[Question]
  [
Imagine a ***W*** by ***H*** grid of squares that wraps toroidally. Items are placed onto the grid as follows.
The first item can be placed on any square, but subsequent items must not be within a [Manhattan distance](http://en.wikipedia.org/wiki/Taxicab_geometry "Wikipedia article") ***R*** of any previous item (also known as a [Von Neumann neighbourhood of range ***R***](http://en.wikipedia.org/wiki/Von_Neumann_neighborhood#von_Neumann_neighborhood_of_range_r "Wikipedia article")). Carefully choosing the positions allows fitting a large number of items onto the grid before there are no more valid positions. However, instead consider the opposite aim: What is the lowest number of items that can be placed and leave no further valid positions?
Here is a radius 5 exclusion zone:
![Radius 5 exclusion zone](https://i.stack.imgur.com/leNg0.png "Radius 5 exclusion zone")
Here is another radius 5 exclusion zone, this time near the edges so the wrapping behaviour is apparent:
![Wrapping radius 5 exclusion zone](https://i.stack.imgur.com/AykgT.png "Radius 5 exclusion zone with toroidal edge wrapping")
### Input
Three integers:
* ***W***: width of grid *(positive integer)*
* ***H***: height of grid *(positive integer)*
* ***R***: radius of exclusion zone *(non-negative integer)*
### Output
An integer ***N***, which is the smallest number of items that can be placed preventing any further valid placements.
### Details
* A radius of zero gives an exclusion zone of 1 square (the one the item was placed on).
* A radius of N excludes the zone that can be reached in N orthogonal steps (remember the edges wrap toroidally).
Your code must work for the trivial case of ***R*** = 0, but does not need to work for ***W*** = 0 or ***H*** = 0.
Your code must also deal with the case where ***R*** > ***W*** or ***R*** > ***H***.
# Time limit and test cases
Your code must be able to deal with all of the test cases, and each test case must complete within 5 minutes. This should be easy (the example JavaScript solution takes a few seconds for each test case). The time limit is mainly to exclude the extreme brute force approach. The example approach is still fairly brute force.
If your code completes within 5 minutes on one machine but not on another that will be close enough.
**Test cases** in the form inputs: output as `W H R : N`
```
5 4 4 : 1
5 4 3 : 2
5 4 2 : 2
5 4 1 : 5
7 5 5 : 1
7 5 4 : 2
7 5 3 : 2
7 5 2 : 4
8 8 8 : 1
8 8 7 : 2
8 8 6 : 2
8 8 5 : 2
8 8 4 : 2
8 8 3 : 4
 7  6  4 : 2
 7  6  2 : 4
11  7  4 : 3
11  9  4 : 4
13 13  6 : 3
11 11  5 : 3
15 14  7 : 2
16 16  8 : 2
```
# Snippet to help visualise and play around with ideas
```
canvas = document.getElementById('gridCanvas');ctx = canvas.getContext('2d');widthSelector = document.getElementById('width');heightSelector = document.getElementById('height');radiusSelector = document.getElementById('radius');itemCount = document.getElementById('itemCount');canvasMaxWidth = canvas.width;canvasMaxHeight = canvas.height;gridlineColour = '#888';emptyColour = '#ddd';itemColour = '#420';hoverCellColour = '#840';exclusionColour = '#d22';validColour = '#8f7';invalidColour = '#fbb';overlapColour = '#600';resetGrid();x = -1;y = -1;function resetGrid() {gridWidth = widthSelector.value;gridHeight = heightSelector.value;radius = radiusSelector.value;numberOfItems = 0;itemCount.value = numberOfItems + ' items placed.';cells = [gridWidth * gridHeight];for (i=0; i<gridWidth*gridHeight; i++) {cells[i] = '#ddd';}cellSize = Math.min(Math.floor(canvasMaxWidth/gridWidth), Math.floor(canvasMaxHeight/gridHeight));pixelWidth = gridWidth * cellSize + 1;canvas.width = pixelWidth;pixelHeight = gridHeight * cellSize + 1;canvas.height = pixelHeight;leaveCanvas();}function checkPosition(event) {bound = canvas.getBoundingClientRect();canvasX = event.clientX - bound.left;canvasY = event.clientY - bound.top;newX = Math.min(Math.floor(canvasX / cellSize), gridWidth-1);newY = Math.min(Math.floor(canvasY / cellSize), gridHeight-1);if (newX != x || newY != y) {x = newX;y = newY;drawGrid();}}function leaveCanvas() {x = -1;y = -1;drawGrid();}function drawGrid() {ctx.fillStyle = gridlineColour;ctx.fillRect(0, 0, pixelWidth, pixelHeight);cellColour = cells[x + gridWidth * y];if (cellColour == itemColour || cellColour == exclusionColour) {zoneColour = invalidColour;} else {zoneColour = validColour;}for (gridX=0; gridX<gridWidth; gridX++) {for (gridY=0; gridY<gridHeight; gridY++) {colour = (cells[gridX + gridWidth * gridY]);if (x >= 0 && y >= 0) {if (x == gridX && y == gridY) {colour = hoverCellColour;} else if (manhattanDistance(x, y, gridX, gridY) <= radius) {if (colour == exclusionColour) {colour = overlapColour;} else if (colour != itemColour) {colour = zoneColour;}}}ctx.fillStyle = colour;ctx.fillRect(gridX * cellSize + 1, gridY * cellSize + 1, cellSize - 1, cellSize - 1);}}}function manhattanDistance(a, b, c, d) {horizontal = Math.min(Math.abs((gridWidth + c - a) % gridWidth), Math.abs((gridWidth + a - c) % gridWidth));vertical = Math.min(Math.abs((gridHeight + d - b) % gridHeight), Math.abs((gridHeight + b - d) % gridHeight));return vertical + horizontal;}function placeItem(event) {bound = canvas.getBoundingClientRect();canvasX = event.clientX - bound.left;canvasY = event.clientY - bound.top;attemptX = Math.min(Math.floor(canvasX / cellSize), gridWidth-1);attemptY = Math.min(Math.floor(canvasY / cellSize), gridHeight-1);colour = cells[attemptX + gridWidth * attemptY];if (colour != itemColour && colour != exclusionColour) {for (a=0; a<gridWidth; a++) {for (b=0; b<gridHeight; b++) {if (manhattanDistance(a, b, attemptX, attemptY) <= radius) {cells[a + gridWidth * b] = exclusionColour;}}}cells[attemptX + gridWidth * attemptY] = itemColour;drawGrid();numberOfItems++;itemCount.value = numberOfItems + ' items placed.';}}
```
```
<canvas id='gridCanvas' width='500' height='300' style='border:none' onmousemove='checkPosition(event)' onmouseleave='leaveCanvas()' onclick='placeItem(event)'></canvas><br><textarea id='itemCount' readonly></textarea><br><button id='reset' onclick='resetGrid()'>Reset</button> with the following values:<br>Width:<input id='width' type='number' value='20' min='1' max='50' maxlength='2' step='1'><br>Height:<input id='height' type='number' value='13' min='1' max='30' maxlength='2' step='1'><br>Radius:<input id='radius' type='number' value='5' min='0' max='60' maxlength='2' step='1'>
```
# Example (ungolfed) solution
*Just an example for small outputs (resulting from radius not much smaller than the width and height). Can handle any of the test cases but will time out and give up for most larger cases.*
```
itemCount = document.getElementById('itemCount')
calculateButton = document.getElementById('calculate')
widthSelector = document.getElementById('width')
heightSelector = document.getElementById('height')
radiusSelector = document.getElementById('radius')
function calculate() {
    calculateButton.disabled = true
    widthSelector.disabled = true
    heightSelector.disabled = true
    radiusSelector.disabled = true
    itemCount.value = 'Calculating...'
    setTimeout(delayedCalculate, 100)
}
function delayedCalculate() {
    gridWidth = widthSelector.value
    gridHeight = heightSelector.value
    radius = radiusSelector.value
    startingMin = gridWidth*gridHeight + 1
    var cells = [gridWidth * gridHeight]
    for (i=0; i<gridWidth*gridHeight; i++) {
        cells[i] = 0
    }
    var gridState = gridWithItemAdded(cells, 0, 0)
    var minimum = minFromHere(gridState) + 1
    itemCount.value = 'Minimum ' + minimum + ' items required.'
    calculateButton.disabled = false
    widthSelector.disabled = false
    heightSelector.disabled = false
    radiusSelector.disabled = false
}
function minFromHere(gridState) {
    var x
    var y
    var newGridState
    var noCells = true
    var min = startingMin
    for (x=0; x<gridWidth; x++) {
        for (y=0; y<gridHeight; y++) {
            if (gridState[x + gridWidth * y] == 0) {
                noCells = false
                newGridState = gridWithItemAdded(gridState, x, y)
                m = minFromHere(newGridState)
                if (m < min) {
                    min = m
                }
            }
        }
    }
    if (noCells) return 0
    return min + 1
}
function gridWithItemAdded(gridState, x, y) {
    var a
    var b
    var grid = gridState.slice()
    for (a=0; a<gridWidth; a++) {
        for (b=0; b<gridHeight; b++) {
            if (manhattanDistance(a, b, x, y) <= radius) {
                grid[a + gridWidth * b] = 1
            }
        }
    }    
    return grid
}
function manhattanDistance(a, b, c, d) {
    horizontal = Math.min(Math.abs((gridWidth + c - a) % gridWidth), Math.abs((gridWidth + a - c) % gridWidth))
    vertical = Math.min(Math.abs((gridHeight + d - b) % gridHeight), Math.abs((gridHeight + b - d) % gridHeight))
    return vertical + horizontal
}
```
```
<textarea id='itemCount' readonly></textarea>
<br>
<button id='calculate' onclick='calculate()'>Calculate</button> with the following values:
<br>
Width:<input id='width' type='number' value='5' min='1' max='50' maxlength='2' step='1'>
<br>
Height:<input id='height' type='number' value='4' min='1' max='30' maxlength='2' step='1'>
<br>
Radius:<input id='radius' type='number' value='1' min='0' max='60' maxlength='2' step='1'>
```
  
[Answer]
# Python 2, ~~216~~ 182 bytes
```
def f(W,H,R):L={(i%W,i/W)for i in range(W*H)};M={(x,y)for x,y in L if min(x,W-x)+min(y,H-y)>R};g=lambda s:min([1+g(s-{((a+x)%W,(b+y)%H)for x,y in L-M})for a,b in s]or[1]);return g(M)
```
Input like `f(16,16,8)`. Uses pretty much the same algorithm as [@trichoplax's sample](https://codegolf.stackexchange.com/a/51611/21487), but with sets. Initially I didn't place the first item at `(0, 0)`, but this made it choke on the last few cases.
