus ten times the height difference
//   (as an absolute value).
static double
CostBetween(long ptAi, long ptBi, const Point3D p[])
{
double dx, dy, dht;
dht   = p[ptAi].ht;
dx      = p[ptAi].thePoint.x;
dy      = p[ptAi].thePoint.y;
dht   -= p[ptBi].ht;
dx      -= p[ptBi].thePoint.x;
dy      -= p[ptBi].thePoint.y;
dht   *= 10.0;
dx      *= dx;
dy      *= dy;
if      (dht < 0.0)
dht = -dht;
return sqrt(dx + dy) + dht;
}

CostOfSegmentPath
// returns cost of whole path (given as list of Segments)
// computed by adding cost between the startingPoint and the
//   endingPoint of each Segment.
static double
CostOfSegmentPath(Segment s[], long sCount,
const Point3D p[], const long pCount)
{
double total = 0.0;
long i;
long ptAi = FindPtIdx(p, pCount,
s[0].startingPoint.x, s[0].startingPoint.y);
for (i = 0; i < sCount; i++)
{
long ptBi = FindPtIdx(p, pCount,
s[i].endingPoint.x, s[i].endingPoint.y);
total += CostBetween(ptAi, ptBi, p);
ptAi = ptBi;
}
}

Search Queue & Paths
#pragma mark == Search Queue & Paths ==
// This structure is the heart of my depth-first-search
// It is both a search queue element and a path element
typedef struct QE {
struct QE *      nextSearchSQ;    // 32K max (+ 2 sentries)
//   short         ptIdx;            // 32K max
unsigned short   nextPathQEi;     // 32K max (+ 2 sentries)
short         refCount;           // path points are shared
double         costSoFar;         // total cost of folowing this path
} SearchQueueElem, *SearchQueue, *Path, **SearchQueuePtr;
#define IdxOfPath(p)      ((p) - gQ)
#define IdxOfSQ(q)         ((q) - gQ)
#define CostOfPath(p)      ((p)->costSoFar)
#define Idx2Path(ptidx)      (&(gQ[ptidx]))
#define Idx2SQ(qidx)      (&(gQ[qidx]))
#define NextSQOf(q)         ((q)->nextSearchSQ)
#define NextSQIdxOf(q)      (IdxOfSQ(NextSQOf(q)))
#define SetNextSQOf(q,nq)   (NextSQOf(q) = (nq))
#define NextPathIdxOf(p)   ((p)->nextPathQEi)
#define NextPathOf(p)      (Idx2Path(NextPathIdxOf(p)))
#define SetNextPathOf(p,np)   (NextPathIdxOf(p) = IdxOfPath(np))
#define RefCountOf(p)      ((p)->refCount)
#define kMaxQE         (kMaxPoints)
#define kLastValidQEi   (kMaxQE - 1)
#define kLastPath      (kMaxQE + 1)
#define kLastQE         (kMaxQE + 2)
static SearchQueueElem gQ[kMaxQE + 3];
// Sentry values (not nil, so I can tell whether an element is
//   in a path or queue, even at the end of it)
static const SearchQueue gkLastSQ = &gQ[kLastQE];
static const Path gkLastPath = &gQ[kLastPath];
//----//----//----//----//----//----//
static void
InitSearchQueue(SearchQueuePtr qp, const long pCount)
{
long i;
for (i = 0; i < pCount; i++)
{
gQ[i].nextSearchSQ = nil;
gQ[i].refCount = 0;
#if MYDEBUG
gQ[i].nextPathQEi = kGarbage;
gQ[i].costSoFar = (double) kGarbage;
#endif
}
*qp = gkLastSQ;
}

//----//----//----//----//----//----//
static void
DeInitSearchQueue(SearchQueuePtr qp)
{
#pragma unused(qp)
}
IsEmptySearchQueue
static inline Boolean
IsEmptySearchQueue(SearchQueuePtr qp)
{
ASSERT(qp != nil);
ASSERT(*qp >= &gQ[0]);
ASSERT(*qp <= gkLastSQ);
ASSERT(*qp != gkLastPath);
return (*qp == gkLastSQ);
}

//----//----//----//----//----//----//
static inline Boolean
{
ASSERT(p != nil);
ASSERT(p != gkLastSQ);
ASSERT(p != gkLastPath);
return (nil != NextSQOf(p));
}

//----//----//----//----//----//----//
// Add to search queue in sorted order (least cost at front)
static void
{
double pCost;
SearchQueue q, lastq;
ASSERT(qp != nil);
ASSERT(*qp != gkLastPath);
ASSERT(p != nil);
ASSERT(p != gkLastSQ);
ASSERT(p != gkLastPath);
if (*qp == gkLastSQ)
{
*qp = p;
SetNextSQOf(p, gkLastSQ);
return;
}
lastq = nil;
q = *qp;
pCost = CostOfPath(p);
while (q != gkLastSQ && CostOfPath(q) < pCost)
{
lastq = q;
q = NextSQOf(q);
}
if (lastq == nil)
else               // insert p after lastq
SetNextSQOf(lastq, p);
SetNextSQOf(p, q);
}

//----//----//----//----//----//----//
// Remove from the queue the least cost path
// Hint: it's at the front!
// Does NOT remove it from path NOR releases it
// DOES set nextSQ to nil
static inline Path
RemoveBestElem(SearchQueuePtr qp)
{
Path p = *qp;
ASSERT(p != nil);
ASSERT(p != gkLastSQ);
ASSERT(p != gkLastPath);
*qp = NextSQOf(p);
SetNextSQOf(p, nil);
return p;
}

//----//----//----//----//----//----//
// Remove from the queue the given path
// Does NOT remove it from path NOR releases it
// DOES set nextSQ to nil
static void
RemoveFromSearchQueue(SearchQueuePtr qp, Path p)
{
Path seek, seekLast;
ASSERT(qp != nil);
ASSERT(*qp != gkLastSQ);
ASSERT(*qp != gkLastPath);
ASSERT(p != nil);
ASSERT(p != gkLastSQ);
ASSERT(p != gkLastPath);
seekLast = nil;
seek = *qp;
while (seek != p)
{
seekLast = seek;
seek = NextSQOf(seek);
ASSERT(seek != nil);
ASSERT(seek != gkLastSQ);
}
if (seekLast == nil)   // p was first in list
*qp = NextSQOf(p);
else   // make the node before p point to the one after p
SetNextSQOf(seekLast, NextSQOf(p));
SetNextSQOf(p, nil);
}

//----//----//----//----//----//----//
static Path
MakeFirstPath(long pointIdx)
{
Path p = Idx2Path(pointIdx);
ASSERT(p >= gQ);
ASSERT(p <= &(gQ[kLastValidQEi]));
SetNextPathOf(p, gkLastPath);
SetNextSQOf(p, nil);
RefCountOf(p) = 64;   // anomoly to prevent loops at beginning
CostOfPath(p) = 0.0;
return p;
}

//----//----//----//----//----//----//
static inline void
AddPath(Path oldp, Path newp, double newTotalCost)
{
ASSERT(oldp != nil);
ASSERT(newp != nil);
ASSERT(newTotalCost > CostOfPath(oldp));
ASSERT(RefCountOf(newp) == 0);
SetNextPathOf(newp, oldp);
CostOfPath(newp) = newTotalCost;
RefCountOf(oldp)++;      // oldp becomes interior node
}

//----//----//----//----//----//----//
static void
ReleasePath(Path p)
{
ASSERT(p != nil);
ASSERT(p != gkLastSQ);
if (p == gkLastPath)
return;
if (RefCountOf(p) == 0)
ReleasePath(NextPathOf(p));
else
RefCountOf(p)-;
}

//----//----//----//----//----//----//
// Returns the number of segments, i.e., jumps between path nodes
static long
PathLength(Path p)
{
long len = -1;
ASSERT(p != nil);
ASSERT(p != gkLastPath);
ASSERT(p != gkLastSQ);
do   {
p = NextPathOf(p);
++len;
} while (p != gkLastPath);
return len;
}