All cases above complete within 10 seconds, well under the limit. In fact, the cases are small enough that I had a little room to be less efficient, allowing me to remove a check which checked for duplicate states.
*(Thanks to @trichoplax for golfing help)*
**Expanded:**
```
def f(W,H,R):
  # All cells
  L={(i%W,i/W)for i in range(W*H)}                 
  # Mask: Complement of exclusion zone around (0, 0) 
  M={(x,y)for x,y in L if min(x,W-x)+min(y,H-y)>R}
  # Place recursively
  g=lambda s:min([1+g(s-{((a+x)%W,(b+y)%H)for x,y in L-M})for a,b in s]or[1])
  return g(M)
```
[Answer]
# Python 3, ~~270~~ ~~262~~ ~~260~~ ~~251~~ ~~246~~ 226
*(Thanks to Sp3000 for:*
* `-~` *instead of* `+1`, *which lets me lose a space after* `return` *on the last line.*
* *losing superfluous parentheses around* `W*H`.
* *lambdas...*
* *putting everything on one line.*
* *python modulo `%` gives positive result for negative numbers, to save another 20 bytes)*
This is the JavaScript example answer from the question ported into Python 3.
To avoid having to pass so many function arguments around, I've moved the two supporting functions inside the calculate function so that they share its scope. I've also condensed each of these functions into a single line to avoid the cost of indentation.
### Explanation
This fairly brute force approach places the first item at (0, 0), and then marks all the excluded squares. It then recursively places an item at all remaining valid squares until all squares are excluded, and returns the minimum number of items required.
### Golfed code:
```
def C(W,H,R):r=range;M=lambda g:min([M(G(g,x,y))for x in r(W)for y in r(H)if g[x+W*y]]or[-1])+1;G=lambda g,x,y:[g[a+W*b]if min((x-a)%W,(a-x)%W)+min((y-b)%H,(b-y)%H)>R else 0for b in r(H)for a in r(W)];return-~M(G([1]*W*H,0,0))
```
### Ungolfed code:
```
def calculate(W, H, R):
    starting_min = W * H + 1
    cells = [0] * (W * H)
    grid_state = grid_with_item_added(cells, 0, 0, W, H, R)
    return min_from_here(grid_state, starting_min, W, H, R) + 1
def min_from_here(grid_state, starting_min, W, H, R):
    no_cells = True
    min = starting_min
    for x in range(W):
        for y in range(H):
            if grid_state[x + W * y] == 0:
                no_cells = False
                new_grid_state = grid_with_item_added(grid_state, x, y, W, H, R)
                m = min_from_here(new_grid_state, starting_min, W, H, R)
                if m < min:
                    min = m
    if no_cells:
        return 0
    else:
        return min + 1
def grid_with_item_added(grid_state, x, y, W, H, R):
    grid = grid_state[:]
    for a in range(W):
        for b in range(H):
            if manhattan_distance(a, b, x, y, W, H) <= R:
                grid[a + W * b] = 1
    return grid
def manhattan_distance(a, b, c, d, W, H):
    horizontal = min(abs(W + c - a) % W, abs(W + a - c) % W)
    vertical = min(abs(H + d - b) % H, abs(H + b - d) % H)
    return horizontal + vertical
if __name__ == '__main__':
    import sys
    arguments = sys.argv[1:]
    if len(arguments) < 3:
        print('3 arguments required: width, height and radius')
    else:
        print(calculate(int(arguments[0]), int(arguments[1]), int(arguments[2])))
```
The ungolfed code defines a function and also includes code to allow it to be called from the command line. The golfed code just defines the function, which is [sufficient for standard code golf questions](http://meta.codegolf.stackexchange.com/questions/2419/default-for-code-golf-program-function-or-snippet).
If you would like to test the golfed code from the command line, here it is with the command line handling included (but not golfed):
### Command line testable golfed code
```
def C(W,H,R):r=range;M=lambda g:min([M(G(g,x,y))for x in r(W)for y in r(H)if g[x+W*y]]or[-1])+1;G=lambda g,x,y:[g[a+W*b]if min((x-a)%W,(a-x)%W)+min((y-b)%H,(b-y)%H)>R else 0for b in r(H)for a in r(W)];return-~M(G([1]*W*H,0,0))
if __name__ == '__main__':
    import sys
    arguments = sys.argv[1:]
    if len(arguments) < 3:
        print('3 arguments required: width, height and radius')
    else:
        print(C(int(arguments[0]), int(arguments[1]), int(arguments[2])))
```
]