//----//----//----//----//----//----//
// Record the segments that make of this (newer) better path
// The path is from end to start, so work backwards
// NOTE: this will fail if the path is size zero
static void
RecordPath(Segment s[], long *sCount, Path path,
const Point3D p[], const Triangle t[], const long tCount)
{
double xE, yE, xS, yS;
long len = *sCount = PathLength(path);
long triNum;
xE = p[IdxOfPath(path)].thePoint.x;
yE = p[IdxOfPath(path)].thePoint.y;
ASSERT(len > 0);
do   {
path = NextPathOf(path);
len-;
xS = p[IdxOfPath(path)].thePoint.x;
yS = p[IdxOfPath(path)].thePoint.y;
triNum = FindTriangleContaining(p, t,
tCount, xS, yS, xE, yE);
ASSERT(triNum != kNoTriangle);
s[len].theTriangleNum   = triNum;
s[len].startingPoint.x   = xS;
s[len].startingPoint.y   = yS;
s[len].endingPoint.x   = xE;
s[len].endingPoint.y   = yE;
xE = xS;
yE = yS;
} while (len > 0);
}

#if MYDEBUG
static long
FindPtInTriangle(const Point3D p[], const PointNum pn[3],
double x, double y)
{
long j;
for (j = 0; j < 3; j++)
{
long pti = PtNumToIdx(pn[j]);
if (p[pti].thePoint.x == x &&
p[pti].thePoint.y == y)
return pn[j];
}
DebugStr("\p couldn't find point in triangle!");
return -1;
}

static void
PrintPath(Segment s[], long sCount, const Point3D p[], const Triangle t[])
{
long i;
for (i = 0; i < sCount; i++)
{
long triNum, ptNumFrom, ptNumTo;
triNum = s[i].theTriangleNum;
ptNumFrom = FindPtInTriangle(p, t[triNum].thePoints,
s[i].startingPoint.x, s[i].startingPoint.y);
ptNumTo   = FindPtInTriangle(p, t[triNum].thePoints,
s[i].endingPoint.x,   s[i].endingPoint.y);
printf("seg %d \t%d \t%d \t%d",
i, triNum, ptNumFrom, ptNumTo);
printf(" cost=%f\n",
CostBetween(PtNumToIdx(ptNumFrom), PtNumToIdx(ptNumTo), p));
}
}
#endif

EvaluateFinishedPath
// Deal with a path to the end point
// If it is a better path,
//      record it, release the old best path, and return the new value
//   otherwise
//      release the path (it wasn't better), and return the old value
// Return the current best path in *bestPath
static double
EvaluateFinishedPath(Segment s[], long *sCount,
const Point3D p[], const Triangle t[], const long tCount,
Path path, Path *bestPath, double bestCost)
{
double newCost;
newCost = CostOfPath(path);
if (newCost < bestCost)
{   // found a better path to end!!
RecordPath(s, sCount, path, p, t, tCount);
#if MYDEBUG
printf("better path:\n");
PrintPath(s, *sCount, p, t);
#endif
bestCost = newCost;
ASSERT(bestPath != nil);
if (*bestPath != nil)
ReleasePath(*bestPath);
*bestPath = path;
}
else
ReleasePath(path);
return bestCost;
}

ExpandSearch
// Expand the search from the given path endpoint 'path'
// Expand by following each node that is neighbor to 'path'
// Expand just one level, then return.
static void
ExpandSearch(SearchQueuePtr qp, const Point3D p[],
Path path, long pathIdx, double bestCost)
{
long follow;
// expand search from partial path ending at 'path'
{
double newCost;
Path followPath = Idx2Path(follow);
// don't go back on yourself (problem with neighbor logic)
if (RefCountOf(followPath) > 0)
{         // it's an interior node,
// should not be in search path
continue;   // skip it
}
newCost = CostOfPath(path) + CostBetween(pathIdx, follow, p);
if (newCost < bestCost)
{   // survived cutoff
if (newCost >= CostOfPath(followPath))
// new path not an improvement
continue;
else   // remove to add in proper (sorted) place
RemoveFromSearchQueue(qp, followPath);
}
}
}

SearchAlternatePath
// Search for a better path from start to end than the
//   through-the-center route already in theSegments
// If found, fill in theSegments and return its length
//   else, just return the old length (sCount)
static long
SearchAlternatePath(Segment s[], long sCount,
const Point3D p[], const long pCount,
const Triangle t[], const long tCount,
long ptIdxStart, long ptIdxEnd)
{
double bestCost = CostOfSegmentPath(s, sCount, p, pCount);
SearchQueue q;
Path pathStartPtr, bestPath = nil;
InitSearchQueue(&q, pCount);
pathStartPtr = MakeFirstPath(ptIdxStart);
do   {
Path path = RemoveBestElem(&q);
long pathIdx = IdxOfPath(path);
if (pathIdx == ptIdxEnd)   // found a path to end!
bestCost = EvaluateFinishedPath(s, &sCount,
p, t, tCount, path, &bestPath, bestCost);
else
ExpandSearch(&q, p, path, pathIdx, bestCost);
} while (!IsEmptySearchQueue(&q));
DeInitSearchQueue(&q);
return sCount;
}

FindAPath
// Find a minimal cost path between the two points
// First:
// Connect the start to the middle
// Connect the middle to the end
// Then:
// Do a breadth-first search from start
// Use the simple through-the-center path as an optimizing cutoff
long /*numSegments*/
FindAPath(
const Point3D thePoints[],
long numPoints,
const Triangle theTriangles[],
long numTriangles,
const Point2D pathStart,
const Point2D pathEnd,
Segment theSegments[]
) {
#pragma unused (numPoints)
double x1, y1, xc, yc;
long numSegments = 0;
long triNumStart, triNumEnd;
PointNum ptIdxStart, ptIdxEnd;
x1 = pathStart.x;
y1 = pathStart.y;
xc = thePoints[gCenterIdx].thePoint.x;
yc = thePoints[gCenterIdx].thePoint.y;
triNumStart = FindTriangleContaining(thePoints, theTriangles,
numTriangles, x1, y1, xc, yc);
if (triNumStart != kNoTriangle)
// triNum == kNoTriangle means pathStart IS center
{
theSegments[0].theTriangleNum   = triNumStart;
theSegments[0].startingPoint   = pathStart;
theSegments[0].endingPoint.x   = xc;
theSegments[0].endingPoint.y   = yc;
numSegments = 1;
}
ptIdxStart = FindPtIdx(thePoints, numPoints, x1, y1);
x1 = pathEnd.x;
y1 = pathEnd.y;
triNumEnd = FindTriangleContaining(thePoints, theTriangles,
numTriangles, x1, y1, xc, yc);
if (triNumEnd != kNoTriangle)
// triNum == kNoTriangle means pathEnd IS center
{
theSegments[numSegments].theTriangleNum   = triNumEnd;
theSegments[numSegments].startingPoint.x   = xc;
theSegments[numSegments].startingPoint.y   = yc;
theSegments[numSegments].endingPoint       = pathEnd;
numSegments++;
}
#if MYDEBUG
printf("starting point: %f, %f\n", pathStart.x, pathStart.y);
printf("ending point:   %f, %f\n", pathEnd.x,   pathEnd.y);
printf("initial path through center:\n");
PrintPath(theSegments, numSegments, thePoints, theTriangles);
#endif
if (numSegments > 1)   // can't improve a path of length one
{
ptIdxEnd = FindPtIdx(thePoints, numPoints, x1, y1);
numSegments = SearchAlternatePath(theSegments, numSegments,
thePoints, numPoints,
theTriangles, numTriangles,
ptIdxStart, ptIdxEnd);
}
// else SearchAlternatePath is unnecessary (and fails!)
return numSegments;
}
```

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Homerun Clash 2: Legends Derby opens for...
Since launching in 2018, Homerun Clash has performed admirably for HAEGIN, racking up 12 million players all eager to prove they could be the next baseball champions. Well, the title will soon be up for grabs again, as Homerun Clash 2: Legends... | Read more »
‘Neverness to Everness’ Is a Free To Pla...
Perfect World Games and Hotta Studio (Tower of Fantasy) announced a new free to play open world RPG in the form of Neverness to Everness a few days ago (via Gematsu). Neverness to Everness has an urban setting, and the two reveal trailers for it... | Read more »
Meditative Puzzler ‘Ouros’ Coming to iOS...
Ouros is a mediative puzzle game from developer Michael Kamm that launched on PC just a couple of months back, and today it has been revealed that the title is now heading to iOS and Android devices next month. Which is good news I say because this... | Read more »

## Price Scanner via MacPrices.net

Amazon is still selling 16-inch MacBook Pros...
Prime Day in July is over, but Amazon is still selling 16-inch Apple MacBook Pros for \$500-\$600 off MSRP. Shipping is free. These are the lowest prices available this weekend for new 16″ Apple... Read more
Walmart continues to sell clearance 13-inch M...
Walmart continues to offer clearance, but new, Apple 13″ M1 MacBook Airs (8GB RAM, 256GB SSD) online for \$699, \$300 off original MSRP, in Space Gray, Silver, and Gold colors. These are new MacBooks... Read more
Apple is offering steep discounts, up to \$600...
Apple has standard-configuration 16″ M3 Max MacBook Pros available, Certified Refurbished, starting at \$2969 and ranging up to \$600 off MSRP. Each model features a new outer case, shipping is free,... Read more
Save up to \$480 with these 14-inch M3 Pro/M3...
Apple has 14″ M3 Pro and M3 Max MacBook Pros in stock today and available, Certified Refurbished, starting at \$1699 and ranging up to \$480 off MSRP. Each model features a new outer case, shipping is... Read more
Amazon has clearance 9th-generation WiFi iPad...
Amazon has Apple’s 9th generation 10.2″ WiFi iPads on sale for \$80-\$100 off MSRP, starting only \$249. Their prices are the lowest available for new iPads anywhere: – 10″ 64GB WiFi iPad (Space Gray or... Read more
Apple is offering a \$50 discount on 2nd-gener...
Apple has Certified Refurbished White and Midnight HomePods available for \$249, Certified Refurbished. That’s \$50 off MSRP and the lowest price currently available for a full-size Apple HomePod today... Read more
The latest MacBook Pro sale at Amazon: 16-inc...
Amazon is offering instant discounts on 16″ M3 Pro and 16″ M3 Max MacBook Pros ranging up to \$400 off MSRP as part of their early July 4th sale. Shipping is free. These are the lowest prices... Read more
14-inch M3 Pro MacBook Pros with 36GB of RAM...
B&H Photo has 14″ M3 Pro MacBook Pros with 36GB of RAM and 512GB or 1TB SSDs in stock today and on sale for \$200 off Apple’s MSRP, each including free 1-2 day shipping: – 14″ M3 Pro MacBook Pro (... Read more
14-inch M3 MacBook Pros with 16GB of RAM on s...
B&H Photo has 14″ M3 MacBook Pros with 16GB of RAM and 512GB or 1TB SSDs in stock today and on sale for \$150-\$200 off Apple’s MSRP, each including free 1-2 day shipping: – 14″ M3 MacBook Pro (... Read more
Amazon is offering \$170-\$200 discounts on new...
Amazon is offering a \$170-\$200 discount on every configuration and color of Apple’s M3-powered 15″ MacBook Airs. Prices start at \$1129 for models with 8GB of RAM and 256GB of storage: – 15″ M3... Read more

## Jobs Board

*Apple* Systems Engineer - Chenega Corporati...
…LLC,** a **Chenega Professional Services** ' company, is looking for a ** Apple Systems Engineer** to support the Information Technology Operations and Maintenance Read more
Solutions Engineer - *Apple* - SHI (United...
**Job Summary** An Apple Solution Engineer's primary role is tosupport SHI customers in their efforts to select, deploy, and manage Apple operating systems and Read more
*Apple* / Mac Administrator - JAMF Pro - Ame...
Amentum is seeking an ** Apple / Mac Administrator - JAMF Pro** to provide support with the Apple Ecosystem to include hardware and software to join our team and Read more
Operations Associate - *Apple* Blossom Mall...
Operations Associate - Apple Blossom Mall Location:Winchester, VA, United States (https://jobs.jcp.com/jobs/location/191170/winchester-va-united-states) - Apple Read more
Cashier - *Apple* Blossom Mall - JCPenney (...
Cashier - Apple Blossom Mall Location:Winchester, VA, United States (https://jobs.jcp.com/jobs/location/191170/winchester-va-united-states) - Apple Blossom Mall Read more